On Twisted Cyclic Algebras and the Chain Equivalence of Kummer Elements

Dissertation

zur Erlangung des Doktorgrades der Naturwissenschaften (Dr. rer. nat.)

der Fakultät für Mathematik der Universität Regensburg

vorgelegt von Wieland Fischer

aus Amberg

Fakultät für Mathematik der Universität Regensburg Regensburg im Juni 1999

Table of Contents

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Chapter I . Twisted Cyclic Extensions . . . . . . . . . . . . . . . . . . . . . 8

§ 1. Twisted Cyclic Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1. Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2. Twisted Cyclic Structures . . . . . . . . . . . . . . . . . . . . . . . . . 12 3. Uniqueness of Twisted Cyclic Structures . . . . . . . . . . . . . . . . . 16

§ 2. C_{0}-Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 1. C_{0}-Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2. Classification of C_{0}-Extensions . . . . . . . . . . . . . . . . . . . . . . . 19 3. μ_{n}-Extensions and Kummer Extensions . . . . . . . . . . . . . . . . . . 25

Chapter II . Twisted Cyclic Algebras . . . . . . . . . . . . . . . . . . . . . . 30

§ 3. Twisted Cyclic Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 1. Twisted Cyclic Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . 30 2. Pairs of Twisted Cyclic Extensions associated to Twisted Cyclic Algebras 32 3. Twisted Cyclic Algebras associated to Pairs of Twisted Cyclic Extensions 38 4. A Cohomological Classification of Twisted Cyclic Algebras . . . . . . . 41

§ 4. Existence of Twisted Cyclic Decompositions of Central Simple Algebras . . 49 1. A Cohomological Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . 49 2. Galois Actions on Ā_{0} . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 3. The Existence of a Twisted Cyclic Decomposition . . . . . . . . . . . . 57

§ 5. A Cohomological Description of Twisted Cyclic Algebras . . . . . . . . . . 59 1. The Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 2. Cocycles Associated to a Twisted Cyclic Algebra . . . . . . . . . . . . . 60 3. The 1-Cocycles Associated to the Pairs (L, ρ_{L}) and (K, ρ_{K}) . . . . . . . 62 4. Proof of the Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

Chapter III . The Notion of Chain Equivalence . . . . . . . . . . . . . . . 67

§ 6. Kummer Elements and Kummer Relation . . . . . . . . . . . . . . . . . . . 67 1. Kummer Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 2. Description of Twisted Cyclic Algebras via Elements in Kummer Relation 70 3. Galois Action on Kummer Elements and Twisted Cyclic Decompositions 72 4. Description of Split Twisted Cyclic Algebras . . . . . . . . . . . . . . . 73

Table of Contents

§ 7. Chains of Twisted Cyclic Extensions . . . . . . . . . . . . . . . . . . . . . 75 1. Definition of Chains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 2. Chains with Fixed Starting and End Points . . . . . . . . . . . . . . . . 75 § 8. Chains of Kummer Elements . . . . . . . . . . . . . . . . . . . . . . . . . . 80

1. Definition of Chains of Kummer Elements . . . . . . . . . . . . . . . . 80 2. The Notion of Chain Equivalence of Kummer Elements . . . . . . . . . 82 3. Connection between K-Chains and c-Chains . . . . . . . . . . . . . . . 83

Chapter IV . Geometry of Kummer Elements and Chains . . . . . . . . . 85

§ 9. The Variety of Kummer Elements . . . . . . . . . . . . . . . . . . . . . . . 85 1. A as a k-Scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 2. W(A) and W(A) as Varieties over an Algebraically Closed Field …. 88 3. W(A) and W(A) as k-Schemes . . . . . . . . . . . . . . . . . . . . . . . 91

§ 10. The Variety of Chains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 1. The “Vectorbundle of Kummer Pairs” . . . . . . . . . . . . . . . . . . . 92 2. Construction of the Varieties of Chains . . . . . . . . . . . . . . . . . . 98 3. Chains with Starting Point . . . . . . . . . . . . . . . . . . . . . . . . . 102

§ 11. The Topological Closure of W^{1}(A) . . . . . . . . . . . . . . . . . . . . . . . 103 1. Some Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 2. The Closure of W^{1}(A) . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

Chapter V . Chain Equivalence for Algebras of Degree 2 and 3 . . . . . . 110

§ 12. The Case of Algebras of Degree 2 . . . . . . . . . . . . . . . . . . . . . . . 111 1. Chain Equivalence for n = ℓ = 2 . . . . . . . . . . . . . . . . . . . . . . 111 2. The Geometric Point of View . . . . . . . . . . . . . . . . . . . . . . . 113

§ 13. The Case of Algebras of Degree 3 and Length 3 . . . . . . . . . . . . . . . 117 1. Preliminaries and Conditions for Chains of Length 2 . . . . . . . . . . . 117 2. Existence of Connecting Chains of Length 3 . . . . . . . . . . . . . . . 120

§ 14. The Case of Algebras of Degree 3 and Length 4 . . . . . . . . . . . . . . . 124

Chapter VI . Relation to the Product Map of Tori . . . . . . . . . . . . . 128

§ 15. A Covering of W(A) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 1. The Variety GL_{1}(A)/D . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 2. The Covering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

§ 16. A Covering of Chain Varieties . . . . . . . . . . . . . . . . . . . . . . . . . 133 1. The Kummer Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 2. The Covering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

§ 17. Reformulation in the Product Map of Tori . . . . . . . . . . . . . . . . . . 140

Introduction

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 A well known theorem about quaternion algebras is the common slot lemma by A. Albert: For a, b, c, d ∈ k^{∗}, non-zero elements in a field such that (a, b) ≅ (c, d) there exists an element e ∈ k^{∗} such that

(a, b) ≅ (a, e) ≅ (c, e) .

For non-zero elements a, b in a field k of characteristic not 2, the symbol (a, b) denotes the central simple k-algebra of degree 2 defined by the presentation

⟨X, Y : X^{2} = a, Y^{2} = b, XY = −Y X⟩ .

There are several other theorems of this form. One example is the chain-p-equivalence of Pfister forms, cf. chapter 4, §1 in Scharlau [SchQF]. Another example is Witt’s chain equivalence theorem of diagonal quadratic forms, cf. chapter 1, (5.2) in Lam [LaQF].

Concerning the first example, one may ask similar questions for central simple algebras of degree > 2. E.g., for some integer n ≥ 2 let k be a field of characteristic prime to n, containing a primitive n-th root of unity ζ. Then for a, b, c, d ∈ k^{∗} such that (a, b)_{ζ} ≅ (c, d)_{ζ}, does there exist elements e_{1}, …, e_{ℓ−1} ∈ k^{∗} such that there is a chain of isomorphisms of length ℓ of the form

(a, b)_{ζ} ≅ (e_{1}, b)_{ζ} ≅ (e_{1}, e_{2})_{ζ} ≅ ⋯ ≅ (e_{ℓ−2}, e_{ℓ−1})_{ζ} ≅ (c, e_{ℓ−1})_{ζ}

if ℓ is odd, or

(a, b)_{ζ} ≅ (a, e_{1})_{ζ} ≅ (e_{2}, e_{1})_{ζ} ≅ ⋯ ≅ (e_{ℓ−2}, e_{ℓ−1})_{ζ} ≅ (c, e_{ℓ−1})_{ζ}

if ℓ is even? Here (a, b)_{ζ} denotes the central simple k-algebra of degree n defined by the presentation

⟨X, Y : X^{n} = a, Y^{n} = b, XY = ζY X⟩ .

If this statement is true, i.e., answered with “yes”, we call it a common slot lemma for central simple k-algebras of degree n and chains of length ℓ.

J.-P. Tignol answered the question for central simple algebras of degree n = 3 and chains of length ℓ = 2, by giving a counter-example: There are fields k and elements a, b, c, d ∈ k^{∗} with (a, b)_{ζ} ≅ (c, d)_{ζ}, such that for no e ∈ k^{∗} the following holds:

(a, b)_{ζ} ≅ (a, e)_{ζ} ≅ (c, e)_{ζ} .

Cf. the appendix in [RoCL].

A common slot lemma for exceptional Jordan algebras over quadratically closed fields was proven by H. P. Petersson & M. L. Racine in [PeSR]. Their methods were used by M. Rost in [RoCL] in order to show a common slot lemma for central simple algebras of degree 3 and chains of length 4. For completeness we reproduce his proof (with minor adaptations) in §14 of this paper.

M. Rost announced a chain lemma for Kummer elements, which we are going to state below.—A Kummer element in a central simple k-algebra A of degree n is an invertible element X in A such that its reduced characteristic polynomial has the form Prd(X, t) = t^{n} − a. Furthermore, a ζ-pair in A is a pair (X, Y) of Kummer elements in A such that XY = ζY X, where ζ denotes a primitive n-th root of unity. We will see in §6 that a ζ-pair (X, Y) generates the algebra A; so with a := X^{n} and b := Y^{n} ∈ k^{∗} we have A ≅ (a, b)_{ζ}.

Now Rost’s chain lemma says: Let n = p be a prime number and k a p′-closed field, i.e., k has no finite algebraic extension of degree prime to p. Let A be a division algebra over k of degree p and ζ ∈ k^{∗} a primitive p-th root of unity. Then for any two Kummer elements Z_{0} and Z_{p} in A, there exist Kummer elements Z_{1}, …, Z_{p−1} in A such that (Z_{i−1}, Z_{i}) are ζ-pairs for i = 1, …, p. For the last fact we are going to say that Z_{0}, …, Z_{p} is a (ζ-)chain of Kummer elements in A of length p.

As a corollary we easily get a common slot lemma for central simple k-algebras of degree p and chains of length p, with the field k from above: Let a, b, c, d ∈ k^{∗} with A := (a, b)_{ζ} ≅ (c, d)_{ζ}, then there are Kummer elements Z_{0} and Z_{p} in A with Z_{0}^{p} = b if (p ≡ 1 mod 4) or Z_{0}^{p} = b^{−1} if (p ≡ 3 mod 4) and Z^{p} = c. We assume that p is odd. Then there are (in the generic case) Kummer elements Z_{1}, …, Z_{p−1} such that Z_{0}, …, Z_{p} is a ζ-chain of Kummer elements in A, and we get

A = ⟨Z_{0}, Z_{1}⟩ = ⟨Z_{1}, Z_{2}⟩ = ⋯ = ⟨Z_{p−1}, Z_{p}⟩ .

In the language of symbols this is

(a, b)_{ζ} ≅ (e_{1}, b)_{ζ} ≅ (e_{1}, e_{2})_{ζ} ≅ ⋯ ≅ (e_{p−2}, e_{p−1})_{ζ} ≅ (c, e_{p−1})_{ζ}

for suitable e_{i} := Z_{i}^{±n} ∈ k^{∗}.

The following paper consists of two parts: In the first part we set the notion of ζ-pairs in a more conceptual frame: If (X, Y) is a ζ-pair in a central simple k-algebra A of degree n, then we get other ζ-pairs by taking powers of the two elements X and Y with exponents prime to n, i.e., (X^{ν}, Y^{μ}) is a ζ^{νμ}-pair. The same we can do more generally for a whole chain of Kummer elements. But the pairs we got by this process do not give anything new. In particular the entries a and b in the associated symbol (a, b)_{ζ} just change to the ν-th and μ-th power. We also observe that X and Y generate maximal commutative subalgebras L = k[X] and K = k[Y] of A, cf. §6, and that these algebras remain unchanged, even if we first applied any operation of taking powers, mentioned above. We call the pairs (L, K) arising in this way a decomposition of A. In chapter II we analyze these kinds of objects—the triples (A, L, K). We even consider objects, which are models (over a subfield of k) of this triple.

It is easy to see (cf. §3) that the powers of Y act on L and mutatis mutandis X on K by conjugation in A. This situation is the subject of chapter I.

In chapter III we give the connection between the decompositions of A and ζ-pairs in A: We specify several notions of chains and show how they are related. This chapter is the link to the second part of the paper.

Here we give a geometric view of the main objects—Kummer elements and chains of Kummer elements: We will find smooth irreducible k-schemes W^{0}(A) and W^{ℓ}(A) which classify by their points the Kummer elements and chains of Kummer elements of length ℓ, up to scaling of the Kummer elements (i.e., we work in projective spaces). By this interpretation, the problem of the chain lemma becomes a question in Algebraic Geometry, where we have much additional structure. For example we may stress dimension theory. The problem, if any two (generic) Kummer elements can be connected by a chain of Kummer elements of length ℓ now has the following form: Let

π: W^{ℓ}(A) ⟶ W^{0}(A) ×_{k} W^{0}(A)

be the morphism of schemes which on the points is just the projection (Z_{0}, …, Z_{ℓ}) ↦ (Z_{0}, Z_{ℓ}) of a chain of Kummer elements of the length ℓ onto the first and last factor. Question: Is the morphism dominant, and—if dim W^{ℓ}(A) = dim W^{0}(A) ×_{k} W^{0}(A)— what is the degree of π?

We will find that dim W^{ℓ}(A) = (ℓ + n)(n − 1) and dim W^{0}(A) = n(n − 1). So one easily can see that for a chain lemma of a central simple algebra A of degree n we need at least ℓ ≥ n. In the case from above, for central simple algebras of degree 3 we recognize that there is no chance connecting 2 arbitrary Kummer elements by a chain of length 2—for dimensional reasons! In his counter-example, Tignol even showed more: He proved that one can find two Kummer elements X and Y such that X can not be connected to any conjugate of Y, by a chain of length 2. In the end of chapter IV, we show an interesting property about the topology of the scheme W^{1}(A) of ζ-pairs, i.e., we will describe the closure of the space.

In chapter V, we explicitly work out three cases: The first case is concerned with the common slot lemma for central simple algebras of degree 2 and chains of length 2; the “classical” common slot lemma, given at the beginning.—But of course we do it in our terminology of Kummer elements. We will see the geometric structure of the chain lemma: The morphism π: W^{2}(A) ⟶ W^{0}(A) ×_{k} W^{0}(A) will be revealed as part of the blowingup morphism along the diagonal. The second case is the chain lemma for central simple algebras of degree 3 and chains of length 3. We will see that for a quadratically closed field k, almost any two Kummer elements X and Y can be connected by exactly two chains of Kummer elements of length 2. The third case is the case of central simple algebras of degree 3 and chains of length 4, already mentioned above.

The last chapter gives a new interpretation of the morphism π: We will construct a more explicit map ω between spaces which are coverings of degree n^{ℓ} and n of W^{ℓ}(A) and W^{0}(A) ×_{k} W^{0}(A) respectively, and via these coverings ω lies over π. So, one may ask the question of dominance and even (for n = ℓ) of the degree anew for the map ω. Finally we will see that our new map ω has a nice interpretation for ℓ = n: Let S and T be the maximal tori in PGL(A) given by the subalgebras generated by Kummer elements X and Y; where (X, Y) is a ζ-pair. Then for ℓ = n odd, the map ω can be interpreted as the multiplication map

S × T × S × T × ⋯ × S × T = (S × T)^{\frac{n+1}{2}} ⟶ PGL(A) .

Acknowledgments: I deeply wish to express my thanks to PD Dr. M. Rost. He took care of me after the tragic death of my former teacher Jürgen Neukirch. I am greatly indebted to him for presenting to me this new interesting subject.

Notations

Throughout the paper k denotes a field and—unless specified in a different way—we denote by k̄ a separable algebraic closure of k. Then Γ = Gal(k̄|k) denotes the absolute Galois group of k. Furthermore we use the following “standard” notations:

E_{ij}  quadratic matrix (of specified size) with entry 1 at the place (i, j)

and 0 otherwise

diag(d_{1}, …, d_{n}) diagonal n × n-matrix with diagonal entries d_{1},…, d_{n}, in this

order, i.e., d_{1}E_{11} + ⋯ + d_{n}E_{nn}

EigVal(f)  set of eigenvalues of some endomorphism f of a vector space
EigVec(f, θ)  set of (non-zero) eigenvectors of f to the eigenvalue θ
EigVec(f)  ∐_{θ∈EigVal(f)} EigVec(f, θ)
∐  sum in the category of sets, i.e., disjoint union
κ_{x}  conjugation with the element x in some domain, whenever this

action is defined, i.e., κ_{x}(y) = xyx^{−1}

ev_{x}  evaluation in x, i.e., ev_{x}(f) = f(x), whenever f is a map which

can be evaluated in x

Prd(X, t)  reduced characteristic polynomial (in the variable t) of

some element X in a central simple algebra

Nrd(X)  reduced norm of some element X in a central simple algebra
Trd(X)  reduced trace of some element X in a central simple algebra

If X is any scheme and ℰ an O_{X}-module on X, then we denote with ℰ (x), for a point x ∈ X, the κ(x)-vector space ℰ_{x} ⊗_{O_{X,x}} κ(x), where in this context κ(x) is the residue field O_{X,x}/𝔪_{x} of x and ℰ_{x} the stalk of ℰ at x. For such an ℰ on a k-scheme X and a k-vector space E, we use the following notations:

A(E) Spec(S(Ě)), the k-Scheme, which has E as its k-rational points
P(E) Proj(S(Ě)), the k-Scheme, which has E − {0}/k^{∗} as its k-rational points
𝕍(ℰ) Spec(S(ℰ))
ℙ(ℰ) Proj(S(ℰ))

Here Ě stands for the k-vector space dual Hom_{k-lin}(E, k) of E, and S is the (naturally graded) symmetric k-algebra on some k-vector space or sheaf of modules on some kscheme X. In a similar way, one has to read the k-Schemes GL_{1}(A) and PGL_{1}(A) for k-algebras A. Note that in this notation, if one interprets E as a sheaf of O_{X}-modules, for X = Spec(k), this implies: A(Ě) = 𝕍(E) and P(Ě) = ℙ(E). In most cases we use the notation [x] for some equivalence class of x, however, sometimes we write ā for (a mod nℤ).

Chapter I

Twisted Cyclic Extensions

A cyclic Galois extension L|k is known to be a Galois extension L|k of finite degree n with a Galois group Gal(L|k) ≅ ℤ/nℤ. We can characterize a cyclic Galois extension in the following way: Let L|k be a separable field extension of degree n and k̄ = L̄ be a separable algebraic closure of L and k. Further we denote by L̃ the Galois closure of L in k̄. If θ is a primitive element of L, i.e., L = k(θ), then θ has n conjugates θ_{1}, …, θ_{n} in k̄ and hence L̃ = k(θ_{1}, …, θ_{n}). Now obviously we can say: L|k is a cyclic Galois extension ⇔ Gal(L̃|k) ≅ ℤ/nℤ. From Galois theory one knows the morphism

\begin{matrix}φ: Gal(k̄|k) & ⟶ S(\{θ_{1}, …, θ_{n}\}) \\ γ & ⟼ γ|_{\{θ_{1},...,θ_{n}\}} \end{matrix}

from the absolute Galois group of k into the permutation group of the conjugates of θ. The kernel of this morphism in Gal(k̄|L̃) and therefore the image is isomorphic to Gal(L̃|k). So we can say: L|k is a cyclic Galois extension ⇔ im(φ) ≅ ℤ/nℤ. In this case im(φ) is a group which acts free on the set {θ_{1}, …, θ_{n}}. The group Gal(k̄|k) acts on S(θ_{1}, …, θ_{n}) by conjugation via φ. However this action is trivial on im(φ). In the following chapter we will treat objects which are slightly more general than the cyclic Galois extensions, namely pairs (L, C) where C ⊆ S({θ_{1}, …, θ_{n}}) is not any more the image of φ but some other cyclic subgroup of order n which is preserved by the action of the absolute Galois group of k.

§ 1. Twisted Cyclic Extensions

1. Preliminaries

Let M be a finite set of n elements, S(M) denotes the symmetric group on M and let C ⊆ S(M) be a transitive cyclic subgroup of order n, i.e., a cyclic subgroup of order n such that M = {π(m) : π ∈ C} for any m ∈ M. About the normalizer subgroup N_{C} := {π ∈ S(M) : πCπ^{−1} = C} one knows:

(1.1) Claim. There is a natural short exact sequence

\begin{matrix}1 → C ⟶^{incl} N_{C} & ⟶^{κ} Aut(C) → 1 \\ π & ⟼ κ_{π} \end{matrix}

where κ_{π}(φ) = πφπ^{−1} is the inner automorphism on S(M) restricted to C, given by conjugation with π.

§ 1. Twisted cyclic extensions

Proof: We may assume M = {1, 2, …, n} and—since C is a transitive subgroup of order n—that C is generated by σ := (1 2 … n), the permutation which maps i to i + 1 modulo n. Let π ∈ ker(κ) be an element of the kernel of κ. Then πφ = φπ for all elements φ ∈ C. Let i = π(n), then we have

π(j) = σ^{j}πσ^{−j}(j) = σ^{j}π(n) = σ^{j}(i) = σ^{i+j}(n) = σ^{i}(j)

for all j, hence π = σ^{i} ∈ C. It remains to show the surjectivity of κ: An automorphism f ∈ Aut(C) = (ℤ/nℤ)^{∗} is given by some (m mod nℤ) ∈ (ℤ/nℤ)^{∗} such that f: C → C, π ↦ π^{m}. Define φ ∈ S_{n} = S(ℤ/nℤ) to be ℤ/nℤ \underrightarrow{∼} ℤ/nℤ, ℓ mod n ↦ m · ℓ mod n, then considering

σ^{m} = (1m 2m … nm) = φ (1 2 … n) φ^{−1} = κ_{φ}(σ)

one can see both—φ ∈ N_{C} and κ_{φ} = f.

The surjective homomorphism κ has sections. There is no canonical section, but we will see that there is indeed a class of sections which are given naturally.

(1.2) Lemma. For each m ∈ M the restriction of κ: N_{C} ⟶ Aut(C) to

I_{m} := \{π ∈ N_{C} : π(m) = m\} \underrightarrow{∼} Aut(C)

is an isomorphism. It gives rise to a section s_{m}: Aut(C) ⟶ N_{C} of κ: N_{C} ⟶ Aut(C), namely its inverse. It can be described by

\begin{matrix}s_{m}(f): M & ⟶ M \\ π(m) & ⟼ (^{f}π)(m) , \end{matrix}

for all f ∈ Aut(C) and π ∈ C.

Proof: First of all, the restriction κ: I_{m} → Aut(C) is injective: If π, π′ ∈ I_{m} are two elements such that κ_{π} = κ_{π′}, we can conclude π′π^{−1} ∈ ker(κ) = C and hence π′ = π. In fact, since π′π^{−1}(m) = m we know π′π^{−1} = 1. Now, in order to show that the morphism is surjective, choose an f ∈ Aut(C). Then the permutation defined by s̃: M → M, π(m) ↦ (^{f}π)(m) for π ∈ C, is a pre-image of f lying in I_{m}: We first claim for any π ∈ C:

s̃ π s̃^{−1} = ^{f}π .

Evaluation with ψ(m) ∈ M for ψ ∈ C gives

(s̃ π s̃^{−1})(ψ(m)) = (s̃ π)(^{f^{−1}}ψ)(m)
= s̃(π ^{f^{−1}}ψ)(m)
= ^{f}(π ^{f^{−1}}ψ)(m)
= (^{f}π ψ)(m)
= (^{f}π)(ψ(m)) .

This proves that s̃ ∈ N_{C}—thus we get s̃ ∈ I_{m}, since s̃(m) = m—and that κs̃ = f.

(1.3) Lemma. For any m ∈ M and π ∈ C we have the formula s_{π(m)}(f) = πs_{m}(f)π^{−1} for every f ∈ Aut(C), i.e.,

s_{π(m)} = κ_{π} ∘ s_{m} ,

where κ_{π} is the conjugation κ_{π}: N_{C} ⟶ N_{C}, φ ⟼ πφπ^{−1}. In other words, we have the commutative diagram

�  _{s_{m}} �  _{s_{π(m)}} �  _{κ_{π}} N_{C} . N_{C} Aut(C)

Proof: Since κ_{π}(I_{m}) = I_{π(m)} we can change the two right hand side objects in the diagram to I_{m} and I_{π(m)} such that we have to show the commutativity of

�  _{s_{m}} �  _{s_{π(m)}} �  _{κ_{π}} I_{π(m)} . I_{m} Aut(C)

Now the two diagonal arrows are bijections, and in order to show the commutativity of the diagram it is equivalent to prove the commutativity of the diagram with the reversed arrows:

�  _{κ} �  _{κ} �  _{κ_{π}} I_{π(m)} . I_{m} Aut(C)

Thus we have to prove κ ∘ κ_{π} = κ on I_{m}. But this is clear. We even have the commutativity on N_{C}

�  _{κ} �  _{κ} �  _{κ_{π}} N_{C} , N_{C} Aut(C)

since π is an element of the kernel of κ: N_{C} ⟶ Aut(C): For any τ ∈ N_{C}

κ ∘ κ_{π}(τ) = κ(πτπ^{−1})
= κ(π)κ(τ)κ(π^{−1})
= κ(τ)

and this completes the proof.

(1.4) Claim. For each m ∈ M there is an isomorphism of groups

\begin{matrix}σ_{m}: C⋊ Aut(C) & \underrightarrow{∼} N_{C} \\ (φ, f) & ⟼ φ s_{m}(f) . \end{matrix}

We have the commutative diagram for any π ∈ C

�_{κ_{(π,id_{C})}} �  _{κ_{π}} �  ^{σ_{m}} �  _{σ_{m}} �  _{σ_{π(m)}} N_{C} . C⋊ Aut(C)  N_{C} C⋊ Aut(C)

Proof: With the last lemmas there is nothing to prove anymore. The commutativity of the right triangle follows from φ s_{π(m)}(f) = φπs_{m}(f)π^{−1} = πφ s_{m}(f)π^{−1} for φ ∈ C and f ∈ Aut(C). To recall the multiplication in the semidirect product we show the homomorphism property:

σ_{m}((φ, f)(φ′, f′))(ψ(m)) = σ_{m}((φ ^{f}φ′, ff′))(ψ(m))
= φ ∘^{f}φ′ ∘ s_{m}(ff′)(ψ(m))
= φ ∘^{f}φ′ ∘ (^{ff′}ψ)(m)
= φ ∘ s_{m}(f)((φ′ ^{f′}ψ)(m))
= σ_{m}(φ, f)(σ_{m}(φ′, f′)(ψ(m)))
= σ_{m}(φ, f) ∘ σ(φ′, f′)(ψ(m))

for (φ, f), (φ′, f′) ∈ C⋊ Aut(C) and ψ ∈ C.

(1.5) Remark. Up to an inner automorphism with an element of C (which is a subgroup of either C⋊ Aut(C) and N_{C}) there is a canonical isomorphism of the groups C⋊ Aut(C) ≅ N_{C}.

(1.6) Lemma. If n is of the form n = 2^{δ}n′, where n′ is an odd and squarefree number and δ = 0, 1, 2, then in (ℤ/nℤ)⋊ Aut(ℤ/nℤ) = (ℤ/nℤ)⋊(ℤ/nℤ)^{∗} one has

ord(m̄, ā) = n ⟹ ā = 1

for all (m̄, ā) ∈ (ℤ/nℤ)⋊(ℤ/nℤ)^{∗}.

Proof: Let (m̄, ā) ∈ (ℤ/nℤ)⋊(ℤ/nℤ)^{∗} and let m̄ = (m mod nℤ), ā = (a mod nℤ). Assume first that n = p is a prime. If a ≡ 1 (mod p), then

(m̄, ā)^{p} = (m̄, 1)^{p} = (0, 1) .

If a ≢ 1 (mod p), then

1 + a + a^{2} + ⋯ + a^{p−2} = \frac{a^{p−1} − 1}{a − 1} ≡ 0 (mod p)

and therefore

(m̄, ā)^{p−1} = (m + am + ⋯ + a^{p−2}m mod pℤ, a^{p−1} mod pℤ) = (0, 1) .

Let first n be (not necessarily odd) squarefree and n = p_{1} ⋯ p_{r} be the prime factorization of n. We define for i = 1, …, r

μ_{i} := \{ \begin{matrix}p_{i}, & if a ≡ 1 (mod p_{i}) \\ p_{i} − 1, & if a ≢ 1 (mod p_{i}) \end{matrix}

and μ := μ_{1} ⋯ μ_{r}. Then

(m̄, ā)^{μ} = (0, 1)

since

(ℤ/nℤ)⋊(ℤ/nℤ)^{∗} = ∏_{i=1}^{r}(ℤ/p_{i}ℤ)⋊(ℤ/p_{i}ℤ)^{∗} .

Now if ā ≠ 1 we have an i such that a ≢ 1 (mod p_{i}), hence μ_{i} = p_{i} − 1 and therefore μ < n. It follows that ord(m̄, ā) < n. In the case of n = 4n′ = p_{1}^{2}p_{2} ⋯ p_{r}, where n′ is odd squarefree and p_{1} = 2 we modify

μ_{i} := \{ \begin{matrix}p_{i}, & if a ≡ 1 (mod p_{i}) and 2 ≤ i \\ p_{i} − 1, & if a ≢ 1 (mod p_{i}) and 2 ≤ i \\ 4, & if a ≡ 1 (mod p_{i}) and 1 = i \\ 2, & if a ≢ 1 (mod p_{i}) and 1 = i \end{matrix}

with μ := μ_{1} ⋯ μ_{r}. Then one observes for i = 1 that in (ℤ/4ℤ)⋊(ℤ/4ℤ)^{∗}

(m̄, ā)^{μ_{1}} = (0, 1)

and therefore we have analogously in (ℤ/nℤ)⋊(ℤ/nℤ)^{∗}

(m̄, ā)^{μ} = (0, 1)

also with μ < n which implies ord(m̄, ā) < n.

(1.7) Corollary. Let π ∈ S(M) be an element of order n and C some transitive cyclic subgroup of S(M). Assume n = m or n = 2m for some squarefree m. Then the condition πCπ^{−1} ⊆ C implies ⟨π⟩ = C.

Proof: The condition says that π ∈ N_{C} ≅ C⋊ Aut(C). It suffices to show the following statement: If π ∈ ℤ/nℤ⋊ Aut(ℤ/nℤ) = ℤ/nℤ⋊(ℤ/nℤ)^{∗} is of order n, then π ∈ ℤ/nℤ, i.e., π = (m, 1) for some m ∈ ℤ/nℤ. But this is exactly (1.6).

2. Twisted Cyclic Structures

We fix a separable algebraic closure k̄ of a field k. Let Γ := Gal(k̄|k) denote the absolute Galois group of k. In the following let L be a separable k-algebra of degree n, i.e., L is a commutative k-algebra with

L̄ := L ⊗_{k} k̄ ≅ k̄^{n}

as k̄-algebra (cf. (18.3) in [BI]). Γ acts (continuously) on L̄ by semilinear automorphisms via Γ × L̄ → L̄, (γ, ℓ ⊗ x) ↦ ℓ ⊗ γ(x). Obviously L̄^{Γ} = L.

If M_{L} is the set of the n primitive idempotents of L̄, then there is a canonical identification

\begin{matrix}Autk̄(L̄) & = S(M_{L}) \\ f & ⟼ f|_{M_{L}} , \end{matrix}

where S(M_{L}) is the symmetric group on the set M_{L}.

(1.8) Remark. By this identification one can write the morphism considered in (1.4) as

\begin{matrix}σ_{e}: C⋊ Aut(C) & ⟶ Autk̄(L) \\ (f, φ) & ⟼ f ∘ s_{e}(φ) , \end{matrix}

where e ∈ M_{L} is a primitive idempotent of L̄ and C ⊆ Autk̄(L) a transitive cyclic subgroup of order n, and s_{e}(φ) ∈ Autk̄(L) is given by

\begin{matrix}s_{e}(φ): L̄ & ⟶ L̄ \\ ∑_{c∈C} x_{c} · c(e) & ⟼ ∑_{c∈C} x_{c} · (^{φ}c)(e) , \end{matrix}

where x_{c} ∈ k̄. Recall that L̄ = ⨁_{e∈M_{L}} k̄ · e = ⨁_{c∈C} k̄ · c(e) .

There is a left action of Γ on the group Autk̄(L) given by

Γ × Autk̄(L) ⟶ Autk̄(L)
(γ, f) ⟼ ^{γ}f := (id_{L} ⊗γ) ∘ f ∘ (id_{L} ⊗γ^{−1}) .

One checks that ^{γ}f again is k̄-linear, even if the composition on the right hand side takes place in the larger group Aut_{k}(L̄).

(1.9) Remark. There is a canonical identification

Aut_{k}(L) \underrightarrow{∼} Autk̄(L)^{Γ}
f ⟼ f ⊗ idk̄
g|_{L⊗k} ⟵ g .

Proof: All we have to show is g(L ⊗ k) ⊆ L ⊗ k, and this is clear since L̄^{Γ} = L: Let γ ∈ Γ and ℓ ∈ L, then g(ℓ ⊗ 1) = (^{γ}g)(ℓ ⊗ 1) = γ(g(ℓ ⊗ 1)).

The action of Γ on Autk̄(L) is already given by inner automorphisms. Define

\begin{matrix}φ_{L}: Γ & ⟶ Autk̄(L) = S(M_{L}) \\ γ & ⟼ φ_{L}(γ) = (id_{L} ⊗γ)|_{M_{L}} , \end{matrix}

i.e., φ_{L}(γ) has the same effect on M_{L} as (id_{L} ⊗γ) but is k-linear.

(1.10) Claim. The action of Γ on Autk̄(L) is given by κ ∘ φ_{L}, i.e.,

\begin{matrix}Γ × Autk̄(L) & ⟶ Autk̄(L) \\ (γ, f) & ⟼ κ_{φ_{L}(γ)}(f) = φ_{L}(γ) ∘ f ∘ φ_{L}(γ)^{−1} . \end{matrix}

Proof:

(φ_{L}(γ) ∘ f ∘ φ_{L}(γ)^{−1})|_{M_{L}} = φ_{L}(γ)|_{M_{L}} ∘ f|_{M_{L}} ∘ φ_{L}(γ)^{−1}|_{M_{L}}
= (id_{L} ⊗γ)|_{M_{L}} ∘ f|_{M_{L}} ∘ (id_{L} ⊗γ^{−1})|_{M_{L}}
= ^{γ}f|_{M_{L}} .

(1.11) Remark. Let L = k(θ)|k be a separable field extension of degree n. We assume L ⊆ k̄, so θ ∈ k̄. Let {θ_{1}, …, θ_{n}} be the set of the n distinct conjugates of θ in k̄. Then we have the canonical identifications

\begin{matrix}M_{L} & = Hom_{k}(L, k̄) \\ e & ⟼ e^{∗} , \end{matrix}

where e^{∗}: L → k̄, ℓ ↦ tr_{L̄|k̄}(eℓ) and

\begin{matrix}Hom_{k}(L, k̄) & = \{θ_{1}, …, θ_{n}\} \\ f & ⟼ f(θ) . \end{matrix}

With this identification of M_{L} with {θ_{1}, … θ_{n}}, we can give the following interpretation of φ_{L}: For γ ∈ Γ the permutation φ_{L}(γ) ∈ S(M_{L}) is given by

\begin{matrix}φ_{L}(γ): \{θ_{1}, …, θ_{n}\} & \underrightarrow{∼} \{θ_{1}, …, θ_{n}\} \\ θ_{i} & ⟼ γθ_{i} . \end{matrix}

So the map φ_{L}: Γ ⟶ S({θ_{1}, …, θ_{n}}) is just the morphism one knows from Galois theory.

(1.12) Definition. We call the image im(φ_{L})(⊆ Autk̄(L)) the Galois group of the k-algebra L and the fixed field k̄^{ker(φ_{L})} the splitting field of L.

(1.13) Example. If L|k is a separable field extension, then the splitting field of L is the Galois closure L̃ of any embedding of L in k̄, and φ_{L} induces the isomorphism Gal(L̃|k) \underrightarrow{∼} im(φ_{L}).

(1.14) Definition. A twisted cyclic structure on the separable k-algebra L is a Γ-invariant subgroup of Autk̄(L) which is a transitive cyclic subgroup of order n. A twisted cyclic extension of k of degree n is a pair (L, C) of a separable k-algebra L of degree n and a twisted cyclic structure C on L.

(1.15) Remark. A transitive cyclic subgroup C ⊆ Autk̄(L) of order n is a twisted cyclic structure on L if and only if im(φ_{L}) ⊆ N_{C}, i.e., if φ_{L}: Γ → Autk̄(L) factorizes in the form Γ ⟶^{τ_{e}} C⋊ Aut(C) ⟶^{σ_{e}} Autk̄(L) (for any e ∈ M_{L}). Here τ_{e} is uniquely defined by the choice of e; and τ_{e} = κ_{(f,id)} ∘ τ_{f(e)} for any f ∈ C, as one can see regarding (1.4).

(1.16) Remark. Let k̃ be some other separable algebraic closure of k and α: k̄ \underrightarrow{∼} k̃ a k-isomorphism, then we have the induced isomorphism

α_{∗}: Autk̄(L ⊗_{k} k̄) ⟶ Autk̃(L ⊗_{k} k̃)
f ⟼ α_{∗}(f) = (id_{L} ⊗α) ∘ f ∘ (id_{L} ⊗α)^{−1} .

which gives a twisted cyclic structure α_{∗}(C) on L if and only if C is one. In this case, i.e., if C is a twisted cyclic structure, the subgroup α_{∗}(C) is independent of the choice of the k-isomorphism α: Let β = α ∘ γ, (γ ∈ Γ) be a second k-isomorphism. Then β_{∗}(C) = (id_{L} ⊗α)(id_{L} ⊗γ)C(id_{L} ⊗γ)^{−1}(id_{L} ⊗α)^{−1} = (id_{L} ⊗α)^{γ}C(id_{L} ⊗α)^{−1} = α_{∗}(C), since ^{γ}C = C.

(1.17) Example. If n = 2, 3 there is a unique twisted cyclic structure on L because S_{2} and S_{3} have exactly one cyclic subgroup of order n, and this one is a normal subgroup.

(1.18) Remark. In general there is not only one twisted cyclic structure on the algebra L. For example if k̄ = k, then the absolute Galois group Γ = 1 is trivial, so is im(φ_{L}) and every transitive cyclic subgroup of S(M_{L}) ≅ S_{n} of order n is a twisted cyclic structure, and there are exactly (n − 1)!/φ(n) transitive cyclic subgroups of order n. (φ is the Euler phi function.)

(1.19) Definition. A twisted cyclic structure C on L is called a cyclic structure on L, if Γ acts trivially on C. Then the pair (L, C) is called a cyclic extension of k of degree n.

(1.20) Remark. A transitive cyclic subgroup C ⊆ Autk̄(L) of order n is a cyclic structure on L, if and only if im(φ_{L}) ⊆ C, because the only elements of Autk̄(L) ≅ S_{n} commuting with the elements of C are the elements of C. In other words: C is a cyclic structure, if and only if one gets the (unique) factorization of φ_{L} in the form

�_{τ_{e}} �  _{σ_{e}} �  ^{φ_{L}} �  C⋊ Aut(C) , C = C⋊{1}  Autk̄(L) Γ

where again e ∈ M_{L} is any primitive idempotent of L̄.

(1.21) Remark. If C is a cyclic structure on L, then one has

C = C^{Γ} ⊆ Autk̄(L)^{Γ} = Aut_{k}(L) .

Therefore, if L is a field, then from # Aut_{k}(L) ≤ n follows C = Aut_{k}(L). Hence L|k is a cyclic Galois extension with Galois group C.

(1.22) Example. If L|k is a cyclic Galois extension of degree n, then im(φ_{L}) = Gal(L|k) is a cyclic structure. It is the only cyclic structure on L, as we have seen in (1.21). In the following we will see that it is in fact the only twisted cyclic structure on L.

(1.23) Definition. A twisted cyclic structure C on L is called a dihedral structure on L, if Γ acts on C only as identity and taking the inverse, i.e., if one has the following factorization:

� ^{φ_{L}} �  ^{κ} � �  ℤ/2ℤ  Aut(C) N_{C} Γ

where the lower right hand arrow takes (0 mod 2ℤ) to id_{C} and (1 mod 2ℤ) to f ↦ f^{−1}.

(1.24) Remark. A transitive cyclic subgroup C ⊆ Autk̄(L) of order n is a dihedral structure on L, if and only if φ_{L}: Γ → Autk̄(L) factorizes in the form

� �  _{σ_{e}} �  ^{φ_{L}} � �  _{τ_{e}}  C⋊ Aut(C) , C⋊{±1}  Autk̄(L) Γ

where {±1} ⊆ (ℤ/nℤ)^{∗} = Aut(C) and e ∈ M_{L}.

(1.25) Example. For n = 3, 4, 6 every twisted cyclic structure on L is dihedral. In fact in this case is Aut(C) = (ℤ/nℤ)^{∗} = ℤ/2ℤ.

(1.26) Example. Let L|k be a separable field extension of degree n = 4 of the form L = k(\sqrt{a}, \sqrt{b}). Then L|k is Galois and Gal(L|k) ≅ ℤ/2ℤ × ℤ/2ℤ. There are three twisted cyclic structures on L (which are already dihedral by (1.25))—the 3 = (n − 1)!/φ(n) (transitive) cyclic subgroups of order 4. The action of Γ on M_{L} can be identified with the action of Γ on M := \{±\sqrt{a} ± \sqrt{b}\}. One observes that im(φ_{L}) ⊆ S(M) is the Klein Four Group

\begin{aligned}U : & = \{σ ∈ S(M) : σ^{2} = 1, sgn σ = 1\} \\ & = \{(rs)(tu) : \{r, s, t, u\} = M\} . \end{aligned}

The claim follows since the Klein Four Group U operates on the three cyclic subgroups by conjugation: One checks this for M = {1, 2, 3, 4}, e.g., consider first

((rs)(tu))(1 2 3 4)((rs)(tu))^{−1} = \{ \begin{matrix}(1 2 3 4) \\ (4 3 2 1) , \end{matrix}

with {r, s, t, u} = {1, 2, 3, 4}. The permutation (1 2 3 4) is a generator of one of the three transitive cyclic subgroups and conjugating it with an element of the group U yields the element itself or its inverse, i.e., U acts by conjugation on the group generated by (1 2 3 4). We get all the other generators of the transitive cyclic subgroups by conjugating (1 2 3 4) by any element of S(M) = S_{4}, and so the conjugation of these generators by an elements of U again yields the element itself or its inverse. Note that the Klein Four Group is a normal subgroup.

3. Uniqueness of Twisted Cyclic Structures

Let’s start with a “counter-example” for n = 6.

(1.27) Example. Let n = 6. In S_{6} we have the elements σ_{1} := (1 2 3 4 5 6), σ_{2} := (1 4 3 6 5 2) and σ_{3} = (1 6 3 2 5 4) of order 6, furthermore φ := (1 6)(2 5)(3 4) of order 2 and finally ψ := (1 3 5)(2 4 6) of order 3. Also we know for i = 1, 2, 3

\begin{aligned}ψ & = σ_{i}^{2} , \\ φσ_{i}φ^{−1} & = σ_{i}^{−1} , \\ φψφ^{−1} & = ψ^{2} . \end{aligned}

Obviously, the subgroup U := ⟨φ, ψ⟩ (of order 6) generated by φ and ψ acts on C_{i} := ⟨σ_{i}⟩ by conjugation, i.e., U ⊆ N_{C_{i}}. Notice that the three groups C_{1}, C_{2} and C_{3} are different! They are the only cyclic subgroups of order 6 with this last property: Let C̃ = ⟨σ̃⟩ ⊆ S_{6} be a transitive cyclic subgroup of order 6 with U ⊆ NC̃ ≅ C̃⋊ Aut(C̃). After some identification NC̃ = C̃⋊ Aut(C̃) we can write ψ = (c, f) ∈ C̃⋊ Aut(C̃). Since ψ = ψ^{4} and Aut(C̃) ≅ ℤ/2ℤ we have f = 1, i.e., ψ = c ∈ C̃. Hence by ord(ψ) = 3 we have ψ = σ̃^{2} or ψ = σ̃^{−2}. Since σ̃ and σ̃^{−1} are the only two generators of C̃, we may assume (1 3 5)(2 4 6) = ψ = σ̃^{2}. Therefore σ̃ has the form σ̃ = (1 a 3 b 5 c) with (a b c) = (2 4 6). There are three possibilities for substituting the variables a, b and c by the numbers 2, 4 and 6 (in the right order!), and they yield the elements σ_{1}, σ_{2} and σ_{3}. Now, set k = ℚ and L := ℚ(ζ_{3}, ^{3} \sqrt{2}) where ζ_{3} is a primitive third root of unity. Then L|k is Galois of degree 6 and the Galois group Gal(L|k) = im(φ_{L}) can easily be identified with U. So L has exactly three twisted cyclic structures. Note that L|k is a non-abelian Galois extension. This is interesting, since for cyclic Galois extensions of fields we have the following:

(1.28) Proposition. A cyclic Galois extension L|k of fields of degree n has a unique twisted cyclic structure—namely the cyclic structure mentioned in (1.22).

Proof: Let C := im(φ_{L}) be our cyclic (hence twisted cyclic) structure from (1.22), and we assume that C′ is another twisted cyclic structure on L. We choose generators: ⟨σ⟩ = C and ⟨τ⟩ = C′. Then by definition, im(φ_{L}) = C ⊆ N_{C′}, i.e., τστ^{−1} ∈ C. So there is an integer i with τστ^{−1} = σ^{i}. But σ^{i} = τστ^{−1} is also a generator of C, hence (i mod nℤ) ∈ (ℤ/nℤ)^{∗}. Chose an integer j such that ij ≡ 1 (mod n). Then τ^{j}στ^{−j} = σ^{ij} = σ. Since C is a transitive subgroup of S(M_{L}) we know by (1.1) that τ^{j} ∈ C, hence τ = τ^{ij} ∈ C. Therefore C = C′.

(1.29) Proposition. Assume n = m or n = 2m for some squarefree m. Let C be a twisted cyclic structure on a separable k-algebra L of degree n, and suppose that there exists an element π ∈ im(φ_{L}) of order n, then C = ⟨π⟩, and C is the only twisted cyclic structure on L.

Proof: After the identification Autk̄(L) = S(M_{L}) one can use (1.7), which says that πCπ^{−1} ⊆ C implies ⟨π⟩ = C.

(1.30) Proposition. Let L|k be a separable field extension of degree n. If the condition (n, φ(n)) = 1 holds—in this case n is squarefree—then there exists at most one twisted cyclic structure on L. (φ denotes the Euler phi function!)

Proof: Assume C is a twisted cyclic structure on L, by (1.29) it is enough to find an element π ∈ im(φ_{L}) of order n. Certainly there are elements π_{i} ∈ im(φ_{L}), i = 1, …, m of order p_{i}, where n = p_{1} ⋯ p_{m}. This follows from the fact that n|# im(φ_{L}). Identifying somehow N_{C} = C⋊ Aut(C) we can write π_{i} = (c_{i}, φ_{i}) for c_{i} ∈ C and φ_{i} ∈ Aut(C). We claim φ_{i} = id_{C} for all i. Otherwise we have 1 ≠ ord(φ_{i}) | ord(c_{1}, φ_{i}) = p_{i}, i.e., ord(φ_{i}) = p_{i}. But ord(φ_{i}) | φ(n), which is prime to n. Therefore π := π_{1} ⋯ π_{m} = (c_{1} ⋯ c_{m}, id_{C}) has order p_{1} ⋯ p_{m} = n.

(1.31) Remark. Example (1.27) shows that (1.30) doesn’t hold without the condition (n, φ(n)) = 1.

§ 2. C_{0}-Extensions

1. C_{0}-Extensions

Again, let k be a field and Γ = Gal(k̄|k) the absolute Galois group of k (for some separable algebraic closure k̄ of k). Now fix a natural number n and a continuous discrete Γ-module C_{0} which is isomorphic to ℤ/nℤ as a group. Let α: Γ → Aut(C_{0}) denote the morphism given by the action of Γ on C_{0}, i.e., α(γ)(c) = ^{γ}c for γ ∈ Γ, c ∈ C_{0}.

(2.1) Definition. Let L be a separable k-algebra of degree n. A C_{0}-structure on L is an injective Γ-equivariant morphism ρ: C_{0} ⟶ Autk̄(L) = S(M_{L}) of groups such that im(ρ) is a transitive cyclic subgroup of order n. A C_{0}-extension is a pair (L, ρ) of a separable k-algebra L of degree n and a C_{0}-structure ρ on L.

