A criterion for a central simple algebra to be split

Giordano Favi, Emmanuel Lequeu

1 september 