(2.2) Remark. If (L, ρ) is a C_{0}-extension, then im(ρ) ⊆ Autk̄(L) is a twisted cyclic structure on L and (L, im(ρ)) is a twisted cyclic extension of k of degree n. On the other hand: If (L, C) is a twisted cyclic extension, it gives rise to the C-extension

(L, ρ: C ⟶^{incl} Autk̄(L)).

(2.3) Example. Let us take a look at the case of a trivial Γ-module: If C_{0} is a trivial Γ-module, then im(ρ) is a cyclic structure on L for any C_{0}-extension (L, ρ). From (1.21) already follows im(ρ) ⊆ Aut_{k}(L) ⊆ Autk̄(L). So the C_{0}-structures on L are already the injective morphisms C_{0} ⟶ Aut_{k}(L) of groups. Therefore after fixing some generator c_{0} of C_{0}, there is a bijection

\{C_{0}-structures on L\} \underrightarrow{∼} \{c ∈ Aut_{k}(L) : ord(c) = n\}
ρ ⟼ ρ(c_{0}) .

Since a cyclic group of order n has exactly φ(n) = #(ℤ/nℤ)^{∗} generators (i.e., elements of order n), there is a φ(n) to 1 correspondence between C_{0}-structures and cyclic structures on L. In the case C_{0} = ℤ/nℤ we have the canonical generator (1 mod nℤ) and instead of writing (L, ρ) we may write (L, ρ(1 mod nℤ)).

2. Classification of C_{0}-Extensions

There is a classification of the C_{0}-extensions of k by the Galois cohomology group H^{1}(k, C_{0}) = H^{1}(Γ, C_{0}).

(2.4) Definition. A morphism f: (L, ρ) ⟶ (L′, ρ′) of C_{0}-extensions of k is a morphism f: L → L′ of k-algebras such that for every c ∈ C_{0}, the diagram

�_{ρ(c)} �  _{ρ′(c)} �  _{f̄} �  ^{f̄}  L̄′ L̄  L̄′ L̄

is commutative. Here f̄ is the morphism

f ⊗ idk̄: L̄ = L ⊗_{k} k̄ ⟶ L′ ⊗_{k} k̄ = L̄′ .

(2.5) Remark. In other words: L̄ and L̄′ become C_{0}-modules via ρ: C_{0} → Autk̄(L) and ρ′: C_{0} → Autk̄(L̄′) and f̄ is a morphism of C_{0}-modules.

(2.6) Remark. The morphisms f and f̄ are then already isomorphisms. In fact f̄ is bijective, since f̄|_{M_{L}}: M_{L} → M_{L′} is bijective: C_{0} acts free on the sets of primitive idempotents M_{L} and M_{L′}.

(2.7) Remark. Recall some well known facts about Galois cohomology; one has the following groups: Let

𝒵^{1}(Γ, C_{0}) = {continuous crossed homomorphisms Γ → C_{0}}

denote the group of the 1-cocycles of Γ with values in C_{0}. On this group we have (an equivalence relation given by) the normal subgroup

ℬ^{1}(Γ, C_{0}) = {f: Γ → C_{0} : there exists a c ∈ C_{0} such that f(γ) = γc − c for all γ ∈ Γ}

of the 1-coboundaries. Then we have the cohomology group H^{1}(Γ, C_{0}) = 𝒵^{1}(Γ, C_{0})/ℬ^{1}(Γ, C_{0}). Let

Hom_{cont}_{,Γ}(Γ, C_{0}⋊Γ) := {f ∈ Hom_{cont}(Γ, C_{0}⋊Γ) : pr_{2} ∘f = id_{Γ}} ,

where pr_{2}: C_{0}⋊Γ → Γ is the projection on the second factor. On this group we have the equivalence relation given by:

f ∼ g :⟺ f = κ_{(c,1_{Γ})} ∘ g for some c ∈ C_{0} .

Here κ_{(c,1_{Γ})} is the conjugation in C_{0}⋊Γ with the element (c, 1_{Γ}). Let

Hom_{cont}_{,α}(Γ, C_{0}⋊ Aut(C_{0})) := {f ∈ Hom_{cont}(Γ, C_{0}⋊ Aut(C_{0})) : pr_{2} ∘f = α} ,

where pr_{2}: C_{0}⋊ Aut(C_{0}) → Aut(C_{0}) is the projection on the second factor. On this group we have the equivalence relation given by:

f ∼ g :⟺ f = κ_{(c,id_{C_{0}} )} ∘ g for some c ∈ C_{0} .

Here κ_{(c,id_{C_{0}} )} is the conjugation in C_{0}⋊ Aut(C_{0}) with the element (c, id_{C_{0}}). Also, there are the identifications of these groups

𝒵^{1}(Γ, C_{0}) \underrightarrow{∼} Hom_{cont}_{,Γ}(Γ, C_{0}⋊Γ)
h ⟼ h̃ ,

where h̃(γ) = (h(γ), γ) and

𝒵^{1}(Γ, C_{0}) \underrightarrow{∼} Hom_{cont}_{,α}(Γ, C_{0}⋊ Aut(C_{0}))
h ⟼ h̃ ,

where h̃(γ) = (h(γ), α(γ)). They respect the equivalence relations, such that canonically

H^{1}(Γ, C_{0}) ≅ Hom_{cont}_{,Γ}(Γ, C_{0}⋊Γ)/∼
≅ Hom_{cont}_{,α}(Γ, C_{0}⋊ Aut(C_{0}))/∼ .

The theorem (2.10) states that there is an identification of isomorphism classes of C_{0}- extensions and the elements of H^{1}(Γ, C_{0}).

First, we want to assign a cohomology class [h_{(L,ρ)}] ∈ H^{1}(Γ, C_{0}) to a C_{0}-extension (L, ρ): Let C := im(ρ) be the twisted cyclic structure associated to ρ. Choose a primitive idempotent e ∈ M_{L} of L̄. Now ρ induces the isomorphism

ρ_{∗}: C_{0}⋊ Aut(C_{0}) \underrightarrow{∼} C⋊ Aut(C) ,

and there is the commutative diagram

�_{ρ_{∗}^{−1}∘τ_{e}} � ^{τ_{e}} �  ^{σ_{e}} �  _{ρ_{∗}} C_{0}⋊ Aut(C_{0})  Autk̄(L) = S(M_{L}) C⋊ Aut(C) Γ

of the factorization of φ_{L}. Now define the crossed homomorphism h_{(L,ρ)}: Γ → C_{0} by the equation h̃_{(L,ρ)} = ρ_{∗}^{−1} ∘τ_{e}, so

(1)  φ_{L} = σ_{e} ∘ ρ_{∗} ∘ h̃_{(L,ρ)} .

The cohomology class [h_{(L,ρ)}] ∈ H^{1}(Γ, C_{0}) is independent of the choice of the primitive idempotent e: Let c_{0} ∈ C_{0}, c := ρ(c_{0}) ∈ C, then by (1.15) τ_{e} = κ_{(c,id)} ∘ τ_{c(e)}, therefore

(ρ_{∗}^{−1} ∘ τ_{e}) = ρ_{∗}^{−1} ∘ κ_{(c_{0},id)} ∘ τ_{c(e)} = κ_{(c_{0},id)} ∘ (ρ_{∗}^{−1} ∘ τ_{c(e)}) .

Furthermore it is obvious that [h_{(L,ρ)}] = [h_{(L′,ρ′)}], if (L, ρ) ≅ (L′, ρ′).

(2.8) Lemma. For the 1-cocycle h_{(L,ρ)} which is given by h̃_{(L,ρ)} := ρ_{∗}^{−1} ∘ τ_{e} we have the explicit description: The diagram

�_{φ_{L}(−)(e)} � ^{h_{(L,ρ)}} �  _{ρ(−)(e)} M_{L} C_{0} Γ

is commutative, i.e., for any γ ∈ Γ,

ρ(h_{(L,ρ)}(γ))(e) = φ_{L}(γ)(e) = (id_{L} ⊗γ)(e) .

Proof: By definition

(ρ_{∗}^{−1} ∘ τ_{e})(γ) = h̃_{(L,ρ)}(γ) = (h_{(L,ρ)}(γ), α(γ))

for any γ ∈ Γ. Therefore

τ_{e}(γ) = (ρ(h_{(L,ρ)}(γ)), ρ_{∗}α(γ)) ,

where ρ_{∗}α(γ) = ρ ∘ α(γ) ∘ ρ^{−1}. Hence

φ_{L}(γ) = (σ_{e} ∘ τ_{e})(γ)
= σ_{e}(ρ(h_{(L,ρ)}(γ)), ρ_{∗}α(γ))
= ρ(h_{(L,ρ)}(γ)) ∘ s_{e}(ρ_{∗}α(γ)) ;

for s_{e} cf. (1.8). Evaluation in e ∈ M_{L} yields

φ_{L}(γ)(e) = ρ(h_{(L,ρ)}(γ)) ∘ s_{e}(ρ_{∗}α(γ))(1_{C}(e))
= ρ(h_{(L,ρ)}(γ))(^{ρ_{∗}α(γ)}1_{C}(e))
= ρ(h_{(L,ρ)}(γ))(e) .

and we are done.

For the other direction we want to assign a C_{0}-extension (L_{h}, ρ_{h}) to a 1-cocycle h: Let h ∈ 𝒵^{1}(Γ, C_{0}), and set M := C_{0}. Then C_{0} is canonically embedded in the symmetric group S(M) viz

\begin{matrix}l: C_{0} & ⟶ S(M) \\ c & ⟼ l_{c} , \end{matrix}

where l_{c} : M → M is the left multiplication with c. Denote C := im(l). Now let Γ act on M in the following way: The cocycle h gives a homomorphism

h̃: Γ ⟶ C_{0}⋊ Aut(C_{0}) ,

as described above. The embedding l yields the embedding (which maps isomorphically to the normalizer N_{C}—cf. (1.4))

\begin{matrix}σ_{1} ∘ l_{∗}: C_{0}⋊ Aut(C_{0}) & ⟶ S(M) \\ (c, φ) & ⟼ l_{c} ∘ φ , \end{matrix}

where l_{c} ∘ φ: M → M, m ↦ c · φ(m) and 1 = 1_{C_{0}} ∈ M. Composing these two maps one gets the (continuous) morphism

(2)  \begin{matrix}ϕ_{h}: Γ & ⟶ S(M) \\ γ & ⟼ l_{h(γ)} ∘ α(γ) , \end{matrix}

which gives an action of Γ on M. Now let Γ act semilinear on \overline{L_{h}} := ⨁_{m∈M} k̄ · m by

\begin{matrix}Γ × ⨁_{m∈M} k̄ · m & ⟶ ⨁_{m∈M} k̄ · m \\ (γ, (x_{m} · m)) & ⟼ (γ(x_{m}) · ϕ_{h}(γ)(m)) \end{matrix}

for x_{m} ∈ k̄. Then L_{h} := \overline{L_{h}}^{Γ} is a separable k-algebra of degree n. (cf. (18.1) in [BI])

(2.9) Remark. Because of L_{h} ⊗_{k} k̄ = \overline{L_{h}}, we can identify M with M_{L_{h}}, and then one has the equality

(3)  φ_{L_{h}} = ϕ_{h}: Γ ⟶ S(M_{L_{h}}) = S(M) ,

since both morphisms describe the Galois action of Γ on the set of the primitive idempotents M = M_{L_{h}}.

Finally we define the C_{0}-structure ρ_{h} as the injection l from above:

(4)  \begin{matrix}ρ_{h}: C_{0} & ⟶ S(M_{L_{h}}) = S(M) \\ c & ⟼ l_{c} . \end{matrix}

We have to prove that ρ_{h} is Γ-equivariant, i.e., for every γ ∈ Γ, the diagram

�_{α(γ)} �  _{κ∘φ_{L_{h}} (γ)=κ_{ϕ_{h}(γ)}} �  _{ρ_{h}} �  ^{ρ_{h}}  S(M) C_{0}  S(M) C_{0}

is commutative. For every c ∈ C_{0} we have to verify

κ_{ϕ_{h}(γ)}(ρ_{h}(c)) = ρ_{h}(α(γ)(c)) ,

i.e., ϕ_{h}(γ) ∘ ρ_{h}(c) = ρ_{h}(^{γ}c) ∘ ϕ_{h}(γ) in S(M). Indeed, evaluation in m ∈ M gives

ϕ_{h}(γ) ∘ ρ_{h}(c)(m) =^{(2),(4)} l_{h(γ)} ∘ α(γ) ∘ l_{c}(m)

= h(γ) ·^{γ}(c · m)

= ^{γ}c · h(γ) ·^{γ}m

= l_{(^{γ}c)} · (h(γ) ·^{γ}m)

=^{(2),(4)} ρ_{h}(^{γ}c) ∘ ϕ_{h}(γ)(m) .

If [h] = [h′], then (L_{h}, ρ_{h}) ≅ (L_{h′}, ρ_{h′}). One can check this directly as in the previous case or it follows with the proof of (2.10).

Now we can state and prove the Theorem.

(2.10) Theorem. There is a canonical bijection

\{C_{0}-extensions of k\}/≅ \underrightarrow{∼} H^{1}(Γ, C_{0})
[(L, ρ)] ⟼ [h_{(L,ρ)}]
[(L_{h}, ρ_{h})] ⟵ [h]

between the set of isomorphism classes of C_{0}-extensions and the first Galois cohomology group.

Proof: We have to prove that the two maps are inverse to each other.

Let h ∈ 𝒵^{1}(Γ, C_{0}) be a 1-cocycle and write ℒ_{h} for (L_{h}, ρ_{h}) We will show: h = h_{ℒ_{h}} . But it is enough to show h̃ = h̃_{ℒ_{h}}: Γ ⟶ C_{0}⋊ Aut(C_{0}). It is even enough to show

σ_{1} ∘ l_{∗} ∘ h̃ = σ_{1} ∘ l_{∗} ∘ h̃_{ℒ_{h}}: Γ ⟶ S(M) = S(M_{L_{h}}) ,

where 1 = 1_{C_{0}} ∈ M. But that follows from

σ_{1} ∘ l_{∗} ∘ h̃_{ℒ_{h}} = σ_{1} ∘ (ρ_{h})_{∗} ∘ h̃_{ℒ_{h}}

=^{(1)} φ_{L_{h}}

=^{(3)} ϕ_{h}

=^{(2)} σ_{1} ∘ l_{∗} ∘ h̃ .

Now let ℒ = (L, ρ) be a C_{0}-extension. h_{ℒ} is defined by the equation

φ_{L} = σ_{e} ∘ ρ_{∗} ∘ h̃_{ℒ}

for some primitive idempotent e ∈ M_{L}. We have to show

(L, ρ) ≅ (L_{h_{ℒ}} , ρ_{h_{ℒ}} ) ,

but that means just

L̄ ≅ \overline{L_{h_{ℒ}}}

as k̄-algebra, Γ-module and C_{0}-module (as described in (2.5)). The bijective map

\begin{matrix}r: M & ⟶ M_{L} \\ c & ⟼ ρ(c)(e) \end{matrix}

(remember M = C_{0}!) gives an isomorphism

\begin{matrix}r: \overline{L_{h_{ℒ}}} = ⨁_{c∈M} k̄ · c & \underrightarrow{∼} ⨁_{m∈M_{L}} k̄ · m = L̄ \\ (x_{c} · c) & ⟼ (x_{c} · r(c)) \end{matrix}

of k̄-algebras. But r is also Γ- and C_{0}-equivariant: First we claim that the diagram

�_{ρ_{∗}} �  ^{σ_{1}} �  _{σ_{e}} �  _{r_{∗}} S(M_{L}) C⋊ Aut(C)  S(M) = S(M_{L_{h_{ℒ}}} ) C_{0}⋊ Aut(C_{0})

with C := im(ρ) and 1 = 1_{C_{0}} ∈ M, is commutative, i.e., for (c, φ) ∈ C_{0}⋊ Aut(C_{0}), we have the identity

σ_{e}(ρ(c), ρ_{∗}(φ)) = r_{∗}(σ_{1}(c, φ)) .

To show this, evaluate the equation in an element m = r(d) ∈ M_{L}, d ∈ M = C_{0}, and the claim follows from

σ_{e}(ρ(c), ρ_{∗}(φ))(m) = σ_{e}(ρ(c), ρ_{∗}(φ))(ρ(d)(e))
= ρ(c) ·^{φ}(ρ(d))(e)
= ρ(c ·^{φ}d)(e)
= r(c ·^{φ}d)
= r(c ·^{φ}(r^{−1}(m)))
= r(σ_{1}(c, φ)(r^{−1}(m)))
= r_{∗}(σ_{1}(c, φ))(m) .

Composing this diagram in the upper left corner with either C_{0} ↪ C_{0}⋊ Aut(C_{0}) and h̃_{ℒ}: Γ → C_{0}⋊ Aut(C_{0}) one gets the commutative diagrams

� _{ρ_{h_{ℒ}}} � _{ρ} �  _{r_{∗}} S(M_{L}) S(M) C_{0}

and � _{ϕ_{h_{ℒ}}} � _{φ_{L}} �  _{r_{∗}} S(M_{L}) . S(M) Γ

But this is just the C_{0}- and Γ-equivariance of r, and we are done.

(2.11) Notation. If (L, C) is a twisted cyclic extension of k, then—by virtue of

(2.10)—the C-extension (L, ρ: C ⟶^{incl} Autk̄(L)) maps to [h_{(L,ρ)}] ∈ H^{1}(Γ, C). We will denote this cohomology class by [L, C] ∈ H^{1}(Γ, C).

(2.12) Remark. If (L, ρ) is a C_{0}-extension and C := im(ρ), then ρ induces an isomorphism

ρ_{∗}: H^{1}(Γ, C_{0}) \underrightarrow{∼} H^{1}(Γ, C) ,

which maps [h_{(L,ρ)}] to [L, C].

3. μ_{n}-Extensions and Kummer Extensions

Now we take a look at the case where C_{0} is μ_{n} := μ_{n}(k̄), the Γ-module of the n-th roots of unity. We assume that the characteristic of k does not divide n, so that μ_{n} is a cyclic group of order n. Let (L, ρ) be a μ_{n}-extension of k, i.e., ρ: μ_{n} → Autk̄(L) is Γ-equivariant. That means

ρ(^{γ}ζ) = κ_{φ_{L}(γ)}(ρ(ζ))
= φ_{L}(γ) ∘ ρ(ζ) ∘ φ_{L}(γ)^{−1}
= (id_{L} ⊗γ) ∘ ρ(ζ) ∘ (id_{L} ⊗γ)^{−1}

for ζ ∈ μ_{n} and γ ∈ Γ (cf. (1.10)), therefore

(5)  ρ(^{γ}ζ) ∘ (id_{L} ⊗γ) = (id_{L} ⊗γ) ∘ ρ(ζ) ,

which takes place in Aut_{k}(L̄) = Aut_{k}(L ⊗_{k} k̄). Define

\overline{K}_{ρ} := \{x ∈ L̄ : ρ(ζ)(x) = ζ · x, for all ζ ∈ μ_{n}\} .

Of course, if one chooses a primitive n-th root of unity ξ ∈ μ_{n}, one has

\overline{K}_{ρ} = \{x ∈ L̄ : ρ(ξ)(x) = ξ · x, \} .

(2.13) Claim. The subset \overline{K}_{ρ}(⊆ L̄) is Γ-invariant with respect to the semilinear action of Γ on L̄ (cf. the beginning of §1, 2.)

Proof: Let x ∈ \overline{K}_{ρ} and γ ∈ Γ. Then for any ζ ∈ μ_{n},

ρ(ζ)((id_{L} ⊗γ)(x)) = (ρ(ζ) ∘ (id_{L} ⊗γ))(x)

=^{(5)} ((id_{L} ⊗γ) ∘ ρ(^{γ^{−1}}ζ))(x)

= (id_{L} ⊗γ)(^{γ^{−1}}ζ · x)

= ζ · (id_{L} ⊗γ)(x) .

and that means (id_{L} ⊗γ)(x) ∈ \overline{K}_{ρ}.

(2.14) Claim. \overline{K}_{ρ} is a one dimensional k̄-vector space. The Γ-semilinear action on \overline{K}_{ρ}

gives a k-vector space K_{ρ} := \overline{K}_{ρ}^{Γ} ⊆ L̄^{Γ} = L of k-dimension one.

Proof: \overline{K}_{ρ} is obviously a vector space, so all we have to prove is dimk̄ \overline{K}_{ρ} = 1. But that is clear, regarding the following explicit description: Since

L̄ = ⨁_{ν=0}^{n−1} k̄ · π^{ν}(e) ,

where e ∈ M_{L} is any primitive idempotent and π := ρ(ξ) ∈ S(M_{L}) = Autk̄(L) for some fixed primitive n-th root of unity ξ ∈ μ_{n}, it is obvious that

\overline{K}_{ρ} = \{∑_{ν=0}^{n−1} αξ^{−ν} · π^{ν}(e) : α ∈ k̄\} ,

because for any y := ∑_{ν=0}^{n−1} α_{ν}ξ^{−ν} ·π^{ν}(e) ∈ L̄, one has ρ(ξ)(y) = ξ·∑_{ν=0}^{n−1} α_{ν−1}ξ^{−ν} ·π^{ν}(e), where one has to read the index ν modulo n. For the last claim cf. again (18.1) in [BI].

(2.15) Claim. If x ∈ K_{ρ} is non-zero, i.e., K_{ρ} = k · x, then one has x^{n} ∈ k^{∗} = k − {0} and L = k(x).

Proof: Let α ∈ k̄ such that x = ∑_{ν=0}^{n−1} αξ^{−ν} · π^{ν}(e) ∈ K_{ρ} − {0}. Then

x^{n} = ∑_{ν=0}^{n−1} α^{n} · π^{ν}(e) = α^{n} · ∑_{ν=0}^{n−1} π^{ν}(e) = α^{n} · 1 .

But of course x^{n} = α^{n} is Γ-invariant, i.e., α^{n} = x^{n} ∈ k̄^{Γ} = k. Because the Vandermonde determinant

det(ξ^{νω})_{ν,ω=0,...,n−1} = ± ∏_{0≤ν<ω≤n}(ξ^{ν} − ξ^{ω}) ≠ 0

is non-zero, the set {1, x, x^{2}, …, x^{n−1}} is k̄-linearly independent and therefore a basis of L̄|k̄ and (since x ∈ L) of L|k.

This claim gives rise to the following

(2.16) Definition. A Kummer structure on a separable k-algebra L of degree n is a one dimensional k-vector subspace K ⊆ L such that for some non-zero element x of K, one has the following properties

(i) x^{n} ∈ k^{∗}, (ii) the morphism of k-algebras k[t]/(t^{n} − x^{n}) → L , t ↦ x is an isomorphism. (In this case these properties hold for any non-zero element x of K.) A Kummer extension is a pair (L, K) of a separable k-algebra L of degree n and a Kummer structure K on L.

(2.17) Proposition. For a separable k-algebra L of degree n there is a bijection between the set of μ_{n}-structures on L and the set of Kummer structures on L, given by

\{Kummer structures on L\} \underrightarrow{∼} \{μ_{n}-structures on L\}
K ⟼ ρ_{K}
K_{ρ} ⟵ ρ ,

where ρ_{K}: μ_{n} → Autk̄(L) is given by ρ_{K}(ζ)(x) = ζ · x for some non-zero element x of K.

Proof: Note that the definition of ρ_{K} is independent of the choice of x and since L̄ = k̄(x), the morphism ρ_{K} is well defined.—The two maps are inverse to each other: First we prove K = K_{ρ_{K}}: Let x ∈ K be non-zero, then

K_{ρ_{K}} = {y ∈ L : ρ_{K}(ζ)(y) = ζ · y} ⊇ k · x = K ,

and both are one dimensional vector spaces. Now for ρ = ρ_{K_{ρ}}: ρ(ζ)(x) = ζ · x = ρ_{K_{ρ}}(ζ)(x) for some x ∈ K_{ρ}.

(2.18) Definition. Two Kummer extension (L, K) and (L′, K′) of k are said to be isomorphic, if there exists an isomorphism f: L \underrightarrow{∼} L′ of k-algebras such that f(K) = K′.

(2.19) Remark. Of course the bijection of (2.17) respects this notion of being isomorphic, i.e.,

(L, K) ≅ (L′, K′) ⟺ (L, ρ_{K}) ≅ (L′, ρ_{K′})

Therefore (2.17) gives us a bijection

{Kummer extensions of degree n}/≅ ⟶ {μ_{n}-extensions of k}/≅
[(L, K)] ⟼ [(L, ρ_{K})]
[(L, K_{ρ})] ⟵ [(L, ρ)] ,

between the isomorphism classes of Kummer extensions and the isomorphism classes of μ_{n}-extensions.

Taking the long exact cohomology sequence to the short exact sequence

1 ⟶ μ_{n} ⟶ k̄^{∗} ⟶^{n} k̄^{∗} ⟶ 1

of Γ-modules and using Hilbert’s Theorem 90 we get the well known isomorphism

\begin{matrix}∂: k^{∗}/k^{∗n} & \underrightarrow{∼} H^{1}(Γ, μ_{n}) \\ a · k^{∗n} & ⟼ [h_{α}] , \end{matrix}

where h_{α}: Γ → μ_{n} is given by h_{α}(γ) = \frac{α}{^{γ}α} for some root α ∈ k̄^{∗} of the polynomial t^{n} −a. With this interpretation of the cohomology group we can describe the composition of the maps mentioned in (2.19) and (2.10):

(2.20) Proposition. The composition of the two maps of (2.19) and (2.10)

{Kummer extensions of degree n}/≅ ⟶ {μ_{n}-extensions}/≅

⟶ H^{1}(Γ, μ_{n})

⟶^{∂^{−1}} k^{∗}/k^{∗n}

is given by the application

[(L, K)] ⟼ K^{∗n} ,

where K^{∗n} = x^{n} · k^{∗n}, if K = k · x.

Proof: We prove this by going a “step back” via ∂, i.e., we show [h_{(L,ρ_{K})}] = [h_{α}], where α is a solution of t^{n} − x^{n} = 0 in k̄^{∗}. (Note first that x^{n} ∈ k^{∗} and second that α = 1 ⊗ α ∈ k̄ = k ⊗_{k} k̄ but x = x ⊗ 1 ∈ L = L ⊗_{k} k.) Going a second step back, we are actually proving the equality

(*)  φ_{L} = σ_{e} ∘ ρ_{K∗} ∘ h̃_{α}: Γ ⟶ S(M_{L}) = Autk̄(L) ,

for a suitable e ∈ M_{L}—cf. formula (1). First observe that

e = \frac{1}{n} ∑_{ν=0}^{n−1} (\frac{x}{α})^{ν}

is an idempotent, since

e^{2} = \frac{1}{n^{2}} ∑_{ν,ω=0}^{n−1} (\frac{x}{α})^{ν+ω} = e .

It is even a primitive idempotent because

∑_{ζ∈μ_{n}} ρ_{K∗}(ζ)(e) = \frac{1}{n} ∑_{ζ∈μ_{n}} ∑_{ν=0}^{n−1} (\frac{ζ · x}{α})^{ν}

= \frac{1}{n} ∑_{ν=0}^{n−1} ∑_{ζ∈μ_{n}} ζ^{ν} · (\frac{x}{α})^{ν}

= \frac{1}{n} ∑_{ζ∈μ_{n}} ζ^{0} · (\frac{x}{α})^{0}

= 1 .

—Note that ρ_{K}(ζ)(e) = \frac{1}{n} ∑_{ν=0}^{n−1} (\frac{ζ·x}{α})^{ν}. Furthermore

L̄ = ⨁_{ν=0}^{n−1} k̄ · x^{ν} (= ⨁_{ν=0}^{n−1} x^{ν}(k ⊗_{k} k̄)) .

Now evaluation of (*) in γ ∈ Γ and then in the element (for an arbitrary ζ ∈ μ_{n})

e′ := ρ_{K}(ζ)(e) = \frac{1}{n} ∑_{ν=0}^{n−1} (\frac{ζ · x}{α})^{ν} ∈ M_{L}

§ 2. C_{0}-extensions

of the primitive idempotents, gives on the left side

φ_{L}(γ)(e′) = (id_{L} ⊗γ)(e′)

= (id_{L} ⊗γ) (\frac{1}{n} ∑_{ν=0}^{n−1} (\frac{ζ · x}{α})^{ν})

= \frac{1}{n} ∑_{ν=0}^{n−1} (\frac{^{γ}ζ · x}{^{γ}α})^{ν}

On the other side of (*) we get

((σ_{e} ∘ ρ_{K∗} ∘ h̃_{α})(γ))(e′) = σ_{e} (ρ_{K} (\frac{α}{^{γ}α}) , ρ_{K∗}(γ)) (ρ_{K}(ζ)(e))

= ρ_{K} (\frac{α}{^{γ}α}) (ρ_{K}(^{γ}ζ)(e))

= ρ_{K} (\frac{α}{^{γ}α}^{γ}ζ) \frac{1}{n} ∑_{ν=0}^{n−1} (\frac{x}{α})^{ν}

= \frac{1}{n} ∑_{ν=0}^{n−1} (^{γ}ζ \frac{α}{^{γ}α} \frac{x}{α})^{ν}

= \frac{1}{n} ∑_{ν=0}^{n−1} (\frac{^{γ}ζ · x}{^{γ}α})^{ν}

This proves the claim.

Chapter II

Twisted Cyclic Algebras

If we assume that the ground field k, of characteristic prime to n, contains the n-th roots of unity, then a cyclic algebra A of degree n is a k-algebra of the form

(a, b)_{ζ} := ⟨X, Y : X^{n} = a, X^{n} = b, XY = ζY X⟩

—the k-algebra generated by the two variables X and Y with the three relations X^{n}−a, Y^{n} − b and XY − ζY X. Here, a, b ∈ k^{∗}, and ζ is a primitive n-th root of unity. After the base extension k̄|k we may assume a = b = 1 and we can identify (a, b)_{ζ} with M_{n}(k̄) in a way that X = diag(ζ, ζ^{2}, …, ζ^{n}) and Y is the invertible matrix which maps the canonical basis vector e_{i} of k̄^{n} to e_{i+1}. In this chapter, we consider the twisted forms of this kind of algebras, i.e., algebras, which become, after base extension, isomorphic to a cyclic algebra, together with two generators fulfilling the three relations from above.

§ 3. Twisted Cyclic Algebras

In this section again fix a field k and some separable algebraic closure k̄, and let be Γ = Gal(k̄|k). Choose a positive integer n ≥ 2 relatively prime to the characteristic of k. We abbreviate μ_{n} = μ_{n}(k̄).

1. Twisted Cyclic Algebras

A twisted cyclic algebra is a twisted version of the standard triple (A_{0}, L_{0}, K_{0}), which is defined as follows: A_{0} := M_{n}(k) is the k-algebra of the n × n matrices, L_{0} and K_{0} are two commutative subalgebras: L_{0} is the subalgebra of the diagonal matrices and K_{0} is the k-algebra generated by the matrix Y_{0}:

K_{0} = k[Y_{0}] = ⨁_{ν=0}^{n−1} k · Y_{0}^{ν} , where Y_{0} := ( \begin{matrix}0 & & & & & 0 & 1 \\ 1 & 0 & & & & & 0 \\ 0 & 1 & 0 \\ & ... & ... & ... \\ & & ... & ... & 0 \\ & & & ... & 1 & 0 \\ & & & & 0 & 1 & 0 \end{matrix} )

is the matrix which maps e_{i} to e_{i+1}. L_{0} and K_{0} are maximal commutative k-subalgebras of A_{0} (of degree n).

§ 3. Twisted cyclic algebras

If μ_{n} ⊆ k then in M_{n}(k̄) define for any ζ ∈ μ_{n},

X_{0}(ζ) := ( \begin{matrix}ζ^{1} \\ & ζ^{2} \\ & & . & . & . \\ & & & & & ζ^{n} \end{matrix} ) .

(3.1) Remark. If μ_{n} ⊆ k, then L_{0} = k[X_{0}], where X_{0} := X_{0}(ζ) for any primitive n-th root of unity ζ ∈ μ_{n}.

(3.2) Notation. For k-algebras A, L, K, A_{0}, L_{0}, K_{0}, … we denote with Ā, L̄, … the k̄-algebras A ⊗_{k} k̄, L ⊗_{k} k̄, …

(3.3) Definition. A twisted cyclic k-algebra (of degree n) is a triple (A, L, K), where A is a central simple k-algebra of degree n, L and K are commutative ksubalgebras such that (Ā, L̄, K̄) ≅ (Ā_{0}, L̄_{0}, K̄_{0}) over k̄, i.e., there exists an isomorphism

α: Ā ⟶ Ā_{0} = M_{n}(k̄)

of k̄-algebras such that α(L̄) = L̄_{0} and α(K̄) = K̄_{0}. The pair (L, K) is called a (twisted cyclic) decomposition of A.

(3.4) Remark. If (A, L, K) is a twisted cyclic k-algebra and k′ is any separable algebraic field extension of k, then (A, L, K)_{k′} := (A ⊗_{k} k′, L ⊗_{k} k′, K ⊗_{k} k′) is a twisted cyclic k′-algebra and vice versa. In §6 we will see that the fact holds even for algebraic field extensions which are not necessarily separable.

(3.5) Lemma. If (A, K, L) is a twisted cyclic k-algebra, then the multiplication map L ⊗_{k} K → A, x ⊗ y ↦ xy is an isomorphism of k-modules.

Proof: Of course it is enough to prove this after base change with k̄|k. Because of the commutative diagram

�_{α} �  _{α} � �  Ā_{0} L̄_{0} ⊗k̄ K̄_{0}  Ā L̄ ⊗k̄ K̄

of k̄-modules, where α is defined as in (3.3) and the horizontal arrows are the multiplication maps, all we have to consider is the case (Ā_{0}, L̄_{0}, K̄_{0}). But it is easy to see that

Ā_{0} = ⨁_{ν=0}^{n−1} L̄_{0} · Y_{0}^{ν}

and so the surjectivity of the map is clear, hence also the injectivity for dimension reasons.

(3.6) Remark. Since L̄ ≅ L̄_{0} ≅ k̄^{n} and K̄ ≅ K̄_{0} ≅ k̄^{n} as k̄-algebras, L and K are separable k-algebras of degree n for any twisted cyclic k-algebra (A, L, K) of degree n.

(3.7) Lemma. If (A, L, K) is a twisted cyclic k-algebra, then (A, K, L) is also one.

Proof: We only have to show it for the case (A_{0}, L_{0}, K_{0}), since then

(Ā, K̄, L̄) ≅ (Ā_{0}, K̄_{0}, L̄_{0}) ≅ (Ā_{0}, L̄_{0}, K̄_{0}) .

So let ζ ∈ μ_{n} be a primitive n-th root of unity, L̄_{0} = k̄[X_{0}], K̄_{0} = k̄[Y_{0}], where X_{0} := X_{0}(ζ). For the inner automorphism

κ_{Z_{0}}: Ā_{0} = M_{n}(k̄) \underrightarrow{∼} Ā_{0} = M_{n}(k̄)

given by the conjugation with the matrix Z_{0} := Z_{0}(ζ) := (ζ^{−ij})_{i,j=1,...,n} ∈ M_{n}(k̄) we have κ_{Z_{0}}(L̄_{0}) = K̄_{0} and κ_{Z_{0}}(K̄_{0}) = L̄_{0}; more precisely: One checks that Z_{0}X_{0}Z_{0}^{−1} = Y_{0} and Z_{0}Y_{0}Z_{0}^{−1} = X_{0}^{−1}. (The second one follows from the first by taking transposes.)

2. Pairs of Twisted Cyclic Extensions associated to Twisted Cyclic Algebras

Define for any twisted cyclic k-algebra (A, L, K) of degree n,

K_{(L,K)} := {X ∈ L̄ : X^{n} = 1, κ_{X}(K̄) = K̄} ,

—the elements of L̄^{∗} of order dividing n which act on K̄ by conjugation. Also

K_{(K,L)} := {Y ∈ K̄ : Y^{n} = 1, κ_{Y} (L̄) = L̄} .

For these sets, we have the following obvious properties:

(3.8) Proposition. Let (A, L, K) be a twisted cyclic k-algebra of degree n. Then (i) K_{(L,K)} is a subgroup of L̄^{∗}. (ii) K_{(L,K)} is Γ-invariant. (iii) μ_{n} ⊆ K_{(L,K)}, where we view (μ_{n} ⊆)k̄ ⊆ L̄ via k̄ ↪ L̄, x ↦ 1 ⊗ x. (iv) The isomorphism α: Ā \underrightarrow{∼} Ā_{0} of k̄-algebras of (3.3) induces an isomorphism

α: K_{(L,K)} \underrightarrow{∼} K_{(L_{0},K_{0})} of groups. The same holds for K_{(K,L)}.

(3.9) Lemma. For the twisted cyclic k-algebra (A_{0}, L_{0}, K_{0}) we have the explicit description:

K_{(L_{0},K_{0})} = {ζX_{0}(ζ′) : ζ, ζ′ ∈ μ_{n}}
= {ξ^{i}X_{0}(ξ)^{j} : i, j = 0, …, n − 1}
K_{(K_{0},L_{0})} = {ζY_{0}^{j} : ζ ∈ μ_{n}, j = 0, …, n − 1}
= {ξ^{i}Y_{0}^{j} : i, j = 0, …, n − 1} ,

where ξ ∈ μ_{n} is any primitive n-th root of unity.

Proof: Let X ∈ K_{(L_{0},K_{0})}, then XK̄_{0}X^{−1} = K̄_{0}, especially

XY_{0}X^{−1} = a_{0}Y_{0}^{0} + a_{1}Y_{0}^{1} + ⋯ + a_{n−1}Y_{0}^{n−1}

for some a_{i} ∈ k̄. Since X is diagonal we must have a_{i} = 0 for i ≠ 1: XY_{0}X^{−1} = aY_{0} for some a ∈ k̄^{∗}, i.e., X = aY_{0}XY_{0}^{−1}, and one observes that X has the form

X = ( \begin{matrix}α_{1} \\ & . & . & . \\ & & & & α_{n} \end{matrix} ) ,

where α_{i} ∈ k̄^{∗} and α_{i+1} = aα_{i} for all i (seen modulo n), i.e., α_{i} = a^{i}α for α := α_{n}, hence a^{n} = 1 (thus a = ζ ∈ μ_{n}) and X = αX_{0}(ζ). Finally X^{n} = α^{n} = 1, and that means α ∈ μ_{n}. On the other hand the elements ζX_{0}(ζ′) obviously lie in K_{(L_{0},K_{0})} The second part follows from the first if one uses the isomorphism

\begin{matrix}κ_{Z_{0}(ξ)}: K_{(L_{0},K_{0})} & \underrightarrow{∼} K_{(K_{0},L_{0})} \\ ξ^{i}X_{0}(ξ)^{j} & ⟼ ξ^{i}Y_{0}^{j} . \end{matrix}

mentioned in (3.7), and (3.8)(iv).

(3.10) Proposition. For any twisted cyclic k-algebra (A, L, K) of degree n, the groups K_{(L,K)}/μ_{n} and K_{(K,L)}/μ_{n} are cyclic groups of order n and they are Γ-modules.

Proof: The last part is clear. The first part is certainly true for (A_{0}, L_{0}, K_{0}) as one can see at the explicit description given in (3.9). But the induced isomorphisms (of groups)

\begin{aligned}α: K_{(L,K)}/μ_{n} & \underrightarrow{∼} K_{(L_{0},K_{0})}/μ_{n} \\ α: K_{(K,L)}/μ_{n} & \underrightarrow{∼} K_{(K_{0},L_{0})}/μ_{n} \end{aligned}

show the general case.

(3.11) Remark. We immediately get for any two generators (of the cyclic groups)

[X] ∈ K_{(L,K)}/μ_{n} and [Y] ∈ K_{(K,L)}/μ_{n}:

K_{(L,K)} = {ζX^{j} : ζ ∈ μ_{n}, j = 0, …, n − 1}

and

K_{(K,L)} = {ζY^{j} : ζ ∈ μ_{n}, j = 0, …, n − 1} .

X and Y are any representatives of the classes [X] and [Y].

(3.12) Corollary. For a twisted cyclic k-algebra (A, L, K) of degree n let [X] ∈ K_{(L,K)}/μ_{n} and [Y] ∈ K_{(K,L)}/μ_{n} be generators of the cyclic groups. Then

L̄ = k̄[X] = ⨁_{ν=0}^{n−1} k̄X^{ν} and K̄ = k̄[Y] = ⨁_{ν=0}^{n−1} k̄Y^{ν} .

Proof: Again, we only have to prove this for our standard triple (A_{0}, L_{0}, K_{0}). In this situation it is clear that

X = ξX_{0}(ζ) and Y = ξ′Y_{0}^{j}

for ζ ∈ μ_{n} primitive and ξ, ξ′ ∈ μ_{n}; and the claim is obvious.

By definition, K_{(K,L)} resp. K_{(L,K)} are sets of elements of K̄ resp. L̄ which act on L̄ resp. K̄ by conjugation. Since μ_{n} lies in the center of Ā we get the maps

\begin{matrix}ρ_{L}: K_{(K,L)}/μ_{n} & ⟶ Autk̄(L) \\ [Y] & ⟼ κ_{Y} \end{matrix}

and

\begin{matrix}ρ_{K}: K_{(L,K)}/μ_{n} & ⟶ Autk̄(K̄) \\ [X] & ⟼ κ_{X} . \end{matrix}

About these maps one knows:

(3.13) Remark. Let (A, K, L) and (A′, L′, K′) be twisted cyclic k-algebras of degree n. Let α: Ā \underrightarrow{∼} Ā′ be an isomorphism of k̄-algebras such that α(L̄) = L̄′ and α(K̄) = K̄′, then we have the commutative diagram

�_{α} �  _{α} �  _{ρ_{L′}} �  ^{ρ_{L}} Autk̄(L̄′) . K_{(K′,L′)}/μ_{n}  Autk̄(L) K_{(K,L)}/μ_{n}

The same holds for ρ_{K} and ρ_{K′}.

(3.14) Proposition. Let (A, L, K) be a twisted cyclic k-algebra. Then (L, ρ_{L}) and (K, ρ_{K}) are K_{(K,L)}/μ_{n}- and K_{(L,K)}/μ_{n}-extensions.

Proof: The map ρ_{L} is Γ-equivariant since for any Y ∈ K_{(K,L)} and γ ∈ Γ

κ_{(id_{K} ⊗γ)Y} = φ_{L}(γ) ∘ κ_{Y} ∘ φ_{L}(γ)^{−1} = (id_{L} ⊗γ) ∘ κ_{Y} ∘ (id_{L} ⊗γ)^{−1} .

In fact we evaluate in any X ∈ L̄:

κ_{(id_{K} ⊗γ)Y} (X) = ((id_{A} ⊗γ)Y )X((id_{A} ⊗γ)Y )^{−1}
= (id_{L} ⊗γ)(Y ((id_{L} ⊗γ)^{−1}X)Y^{−1})
= ((id_{L} ⊗γ) ∘ κ_{Y} ∘ (id_{L} ⊗γ)^{−1})(X) .

ρ_{L} is injective: Regarding (3.13) it is enough to consider (A_{0}, L_{0}, K_{0}). Therefore Y_{0} is a generator of K_{(K_{0},L_{0})}/μ_{n}. Now if ρ_{L}([Y_{0}]^{j}) = idL̄, then X_{0} = Y_{0}^{j}X_{0}Y_{0}^{−j} = ζ^{−j}X_{0} (X_{0} = X_{0}(ζ) and ζ some primitive n-th root of unity) therefore n|j. It is easy to see that im(ρ_{L}) is a transitive subgroup.—Analogously one proves the same fact about about ρ_{K}.

(3.15) Proposition. For a twisted cyclic k-algebra (A, L, K) one has the isomorphism

\begin{matrix}𝔠_{(A,L,K)}: K_{(K,L)}/μ_{n} ⊗_{ℤ} K_{(L,K)}/μ_{n} & \underrightarrow{∼} μ_{n} ⊆ k̄^{∗} \\ [Y] ⊗ [X] & ⟼ XY X^{−1}Y^{−1} . \end{matrix}

of Γ-modules.

Proof: Let X ∈ K_{(L,K)} and Y ∈ K_{(K,L)}, then XY X^{−1}Y^{−1} ∈ K̄^{∗} ∩ L̄^{∗} = k̄^{∗}. Therefore 𝔠 := 𝔠_{(A,L,K)} is well defined as a map going to k̄^{∗}. Obviously 𝔠 is Γ-equivariant. 𝔠 is a homomorphism of groups: e.g., let [X] ∈ K_{(L,K)}/μ_{n} and [Y] ∈ K_{(K,L)}/μ_{n} be generators and a, b ∈ ℤ, then

𝔠([Y^{a+b}] ⊗ [X]) = XY^{a+b}X^{−1}Y^{−a−b}
= XY^{a}X^{−1}(XY^{b}X^{−1}Y^{−b})Y^{−a}
= (XY^{a}X^{−1}Y^{−a})(XY^{b}X^{−1}Y^{−b})
= 𝔠([Y^{a}] ⊗ [X])𝔠([Y^{b}] ⊗ [X]) .

The object on the left hand side of the arrow is as an abelian group isomorphic to ℤ/nℤ ⊗_{ℤ} ℤ/nℤ = ℤ/nℤ, and therefore 𝔠 factors through μ_{n}, the subgroup of k̄^{∗} of the elements of order dividing n. In order to show that 𝔠 is an isomorphism one has to show that 𝔠 is surjective. Let ζ ∈ μ_{n}. With the isomorphism α: Ā \underrightarrow{∼} Ā_{0} of (3.3) we have the commutative diagram

�  _{𝔠_{(A,L,K)}} �_{α} �  _{𝔠_{(A_{0},L_{0},K_{0})}} K_{(K_{0},L_{0})}/μ_{n} ⊗ K_{(L_{0},K_{0})}/μ_{n}  μ_{n} K_{(K,L)}/μ_{n} ⊗ K_{(L,K)}/μ_{n}

All we have to show is ζ ∈ im(𝔠_{(A_{0},L_{0},K_{0})}). But X_{0}(ζ)Y_{0}X_{0}(ζ)^{−1}Y_{0}^{−1} = ζ.

Now, one easily gets the following descriptions of the ρ’s.

(3.16) Corollary. For a twisted cyclic k-algebra (A, L, K) we know about ρ_{L} and ρ_{K} the following description: Let X ∈ K_{(L,K)} and Y ∈ K_{(K,L)}, then

ρ_{L}([Y])(X) = ξ^{−1}X and ρ_{K}([X])(Y) = ξY ,

where ξ = 𝔠_{(A,L,K)}([Y] ⊗ [X]) ∈ μ_{n}.

(3.17) Remark. For any twisted cyclic k-algebra (A, L, K) there is a commutative diagram

�  ^{𝔠_{(A,L,K)}} �_{switch} �  _{𝔠_{(A,K,L)}} �  _{inverse} μ_{n} . K_{(L,K)}/μ_{n} ⊗ K_{(K,L)}/μ_{n}  μ_{n} K_{(K,L)}/μ_{n} ⊗ K_{(L,K)}/μ_{n}

Starting from a twisted cyclic k-algebra (A, L, K) we got two twisted cyclic extensions (L, C) and (K, D), where we denote C := C_{(A,L,K)} := im(ρ_{L}) and D := D_{(A,L,K)} := im(ρ_{K}). Observe the (almost) tautology C_{(A,L,K)} = D_{(A,K,L)} and D_{(A,L,K)} = C_{(A,K,L)}. Additionally we got an isomorphism 𝔠_{(A,L,K)}: K_{(K,L)}/μ_{n} ⊗ K_{(L,K)}/μ_{n} \underrightarrow{∼} μ_{n}. This factorizes through ρ_{L} ⊗ ρ_{K} in the form

�_{ρ_{L}⊗ρ_{K}} �  _{𝔠_{(A,K,L)}} �  _{𝔠_{(A,K,L)}}  μ_{n} . C ⊗ D K_{(K,L)}/μ_{n} ⊗ K_{(L,K)}/μ_{n}

(3.18) Definition. A pair of twisted cyclic extensions (of degree n) is a tuple ((L, C), (K, D), 𝔠), where (L, C) and (K, D) are twisted cyclic extensions (of degree n) and 𝔠: C ⊗_{ℤ} D \underrightarrow{∼} μ_{n} is an isomorphism of Γ-modules.

Now we can say: There is a map from the twisted cyclic k-algebras to the pairs of twisted cyclic extensions of k (of degree n)

(A, L, K) ⟼ ((L, C_{(A,L,K)}), (K, D_{(A,L,K)}), 𝔠_{(A,L,K)}) .

In the next section we will ask the question, if every pair of twisted cyclic extensions comes from a twisted cyclic algebra. If we have a pair of twisted cyclic extension, associated to a twisted cyclic algebra, we immediately get back A as a k-module, namely L ⊗_{k} K (cf. (3.5)). How about the multiplication, i.e., what are the relations of the elements of L and K by multiplication? At least we know it for the elements K_{(L,K)} ⊆ L̄ and K_{(K,L)} ⊆ K̄, since 𝔠_{(A,L,K)} gives us the rule, and that is enough. But since we have only C and D, we first have to get back the groups K_{(L,K)} and K_{(K,L)}.

(3.19) Lemma. Let (A, L, K) be a twisted cyclic k-algebra and let C and D be the associated twisted cyclic structures on L and K. Then we have the following description:

K_{(L,K)} = {ζ′ ∑_{ν=0}^{n−1} ζ^{ν}c^{ν}(e) : ζ′, ζ ∈ μ_{n}}

for any generator c ∈ C and any primitive idempotent e ∈ M_{L};

K_{(K,L)} = {ζ′ ∑_{ν=0}^{n−1} ζ^{ν}d^{ν}(f) : ζ′, ζ ∈ μ_{n}}

for any generator d ∈ D and any primitive idempotent f ∈ M_{K}.

Proof: First observe that the sets on the right side are independent of the choices! Let α: Ā → Ā_{0} be an isomorphism of k̄-algebras as in (3.3). Then α maps bijectively K_{(A,L,K)} to K_{(A_{0},L_{0},K_{0})}, im(ρ_{L}) to im(ρ_{L_{0}}), M_{L} to M_{L_{0}} …. So we just have to prove the claim in this case: It is easy to see that κ_{Y_{0}} ∈ Autk̄(L) generates C = C_{(A_{0},L_{0},K_{0})} and that E_{11} (we denote with E_{ij} the n × n-matrix which has just one non-zero entry: a “1”at the place (i, j)) is a primitive idempotent of L̄_{0}, in fact M_{L_{0}} = {E_{11}, E_{22}, …, E_{nn}}. Furthermore

κ_{Y_{0}^{ν}} (E_{11}) = Y_{0}^{ν}E_{11}Y_{0}^{−ν} = E_{1+ν,1+ν} .

thus

ζ′ ∑_{ν=0}^{n−1} ζ^{ν}κ_{Y_{0}}^{ν}(E_{11}) = ζ′ ∑_{ν=0}^{n−1} ζ^{ν}E_{1+ν,1+ν} = ζ′ζ^{−1}X_{0}(ζ) .

Now compare with (3.9). The second part is proved analogously.

Now we can give an intrinsic description of (the inverse of) ρ_{L} and ρ_{K} without using the embeddings of L and K in A.

(3.20) Proposition. Let (A, L, K) be a twisted cyclic k-algebra and ρ_{L}, ρ_{K}, 𝔠 = 𝔠_{(A,L,K)} the associated data. Then the inverse maps of ρ_{K} and ρ_{L} can be described in the following way:

\begin{matrix}λ_{K}: D & ⟶ K_{(L,K)}/μ_{n} \\ d & ⟼ [∑_{ν=0}^{n−1} ζ^{ν}c^{ν}(e)] , \end{matrix}

where c ∈ C is any generator of C, ζ := 𝔠(c ⊗ d) and e ∈ M_{L};

\begin{matrix}λ_{L}: C & ⟶ K_{(K,L)}/μ_{n} \\ c & ⟼ [∑_{ν=0}^{n−1} ζ^{−ν}d^{ν}(f)] , \end{matrix}

where d ∈ D is any generator of D, ζ := 𝔠(c ⊗ d) and f ∈ M_{K}.

Proof: First observe that the maps are well defined and independent of the various choices, e.g., let’s prove it for λ_{K}. Let c′ be another generator of C, c′ = c^{ω} for some integer ω with (ω, n) = 1. Then ζ′ := 𝔠(c′ ⊗ d) = ζ^{ω}.

[∑_{ν=0}^{n−1} ζ′c^{′ν}(e)] = [∑_{ν=0}^{n−1} ζ^{ων}c^{ων}(e)] = [∑_{ν=0}^{n−1} ζ^{ν}c^{ν}(e)] .

Let e′ = c^{ω}(e) ∈ M_{L} be any other primitive idempotent, then

[∑_{ν=0}^{n−1} ζ^{ν}c^{ν}(e′)] = [∑_{ν=0}^{n−1} ζ^{ν}c^{ν+ω}(e)]

= [ζ^{−ω} ∑_{ν=0}^{n−1} ζ^{ν+ω}c^{ν+ω}(e)]

= [ζ^{−ω} ∑_{ν=0}^{n−1} ζ^{ν}c^{ν}(e)]

= [∑_{ν=0}^{n−1} ζ^{ν}c^{ν}(e)] .

The same holds for λ_{L}. Also λ_{K} is a homomorphism. Let α, β ∈ ℤ

λ_{K}(d^{α}d^{β}) = [∑_{ν=0}^{n−1} ζ^{(α+β)ν}c^{ν}(e)]

= [∑_{ν=0}^{n−1} ζ^{αν}ζ^{βν}c^{ν}(e)]

= [∑_{ν=0}^{n−1} ζ^{αν}c^{ν}(e)] [∑_{ν=0}^{n−1} ζ^{βν}c^{ν}(e)]

= λ_{K}(d^{α})λ_{K}(d^{β}) ; the proof for λ_{L} is similar. In order to show that λ_{K} = ρ_{K}^{−1} we only need to prove λ_{K}(ρ_{K}([X])) = [X] for X ∈ K_{(L,K)}. Again by reasons of “functoriality”(cf. (3.13)) it is enough to consider the case (A_{0}, L_{0}, K_{0}), so we will show λ_{K_{0}}(ρ_{K_{0}}([X_{0}])) = [X_{0}] for any X_{0} = X_{0}(ζ) ∈ K_{(L,K)}:

λ_{K_{0}}(ρ_{K_{0}}([X_{0}])) = λ_{K_{0}}(κ_{X_{0}})

= [∑_{ν=0}^{n−1} ζ^{ν}Y_{0}^{ν}E_{11}Y_{0}^{−ν}]

= [∑_{ν=0}^{n−1} ζ^{ν}E_{1+ν,1+ν}]

= [ζ^{−1} ∑_{ν=0}^{n−1} ζ^{ν}E_{νν}]

= [ζ^{−1}X_{0}] = [X_{0}] ;

note that ζ = 𝔠_{(A_{0},L_{0},K_{0})}([Y_{0}] ⊗ [X_{0}]). Using (3.7) together with (3.17) the claim for λ_{L} follows from the proven one. Alternatively one can prove it directly like above.

3. Twisted Cyclic Algebras associated to Pairs of Twisted Cyclic Extensions

Now we want to go the other direction. Given a pair of twisted cyclic extensions ((L, C), (K, D), 𝔠) of degree n we ask: Is there a central simple k-algebra A of degree n together with embeddings L ↪ A, K ↪ A such that (A, L, K) is a twisted cyclic k-algebra and the associated pair ((L, C_{(A,L,K)}), (K, D_{(A,L,K)}), 𝔠_{(A,L,K)}) is the given one? If there is one it must be L ⊗_{k} K by (3.5) and C, D, 𝔠 will give us the right multiplication on it.

(3.21) Lemma. Given a pair ((L, C), (K, D), 𝔠) of twisted cyclic extensions of degree n then the following maps are well defined injective morphisms of Γ-modules, independent of the various choices:

\begin{matrix}λ_{K}: D & ⟶ L̄^{∗}/μ_{n} \\ d & ⟼ [∑_{ν=0}^{n−1} ζ^{ν}c^{ν}(e)] , \end{matrix}

where c ∈ C is any generator of C, ζ := 𝔠(c ⊗ d) and e ∈ M_{L};

\begin{matrix}λ_{L}: C & ⟶ K̄^{∗}/μ_{n} \\ c & ⟼ [∑_{ν=0}^{n−1} ζ^{−ν}d^{ν}(f)] , \end{matrix}

where d ∈ D is any generator of D, ζ := 𝔠(c ⊗ d) and f ∈ M_{K}.

Proof: The independence of the choices are proved exactly the same way as in the proof of (3.20).

The injectivity and the fact that the maps go to the units of L̄ and K̄: Taking the identification L̄ ≅ k̄^{n} of k̄-algebras, observe that the matrix, given by (representatives of) the vectors of λ_{K}(d), d ∈ D is (up to some elementary row and column operations) the Vandermonde matrix (ζ^{ij})_{i,j=0,...,n−1} for some primitive n-th root of unity ζ, and this matrix is invertible.

Now define

K_{(K,L),𝔠} := im(λ_{L}) and K_{(L,K),𝔠} := im(λ_{K}) .

They are cyclic groups of order n. Again with the Vandermonde determinant one sees

(3.22) Lemma. If c ∈ C is a generator of C, then the element λ_{L}(c) (better: any representative) is a generator of K_{(K,L),𝔠} and it is a generator of K̄ as a k̄-algebra, i.e., K̄ = k̄[λ_{L}(c)]. Mutatis mutandis for d ∈ D and L̄.

(3.23) Theorem. Let ((L, C), (K, D), 𝔠) be a pair of twisted cyclic extensions of k. Then there exists a—up to a unique k-isomorphism—unique twisted cyclic k-algebra (A, L, K) such that C = C_{(A,L,K)}, D = D_{(A,L,K)}, 𝔠 = 𝔠_{(A,L,K)}.

Proof: Existence: Define A to be L⊗_{k}K as k-module. We have to give a multiplication on A. But this is the same as giving a—the Γ-action respecting—multiplication on Ā = L̄ ⊗k̄ K̄. Choose any primitive idempotents e ∈ M_{L} and f ∈ M_{K} and some generators c ∈ C and d ∈ D, then (because 𝔠 is an isomorphism) ζ := 𝔠(c ⊗ d) ∈ μ_{n} is a primitive n-th root of unity. Set

ℓ_{K} := ∑_{ν=0}^{n−1} ζ^{ν}c^{ν}(e) and ℓ_{L} := ∑_{ν=0}^{n−1} ζ^{−ν}d^{ν}(f) ,

then λ_{K}(d) = [ℓ_{K}] ∈ L̄^{∗}/μ_{n}, and λ_{L}(c) = [ℓ_{L}] ∈ K̄^{∗}/μ_{n} and from (3.22) we know

L̄ = k̄[ℓ_{K}] = ⨁_{ν=0}^{n−1} k̄ · ℓ_{K}^{ν} and K̄ = k̄[ℓ_{L}] = ⨁_{ν=0}^{n−1} k̄ · ℓ_{L}^{ν} .

There is a well defined isomorphism

\begin{matrix}L̄ ⊗k̄ K̄ & \underrightarrow{∼} Ā_{0} \\ ℓ_{K}^{i} ⊗ ℓ_{L}^{j} & ⟼ X_{0}(ζ)^{i}Y_{0}^{j} \end{matrix}

of k̄-modules. The two restrictions L̄ ↪ L̄⊗k̄ K̄ → Ā_{0} and K̄ ↪ L̄⊗k̄ K̄ → Ā_{0} are monomorphisms of k̄-algebras, and we can define (by structure transport via this isomorphism) on L̄ ⊗k̄ K̄ a k̄-algebra structure such that L̄ and K̄ are k̄-subalgebras. The semilinear action of Γ on L̄ ⊗k̄ K̄ is even an action on the new k̄-algebra: For that, all we have to consider is that Γ respects the multiplication of the pair ℓ_{K} and ℓ_{L}. We know ℓ_{K}ℓ_{L}ℓ_{K}^{−1}ℓ_{L}^{−1} = ζ since X_{0}(ζ)Y_{0}X_{0}(ζ)^{−1}Y_{0}^{−1} = ζ. For an element γ ∈ Γ we have ^{γ}c = c^{i} and ^{γ}d = d^{j} for some integers i, j which are prime to n. Hence

γζ = ^{γ}𝔠(c ⊗ d) = 𝔠(^{γ}c ⊗^{γ}d) = 𝔠(c^{i} ⊗ d^{j}) = ζ^{ij} ,

and this implies ^{γ}ℓ_{K}^{γ}ℓ_{L}^{γ}ℓ_{K}^{−1}^{γ}ℓ_{L}^{−1} = ^{γ}ζ. In fact

γℓ_{K} = ∑_{ν=0}^{n−1}(^{γ}ζ)^{ν}(^{γ}c)^{ν}(γe)

= ∑_{ν=0}^{n−1} ζ^{ijν}c^{iν}(γe)

= ∑_{ν=0}^{n−1} ζ^{jν}c^{ν}(γe)

= (∑_{ν=0}^{n−1} ζ^{ν}c^{ν}(γe))^{j}

= (∑_{ν=0}^{n−1} ζ^{ν}c^{ν+p}(e))^{j}

= (ζ^{−p} ∑_{ν=0}^{n−1} ζ^{ν}c^{ν}(e))^{j}

= ζ^{−pj}ℓ_{K}^{j} ,

where p is an integer such that γe = c^{p}(e). Analogously ^{γ}ℓ_{L} = ζ^{qi}ℓ_{L}^{i} for some integer q such that γf = d^{q}(f). So from ℓ_{K}ℓ_{L} = ζℓ_{L}ℓ_{K} follows ℓ_{K}^{j}ℓ_{L}^{i} = ζ^{ij}ℓ_{L}^{i}ℓ_{K}^{j}, i.e.,

γℓ_{K}^{γ}ℓ_{L}^{γ}ℓ_{K}^{−1}^{γ}ℓ_{L}^{−1} = ℓ_{K}^{j}ℓ_{L}^{i}ℓ_{K}^{−j}ℓ_{L}^{−i} = ζ^{ij} = ^{γ}ζ

Now we have a k-algebra structure on A. Finally we have to show that C = C_{(A,L,K)}, D = D_{(A,L,K)} and 𝔠 = 𝔠_{(A,L,K)}. Claim: ρ_{L}(ℓ_{L}) = c, i.e., κ_{ℓ_{L}} = c: L̄ → L̄. It suffices to prove it for evaluation in ℓ_{K} ∈ L̄:

κ_{ℓ_{L}}(ℓ_{K}) = ℓ_{L}ℓ_{K}ℓ_{L}^{−1} = ζ^{−1}ℓ_{K}

= ζ^{−1} ∑_{ν=0}^{n−1} ζ^{ν}c^{ν}(e)

= ∑_{ν=0}^{n−1} ζ^{ν−1}c^{ν}(e)

= ∑_{ν=0}^{n−1} ζ^{ν}c^{ν+1}(e)

= c ∑_{ν=0}^{n−1} ζ^{ν}c^{ν}(e)

= c ℓ_{K} ;

analogously ρ_{K}(ℓ_{K}) = d.—The claim 𝔠 = 𝔠_{(A,L,K)} is obvious. Uniqueness: Let (A, L, K) and (B, L, K) be two twisted cyclic k-algebras such that C_{(A,L,K)} = C_{(B,L,K)}, D_{(A,L,K)} = D_{(B,L,K)} and 𝔠_{(A,L,K)} = 𝔠_{(B,L,K)}: C ⊗ D → μ_{n}. Then the isomorphisms of (3.5) yield an isomorphism δ of k-modules such that

�  _{α} �  _{β} �  _{δ} B A L ⊗ K

commutes. Here α and β denote the multiplication morphisms. We have to show that δ is a morphism of k-algebras. We verify it after the base extension k̄|k. Let ℓ_{K} ∈ L̄ and ℓ_{L} ∈ K̄ be—as above—elements representing λ_{K}(d), λ_{L}(c), then we know

Ā = ⨁_{i,j=0}^{n−1} k̄ · α(ℓ_{K})^{i}α(ℓ_{L})^{j} and B̄ = ⨁_{i,j=0}^{n−1} k̄ · β(ℓ_{K})^{i}β(ℓ_{L})^{j} .

Now we have to show only

δ(α(ℓ_{K})α(ℓ_{L})α(ℓ_{K})^{−1}α(ℓ_{L})^{−1}) = β(ℓ_{K})β(ℓ_{L})β(ℓ_{K})^{−1}β(ℓ_{L})^{−1} .

But the two groups K_{(L,K)} derived from the two twisted cyclic algebras are equal (cf. (3.19)), and also the two groups K_{(K,L)}. Also ℓ_{K} ∈ K_{(L,K)}, ℓ_{L} ∈ K_{(K,L)}, hence

α(ℓ_{K})α(ℓ_{L})α(ℓ_{K})^{−1}α(ℓ_{L})^{−1} = 𝔠_{(A,L,K)}([ℓ_{L}] ⊗ [ℓ_{K}])
= 𝔠_{(B,L,K)}([ℓ_{L}] ⊗ [ℓ_{K}])
= β(ℓ_{K})β(ℓ_{L})β(ℓ_{K})^{−1}β(ℓ_{L})^{−1} .

(3.24) Notation. In the following we will denote the algebra A constructed in the proof of (3.23) with A((L, C), (K, D), 𝔠).

4. A Cohomological Classification of Twisted Cyclic Algebras

We want to classify the twisted cyclic k-algebras of degree n. Since the twisted cyclic k-algebras are exactly the objects which become—after the base field extension k̄|k—isomorphic to our standard model (A_{0}, L_{0}, K_{0}), the set of isomorphism classes of twisted cyclic k-algebras of degree n is the first Galois cohomology group over k with values in the automorphism group of our standard model. Cf. chapter X, §2 in [SeLF] or chapter VII, 29.A in [BI].

(3.25) Definition. An isomorphism of twisted cyclic k-algebras f: (A, L, K) \underrightarrow{∼} (A′, L′, K′) is an isomorphism f: A \underrightarrow{∼} A′ of k-algebras such that f(L) = L′ and f(K) = K′. The set of the automorphisms of a twisted cyclic k-algebra (A, L, K) will be denoted by Aut(A, L, K).

(3.26) Remark. As we have seen earlier, an isomorphism f: (A, L, K) \underrightarrow{∼} (A′, L′, K′) of twisted cyclic k-algebras induces isomorphisms

f: K_{(L,K)} \underrightarrow{∼} K_{(L′,K′)} and f: K_{(K,L)} \underrightarrow{∼} K_{(K′,L′)}

as well as

f: K_{(L,K)}/μ_{n} \underrightarrow{∼} K_{(L′,K′)}/μ_{n} and f: K_{(K,L)}/μ_{n} \underrightarrow{∼} K_{(K′,L′)}/μ_{n} .

So in the case of an automorphism f ∈ Aut(A, L, K) we get the induced automorphisms

f_{(L,K)} := f: K_{(L,K)}/μ_{n} \underrightarrow{∼} K_{(L,K)}/μ_{n} and f_{(K,L)} := f: K_{(K,L)}/μ_{n} \underrightarrow{∼} K_{(K,L)}/μ_{n} ,

i.e., f_{(L,K)} ∈ Aut(K_{(L,K)}/μ_{n}) and f_{(K,L)} ∈ Aut(K_{(K,L)}/μ_{n}). From (3.10) we know that K_{(L,K)}/μ_{n} and K_{(K,L)}/μ_{n} are cyclic groups of order n. Therefore we have the canonical identifications

\begin{matrix}(ℤ/nℤ)^{∗} & = Aut(K_{(L,K)}/μ_{n}) \\ (ν mod nℤ) & ⟼ ([X] ↦ [X^{ν}]) \end{matrix}

and

\begin{matrix}(ℤ/nℤ)^{∗} & = Aut(K_{(K,L)}/μ_{n}) \\ (ν mod nℤ) & ⟼ ([Y] ↦ [Y^{ν}]) \end{matrix}

where here [X] and [Y] stand for (X mod μ_{n}) and (Y mod μ_{n}). So we will view f_{(L,K)} and f_{(K,L)} also as elements of (ℤ/nℤ)^{∗}.

(3.27) Lemma. With these identifications we have

f_{(L,K)} · f_{(K,L)} = (1 mod nℤ) ∈ (ℤ/nℤ)^{∗}

for any automorphism f ∈ Aut(A, L, K) of a twisted cyclic k-algebra (A, L, K) of degree n.

Proof: We may assume that k = k̄ and hence that (A, L, K) = (A_{0}, L_{0}, K_{0}) is our standard triple. Regarding f_{(L,K)} = (ν mod nℤ) and f_{(K,L)} = (ν′ mod nℤ) as elements of (ℤ/nℤ)^{∗}, then

\begin{matrix}f_{(L,K)}: K_{(L,K)}/μ_{n} & \underrightarrow{∼} K_{(L,K)}/μ_{n} \\ [X_{0}(ξ)] & ⟼ [X_{0}(ξ)^{ν}] \end{matrix}

for any ξ ∈ μ_{n}, and

\begin{matrix}f_{(K,L)}: K_{(K,L)}/μ_{n} & \underrightarrow{∼} K_{(K,L)}/μ_{n} \\ [Y_{0}^{m}] & ⟼ [Y_{0}^{mν′}] \end{matrix}

for any m ∈ ℤ. Now for any primitive n-th root ζ ∈ μ_{n} of unity

ζ = f(ζ)
= f(X_{0}(ζ)Y_{0}X_{0}(ζ)^{−1}Y_{0}^{−1})
= X_{0}(ζ)^{ν}Y_{0}^{ν′}X_{0}(ζ)^{−ν}Y_{0}^{−ν′}
= ζ^{νν′} .

Therefore νν′ ≡ 1 (mod n).

Now we assume that μ_{n} = μ(k̄) ⊆ k.

(3.28) Lemma. If ζ ∈ μ_{n} is a primitive n-th root of unity, then for an automorphism f ∈ Aut(A_{0}, L_{0}, K_{0}), the following conditions are equivalent

(i) f(Y_{0}) ∈ μ_{n} · Y_{0} (ii) f(X_{0}(ζ)) ∈ μ_{n} · X_{0}(ζ) (iii) f(X_{0}(ξ)) ∈ μ_{n} · X_{0}(ξ) for all ξ ∈ μ_{n} (iv) f_{(L_{0},K_{0})} = (1 mod nℤ) is the identity

(v) f_{(K_{0},L_{0})} = (1 mod nℤ) is the identity.

Proof: The equivalence of (i) and (v) as well as (ii) and (iv) is trivial since [X_{0}(ζ)] and [Y_{0}] are generators of the cyclic groups K_{(L_{0},K_{0})}/μ_{n} and K_{(K_{0},L_{0})}/μ_{n}. The equivalence of (iv) and (v) is just (3.27). The equivalence of (ii) and (iii) is clear.

(3.29) Lemma. There is an injective morphism

\begin{matrix}μ_{n} × (ℤ/nℤ) & ⟶ Aut(A_{0}, L_{0}, K_{0}) \\ (ξ, m̄) & ⟼ κ_{(ξ,m̄)} \end{matrix}

where

\begin{matrix}κ_{(ξ,m̄)} := κ_{X_{0}(ξ)Y_{0}^{m}}: A_{0} & \underrightarrow{∼} A_{0} \\ Z & ⟼ X_{0}(ξ)Y_{0}^{m}ZY_{0}^{−m}X_{0}(ξ)^{−1} . \end{matrix}

—We write m̄ for (m mod nℤ). The image of this injection is the set of all automorphisms f ∈ Aut(A_{0}, L_{0}, K_{0}) which fulfill the conditions of (3.28).

Proof: The map is a morphism since

κ_{X_{0}(ξ)Y_{0}^{m}} ∘ κ_{X_{0}(ξ′)Y_{0}^{m′}} = κ_{X_{0}(ξξ′)Y_{0}^{(m+m′)}}

for ξ, ξ′ ∈ μ_{n} and m, m′ ∈ ℤ. This follows from the fact that

X_{0}(ξ′)Y_{0}^{m} = ξ^{′m}Y_{0}^{m}X_{0}(ξ′)

and hence κ_{X_{0}(ξ′)} ∘ κ_{Y_{0}^{m}} = κ_{Y_{0}^{m}} ∘ κ_{X_{0}(ξ′)}. Injectivity: Let (ξ, m̄) ∈ μ_{n} × (ℤ/nℤ) such that κ_{(ξ,m̄)} is the identity. Then

X_{0}(ζ) = κ_{(ξ,m̄)}(X_{0}(ξ))
= X_{0}(ξ)Y_{0}^{m}X_{0}(ζ)Y_{0}^{−m}X_{0}(ξ)^{−1}
= ζ^{−m}X_{0}(ζ)

for any ζ ∈ μ_{n}, hence m ≡ 0 (mod n).

Y_{0} = κ_{(ξ,m̄)}(Y_{0})
= X_{0}(ξ)Y_{0}^{m}Y_{0}Y_{0}^{−m}X_{0}(ξ)^{−1}
= X_{0}(ξ)Y_{0}X_{0}(ξ)^{−1}
= ξY_{0} ,

hence ξ = 1. For the last statement, first note that for any (ξ, m̄) ∈ μ_{n} × (ℤ/nℤ), the induced morphism

(κ_{(ξ,m̄)})_{(L_{0},K_{0})}: K_{(L_{0},K_{0})}/μ_{n} \underrightarrow{∼} K_{(L_{0},K_{0})}/μ_{n}

is the identity, since

κ_{(ξ,m̄)}(X_{0}(ζ)) = ζ^{−m}X_{0}(ζ)

for any ζ ∈ μ_{n}. Cf. (3.28). On the other hand let f ∈ Aut(A_{0}, L_{0}, K_{0}) such that the conditions of (3.28) hold for f. Then f(Y_{0}) = ξY_{0} for some ξ ∈ μ_{n}. Choose a primitive n-th root of unity ζ ∈ μ_{n}, then we also have by (3.28) f(X_{0}(ζ)) = ζ^{−m}X_{0}(ζ) for some m ∈ ℤ. Obviously now f = κ_{(ξ,m̄)}, since both automorphisms are equal on the generators of A_{0},—X_{0}(ζ) and Y_{0}.

As a consequence we get

(3.30) Proposition. There are two short exact sequences

\begin{matrix}0 ⟶ μ_{n} × (ℤ/nℤ) ⟶ Aut(A_{0}, L_{0}, K_{0}) & ⟶ Aut(K_{(L_{0},K_{0})}/μ_{n}) ⟶ 1 \\ f & ⟼ f_{(L_{0},K_{0})} \end{matrix}

and

\begin{matrix}0 ⟶ μ_{n} × (ℤ/nℤ) ⟶ Aut(A_{0}, L_{0}, K_{0}) & ⟶ Aut(K_{(K_{0},L_{0})}/μ_{n}) ⟶ 1 \\ f & ⟼ f_{(K_{0},L_{0})} . \end{matrix}

The surjective morphisms on the right hand side have the following sections:

Aut(K_{(L_{0},K_{0})}/μ_{n}) = (ℤ/nℤ)^{∗} ⟶ Aut(A_{0}, L_{0}, K_{0})
ν̄ = (ν mod nℤ) ⟼ fν̄

and

Aut(K_{(K_{0},L_{0})}/μ_{n}) = (ℤ/nℤ)^{∗} ⟶ Aut(A_{0}, L_{0}, K_{0})
ν̄ = (ν mod nℤ) ⟼ gν̄

where fν̄ and gν̄ are given by

\begin{matrix}fν̄: A_{0} & \underrightarrow{∼} A_{0} \\ X_{0}(ξ) & ⟼ X_{0}(ξ)^{ν} \\ Y_{0} & ⟼ Y_{0}^{(ν^{−1})} \end{matrix}

and

\begin{matrix}gν̄: A_{0} & \underrightarrow{∼} A_{0} \\ X_{0}(ξ) & ⟼ X_{0}(ξ)^{(ν^{−1})} \\ Y_{0} & ⟼ Y_{0}^{ν} \end{matrix}

for any ξ ∈ μ_{n}; and ν^{−1} denotes any integer such that νν^{−1} ≡ 1 (mod n).

Proof: The first part of the proposition—except for the surjectivity in the sequences— is just (3.28) and (3.29).

The surjectivity follows with the sections: For the fact that fν̄ and gν̄ are well defined automorphisms of (A_{0}, L_{0}, K_{0}) observe that they conserve the relation of the two generators X_{0}(ζ) (for any primitive n-th root ζ ∈ μ_{n}), Y_{0} and their images, e.g., for fν̄:

X_{0}(ζ)^{ν}Y_{0}^{(ν^{−1})}X_{0}(ζ)^{−ν}Y_{0}^{−(ν^{−1})} = ζ^{νν^{−1}} = ζ = X_{0}(ζ)Y_{0}X_{0}(ζ)^{−1}Y_{0}^{−1} .

It is obvious that the applications ν̄ ↦ fν̄ and ν̄ ↦ gν̄ are morphisms and sections.

(3.31) Remark. For every ν̄ ∈ (ℤ/nℤ)^{∗} the automorphisms fν̄ and gν̄ are inverse to each other, i.e., fν̄ ∘ gν̄ = gν̄ ∘ fν̄ = id_{A_{0}}.

In our situation of (3.30), the right hand side objects of the short exact sequences act on the left hand side objects, in a well known manner: E.g., in the first case, one takes a preimage φ ∈ Aut(A_{0}, L_{0}, K_{0}) of an element ν̄ ∈ Aut(K_{(L_{0},K_{0})}/μ_{n}) = (ℤ/nℤ)^{∗} (one may take φ = fν̄) and, after identification of μ_{n} × (ℤ/nℤ) with its image in Aut(A_{0}, L_{0}, K_{0}), the action of ν̄ on μ_{n} × (ℤ/nℤ) is given by conjugation with φ. One knows that this is in fact a well defined action.

How do these actions look like in our case?

(3.32) Lemma. The actions of the groups Aut(K_{(L_{0},K_{0})}/μ_{n}) and Aut(K_{(K_{0},L_{0})}/μ_{n}) on the group μ_{n} × (ℤ/nℤ) from above can be described in the following way: After the identification Aut(K_{(L_{0},K_{0})}/μ_{n}) = (ℤ/nℤ)^{∗}, the action of this group is

\begin{matrix}(ℤ/nℤ)^{∗} × (μ_{n} × (ℤ/nℤ)) & ⟶ μ_{n} × (ℤ/nℤ) \\ (ν̄, (ξ, m̄)) & ⟼ (ξ^{ν}, ν̄^{−1}m̄) . \end{matrix}

After the identification Aut(K_{(K_{0},L_{0})}/μ_{n}) = (ℤ/nℤ)^{∗}, the action of this group is

\begin{matrix}(ℤ/nℤ)^{∗} × (μ_{n} × (ℤ/nℤ)) & ⟶ μ_{n} × (ℤ/nℤ) \\ (ν̄, (ξ, m̄)) & ⟼ (ξ^{(ν^{−1})}, ν̄m̄) \end{matrix}

where ν^{−1} is any integer such that νν^{−1} ≡ 1 (mod n).

Proof: For the first case we choose φ = fν̄ and we have to prove

fν̄ ∘ κ_{(ξ,m̄)} = κ_{(ξ^{ν},ν̄^{−1}m̄)} ∘ fν̄ .

It is enough to verify this equation after evaluation in X_{0}(ζ) (for any ζ ∈ μ_{n}) and Y_{0}:

(fν̄ ∘ κ_{(ξ,m̄)})(X_{0}(ζ)) = fν̄(κ_{(ξ,m̄)}(X_{0}(ζ)))

= fν̄(ζ^{−m}X_{0}(ζ))

= ζ^{−m}X_{0}(ζ)^{ν} = ζ^{−m}X_{0}(ζ^{ν})

= κ_{(ξ^{ν},ν̄^{−1}m̄)}(X_{0}(ζ^{ν}))

= (κ_{(ξ^{ν},ν̄^{−1}m̄)} ∘ fν̄)(X_{0}(ζ)) and

(fν̄ ∘ κ_{(ξ,m̄)})(Y_{0}^{ν}) = fν̄(κ_{(ξ,m̄)}(Y_{0}^{ν}))
= fν̄(ξ^{ν}Y_{0}^{ν})
= ξ^{ν}Y_{0}
= κ_{(ξ^{ν},ν̄^{−1}m̄)}(Y_{0})
= κ_{(ξ^{ν},ν̄^{−1}m̄)}(fν̄(Y_{0}^{ν}))
= (κ_{(ξ^{ν},ν̄^{−1}m̄)} ∘ fν̄)(Y_{0}^{ν}) .

The description of the other action follows from the fact that gν̄ = f −1 ; ν̄ for every ν̄ ∈ (ℤ/nℤ)^{∗}.

From the last points now follows:

(3.33) Proposition. With these actions we can construct the following semidirect products and get the isomorphisms

\begin{matrix}(μ_{n} × (ℤ/nℤ))⋊ Aut(K_{(L_{0},K_{0})}/μ_{n}) = (μ_{n} × (ℤ/nℤ))⋊(ℤ/nℤ)^{∗} & \underrightarrow{∼} Aut(A_{0}, L_{0}, K_{0}) \\ ((ξ, m̄), ν̄) & ⟼ κ_{(ξ,m̄)} ∘ fν̄ \end{matrix}

and

\begin{matrix}(μ_{n} × (ℤ/nℤ))⋊ Aut(K_{(K_{0},L_{0})}/μ_{n}) = (μ_{n} × (ℤ/nℤ))⋊(ℤ/nℤ)^{∗} & \underrightarrow{∼} Aut(A_{0}, L_{0}, K_{0}) \\ ((ξ, m̄), ν̄) & ⟼ κ_{(ξ,m̄)} ∘ gν̄ . \end{matrix}

Now let again k be a field not necessarily with μ_{n} ⊆ k, and let (A, L, K) be a twisted cyclic k-algebra of degree n. We still have a separable algebraic closure k̄ of k and Γ = Gal(k̄|k). Then Γ acts on Aut(Ā, L̄, K̄) via conjugation

Γ × Aut(Ā, L̄, K̄) ⟶ Aut(Ā, L̄, K̄)
(γ, f) ⟼ (id_{A} ⊗γ) ∘ f ∘ (id_{A} ⊗γ)^{−1} .

In an analogous way Γ acts by conjugation on Aut(K_{(L,K)}/μ_{n}) and Aut(K_{(K,L)}/μ_{n}). On μ_{n} × (ℤ/nℤ) the group Γ may act in the obvious way.

(3.34) Lemma. The actions of Γ on the groups Aut(K_{(L,K)}/μ_{n}) and Aut(K_{(K,L)}/μ_{n}) are trivial, so their identification with (ℤ/nℤ)^{∗} are even Γ-isomorphisms.

Proof: Let f ∈ Aut(K_{(L,K)}/μ_{n}), i.e., there is a ν̄ ∈ (ℤ/nℤ)^{∗} such that

\begin{matrix}f: K_{(L,K)}/μ_{n} & \underrightarrow{∼} K_{(L,K)}/μ_{n} \\ [X] & ⟼ [X^{ν}] . \end{matrix}

Then for any γ ∈ Γ

(id_{A} ⊗γ) ∘ f ∘ (id_{A} ⊗γ)^{−1}([X]) = (id_{A} ⊗γ) ∘ f([(id_{A} ⊗γ)^{−1}X])
= (id_{A} ⊗γ)([(id_{A} ⊗γ)^{−1}X^{ν}])
= [X^{ν}]
= f([X]) .

Analogous for the other action.

§ 3. Twisted cyclic algebras

(3.35) Lemma. The injective morphism

\begin{matrix}μ_{n} × (ℤ/nℤ) & ⟶ Aut(Ā_{0}, L̄_{0}, K̄_{0}) \\ (ξ, m̄) & ⟼ κ_{(ξ,m̄)} \end{matrix}

is Γ-equivariant. Proof: For γ ∈ Γ, ξ ∈ μ_{n}, m̄ ∈ ℤ/nℤ we have to show

(id ⊗γ) ∘ κ_{(ξ,m̄)} ∘ (id ⊗γ)^{−1} = κ_{(γξ,m̄)} .

For any Z ∈ Ā_{0} we have

((id ⊗γ) ∘ κ_{(ξ,m̄)} ∘ (id ⊗γ)^{−1})(Z) = (id ⊗γ)(X_{0}(ξ)Y_{0}^{m}((id ⊗γ)^{−1}Z)Y_{0}^{−m}X_{0}(ξ)^{−1})
= (id ⊗γ)(X_{0}(ξ)Y_{0}^{m}) · Z · (id ⊗γ)(Y_{0}^{−m}X_{0}(ξ)^{−1})
= X_{0}(γξ)Y_{0}^{m}ZY_{0}^{−m}X_{0}(γξ)^{−1}
= κ_{(γξ,m̄)}(Z) ,

and we are done. Now we apply the former propositions to (Ā_{0}, L̄_{0}, K̄_{0}) and we get (3.36) Theorem. The short exact sequences from (3.30)

0 ⟶ μ_{n} × (ℤ/nℤ) ⟶ Aut(Ā_{0}, L̄_{0}, K̄_{0}) ⟶ Aut(K_{(L_{0},K_{0})}/μ_{n}) ⟶ 1 and

0 ⟶ μ_{n} × (ℤ/nℤ) ⟶ Aut(Ā_{0}, L̄_{0}, K̄_{0}) ⟶ Aut(K_{(K_{0},L_{0})}/μ_{n}) ⟶ 1

are sequences of Γ-groups and the morphisms are Γ-equivariant. The two induced isomorphisms from (3.33) β: (μ_{n} × (ℤ/nℤ))⋊ Aut(K_{(L_{0},K_{0})}/μ_{n}) = (μ_{n} × (ℤ/nℤ))⋊(ℤ/nℤ)^{∗} \underrightarrow{∼} Aut(A_{0}, L_{0}, K_{0}) and β′: (μ_{n} × (ℤ/nℤ))⋊ Aut(K_{(K_{0},L_{0})}/μ_{n}) = (μ_{n} × (ℤ/nℤ))⋊(ℤ/nℤ)^{∗} \underrightarrow{∼} Aut(A_{0}, L_{0}, K_{0}) are isomorphisms of Γ-groups. Now, by the general theory of Galois cohomology we know that the first cohomology group over k with coefficients in Aut(Ā_{0}, L̄_{0}, K̄_{0}) classify the twisted objects of (A_{0}, L_{0}, K_{0}): (3.37) Remark. There is an isomorphism, i.e., bijection,

\{Twisted cyclic k-algebras of degree n\}/≅ \underrightarrow{∼} H^{1}(k, Aut(Ā_{0}, L̄_{0}, K̄_{0}))
(A, L, K) ⟼ [c_{γ}] ,

where a representing cocycle Γ → Aut(Ā_{0}, L̄_{0}, K̄_{0})), γ ↦ c_{γ} is given by

c_{γ} = α ∘ (id_{A} ⊗γ) ∘ α^{−1} ∘ (id_{A_{0}} ⊗γ)^{−1} ;

α is any isomorphism as in the definition (3.3). Now we fix (e.g.) the first action of (ℤ/nℤ)^{∗} on μ_{n} × (ℤ/nℤ) described in (3.32) and the theorem (3.36) yields the

§ 3. Twisted cyclic algebras

(3.38) Corollary. There is an isomorphism

\{Twisted cyclic k-algebras of degree n\}/≅ \underrightarrow{∼} H^{1}(k, (μ_{n} × (ℤ/nℤ))⋊(ℤ/nℤ)^{∗})
(A, L, K) ⟼ [β^{−1}c_{γ}] .

§ 4. Existence of Twisted Cyclic Decompositions of Central Simple Algebras

In §3, 3., we considered the question if—expressed in a sloppy manner—we can fill in the gap in (?, L, K). Now we want to fill the gap in (A, L,?), i.e., we ask if we can complete L to a twisted cyclic decomposition of A.

The way we are going to find some K is the following: We transfer our situation to Ā_{0}, i.e., instead of looking for a K in A, we are looking for a Γ-invariant K̄ in Ā—but for computational reasons we work in Ā_{0}. And all we have to do is pushing the action of Γ on Ā to an action of Γ on Ā_{0}, which is not necessarily the canonical action. The main ingredient however is a cohomological one, which we will discuss in the first part.

1. A Cohomological Lemma

Let (L, C) be a twisted cyclic extension of k of degree n. As always Γ denotes the absolute Galois group of k—with respect to a chosen separable algebraic closure k̄ of k. We assume that L is a field. There is a canonical morphism of Γ-modules

\begin{matrix}Ξ: Hom_{ℤ}(C, μ_{n}) & ⟶ L̄^{∗}/k̄^{∗} \\ u & ⟼ [∑_{c∈C} u(c) · c(e)] \end{matrix}

for any primitive idempotent e ∈ M_{L} of L̄: Γ acts on the left group in the usual way. Ξ is independent of the choice of e: Let e′ = c′(e) be any other primitive idempotent, c′ ∈ C, then

∑_{c∈C} u(c) · c(e′) = ∑_{c∈C} u(c) · cc′(e)

= ∑_{c∈C} u(cc′^{−1}) · c(e)

= u(c′)^{−1} ∑_{c∈C} u(c) · c(e) .

Ξ is Γ-equivariant. Let γ ∈ Γ, then giving heed to (1.10)

Ξ(^{γ}u) = Ξ(γ ∘ u ∘ κ_{φ_{L}(γ)}^{−1})

= [∑_{c∈C} γu(φ_{L}(γ)^{−1}cφ_{L}(γ)) · c(e)]

= [ ∑_{c′∈C} γ(u(c′)) · (id_{L} ⊗γ)(c′((id_{L} ⊗γ)^{−1}e))]

= (id_{L} ⊗γ)(Ξ(u)) .

Taking cohomology, this morphism induces the morphism

Ξ_{∗} := H^{1}(Γ, Ξ): H^{1}(Γ, Hom_{ℤ}(C, μ_{n})) ⟶ H^{1}(Γ, L̄^{∗}/k̄^{∗}) .

§ 4. Existence of twisted cyclic decompositions

(4.1) Lemma. The cohomology group H^{1}(Γ, L̄^{∗}/k̄^{∗}) is isomorphic to the relative Brauer group Br(L|k), and therefore it is an n-torsion group.

Proof: First we may assume that L is embedded in k̄, i.e., k ⊆ L ⊆ k̄. Let Γ′ := Gal(k̄|L) be the absolute Galois group of L which is an open subgroup of Γ of index n. Note: L̄ does not denote an algebraic closure of L but still the base extension L ⊗_{k} K̄. Claim: L̄^{∗} is an induced module for Γ′ ⊆ Γ. The morphism of Γ-modules

\begin{matrix}ψ: L̄^{∗} = (L ⊗_{k} k̄)^{∗} & \underrightarrow{∼} Map_{Γ′}(Γ, k̄^{∗}) \\ ∑_{ν} l_{ν} ⊗ x_{ν} & ⟼ (τ ↦ ∑_{ν} l_{ν}τ(x_{ν})) \end{matrix}

is an isomorphism. In fact it is the composition of the obviously bijective morphisms of Γ-modules

\begin{matrix}L̄^{∗} = (L ⊗_{k} k̄)^{∗} & \underrightarrow{∼} ∏_{σ̄∈Γ/Γ′} k̄^{∗} \\ ∑_{ν}(l_{ν} ⊗ x_{ν}) & ⟼ (∑_{ν} σ(l_{ν})x_{ν})σ̄ \end{matrix}

and

\begin{aligned}∏_{σ̄∈Γ/Γ′} k̄^{∗} & \underrightarrow{∼} Map_{Γ′}(Γ, k̄^{∗}) \\ (aσ̄)σ̄ & ⟼ (τ ↦ τa_{}_{τ̄^{−1}}) \\ (σw(σ^{−1}))σ̄ & ⟵ w . \end{aligned}

Here σ̄ stands for (σ mod Γ′) and σ is any representative. Applying cohomology to the short exact sequence

1 ⟶ k̄^{∗} ⟶ L̄^{∗} ⟶ L̄^{∗}/k̄^{∗} ⟶ 1

of Γ-modules yields, as part of the long exact cohomology sequence, the exact sequence

H^{1}(Γ, L̄^{∗}) ⟶ H^{1}(Γ, L̄^{∗}/k̄^{∗}) ⟶^{δ} H^{2}(Γ, k̄^{∗}) ⟶ H^{2}(Γ, L̄^{∗}) .

Hilbert’s Theorem 90 and Shapiro’s lemma—via ψ—(cf. [ShPG], II, §2, Theorem 8, p. 31) shows

H^{1}(Γ, L̄^{∗}) ≅ H^{1}(Γ, Map_{Γ′}(Γ, k̄^{∗})) ≅ H^{1}(Γ′, k̄^{∗}) = 1 .

Therefore δ identifies H^{1}(Γ, L̄^{∗}/k̄^{∗}) with the kernel of the right arrow of the latter sequence. We have the commutative diagram

� �  _{(ϵk̄^{∗})_{∗}} �  _{ψ_{∗}} H^{2}(Γ, Map_{Γ′}(Γ, k̄^{∗})) , H^{2}(Γ, L̄^{∗}) H^{2}(Γ, k̄^{∗})

where ϵk̄^{∗} is the composition

k̄^{∗} = (k ⊗_{k} k̄)^{∗} ⟶ (L ⊗_{k} k̄)^{∗} ⟶^{ψ} Map_{Γ′}(Γ, k̄^{∗}) ,

ϵk̄^{∗}(x)(τ) = τ(x) for x ∈ k̄^{∗} and τ ∈ Γ,—in concordance with the notation in [ShPG], II, §3, Proposition 7, p. 33—gives the commutative diagram

�  ^{(ϵk̄^{∗})_{∗}} �  _{res} �  _{shapiro} H^{2}(Γ′, k̄^{∗}) H^{2}(Γ, Map_{Γ′}(Γ, k̄^{∗})) H^{2}(Γ, k̄^{∗})

Therefore our kernel is equal to the kernel of the restriction, given in the last diagram. This kernel is known to be the relative Brauer group Br(L|k) and this is n-torsion since the composition

H^{2}(Γ, k̄^{∗}) ⟶^{res} H^{2}(Γ′, k̄^{∗}) ⟶^{cor} H^{2}(Γ, k̄^{∗})

restricted to Br(L|k) is on the one hand zero, and on the other hand it is the multiplication with n.

(4.2) Lemma. Assume that Γ acts trivially on C, i.e., (L, C) is a cyclic extension, then

Ξ_{∗}: H^{1}(Γ, Hom_{ℤ}(C, μ_{n})) ⟶ H^{1}(Γ, L̄^{∗}/k̄^{∗})

is surjective.

Proof: In this case we have the isomorphism ev_{π}: Hom_{ℤ}(C, μ_{n}) \underrightarrow{∼} μ_{n}, u ↦ u(π) of Γ-modules; here we have chosen a generator π ∈ C of the cyclic group C(⊆ Autk̄(L)). We will show the surjectivity of the composition

Ξ_{∗}^{π} := Ξ_{∗} ∘ (ev_{π}^{−1})_{∗}: H^{1}(Γ, μ_{n}) ⟶ H^{1}(Γ, L̄^{∗}/k̄^{∗}) ,

where this morphism is induced by

\begin{matrix}Ξ^{π} := Ξ ∘ ev_{π}^{−1}: μ_{n} & ⟶ L̄^{∗}/k̄^{∗} \\ ζ & ⟼ [∑_{ν=0}^{n−1} ζ^{ν}π^{ν}(e)] \end{matrix}

for any primitive idempotent e ∈ M_{L}, which we want to fix. Let us apply cohomology to the two short exact sequences

1 ⟶ μ_{n} ⟶ k̄^{∗} ⟶^{n} k̄^{∗} ⟶ 1

and

\begin{matrix}1 ⟶ L̄^{∗}/k̄^{∗} & ⟶^{1−π} L̄^{∗} ⟶^{N} k̄^{∗} ⟶ 1 , \\ l & ⟼ \frac{l}{π(l)} \end{matrix}

where N is the morphism

\begin{matrix}N: L̄^{∗} = ⨁_{e′∈M_{L}} k̄^{∗} · e′ & ⟶ k̄^{∗} \\ (a_{e′}) & ⟼ ∏_{e′∈M_{L}} a_{e′} , \end{matrix}

induced by the Norm map N_{L|k}. Then we get the exact sequences

k^{∗} ⟶^{n} k^{∗} ⟶^{δ} H^{1}(Γ, μ_{n}) ⟶ H^{1}(Γ, k̄^{∗})

§ 4. Existence of twisted cyclic decompositions

and

L^{∗} ⟶^{N_{L|k}} k^{∗} ⟶^{δ} H^{1}(Γ, L̄^{∗}/k̄^{∗}) ⟶ H^{1}(Γ, L̄^{∗}) .

But H^{1}(Γ, k̄^{∗}) = 1 by Hilbert’s Theorem 90 and H^{1}(Γ, L̄^{∗}) = 1 as we have seen in the proof of (4.1). Therefore we get isomorphisms

\begin{matrix}δ: k^{∗}/k^{∗n} & \underrightarrow{∼} H^{1}(Γ, μ_{n}) \\ (a mod k^{∗n}) & ⟼ [h_{α}] , \end{matrix}

where h_{α}(γ) = \frac{α}{γ(α)}, with some α ∈ k̄^{∗} such that α^{n} = a, and

\begin{matrix}δ: k^{∗}/ N_{L|k}(L^{∗}) & \underrightarrow{∼} H^{1}(Γ, L̄^{∗}/k̄^{∗}) \\ (a mod N_{L|k}(L^{∗})) & ⟼ [h_{β}] , \end{matrix}

where for γ ∈ Γ the element h_{β}(γ) ∈ L̄^{∗}/k̄^{∗} is given by (1 − π)h_{β}(γ) = \frac{β}{γ(β)}, with some β ∈ L̄^{∗} such that N(β) = a. For example one may choose β = ∑_{c∈C} α · c(e) for some α ∈ k̄^{∗} with α^{n} = a. Then \frac{β}{γ(β)} = ∑_{c∈C}(\frac{α}{γ(α)}) · c(e) and therefore

h_{β}(γ) = [∑_{ν=0}^{n−1} (\frac{α}{γ(α)})^{ν} π^{ν}(e)] .

The diagram

�  ^{Ξ_{∗}^{π}} �  _{proj} �_{δ} �  _{δ} k^{∗}/ N_{L|k} L^{∗} k^{∗}/k^{∗n}  H^{1}(Γ, L̄^{∗}/k̄^{∗}) H^{1}(Γ, μ_{n})

is commutative: For any a, α and β as above,

Ξ^{π} ∘ h_{α}(γ) = Ξ^{π} (\frac{α}{γ(α)})

= [∑_{ν=0}^{n−1} (\frac{α}{γ(α)})^{ν} π^{ν}(e)]

= h_{β}(γ) .

The claim follows immediately. (4.3) Proposition. If (n, φ(n)) = 1, then the morphism

Ξ_{∗}: H^{1}(Γ, Hom_{ℤ}(C, μ_{n})) ⟶ H^{1}(Γ, L̄^{∗}/k̄^{∗})

is surjective. Proof: For any e ∈ M_{L} we have the factorization from (1.15)

φ_{L}: Γ ⟶^{τ_{e}} C⋊ Aut(C) ⟶^{σ_{e}} Autk̄(L) .

Set Γ′ := τ_{e}^{−1}(C⋊{1}) (this group is independent of the choice of e), then we see that Γ′ acts trivially on C. The field extension k′ := k̄^{Γ′}|k has the degree n′ := (Γ : Γ′), which divides φ(n) and thus is prime to n. Therefore (4.2) says that

Ξ′_{∗}: H^{1}(Γ′, Hom_{ℤ}(C, μ_{n})) ⟶ H^{1}(Γ′, L̄^{∗}/k̄^{∗})

is surjective. Consider that in this case—the situation of k′ as base field—we have (L′ := L ⊗_{k} k′, C) as (twisted) cyclic extension of degree n, k̄′ = k̄, L̄′ = L′ ⊗_{k′} ⊗k̄ = L ⊗_{k} k̄ = L̄ and M_{L′} = M_{L}, especially L̄^{′∗}/k̄^{′∗} = L̄^{∗}/k̄^{∗}. Now there is the commutative diagram

�  ^{Ξ′_{∗}} �  _{Ξ_{∗}} �_{cor} �  _{cor} H^{1}(Γ, L̄^{∗}/k̄^{∗}) . H^{1}(Γ, Hom_{ℤ}(C, μ_{n}))  H^{1}(Γ′, L̄^{∗}/k̄^{∗}) H^{1}(Γ′, Hom_{ℤ}(C, μ_{n}))

Since H^{1}(Γ, L̄^{∗}/k̄^{∗}) is n-torsion the multiplication with n′ is surjective, but this is also cor ∘ res, therefore cor is surjective. Since now cor ∘Ξ′_{∗} = Ξ_{∗} ∘ cor is surjective, this is also true for Ξ_{∗}.

2. Galois Actions on Ā_{0}

In this section we are interested in the following problem: Given a k-algebra A such that Ā ≅ Ā_{0} as k̄-algebras. How does the action (id_{A} ⊗γ) of γ ∈ Γ on Ā look like, after “transporting” it to Ā_{0}? Especially in the case when we have additional structures like twisted cyclic decompositions of A.

(4.4) Notation. Let σ ∈ S_{n} be a permutation, then we denote by E_{σ} the “permutational matrix” in GL_{n}(k), which is defined by the condition that it maps the i-th canonical basis vector e_{i} ∈ k^{n} to e_{σ(i)}. For example, if π_{0} := (1 2 … n) ∈ S_{n}, then we have Y_{0} = E_{π_{0}} and more general Y_{0}^{ν} = E_{π_{0}^{ν}} for any ν ∈ ℤ.

(4.5) Lemma. Let X ∈ GL_{n}(k) be an invertible matrix such that the inner automorphism κ_{X}: A_{0} \underrightarrow{∼} A_{0} maps L_{0} to L_{0}. Then X has the form

X = D · E_{σ} ,

where D is a diagonal matrix, i.e., D ∈ L_{0}, and σ ∈ S_{n} is uniquely defined by

κ_{X}(E_{ii}) = E_{σ(i),σ(i)} .

This decomposition of X into a product of a diagonal matrix and a permutational matrix is therefore unique.

Proof: Since κ_{X} is also an automorphism of the k-algebra L_{0} it permutes the primitive idempotents E_{11}, …, E_{nn}. Hence define σ ∈ S_{n} by κ_{X}(E_{ii}) = E_{σ(i),σ(i)}. So we have X ·E_{ii} = E_{σ(i),σ(i)} ·X for all i = 1, 2, …, n, and the image of the right-hand-side endomorphism is k · e_{σ(i)} whereas the image of the left-hand-side endomorphism is X(k · e_{i}). Therefore X must map k · e_{i} onto k · e_{σ(i)}. So X ∘ E_{σ^{−1}} maps k · e_{σ(i)} onto itself, and that means that it is a diagonal matrix.

(4.6) Lemma. Additionally to the situation in (4.5) we assume that κ_{X} maps K_{0} to K_{0}. Then D has the form

D = η · X_{0}(ζ)

for a n-th root of unity ζ ∈ μ_{n}(k) and η ∈ k^{∗}. Moreover σ ∈ N_{⟨π_{0}⟩} ⊆ S_{n} is an element of the normalizer subgroup of ⟨π_{0}⟩, where again π_{0} = (1 2 … n) ∈ S_{n}.

Proof: Since κ_{X}(Y_{0}) = XY_{0}X^{−1} = DE_{σ}E_{π_{0}}E_{σ}^{−1}D^{−1} = DE_{σπ_{0}σ^{−1}}D^{−1} ∈ K_{0} = k[Y_{0}] there exists a λ ∈ k^{∗} and (i mod nℤ) ∈ ℤ/nℤ such that κ_{X}(Y_{0}) = λY_{0}^{i}, i.e., E_{σπ_{0}σ^{−1}} = Y_{0}^{i} or equivalently σπ_{0}σ^{−1} = π_{0}^{i} —hence the last claim follows. Since the order of Y_{0} is n, the same holds for the order of κ_{X}(Y_{0}) and that means (i mod nℤ) ∈ (ℤ/nℤ)^{∗}. Furthermore (λY_{0}^{i})^{n} = κ_{X}(Y_{0})^{n} = E we have λ ∈ μ_{n}(k). Let (j mod nℤ) ∈ (ℤ/nℤ)^{∗}, such that ji ≡ 1 (mod nℤ). We have

XE_{σ}^{−1}e_{n} = De_{n} = ηe_{n}

for some η ∈ k^{∗}. Claim: D = ηX_{0}(ζ) with ζ = λ^{j}. All we have to check is De_{ν} = ηζ^{ν}e_{ν}. For e_{n} this is the definition of η. Now we know

XY_{0} = λY_{0}^{i}X and E_{σ}Y_{0}E_{σ}^{−1} = Y_{0}^{i}

and the latter can be written as E_{σ}Y_{0}^{j}E_{σ}^{−1} = Y_{0}, i.e., E_{σ}^{−1}Y_{0} = Y_{0}^{j}E_{σ}^{−1}. Therefore

De_{ν} = (XE_{σ}^{−1})(Y_{0}^{ν}e_{n})
= XE_{σ}^{−1}Y_{0}^{ν}e_{n}
= XY_{0}^{νj}E_{σ}^{−1}e_{n}
= λ^{νj}Y_{0}^{νji}XE_{σ}^{−1}e_{n}
= ζ^{ν}Y_{0}^{ν}ηe_{n}
= ηζ^{ν}e_{ν} .

(4.7) Remark. Of course both lemmas also go into the other direction, i.e., if X = D · E_{σ} or X = ηX_{0}(ζ)E_{σ} with σ ∈ N_{⟨π_{0}⟩}, then κ_{X} leaves L_{0}, or both L_{0} and K_{0} respectively, invariant.

Now let A be a central simple k-algebra of degree n and L ⊆ A a separable commutative k-subalgebra of k-dimension n. Then L is a maximal commutative k-subalgebra of A. We fix an isomorphism α: Ā \underrightarrow{∼} Ā_{0} of k̄-algebras with α(L̄) = L̄_{0}. Such an isomorphism always exist: If α′: Ā \underrightarrow{∼} Ā_{0} is any isomorphism of k̄-algebras, then the set α(M_{L}) is a set of—as projectors—diagonalizable and commuting endomorphisms in Ā_{0} = M_{n}(k̄). Therefore they are simultaneously diagonalizable by say G ∈ GL_{n}(k̄). Now define α := κ_{G} ∘ α′.

(4.8) Claim. We endow the group PGL_{n}(k̄) = GL_{n}(k̄)/k̄^{∗} with the canonical Γ-action (or we just say PGL_{n}(k̄) := Ā_{0}^{∗}/k̄^{∗}.) For every γ ∈ Γ there exists a unique element

(X_{γ} mod k̄^{∗}) ∈ PGL_{n}(k̄)

such that the diagram

�  ^{(id_{A} ⊗γ)} � _{κ_{Xγ} ∘(id_{A_{0}} ⊗γ)} �_{α} �  _{α} Ā_{0} Ā_{0}  Ā Ā

is commutative, i.e., α_{∗}(id_{A} ⊗γ) := α ∘ (id_{A} ⊗γ) ∘ α^{−1} = κ_{X_{γ}} ∘ (id_{A_{0}} ⊗γ). Moreover (X_{γ} mod k̄^{∗}) defines a 1-cocycle

(X_{γ} mod k̄^{∗}) ∈ 𝒵^{1}(Γ, PGL_{n}(k̄)) .

Proof: Since the morphism α_{∗}(id_{A} ⊗γ) ∘ (id_{A_{0}} ⊗γ)^{−1}: Ā_{0} \underrightarrow{∼} Ā_{0} is obviously k̄-linear we get by Skolem-Noether’s theorem a unique element (X_{γ} mod k̄^{∗}) ∈ PGL_{n}(k̄) such that κ_{X_{γ}} = α_{∗}(id_{A} ⊗γ)∘(id_{A_{0}} ⊗γ)^{−1}, hence κ_{X_{γ}} ∘(id_{A_{0}} ⊗γ) = α_{∗}(id_{A} ⊗γ). Let γ, γ′ ∈ Γ, then

κ_{X_{γγ′}} = α ∘ (id_{A} ⊗γγ′) ∘ α^{−1} ∘ (id_{A_{0}} ⊗γγ′)^{−1}
= α ∘ (id_{A} ⊗γ) ∘ α^{−1} ∘ α ∘ (id_{A} ⊗γ′) ∘ α^{−1} ∘ (id_{A_{0}} ⊗γ′)^{−1} ∘ (id_{A_{0}} ⊗γ)^{−1}
= α ∘ (id_{A} ⊗γ) ∘ α^{−1} ∘ κ_{X_{γ′}} ∘ (id_{A_{0}} ⊗γ)^{−1}
= κ_{X_{γ}} ∘ (id_{A_{0}} ⊗γ) ∘ κ_{X_{γ′}} ∘ (id_{A_{0}} ⊗γ)^{−1}
= κ_{X_{γ}} ∘ κ_{(id_{A_{0}} ⊗γ)(X_{γ′})}
= κ_{X_{γ}(id_{A_{0}} ⊗γ)(X_{γ′})} ,

hence X_{γγ′} ≡ X_{γ}(id_{A_{0}} ⊗γ)(X_{γ′}) (mod k̄^{∗}), and (X_{γ} mod k̄^{∗}) is a crossed homomorphism.

(4.9) Notation. If there is no way of misunderstanding we will write γ instead of (id_{A} ⊗γ) and (id_{A_{0}} ⊗γ), and we will write α_{∗}(γ) for α_{∗}(id_{A} ⊗γ).

(4.10) Remark. This is the first step (of two steps) in the definition of a morphism— which will be the opposite of the standard morphism—from the Brauer group Br(k) to the cohomology group H^{2}(k, k̄^{∗}); cf. [SeLF], X, §2 and §5.

(4.11) Notation. The set of the primitive idempotents of L_{0} is

M_{L_{0}} = {E_{11}, E_{22}, …, E_{nn}} .

We identify this set with the set {1, …, n} via i ↦ E_{ii}, such that we also can easily identify S(M_{L_{0}}) with S_{n}. The isomorphism α: L̄ \underrightarrow{∼} L̄_{0} gives a bijection α: M_{L} \underrightarrow{∼} M_{L_{0}} = \{1, …, n\}. This gives rise to the composition

Γ ⟶^{φ_{L}} S(M_{L}) ⟶^{α_{∗}} S(M_{L_{0}}) = S_{n} ,

and we will denote for any γ ∈ Γ,

γ̃ := α_{∗}(φ_{L}(γ)) ∈ S_{n} .

§ 4. Existence of twisted cyclic decompositions

In other words we have the commutative diagram

� ^{φ_{L}(γ)=(id_{A} ⊗γ)} � �  _{γ̃} �_{α} �  _{α} � �  {1, …, n} {1, …, n}  M_{L_{0}} M_{L_{0}}  M_{L} M_{L}

and the middle horizontal arrow is defined by E_{ii} ↦ E_{γ̃(i),γ̃(i)}. We have

E_{γ̃(i),γ̃(i)} = α_{∗}(id_{A} ⊗γ)(E_{ii}) = α_{∗}(id_{A} ⊗γ)(id_{A_{0}} ⊗γ)^{−1}(E_{ii}) = κ_{X_{γ}} (E_{ii}) . (4.12) Remark. Since both α_{∗}(id_{A} ⊗γ) and (id_{A_{0}} ⊗γ) leave L̄_{0} invariant we know by (4.5) about the matrix X_{γ}, that it has the form X_{γ} = D_{γ} · Eγ̃, where D_{γ} is a diagonal matrix and (D_{γ} mod k̄^{∗}) ∈ L̄_{0}^{∗}/k̄^{∗} is unique. But we can say more about D_{γ}: (4.13) Proposition. There is a unique 1-cocycle

(D̃_{γ} mod k̄^{∗}) ∈ 𝒵^{1}(Γ, L̄^{∗}/k̄^{∗})

such that

D_{γ} ≡ α(D̃_{γ}) (mod k̄^{∗}) .

Proof: Since α: Ā \underrightarrow{∼} Ā_{0} yields am isomorphism α: L̄^{∗}/k̄^{∗} \underrightarrow{∼} L̄_{0}^{∗}/k̄^{∗} just define

D̃_{γ} mod k̄^{∗} := α^{−1}(D_{γ} mod k̄^{∗}) .

All we have to prove is that (D̃_{γ} mod k̄^{∗}) is a 1-cocycle, i.e., for γ, γ′ ∈ Γ,

D̃_{γγ′} ≡ D̃_{γ} · (id_{A} ⊗γ)(D̃_{γ′}) (mod k̄^{∗}) .

Applying α_{∗} this is equivalent to showing

D_{γγ′} ≡ D_{γ} · α_{∗}(id_{A} ⊗γ)(D_{γ′}) (mod k̄^{∗}) .

We know from (4.8) and (4.12)

D_{γγ′}Eγ̃_{}_{γ̃′} ≡ X_{γγ′}
≡ X_{γ}(id_{A_{0}} ⊗γ)(X_{γ′})
≡ D_{γ}Eγ̃(id_{A_{0}} ⊗γ)(D_{γ′}E_{}_{γ̃′})
≡ D_{γ}Eγ̃(id_{A_{0}} ⊗γ)(D_{γ′})E_{}_{γ̃′} (mod k̄^{∗}) ;

and this is equivalent to

D_{γγ′} ≡ D_{γ}κ_{Eγ̃} ((id_{A_{0}} ⊗γ)(D_{γ′})) (mod k̄^{∗}) .

But the latter equivalence shows that κ_{Eγ̃} ((id_{A_{0}} ⊗γ)(D_{γ′})) is diagonal, so it is equal to

κ_{D_{γ}} ∘ κ_{Eγ̃} ((id_{A_{0}} ⊗γ)(D_{γ′})) = α_{∗}(id_{A} ⊗γ)(D_{γ′}) .

Replacing this in the last equation we get our cocycle condition.

3. The Existence of a Twisted Cyclic Decomposition

Again, let A be a central simple k-algebra of degree n, let (L, C) be a twisted cyclic extension of k of degree n and assume that L ⊆ A is a subalgebra of A. Choose a generator π ∈ C of C and an isomorphism α: Ā \underrightarrow{∼} Ā_{0} of k̄-algebras such that α(L̄) = L̄_{0}. We may assume that α_{∗}(π) = α ∘ π ∘ α^{−1} ∈ S(M_{L_{0}}) = S_{n} is the permutation π_{0} := (1 2 … n) ∈ S_{n}, i.e., (α ∘ π ∘ α^{−1})(E_{ii}) = E_{i+1,i+1}. The index is viewed modulo n. If this is not already the case, then there exists a σ ∈ S_{n} such that σα_{∗}(π)σ^{−1} = π_{0}. If we change α to κ_{E_{σ}} ∘ α, then we have our desired property.

(4.14) Lemma. In this situation, the composition

Hom_{ℤ}(C, μ_{n}) ⟶^{Ξ} L̄^{∗}/k̄^{∗} ⟶^{α} L̄_{0}^{∗}/k̄^{∗}

can be described as

(α ∘ Ξ)(u) = [X_{0}(u(π))] .

Proof: We choose for the primitive idempotent e ∈ M_{L} the element e := α^{−1}(E_{nn}). Then

(α ∘ Ξ)(u) = α [∑_{ν=0}^{n−1} u(π^{ν})π^{ν}(e)]

= [∑_{ν=0}^{n−1} u(π)^{ν}α_{∗}(π)^{ν}(E_{nn})]

= [∑_{ν=0}^{n−1} u(π)^{ν}E_{π_{0}^{ν}(n),π_{0}^{ν}(n)}]

= [∑_{ν=0}^{n−1} u(π)^{ν}E_{ν,ν}]

= [X_{0}(u(π))] .

(4.15) Theorem. Assume that L is a field and (n, φ(n)) = 1. Then there exists a separable k-subalgebra K of A of degree n such that (A, L, K) is a twisted cyclic k-algebra of degree n. Moreover C = C_{(A,L,K)}.

Proof: Let (X_{γ} mod k̄^{∗}) ∈ 𝒵^{1}(Γ, PGL_{n}(k̄)) be the 1-cocycle constructed in (4.8). (This, of course, depends on the choice of α—we take the α from above.) Then, with D_{γ} := X_{γ}E−1 ; γ̃, by (4.13) there exists a 1-cocycle (D̃_{γ} mod k̄^{∗}) ∈ 𝒵^{1}(Γ, L̄^{∗}/k̄^{∗}) such that D_{γ} ≡ α(D̃_{γ}) (mod k̄^{∗}). Because of (4.3) there is a 1-cocycle

u_{γ} ∈ 𝒵^{1}(Γ, Hom_{ℤ}(C, μ_{n}))

such that

[Ξ ∘ u_{γ}] = [D̃_{γ}] ∈ H^{1}(Γ, L̄^{∗}/k̄^{∗}) ,

i.e., there is a Δ ∈ L̄^{∗} such that

D̃_{γ} ≡ (Ξ ∘ u_{γ}) ·^{γ}Δ · Δ^{−1} (mod k̄^{∗})

for all γ ∈ Γ. Applying α and defining Δ_{0} := α(Δ) we get with (4.14)

D_{γ} ≡ X_{0}(u_{γ}(π)) · α_{∗}(id_{A} ⊗γ)(Δ_{0}) · Δ_{0}^{−1} (mod k̄^{∗}) ,

For all γ ∈ Γ. But

α_{∗}(id_{A} ⊗γ) = Eγ̃ · (id_{A_{0}} ⊗γ)(Δ_{0}) · E−1 ; γ̃ (mod k̄^{∗}) ,

since Δ_{0} ∈ L̄_{0} is diagonal. Therefore

Δ_{0} · D_{γ} · Eγ̃ ≡ X_{0}(u_{γ}(π)) · Eγ̃ · (id_{A_{0}} ⊗γ)(Δ_{0}) (mod k̄^{∗}) ;

in other words: The diagram

� ^{κ_{(Dγ Eγ̃)}∘(id_{A_{0}} ⊗γ)} � _{κ_{(X_{0}(uγ (π))Eγ̃)}∘(id_{A_{0}} ⊗γ)} �_{κ_{Δ_{0}}} �  _{κ_{Δ_{0}}} Ā_{0} Ā_{0}  Ā_{0} Ā_{0}

is commutative. If we replace α by κ_{Δ_{0}} ∘ α, then

X_{γ} = X_{0}(u_{γ}(π)) · Eγ̃

and (4.7) says that κ_{X_{γ}}, and also κ_{X_{γ}} ∘ (id_{A_{0}} ⊗γ), leaves L̄_{0} and K̄_{0} invariant. Define K̄ := α^{−1}(K̄_{0}), then K̄ ⊆ Ā is Γ-invariant. Define K to be K̄^{Γ}, and we are done. The last statement of the theorem follows, e.g., from the uniqueness of C, cf. (1.30).

§ 5. A Cohomological Description of Twisted Cyclic Algebras

In this section we fix a twisted cyclic k-algebra (A, L, K) of degree n—or equivalently— a pair of twisted cyclic extensions ((L, C), (K, D), 𝔠) of degree n (which is associated to this algebra.) The well known isomorphism between the Brauer group Br(k) of k and the cohomology group H^{2}(k, k̄^{∗}) gives us a two-dimensional cohomology class [A], associated to the central simple k-algebra A. On the other hand we constructed in §2 the one-dimensional cohomology classes [L, C] and [K, D]. In this rather technical section we will show how these three classes are related.

1. The Theorem

In §2 we constructed the cohomology classes

[L, C] ∈ H^{1}(Γ, C) and [K, D] ∈ H^{1}(Γ, D) .

Their cup-product is a two-dimensional cohomology class

[L, C] ∪ [K, D] ∈ H^{2}(Γ, C ⊗_{ℤ} D) .

The isomorphism 𝔠: C ⊗_{ℤ} D \underrightarrow{∼} μ_{n} of Γ-modules induces the isomorphism

𝔠_{∗}: H^{2}(Γ, C ⊗_{ℤ} D) \underrightarrow{∼} H^{2}(Γ, μ_{n})

of the cohomology groups, and the right-hand-side group can canonically be identified with the n-torsion part of H^{2}(Γ, k̄^{∗}). (Apply cohomology to the short exact sequence 1 → μ_{n} ⟶ k̄^{∗} ⟶^{n} k̄^{∗} → 1 and use Hilbert’s Theorem 90 to get the exact sequence 1 ⟶^{δ} H^{2}(Γ, μ_{n}) ⟶ H^{2}(Γ, k̄^{∗}) ⟶^{n} H^{2}(Γ, k̄^{∗}).) Therefore

𝔠_{∗}([L, C] ∪ [K, D]) ∈ H^{2}(Γ, k̄^{∗}) .

If we use the well known identification H^{2}(Γ, k̄^{∗}) = Br(k̄)—we will use the construction of Serre in [SeLF], X, which is the opposite of the standard morphism, constructed by means of “crossed products”—we can see this class as an element of the Brauer group of k. On the other hand we have the element [A] ∈ Br(k). They are connected in the following way:

(5.1) Theorem. Let ((L, C), (K, D), 𝔠) be a pair of twisted cyclic extensions of degree n and assume that Γ acts transitively on M_{L}, then

𝔠_{∗}([L, C] ∪ [K, D]) = −[A((L, C), (K, D), 𝔠)] .

(5.2) Remark. The additional assumption is true, e.g., if L is a field. Of course for reasons of symmetry, one can alternatively assume that Γ acts transitively on M_{K}.

We will show the theorem by a—lengthy—direct computation with cocycles. In order to do that we first have to compute a 2-cocycle which represents the cohomology class [A] ∈ Br(k) = H^{2}(Γ, k̄^{∗}) and two 1-cocycles representing the cohomology classes [L, C] ∈ H^{1}(Γ, C) and [K, D] ∈ H^{1}(Γ, D).

Recall that the construction given in [SeLF], X, §2 and §5 has two steps: Starting with a central simple algebra of degree n, one first gets (as shown in (4.8)) a cohomology class in H^{1}(Γ, PGL_{n}(k̄)). By virtue of the connecting morphism δ: H^{1}(Γ, PGL_{n}(k̄)) ⟶ H^{2}(Γ, k̄^{∗}) that appears in the long exact cohomology sequence which is associated to the short exact sequence

1 ⟶ k̄^{∗} ⟶ GL_{n}(k̄) ⟶ PGL_{n}(k̄) ⟶ 1

of (non-abelian) Γ-groups, we get the cohomology class [A].

Actually, we will not go all the way but compare the cohomology class of A with the cup-product in the left-hand group H^{1}(Γ, PGL_{n}(k̄)). In order to do that we have to find a pre-image of the cup-product.

Moreover we are not working with the classes [L, C] and [K, D] but—and this is equivalent because of (2.12) and the remark before (3.18)—with the more explicit [h_{(L,ρ_{L})}] ∈ H^{1}(Γ, K_{(K,L)}/μ_{n}) and [h_{(K,ρ_{K})}] ∈ H^{1}(Γ, K_{(L,K)}/μ_{n}) constructed in §2, 2.

2. Cocycles Associated to a Twisted Cyclic Algebra

For our twisted cyclic k-algebra (A, L, K) choose an isomorphism α: Ā \underrightarrow{∼} Ā_{0} of k̄- algebras as in (3.3). We adopt the notations of (4.11) and set again π_{0} := (1 2 … n) ∈ S_{n} = S(M_{L_{0}}). Let Y := α^{−1}(Y_{0}) ∈ K_{(K,L)} and e := α^{−1}(E_{nn}) ∈ M_{L}. Then define π := ρ_{L}([Y]) ∈ C. Since Y and Y_{0} have order n the cyclic group C is generated by π.

(5.3) Lemma. The morphism

α_{∗}: S(M_{L}) ⟶ S(M_{L_{0}}) = S_{n}

maps π to π_{0}.

Proof: Let i = 1, …, n, then

α_{∗}(π)^{i}(E_{nn}) = α(π^{i}(e))
= α(Y^{i}eY^{i})
= Y_{0}^{i}E_{nn}Y_{0}^{−i}
= E_{ii} .

This implies our claim.

Now we want to describe the cohomology class in H^{1}(Γ, PGL_{n}(k̄)) assigned to the algebra A, which is mapped to [A] by the morphism δ: H^{1}(Γ, PGL_{n}(k̄)) ⟶ H^{2}(Γ, k̄^{∗}) mentioned above.

We have done this already in §4, 2.: The cohomology class is given by

[X_{γ} mod k̄^{∗}] ∈ H^{1}(Γ, PGL_{n}(k̄)) ,

where X_{γ} is the cocycle we got in (4.8). Now (4.6) tells us that we can write

X_{γ} = X_{0}(ξ_{γ}) · Eγ̃ ,

where ξ_{γ} ∈ μ_{n} for every γ ∈ Γ. Let us denote the pre-image of X_{0}(ξ_{γ}) via α by

X(ξ_{γ}) := α^{−1}(X_{0}(ξ_{γ})) ∈ K_{(L,K)} .

For the proof we want to write PGL_{n}(k̄) in a different way: Define the k-vector space V to be

V := ⨁_{e′∈M_{L}} k · e′

and V̄ := V ⊗_{k} k̄ with the canonical Γ-action: Γ does not permute the components. So the linear map

\begin{matrix}β: k̄^{n} & ⟶ V̄ \\ e_{i} & ⟼ 1 · π^{i}(e) \end{matrix}

for i = 1, ⋯, n, is an isomorphism of Γ-modules and it induces an isomorphism

β_{∗}: PGL_{n}(k̄) = (GL_{n}(k) ⊗_{k} k̄)/k̄^{∗} \underrightarrow{∼} PGL(V̄) = (GL(V) ⊗_{k} k̄)/k̄^{∗} .

of Γ-groups. Let now [q_{γ} mod k̄^{∗}] denote the image of [X_{γ} mod k̄^{∗}] under the induced isomorphism

β_{∗}: H^{1}(Γ, PGL_{n}(k̄)) \underrightarrow{∼} H^{1}(Γ, PGL(V̄)) ,

i.e., q_{γ} ∈ GL(V̄) is some automorphism of V̄ for every γ ∈ Γ. In order to describe q_{γ} better let us make some conventions on notations:

(5.4) Notation. If e′, e′′ ∈ M_{L} are some primitive idempotents of L̄, then there a unique c ∈ C such that e′′ = c(e′). We will write

\frac{e′′}{e′} := c .

Furthermore, if c = π^{ν} for some ν̄ ∈ ℤ/nℤ, then we will write

log_{π}(c) := ν̄ .

(5.5) Lemma. A representative q_{γ} of the class [q_{γ} mod k̄^{∗}] ∈ H^{1}(Γ, PGL(V̄)) is given by

\begin{matrix}q_{γ}: ⨁_{e′∈M_{L}} k̄ · e′ & ⟶ ⨁_{e′∈M_{L}} k̄ · e′ \\ e′ & ⟼ ξ_{γ}^{log_{π}( \frac{γ(e′)}{e})} · γ(e′) , \end{matrix}

where γ(e′) is an abbreviation for (id_{A} ⊗γ)(e′) = φ_{L}(γ)(e′), which we will use in the following in connection with the action on the idempotents; cf. (4.9).

Proof: We denote the linear morphism defined in the lemma by q′_{γ}. We have to show for γ ∈ Γ,

β_{∗}^{−1}(q′_{γ}) ≡ D_{γ}Eγ̃ (mod k̄^{∗}) ,

and we will show

q′_{γ}(β(e_{i})) = β(D_{γ}Eγ̃(e_{i}))

for i = 1, …, n, and e_{i} is still the i-th basis vector of k̄^{n}. First we claim: γ̃(i) = log_{π}(\frac{γπ^{i}(e)}{e}). Since α maps e to E_{nn} ∈ M_{L_{0}} or—equivalently in our convention of identifying M_{L_{0}} with the set {1, …, n}—to n ∈ {1, …, n} and since α_{∗}: S(M_{L}) → S_{n} maps γπ^{i} to γ̃π_{0}^{i} we have

α(γπ^{i}(e)) = α_{∗}(γπ^{i})α(e) = γ̃π_{0}^{i}(n) = γ̃(i) = π_{0}^{γ̃(i)}(n) = α(π^{γ̃(i)}(e)) .

Hence γπ^{i}(e) = π^{γ̃(i)}(e)—the claim. Now

β(D_{γ}Eγ̃e_{i}) = β(X_{0}(ξ_{γ})e_{γ̃(i)})
= ξ_{γ}^{γ̃(i)} · β(e_{γ̃(i)})
= ξ_{γ}^{γ̃(i)} · π^{γ̃(i)}(e)
= ξ_{γ}^{γ̃(i)} · γπ^{i}(e)
= ξ_{γ}^{γ̃(i)} · γ(β(e_{i}))
= q′_{γ}(β(e_{i})) ,

since γ̃(i) = log_{π}(\frac{γπ^{i}(e)}{e}) = log_{π}(\frac{γβ(e_{i})}{e}).

(5.6) Remark. If—in the description of the map q_{γ}—one replaces the idempotent e by any other primitive idempotent, then one gets the same 1-cocycle in 𝒵^{1}(Γ, PGL(V̄)). If one changes π to some other generator of C one gets a different cocycle, but the new one still represents the same cohomology class.

3. The 1-Cocycles Associated to the Pairs (L, ρ_{L}) and (K, ρ_{K})

We still keep the notations from above. For the 1-cocycle h_{L} := h_{(L,ρ_{L})} which represents the cohomology class [h_{(L,ρ_{L})}] ∈ H^{1}(Γ, K_{(K,L)}/μ_{n}), constructed in §2, we know by (2.8) that it is given by the formula

(id_{A} ⊗γ)(e) = ρ_{L}(h_{L}(γ))(e)

for all γ ∈ Γ.

(5.7) Lemma. A 1-cocycle h_{L}: Γ ⟶ K_{(K,L)}/μ_{n} representing the class [h_{(L,ρ_{L})}] is given by

h_{L}(γ) = (Y^{log_{π}( \frac{γ(e)}{e})} mod μ_{n})

for all γ ∈ Γ.

Proof: We have to check the formula from above. It suffices to prove this formula after applying α to it, i.e., we have to prove

α_{∗}(id_{A} ⊗γ)(αe) = ρ_{L_{0}}(Y_{0}^{log_{π}( \frac{γ(e)}{e})} mod μ_{n})(αe) .

First we claim

log_{π}(\frac{γ(e)}{e}) = γ̃(n) .

Let γ(e) = π^{i}(e) for some i = 1, …, n, then

γ̃(n) = α_{∗}(γ(e)) = α_{∗}(π^{i}(e)) = π_{0}^{i}(n) = i .

Now with (4.8)

α_{∗}(id_{A} ⊗γ)(αe) = κ_{X_{γ}} ∘ (id_{A_{0}} ⊗γ)(E_{nn})

= X_{0}(ξ_{γ})Eγ̃E_{nn}E−1 ; γ̃X_{0}(ξ_{γ})^{−1}

= X_{0}(ξ_{γ})E_{γ̃(n),γ̃(n)}X_{0}(ξ_{γ})^{−1}

= E_{γ̃(n),γ̃(n)}

= Y_{0}^{γ̃(n)}E_{nn}Y_{0}^{−γ̃(n)}

= ρ_{L_{0}}(Y_{0}^{γ̃(n)} mod μ_{n})(E_{nn}) .

(5.8) Lemma. The action of Γ on Y = α^{−1}(Y_{0}) can be described as follows:

(id_{A} ⊗γ)(Y) = (ξ_{γ} · Y)^{log_{π}( \frac{γπ(e)}{γ(e)} )}

for all γ ∈ Γ.

Proof: Again the formula will be verified after we applied α. Let φ_{L}(γ)πφ_{L}(γ)^{−1} = π^{i} for some i = 1, …, n, then γ̃π_{0}γ̃^{−1} = π_{0}^{i} and γπ(e) = γπγ^{−1}(γe) = π^{i}γ(e), hence i = log_{π}(\frac{γπ(e)}{γ(e)}). Now since

X_{0}(ξ_{γ})E_{π_{0}}X_{0}(ξ_{γ})^{−1} = X_{0}(ξ_{γ})Y_{0}X_{0}(ξ_{γ})^{−1} = ξ_{γ}Y_{0}

we have

α_{∗}(id_{A} ⊗γ)(Y_{0}) = κ_{X_{γ}} ∘ (id_{A_{0}} ⊗γ)E_{π_{0}}

= X_{0}(ξ_{γ})Eγ̃E_{π_{0}}E−1 ; γ̃X_{0}(ξ_{γ})^{−1}

= X_{0}(ξ_{γ})E_{γ̃π_{0}γ̃^{−1}}X_{0}(ξ_{γ})^{−1}

= X_{0}(ξ_{γ})E_{π_{0}^{i}}X_{0}(ξ_{γ})^{−1}

= X_{0}(ξ_{γ})E_{π_{0}}^{i}X_{0}(ξ_{γ})^{−1}

= (ξ_{γ}Y_{0})^{i} .

Like in the first case we can describe a 1-cocycle h_{K} := h_{(K,ρ_{K})} which represents the cohomology class [h_{(K,ρ_{K})}] ∈ H^{1}(Γ, K_{(L,K)}/μ_{n}) by the formula

(id_{A} ⊗γ)(g) = ρ_{K}(h_{K}(γ))(g) ,

where g ∈ M_{K} is any primitive idempotent of K̄.

Recall the isomorphism κ_{Z_{0}}: L̄_{0} \underrightarrow{∼} K̄_{0} defined in the proof of (3.7). This induces a bijection κ_{Z_{0}}: M_{L_{0}} ⟶ M_{K_{0}} and explicit computation yields

M_{K_{0}} = \{\frac{1}{n} ∑_{ν=0}^{n−1} ζ^{ν}Y_{0}^{ν} : ζ ∈ μ_{n}\} .

We will choose for g the easiest case ζ = 1, i.e., g := α^{−1}(\frac{1}{n} ∑_{ν=0}^{n−1} Y_{0}^{ν}) = \frac{1}{n} ∑_{ν=0}^{n−1} Y^{ν}.

(5.9) Lemma. A 1-cocycle h_{K}: Γ ⟶ K_{(L,K)}/μ_{n} representing the class [h_{(K,ρ_{K})}] is given by

h_{K}(γ) = (X(ξ_{γ}) mod μ_{n})

for all γ ∈ Γ.

Proof: We have X(ξ_{γ})Y X(ξ_{γ})^{−1} = ξ_{γ}Y and therefore

(id_{A} ⊗γ)(g) = (id_{A} ⊗γ)(\frac{1}{n} ∑_{ν=0}^{n−1} Y^{ν})

= \frac{1}{n} ∑_{ν=0}^{n−1}(id_{A} ⊗γ)Y^{ν}

= \frac{1}{n} ∑_{ν=0}^{n−1}(ξ_{γ} · Y)^{ν log_{π}( \frac{γπ(e)}{γ(e)} )}

= \frac{1}{n} ∑_{ν=0}^{n−1}(ξ_{γ} · Y)^{ν}

= \frac{1}{n} ∑_{ν=0}^{n−1}(X(ξ_{γ})Y X(ξ_{γ})^{−1})^{ν}

= X(ξ_{γ})gX(ξ_{γ})^{−1}

= ρ_{K}(X(ξ_{γ}) mod μ_{n})(g) .

Note that

i := log_{π}(\frac{γπ(e)}{γ(e)}) = log_{π}(\frac{γπγ^{−1}(e)}{e}) ∈ (ℤ/nℤ)^{∗} ,

since γπγ^{−1} = π^{i} has order n.

4. Proof of the Theorem

We now want to prove the Theorem, i.e., the equation

𝔠([h_{L}] ∪ [h_{K}]) = −[A] .

The cohomology class

[h_{L}] ∪ [h_{K}] ∈ H^{2}(Γ, K_{(K,L)}/μ_{n} ⊗_{ℤ} K_{(L,K)}/μ_{n})

has the 2-cocycle b_{σ,τ} := h_{L}(σ) ⊗ (id_{A} ⊗σ)h_{K}(τ) as a representative. Because of the anticommutativity of the cup-product of one-dimensional cohomology classes we will actually show

(𝔠 ∘ switch)([h_{K}] ∪ [h_{L}]) = [A] .

A 2-cocycle which represents the left side is then

\begin{aligned}a_{σ,τ} & = 𝔠(((id_{A} ⊗σ)h_{L}(τ)) ⊗ h_{K}(σ)) \\ & = ĥ_{K}(σ) · (id_{A} ⊗σ)(ĥ_{L}(τ)) · ĥ_{K}(σ)^{−1} · (id_{A} ⊗σ)(ĥ_{L}(τ))^{−1} ∈ μ_{n} , \end{aligned}

Where ĥ_{K}(σ) and ĥ_{L}(τ) are representatives of h_{K}(σ) and h_{L}(τ) respectively. In order to compare [a_{σ,τ}] ∈ H^{2}(Γ, k̄^{∗}) with [q_{γ} mod k̄^{∗}] ∈ H^{1}(Γ, PGL(V̄)) we first take a preimage of [a_{σ,τ}] under the composition

H^{1}(Γ, PGL(V̄)) ⟶^{β_{∗}^{−1}} H^{1}(Γ, PGL_{n}(k̄^{∗})) ⟶^{δ} H^{2}(Γ, k̄^{∗}) .

(5.10) Lemma. A pre-image [p_{γ} mod k̄^{∗}] ∈ H^{1}(Γ, PGL(V̄)) of [a_{σ,τ}] under δ ∘ β_{∗}^{−1} is given by

\begin{matrix}p_{γ}: ⨁_{e′∈M_{L}} k̄e′ & ⟶ ⨁_{e′∈M_{L}} k̄e′ \\ τ(e) & ⟼ a_{γ,τ} · γτ(e) \end{matrix}

for τ ∈ Γ.

(5.11) Remark. p_{γ} is well defined: If for two elements τ, τ′ ∈ Γ we suppose τ(e) = τ′(e), then a_{γ,τ} = a_{γ,τ′}, since the assumption implies (id_{A} ⊗τ)(e) = (id_{A} ⊗τ′)(e) and thus h_{L}(τ) = h_{L}(τ′), by (2.8). Moreover M_{L} = {τ(e) : τ ∈ Γ} since in (5.1) we assumed that Γ acts transitively on the set M_{L} of the primitive idempotents of L̄.

Proof: (of (5.10)). We have to show that a_{σ,τ} is cohomologous to p_{σ} ∘^{σ}p_{τ} ∘ p_{στ}^{−1}. In fact we will show that they are equal: Let σ, τ, ω ∈ Γ, then

p_{σ} ∘^{σ}p_{τ} (ω(e)) = p_{σ}(σ(a_{τ,ω}) · (τω)(e))
= a_{σ,τω} · σ(a_{τ,ω}) · (στω)(e)
= a_{σ,τ} · a_{στ,ω} · (στ)(ω(e))
= a_{σ,τ} · p_{στ} (ω(e)) .

In the third equality we used the cocycle property of a_{σ,τ}.

§ 5. A cohomological description of twisted cyclic algebras

(5.12) Lemma. For any σ, τ ∈ Γ we have the formula

log_{π}(\frac{τ(e)}{e}) · log_{π}(\frac{σπ(e)}{σ(e)}) = log_{π}(\frac{στ(e)}{σ(e)}) .

Proof: Let φ_{L}(σ)πφ_{L}(σ)^{−1} = π^{i}, i.e., log_{π}(\frac{σπ(e)}{σ(e)}) = i and set j := log_{π}(\frac{τ(e)}{e}). Then τ(e) = π^{j}(e) and we get στ(e) = σπ^{j}(e) = π^{ij}σ(e). We come to the proof of our theorem: Proof: (of (5.1)). Now what remains to show is

[q_{γ} mod k̄^{∗}] = [p_{γ} mod k̄^{∗}] ∈ H^{1}(Γ, PGL(V̄)) .

We will show for every γ ∈ Γ,

ξ_{γ}^{log_{π}( \frac{e}{γ(e)} )} · q_{γ} = p_{γ}

for the representatives from above. Let γ, τ ∈ Γ, then

ξ_{γ}^{log_{π}( \frac{e}{γ(e)} )} · q_{γ}(τ(e)) = ξ_{γ}^{log_{π}( \frac{e}{γ(e)} )} · ξ_{γ}^{log_{π}( \frac{γτ(e)}{e})} · γτ(e) = ξ_{γ}^{log_{π}( \frac{γτ(e)}{γ(e)} )} · γτ(e) by (5.5). And

p_{γ}(τ(e)) = a_{γ,τ} · γτ(e) ,

where

a_{γ,τ} = ĥ_{K}(γ) · (id_{A} ⊗γ)(ĥ_{L}(τ)) · ĥ_{K}(γ)^{−1} · (id_{A} ⊗γ)(ĥ_{L}(τ))^{−1}
= X(ξ_{γ}) · Y^{log_{π}( \frac{γτ(e)}{γ(e)} )} · X(ξ_{γ})^{−1} · Y^{− log_{π}( \frac{γτ(e)}{γ(e)} )}
= ξ_{γ}^{log_{π}( \frac{γτ(e)}{γ(e)} )} ,

where for the representatives ĥ_{K}(γ) and ĥ_{L}(τ) we used X(ξ_{γ}) and Y^{log_{π}( \frac{τ(e)}{e})} respectively. Further we used in the second equation

(id_{A} ⊗γ)(ĥ_{L}(τ)) = (id_{A} ⊗γ)(Y^{log_{π}( \frac{τ(e)}{e})})
= (ξ_{γ} · Y)^{log_{π}( \frac{γπ(e)}{γ(e)} )·log_{π}( \frac{τ(e)}{e})}
= (ξ_{γ} · Y)^{log_{π}( \frac{γτ(e)}{γ(e)} )}
= Y^{log_{π}( \frac{γτ(e)}{γ(e)} )} .

We are done.

Chapter III

The Notion of Chain Equivalence

§ 6. Kummer Elements and Kummer Relation

In §3 we have seen that a twisted cyclic k-algebra (A, L, K) gives rise to elements X ∈ K_{(L,K)} and Y ∈ K_{(K,L)} which are in a—what we are going to call—Kummer relation, i.e., XY = ζY X, where ζ ∈ μ_{n} = μ_{n}(k̄) is a primitive n-th root of unity. Since these elements generate L̄ and K̄, they already give rise to our twisted cyclic decomposition of A. We are going to use these Kummer elements in order to describe twisted cyclic decompositions.

In this section we fix a field k and a positive integer n ≥ 2 which is prime to the characteristic of k, and ζ denotes a primitive n-th root of unity in k̄. Let A be a central simple k-algebra of degree n.

1. Kummer Elements

For every element X ∈ A we have its reduced characteristic polynomial

Prd(X, t) = t^{n} − Srd_{1}(X)t^{n−1} + ⋯ + (−1)^{n} Srd_{n}(X) ∈ k[t] ,

especially the reduced trace Trd(X) = Srd_{1}(X) ∈ k and the reduced norm Nrd(X) = Srd_{n}(X) ∈ k. If A is a matrix algebra, then Nrd(X) and Trd(X) are just determinant and trace of the matrix X. (Cf. [SchQH], chap. 8, §8, p. 296, or [BI]). The condition Nrd(X) ≠ 0 is equivalent to X ∈ A^{∗}. If A = M_{n}(k), then Srd_{ν}(X) is a homogeneous polynomial of degree ν in the n^{2} entries of the matrix X. The coefficients lie in ℤ (or better: in the image of ℤ in k).

(6.1) Definition. A Kummer Element in A is an element X ∈ A such that

Prd(X, t) = t^{n} − a ,

where a ∈ k^{∗}; or equivalently X ∈ A^{∗}, Srd_{ν}(X) = 0 for all ν = 1, …, n − 1.

(6.2) Notation. We denote the set of all Kummer elements of A by

W(A) := {X ∈ A^{∗} : Prd(X, t) = t^{n} − a} .

Since W(A) is stable under scaling by elements of k^{∗}, we also call the “projective version”

W(A) := W(A)/k^{∗} = {[X] ∈ A^{∗}/k^{∗} : X ∈ W(A)}

the set of Kummer elements or more accurate Kummer lines.

§ 6. Kummer elements and Kummer relation

(6.3) Example. In A = M_{n}(k̄), we have the Kummer elements X_{0}(ζ), where ζ is any primitive n-th root of unity, and Y_{0}:

Prd(X_{0}(ζ), t) = Prd(Y_{0}, t) = t^{n} − 1 .

In fact, we have

W(M_{n}(k̄)) = {[GX_{0}(ζ)G^{−1}] : G ∈ GL_{n}(k̄)} ,

since an X ∈ W(A), scaled such that det(X) = 1, has eigenvalues EigVal(X) = μ_{n}(k̄) and therefore is similar to X_{0}(ζ) or Y_{0}.

(6.4) Remark. For a Kummer element X ∈ W(A) the reduced characteristic polynomial Prd(X, t) = t^{n} − a has n distinct roots. Hence Prd(X, t) is also the minimal polynomial of X.

(6.5) Lemma. We assume that (n!) is prime to the characteristic of k. For any X ∈ A the following conditions are equivalent:

(i) Prd(X, t) = t^{n} − a,

(ii) Trd(X) = Trd(X^{2}) = … = Trd(X^{n−1}) = 0.

Proof: We may assume k = k̄ and A = M_{n}(k̄). Furthermore we can assume that X is a triangular matrix with the diagonal elements α_{1}, …, α_{n} ∈ k. Then (i) is equivalent to

σ_{ν}(α_{1}, …, α_{n}) = 0 for all ν = 1, …, n − 1 ,

where σ_{ν}(T_{1}, …, T_{n}) ∈ ℤ[T_{1}, …, T_{n}] is the elementary symmetric polynomial of degree ν. The condition (ii) is equivalent to

τ_{ν}(α_{1}, …, α_{n}) = 0 for all ν = 1, …, n − 1 ,

where τ_{ν}(T_{1}, …, T_{n}) := T_{1}^{ν} + ⋯ + T_{n}^{ν} ∈ ℤ[T_{1}, …, T_{n}]. The equivalence of these conditions immediately follows from Newton’s formula (cf. [WaAL],§33, Aufgabe 1)

τ_{ν} − τ_{ν−1}σ_{1} + τ_{ν−2}σ_{2} − ⋯ + (−1)^{ν−1}τ_{1}σ_{ν−1} + (−1)^{ν}νσ_{ν} = 0

for 1 ≤ ν ≤ n, using the fact that ν = 1, …, n − 1 is invertible in k.

(6.6) Proposition. If n = 2, 3, then W(A) ≠ ∅.

Proof: n = 2: We may assume A is a division algebra. (Otherwise A ≅ M_{2}(k) and Y_{0} is a Kummer element.) Take any X′ ∈ A − k · 1_{A} and set

X := X′ − \frac{Trd(X′)}{2} · 1_{A} .

Then X ∈ W(A), since Trd(X) = 0. n = 3: Cf. (19.2) in [BI].

(6.7) Lemma. If X, Y ∈ GL_{n}(k̄) are invertible matrices such that XY = ζY X, then there are α, β ∈ k̄^{∗} such that we have for the eigenvalues of X and Y:

EigVal(X) = α · μ_{n} and EigVal(Y) = β · μ_{n} ,

and α^{n} = (−1)^{n−1}det(X), β^{n} = (−1)^{n−1}det(Y).

Proof: From Y^{−1}XY = ζX we see

ζ · EigVal(X) = EigVal(ζX) = EigVal(Y^{−1}XY) = EigVal(X) .

Let α ∈ EigVal(X) − {0}, then ζα, ζ^{2}α, … ∈ EigVal(X). We have

det(X) = ∏_{ξ∈μ_{n}} ξα

= ( ∏_{ξ∈μ_{n}} ξ)α^{n}

= (−1)^{n−1}α^{n} .

The same holds for Y.

(6.8) Proposition. If X, Y ∈ A^{∗} are invertible elements such that XY = ζY X, then X, Y ∈ W(A) are Kummer elements.

Proof: Since the reduced characteristic polynomial remains stable under base field extension we may assume that k = k̄ and A = M_{n}(k̄). Because of (6.7) there is an α ∈ k̄^{∗} such that X is similar to αX_{0}(ζ) ∈ W(A). The same holds for Y.

(6.9) Corollary. If X, Y ∈ A^{∗} are invertible elements such that XY = ζY X, then also XY ∈ W(A) is a Kummer element.

Proof: XY = ζY X implies (XY)Y = ζY (XY).

(6.10) Lemma. If X, Y ∈ W(A) such that XY = ζY X, then for X′, Y′ ∈ A:

\begin{aligned}XY′ = ζY′X & ⟺ Y′ ∈ k[X] · Y = Y · k[X] \\ X′Y = ζY X′ & ⟺ X′ ∈ k[Y] · X = X · k[Y] \end{aligned}

Proof: The direction “⇐” is clear since the elements of k[X] commute with X and the elements of k[Y] commute with Y. “⇒”: Let XY′ = ζY′X, then

Y^{−1}Y′X = ζY^{−1}XY′ = XY^{−1}Y′ ,

i.e., Y^{−1}Y′ commutes with X. Therefore Y^{−1}Y′ ∈ k[X], since k[X] is a maximal commutative subalgebra of A.—Analogous for X′.

2. Description of Twisted Cyclic Algebras via Elements in Kummer Relation

We start with an important technical lemma:

(6.11) Lemma. Let X, Y ∈ GL_{n}(k) such that det(X) = det(Y) = (−1)^{n−1} and XY = ζY X. Then the pair (X, Y) is similar to (X_{0}(ζ), Y_{0}), i.e., there exists a matrix G ∈ GL_{n}(k) such that

GXG^{−1} = X_{0}(ζ) and GY G^{−1} = Y_{0} .

Proof: Because of det(X) = det(Y) = (−1)^{n−1} and (6.7) we know EigVal(X) = EigVal(Y) = μ_{n}. Therefore there exists a G_{1} ∈ GL_{n}(k) such that G_{1}XG_{1}^{−1} = X_{0}(ζ).— Note that ζ ∈ k because ζ = XY X^{−1}Y^{−1}. Since conjugation by G_{1} does not change our assumption, we may for the proof assume that actually X = X_{0}(ζ). Since

XY = ζY X and XY_{0} = ζY_{0}X ,

we know by (6.10) that Y^{−1}Y_{0} = D ∈ k[X_{0}(ζ)] = L_{0} is a diagonal matrix. Let D = diag(d_{1}, …, d_{n}) for d_{1}, …, d_{n} ∈ k^{∗}, then we know, since det(Y^{−1}Y_{0}) = 1, that d_{1} ⋯ d_{n} = 1. Define (we now use the “Hilbert’s Theorem 90-trick”)

G := diag(1, d_{1}, d_{1}d_{2}, …, d_{1} ⋯ d_{n−1}) .

Then

Y_{0}^{−1}GY_{0} = diag(d_{1}, d_{1}d_{2}, …, d_{1} ⋯ d_{n}) ,

hence

G^{−1}(Y_{0}^{−1}GY_{0}) = diag(d_{1}, d_{2}, …, d_{n}) = Y^{−1}Y_{0} ,

i.e.,

GY G^{−1} = Y_{0} .

But since G ∈ L_{0} is diagonal, we still have GXG^{−1} = X = X_{0}(ζ) and we are done.

As an immediate consequence we get:

(6.12) Corollary. Let X, Y ∈ GL_{n}(k) with XY = ζY X, then there exists a G ∈ GL_{n}(k̄) and α, β ∈ k̄^{∗} such that

GXG^{−1} = αX_{0}(ζ) and GXG^{−1} = βY_{0} .

In this case α^{n} = (−1)^{n−1}det(X), β^{n} = (−1)^{n−1}det(Y).

(6.13) Definition. Let A be a central simple k-algebra of degree n and X, Y ∈ A^{∗}. Then we say (X, Y) is a ζ-pair, or X and Y are in (ζ-)Kummer relation if

XY = ζY X ,

where ζ is a primitive n-th root of unity. We will also denote this fact by the symbol

X ⟶^{ζ} Y .

(6.14) Proposition. Let A be a central simple k-algebra of degree n and (X, Y) a ζ-pair in A. Then X and Y generate A, we even have

A = ⨁_{i,j=0,...,n−1} k · X^{i}Y^{j} .

If there are elements α, β ∈ k^{∗} in the ground field such that α^{n} = (−1)^{n−1}Nrd(X) = X^{n} and β^{n} = (−1)^{n−1}Nrd(Y) = Y^{n}, then the morphism of k-algebras, given by

\begin{matrix}A & \underrightarrow{∼} A_{0} \\ X^{i}Y^{j} & ⟼ (αX_{0}(ζ))^{i}(βY_{0})^{j} \end{matrix}

for i, j = 1, …, n − 1, is an isomorphism.

Proof: There is an isomorphism of k̄-algebras

Ā = A ⊗_{k} k̄ \underrightarrow{∼} Ā_{0} = M_{n}(k̄) ,

which maps X to αX_{0}(ζ) and Y to βY_{0}: Take any isomorphism φ: Ā → Ā_{0}. Since (X, Y) is a ζ-pair (φ(X), φ(Y)) is also one. Now, take the G ∈ GL_{n}(k̄) of (6.12) and compose φ with κ_{G}. Obviously

Ā_{0} = ⨁_{i,j=0,...,n−1} k̄ · X_{0}(ζ)^{i}Y_{0}^{j} = ⨁_{i,j=0,...,n−1} k̄ · (αX_{0}(ζ))^{i}(βY_{0})^{j} ,

therefore

Ā = ⨁_{i,j=0,...,n−1} k̄ · X^{i}Y^{j} .

Now, in order to show the first part of the proposition, it is enough to prove it after the (faithfully flat) base extension k̄|k; what we have done above. The second part also follows, since the isomorphism κ_{G} ∘ φ is already defined over k.

Now we can characterize twisted cyclic algebras by ζ-pairs:

(6.15) Proposition. Let A be a central simple k-algebra of degree n and L, K ⊆ A two commutative k-subalgebras of degree n. Then the following conditions are equivalent:

(i) (L, K) is a twisted cyclic decomposition of A, (ii) There exist X ∈ L̄ and Y ∈ K̄ such that (X, Y) is a ζ-pair. In this case

L̄ = k̄[X] and K̄ = k̄[Y] .

and

\begin{aligned}X ∈ K_{(L,K)} & = \{ζ^{i}X^{j} : i, j = 0, …, n − 1\} \\ Y ∈ K_{(K,L)} & = \{ζ^{i}Y^{j} : i, j = 0, …, n − 1\} . \end{aligned}

Proof: Let (A, L, K) be a twisted cyclic k-algebra and α: Ā \underrightarrow{∼} Ā_{0} a Γ-isomorphism as in the definition (3.3). Then take X := α^{−1}(X_{0}(ζ)) and Y := α^{−1}(Y_{0}). The other direction follows with the isomorphism of (6.14) in the case k = k̄.

In (6.6) we said something about the existence of Kummer elements. Now we ask for the existence of ζ-pairs.

(6.16) Proposition. Assume that μ_{n} = μ_{n}(k̄) ⊆ k. If A is a division algebra and X ∈ W(A) a Kummer element in A, then there exist Kummer elements Y ∈ W(A) such that (X, Y) is a ζ-pair.

Proof: k[X] is a commutative subalgebra of A of dimension n. Therefore it is a field, in particular simple. The automorphism k[X] \underrightarrow{∼} k[X], X ⟼ ζ^{−1}X can be extended by Skolem-Noether to an inner automorphism κ_{Y} for some Y ∈ A^{∗}, i.e., we have Y XY^{−1} = ζ^{−1}X.

3. Galois Action on Kummer Elements and Twisted Cyclic Decompositions

Let k′|k be a Galois extension and A a central simple k-algebra. Then G := Gal(k′|k) acts on the central simple k′-algebra A_{k′} = A ⊗_{k} k′. This action gives rise to actions on the set of Kummer elements and twisted cyclic decompositions.

(6.17) Remark. For σ ∈ G and X ∈ A_{k′} we have the identity

Prd(σX, t) = σ Prd(X, t) ,

where σ acts on the coefficients of the polynomial. Therefore, if X ∈ W(A_{k′}) is a Kummer element, then σX ∈ W(A_{k′}) is also one. The same holds for W(A_{k′}).

(6.18) Proposition. The map

W(A) ⟶ W(A_{k′}) , X ⟼ X

has Galois descent, i.e., it induces the bijection

W(A) \underrightarrow{∼} W(A_{k′})^{G} .

Proof: This follows from the Galois descent of the map A ⟶ A_{k′}.

The Galois group G acts in the following way on the set of twisted cyclic decompositions of A_{k′}.

(6.19) Lemma. Let (L′, K′) be a twisted cyclic decomposition of the central simple k′- algebra A_{k′}. Then for σ ∈ G, the pair (σL′, σK′) is also a twisted cyclic decomposition of A_{k′}.

Proof: Because of (6.15) there are X ∈ L̄′ = L′ ⊗_{k′} k̄ and Y ∈ K̄′ = K′ ⊗_{k′} k̄ such that (X, Y) is a ζ-pair and L̄′ = k̄[X] and K̄′ = k̄[Y]. But this implies that (σX, σY) is a σζ-pair and \overline{σL′} = k̄[σX], \overline{σK′} = k̄[σY]. Again with (6.15) we are done.

(6.20) Proposition. The map

{twisted cyclic decompositions of A} ⟶ {twisted cyclic decompositions of A_{k′}}
(L, K) ⟼ (L ⊗_{k} k′, K ⊗_{k} k′)

has Galois descent.

Proof: The map is obviously injective and well defined (cf. (3.4)). Let (L′, K′) be a decomposition which is invariant under G. Then G acts semilinear on L′ and K′. Therefore we have for L := L^{′G} and K := K^{′G},

L′ = L ⊗_{k} k′ and K′ = K ⊗_{k} k′ ;

cf. (18.1) in [BI].

4. Description of Split Twisted Cyclic Algebras

We will give a criterion for a triple (A, L, K) of k-algebras to be isomorphic to the standard one. This will be used to prove the last remark of (3.4).

(6.21) Lemma. We assume μ_{n} ⊆ k. Let (A, L, K) be a triple, where A is a k-algebra, L and K are commutative k-subalgebras of A. Then the condition

(A, L, K) ≅ (A_{0}, L_{0}, K_{0})

is equivalent to the following set of two conditions:

(i) A ≅ A_{0}, L ≅ L_{0}, K ≅ K_{0} as k-algebras; (ii) One can number the primitive idempotents {e_{1}, …, e_{n}} of L and {f_{1}, …, f_{n}}

of K in a way such that the elements X := ζ^{1}e_{1} + ζ^{2}e_{2} + ⋯ + ζ^{n}e_{n} ∈ L and Y := ζ^{1}f_{1} + ζ^{2}f_{2} + ⋯ + ζ^{n}f_{n} ∈ K are in a Kummer relation.

Note that the point (i) implies that L ≅ k^{n} and K ≅ k^{n} and therefore they have n primitive idempotents.

Proof: If (A, L, K) ≅ (A_{0}, L_{0}, K_{0}), the point (i) is clear. The point (ii) follows, since it is true for the case (A, L, K) = (A_{0}, L_{0}, K_{0}); this we can see putting together (3.9) and (3.19). The other direction follows with (6.14): Since X^{n} = Y^{n} = 1, we can set in this proposition α = β = 1, and that gives the isomorphism (A, k[X], k[Y]) ≅ (A_{0}, L_{0}, K_{0}). But for dimension reasons the inclusions k[X] ⊆ L and k[Y] ⊆ K are actually equalities.

(6.22) Proposition. Let (A, L, K) be a triple, where A is a k-algebra, L and K are commutative k-subalgebras of A and k′|k is an algebraic (not necessarily separable) field extension. Then we have the equivalence: (A, L, K) is a twisted cyclic k-algebra of degree n ⟺ (A, L, K)_{k′} is a twisted cyclic k′-algebra of degree n.

Proof: The direction “⇒”is clear. For the other direction we may assume that k′ = k̃ is an algebraic closure of k.

Furthermore—because of (3.4)—we may assume that k = k̄ is separably closed. The separable closure of k̃ is of course k̃, therefore (Ã, L̃, K̃) := (A, L, K)k̃ is isomorphic to (Ã_{0}, L̃_{0}, K̃_{0}) := (A_{0}, L_{0}, K_{0})k̃, i.e., Ã is isomorphic to a matrix algebra. Thus A is a central simple k-algebra; and hence, A is already split, since k is its own separable closure. So A ≅ M_{n}(k) = A_{0}. Since L̃_{0} ≅ k̃^{n} and K̃_{0} ≅ k̃^{n} are diagonalizable, the same holds for L̃ and K̃, i.e., L̃ and K̃ are separable k̃-algebras of degree n. Since

n = Homk̃(L̃, k̃) = Hom_{k}(L, k̃) and n = Homk̃(K̃, k̃) = Hom_{k}(K, k̃) ,

L and K are separable k-algebras. And since k is separably closed, we get

L ≅ k^{n} ≅ L_{0} and K ≅ k^{n} ≅ K_{0} ;

in other words: The primitive idempotents e_{1}, …, e_{n} of L̃ and f_{1}, …, f_{n} of K̃ already lie in L and K respectively. Now

L = k · e_{1} ⊕ ⋯ ⊕ k · e_{n} and K = k · f_{1} ⊕ ⋯ ⊕ k · f_{n}

as well as

L̃ = k̃ · e_{1} ⊕ ⋯ ⊕ k̃ · e_{n} and K̃ = k̃ · f_{1} ⊕ ⋯ ⊕ k̃ · f_{n} .

Since (Ã, L̃, K̃) ≅ (Ã_{0}, L̃_{0}, K̃_{0}), we may, by (6.21), assume that the elements X := ζ^{1}e_{1} + ζ^{2}e_{2} + ⋯ + ζ^{n}e_{n} and Y := ζ^{1}f_{1} + ζ^{2}f_{2} + ⋯ + ζ^{n}f_{n} are in a Kummer relation. But X and Y already lie in L and K respectively. Using the other direction of (6.21) we get the claim.

§ 7. Chains of Twisted Cyclic Extensions

In this section we establish the first version of chains, give some elementary properties and classify these by virtue of certain “structure morphisms”. We fix a field k and positive integers n ≥ 2 and ℓ such that n is prime to the characteristic of k. Let A be a central simple k-algebra of degree n.

1. Definition of Chains

(7.1) Definition. A k-chain in A of length ℓ is an (ℓ + 1)-tuple (L_{0}, …, L_{ℓ}) of separable commutative k-subalgebras of A such that (L_{i}, L_{i+1}) is a twisted cyclic decomposition of A for all i = 0, …, ℓ − 1.

(7.2) Notation. We denote the set of all k-chains in A of length ℓ by the expression

Chain_{ℓ}(A) .

(7.3) Remark. If k′|k is a field extension, then we have a natural injective map

Chain_{ℓ}(A) ⟶ Chain_{ℓ}(A ⊗_{k} k′)
(L_{0}, …, L_{ℓ}) ⟼ (L_{0}, …, L_{ℓ})_{k′} ,

where (L_{0}, …, L_{ℓ})_{k′} := (L_{0} ⊗_{k} k′, …, L_{ℓ} ⊗_{k} k′); cf. (3.4).

Furthermore if k′|k is a Galois extension, then Gal(k′|k) acts on Chain_{ℓ}(A ⊗_{k} k′), in the way of (6.19). Analogously to (6.20) we have:

(7.4) Proposition. If k′|k is a Galois extension, then the map

Chain_{ℓ}(A) ⟶ Chain_{ℓ}(A_{k′}) ,

has Galois descent.

2. Chains with Fixed Starting and End Points

We fix two separable commutative k-subalgebras L, K ⊆ A of degree n. Since we are interested in the question if two separable commutative subalgebras can be connected by chains, we make the following definition.

(7.5) Definition. Let k′|k be a field extension. A k′-chain in A from L to K of length ℓ is a k′-chain (L′_{0}, …, L′_{ℓ}) ∈ Chain_{ℓ}(A_{k′}) such that

L′_{0} = L_{k′} and L′_{ℓ} = K_{k′} .

(7.6) Notation. We make the following notations

Chain_{ℓ}(L, K; A)_{k′} := {(L′_{0}, …, L′_{ℓ}) ∈ Chain_{ℓ}(A_{k′}) : L′_{0} = L_{k′}, L′_{ℓ} = K_{k′}}

and

Chain_{ℓ}(L, −; A)_{k′} := {(L′_{0}, …, L′_{ℓ}) ∈ Chain_{ℓ}(A_{k′}) : L′_{0} = L_{k′}} .

If k′ = k, we leave the index “k”.

(7.7) Remark. Obviously

Chain_{ℓ}(L, K; A)_{k′} = Chain_{ℓ}(L_{k′}, K_{k′}; A_{k′}) .

If k′|k is a Galois extension, then Gal(k′|k) acts on this set, since the Galois action leaves L_{k′} and K_{k′} fixed. An easy consequence of (7.4) is:

(7.8) Proposition. We have the natural injective map

Chain_{ℓ}(L, K; A) ⟶ Chain_{ℓ}(L, K; A)_{k′}
(L_{0}, …, L_{ℓ}) ⟼ (L_{0}, …, L_{ℓ})_{k′} .

If k′|k is a Galois extension, then this map has Galois descent.

We remind that we wrote Γ = Gal(k̄|k) for the absolute Galois group of k. Let 𝒞_{n} be the category of the Γ-modules C which are cyclic groups of order n. Then we have on 𝒞_{n} the following involutionary endo-functor

\begin{matrix}T := Hom_{ℤ}(−, μ_{n}): 𝒞_{n} & ⟶ 𝒞_{n} \\ C & ⟼ Hom_{ℤ}(C, μ_{n}) , \end{matrix}

where μ_{n} = μ_{n}(k̄). The natural transformation φ: id \underrightarrow{∼} T ∘ T is given by

φ_{C}: C ⟶ Hom_{ℤ}(Hom_{ℤ}(C, μ_{n}), μ_{n})
c ⟼ ev_{c} ,

the evaluation morphism. The following is well known.

(7.9) Lemma. For any C, D ∈ 𝒞_{n} we have the identity

Isom_{Γ}(C ⊗_{ℤ} D, μ_{n}) = Isom_{Γ}(C, T (D)) .

Let (L_{0}, …, L_{ℓ}) ∈ Chain_{ℓ}(A) be a k-chain, then (A, L_{i}, L_{i+1}) is twisted cyclic k-algebra for i = 0, …, ℓ − 1. This induces the twisted cyclic structures

C_{i} := C_{(A,L_{i},L_{i+1})} on L_{i}

and

D_{i+1} := D_{(A,L_{i},L_{i+1})} on L_{i+1}

and the isomorphism of Γ-modules

𝔠_{i} := 𝔠_{(A_{i},L_{i},L_{i+1})}: C_{i} ⊗_{ℤ} D_{i+1} \underrightarrow{∼} μ_{n} = μ_{n}(k̄) ;

cf. §3, 2.

(7.10) Definition. A coherent chain in A is a chain (L_{0}, …, L_{ℓ}) ∈ Chain_{ℓ}(A) with

C_{i} = D_{i} for all i = 1, …, ℓ − 1 .

(7.11) Notation. We write for the set of coherent chains of length ℓ,

c-Chain_{ℓ}(A) := {(L_{0}, …, L_{ℓ}) ∈ Chain_{ℓ}(A) : (L_{0}, …, L_{ℓ}) is coherent}

and

c-Chain_{ℓ}(L, K; A) := c-Chain_{ℓ}(A) ∩ Chain_{ℓ}(K, L; A) ,

and analogously for all other notations of this kind.

If (L_{0}, …, L_{ℓ}) ∈ c-Chain_{ℓ}(A) is a coherent chain, then we can compose the Γ-morphisms, given by (7.9):

T^{i}(𝔠_{i}): T^{i}(C_{i}) \underrightarrow{∼} T^{i+1}(D_{i+1}) = T^{i+1}(C_{i+1}) ,

and we get

C_{0} \underrightarrow{∼} T^{1}(C_{1}) \underrightarrow{∼} T^{2}(C_{2}) \underrightarrow{∼} ⋯ \underrightarrow{∼} T^{ℓ−1}(C_{ℓ−1}) \underrightarrow{∼} T^{ℓ}(D_{ℓ}) ,

which we will denote by

ρ_{(L_{0},...,L_{ℓ})}: C_{0} \underrightarrow{∼} T^{ℓ}(D_{ℓ}) .

Thus we can classify coherent chains by means of this isomorphism.

(7.12) Notation. For C, D ∈ 𝒞_{n} and ρ ∈ Isom_{Γ}(C, T^{ℓ}(D)) we define

Chain_{ℓ}^{ρ}(L, K; A) :=
{(L_{0}, …, L_{ℓ}) ∈ c-Chain_{ℓ}(L, K; A) : C_{0} = C, D_{ℓ} = D, ρ_{(L_{0},...,L_{ℓ})} = ρ} .

(7.13) Remark. Obviously we have the disjoint union

c-Chain_{ℓ}(L, K; A) = ∐_{C,D} ∐_{ρ} Chain_{ℓ}^{ρ}(L, K; A) ,

where C and D run through the set of twisted cyclic structures of L and K respectively and ρ runs through the set Isom_{Γ}(C, T^{ℓ}(D)).

(7.14) Proposition. If A is a division algebra and (n, φ(n)) = 1, then

c-Chain_{ℓ}(A) = Chain_{ℓ}(A) .

Proof: This follows immediately from the uniqueness of twisted cyclic structures, proven in (1.30).

(7.15) Corollary. If A is a division algebra and (n, φ(n)) = 1, and if we assume that Chain_{ℓ}(L, K; A) ≠ ∅, then there are unique twisted cyclic structures C and D on L and K respectively, and we have the disjoint union

Chain_{ℓ}(L, K; A) = ∐_{ρ∈Isom_{Γ}(C,T^{ℓ}(D))} Chain_{ℓ}^{ρ}(L, K; A) .

(7.16) Definition. Let C and D be some twisted cyclic structures on L and K respectively and let ρ ∈ Isom_{Γ}(C, T^{ℓ}(D)), then we say that L and K are (ℓ, ρ)-chain equivalent over k′, if the set

Chain_{ℓ}^{ρ}(L, K; A)_{k′} = Chain_{ℓ}^{ρ}(L_{k′}, K_{k′}; A_{k′}) ≠ ∅

is non-empty. (Note that C and D are also twisted cyclic structures on L_{k′} and K_{k′}.)

(7.17) Proposition. Let k′|k be an algebraic field extension, then we have for any (L_{0}, …, L_{ℓ}) ∈ Chain_{ℓ}(A) the equivalence

(L_{0}, …, L_{ℓ}) is coherent ⟺ (L_{0}, …, L_{ℓ})_{k′} is coherent .

In particular, there is the injective map

c-Chain_{ℓ}(A) ⟶ c-Chain_{ℓ}(A_{k′})
(L_{0}, …, L_{ℓ}) ⟼ (L_{0}, …, L_{ℓ})_{k′} ,

which has Galois descent, if k′|k is a Galois extension.

Proof: One only has to observe that the associated twisted cyclic structures C_{i} and D_{i} do not change under base extension.

Finally, we want to state a useful lemma, which we will need later.

(7.18) Lemma. Let (L_{0}, …, L_{ℓ}) ∈ Chain_{ℓ}(A) be a chain. Then (L_{0}, …, L_{ℓ}) is coherent if and only if

K_{(L_{i},L_{i−1})} = K_{(L_{i},L_{i+1})}

for i = 1, …, ℓ − 1.—More precisely: The last condition for i is equivalent to the condition C_{i} = D_{i} for i.

Proof: We fix an i ∈ {1, …, ℓ − 1}. Let d ∈ D_{i} and c ∈ C_{i} be generators of the cyclic groups and e ∈ M_{L_{i}} a primitive idempotent of L̄_{i}. Then (3.19) says

K_{(L_{i},L_{i−1})} = {ζ′ ∑_{ν=0}^{n−1} ζ^{ν}d^{ν}(e) : ζ′, ζ ∈ μ_{n}}

and

K_{(L_{i},L_{i+1})} = {ζ′ ∑_{ν=0}^{n−1} ζ^{ν}c^{ν}(e) : ζ′, ζ ∈ μ_{n}} .

If C_{i} = D_{i}, then we may choose c = d and the claim K_{(L_{i},L_{i−1})} = K_{(L_{i},L_{i+1})} is clear. If K_{(L_{i},L_{i−1})} = K_{(L_{i},L_{i+1})}, then by the following lemma

D_{i} = D_{(A,L_{i−1},L_{i})}
= C_{(A,L_{i},L_{i−1})}
= C_{(A,L_{i},L_{i+1})}
= C_{i} ,

where the first and last equality are the definitions, the second equality is the tautological one and the third equality follows from (7.19).

(7.19) Lemma. Let (A, L, K) be a twisted cyclic k-algebra of degree n, further [X] ∈ K_{(L,K)}/μ_{n} and [Y] ∈ K_{(K,L)}/μ_{n} generators of the cyclic groups. Then

C_{(A,L,K)} = \{(L̄ \underrightarrow{∼} L̄, X ⟼ ξX) : ξ ∈ μ_{n}\}

and

D_{(A,L,K)} = \{(K̄ \underrightarrow{∼} K̄, Y ⟼ ξY) : ξ ∈ μ_{n}\} .

Therefore C and D depend solely on K_{(L,K)} ⊆ L̄^{∗} and K_{(K,L)} ⊆ K̄^{∗} respectively

Proof: This is just a corollary of (3.12) and (3.16): The k̄-automorphisms are defined by the image of X and Y respectively. (3.16) shows that for these images there are only the n possibilities of multiples by elements of μ_{n}.

§ 8. Chains of Kummer Elements

If we want to consider k̄-chains in a central simple algebra A, we can make use of the interpretation of twisted cyclic decompositions we got in (6.15). We fix a field k and positive integers n ≥ 2 and ℓ such that n is prime to the characteristic of k. Let A be a central simple k-algebra of degree n.

1. Definition of Chains of Kummer Elements

First we define the chains of Kummer elements:

(8.1) Definition. A chain of Kummer elements in A of length ℓ is a ℓ + 1-tuple ([X_{0}], …, [X_{ℓ}]) ∈ W(A)^{ℓ+1} of Kummer elements such that consecutive elements are in Kummer relation, i.e.,

X_{i−1}X_{i} = ζ_{i}X_{i}X_{i−1}

for i = 1, …, ℓ and primitive n-th roots of unity ζ_{i} ∈ μ_{n} = μ_{n}(k̄).

Note that the condition X_{i−1} ⟶^{ζ_{i}} X_{i} is independent of the choice of representatives X_{i} ∈ [X_{i}].

(8.2) Notation. We denote the set of chains of Kummer elements in A of length ℓ by the symbol

K-Chain_{ℓ}(A) .

Let Z := (ζ_{1}, …, ζ_{ℓ}) be a ℓ-tuple of primitive n-th root of unity. Then we write

K-Chain_{ℓ}^{Z}(A) := {([X_{0}], …, [X_{ℓ}]) ∈ K-Chain_{ℓ}(A) : X_{i−1}X_{i} = ζ_{i}X_{i}X_{i−1}, i = 1, …, ℓ} .

If Z = (ζ, …, ζ), then we write just ζ instead of Z:

K-Chain_{ℓ}^{ζ}(A) := K-Chain_{ℓ}^{Z}(A) .

The notations K-Chain_{ℓ}([X], [Y]; A), K-Chain_{ℓ}^{Z}([X], [Y]; A), … are to be read in the same obvious way like the corresponding ones in §7.

(8.3) Remark. Clearly, for Kummer elements [X], [Y] ∈ W(A),

K-Chain_{ℓ}(A) = ∐_{Z} K-Chain_{ℓ}^{Z}(A) ,

and herein

K-Chain_{ℓ}([X], [Y]; A) = ∐_{Z} K-Chain_{ℓ}^{Z}([X], [Y]; A) ,

where Z runs through the ℓ-tuples of primitive n-th roots of unity.

How are the different sets on the right hand side connected mutually?

(8.4) Proposition. Let Z = (ζ_{1}, …, ζ_{ℓ}) and Z′ = (ζ′_{1}, …, ζ′_{ℓ}) be two ℓ-tuples of primitive n-th roots of unity. Let ν_{i} ∈ (ℤ/nℤ)^{∗} for i = 1, …, ℓ, be the unique elements such that ζ′_{i} = ζ_{i}^{ν_{i}}. Define ϵ_{0} := 1 and for i = 1, …, ℓ

ϵ_{i} := \frac{ν_{i}}{ν_{i−1}} \frac{ν_{i−2}}{ν_{i−3}} ⋯ = ∏_{j=1}^{i} ν_{j}^{(−1)^{i−j}} ,

then the map

ψ_{Z}^{Z′}: K-Chain_{ℓ}^{Z}(A) \underrightarrow{∼} K-Chain_{ℓ}^{Z′}(A)
([X_{0}], …, [X_{ℓ}]) ⟼ ([X_{0}^{ϵ_{0}}], [X_{1}^{ϵ_{1}}], …, [X_{ℓ}^{ϵ_{ℓ}}])

is a bijection. Furthermore we have the functoriality

ψ_{Z}^{Z} = id and ψ_{Z}^{Z′′} = ψ_{Z′}^{Z′′} ∘ ψ_{Z}^{Z′} ,

where Z′′ is a third ℓ-tuple of primitive n-th roots of unity. In the special case of Z = (ζ, …, ζ) and Z′ = (ζ′, …, ζ′), where ζ′ = ζ^{ν}, we get

ϵ_{i} = \{ \begin{matrix}ν, & if i is odd \\ 1, & if i is even. \end{matrix}

Proof: The map ψ_{Z}^{Z′} is well defined: First note that [X_{i}^{n}] = [1]; moreover X_{i−1} ⟶^{ζ_{i}} X_{i}

implies X_{i−1}^{ϵ_{i−1}} ⟶^{ζ′_{i}} X_{i}^{ϵ_{i}}, since ϵ_{i−1}ϵ_{i} = ν_{i} and

X_{i−1}^{ϵ_{i−1}}X_{i}^{ϵ_{i}}X_{i−1}^{−ϵ_{i−1}}X_{i}^{−ϵ_{i}} = ζ_{i}^{ϵ_{i−1}ϵ_{i}} = ζ_{i}^{ν_{i}} = ζ′_{i} .

The functoriality is obvious and it implies the bijectivity by the standard argument.

(8.5) Corollary. Let [X], [Y] ∈ W(A) be two Kummer elements. In the situation of (8.4) the map ψ_{Z}^{Z′} induces the bijections

ψ_{Z}^{Z′}: K-Chain_{ℓ}^{Z}([X], −; A) \underrightarrow{∼} K-Chain_{ℓ}^{Z′}([X], −; A)

and

ψ_{Z}^{Z′}: K-Chain_{ℓ}^{Z}([X], [Y]; A) \underrightarrow{∼} K-Chain_{ℓ}^{Z′}([X], [Y^{ϵ_{ℓ}}]; A) .

In particular, we have the bijective map

ψ_{Z}^{Z′}: ∐_{i∈(ℤ/nℤ)^{∗}} K-Chain_{ℓ}^{Z}([X], [Y^{i}]; A) \underrightarrow{∼} ∐_{i∈(ℤ/nℤ)^{∗}} K-Chain_{ℓ}^{Z′}([X], [Y^{i}]; A) .

Now we see that for many questions, it is enough to consider the case Z = (ζ, …, ζ). For these cases, we have the addendum:

(8.6) Corollary. Let ζ and ζ′ = ζ^{ν} be two primitive n-th roots of unity, such that ν ∈ (ℤ/nℤ)^{∗}. If ℓ is even, we have the bijective map

ψ_{ζ}^{ζ′}: K-Chain_{ℓ}^{ζ}([X], [Y]; A) \underrightarrow{∼} K-Chain_{ℓ}^{ζ′}([X], [Y]; A)
([X_{0}], …, [X_{ℓ}]) ⟼ ([X_{0}], [X_{1}^{ν}], [X_{2}], …, [X_{ℓ−1}^{ν}], [X_{ℓ}]) .

If ℓ is odd, we have the bijective map

ψ_{ζ}^{ζ′}: K-Chain_{ℓ}^{ζ}([X], [Y]; A) \underrightarrow{∼} K-Chain_{ℓ}^{ζ′}([X], [Y^{ν}]; A)
([X_{0}], …, [X_{ℓ}]) ⟼ ([X_{0}], [X_{1}^{ν}], [X_{2}], …, [X_{ℓ−1}], [X_{ℓ}^{ν}]) .

(8.7) Notation. For any ℓ-tuple Z of primitive n-th roots of unity and Kummer elements [X], [Y] ∈ W(A), we define

K̃-Chain_{ℓ}^{Z}([X], [Y]; A) := ∐_{i∈(ℤ/nℤ)^{∗}} K-Chain_{ℓ}^{Z}([X], [Y^{i}]; A) .

(8.8) Remark. By virtue of the bijections ψ_{Z}^{Z′}, we can canonically identify these sets for different ℓ-tuples Z and Z′.

2. The Notion of Chain Equivalence of Kummer Elements

Analogously to §7, 2. we make the following notations and definitions:

(8.9) Notation. Let k′|k be a field extensions, and let [X], [Y] ∈ W(A) be two Kummer elements, then we denote with

K-Chain_{ℓ}([X], [Y]; A)_{k′} := K-Chain_{ℓ}([X], [Y]; A_{k′})

the set of all k′-chains from [X] to [Y] in A of length ℓ.

(8.10) Definition. Two Kummer elements [X], [Y] ∈ W(A) are called (ℓ, Z, ν)-chain equivalent (or -related) over k′, if

K-Chain_{ℓ}^{Z}([X], [Y^{ν}]; A)_{k′} ≠ ∅ ,

where ν ∈ (ℤ/nℤ)^{∗} and Z an ℓ-tuple of primitive n-th roots of unity. If Z = (ζ, …, ζ) we write (ℓ, ζ, ν) instead of (ℓ, Z, ν).

(8.11) Remark. In the situation of (8.5) we can see:

[X], [Y] are (ℓ, Z, ν)-related ⟺ [X], [Y] are (ℓ, Z′, ϵ_{ℓ} · ν)-related .

Hence we always can reduce the question of being (ℓ, Z, ν)-related to the question of being (ℓ, ζ, ν′)-related, for suitable ν′.

(8.12) Lemma. Let [X], [Y] ∈ W(A) be Kummer elements, ζ, ζ′ primitive n-th roots of unity and ν, ν′ ∈ (ℤ/nℤ)^{∗}. Then we have the following equivalences:

If ℓ is even, then

[X], [Y] are (ℓ, ζ, ν)-related ⟺ [X], [Y] are (ℓ, ζ′, ν)-related .

If ℓ is odd and ζ^{ν′} = ζ^{′ν}, then

[X], [Y] are (ℓ, ζ, ν)-related ⟺ [X], [Y] are (ℓ, ζ′, ν′)-related .

Proof: The first part follows directly with (8.6). The second part has to be shown just in one direction: For Z = (ζ, …, ζ) and Z′ = (ζ′, …, ζ′) with the notations of (8.5) we have ζ′ = ζ^{ϵ_{ℓ}}, therefore ζ^{ν′} = ζ^{′ν} = ζ^{ϵ_{ℓ}·ν}, i.e., ν′ = ϵ_{ℓ} · ν ∈ (ℤ/nℤ)^{∗}. But (8.11) tells us that: [X], [Y] are (ℓ, ζ, ν)-related ⟺ [X], [Y] are (ℓ, ζ′, ϵ_{ℓ} · ν)-related .

(8.13) Remark. If [X] and [Y] are (ℓ, ζ, ν)-related, then they are (ℓ + 1, ζ, ν)-related: This follows from the fact that a chain of Kummer elements can be expanded at any link: If (X_{i−1}, X_{i}) is a ζ-pair, then (X_{i−1}, X_{i−1}X_{i}) and (X_{i−1}X_{i}, X_{i}) are also ζ-pairs. Cf. §6, 1.

3. Connection between K-Chains and c-Chains

To every chain of Kummer elements we can assign a chain of twisted cyclic extensions:

(8.14) Theorem. Let Z be a ℓ-tuple of primitive n-th roots of unity and [X], [Y] ∈ W(A) any Kummer elements. Then we have a canonical bijection

ψ_{Z}: K̃-Chain_{ℓ}^{Z}([X], [Y]; A)k̄ \underrightarrow{∼} c-Chain_{ℓ}(k[X], k[Y]; A)k̄
([X_{0}], …, [X_{ℓ}]) ⟼ (k̄[X_{0}], …, k̄[X_{ℓ}])

and for another Z′, we have the compatibility relation

ψ_{Z′} ∘ ψ_{Z}^{Z′} = ψ_{Z} .

Proof: The map is well defined, since the algebras k̄[X_{i}] are separable commutative k̄-subalgebras of Ā = Ak̄. Because of (6.15) the pairs (k̄[X_{i−1}], k̄[X_{i}]) are twisted cyclic decompositions of Ā. Furthermore one observes

K_{(k̄[X_{i}],k̄[X_{i−1}])} = {[X_{i}^{0}], …, [X_{i}^{n−1}]}

and

K_{(k̄[X_{i}],k̄[X_{i+1}])} = {[X_{i}^{0}], …, [X_{i}^{n−1}]} ,

therefore by (7.18) the chain (k̄[X_{0}], …, k̄[X_{ℓ}]) is coherent. The last formula is clear. Injectivity of ψ_{Z}: Let be given two chains ([X_{0}], …, [X_{ℓ}]) and ([X′_{0}], …, [X′_{ℓ}]) out of K̃-Chain_{ℓ}^{Z}([X], [Y]; A)k̄ which are mapped to the same coherent chain via ψ_{Z}. Let i be the least integer such that [X_{i}] ≠ [X′_{i}]. Since [X_{0}] = [X] = [X′_{0}] we know i ≥ 1.

Furthermore we know that (X_{i−1}, X_{i}) and (X_{i−1}, X′_{i}) = (X′_{i−1}, X′_{i}) are ζ_{i}-pairs, where Z = (ζ_{1}, …, ζ_{ℓ}). Since

k̄[X_{i−1}] = k̄[X′_{i−1}] and k̄[X_{i}] = k̄[X′_{i}]

the sets

K_{(k̄[X_{i}],k̄[X_{i−1}])} = K_{(k̄[X′_{i}],k̄[X′_{i−1}])}

are equal and [X_{i}] as well as [X′_{i}] are the unique element of this set which is in ζ_{i}-relation with [X_{i−1}] = [X′_{i−1}]. Hence [X_{i}] = [X′_{i}], and we have shown injectivity. Surjectivity of ψ_{Z}: Let (L̄_{0}, …, L̄_{ℓ}) ∈ c-Chain_{ℓ}(k[X], k[Y]; A)k̄ be a coherent chain. For i = 1, …, ℓ choose a generator [X_{i}] of the cyclic group K_{(L̄_{i},L̄_{i−1})} (= K_{(L̄_{i},L̄_{i+1})} for i < ℓ). Then L̄_{i} = k̄[X_{i}] and ([X_{0}], …, [X_{ℓ}]) ∈ K̃-Chain_{ℓ}^{Z′}([X], [Y]; A)k̄ for [X_{0}] := [X] and some ℓ-tuple Z′ of primitive n-th root of unity. This chain maps to (L̄_{0}, …, L̄_{ℓ}) via ψ_{Z′}. Because of ψ_{Z′} = ψ_{Z} ∘ ψ_{Z′}^{Z}, our surjectivity follows.

Chapter IV

Geometry of Kummer Elements and Chains

The objects we considered in the last chapter have, in a natural way, a geometric structure: We will see that W(A), K-Chain_{ℓ}(A), K-Chain_{ℓ}([X], −; A), … can be identified with the k-rational points of smooth varieties.

§ 9. The Variety of Kummer Elements

In this section we fix a field k and a positive integer n ≥ 2 which is not divisible by the characteristic of k. Let A be a central simple k-algebra of degree n and ζ ∈ μ_{n} = μ_{n}(k̄) a primitive n-th root of unity.

1. A as a k-Scheme

Since A is an n^{2}-dimensional k-vector space, it gives rise, in the well known way, to an n^{2}-dimensional affine space over k, which is isomorphic to 𝔸_{k}^{n^{2}} as k-scheme.

(9.1) Definition. Let Ǎ := Hom_{k-lin}(A, k) be the (k-vector space) dual of A and S(Ǎ) its associated symmetric algebra, then we define the k-schemes

A(A) := Spec(S(Ǎ)) and P(A) := Proj(S(Ǎ)) .

(9.2) Remark. For every field extension k′|k we have the canonical identification of the k′-rational points of A(A) with the elements of A_{k′} = A ⊗_{k} k′:

A(A)(Spec k′) = Hom_{k-alg}(S(Ǎ), k′)
= Hom_{k-lin}(Ǎ, k′)
= Hom_{k-lin}(Ǎ, k) ⊗ k′
= A_{k′} .

Analogously

P(A)(Spec k′) = (A_{k′} − {0})/k^{′∗} .

If we choose a k-basis a_{1}, …, a_{n^{2}} of A, then the isomorphism of the (graded) k-algebras

\begin{matrix}k[X_{1}, …, X_{n^{2}}] & \underrightarrow{∼} S(Ǎ) \\ X_{i} & ⟼ ǎ_{i} \end{matrix}

§ 9. The variety of Kummer elements

for i = 1, …, n^{2} and the dual basis ǎ_{1}, …, ǎ_{n^{2}} of a_{1}, …, a_{n^{2}}, induces isomorphisms A(A) \underrightarrow{∼} 𝔸_{k}^{n^{2}} and P(A) \underrightarrow{∼} ℙ_{k}^{n^{2}−1}. Additionally we have a multiplication morphism on the affine scheme: The multiplication induces a morphism A ⊗_{k} A → A, x ⊗ y ↦ xy of k-modules, and after dualizing we get

Ǎ ⟶ Ǎ ⊗_{k} Ǎ .

If we compose this morphism with the double tensor product of the canonical injections Ǎ ⊗_{k} Ǎ → S(Ǎ) ⊗_{k} S(Ǎ), we get the k-linear morphism

Ǎ ⟶ S(Ǎ) ⊗_{k} S(Ǎ) ;

by the universal property of the symmetric algebras. This factorizes in the form

� � �  _{ψ} �  S(Ǎ) ⊗_{k} S(Ǎ) , S(Ǎ)  Ǎ ⊗_{k} Ǎ Ǎ

where ψ is unique. Then it induces the morphism of k-schemes

Spec(ψ): A(A) ×_{k} A(A) ⟶ A(A) ,

which is on the k′-rational points just the multiplication on A_{k′}.

In §6, 1., we defined the maps

Srd_{ν}: A ⟶ k ,

ν = 1, …, n, in particular the reduced norm Nrd = Srd_{n} and trace Trd = Srd_{1}. Furthermore we define the maps

\begin{matrix}T^{ν}: A & ⟶ k \\ X & ⟼ Trd(X^{ν}) . \end{matrix}

(9.3) Lemma. Assume k is algebraically closed. Then the maps Srd_{ν} and T^{ν} are regular functions on the scheme A(A), i.e., there are uniquely defined (homogeneous) elements in S(Ǎ) = O(A(A))—which we also denote by Srd_{ν} and T^{ν}—such that they represent these maps. Srd_{ν} and T^{ν} are homogeneous of degree ν.

Proof: Choose an isomorphism α: A \underrightarrow{∼} A_{0} = M_{n}(k) of k-algebras. This also induces an isomorphism A(A) \underrightarrow{∼} A(A_{0}) of k-schemes as well as the graded isomorphism of the global section rings S(Ǎ_{0}) \underrightarrow{∼} S(Ǎ). Therefore we may assume that A = A_{0}. We choose the canonical basis E_{ij}, i, j = 1, …, n, of the k-vector space A_{0}, and we denote with X_{ij}, i, j = 1, …, n, its dual basis. Then

S(Ǎ_{0}) = k[X_{11}, X_{12}, …, X_{nn}] ,

and it is clear that Srd_{ν} is a homogeneous polynomial in the X_{ij} of degree ν, e.g.,

Srd_{1} = Trd = X_{11} + X_{22} + ⋯ + X_{nn} .

The ν-th power of the matrix (X_{ij}) is given by n^{2} homogeneous polynomials of degree ν. Therefore Trd((X_{ij})^{ν}) is also a homogeneous polynomial of degree ν.

(9.4) Remark. In the proof we have of course Srd_{ν}, T^{ν} ∈ ℤ[X_{11}, X_{12}, …, X_{nn}].

Now let k again be an arbitrary field of characteristic prime to n.

(9.5) Lemma. The regular functions of (9.3) are already defined over the base field:

Srd_{ν}, T^{ν} ∈ S(Ǎ) .

Proof: First of all

Srd_{ν}, T^{ν} ∈ S(Ǎ) ⊗_{k} k̄ = O(A(A) ×_{k} k̄) ,

where k̄ is a separable closure. In fact this is true for A = A_{0}, as we have seen above, and there are already k̄-isomorphisms α: Ā \underrightarrow{∼} Ā_{0}, hence α_{∗}: A(Ā) \underrightarrow{∼} A(Ā_{0}), because A is a central simple k-algebra. Now all we have to show is that for f := Srd_{ν}, T^{ν} and γ ∈ Γ = Gal(k̄|k), the diagram

� ^{f} �_{id ⊗γ} � _{f} �  _{γ} k̄ Ā  k̄ Ā

is commutative. This follows from the commutative diagram

�  _{α} �  ^{f} �_{id ⊗γ} �  _{κ_{Xγ} ∘(id ⊗γ)} �  _{γ} �  _{α} �  _{f} k̄ . M_{n}(k̄) Ā  k̄ M_{n}(k̄) Ā

For the left part cf. §4,2. The commutativity of the right part is clear, since Srd_{ν} and T^{ν} are invariant under any conjugation κ_{X_{γ}}, and commute with the Galois action.

(9.6) Remark. T^{ν} ∈ S(Ǎ) already follows from Trd ∈ S(Ǎ) since the morphism “taking the ν-th power” A(A) → A(A) is a morphism, which is defined over k. The krationality of Trd is clear, since Trd is linear, so the k-rationality only has to be checked on a k-basis of A and this is clear.

(9.7) Notation. We denote the distinguished open subschemes of A(A) and P(A), associated to Nrd, by

GL_{1}(A) := Spec S(Ǎ)_{Nrd} = D(Nrd) ⊆ A(A)

and

PGL_{1}(A) := Spec(S(Ǎ))_{(Nrd)} = D_{+}(Nrd) ⊆ P(A) .

(9.8) Remark. For every field extension k′|k we have the identifications of the k′- rational points GL_{1}(A)(Spec k′) = (A ⊗_{k} k′)^{∗} and PGL_{1}(A)(Spec k′) = A_{k′}^{∗}/k^{′∗}. Moreover there are morphisms of k-schemes

GL_{1}(A) ×_{k} A(A) ⟶ A(A)

and

GL_{1}(A) ×_{k} P(A) ⟶ P(A) ,

which are on the k′-rational points just the conjugation:

\begin{matrix}A_{k′}^{∗} × A_{k′} & ⟶ A_{k′} \\ (g, a) & ⟼ gag^{−1} . \end{matrix}

2. W(A) and W(A) as Varieties over an Algebraically Closed Field

We want to see W(A) and W(A) as subvarieties of A(A) and P(A) respectively. First we want to consider this in the algebraically closed setting. So in this subsection let k be algebraically closed. In the last subsection we have seen that the maps Srd_{ν}, T^{ν}: A → k are homogeneous elements of the global section ring O(A(A)). So we can make the following definition

(9.9) Definition. We define the subvarieties

W̄(A) := V (Srd_{1}, ⋯, Srd_{n−1}) ⊆ A(A)
W̄(A) := V_{+}(Srd_{1}, ⋯, Srd_{n−1}) ⊆ P(A)
W(A) := W̄(A) ∩ D(Nrd) ⊆ A(A)
W(A) := W̄(A) ∩ D_{+}(Nrd) ⊆ A(A) ,

where D(Nrd) and D_{+}(Nrd) are the distinguished open subsets associated to Nrd.

(9.10) Remark. The closed, i.e., k-rational points of W(A) and W(A) are exactly the Kummer elements and Kummer lines of A. Hence we used the same symbols as in the sections above.

In the language of schemes we can define the varieties in the following way:

(9.11) Lemma. Let I := (Srd_{1}, …, Srd_{n−1}) ⊆ S(Ǎ) be the graded Ideal generated by the elements Srd_{1},…, Srd_{n−1} and let \sqrt{I} denote the radical of I. Then

W̄(A) := Spec(S(Ǎ)/\sqrt{I})
W̄(A) := Proj(S(Ǎ)/\sqrt{I})
W(A) := Spec(S(Ǎ)/\sqrt{I})_{Nrd}
W(A) := Spec(S(Ǎ)/\sqrt{I})_{(Nrd)} .

We will see that I is actually radical in S(Ǎ)_{Nrd}.

(9.12) Claim. If (n!) is prime to the characteristic of k, then T^{1}, …, T^{n−1} is another set of generators of the Ideal I:

(T^{1}, …, T^{n−1}) = (Srd_{1}, …, Srd_{n−1}) ⊆ S(Ǎ) .

Proof: Like in (6.5) this follows from the formula

T^{ν} − T^{ν−1} Srd_{1} +T^{ν−2} Srd_{2} − ⋯ + (−1)^{ν−1}T^{1} Srd_{ν−1} +(−1)^{ν}ν Srd_{ν} = 0 ,

ν = 1, …, n − 1, which follows from Newton’s formula: In the case A = A_{0} = M_{n}(k) one diagonalizes the generic matrix, and then the formula directly follows.

(9.13) Remark. The morphisms GL_{1}(A) × A(A) → A(A) and GL_{1}(A) × P(A) → P(A), mentioned in (9.8) induce morphisms

GL_{1}(A) × W(A) ⟶ W(A) and GL_{1}(A) × W(A) ⟶ W(A) ,

i.e., GL_{1}(A) acts on W(A) and W(A) by conjugation. The last action is transitive, cf. (6.3), i.e., for any closed point [X] ∈ W(A) the morphism

\begin{matrix}GL_{1}(A) & ⟶ W(A) \\ G & ⟼ [GXG^{−1}] \end{matrix}

is surjective. The first action is not transitive—one misses the scaling; however the morphism

\begin{matrix}GL_{1}(A) × (𝔾_{m})_{k} × W(A) & ⟶ W(A) \\ (G, g, X) & ⟼ gGXG^{−1} \end{matrix}

gives a transitive action.

(9.14) Proposition. W(A) and W(A) are irreducible topological spaces.

Proof: This is a corollary of (9.13) since GL_{1}(A) (which is open, dense in A(A) ≅ 𝔸^{n^{2}}) and GL_{1}(A) × (𝔾_{m})_{k} are irreducible and they map surjectively to W(A) and W(A) respectively, by the action on some fixed element [X] ∈ W(A) or X ∈ W(A).

(9.15) Corollary. The open subsets W(A) ⊆ W̄(A) and W(A) ⊆ W̄(A) are dense, hence W̄(A) and W̄(A) are irreducible topological spaces.

Proof: We may assume A = A_{0} = M_{n}(k). Let X ∈ W̄(A)−W(A) or equivalently [X] ∈ W̄(A)−W(A), i.e., Prd(X, t) = t^{n}. Therefore X is nilpotent and (after conjugation with a suitable element in GL_{1}(A)) we may assume that X is a strictly (upper) triangular matrix: The diagonal entries are zero. For every λ ∈ k^{∗} and any primitive n-th root ζ ∈ μ_{n} of unity we have

X + λX_{0}(ζ) ∈ W(A) and [X + λX_{0}(ζ)] ∈ W(A) ,

since Prd(X + λX_{0}(ζ), t) = t^{n} − λ^{n}. Hence X lies in the closure of W(A). The claim follows.

As a further consequence of the transitive action we immediately get (9.16) Proposition. The varieties W(A) and W(A) are non-singular.

Now we are going to give another proof of the non-singularity of W(A) and W(A). This will be done by direct computation, using the Jacobi criterion.

(9.17) Lemma. Let A = A_{0} = M_{n}(k) and let Srd_{ν} ∈ k[X_{11}, X_{12}, …, X_{nn}] be the homogeneous polynomials of degree ν from above. Then the partial derivatives of Srd_{ν} in the point X = X_{0}(ζ) in the coordinate X_{ij} is given by

\frac{∂ Srd_{ν}}{∂X_{ij}} (X_{0}(ζ)) = \{ \begin{matrix}0, & if i ≠ j, \\ (−1)^{ν−1}ζ^{i(ν−1)}, & if i = j . \end{matrix}

Proof: Let X := (X_{ij})_{i,j=1,...,n}. We have

det(t − X) = t^{n} − t^{n−1} Srd_{1}(X) + ⋯ + (−1)^{n} Srd_{n}(X) ;

therefore we know

\frac{∂ det(t − X)}{∂X_{ij}} = −t^{n−1} \frac{∂ Srd_{1}(X)}{∂X_{ij}} + t^{n−2} \frac{∂ Srd_{2}(X)}{∂X_{ij}} − ⋯ + (−1)^{n}t^{0} \frac{∂ Srd_{n}(X)}{∂X_{ij}} .

On the other hand we have

\frac{∂ det(t − X)}{∂X_{ij}} |_{X} = \frac{∂ det(t − (X + λE_{ij}))}{∂λ}|_{λ=0}

= \{ \begin{matrix}\frac{∂}{∂λ} [∏_{ν=1}^{n}(t − ζ^{ν})]|_{λ=0} = 0, & if i ≠ j \\ \frac{∂}{∂λ} [(t − ζ^{i} − λ) ∏_{{ν=1 \atop ν≠i}}^{n}(t − ζ^{ν})]|_{λ=0} = −\frac{t^{n} − 1}{t − ζ^{i}}, & if i = j , \end{matrix}

but

\frac{t^{n} − 1}{t − ζ^{i}} = \frac{t^{n} − (ζ^{i})^{n}}{t − ζ^{i}} = t^{n−1} + t^{n−2}ζ^{i} + t^{n−3}ζ^{2i} + ⋯ + ζ^{(n−1)i} .

Comparing the last with the first formula completes the proof.

(9.18) Theorem. The varieties W(A) and W(A) are non-singular k-varieties of dimension n^{2} − (n − 1) and n^{2} − n respectively.

Proof: We may assume A = A_{0}. Because of the transitive actions on W(A) and W(A) mentioned in (9.13), all we have to prove (by virtue of the Jacobi criterion) is that the (n − 1) × n^{2}-matrix

(\frac{∂ Srd_{ν}}{∂X_{ij}} )_{{ν=1,...,n−1 \atop i,j=1,...,n}} |_{X_{0}(ζ)} = (\frac{∂ Srd_{ν}}{∂X_{ij}} (X_{0}(ζ)))_{{ν=1,...,n−1 \atop i,j=1,...,n}}

has rank n − 1. Again, ζ ∈ μ_{n} is a primitive n-th root of unity. But by lemma (9.17) this matrix has a (n − 1) × n minor of the form

((−1)^{ν−1}ζ^{i(ν−1)})_{{ν=1,...,n−1 \atop i=1,...,n}} .

We even know that besides this minor there are only trivial (zero) entries in the matrix. But this minor is a (modified) minor of the Vandermonde matrix (ζ^{ij})_{i,j=1,...,n} which has full rank n. Therefore our minor has rank (n − 1) and we are done.

We immediately get from the proof:

(9.19) Corollary. The ideal I = (Srd_{1}, …, Srd_{n−1}) ⊆ O(W(A)) = S(Ǎ)_{Nrd} is already the ideal of vanishing functions. Since W(A) is irreducible, I is a prime ideal, especially \sqrt{I} = I in S(Ǎ)_{Nrd}.

3. W(A) and W(A) as k-Schemes

Now let k again be an arbitrary field of characteristic prime to n. Then we define the following k-schemes:

(9.20) Definition. For a central simple k-algebra A of degree n we define the ksubschemes of A(A) and P(A)

W^{0}(A) := Spec(S(Ǎ)/(Srd_{1}, …, Srd_{n−1}))_{Nrd} ⊆ A(A)

and

W^{0}(A) := Spec(S(Ǎ)/(Srd_{1}, …, Srd_{n−1}))_{(Nrd)} ⊆ P(A)

(9.21) Remark. For any field extension k′|k we have

W^{0}(A_{k′}) = W^{0}(A) ×_{Spec(k)} Spec(k′) and W^{0}(A_{k′}) = W^{0}(A) ×_{Spec(k)} Spec(k′) .

(9.22) Proposition. W^{0}(A) and W^{0}(A) are smooth integral k-schemes of dimension n^{2}−(n−1) and n^{2}−n respectively. For any field extension k′|k we have the identification of k′-rational points and Kummer elements

W^{0}(A)(Spec k′) = W(A_{k′}) and W^{0}(A)(Spec k′) = W(A_{k′}) ,

which is induced by the identification of (9.2).

Proof: The first part only has to be proven over an algebraic closure of k—what we have done in the last subsection, since there W^{0}(A) = W(A) and W^{0}(A) = W(A). The second part is clear.

§ 10. The Variety of Chains

We fix a field k and a positive integer n ≥ 2 which is not divisible by the characteristic of k. Let A be a central simple k-algebra of degree n and ζ ∈ μ_{n} = μ_{n}(k̄) a primitive n-th root of unity lying in k. Furthermore we choose a k-basis a_{1}, a_{2}, …, a_{n^{2}} of the vector space A, and we denote its dual basis by ǎ_{1}, …, ǎ_{n^{2}}.

1. The “Vectorbundle of Kummer Pairs”

We want to assign to the set

{(X, Y) ∈ W(A) × A : XY − ζY X = 0}

of Kummer pairs (in a slight broader sense, i.e., we don’t demand Y to be invertible) the structure of a scheme—and more: of a vector bundle over the k-scheme W^{0}(A). Because of (6.10) for any Kummer element X ∈ W(A), the set {Y ∈ A : XY −ζY X = 0} is a vector space of dimension n: Using the basis a_{1}, a_{2}, …, a_{n^{2}} one can interpret the equation XY − ζY X = 0 (in Y) as n^{2} linear equations in the entries y_{1}, y_{2}, …, y_{n^{2}} of Y = y_{1} · a_{1} + y_{2} · a_{2} + ⋯ + y_{n^{2}} · a_{n^{2}} for y_{i} ∈ k. If k is algebraically closed it is clear by (6.10) that our set is an n-dimensional vector subspace of A. Therefore the gradients of the n^{2} linear equations span an (n^{2} − n)-dimensional vector space. That means, one can leave out n of the n^{2} equations. Since the gradients do not change under base field extension, everything holds for a general field k.

We will write

R := O(W^{0}(A)) = (S(Ǎ)/(Srd_{1}, …, Srd_{n−1}))_{Nrd}

for the global section ring of the affine scheme W^{0}(A)—cf. §9.

The bilinear map A × A → A, (x, y) ↦ xy − ζyx induces, after dualizing, a morphism

φ: Ǎ ⟶ Ǎ ⊗_{k} Ǎ

of k-modules. If we compose φ with the morphism Ǎ ⊗_{k} Ǎ → R ⊗_{k} S(Ǎ), a ⊗ b ↦ a ⊗ b, of k-modules, we get a morphism Ǎ → R ⊗_{k} S(Ǎ) which induces—by the universal property of the symmetric algebra—a morphism

ϕ: S(Ǎ) ⟶ R ⊗_{k} S(Ǎ)

of graded k-algebras. We set f_{i} := ϕ(ǎ_{i}) = φ(ǎ_{i}). Applying the functor Spec to it, we get the morphism

Spec(ϕ): W^{0}(A) ×_{k} A(A) ⟶ A(A) ,

which on the rational points is just the map

(x, y) ⟼ xy − ζyx .

So the fibre over the k-rational point zero is our choice for the scheme of Kummer pairs (in the broader sense).

(10.1) Lemma. The fibre of Spec(ϕ) over zero is given by

Spec(R ⊗_{k} S(Ǎ))/(im φ) = Spec(R ⊗_{k} S(Ǎ))/(ϕ(S(Ǎ)_{+})) ,

where (im φ) is the ideal in R ⊗_{k} S(Ǎ) generated by the elements of im(φ)—or equivalently by the elements of ϕ(S(Ǎ)_{+}), which are of degree > 0.

(10.2) Remark. The ideal (ϕ(S(Ǎ)_{+})) = (im φ) is a graded ideal, already generated by the n^{2} elements f_{1}, f_{2}, …, f_{n^{2}}, which are of degree 1. The grading of R ⊗_{k} S(Ǎ) is given by the grading of the symmetric algebra S(R ⊗_{k} Ǎ) = R ⊗_{k} S(Ǎ).

Proof: (of Lemma (10.1)). The fibre is given by the spectrum of

(R ⊗_{k} S(Ǎ)) ⊗_{S(Ǎ)} k ,

where the structure morphisms of the two S(Ǎ)-algebras involved in the tensor product are ϕ and the projection S(Ǎ) → k with kernel S(Ǎ)_{+}. The last ring morphism is the one which corresponds to the k-rational point 0 in A(A). Therefore

\begin{aligned}(R ⊗_{k} S(Ǎ)) ⊗_{S(Ǎ)} k & = (R ⊗_{k} S(Ǎ)) ⊗_{S(Ǎ)} (S(A)/S(A)_{+}) \\ & = R ⊗_{k} S(Ǎ)/(ϕ(S(Ǎ)_{+})) , \end{aligned}

where (ϕ(S(Ǎ)_{+})) = (im φ) since the ideal S(Ǎ)_{+} is generated by the elements of Ǎ, which are just the elements of S(Ǎ) of degree 1.

(10.3) Definition. We define the k-scheme

E(W^{0}(A)) := Spec(R ⊗_{k} S(Ǎ))/(im φ) = Spec(R ⊗_{k} S(Ǎ))/(ϕ(S(Ǎ)_{+})) .

(10.4) Remark. Let k′|k be any field extension. Then we have the identity for the k′-rational points

E(W^{0}(A))(Spec k′) = {(X, Y) ∈ W(A_{k′}) × A_{k′} : XY − ζY X = 0} .

(10.5) Remark. The scheme E(W^{0}(A)) is equipped with the two morphisms

π_{0}: E(W^{0}(A)) ⟶ W^{0}(A) ,

induced by R → (R ⊗_{k} S(Ǎ))/(im φ), r ↦ r ⊗ 1 and

π_{1}: E(W^{0}(A)) ⟶ A(A) ,

induced by S(Ǎ) → (R ⊗_{k} S(Ǎ))/(im φ), s ↦ 1 ⊗ s, which are on the rational points just the projections on the first and second factor.

Now we want to show that

π_{0}: E(W^{0}(A)) ⟶ W^{0}(A) ,

is a vector bundle. For this aim, we will define a locally free sheaf ℱ of O_{W^{0}(A)}-modules for which it is the vector bundle, associated to ℱ:

E(W^{0}(A)) = 𝕍(ℱ) (:= Spec S(ℱ)) .

Let σ: W^{0}(A) → Spec(k) be the structure morphism and

ℳ := O_{W^{0}(A)} ⊗_{k} Ǎ = σ^{∗}(Ǎ) .

This is an O_{W^{0}(A)}-algebra, which is a free O_{W^{0}(A)}-module of rank n^{2}. ℳ is the sheaf of O_{W^{0}(A)}-modules associated to the R-module M := R ⊗_{k} Ǎ. Let N := R · im(φ) ⊆ M be the R-submodule of M, generated by im(φ); or say

N := R · f_{1} + R · f_{2} + ⋯ + R · f_{n^{2}} .

Let 𝒩 be the sheaf of O_{W^{0}(A)}-modules associated to the R-module N. 𝒩 is a sheaf of O_{W^{0}(A)}-submodules of ℳ. We define the sheaf of O_{W^{0}(A)}-modules ℱ to be

ℱ := ℳ /𝒩 ;

or in other words: ℱ is the sheaf of O_{W^{0}(A)}-modules associated to the R-module M/N. We are going to show first that M/N is a projective R-module of constant rank n.

(10.6) Lemma. Let 𝔭 ∈ Spec(R) = W^{0}(A) be a prime ideal and κ(𝔭) its residue field. Then the vectors f_{1}, f_{2}, …, f_{n^{2}} ∈ M ⊗_{R} κ(𝔭) = Ǎ_{κ(𝔭)} ≅ κ(𝔭)^{n^{2}} span an (n^{2} − n)- dimensional vector subspace.

Proof: The elements f_{1}, f_{2}, …, f_{n^{2}} ∈ Ǎ_{κ(𝔭)} describe the equation XY − ζY X = 0 for the Kummer element X = 𝔭 in W(A_{κ(𝔭)}) ⊆ A_{κ(𝔭)}. We have seen that the space of solutions of this set of linear equations in A_{κ(𝔭)} is n-dimensional. Hence the linear equations f_{1}, f_{2}, …, f_{n^{2}} span a vector subspace of codimension n.

(10.7) Lemma. Let a, b and m be positive integers with a ≤ m, T a topological space, P ∈ T a point and Ω a field with some topology such that {0} is closed in Ω. If for j = 1, 2, …, a+b, the maps F_{j} = (F_{ij})_{i=1,...,m}: T ⟶ Ω^{m} are continuous functions (Ω^{m} is equipped with the product topology) such that the rank of the m×(a+b) matrices

(F_{1}(Q), F_{2}(Q), …, F_{a+b}(Q)) = (F_{ij}(Q))_{{i=1,...,m \atop j=1,...,a+b}}

for all Q ∈ T and the (m × a) matrix

(F_{1}(P), F_{2}(P), …, F_{a}(P)) = (F_{ij}(P))_{{i=1,...,m \atop j=1,...,a}}

is a, then there exists an open neighbourhood U ⊆ T of P and continuous functions α_{μ}^{(ν)}: U ⟶ Ω for μ = 1, …, a and ν = a + 1, …, a + b such that

(i) α_{μ}^{(ν)} ∈ ω(F_{ij}|_{U} : i = 1, …, m; j = 1, …, a + b), where ω is the prime field of Ω, i.e., ℚ or 𝔽_{p}; in other words, the α_{μ}^{(ν)} are quotients of polynomials in the F_{ij}|_{U} with coefficients in ℤ.

(ii) In U we have: F_{ν} = α_{1}^{(ν)}F_{1} + α_{2}^{(ν)}F_{2} + ⋯ + α_{a}^{(ν)}F_{a} for ν = a + 1, …, a + b.

Proof: After re-indexing in i, we may assume that

det(F_{ij}(P))_{i,j=1,...,a} ≠ 0

and hence there is an open neighbourhood U ⊆ T of P with

det(F_{ij}(Q))_{i,j=1,...,a} ≠ 0

for all Q ∈ U. Since for ν = a + 1, …, a + b, we have

rank(F_{1}(Q), F_{2}(Q), …, F_{a}(Q)) = rank(F_{1}(Q), F_{2}(Q), …, F_{a}(Q), F_{ν}(Q)) = a ,

there are uniquely defined elements α_{μ}^{(ν)}(Q) ∈ Ω for μ = 1, …, a such that

F_{ν}(Q) = α_{1}^{(ν)}(Q)F_{1}(Q) + α_{2}^{(ν)}(Q)F_{2}(Q) + ⋯ + α_{a}^{(ν)}(Q)F_{a}(Q) .

In this way we get maps α_{μ}^{(ν)}: U → Ω with property (ii). It remains to prove property (i). (Continuity is then for free!): By Cramer’s Rule we have

\begin{aligned}α_{μ}^{(ν)}(Q) & det(F_{ij}(Q))_{i,j=1,...,a} = \\ & = det(F_{i1}(Q), …, F_{i,μ−1}(Q), F_{iν}(Q), F_{i,μ+1}(Q), …, F_{ia}(Q))_{i=1,...,a} . \end{aligned}

On U the left determinant is invertible, hence we get property (i).

Putting together the last two lemmas yields the

(10.8) Proposition. Let 𝔭 ∈ Spec(R) be a maximal prime ideal. Then—if necessary, after re-indexing of the f_{1}, f_{2}, …, f_{n^{2}}—we have

N_{𝔭} = R_{𝔭} · f_{1} + … + R_{𝔭} · f_{n^{2}} = R_{𝔭} · f_{1} + … + R_{𝔭} · f_{n^{2}−n} ⊆ M_{𝔭} .

Proof: We use lemma (10.7): Let T = W^{0}(A)(Spec Ω) = Hom(R, Ω) be the topological space (with the Zariski topology) of the geometric points of W^{0}(A), where Ω is an algebraic closure of k, and set m := n^{2}, a := n^{2} − n, b := n. Using the identification (via the ǎ_{i}’s) Ǎ = k^{n^{2}} ⊆ Ǎ_{Ω} = Ω^{n^{2}} we define for j = 1, …, n^{2}

\begin{matrix}F_{j} = f_{j}: T & ⟶ M ⊗_{R} Ω = Ω^{n^{2}} \\ x & ⟼ f_{j}(x) , \end{matrix}

where f_{j}(x) is the image of f_{j} under the morphism id_{M}⊗x: M = M ⊗_{R} R → M ⊗_{R} Ω. Then (10.6) shows that the prerequisites of (10.7) hold true for every point x ∈ T with 𝔭 = ker(x). Note that, since R is an integral Jacobson ring, we can identify the elements of R and M with their induced functions on T. Now, (i) tells us that the functions α_{μ}^{(ν)} are elements of R_{𝔭} and (ii) gives the claim, i.e., the equation (ii) first holds true for functions on T, but that implies that it holds for the elements of M.

(10.9) Proposition. The R-module M/N is projective of rank n, in other words: For all 𝔭 ∈ Spec(R) the R_{𝔭}-module (M/N)_{𝔭} is free of rank n.

Proof: We only have to prove the second part for a maximal prime ideal 𝔭. We will show: N_{𝔭} is a free direct summand of M_{𝔭} of rank (n^{2} − n) with a free complementary summand. Then we are done. But this is now easy to show. By (10.8) the R_{𝔭}-module N_{𝔭} is generated by some (n^{2} − n) elements f_{1}, f_{2}, …, f_{n^{2}−n} such that

f_{1}(𝔭), f_{2}(𝔭), …, f_{n^{2}−n}(𝔭) ∈ M(𝔭) = M_{𝔭} ⊗_{R_{𝔭}} κ(𝔭)

are linearly independent. We complete this set of vectors by n vectors to a basis of M(𝔭). We choose liftings g_{1}, g_{2}, …, g_{n} ∈ M_{𝔭} of them. We know from Commutative Algebra that f_{1}, f_{2}, …, f_{n^{2}−n}, g_{1}, g_{2}, …, g_{n} is a basis of the R_{𝔭}-module M_{𝔭}. Hence

M_{𝔭} = N_{𝔭} ⊕ ⟨g_{1}, g_{2}, …, g_{n}⟩ = N_{𝔭} ⊕ R_{𝔭}g_{1} ⊕ R_{𝔭}g_{2} ⊕ ⋯ ⊕ R_{𝔭}g_{n} .

Now we are done.

(10.10) Corollary. The sheaves ℱ, ℳ and 𝒩 of O_{W^{0}(A)}-modules are locally free of rank n, n^{2} and (n^{2} − n) respectively.

(10.11) Theorem. The morphism

π_{0}: E(W^{0}(A)) ⟶ W^{0}(A)

is a vector bundle of rank n; to be precise

E(W^{0}(A)) = 𝕍(ℱ) (= Spec S(ℱ)) .

Proof: Observe

E(W^{0}(A)) = Spec(R ⊗_{k} S(Ǎ))/(im φ)
= Spec(R ⊗_{k} S(Ǎ))/(f_{1}, …, f_{n})
= Spec(S(R ⊗_{k} Ǎ/⟨f_{1}, …, f_{n}⟩))
= Spec(S(M/N))
= Spec(S(ℳ /𝒩 ))
= Spec(S(ℱ))
= 𝕍(ℱ) .

With (9.22) this theorem yields:

(10.12) Corollary. The morphism

π_{0}: E(W^{0}(A)) ⟶ W^{0}(A)

is smooth of relative dimension n, hence E(W^{0}(A)) is a smooth integral k-scheme of the dimension n^{2} + 1 = dim(W^{0}(A)) + n.

Now we take degree zero components of R, M and N:

R_{0} = (S(Ǎ)/(Srd_{1}, …, Srd_{n−1}))_{(Nrd)}
M_{0} = R_{0} ⊗_{k} Ǎ
N_{0} = R_{0} · im(φ) = R_{0} · f_{1} + ⋯ + R_{0} · f_{n^{2}} .

Since W^{0}(A) = Spec(R_{0}), the R_{0}-modules M_{0} and N_{0} induce their associated sheaves of O_{W^{0}(A)}-modules ℳ_{0}, 𝒩_{0} and ℱ_{0} := ℳ_{0}/𝒩_{0}. Obviously, we have the same property:

(10.13) Remark. The sheaves ℱ_{0}, ℳ_{0} and 𝒩_{0} of O_{W^{0}(A)}-modules are locally free of rank n, n^{2} and n^{2} − n respectively.

(10.14) Definition. We define the following affine and projective bundles of rank n over W^{0}(A)

E(W^{0}(A)) := 𝕍(ℱ_{0}) ⊆ W^{0}(A) ×_{k} A(A)

and

E(W^{0}(A)) := ℙ(ℱ_{0}) ⊆ W^{0}(A) ×_{k} P(A) .

These bundles are just the once and twice projectivised versions of E(W^{0}(A)), and like in the former case each of the k-schemes comes with two morphisms

π_{0}: E(W^{0}(A)) ⟶ W^{0}(A) and π_{0}: E(W^{0}(A)) ⟶ W^{0}(A)

which are the bundle morphisms, and

π_{1}: E(W^{0}(A)) ⟶ W^{0}(A) ×_{k} A(A) ⟶^{pr_{2}} A(A)

as well as

π_{1}: E(W^{0}(A)) ⟶ W^{0}(A) ×_{k} P(A) ⟶^{pr_{2}} P(A) .

The inclusions are induced by (applying the functor 𝕍 to) the canonical epimorphism ℳ_{0} → → ℱ_{0} of sheaves of O_{W^{0}(A)}-modules. Again from (10.13) follows:

(10.15) Proposition. The morphisms

π_{0}: E(W^{0}(A)) ⟶ W^{0}(A) and π_{0}: E(W^{0}(A)) ⟶ W^{0}(A)

are smooth of relative dimension n and (n − 1) respectively. Also E(W^{0}(A)) and E(W^{0}(A)) are smooth integral k-schemes of dimension

dim W^{0}(A) + n = n^{2} and dim W^{0}(A) + (n − 1) = n^{2} − 1 .

For any field extension k′|k we have the canonical identifications

E(W^{0}(A))(Spec k′) = {([X], Y ) ∈ W(A_{k′}) × A_{k′} : XY − ζY X = 0}

and (here with ℙA_{k′} = (A_{k′} − 0)/k^{′∗})

E(W^{0}(A))(Spec k′) = {([X], [Y]) ∈ W(A_{k′}) × ℙA_{k′} : XY − ζY X = 0} .

We take the open subschemes of E(W^{0}(A)) and E(W^{0}(A)) of pairs of invertible elements (X, Y) and ([X], [Y]):

(10.16) Definition. We define the open subschemes

W^{1}(A) := E(W^{0}(A)) ∩ (W^{0}(A) ×_{k} D(Nrd))

and

W^{1}(A) := E(W^{0}(A)) ∩ (W^{0}(A) ×_{k} D_{+}(Nrd)) .

(10.17) Remark. By virtue of (6.8) we also can write

W^{1}(A) := E(W^{0}(A)) ∩ (W^{0}(A) ×_{k} W^{0}(A))

and

W^{1}(A) := E(W^{0}(A)) ∩ (W^{0}(A) ×_{k} W^{0}(A)) .

We still have the projection morphisms, induced by π_{0}:

π_{0}^{1}: W^{1}(A) ⟶ W^{0}(A) and π_{0}^{1}: W^{1}(A) ⟶ W^{0}(A) .

(10.18) Proposition. The fibres of the morphisms

π_{0}^{1}: W^{1}(A) ⟶ W^{0}(A) and π_{0}^{1}: W^{1}(A) ⟶ W^{0}(A) .

are non-empty, hence they have (full) dimension n and n − 1 respectively, and they are dense in the corresponding fibres of π_{0}: E(W^{0}(A)) → W^{0}(A) and π_{0}: E(W^{0}(A)) → W^{0}(A).

Proof: One may assume that k is algebraically closed. Then the density and nonemptiness follows from the fact that for any Kummer element X ∈ W(A), one can find an invertible element Y ∈ A which is in ζ-relation with X. All fibres are irreducible, since they are open subsets of vector spaces.

(10.19) Theorem. The morphisms

π_{0}^{1}: W^{1}(A) ⟶ W^{0}(A) and π_{0}^{1}: W^{1}(A) ⟶ W^{0}(A) .

are smooth of relative dimension n and n − 1 respectively. W^{1}(A) and W^{1}(A) are smooth integral schemes of dimension (n^{2} + 1) and (n^{2} − 1) respectively. For any field extension k′|k we have the canonical identifications

W^{1}(A)(Spec k′) = {(X, Y) ∈ W(A_{k′}) × W(A_{k′}) : XY − ζY X = 0}

and

W^{1}(A)(Spec k′) = K-Chain_{1}^{ζ}(A_{k′}) .

(10.20) Remark. Regarding (10.17) there are also morphisms

π_{1}^{1}: W^{1}(A) ⟶ W^{0}(A) and π_{1}^{1}: W^{1}(A) ⟶ W^{0}(A)

induced by the restriction of the morphisms

π_{1}: E(W^{0}(A)) ⟶ A(A) and π_{1}: E(W^{0}(A)) ⟶ P(A)

which then factorize through W^{0}(A) and W^{0}(A).

2. Construction of the Varieties of Chains

In the last subsection we constructed the k-schemes

E^{1}(A) := E(W^{0}(A)) ⊆ W^{0}(A) ×_{k} A(A)
E^{1}(A) := E(W^{0}(A)) ⊆ W^{0}(A) ×_{k} P(A)
W^{1}(A) ⊆ W^{0}(A) ×_{k} W^{0}(A)
W^{1}(A) ⊆ W^{0}(A) ×_{k} W^{0}(A) ,

and all come with their projections π_{0}^{1} and π_{1}^{1} to their first and second factor. They represent chains in A of the length one (or better: pairs of elements in the relation XY − ζY X = 0). Now we want to make the analogous construction for chains of higher length. Again, ℓ denotes a positive integer. We are going to construct k-schemes

\begin{aligned}E^{ℓ}(A) & ⊆ (W^{0}(A))^{ℓ} ×_{k} A(A) \\ E^{ℓ}(A) & ⊆ (W^{0}(A))^{ℓ} ×_{k} P(A) \\ W^{ℓ}(A) & ⊆ (W^{0}(A))^{ℓ+1} \\ W^{ℓ}(A) & ⊆ (W^{0}(A))^{ℓ+1} , \end{aligned}

together with the projections π_{0}^{ℓ}, …, π_{ℓ}^{ℓ} to their ℓ+1 factors, with the obvious properties. This will be done inductively.

(10.21) Definition. Given the objects E^{ℓ}(A), E^{ℓ}(A), W^{ℓ}(A) and W^{ℓ}(A) with their projection morphisms π_{0}^{ℓ}, … π_{ℓ}^{ℓ}, the analogous objects for the index (ℓ + 1) are defined in the following way:

\begin{aligned}E^{ℓ+1}(A) & := W^{ℓ}(A) ×_{W^{0}(A)} E^{1}(A) \\ E^{ℓ+1}(A) & := W^{ℓ}(A) ×_{W^{0}(A)} E^{1}(A) , \end{aligned}

i.e., the diagrams

�  ^{(π_{ℓ}^{ℓ+1},π_{ℓ+1}^{ℓ+1})} �_{(π_{0}^{ℓ+1},...,π_{ℓ}^{ℓ+1})} �  _{π_{0}^{1}} �  _{π_{ℓ}^{ℓ}}  W^{0}(A) W^{ℓ}(A)  E^{1}(A) E^{ℓ+1}(A)

and

�  ^{(π_{ℓ}^{ℓ+1},π_{ℓ+1}^{ℓ+1})} �_{(π_{0}^{ℓ+1},...,π_{ℓ}^{ℓ+1})} �  _{π_{0}^{1}} �  _{π_{ℓ}^{ℓ}}  W^{0}(A) W^{ℓ}(A)  E^{1}(A) E^{ℓ+1}(A)

are cartesian. They also explain the projection morphisms π_{0}^{ℓ+1}, …, π_{ℓ+1}^{ℓ+1}. The k-schemes W^{ℓ+1}(A) and W^{ℓ+1}(A) are the open subschemes defined by

W^{ℓ+1}(A) := E^{ℓ+1}(A) ∩ (W^{0}(A)^{ℓ+1} ×_{k} D(Nrd))
W^{ℓ+1}(A) := E^{ℓ+1}(A) ∩ (W^{0}(A)^{ℓ+1} ×_{k} D_{+}(Nrd))

together with the restrictions of the projection morphisms.

(10.22) Remark. The upper two diagrams are embedded in the cartesian diagrams

�  ^{pr^{ℓ+1,ℓ+2}} � � �  _{pr_{ℓ}}  W^{0}(A) W^{0}(A)^{ℓ+1}  W^{0}(A) ×_{k} A(A) W^{0}(A)^{ℓ+1} ×_{k} A(A)

and

�  ^{pr^{ℓ+1,ℓ+2}} � � �  _{pr_{ℓ}}  W^{0}(A) , W^{0}(A)^{ℓ+1}  W^{0}(A) ×_{k} P(A) W^{0}(A)^{ℓ+1} ×_{k} P(A)

where the vertical arrows are the obvious projections. We count the factors starting with zero!

(10.23) Remark. One can define the objects W^{ℓ+1}(A) and W^{ℓ+1}(A) by the following cartesian diagrams

�  ^{(π_{ℓ}^{ℓ+1},π_{ℓ+1}^{ℓ+1})} �_{(π_{0}^{ℓ+1},...,π_{ℓ}^{ℓ+1})} �  _{π_{0}^{1}} �  _{π_{ℓ}^{ℓ}}  W^{0}(A) W^{ℓ}(A)  W^{1}(A) W^{ℓ+1}(A)

and

�  ^{(π_{ℓ}^{ℓ+1},π_{ℓ+1}^{ℓ+1})} �_{(π_{0}^{ℓ+1},...,π_{ℓ}^{ℓ+1})} �  _{π_{0}^{1}} �  _{π_{ℓ}^{ℓ}}  W^{0}(A) W^{ℓ}(A)  W^{1}(A) W^{ℓ+1}(A)

This follows from the fact that the definition of the objects W^{ℓ+1}(A) and W^{ℓ+1}(A) is equivalent to saying that the diagrams

�  ^{(π_{ℓ}^{ℓ+1},π_{ℓ+1}^{ℓ+1})} �_{incl} �  _{incl} �  _{(π_{ℓ}^{ℓ+1},π_{ℓ+1}^{ℓ+1})} E^{1}(A) E^{ℓ+1}(A)  W^{1}(A) W^{ℓ+1}(A)

and

�  ^{(π_{ℓ}^{ℓ+1},π_{ℓ+1}^{ℓ+1})} �_{incl} �  _{incl} �  _{(π_{ℓ}^{ℓ+1},π_{ℓ+1}^{ℓ+1})} E^{1}(A) E^{ℓ+1}(A)  W^{1}(A) W^{ℓ+1}(A)

are cartesian. Now we only have to compose these diagrams with the ones from (10.21).

(10.24) Remark. Like in (10.17), we can write

\begin{aligned}W^{ℓ+1}(A) & = E^{ℓ+1}(A) ∩ (W^{0}(A))^{ℓ+2} \\ W^{ℓ+1}(A) & = E^{ℓ+1}(A) ∩ (W^{0}(A))^{ℓ+2} , \end{aligned}

i.e., the last projections π_{ℓ+1}^{ℓ+1} factorize through W^{0}(A) ↪ A(A) and W^{0}(A) ↪ P(A) respectively. To see this, one just looks at the diagrams in (10.23) and uses (10.20).

(10.25) Remark. If we define

ℱ_{W^{ℓ}(A)} := (π_{ℓ}^{ℓ})^{∗}(ℱ) and ℱ_{W^{ℓ}(A)} := (π_{ℓ}^{ℓ})^{∗}(ℱ_{0})

to be the pre-images of the O_{W^{0}(A)}-module ℱ and the O_{W^{0}(A)}-module ℱ_{0} under the morphisms π_{ℓ}^{ℓ}: W^{ℓ}(A) → W^{0}(A) and π_{ℓ}^{ℓ}: W^{ℓ}(A) → W^{0}(A), then one has directly

E^{ℓ+1}(A) = 𝕍(ℱ_{W^{ℓ}(A)}) and E^{ℓ+1}(A) = ℙ(ℱ_{W^{ℓ}(A)}) .

The morphisms π_{0}^{ℓ+1}, …, π_{ℓ}^{ℓ+1} are the old morphisms π_{0}^{ℓ}, …, π_{ℓ}^{ℓ} composed with the bundle morphism. And the new projection morphism π_{ℓ+1}^{ℓ+1} is then given in the same way as in the remark after (10.14).

(10.26) Theorem. The morphisms

(π_{0}^{ℓ+1}, …, π_{ℓ}^{ℓ+1}): E^{ℓ+1}(A) ⟶ W^{ℓ}(A) and (π_{0}^{ℓ+1}, …, π_{ℓ}^{ℓ+1}): E^{ℓ+1}(A) ⟶ W^{ℓ}(A)

as well as

(π_{0}^{ℓ+1}, …, π_{ℓ}^{ℓ+1}): W^{ℓ+1}(A) ⟶ W^{ℓ}(A) and (π_{0}^{ℓ+1}, …, π_{ℓ}^{ℓ+1}): W^{ℓ+1}(A) ⟶ W^{ℓ}(A)

are smooth of relative dimension n and (n − 1) respectively. Therefore E^{ℓ}(A) and E^{ℓ}(A) as well as the objects W^{ℓ}(A) and W^{ℓ}(A) are smooth integral k-schemes of dimension

n^{2} + 1 + (ℓ − 1) · n = (ℓ + n − 1)n + 1 and n^{2} − 1 + (ℓ − 1)(n − 1) = (ℓ + n)(n − 1)

respectively. For any field extension k′|k we have the canonical identifications, induced by the morphisms (π_{0}^{ℓ}, …, π_{ℓ}^{ℓ}):

W^{ℓ}(A)(Spec k′) = {(X_{0}, …, X_{ℓ}) ∈ W(A_{k′})^{ℓ+1} : X_{i−1}X_{i} − ζX_{i}X_{i−1} = 0; i = 1, …, ℓ}
W^{ℓ}(A)(Spec k′) = K-Chain_{ℓ}^{ζ}(A_{k′}) .

Proof: The smoothness follows from the cartesian diagrams in (10.21) and (10.23), since it is stable under base extension. The dimension formulas follow by induction.

§ 10. The variety of chains

3. Chains with Starting Point Let X ∈ W(A) be a Kummer element, or equivalently a k-rational point

X: Spec(k) ⟶ W^{0}(A) and [X]: Spec(k) ⟶ W^{0}(A) .

(10.27) Definition. We define the fibres

\begin{aligned}E^{ℓ}(X; A) & := X ×_{W^{0}(A)} E^{ℓ}(A) \\ E^{ℓ}([X]; A) & := [X] ×_{W^{0}(A)} E^{ℓ}(A) \\ W^{ℓ}(X; A) & := X ×_{W^{0}(A)} W^{ℓ}(A) \\ W^{ℓ}([X]; A) & := [X] ×_{W^{0}(A)} W^{ℓ}(A) , \end{aligned}

where the morphisms in the second factor of the product are the projections π_{0}^{ℓ}. (10.28) Proposition. The k-schemes E^{ℓ}(X; A) and E^{ℓ}([X]; A) as well as W^{ℓ}(X; A) and W^{ℓ}([X]; A) are smooth of dimension (ℓ n) and ℓ(n − 1). For a field extension k′|k we have the identification

W^{ℓ}([X]; A)(Spec k′) = K-Chain_{ℓ}^{ζ}([X], −; A)_{k′} .

Proof: Just note that the composition

π_{0}^{ℓ}: E^{ℓ}(A) ⟶ W^{ℓ−1}(A) ⟶ W^{ℓ−2}(A) ⟶ ⋯ ⟶ W^{0}(A)

is smooth of relative dimension (ℓ n). Analogously for the other cases. (10.29) Remark. Like in (10.25) we can see that E^{ℓ+1}(X; A) and E^{ℓ+1}([X]; A) are vector bundles over W^{ℓ}(X; A) and W^{ℓ}([X]; A). Therefore, by induction one observes that all the schemes are irreducible, hence integral.

§ 11. The Topological Closure of W^{1}(A)

We fix a field k and a positive integer n ≥ 2 which is not divisible by the characteristic of k. Let A be a central simple k-algebra of degree n and ζ ∈ μ_{n} = μ_{n}(k̄) a primitive n-th root of unity. For simplicity we assume that k is algebraically closed.

The variety W^{1}(A) is given in D_{+}(Nrd) × D_{+}(Nrd) (⊆ P(A)^{2}) by the homogeneous equation XY − ζY X = 0. We will show that its topological closure in P(A)^{2} is the subvariety of W̄(A) × W̄(A) which is given by the same equation. This is by no means obvious: We have to show that, given elements [X], [Y] ∈ W̄(A) with XY − ζY X = 0, there are elements [Y′] ∈ W(A) in any neighbourhood of [Y] such that XY′ − ζY′X = 0. In order to achieve this, we need some linear algebra.

1. Some Linear Algebra

In this subsection n is allowed to be any positive integer, and k may be an arbitrary field, if we assume that all eigenvalues that appear already lie in k.

(11.1) Lemma. Let W = W_{n} ⊇ W_{n−1} ⊇ ⋯ ⊇ W_{0} = W_{−1} = 0 be a finite dimensional filtered k-vector space. We set W̄_{i} := W_{i}/W_{i−1}, m_{i} := dim_{k} W_{i} and m̄_{i} := dim_{k} W̄_{i} for i = 0, …, n. Let f ∈ End_{filt}(W) be an endomorphism of the filtered k-vector space W, i.e., f(W_{i}) ⊆ W_{i} for all i, then we can make the following statement about the set of eigenvalues: If

dim_{k} W = # (⋃_{i=1}^{n} EigVal(f|W̄_{i})) ,

—that means, if # EigVal(f|W̄_{i}) = m̄_{i} and the sets EigVal(f|W̄_{i}) are disjoint—then

EigVal(f) = ∐_{i=1}^{n} EigVal(f|W̄_{i}) .

is the disjoint sum. To be precise: For every eigenvector v̄ ∈ EigVec(f|W̄_{i}, θ) for the eigenvalue θ (i.e., if (f|W̄_{i} − θ)v = 0), there exists a lifting v ∈ W_{i} of v̄ such that (f − θ)v = 0.

Proof: This is certainly true for n = 1. By induction on n, we may assume that v̄ ∈ EigVec(f|W̄_{n}, θ) and θ ∈ EigVal(f|W̄_{n}), i.e., (f|W̄_{n} − θ)v̄ = 0. Let v_{1} ∈ W be any lifting of v̄. Then

v_{2} := (f − θ)v_{1} ∈ W_{n−1} .

By induction θ ∉ EigVal(f|_{W_{n−1}}), therefore (f − θ)|_{W_{n−1}} is an automorphism of W_{n−1}, and hence there is a v_{3} ∈ W_{n−1} such that

(f − θ)v_{3} = −v_{2} .

For v := v_{1} + v_{3} we have

(f − θ)v = 0 ,

and v is a lifting of v̄.

Now let V be an n-dimensional k-vector space. Fix an element λ ∈ k^{∗} and a non trivial nilpotent endomorphism g ∈ End_{k}(V), i.e., g^{n} = 0 and g ≠ 0. We set

U_{λ} := {f ∈ End_{k}(V) : f ∘ g = λ · g ∘ f} .

We will consider V as an R-module, where R := k[t]/(t^{n}) and t acts on V as g. Then g induces a filtration

V = V_{n} ⊇ V_{n−1} ⊇ ⋯ ⊇ V_{0} = V_{−1} = 0 ,

with V_{i} := ker(g^{i}) for i = 0, 1, …, n . Moreover we take the R-module

W := V/tV = V/ im(g)

with its induced filtration

W = W_{n} ⊇ W_{n−1} ⊇ ⋯ ⊇ W_{0} = W_{−1} = 0 ,

where

W_{i} = (V_{i} + tV)/tV = V_{i}/(V_{i} ∩ tV) = V_{i}/tV_{i+1} ;

note that V_{i} ∩ tV = tV_{i+1}! Furthermore we set

n_{i} := dim_{k} V_{i} , n := dim_{k} V , m_{i} := dim_{k} W_{i} , m := dim_{k} W

and

W̄_{i} := W_{i}/W_{i−1} , m̄_{i} := dim_{k} W̄_{i} .

(11.2) Lemma. We have the formula: n = ∑_{i=1}^{n} m̄_{i} · i .

Proof: Writing the nilpotent g in its Jordan canonical form, one observes that, as R-module,

V ≅ ⨁_{ν=1}^{r} k[t]/(t^{l_{ν}}) ,

l_{ν} ≥ 1 for all ν = 1, …, r. Therefore

V_{i} ≅ ⨁_{{ν:l_{ν}≤i}} k[t]/(t^{l_{ν}}) ⊕ ⨁_{{ν:l_{ν}>i}} t^{(l_{ν}−i)}k[t]/(t^{l_{ν}})

and thus

W_{i} = V_{i}/tV_{i+1} ≅ ⨁_{{ν:l_{ν}≤i}} k[t]/(t) = ⨁_{{ν:l_{ν}≤i}} k .

Now it follows

m̄_{i} = #{ν : l_{ν} = i}

and

∑_{i=1}^{n} m̄_{i} · i = ∑_{i=1}^{n} #{ν : l_{ν} = i} · i = ∑_{ν=1}^{r} l_{ν} = n .

Note that m̄_{i} is the number of the Jordan blocks of g of the size i × i.

In this situation let us fix an f ∈ U_{λ}. Then f ∈ End_{filt}(V) and the induced endomorphism f ∈ End_{filt}(W) respect the filtrations. We further set

EigVal(f|W̄_{i}) := {θ_{ij} : j = 1, …, ϵ(i)}

with ϵ(i) := # EigVal(f|W̄_{i}) and then

ẼigVal(f|W̄_{i}) := {θ_{ij}λ^{ν} : j = 1, …, ϵ(i); ν = 0, …, i − 1} .

We know

# EigVal(f|W̄_{i}) ≤ dim_{k} W̄_{i} = m̄_{i}

and therefore

#ẼigVal(f|W̄_{i}) ≤ dim_{k} W̄_{i} = m̄_{i} · i .

(11.3) Proposition. If n = #(⋃_{i=1}^{n} ẼigVal(f|W̄_{i})), then

(i) EigVal(f) = ∐_{i=1}^{n} ẼigVal(f|W̄_{i}), (ii) For every eigenvalue θ ∈ EigVal(f|W̄_{i}) and eigenvector v̄ ∈ EigVec(f|W̄_{i}, θ) for θ,

there exists a lifting v ∈ V_{i} of v̄ such that (f − θ)v = 0.

Proof: (ii) implies (i), in the following way: Let θ and v be as in (ii). Then v ∈ V_{i} − V_{i−1}, since v̄ ≠ 0. Therefore the vectors v, gv, g^{2}v, …, g^{i−1}v are all non-zero, and for ν = 0, …, i − 1,

(f − λ^{ν}θ)g^{ν}v = g^{ν}(λ^{ν}f − λ^{ν}θ)v = g^{ν}λ^{ν}(f − θ)v = 0 ,

hence

⋃_{i=1}^{n} ẼigVal(f|W̄_{i}) ⊆ EigVal(f) .

Since the left hand side is already a set of order n, we have equality. Because of (11.2) we know that this union is disjoint and more: All the sets

EigVal(f|W̄_{i}) · λ^{ν}

for i = 1, …, n and ν = 0, …, i − 1, are pairwise disjoint sets of order ϵ(i) = m̄_{i}. Now we prove (ii): Let 0 ≤ j < i. From (11.1), applied to the space W_{i}/W_{j}, follows immediately

(1)  EigVal(f|_{W_{i}/W_{j}}) = ∐_{j+1≤ν≤i} EigVal(f|W̄_{ν}) .

Let v̄ ∈ EigVec(f|W̄_{i}, θ) be an eigenvector to the eigenvalue θ := θ_{ij} ∈ EigVal(f|W̄_{i}). By (11.1) this vector has a lifting

(v′ mod tV_{i+1}) ∈ W_{i} = V_{i}/tV_{i+1} ,

with v′ ∈ V_{i}, which is an eigenvector of f|_{W_{i}} to the eigenvalue θ, i.e.,

(2)  (f − θ)v′ = tv′_{i+1}

for some v′_{i+1} ∈ V_{i+1}. Now, for every q = 0, 1, 2, …, we are going to recursively construct a triple of vectors

v_{i+q} ∈ V_{i+q} , v′_{i+q+1} ∈ V_{i+q+1} , v′′_{q} ∈ V_{q}

(one may read V_{p} = V for q > n) in the following way: For q = 0 we set

v_{i} = v′ , v′_{i+1} from above , v′′ = 0 .

For q > 0 we get them from the vectors for q − 1. First we claim:

\frac{θ}{λ^{q}} ∉ EigVal(f|_{W_{i+q}/W_{q}}) =^{(1)} ⋃_{l=q+1}^{i+q} EigVal(f|W̄_{l}) .

Otherwise for some l with q ≤ l − 1,

θ_{ij} = θ = θ_{l,j′}λ^{q} ∈ EigVal(f|W̄_{l}) · λ^{q} ⊆ ẼigVal(f|W̄_{l}) ,

which is a contradiction to the disjointness of the sets EigVal(f|W̄_{i}) and EigVal(f|W̄_{l})·λ^{q}: For i ≠ l this is clear, for i = l observe that 0 < q ≤ l − 1 = i − 1. Therefore (f − \frac{θ}{λ^{q}})|_{W_{i+q}/W_{q}} is an automorphism, and there is a v_{i+q} ∈ V_{i+q} with

λ^{q}(f − \frac{θ}{λ^{q}})v_{i+q} ≡ −v′_{i+q} (mod V_{q} + tV_{i+q+1})

—note that W_{i+q}/W_{q} = V_{i+q}/(V_{q} + tV_{i+q+1}). So there are vectors v′_{i+q+1} ∈ V_{i+q+1} and v′′_{q} ∈ V_{q} such that

(3)  (λ^{q}f − θ)v_{i+q} + v′_{i+q} = v′′_{q} + tv_{i+q+1} .

Now we set

(4)  v := v_{i} + tv_{i+1} + t^{2}v_{i+2} + ⋯ ∈ V_{i} ;

this sum is finite, it stops after at most n steps. We have

(f − θ)v =^{(4)} (f − θ)v_{i} + (f − θ)tv_{i+1} + (f − θ)t^{2}v_{i+2} + ⋯

= (f − θ)v_{i} + t(λf − θ)v_{i+1} + t^{2}(λ^{2}f − θ)v_{i+2} + …

=^{(3)} (f − θ)v_{i} + t(−v′_{i+1} + tv′_{i+2} + v′′_{1}) + t^{2}(−v′_{i+2} + tv′_{i+3} + v′′_{2}) + ⋯

=^{(2)} tv′_{i+1} + t(−v′_{i+1} + tv′_{i+2} + v′′_{1}) + t^{2}(−v′_{i+2} + tv′_{i+3} + v′′_{2}) + ⋯

= tv′_{i+1} − tv′_{i+1} + t^{2}v′_{i+2} − t^{2}v′_{i+2} + t^{3}v′_{i+3} − ⋯

= 0

and we are done.

(11.4) Lemma. The canonical map

φ: U_{λ} ⟶ End_{filt}(W) ⟶ ⨁_{i=1}^{n} End(W̄_{i})

is surjective.

Proof: We choose a basis {v_{ijν} : i = 1, …, n; j = 1, …, m̄_{i}; ν = 0, …, i − 1} of V such that the associated matrix of g has Jordan canonical form. The set {v_{ij0}, …, v_{ij(i−1)}} corresponds to a Jordan block (i.e., elementary Jordan matrix) of size i × i. Setting v_{iji} := 0, we can write

g(v_{ijν}) = v_{ij(ν+1)}

for ν = 0, …, i − 1. We have

V_{i} = ⨁_{1≤l≤n ; {j=1,...,m̄_{l} \atop max\{l−i,0\}≤ν≤l−1}} k · v_{ljν} ,

and

W_{i} = ⨁_{{1≤l≤i \atop j=1,...,m̄_{l}}} k · v_{lj0}  and  W̄_{i} = ⨁_{j=1,...,m̄_{i}} k · v_{ij0} .

Let now

f̄_{i} = (y_{ijβ})_{j,β=1,...,m̄_{i}} ∈ Mm̄_{i}(k) = End(W̄_{i})

be any endomorphism of W̄_{i} for i = 1, …, n, with respect to {v_{ij0} : j = 1, …, m̄_{i}}, the basis of W̄_{i} from above. We define f ∈ U_{λ} ⊆ M_{n}(k) in the following way:

f(v_{ijν}) := λ^{ν} ∑_{β=1}^{m̄_{i}} y_{ijβ} · v_{iβν}

for i = 1, …, n, j = 1, …, m̄_{i}, ν = 0, …, i − 1. Then

f(g(v_{ijν})) = f(v_{ij(ν+1)}) = λ^{ν+1} ∑_{β=1}^{m̄_{i}} y_{ijβ} · v_{iβ(ν+1)}

and

g(f(v_{ijν})) = g (λ^{ν} ∑_{β=1}^{m̄_{i}} y_{ijβ} · v_{iβν} ) = λ^{ν} ∑_{β=1}^{m̄_{i}} y_{ijβ} · v_{iβ(ν+1)} ,

therefore f ∘ g = λ · g ∘ f, i.e., f ∈ U_{λ}. Restricted to

W̄_{i} = ⨁_{j=1}^{m̄_{i}} k · v_{ij0} .

we see

f(v_{ij0}) = ∑_{β=1}^{m̄_{i}} y_{ijβ} · v_{iβ0} = f̄_{i}(v_{ij0}) .

Since the f̄_{i} were chosen arbitrarily we are done.

(11.5) Lemma. Let W be a k-vector space of dimension m, and let f ∈ End_{k}(W) be nilpotent. For any set of elements θ_{1}, …, θ_{m} ∈ k, there exists an endomorphism f̃ ∈ End_{k}(W) such that for any α ∈ k

EigVal(f + αf̃) = {αθ_{1}, …, αθ_{m}} .

Proof: Let v_{1}, …, v_{m} ∈ W be a basis of W such that the matrix Y associated to f with respect to this basis is strictly (upper or lower) triangular, e.g., take the Jordan canonical form. Then we set for Ỹ —the matrix representing f̃—the diagonal matrix with the entries θ_{1}, …, θ_{m}. Then Y + αỸ is a triangular matrix with the diagonal elements αθ_{1}, …, αθ_{m}.

Let again V be an n-dimensional k-vector space and g, V_{i}, W_{i}, … as above. We choose numbers θ_{ij} ∈ k^{∗}, i = 1, …, n; j = 1, …, m̄_{i} such that all the numbers θ_{ij}λ^{ν} for i = 1, …, n; j = 1, …, m̄_{i}; ν = 0, …, i − 1, are pairwise distinct.

(11.6) Proposition. Let f ∈ End_{k}(V) be nilpotent and f ∈ U_{λ}. Then there exists an f̃ ∈ U_{λ} such that

f + αf̃ ∈ U_{λ}

for all α ∈ k and

EigVal(f + αf̃) = {αθ_{ij}λ^{ν} : i = 1, …, n; j = 1, …, m̄_{i}; ν = 0, …, i − 1} .

Proof: Because of (11.3), it is enough to show that there is an f̃ ∈ U_{λ} such that

EigVal((f + αf̃)|W̄_{i}) = {αθ_{ij} : j = 1, …, m̄_{i}} .

for i = 1, …, n. By (11.5) we choose f̃_{i} ∈ End(W̄_{i}) such that

EigVal(f|W̄_{i} + αf̃_{i}) = {αθ_{ij} : j = 1, …, m̄_{i}} .

Since the map φ in (11.4) is surjective we can find an element f̃ ∈ U_{λ} which maps to the tuple (f̃_{i}). We know

EigVal((f + αf̃)|W̄_{i}) = EigVal(f|W̄_{i} + αf̃_{i})
= {αθ_{ij} : j = 1, …, m̄_{i}} .

Therefore we are done.

2. The Closure of W^{1}(A)

In W̄(A) × W̄(A) ⊆ P(A) × P(A) we have the closed set

W̄^{1} := {([X], [Y]) ∈ W̄(A) × W̄(A) : XY − ζY X = 0} .

This set contains the open subset

W^{1} := {([X], [Y]) ∈ W(A) × W(A) : XY − ζY X = 0} .

(11.7) Theorem. W̄^{1} is the (Zariski) topological closure of W^{1} in P(A)^{2}.

Proof: All we have to show is that W^{1} is dense in W̄^{1}. We first show that W^{1} is dense in

Q := {([X], [Y]) ∈ W̄(A) × W(A) : XY − ζY X = 0} .

So let ([X], [Y]) be an element in Q. We may assume that [X] ∉ W(A). Then take any [X̃] ∈ W(A) with X̃Y − ζY X̃ = 0; which always exists. Now

([X + αX̃], [Y]) ∈ W^{1}

for almost all α ∈ k^{∗}, since Nrd(X + αX̃) ≠ 0 for almost all α. Therefore in any neighbourhood of ([X], [Y]) are elements of W^{1}. Now we prove that Q is dense in W̄^{1}. Let ([X], [Y]) ∈ W̄^{1}. We may assume that [Y] ∉ W(A). For reasons of symmetry we may assume that also [X] ∉ W(A), i.e., X and Y are nilpotent. Otherwise we had the former case with switched components. We further assume A = M_{n}(k). In the situation of subsection 1., we set V = k^{n}, λ = ζ^{−1}, g = X and f = Y. Then Y ∈ U_{λ}. Because of (11.6) there exists for the case

{θ_{ij}λ^{ν}} = {ζ^{0}, …, ζ^{n−1}}

—which is easily realized—a Ỹ ∈ U_{λ} such that

EigVal(Y + αỸ) = {αζ^{0}, …, αζ^{n−1}}

for all α ∈ k^{∗}, i.e., Y + αỸ ∈ W(A). So we may find a pair

([X], [Y + αỸ]) ∈ Q

in any neighbourhood of ([X], [Y]).

Chapter V

Chain Equivalence for Algebras of Degree 2 and 3

Let k be a field and n ≥ 2 a positive integer which is not divisible by the characteristic of k. Let A be a central simple k-algebra of degree n, and ζ ∈ μ_{n} = μ_{n}(k̄) denotes a primitive n-th root of unity. We assume that ζ ∈ k.

If X and Y are Kummer elements of A, we are interested in the question if they are (ℓ, ζ, ν)-related for a ν ∈ (ℤ/nℤ)^{∗} and a positive integer ℓ, i.e., if the set

K-Chain_{ℓ}^{ζ}([X], [Y^{ν}]; A) ≠ ∅

is non-empty for some ℓ and ν. If we are interested in this question for generic [X] and [Y] ∈ W^{0}(A), then we only have to consider the case ν = 1. That means the following: If we have shown—for some fixed ℓ and ζ—that K-Chain_{ℓ}^{ζ}([X], [Y]; A) ≠ ∅ (or has a certain constant order c) for generic [X] and [Y] ∈ W^{0}(A), then it is clear that K-Chain_{ℓ}^{ζ}([X], [Y^{ν}]; A) ≠ ∅ (or has the order c) for generic [X] and [Y] ∈ W^{0}(A), since [Y] ↦ [Y^{ν}] is an automorphism of W^{0}(A). Moreover we then can say, that for another primitive n-th root ζ′ the set K-Chain_{ℓ}^{ζ′}([X], [Y^{ν}]; A) ≠ ∅ (or has the order c) for generic [X] and [Y] ∈ W^{0}(A). This follows from (8.6). We may formulate the question in the language of schemes: Let ℓ be a positive integer. Is the morphism of projection to the first and last factor

(π_{0}^{ℓ}, π_{ℓ}^{ℓ}): W^{ℓ}(A) ⟶ W^{0}(A) ×_{k} W^{0}(A)

surjective on the k-rational points, or is it dominant? If yes, then how big are the fibres? In (10.26) we saw

dim W^{ℓ}(A) = (ℓ + n)(n − 1)

and from (9.22) we know

dim W^{0}(A) ×_{k} W^{0}(A) = 2n(n − 1) .

Hence for dominance we need at least ℓ ≥ n.

It is announced by M. Rost that for prime numbers n, the morphism (π_{0}^{n}, π_{n}^{n}) is dominant of degree prime to n. We are going to show this for the cases n = 2 and 3; we even will show that the degree is actually 1 and 2 respectively. It is clear that dominance and degree are independent under base change k′|k. So one may treat and prove the questions of degree and dominance in the case (n, ℓ) for just one special central simple k-algebra A and then automatically get the results for every central simple k′-algebra, where k′ is any other field of the same characteristic as k. Especially one may consider the extreme cases of A being a division algebra or a matrix algebra.—Of course surjectivity depends on the base field and may change with it.

§ 12. The case of algebras of degree 2

The group scheme GL_{1}(A) acts on the schemes W^{ℓ}(A) and W^{0}(A) ×_{k} W^{0}(A) by conjugation in every factor and the projection morphisms π_{0}^{ℓ}: W^{ℓ}(A) → W^{0}(A) as well as pr_{1}: W^{0}(A)×_{k} W^{0}(A) → W^{0}(A) are GL_{1}(A)-morphisms. Since GL_{1}(A) acts transitively on W^{0}(A) we may ask our questions in the following formulation: Is the morphism

π_{ℓ}^{ℓ}: W^{ℓ}([X]; A) ⟶ W^{0}(A)

dominant and what is the degree (which is the same as the degree above)?

§ 12. The Case of Algebras of Degree 2

We fix a field k of characteristic ≠ 2. Let A be a central simple k-algebra of degree 2 and ζ = −1.

1. Chain Equivalence for n = ℓ = 2

We are going to prove that (almost) any two Kummer elements [X], [Y] ∈ W(A) are connected by one chain of length two. We will split up this problem into the two cases that A is a division algebra or a matrix algebra.

(12.1) Lemma. For [X] and [X′] ∈ W(A) the condition XX′ − X′X = 0 implies [X] = [X′].

Proof: X′ commutes with X, therefore X′ ∈ k[X], i.e., X′ = αX + β, where α, β ∈ k. But

0 = Trd(X′) = Trd(αX + β) = Trd(β) = 2β ,

hence X′ = αX.

(12.2) Notation. For any central simple k-algebra A of (arbitrary) degree n, with characteristic of k not dividing n and any primitive n-th root ζ of unity we denote

u(ζ, [X]) := u(ζ, [X]; A) := {[Y] ∈ W(A) : XY − ζY X = 0}

for any [X] ∈ W(A). In other words

{[X]} × u(ζ, [X]) = W^{1}([X]; A)(Spec k) .

Now we first assume that A is a division algebra of degree 2.

(12.3) Proposition. If A is a division algebra and [X], [Y] ∈ W(A) are Kummer elements, then K-Chain_{2}^{ζ}([X], [Y]; A) ≠ ∅. To be precise:

K-Chain_{2}^{ζ}([X], [Y]; A) = \{ \begin{matrix}\{([X], [XY − Y X], [Y])\}, & if [X] ≠ [Y] \\ \{[X]\} × u(ζ, [X]) × \{[Y]\}, & if [X] = [Y] \end{matrix}

Proof: The second case is clear, since ζ = ζ^{−1} = −1; and u(ζ, [X]) ≠ ∅ because of (6.16). If [X] ≠ [Y], then by (12.1) the commutator XY − Y X ≠ 0 is invertible, and we obviously have X ⟶^{ζ} (XY − Y X) ⟶^{ζ} Y, e.g., X(XY − Y X) + (XY − Y X)X = X^{2}Y − XY X + XY X − Y X^{2} = X^{2}Y − Y X^{2} = 0. In particular [XY − Y X] ∈ W(A)—cf. (6.8). All we have to show now is that there is only one [Z] ∈ W(A) with X ⟶^{ζ} Z ⟶^{ζ} Y. In this case, since ζ = ζ^{−1}, we have

[Z] ∈ u(ζ, [X]) ∩ u(ζ, [Y])

and since u(ζ, [X]) and u(ζ, [Y]) are one-dimensional linear subspaces of P(A) they intersect in exactly one point or they are equal. But only the first case is true: Otherwise we got k[X]Z = k[Y]Z, hence k[X] = k[Y]. By (12.1) we had [X] = [Y].

Now we assume that A is a matrix algebra. First let A = A_{0} = M_{2}(k)

(12.4) Remark. From the definition of Kummer elements follows:

W(A_{0}) = \{( \begin{matrix}a & b \\ c & −a \end{matrix} ) : a^{2} + bc ≠ 0\} .

(12.5) Lemma. For X_{0} = ({−1 \atop 0} {0 \atop 1}) we have

u(ζ, [X_{0}]) = \{[( \begin{matrix}0 & β \\ γ & 0 \end{matrix} )] : βγ ≠ 0\} .

Proof: Consider the equivalences

\begin{aligned}( \begin{matrix}α & β \\ γ & δ \end{matrix} ) ∈ u(ζ, [X_{0}]) & ⟺ ( \begin{matrix}−1 & 0 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) ( \begin{matrix}−1 & 0 \\ 0 & 1 \end{matrix} ) = − ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) ∈ GL_{2}(k) \\ & ⟺ α = δ = 0, βγ ≠ 0 . \end{aligned}

(12.6) Proposition. Let X_{0} = ({−1 \atop 0} {0 \atop 1}) and Y = ({a \atop c} {b \atop −a}) ∈ W(A_{0}) be two Kummer elements. Then

K-Chain_{2}^{ζ}([X_{0}], [Y]; A_{0}) =
= \{ \begin{matrix}\{([X_{0}], [X_{0}Y − Y X_{0}], [Y])\}, & if bc ≠ 0 \\ \{[X_{0}]\} × u(ζ, [X_{0}]) × \{[Y]\}, & if b = c = 0, i.e., [X_{0}] = [Y] \\ ∅, & if (b ≠ 0, c = 0) or (b = 0, c ≠ 0) . \end{matrix}

Proof: Because of (12.5) we just have to solve the equation

( \begin{matrix}0 & β \\ γ & 0 \end{matrix} ) ( \begin{matrix}a & b \\ c & −a \end{matrix} ) = − ( \begin{matrix}a & b \\ c & −a \end{matrix} ) ( \begin{matrix}0 & β \\ γ & 0 \end{matrix} )

for γβ ≠ 0. But this equation is equivalent to βc + γb = 0. If b = c = 0 all β, γ ∈ k^{∗} do the job. If cb ≠ 0, then \frac{β}{γ} = \frac{b}{−c}, which gives us the one solution. If (b ≠ 0, c = 0) or (b = 0, c ≠ 0) the equation is not solvable for βγ ≠ 0.

Now let A be any algebra isomorphic to A_{0} = M_{2}(k).

(12.7) Corollary. Let X, Y ∈ W(A) be any Kummer elements. Then

K-Chain_{2}^{ζ}([X], [Y]; A) =
= \{ \begin{matrix}\{([X], [XY − Y X], [Y])\}, & if det(XY − Y X) ≠ 0 \\ \{[X]\} × u(ζ, [X]) × \{[Y]\}, & if XY − Y X = 0, i.e., [X] = [Y] \\ ∅, & otherwise \end{matrix}

Proof: The first two cases are clear. Note that the uniqueness in the first case can be proved after base field extension such that [X] is conjugate to [X_{0}]. The last case may also be shown after base extension: Then we again may assume that [X] = [X_{0}] and Y = ({a \atop c} {b \atop −a}) ∈ W(A). But the condition det(X_{0}Y − Y X_{0}) = 0 and X_{0}Y − Y X_{0} ≠ 0 just means (b ≠ 0, c = 0) or (b = 0, c ≠ 0): Observe X_{0}Y − Y X_{0} = ({0 \atop 2c} {−2b \atop 0}).

2. The Geometric Point of View

In §9 and §10 we gave a description of W(A) and K-Chain_{2}^{ζ}(A) as the k-schemes W^{0}(A) and W^{2}(A). The projection onto the first and last component

(π_{0}^{2}, π_{2}^{2}): W^{2}(A) ⟶ W^{0}(A) ×_{k} W^{0}(A)

is a morphism of k-schemes. In fact, we will show that it is—cum grano salis—a blowing up in the diagonal.

We assume that k is algebraically closed and A = A_{0} = M_{2}(k) is a matrix algebra.

(12.8) Remark. Since W̄(A) is the subvariety of P(A) given by the polynomial Trd, we can write

W̄(A) = \{[( \begin{matrix}a & b \\ c & −a \end{matrix} )] : (a, b, c) ≠ (0, 0, 0)\} .

One can identify W̄(A) with ℙ_{k}^{2} via

\begin{matrix}ψ: W̄(A) & \underrightarrow{∼} ℙ_{k}^{2} \\ [({a \atop c} {b \atop −a})] & ⟼ (2a : b − c : b + c) \end{matrix}

or

\begin{matrix}ψ′: W̄(A) & \underrightarrow{∼} ℙ_{k}^{2} \\ [({a \atop c} {b \atop −a})] & ⟼ (2a : c − b : b + c) . \end{matrix}

The inverse of, e.g., ψ is given by:

ψ^{−1}(α : β : γ) = [( \frac{γ−β}{2} −\frac{α}{2}^{\frac{α}{2} \frac{β+γ}{2}} )] .

§ 12. The case of algebras of degree 2

(12.9) Lemma. The diagram

�  ^{(pr_{1},pr_{3})} �_{φ:=ψ×ψ′×ψ} �  _{p_{13}:=(pr_{1},pr_{3})} �  _{ψ×ψ} ℙ_{k}^{2} × ℙ_{k}^{2} B := {([u], [v], [w]) ∈ (ℙ_{k}^{2})^{3} : v(u) = 0, v(w) = 0}  W̄(A) × W̄(A) {([X], [Y], [Z]) ∈ W̄(A)^{3} : XY + Y X = 0, XZ + ZX = 0}

is commutative with vertical isomorphisms. For [v] = (v_{0} : v_{1} : v_{2}), [u] = (u_{0} : u_{1} : u_{2}) the expression v(u) = 0 stands for v_{0}u_{0} + v_{1}u_{1} + v_{2}u_{2} = 0. Proof: The relations XY + Y X = 0 and Y Z + ZY = 0 are transformed by φ into v(u) = 0 and v(w) = 0, where [u] = ψ([X]), [v] = ψ′([Y]) and [w] = ψ([Z]): Set X = ({a \atop c} {b \atop −a}) and Y = ({a′ \atop c′} {b′ \atop −a′} ), then

XY + Y X = ({2aa′+bc′+cb′ \atop 0} {0 \atop 2aa′+bc′+cb′} ) = 0
⟺ 4aa′ + 2bc′ + 2cb′ = 0
⟺ (2a)(2a′) + (b − c)(c′ − b′) + (b + c)(b′ + c′) = 0
⟺ v(u) = 0 .

Analogous for the other relation. The rest is clear. The projection morphism p_{13}: B ⟶ ℙ_{k}^{2} ×ℙ_{k}^{2} is the blowing up of ℙ_{k}^{2} ×ℙ_{k}^{2} in its diagonal. In order to show that, we need some lemmas. (12.10) Lemma. The projection morphism onto the first and third factor

p_{13} := (pr_{1}, pr_{3}): {(x, [y], z) ∈ 𝔸_{k}^{2} × ℙ_{k}^{1} × 𝔸_{k}^{2} : x ∈ [y]} ⟶ 𝔸_{k}^{2} × 𝔸_{k}^{2}

is the blowing up of 𝔸_{k}^{2} × 𝔸_{k}^{2} in {0} × 𝔸_{k}^{2}. Here the relation x ∈ [y] means y_{1}x_{2} = y_{2}x_{1} for x = (x_{1}, x_{2}) and [y] = (y_{1} : y_{2}). Proof: By definition, the projection morphism on the first factor

pr_{1}: {(x, [y]) ∈ 𝔸_{k}^{2} × ℙ_{k}^{1} : x ∈ [y]} ⟶ 𝔸_{k}^{2}

is the blowing up of 𝔸_{k}^{2} in {0}, and the morphism of the claim is the base extension with 𝔸_{k}^{2} ⟶ Spec k. (12.11) Lemma. The projection morphism onto the first and third factor

p_{13}: {(x, [y], z) ∈ 𝔸_{k}^{2} × ℙ_{k}^{1} × 𝔸_{k}^{2} : x − z ∈ [y]} ⟶ 𝔸_{k}^{2} × 𝔸_{k}^{2}

is the blowing up of 𝔸_{k}^{2} × 𝔸_{k}^{2} in its diagonal. Proof: This follows from (12.10) by the following coordinate transformation

�  ^{p_{13}} � �  _{p_{13}} �  𝔸_{k}^{2} × 𝔸_{k}^{2} , {(u, [v], w) ∈ 𝔸_{k}^{2} × ℙ_{k}^{1} × 𝔸_{k}^{2} : u ∈ [v]}  𝔸_{k}^{2} × 𝔸_{k}^{2} {(x, [y], z) ∈ 𝔸_{k}^{2} × ℙ_{k}^{1} × 𝔸_{k}^{2} : x − z ∈ [y]}

where the left vertical arrow is given by (x, [y], z) ↦ (u, [v], w), with u = x − z, w = z,

[v] = [y]; and the right vertical arrow is given by (x, z) ↦ (u, w). with u = x−z, w = z. It induces an isomorphism between the diagonal of 𝔸_{k}^{2} × 𝔸_{k}^{2} and {0} × 𝔸_{k}^{2}.

(12.12) Proposition. The following morphism is an isomorphism

\begin{matrix}\{(x, [y], z) ∈ 𝔸_{k}^{2} × ℙ_{k}^{1} × 𝔸_{k}^{2} : x − z ∈ [y]\} & \underrightarrow{∼} \\ \{(u, [v] & , w) ∈ 𝔸_{k}^{2} × ℙ_{k}^{2} × 𝔸_{k}^{2} : v(u) = 0, v(w) = 0\} \end{matrix}

defined by

\begin{matrix}((x_{1}, x_{2}), (y_{1} : y_{2}), (z_{1}, z_{2})) & ⟼ ((x_{1}, x_{2}), (y_{1}z_{2} − y_{2}z_{1} : y_{2} : −y_{1}), (z_{1}, z_{2})) \\ ((u_{1}, u_{2}), (v_{2} : −v_{1}), (w_{1}, w_{2})) & ⟵ ((u_{1}, u_{2}), (v_{0} : v_{1} : v_{2}), (w_{1}, w_{2})) \end{matrix}

Here the expression v(u) = 0 means v_{0} + v_{1}u_{1} + v_{2}u_{2} = 0, where [v] = (v_{0} : v_{1} : v_{2}) and u = (u_{1}, u_{2}).

Proof: We only have to prove that the two mappings are well-defined, since they are (almost) obviously inverse to each other. Let x = (x_{1}, x_{2}), z = (z_{1}, z_{2}) ∈ 𝔸_{k}^{2} and y = (y_{1} : y_{2}) ∈ ℙ_{k}^{1}. Then

x − z ∈ [y] ⟹ (x_{1} − z_{1})y_{2} = y_{1}(x_{2} − z_{2})
⟹ (x_{1} − z_{1})y_{2} + (x_{2} − z_{2})(−y_{1}) = 0
⟹ v(u) = v_{0} + v_{1}u_{1} + v_{2}u_{2} = (y_{1}z_{2} − y_{2}z_{1}) + x_{1}y_{2} + x_{2}(−y_{1}) = 0

and

x − z ∈ [y] ⟹ v(w) = v_{0} + v_{1}w_{1} + v_{2}w_{2} = (y_{1}z_{2} − y_{2}z_{1}) + z_{1}y_{2} + z_{2}(−y_{1}) = 0

for u = x, [v] = (y_{1}z_{2} − y_{2}z_{1} : y_{2} : −y_{1}) and w = z. On the other hand, let u = (u_{1}, u_{2}), w = (w_{1}, w_{2}) ∈ 𝔸_{k}^{2} and [v] = (v_{0} : v_{1} : v_{2}) ∈ ℙ_{k}^{2}, then v(u) = 0 and v(w) = 0 implies

0 = v(u) − v(w) = (u_{1} − w_{1})v_{1} + (u_{2} − w_{2})v_{2}

hence (u_{1} − w_{1})(−v_{1}) = (u_{2} − w_{2})(v_{2}), i.e., x − z ∈ [y] for x = u, [y] = (v_{2} : −v_{1}) and z = w. But it also implies v_{0} = −v_{1}w_{1} − v_{2}w_{2} hence

v = (v_{0} : v_{1} : v_{2})
= (−v_{1}w_{1} − v_{2}w_{2} : v_{1} : v_{2})
= (v_{2}w_{2} + v_{1}w_{1} : −v_{1} : v_{2})
= (y_{1}z_{2} − y_{2}z_{1} : y_{2} : −y_{1})

what we need for proving that one composition of the two mappings is the identity.

(12.13) Corollary. The morphism p_{13}: B ⟶ ℙ_{k}^{2} × ℙ_{k}^{2} is the blowing up of ℙ_{k}^{2} × ℙ_{k}^{2} in its diagonal.

Proof: All we have to show is that the map, restricted to the open sets of an open covering of ℙ_{k}^{2} × ℙ_{k}^{2} is a blowing up. Therefore we take a neighbourhood of a point ([u], [w]) ∈ ℙ_{k}^{2}×ℙ_{k}^{2}, [u] = (u_{0} : u_{1} : u_{2}), [w] = (w_{0} : w_{1} : w_{2}). Applying a suitable unitary transformation on all factors ℙ_{k}^{2} of the map p_{13}—it leaves the conditions v(u) = v(w) = 0 invariant—we may assume u_{0} = w_{0} = 1, and we consider the open neighbourhood

\begin{matrix}𝔸_{k}^{2} × 𝔸_{k}^{2} & ⟶ ℙ_{k}^{2} × ℙ_{k}^{2} \\ ((u_{1}, u_{2}), (w_{1}, w_{2})) & ⟼ ((1 : u_{1} : u_{2}), (1 : w_{1} : w_{2})) \end{matrix}

the rest is just (12.12) with (12.11).

Now we know with (12.9) that the map

p_{13}: {([X], [Y], [Z]) ∈ W̄(A)^{3} : XY + Y X = 0, XZ + ZX = 0} ⟶ W̄(A)^{2}

is a blowing up in the diagonal. Restricting (or taking base extension) to the open subscheme W(A) × W(A) we get the blowing up of W(A) × W(A) in its diagonal. Additionally, for any [X_{0}] ∈ W(A) we may take base extension with {[X_{0}]} × W(A) ↪ W(A) × W(A). In all we get

(12.14) Theorem. The morphisms

{([X], [Y], [Z]) ∈ W(A) × W̄(A) × W(A) : XY + Y X = 0, XZ + ZX = 0} ⟶^{p_{13}} W(A)^{2}

and

{([X], [Y], [Z]) ∈ {[X_{0}]} × W̄(A) × W(A) : XY + Y X = 0, XZ + ZX = 0} ⟶^{pr_{3}} W(A)

are blowing ups in the diagonal and in {[X_{0}]} respectively. The left varieties are the (Zariski) closures of W^{2}(A) in W(A)×W̄(A)×W(A) and W^{2}([X_{0}]; A) in W̄(A)×W(A).

(12.15) Remark. Comparing (12.14) with (12.7) one observes that the pre-image (under pr_{3}) of the point (b ≠ 0, c = 0) and (b = 0, c ≠ 0) lie in (W̄(A) − W(A)) × W(A).

§ 13. The Case of Algebras of Degree 3 and Length 3

We fix a field of characteristic prime to 6 with a primitive 3-rd root ζ ∈ μ_{3}(k) of unity lying in k. Let A be a central simple k-algebra of degree 3. We choose a ζ-pair (X, Y), i.e., X, Y ∈ W(A) such that XY − ζY X = 0.

1. Preliminaries and Conditions for Chains of Length 2

From (6.14) we know

A = ⨁_{0≤i,j≤2} k · X^{i}Y^{j}

or, if we set L := k[X] = k ⊕ kX ⊕ kX^{2},

A = L ⊕ Y L ⊕ LY^{−1} .

(13.1) Notation. Every element T ∈ A has a unique presentation in the form

T = α + Y β + γY^{−1}

with α, β, γ ∈ L. In this section, if we write T, we will always work with this presentation. Y acts on L via conjugation:

\begin{matrix}κ_{Y}: L & ⟶ L \\ X & ⟼ Y XY^{−1} = ζ^{−1}X . \end{matrix}

We denote this action by α ↦ \overline{α}.

(13.2) Lemma. For α ∈ L, we have the equivalences

\begin{aligned}α = ζ\overline{\overline{α}} ⟺ \overline{α} = ζα & ⟺ α ∈ kX^{2} , \\ \overline{\overline{α}} = ζα ⟺ α = ζ\overline{α} & ⟺ α ∈ kX , \\ α = \overline{α} & ⟺ α ∈ k . \end{aligned}

Furthermore, if we write α = α_{0} + α_{1}X + α_{2}X^{2} for α_{0}, α_{1}, α_{2} ∈ k, then

Trd(α) = Trd(\overline{α}) = 3α_{0} .

Proof: Observe for α = α_{0} + α_{1}X + α_{2}X^{2} that \overline{α} = α_{0} + ζ^{−1}α_{1}X + ζα_{2}X^{2}.

(13.3) Lemma. For T = α + Y β + γY^{−1} ∈ A we have

Trd(T) = Trd(α)
Trd(T^{2}) = Trd(α^{2}) + 2 Trd(βγ) .

Proof: Note that X, Y, X^{2}, Y^{2}, Y X, Y X^{2}, XY^{2}, X^{2}Y^{2} are Kummer elements (cf. (6.9)) hence their reduced trace is zero.

Trd(T) = Trd(α) + Trd(Y β) + Trd(γY^{−1})
= Trd(α) ,
Trd(T^{2}) = Trd(α^{2} + Y βγY^{−1} + γY^{−1}Y β)
= Trd(α^{2}) + Trd(\overline{βγ}) + Trd(γβ)
= Trd(α^{2}) + Trd(βγ) + Trd(βγ) .

(13.4) Lemma. For invertible T ∈ A^{∗}, the following conditions are equivalent: (i) T ∈ W(A) is a Kummer element, (ii) Trd(α) = 0 and Trd(α^{2}) = −2 Trd(βγ) .

Proof: Since the characteristic of k is prime to (3!) we know from (6.5) that (i) is equivalent to Trd(T) = Trd(T^{2}) = 0, and this is by (13.3) equivalent to (ii).

Any Kummer element X_{1} ∈ W(A) which is in ζ-relation with X_{0} := X, i.e., X_{0}X_{1} − ζX_{1}X_{0} = 0, has the form X_{1} = Y λ for a λ ∈ L^{∗}. What are the conditions for an element T ∈ A to be the third entry in a ζ-chain ([X_{0}], [X_{1}], [T]) ∈ K-Chain_{2}^{ζ}(A)?

(13.5) Lemma. Let X_{1} := Y λ with λ ∈ L and T = α + Y β + γY^{−1} ∈ A, then the following conditions are equivalent:

(i) X_{1}T − ζT X_{1} = 0

(ii) αλ = ζ\overline{\overline{α}}λ, \overline{γ}\overline{λ} = ζγλ, β\overline{\overline{λ}} = ζ\overline{\overline{β}}λ.

Proof: Observe

X_{1}T − ζT X_{1} = (Y λ)(α + Y β + γY^{−1}) − ζ(α + Y β + γY^{−1})(Y λ)
= Y λα + Y λY β + Y λγY^{−1} − ζαY λ − ζY βY λ − ζγλ
= (\overline{λ}\overline{γ} − ζγλ) + Y (λα − ζ\overline{\overline{α}}λ) + Y^{2}(\overline{\overline{λ}}β − ζ\overline{\overline{β}}λ) .

The coefficients are zero if and only if (i) holds.

(13.6) Proposition. If for T = α + Y β + γY^{−1} ∈ W(A) the conditions (I) Trd(αX^{2}) = 0 (II) Trd(βγX) = 0 (III’) β ∈ L^{∗} or (III’’) γ ∈ L^{∗} hold, then there exists an X_{1} ∈ W(A) with

X_{0}X_{1} − ζX_{1}X_{0} = 0 and X_{1}T − ζT X_{1} = 0 .

Proof: With (13.2) and (13.4) we know for α = α_{0}+α_{1}X +α_{2}X^{2} that α_{0} = \frac{1}{3} Trd(α) = \frac{1}{3} Trd(T) = 0.

Condition (I) implies that α_{1} = \frac{1}{3} Trd(αX^{−1}) = Trd(αX^{2})/(3X^{3}) = 0. Therefore α = α_{2}X^{2}, hence α^{2} = α_{2}^{2}X^{4} and Trd(α^{2}) = 0. Because of (13.4) we get Trd(βγ) = 0. Together with condition (II) we get βγ ∈ kX. Now we assume that (III’) holds true: In this case set

λ := βX and X_{1} := Y λ ∈ W(A) ,

and then we have X_{0}X_{1} − ζX_{1}X_{0} = 0. Moreover the equations of (13.5)(ii) are valid: From α = α_{2}X^{2} follows α = ζ\overline{\overline{α}}, hence αλ = ζ\overline{\overline{α}}λ.

\frac{λ}{β} = X implies \overline{\overline{(\frac{λ}{β})}} = ζ(\frac{λ}{β}), i.e., β\overline{\overline{λ}} = ζ\overline{\overline{β}}λ. Finally, βγ ∈ kX shows λγ ∈ kX^{2}, hence \overline{γ}\overline{λ} = ζγλ. If we assume (III’’) then set

λ := γ^{−1}X^{2} and X_{1} := Y λ ∈ W(A) ,

and the rest follows analogously.

(13.7) Proposition. Let T = α + Y β + γY^{−1} ∈ A such that there is an X_{1} ∈ W(A) with

X_{0}X_{1} − ζX_{1}X_{0} = 0 and X_{1}T − ζT X_{1} = 0 .

Then

(I) Trd(αX^{2}) = 0, i.e., α_{1} = 0

(II) Trd(βγX) = 0.

Proof: Because of the first Kummer relation, X_{1} = Y λ for a λ ∈ L^{∗}. The situation, applied to (13.5), gives us

αλ = ζ\overline{\overline{α}}λ, \overline{γ}\overline{λ} = ζγλ, β\overline{\overline{λ}} = ζ\overline{\overline{β}}λ .

Since λ is invertible, we can divide it out of the first equation and get α = ζ\overline{\overline{α}}, hence by (13.2) α ∈ kX^{2}, which implies (I). Conjugating the third equation by Y we get \overline{β}λ = ζβ\overline{λ}; multiplying this one with the second equation yields \overline{β}λ\overline{γ}\overline{λ} = ζ^{2}β\overline{λ}γλ, i.e., \overline{γ}\overline{β} = ζ^{−1}γβ, hence we get γβ ∈ kX, which shows (II).

(13.8) Proposition. If for a T = α + Y β + γY^{−1} ∈ A^{∗}, with β ∈ L^{∗} or γ ∈ L^{∗}, there exists an X_{1} ∈ W(A) such that ([X_{0}], [X_{1}], [T]) ∈ K-Chain_{2}^{ζ}(A), then [X_{1}] is unique with this property.

Proof: Any X_{1} with X_{0} ⟶^{ζ} X_{1} ⟶^{ζ} T has the form X_{1} = Y λ for a λ ∈ L^{∗}. Then if

β ∈ L^{∗}, we know from the proof of (13.7) that \overline{\overline{(\frac{λ}{β})}} = ζ(\frac{λ}{β}), i.e., \frac{λ}{β} ∈ kX or λ ∈ kβX. But that means λ is unique up to scaling with an element of k^{∗}. Therefore [Y λ] is unique. Analogous if γ ∈ L^{∗}.

(13.9) Example. There are exactly two chains of length three connecting [X] with [Y], namely

([X], [XY], [XY^{2}], [Y]) and ([X], [X^{2}Y], [XY], [Y]) .

Indeed, one get the chains by expanding the ζ-pair ([X], [Y]) to ([X], [XY], [Y]) and then again expanding in the first or second link—cf. (6.9). But they are the only ones: Let ([X], [X_{1}], [T], [Y]) ∈ K-Chain_{3}^{ζ}(A) be a ζ-chain. Then we know for T = α + Y β + γY^{−1} from (6.10) that T ∈ k[Y] · X, i.e.,

α_{0} = α_{2} = β_{0} = β_{2} = γ_{0} = γ_{2} = 0

or T = α_{1}X + Y β_{1}X + γ_{1}XY^{−1}. Furthermore we know from (13.7) that (I) α_{1} = 0 and (II) β_{1}γ_{1} = 0. This has two (projective) solutions (β_{1} = 0, γ_{1} ≠ 0) and (β_{1} ≠ 0, γ_{1} = 0) which lead to the two chains from above. Note that the two solution-chains are already k-rational!

2. Existence of Connecting Chains of Length 3

Now we formulate the main theorem of this section which roughly says that almost any two Kummer elements can be connected by exactly two chains of length three.

(13.10) Theorem. The morphism

π_{3}^{3}: W^{3}([X]; A) ⟶ W^{0}(A)

is a dominant morphism of degree 2. In other words, there is an open non-empty, therefore dense, subset U ⊆ W^{0}(A) with the following property: If [Z] ∈ U then (π_{3}^{3})^{−1}([Z]) is a set of two points of degree (of the residue field extension) 1 or one point of degree 2.

(13.11) Remark. We can formulate the theorem in the following way: Let k′|k be a field extension in some algebraic closure k̄ of k and let [Z] ∈ W(A_{k′}) ∩ U be a k′- rational point, i.e., Kummer element, lying in U. Then there exists a field extension l|k′ of degree one or two such that

K-Chain_{3}^{ζ}([X], [Z]; A_{l}) = K-Chain_{3}^{ζ}([X], [Z]; Ak̄) ,

and this is a set of two elements.

At the beginning of this chapter we mentioned that the degree of (π_{3}^{3})—and therefore the whole theorem—does not depend on the special choice of the central simple algebra A (of degree 3) and the field k for a fixed characteristic. So for the proof we choose and fix A and k in a way such that A is a division algebra. (In particular k is not separably closed!)

Let

\begin{matrix}f_{1}: A & ⟶ k \\ T & ⟼ Trd(αX^{2}) \end{matrix}

and

\begin{matrix}f_{2}: A & ⟶ k \\ T & ⟼ Trd(βγX) , \end{matrix}

where T = α + Y β + γY^{−1}. For a 3-dimensional vector subspace E ⊆ A we say that the system (f_{1}, f_{2}) is non degenerate in E if the following two conditions hold:

(i) f_{1}|_{E} ≠ 0 (ii) The quadratic form f_{2} restricted to E ∩ ker(f_{1}) is non degenerate. Note, that by (i) the vector space E′ := E ∩ ker(f_{1}) has dimension 2. Condition (ii) means that the quadratic form q := f_{2}|_{E′} is non degenerate.

Let 𝒜 := A ⊗_{k} O_{W^{0}(A)}—so, 𝒜 is the dual to the sheaf of O_{W^{0}(A)}-modules ℳ, which we know from §10—, and let ℰ be the the O_{W^{0}(A)}-submodule of 𝒜 of rank 3, which is given by the equation

T Z − ζZT = 0 where T ∈ A and [Z] ∈ W^{0}(A) .

—Cf. §10. There we can see that ℰ is the dual of ℱ.

(13.12) Lemma. There exists an open and dense subscheme U ⊆ W^{0}(A) such that for each point u ∈ U, the system (f_{1}, f_{2}) is non degenerate in ℰ (u) = ℰ_{u} ⊗_{O_{W^{0}(A),u}} κ(u) ⊆ A ⊗_{k} κ(u).

Proof: The non degeneracy condition is an open condition and the variety W^{0}(A) is irreducible by (9.22). Therefore it suffices to find one point u ∈ W^{0}(A) such that (f_{1}, f_{2}) is non degenerate in ℰ (u). We put u = [Y]. Then ℰ (u) = X · k[Y] and we have T = α_{1}X + Y β_{1}X + γ_{1}XY^{−1} for α_{1}, β_{1}, γ_{1} ∈ κ(u). The system (f_{1}, f_{2}) restricted to ℰ (u) is given by

(α_{1}, β_{1}, γ_{1}) ⟼ (3X^{3}α_{1}, 3X^{3}β_{1}γ_{1})

which is obviously non degenerate.

Let V ⊆ P(A) ×_{k} W^{0}(A) be the subscheme defined by the equations

(1) T Z − ζZT = 0

(2) Trd(αX^{2}) = 0

(3) Trd(βγX) = 0. where [T] = [α + Y β + γY^{−1}] ∈ P(A) and [Z] ∈ W^{0}(A). As we have seen in §10, equation (1) defines the 2-dimensional projective bundle ℙ(ℱ) = Proj(S(ℰ̌)) on W^{0}(A). Thus V is the subscheme of ℙ(ℱ) given by the system (f_{1}, f_{2}). Let p: V → W^{0}(A) be the restriction of the projection P(A) ×_{k} W^{0}(A) ⟶ W^{0}(A) and let η ∈ W^{0}(A) be the generic point of W^{0}(A), with residue field F := κ(η) and V_{η} = p^{−1}(η) the fiber over η.

(13.13) Proposition. We have the following facts:

(i) One has V_{η} = Spec(K) where the F -algebra K is a separable extension of F of degree 2.

(ii) The F -algebra A_{F} is a division algebra. (iii) If K is a field, the K-algebra A_{K} is a division algebra.

Proof: (i) follows from (13.12). For (ii) note that A is a division algebra and that W^{0}(A) has a smooth k-rational point, e.g., [X] or [Y]. (iii) follows since [K : F] = 2 is prime to the index of A.

Let S ⊆ P(A) ×_{k} W^{0}(A) be the closed subscheme given by the equations

Nrd(β) = Nrd(γ) = 0 where [T] = [α + Y β + γY^{−1}] ∈ P(A) .

(13.14) Proposition. V_{η} ∩ S = ∅.

Proof: Let x ∈ V_{η} and let H := κ(x) be the residue field. Thus H = K if K is a field or H = F if K = F ⊕ F. The point x ∈ V_{η} is of the form

([T_{x}], [Z_{η}]) ∈ P(A)(Spec H) × W^{0}(A)(Spec F)

where [Z_{η}] ∈ W^{0}(A)(Spec F) is the generic element, i.e., the F -rational point localized in the generic point η and [T_{x}] = [α + Y β + γY^{−1}] with α, β, γ ∈ H[X]. If x ∈ V_{η} ∩ S then Nrd(β) = Nrd(γ) = 0. Since A_{H} is a division algebra by (13.13)(ii) and (iii), we must have even β = γ = 0. Further, since Trd(α) = 0 and Trd(αX^{2}) = 0 (the first equation holds, because [T] is a Kummer element and the second by the equation (2)) we have [T_{x}] = [X^{2}]. But then

T_{x}Z_{η} = ζZ_{η}T_{x}

implies

X^{2}Z_{η} = ζZ_{η}X^{2} .

This would mean that the generic element of W^{0}(A) stands in Kummer relation to X^{2}. This is a contradiction.

Let

g := (π_{2}^{3}, π_{3}^{3}): W^{3}([X]; A) ⟶ P(A) ×_{k} P(A) .

By proposition (13.7) the image of g is contained in V. Therefore we have a commutative diagram

�_{g} �  _{π_{3}^{3}} �  _{p}  W^{0}(A) . V W^{3}([X]; A)

By proposition (13.6) and (13.14) we have

V_{η} ⊆ im(g) .

Hence there exists a point y ∈ W^{3}([X]; A) such that g(y) ∈ V_{η}, i.e., y maps under π_{3}^{3} to the generic point of W^{0}(A).

Since dim W^{3}([X]; A) = dim W^{0}(A), the point must be the generic point of W^{3}([X]; A). Hence

g^{−1}(V_{η}) = {y}

consists of only one point. It follows that

V_{η} = {g(y)}

is irreducible and V_{η} = Spec(K) where K|F is a field extension. Moreover by proposition (13.8) the morphism, i.e., field extension, Spec(K) = κ(g(y)) ↪ κ(y) must be of degree one, hence an isomorphism. This proves our theorem.

In the last section we proved that almost any two Kummer elements can be connected by a chain of length 3, if necessary, after a base field extension of degree 2. For chains of length 4 this is true for any two Kummer elements and over the base field itself if A is a skew field.

We fix a field of characteristic prime to 3 with a primitive 3-rd root ζ ∈ μ_{3}(k) of unity lying in k. Let A be a central simple k-algebra of degree 3. Moreover, we assume that A is a division algebra. We choose a ζ-pair (X, Y), i.e., X, Y ∈ W(A) such that XY − ζY X = 0. From above we know

A = L ⊕ Y L ⊕ LY^{−1} ,

with L = k[X].

(14.1) Notation. For any Kummer element Z ∈ W(A) we set

e(ζ, Z) := {Z′ ∈ A : ZZ′ − ζZ′Z = 0} .

For example

e(ζ, X) = k[X]Y = Y k[X] = Y L

and

e(ζ^{2}, X) = k[X]Y^{−1} = Y^{−1}k[X] = LY^{−1} .

(14.2) Theorem. If A is a division algebra then any two Kummer elements of A can be connected by a ζ-chain in A of length 4, i.e., the map induced by (π_{0}^{4}, π_{4}^{4})

K-Chain_{4}^{ζ}(A) ⟶ W(A) × W(A)

is surjective.

For completeness we reproduce the proof with minor changes given in M. Rost [RoCL].

Additionally to the elements X and Y, we take an arbitrary Kummer element Z ∈ W(A) and we will show that there exist invertible elements X_{1}, X_{2}, X_{3} ∈ A^{∗} = A − {0} such that the following conditions hold:

(1) X_{1} ∈ e(ζ, X)

(2) X_{1}X_{2} − ζX_{2}X_{1} = 0

(3) X_{2}X_{3} − ζX_{3}X_{2} = 0

(4) X_{3} ∈ e(ζ^{2}, Z)

(5) X_{2} ∈ X^{2}k ⊕ e(ζ^{2}, X) = X^{2}k ⊕ LY^{−1}

(6) X_{3} ∈ e(ζ, X) ⊕ e(ζ^{2}, X) = Y L ⊕ LY^{−1} (14.3) Remark. The conditions (1)–(4) just mean that

X ⟶^{ζ} X_{1} ⟶^{ζ} X_{2} ⟶^{ζ} X_{3} ⟶^{ζ} Z

is a ζ-chain.

(14.4) Lemma. dim_{k} (e(ζ^{2}, Z) ∩ (e(ζ, X) ⊕ e(ζ^{2}, X))) ≥ 1.

Proof: This follows from

dim_{k} e(ζ^{2}, Z) = 3 and dim_{k}(e(ζ, X) ⊕ e(ζ^{2}, X)) = 6

and the fact that these vector spaces lie in the 8-dimensional vector subspace of the A of trace zero elements. Hence the vector spaces have non-trivial intersection.

Now let X_{3} be a non-zero element of e(ζ^{2}, Z) ∩ (e(ζ, X) ⊕ e(ζ^{2}, X)), i.e., the points (4) and (6) hold for X_{3} and it remains to find X_{1} and X_{2} such that (1)–(3) and (5) are valid. Now, X_{3} has the form

X_{3} = Y μ′ + μ′′Y^{−1}

for some μ′, μ′′ ∈ L.

In two trivial cases it is easy to find the remaining elements.

Case I : If μ′ = 0, then X_{3} = μY^{−1} for μ ∈ L^{∗}. Then the elements

X_{1} := X_{3}^{−1} = Y μ^{−1} and X_{2} := X^{2}

fulfill the remaining conditions. In fact we have: X_{1} = Y μ^{−1} ∈ Y L = e(ζ, X) hence (1), X_{1} = Y μ^{−1} ∈ Y L = e(ζ^{−1}, X^{2}) = e(ζ^{−1}, X_{2}) hence (2), X_{3} = μY^{−1} ∈ LY^{−1} = e(ζ, X^{2}) hence (3), X_{2} ∈ kX^{2} hence (5).

Case II : If μ′′ = 0, then X_{3} = Y μ for μ ∈ L^{∗}, hence X_{3} ∈ e(ζ, X). Then the elements

X_{1} := X_{3}X = Y (μX) and X_{2} := X_{3}^{2}X = X_{3}X_{1}

fulfill the remaining conditions. In fact: X_{1} ∈ Y L ∈ e(ζ, X) hence (1), X_{1}X_{2} = X_{3}XX_{3}^{2}X = ζX_{3}X_{3}XX_{3}X = ζX_{2}X_{1} hence (2), X_{2}X_{3} = X_{3}^{2}XX_{3} = ζX_{3}^{2}X_{3}X = ζX_{3}X_{2} hence (3), X_{2} = Y μY μX ∈ Y^{2}L = LY^{−1} = e(ζ^{2}, X) hence (5).

For the rest we may assume that μ′, μ′′ ∈ L^{∗}. Since the only property of Y we need is X ⟶^{ζ} Y, we may change Y to (Y μ′′). Then we have

X_{3} = Y μ + Y^{−1}

for μ = μ′μ′′ ∈ L^{∗}.

(14.5) Lemma. Let T = Y μ + Y^{−1} for any μ = m_{0} + m_{1}X + m_{2}X^{2} ∈ L for m_{i} ∈ k. Furthermore let c_{2} be the second coefficient of the reduced characteristic polynomial of T. Then c_{2} = −3m_{0}.

Proof: Trd(T) = Trd(Y μ)+Trd(Y^{−1}) = 0 and Trd(T^{2}) = 2 Trd(μ) = 6m_{0}—cf. (13.3). Since 2c_{2} = Trd(T)^{2} − Trd(T^{2}), it follows that 2c_{2} = −6m_{0}. This proves the claim for characteristic ≠ 2. For characteristic 2, consider c_{2} = −3m_{0} as a polynomial identity in the variables m_{i}. It suffices to verify this identity for a standard ζ-pair in M_{3}(ℤ[ζ]). This follows from the characteristic 0 case.

This lemma gives us for X_{3} := T

μ = m_{1}X + m_{2}X^{2} and X_{3} = Y (m_{1}X + m_{2}X^{2}) + Y^{−1}

for some m_{1}, m_{2} ∈ k.

Case III : If m_{1} = 0, then X_{3} = m_{2}Y X^{2} + Y^{−1}. Then the elements

X_{1} := Y and X_{2} := (Y X)^{−1} = X^{−1}Y^{−1}

fulfill the remaining points. In fact:

(1) is clear. X_{1}X_{2} = Y X^{−1}Y^{−1} = ζX^{−1}Y Y^{−1} = ζX_{2}X_{1} hence (2). The point (3) follows from

X_{2}X_{3} = (X^{−1}Y^{−1})(m_{2}Y X^{2} + Y^{−1})
= m_{2}X + X^{−1}Y^{−2}
= ζm_{2}Y XY^{−1} + ζY^{−1}X^{−1}Y^{−1}
= ζX_{3}X_{2} .

Point (5) is obvious since X^{−1}Y^{−1} ∈ LY^{−1} ∈ e(ζ^{2}, X). Finally we have

Case IV : If m_{1} ≠ 0, let

b := Y^{−3} ∈ k^{∗}, c := ζ^{−1}m_{1}b/ N_{L|k}(μ) ∈ k^{∗}, λ := cμX ∈ L^{∗}

and set

X_{1} := Y λ and X_{2} := X^{2}(1 + (Y λ)^{−1}) .

With these settings, (1) and (5) are obvious, and (2) is easily seen: X ⟶^{ζ} Y λ, i.e., Y λ ⟶^{ζ} X^{2} and therefore e(ζ, Y λ) = X^{2}k[(Y λ)]. It remains to verify (3):

X^{2}(1 + (Y λ)^{−1})(Y μ + Y^{−1}) = ζ(Y μ + Y^{−1})X^{2}(1 + (Y λ)^{−1}) .

This is equivalent to either of the following lines:

X^{2}(1 + (Y λ)^{−1})(Y μ + Y^{−1}) = ζX^{2}(ζY μ + ζ^{2}Y^{−1})(1 + (Y λ)^{−1}) ,

i.e.,

(1 + (Y λ)^{−1})(Y μ + Y^{−1}) = (ζ^{2}Y μ + Y^{−1})(1 + (Y λ)^{−1}) ,

§ 14. The case of algebras of degree 3 and length 4

i.e.,

Y μ + Y^{−1} + λ^{−1}μ + λ^{−1}Y^{−2} = ζ^{2}Y μ + Y^{−1} + ζ^{2}Y μλ^{−1}Y^{−1} + Y^{−1}λ^{−1}Y^{−1} .

We use the identity λ^{−1}μ = ζ^{2}Y μλ^{−1}Y^{−1}, which follows from μλ^{−1} = λ^{−1}μ = c^{−1}X^{−1}, in order to get the equivalent version

Y μ + λ^{−1}Y^{−2} = ζ^{2}Y μ + Y^{−1}λ^{−1}Y^{−1} .

Now with Y^{3} = b^{−1} and the notation \overline{λ} := Y λY^{−1} from (13.1) we reformulate our problem to proving the equation

Y μ + Y \overline{\overline{λ}}^{−1}b = ζ^{2}Y μ + Y b\overline{λ}^{−1}

or equivalently

μ + \overline{\overline{λ}}^{−1}b = ζ^{2}μ + b\overline{λ}^{−1} .

This immediately follows from (14.6) Lemma. μ \overline{λ} \overline{\overline{λ}} + \overline{λ} b = ζ^{2} μ \overline{λ} \overline{\overline{λ}} + b \overline{\overline{λ}}. Proof: We have to show

(1 − ζ^{2}) μ \overline{λ} \overline{\overline{λ}} = b (\overline{\overline{λ}} − \overline{λ}) .

Since

N_{L|k}(μ) = μ \overline{μ} \overline{\overline{μ}} and λ = c μ X

we have for the left hand side

(1 − ζ^{2})μ \overline{λ} \overline{\overline{λ}} = (1 − ζ^{2})μ c \overline{μ} \overline{X} c \overline{\overline{μ}} \overline{\overline{X}}
= (1 − ζ^{2})(μ \overline{μ} \overline{\overline{μ}} c) c X^{2}
= (1 − ζ^{2})ζ^{2} m_{1} b c X^{2} .

For the right hand side we have

b (\overline{\overline{λ}} − \overline{λ}) = b (c \overline{\overline{μ}} \overline{\overline{X}} − c \overline{μ} \overline{X})
= b c X (ζ \overline{\overline{μ}} − ζ^{2} \overline{μ})
= b c X (ζ(m_{1}ζX − m_{2}ζ^{2}X^{2}) − ζ^{2}(m_{1}ζ^{2}X + m_{2}ζX^{2}))
= b c X (ζ^{2}m_{1}X − ζm_{1}X)
= b c X (ζ^{2} − ζ)m_{1}X

and we are done.

Chapter VI

Relation to the Product Map of Tori

Out interest lies in the degree of the morphism

π_{n}^{n}: W^{n}([X]; A) ⟶ W^{0}(A)

for a central simple k-algebra of degree n and [X] ∈ W(A), or equivalently in the degree of

(π_{0}^{n}, π_{n}^{n}): W^{n}(A) ⟶ W^{0}(A) ×_{k} W^{0}(A) .

Cf. chapter V. There we mentioned that it does not depend on k and the special form of A but only on n.

Now, in the whole chapter, k is assumed to be algebraically closed (and therefore A ≅ M_{n}(k)).

We are going to construct an explicit map which has—up to the factor (n^{n−1})—the same degree as π_{n}^{n}. Later we will see that this morphism can be interpreted as a multiplication map of tori.

§ 15. A Covering of W(A)

If we projectivise the variety GL_{n} in another way as the usual, namely regarding the column vectors in the matrices, we will see that we can get an n-fold covering of the variety of Kummer lines.

We fix an algebraically closed field k and a positive integer n ≥ 2 not divisible by the characteristic of k. Let ζ ∈ μ_{n}(k) be a primitive n-th root of unity and A a central simple k-algebra of degree n. Furthermore we fix a ζ-pair (X, Y) in A, i.e., Kummer elements X, Y ∈ W(A) such that XY − ζY X = 0, and we set L := k[X], K := k[Y]. Let D := R_{L|k}(𝔾_{m}) be the torus of L-units.

(15.1) Example. Since k is algebraically closed we have essentially only the one case: A = M_{n}(k), X = X_{0}(ζ) and Y = Y_{0}. Then D is the torus of invertible diagonal matrices.

1. The Variety GL_{1}(A)/D

§ 15. A covering of W(A)

The torus D acts on GL_{1}(A) by the right multiplication. Now we are going to divide out this action (i.e., subgroup) and fix the following (15.2) Notation. The variety

GL_{1}(A)/D

is defined to be the units of A modulo the units of L. The canonical epimorphism is denoted by

\begin{matrix}\{\}: GL_{1}(A) & ⟶ GL_{1}(A)/D \\ v & ⟼ \{v\} := vD . \end{matrix}

(15.3) Remark. In our standard case of (15.1) we easily see that GL_{1}(A)/D = GL_{n,k} /D =: P^{n}GL_{n,k} is the open subvariety of (ℙ_{k}^{n−1})^{n} = ℙ_{k}^{n−1} ×_{k} ⋯ ×_{k} ℙ_{k}^{n−1}, defined by the condition

([v_{1}], …, [v_{n}]) ∈ P^{n}GL_{n,k} ⟺ (v_{1}, …, v_{n}) ∈ GL_{n,k} .

The v_{i}’s are column vectors!

Therefore we know

(15.4) Remark. GL_{1}(A)/D is a smooth k-variety of dimension n(n − 1).

There are several operations on GL_{1}(A)/D: The multiplication on GL_{1}(A) induces on GL_{1}(A)/D the left-action

\begin{matrix}GL_{1}(A) × GL_{1}(A)/D & ⟶ GL_{1}(A)/D \\ (g, \{v\}) & ⟼ g\{v\} := \{gv\} . \end{matrix}

This action is transitive. Another action is given by multiplying with Y:

(15.5) Lemma. The right multiplication with Y induces a (right-)action of ℤ/nℤ on GL_{1}(A)/D given by

\begin{matrix}GL_{1}(A)/D × (ℤ/nℤ) & ⟶ GL_{1}(A)/D \\ (\{v\}, (m mod nℤ)) & ⟼ σ^{m}\{v\} := \{vY^{m}\} . \end{matrix}

This action is fixed point free, especially it is faithful.

Proof: Since Y normalizes D, i.e., DY = Y D, the action is well defined. If {v} is a fixed point, i.e., {vY^{m}} = {v} = {v1_{A}}, then by the upper left-action we see that {Y^{m}} = {1_{A}}, i.e., Y^{m} ∈ D. But since K ∩ L we have m ≡ 0 (mod n).

(15.6) Remark. These two actions respect each other, i.e., for g ∈ GL_{1}(A) and m ∈ ℤ we have

gσ^{m}{v} = {gvY^{m}} = σ^{m}g{v} ,

(15.7) Example. In our standard situation of (15.1) one has

σ^{m}([v_{1}], …, [v_{n}]) = ([v_{1+m}], …, [v_{n+m}]) = ([v_{σ^{m}(1)}], …, [v_{σ^{m}(n)}])

for m ∈ ℤ, ([v_{1}], …, [v_{n}]) ∈ P^{n}GL_{n,k}, and the indices are viewed modulo n. Further we view σ on the right hand side as the permutation σ = (1 2 … n) ∈ S_{n}.

This follows from

(v_{1}, …, v_{n}) · Y_{0} = (v_{2}, …, v_{n}, v_{1})

in GL_{n}(k).

2. The Covering

In (6.3) we mentioned that every Kummer element (or better: Kummer line) of M_{n}(k) is conjugate to [X_{0}(ζ)], or

\begin{matrix}κ(X): GL_{1}(A) & ⟶ W(A) \\ g & ⟼ [gXg^{−1}] = [κ_{g}(X)] \end{matrix}

is a surjective morphism. In other words GL_{1}(A) acts transitively on W(A) by conjugation. Since the elements of GL_{1}(A) which commute with X are exactly the elements of D, we immediately get the

(15.8) Claim. The morphism κ(X): GL_{1}(A) → W(A) of k-varieties induces the morphism

\begin{matrix}M := M_{X}: GL_{1}(A)/D & ⟶ W(A) \\ \{v\} & ⟼ M\{v\} := [vXv^{−1}] , \end{matrix}

such that κ(X) = M ∘ {}. This morphism is surjective.

(15.9) Remark. For any v ∈ GL_{1}(A) we have Nrd(X) = Nrd(vXv^{−1}).

(15.10) Lemma. For any v, w ∈ GL_{1}(A) the following conditions are equivalent

(i) vXv^{−1} = wXw^{−1}

(ii) {v} = {w}.

In other words, the lifted morphism

\begin{matrix}M̃: GL_{1}(A)/D & ⟶ GL_{1}(A) \\ \{v\} & ⟼ vXv^{−1} \end{matrix}

is a well defined morphism and it is an injective map.

Proof: vXv^{−1} = wXw^{−1} ⟺ w^{−1}vX = Xw^{−1}v ⟺ w^{−1}v ∈ L.

On W(A) we also have an action of GL_{1}(A), namely the conjugation. The morphism M respects the action of GL_{1}(A):

§ 15. A covering of W(A)

(15.11) Lemma. For any g ∈ GL_{1}(A) the diagram

�  ^{M} �_{g·} �  _{κ_{g}} �  _{M} W(A) GL_{1}(A)/D  W(A) GL_{1}(A)/D

is commutative, i.e., M is GL_{1}(A)-equivariant. Proof: This is clear since for any v ∈ GL_{1}(A),

Mg{v} = M{gv} = [gvXv^{−1}g^{−1}] = κ_{g}[vXv^{−1}] = κ_{g}M{v} .

About the action of ℤ/nℤ on GL_{1}(A)/D we know the following (15.12) Lemma. For any v ∈ GL_{1}(A) and m ∈ ℤ,

M̃(σ^{m}{v}) = ζ^{−m}M̃{v}

hence

M(σ^{m}{v}) = M{v} .

In other words M is (ℤ/nℤ)-invariant. Proof: M̃(σ^{m}{v}) = M̃{vY^{m}} = vY^{m}XY^{−m}v^{−1} = ζ^{−m}vXv^{−1} = ζ^{−m}M̃{v}. Patching together these lemmas we get (15.13) Proposition. For {v}, {w} ∈ GL_{n}(A)/D the following conditions are equivalent

(i) M{v} = M{w} (ii) {v} = σ^{m}{w} for some m ∈ ℤ. Then (m mod nℤ) ∈ ℤ/nℤ is unique. Proof: (i) is equivalent to M̃{v} = λM̃{w} for some λ ∈ 𝔾_{m}_{,k}. With remark (15.9) we see that this is equivalent to

M̃{v} = ζ^{−m}M̃{w}

for some unique (m mod nℤ) ∈ ℤ/nℤ. By (15.12) this is equivalent to

M̃{v} = M̃σ^{m}{w}

and this is equivalent to (ii), by virtue of (15.10).

(15.14) Theorem. The action of (ℤ/nℤ) on GL_{1}(A)/D induces free actions on the fibres of the morphism

M = M_{X}: GL_{1}(A)/D ⟶ W(A) .

(15.15) Corollary. deg(M_{X}) = n.

(15.16) Remark. We get

dim W(A) = dim GL_{1}(A)/D = n^{2} − n .

Furthermore we get the morphism

(GL_{1}(A)/D)/(ℤ/nℤ) ⟶ W(A)

which is bijective. The two varieties are birational equivalent and since GL_{1}(A) acts transitively on both varieties—compatible with the morphism—, it is an isomorphism; which again shows smoothness.

(15.17) Example. In our case of (15.1), the morphism

M_{X_{0}(ζ)}: P^{n}GL_{n,k} ⟶ W(A)

can be interpreted in the following way: For ([v_{1}], …, [v_{n}]) ∈ P^{n}GL_{n,k} the element M̃([v_{1}], …, [v_{n}]) ∈ GL_{n,k} is the (unique) matrix with eigenvalues ζ^{1}, …, ζ^{n} to the eigenvectors v_{1}, …, v_{n} respectively.

§ 16. A Covering of Chain Varieties

By virtue of the covering M = M_{X}: GL_{1}(A)/D → W(A) we will also find a covering of the chain variety W^{ℓ}([X]; A) for any positive integer ℓ. In order to do that we first have to find out how the ζ-relation is reflected in GL_{1}(A)/D.

We use the same assumptions as in §15. Furthermore we assume that X^{n} = Y^{n} ∈ k^{∗}. In this case there exists by Skolem-Noether an element ϕ ∈ GL_{1}(A) = A^{∗} such that

ϕXϕ^{−1} = Y and ϕY ϕ^{−1} = X^{−1} .

Just take the automorphism of A which maps the ζ-pair (X, Y) to the ζ-pair (Y, X^{−1}). As a consequence we can write

Y = M̃{ϕ} = M̃_{X}{ϕ} .

(16.1) Example. In the situation of (15.1) take ϕ to be the matrix (ζ^{−ij})_{i,j=1,...,n} as one can see in the proof of (3.7).

1. The Kummer Relation

(16.2) Lemma. Let d_{1}, d_{2} ∈ D, then the condition ϕd_{1}ϕ^{−1} = d_{2}, i.e., ϕd_{1} = d_{2}ϕ implies the condition [d_{1}] = [d_{2}] = [1_{A}] in P(D) ⊆ P(A).

Proof: If ϕd_{1}ϕ^{−1} = d_{2} then d_{1} ∈ L and d_{1} = ϕ^{−1}d_{2}ϕ ∈ ϕ^{−1}Lϕ = K. Since L ∩ K = k, the claim follows.

(16.3) Corollary. For d_{1}, d_{2} ∈ D, the following conditions are equivalent (i) [d_{1}] = [d_{2}] in P(D) (ii) {d_{1}ϕ} = {d_{2}ϕ} (iii) {vd_{1}ϕ} = {vd_{2}ϕ} for any v ∈ GL_{1}(A).

Proof: The direction “(i)⇒(ii)” is clear; so is the equivalence of (ii) and (iii). Now assume (ii). This means there exists a d ∈ D with d_{1}ϕ = d_{2}ϕd or equivalently

d_{2}^{−1}d_{1}ϕ = ϕd

and this implies [d_{2}^{−1}d_{1}] = [1_{A}], i.e., [d_{1}] = [d_{2}].

(16.4) Remark. The corollary (16.3) just says that for any fixed v ∈ GL_{1}(A) the morphism

\begin{matrix}P(D) & ⟶ GL_{1}(A)/D \\ [d] & ⟼ \{vdϕ\} \end{matrix}

is injective.

§ 16. A covering of chain varieties

(16.5) Lemma. Let v, w ∈ GL_{1}(A). Then the following conditions are equivalent (i) (M̃{v}, M̃{w}) is a ζ-pair (ii) {w} = {vdϕ} for some d ∈ D. In this case, [d] is unique in P(D).

Proof: “(i)⇒(ii)”: By (6.14) there is a G ∈ GL_{1}(A) such that

GM̃{v}G^{−1} = X and GM̃{w}G^{−1} = Y .

Therefore

M̃{Gv} = M̃{1_{A}} and M̃{Gw} = M̃{ϕ} .

By (15.10) we get

{Gv} = {1_{A}} and {Gw} = {ϕ} ,

i.e.,

Gv = d_{1} and Gw = ϕd_{2}

for some d_{1}, d_{2} ∈ D. Combining the two equations gives

w = G^{−1}ϕd_{2} = vd_{1}^{−1}ϕd_{2} ,

hence {w} = {vdϕ} for d := d_{1}^{−1}. “(ii)⇒(i)”: Set G := vd. Then

M̃{v} = GXG^{−1} and M̃{w} = GY G^{−1} .

The first equation is clear, the second is

GY G^{−1} = GM̃{ϕ}G^{−1} = M̃{Gϕ} = M̃{vdϕ} = M̃{w} .

Uniqueness: {vd_{1}ϕ} = {vd_{2}ϕ}, then multiplying with v^{−1} we get {d_{1}ϕ} = {d_{2}ϕ} and (16.3) yields the claim.

The injective morphism

\begin{matrix}ω̂_{v}: P(D) & ⟶ GL_{1}(A)/D \\ [d] & ⟼ \{vdϕ\} \end{matrix}

parametrizes all elements lying (via M̃) over Kummer elements which are in ζ-relation with M̃{v}. There are two operations of (ℤ/nℤ) on P(D): The first one is given in the following way. Let m ∈ ℤ, then

\begin{matrix}P(D) & ⟶ P(D) \\ [d] & ⟼ [X^{−m}d] = [dX^{−m}] \end{matrix}

gives the action of (m mod nℤ) on P(D).

§ 16. A covering of chain varieties

(16.6) Lemma. For m ∈ ℤ, the following diagram is commutative.

�  ω̂_{v} � �  _{σ^{m}} �  _{ω̂_{v}} GL_{1}(A)/D bf P(D)  GL_{1}(A)/D P(D)

where the left vertical arrow is the action of (m mod nℤ) on P(D). Proof: X^{−1}ϕ = ϕY implies {vdX^{−m}ϕ} = {vdϕY^{m}}. The second action is given in the following way. Let m ∈ ℤ then

\begin{matrix}P(D) & ⟶ P(D) \\ [d] & ⟼ [Y^{−m}dY^{m}] \end{matrix}

gives the action of (m mod nℤ) on P(D). (16.7) Lemma. For m ∈ ℤ, the following diagram is commutative.

�  ω̂_{v} � �  _{id} � _{ω̂_{vY ^{m}}} GL_{1}(A)/D P(D)  GL_{1}(A)/D P(D)

where the left vertical arrow is the second action of (m mod nℤ) on P(D). Proof: {vY^{m}Y^{−m}dY^{m}ϕ} = {vdY^{m}ϕ} = {vdϕX^{m}} = {vdϕ}. (16.8) Claim. Both actions are compatible, i.e., they commute. Therefore we get an action of (ℤ/nℤ × ℤ/nℤ) on P(D) described by

(ℤ/nℤ × ℤ/nℤ) × P(D) ⟶ P(D)
((m mod nℤ, p mod nℤ), [d]) ⟼ ^{(m,p)}[d] := [Y^{−m}X^{−p}dY^{m}]

Proof:

(m,0)(0,p)[d] = [Y^{−m}X^{−p}dY^{m}]
= [X^{−p}Y^{−m}dY^{m}ζ^{−pm}]
= ^{(0,p)(m,0)}[d] .

(16.9) Example. In the standard situation of (15.1) we get for d = diag(d_{1}, …, d_{n}) (m,p)[diag(d_{1}, …, d_{n})] = [diag(ζ^{−1p}d_{1+m}, ζ^{−2p}d_{2+m}, …, ζ^{−np}d_{n+p})] .

We choose an element v ∈ GL_{1}(A), then the morphism

\begin{matrix}ω_{v}^{(1)}: P(D) & ⟶^{ω̂_{v}} GL_{1}(A)/D ⟶^{M} W(A) \\ [d] & ⟼ \{vdϕ\} ⟼ M\{vdϕ\} \end{matrix}

has as its image the elements of u(ζ, M{v}), the Kummer elements which are in a ζ-relation with M{v}. This follows from (16.5).

(16.10) Proposition. The morphism

ω_{v}^{(1)}: P(D) ⟶ u(ζ, M{v})

is surjective and the first action of (ℤ/nℤ) on P(D) induces a free action an each fibre.

Proof: The surjectivity follows from (16.5). The fact that the action induces actions on the fibres follows from (16.6) together with (15.12). Free action: Let

ω_{v}^{(1)}([d_{1}]) = ω_{v}^{(1)}([d_{2}])

for [d_{1}], [d_{2}] ∈ P(D), then

M{vd_{1}ϕ} = M{vd_{2}ϕ} .

With (15.13)

{vd_{1}ϕ} = σ^{m}{vd_{2}ϕ} = {vd_{2}ϕY^{m}} = {vd_{2}X^{−m}ϕ}

for unique (m mod nℤ) ∈ ℤ/nℤ. Hence [d_{1}] = [d_{2}X^{−m}] for the unique (m mod nℤ) ∈ ℤ/nℤ, by (16.3).

2. The Covering

We consider the following morphism of k-varieties:

(16.11) Notation. For any v ∈ GL_{1}(A) and any positive integer ℓ we define the morphism

\begin{matrix}ω̂_{v}^{(ℓ)}: P(D)^{ℓ} = P(D) × ⋯ × P(D) & ⟶ GL_{1}(A)/D \\ ([d_{1}], …, [d_{ℓ}]) & ⟼ \{vd_{1}ϕd_{2}ϕ ⋯ d_{ℓ}ϕ\} . \end{matrix}

There is an action of (ℤ/nℤ)^{ℓ} on P(D)^{ℓ} given in the following way:

\begin{matrix}(ℤ/nℤ)^{ℓ} × P(D)^{ℓ} & ⟶ P(D)^{ℓ} \\ ((ā_{1}, …, ā_{ℓ}), ([d_{1}], …, [d_{ℓ}])) & ⟼ (^{(b_{1},c_{1})}[d_{1}], …,^{(b_{ℓ},c_{ℓ})}[d_{ℓ}]) \end{matrix}

where b_{i} := a_{i−1} and c_{i} := a_{i} + a_{i−2} for i = 1, …, ℓ and a_{0} = a_{−1} = 0.

§ 16. A covering of chain varieties

(16.12) Remark. The action of e_{1} = (1, 0, …, 0) on ([d_{1}], …, [d_{ℓ}]) yields

([d_{1}X^{−1}], [Y^{−1}d_{2}Y], [X^{−1}d_{3}], [d_{4}], …) .

The action of e_{i} = (0, …, 0, 1, 0, …, 0) on ([d_{1}], …, [d_{ℓ}]) yields for i ≤ ℓ − 2

(…, [d_{i−1}], [d_{i}X^{−1}], [Y^{−1}d_{i+1}Y], [X^{−1}d_{i+2}], [d_{i+3}], …) .

The action of e_{ℓ−1} = (0, …, 0, 1, 0) on ([d_{1}], …, [d_{ℓ}]) yields

(…, [d_{ℓ−1}], [d_{ℓ−1}X^{−1}], [Y^{−1}d_{ℓ}Y]) .

The action of e_{ℓ} = (0, …, 0, 1) on ([d_{1}], …, [d_{ℓ}]) yields (…, [d_{ℓ−1}], [d_{ℓ}X^{−1}]) .

On GL_{1}(A)/D the group (ℤ/nℤ)^{ℓ} may act in the following way

\begin{matrix}(ℤ/nℤ)^{ℓ} × GL_{1}(A)/D & ⟶ GL_{1}(A)/D \\ ((ā_{1}, …, ā_{ℓ}), \{v\}) & ⟼ \{vY^{a_{ℓ}}\} = σ^{a_{ℓ}}\{v\} . \end{matrix}

(16.13) Lemma. With these actions, the morphism ω̂_{v}^{(ℓ)}: P(D)^{ℓ} ⟶ GL_{1}(A)/D is (ℤ/nℤ)^{ℓ}-equivariant.

Proof: One observes that the action of e_{i} = (0, …, 1, …, 0) commutes with ω̂_{v}^{(ℓ)}: Note that for i ≤ ℓ − 2

d_{i}X^{−1}ϕY^{−1}d_{i+1}Y ϕX^{−1}d_{i+2} = d_{i}ϕd_{i+1}ϕd_{i+2}

and

{v′d_{ℓ−1}X^{−1}ϕY^{−1}d_{ℓ}Y ϕ} = {v′d_{ℓ−1}ϕd_{ℓ}ϕX} = {v′d_{ℓ−1}ϕd_{ℓ}ϕ}

for any v′ ∈ GL_{1}(A). Further

{v′d_{ℓ}X^{−1}ϕ} = {v′d_{ℓ}ϕY } = σ{v′d_{ℓ}ϕ} .

We used X^{−1}ϕY^{−1} = ϕ and Y ϕX^{−1} = ϕ.

(16.14) Notation. Putting together ω̂_{v}^{(1)}, …, ω̂_{v}^{(ℓ)} we define

\begin{matrix}Ω̂_{v}^{(ℓ)} := (ω̂_{v}^{(1)}, …, ω̂_{v}^{(ℓ)}): P(D)^{ℓ} & ⟶ (GL_{1}(A)/D)^{ℓ} \\ ([d_{1}], …, [d_{ℓ}]) & ⟼ (\{vd_{1}ϕ\}, \{vd_{1}ϕd_{2}ϕ\}, …, \{vd_{1}ϕ … d_{ℓ}ϕ\}) \end{matrix}

The group (ℤ/nℤ)^{ℓ} may act on (GL_{1}(A)/D)^{ℓ} in the following way

\begin{matrix}(ℤ/nℤ)^{ℓ} × (GL_{1}(A)/D)^{ℓ} & ⟶ (GL_{1}(A)/D) \\ ((ā_{1}, …, ā_{ℓ}), (\{v_{1}\}, …, \{v_{ℓ}\})) & ⟼ (σ^{a_{1}}\{v_{1}\}, …, σ^{a_{ℓ}}\{v_{ℓ}\}) . \end{matrix}

(16.15) Lemma. With these actions, the morphism Ω̂_{v}^{(ℓ)}: P(D)^{ℓ} ⟶ (GL_{1}(A)/D)^{ℓ} is (ℤ/nℤ)^{ℓ}-equivariant. Proof: This is a direct consequence of (16.13).

(16.16) Proposition. The morphism Ω̂_{v}^{(ℓ)}: P(D)^{ℓ} ⟶ (GL_{1}(A)/D)^{ℓ} is injective.

Proof: We get this from (16.3) by induction:

{vd_{1}ϕ} = {vd′_{1}ϕ} ⟹ [d_{1}] = [d′_{1}]

then

{vd_{1}ϕd_{2}ϕ} = {vd′_{1}ϕd′_{2}ϕ} = {vd_{1}ϕd′_{2}ϕ} ⟹ [d_{2}] = [d′_{2}]

and so on.

Now we compose Ω̂_{v}^{(ℓ)} with (ℓ-times) the morphism M = M_{X}: GL_{1}(A)/D → W(A) in order to get

Ω_{v}^{(ℓ)} := M^{ℓ} ∘ Ω̂_{v}^{(ℓ)}: P(D)^{ℓ} ⟶ {M{v}} × W(A)^{ℓ}

(we just added a trivial first component) and by (16.5) we see that this morphism factorizes through

W^{ℓ}(M{v}; A) ⟶ {M{v}} × W(A)^{ℓ} .

(16.17) Theorem. The morphism

\begin{matrix}Ω_{v}^{(ℓ)}: P(D)^{ℓ} & ⟶ W^{ℓ}(M\{v\}; A) \\ ([d_{1}], …, [d_{ℓ}]) & ⟼ (M\{v\}, M\{vd_{1}ϕ\}, M\{vd_{1}ϕd_{2}ϕ\}, …, M\{vd_{1}ϕ ⋯ d_{ℓ}ϕ\}) \end{matrix}

is surjective and the action of (ℤ/nℤ)^{ℓ} on P(D)^{ℓ} induces free actions on the every fibre of the map.

Proof: The surjectivity is now easily shown: Let (M{v}, M{v_{1}}, …, M{v_{ℓ}}) ∈ W^{ℓ}(M{v}; A), then by (16.5) we see inductively that there exist [d_{1}], …, [d_{ℓ}] ∈ P(D) such that {v_{i}} = {v_{i−1}d_{i}ϕ} for i = 1, …, ℓ with v_{0} := v, and therefore

Ω_{v}^{(ℓ)}([d_{1}], …, [d_{ℓ}]) = (M{v}, M{v_{1}}, …, M{v_{ℓ}}) .

Now, the fibre of

M^{ℓ}: (GL_{1}(A)/D)^{ℓ} ⟶ W^{ℓ}(M{v}; A)

over the given point is by (15.13) the orbit

{(σ^{a_{1}}{v_{1}}, …, σ^{a_{ℓ}}{v_{ℓ}}) : ā_{1}, …, ā_{ℓ} ∈ ℤ/nℤ} = (ℤ/nℤ)^{ℓ} · ({v_{1}}, …, {v_{ℓ}}) .

But this set of n^{ℓ} elements is exactly what we get if we let (ℤ/nℤ)^{ℓ} act on the tuple ([d_{1}], …, [d_{ℓ}]) ∈ P(D)^{ℓ} and then map it via Ω̂_{v}^{(ℓ)} to (GL_{1}(A)/D)^{ℓ}. This proves the claim.

(16.18) Remark. As a consequence we get again statements like the irreducibility of W^{ℓ}([X]; A) and the dimension formula: dim W^{ℓ}([X]; A) = dim P(D)^{ℓ} = ℓ(n − 1).

Now we go back to our problem at the beginning, the case ℓ = n.

§ 16. A covering of chain varieties

We have the commutative diagram

�  ^{Ω_{v}^{(n)}} �_{ω̂_{v}^{(n)}} �  _{M_{X}} �  _{π_{n}^{n}} W(A) GL_{1}(A)/D  W^{n}(M{v}; A) P(D)^{n}

therefore

deg π_{n}^{n} · deg Ω_{v}^{(n)} = deg ω̂_{v}^{(n)} · deg M_{X} .

By (16.17) and (15.15) we get

deg π_{n}^{n} · n^{n} = deg ω̂_{v}^{(n)} · n

(16.19) Corollary. For any v ∈ GL_{1}(A), the morphism

\begin{matrix}ω̂_{v}^{(n)}: P(D)^{n} & ⟶ GL_{1}(A)/D \\ ([d_{1}], …, [d_{n}]) & ⟼ \{vd_{1}ϕd_{2}ϕ ⋯ d_{n}ϕ\} \end{matrix}

has degree

deg ω̂_{v}^{(n)} = n^{n−1} · deg π_{n}^{n} .

§ 17. Reformulation in the Product Map of Tori

First of all, we are going to “de-projectivise” the morphism ω̂_{v}^{(n)} for v = 1 = 1_{A} ∈ A^{∗}. Then we will reformulate the map a second time such that it can be interpreted as the product map of tori.

We keep the notations and assumptions of §16.

(17.1) Notation. We denote with

D̄ = D ×_{𝔾_{m}} ⋯ ×_{𝔾_{m}} D

the quotient of D^{n+1} by the group action

\begin{matrix}𝔾_{m}_{,k}^{n} × D^{n+1} & ⟶ D^{n+1} \\ ((t_{1}, …, t_{n}), (d_{1}, …, d_{n+1})) & ⟼ (t_{1}, …, t_{n}) · (d_{1}, …, d_{n+1}) := (d′_{1}, …, d′_{n+1}) \end{matrix}

where

\begin{aligned}d′_{1} & = d_{1}t_{1}^{−1} \\ d′_{i} & = t_{i−1}d_{i}t_{i}^{−1} for i = 2, …, n \\ d′_{n+1} & = t_{n}d_{n+1} . \end{aligned}

We define Ψ_{n} to be the morphism

\begin{matrix}Ψ_{n}: D̄ & ⟶ GL_{1}(A) \\ (d_{1}, …, d_{n+1}) & ⟼ d_{1}ϕd_{2}ϕ ⋯ d_{n}ϕd_{n+1} . \end{matrix}

(17.2) Remark. The following diagram is commutative:

�  ^{Ψ_{n}} �_{p} �  _{{}} �  _{ω̂_{1}^{(n)}} GL_{1}(A)/D P(D)^{n}  GL_{1}(A) D̄

where

\begin{matrix}p: D̄ & ⟶ P(A)^{n} \\ (d_{1}, …, d_{n+1}) & ⟼ ([d_{1}], …, [d_{n}]) . \end{matrix}

Note that

p^{−1}([d_{1}], …, [d_{n}]) = {(d_{1}, …, d_{n}, d) : d ∈ D} = {(d_{1}, …, d_{n})} × D .

D acts on D̄ and GL_{1}(A) in the following way:

\begin{matrix}D × D̄ & ⟶ D̄ \\ (d, (d_{1}, …, d_{n})) & ⟼ (d_{1}, …, d_{n}d) \end{matrix}

and

\begin{matrix}D × GL_{1}(A) & ⟶ GL_{1}(A) \\ (d, g) & ⟼ gd \end{matrix}

The following is clear:

§ 17. Reformulation in the product map of tori

(17.3) Lemma. With these actions, the morphisms p and {} are D-invariant and Ψ_{n} is D-equivariant. These actions induce free actions an all fibres of p and {}. From this lemma one instantaneously gets (17.4) Proposition. deg Ψ_{n} = deg ω̂_{1_{A}}^{(n)} = n^{n−1} · deg π_{n}^{n}. Since ϕ^{2}Xϕ^{−2} = X^{−1}, conjugation with ϕ^{2} is an automorphism of D. (17.5) Lemma. For any i = 1, …, n, the following diagram is commutative

� ^{Ψ_{n}} �_{β_{i}} �  _{ϕ^{−2}·} � _{α_{i}} GL_{1}(A) D̄  GL_{1}(A) D̄

where

\begin{matrix}β_{i}: D̄ & ⟶ D̄ \\ (d_{1}, …, d_{n+1}) & ⟼ (ϕ^{−2}d_{1}ϕ^{2}, …, ϕ^{−2}d_{i}ϕ^{2}, d_{i+1}, …, d_{n+1}) \end{matrix}

and

\begin{matrix}α_{i}: D̄ & ⟶ GL_{1}(A) \\ (d_{1}, …, d_{n}) & ⟼ d_{1}ϕd_{2} ⋯ ϕd_{i}ϕ^{−1}d_{i+1}ϕ ⋯ ϕd_{n+1} . \end{matrix}

Therefore: deg Ψ_{n} = deg α_{i}. Proof: Clear. In this way we can change ϕ to ϕ^{−1} in any place without changing the degree of the map. Let’s assume that n is an odd number. (17.6) Proposition. The morphism

\begin{matrix}Ψ′_{n}: D̄ & ⟶ GL_{1}(A) \\ (d_{1}, ⋯, d_{n+1}) & ⟼ (d_{1}(ϕd_{2}ϕ^{−1})d_{3}(ϕd_{4}ϕ^{−1}) ⋯ d_{n}(ϕd_{n+1}ϕ^{−1})) \end{matrix}

has degree: deg Ψ′_{n} = deg Ψ_{n}. Proof: One changes the ϕ to ϕ^{−1} at the places after d_{2}, d_{4}, …, d_{n−1} and multiply additionally with ϕ^{−1} from the right. Now projectivising again the morphism and writing S and T for the projective tori associated to L and K = ϕLϕ^{−1} we get

§ 17. Reformulation in the product map of tori

(17.7) Corollary. The multiplication morphism

\begin{matrix}Φ_{n}: S × T × S × T × ⋯ × S × T = (S × T)^{\frac{n+1}{2}} & ⟶ PGL_{1}(A) \\ ([d_{1}], [d_{2}], …, [d_{n+1}]) & ⟼ d_{1}d_{2} ⋯ d_{n+1} \end{matrix}

has degree

deg Φ_{n} = deg Ψ_{n} = n^{n−1} · deg π_{n}^{n} .

(17.8) Remark. If n is even, then one analogously gets: The multiplication morphism

\begin{matrix}Φ_{n}: S × T × S × T × ⋯ × S × T × S = (S × T)^{\frac{n}{2}} × S & ⟶ PGL_{1}(A) \\ ([d_{1}], [d_{2}], …, [d_{n+1}]) & ⟼ d_{1}d_{2} ⋯ d_{n+1} \end{matrix}

has degree

deg Φ_{n} = deg Ψ_{n} = n^{n−1} · deg π_{n}^{n} .

References

[BI] M.-A. Knus & A. S. Merkurjev & H. M. Rost & J.-P. Tignol. The Book of Involutions, American Mathematical Society Colloquium Publications, vol. 44, American Mathematical Society, Providence, RI, 1998

[LaQF] T. Y. Lam. The Algebraic Theory of Quadratic Forms, W. A. Benjamin, Inc., Reading, Massachusetts, 1973

[PeSR] H. P. Petersson & M. L. Racine. An Elementary Approach to the Serre-Rost Invariant of Albert Algebras, Indag. Mathem., N. S. 7 (1996), 343-365

[RoCL] M. Rost. The Chain Lemma for Kummer Elements of degree 3, C. R. Acad. Sci. Paris, t. 328, Série I, p. 185-190 (1999)

[SchQH] W. Scharlau. Quadratic and Hermitian Forms, Springer Verlag, Berlin Heidelberg New York, 1985

[SeCG] J.-P. Serre. Cohomologie Galoisienne, Lecture Notes in Mathematics, vol. 5, Springer-Verlag, Berlin Heidelberg New York, 1964

[SeLF] J.-P. Serre. Local Fields, Springer Verlag, Berlin Heidelberg New York, 1979 [ShPG] S. S. Shatz. Profinite Groups, Arithmetic and Geometry, Annals of Mathe-

matics Studies 67, Princeton University Press, Princeton, 1972

[WaAL] B. L. van der Waerden. Algebra I, Springer Verlag, Berlin Heidelberg New York, 1971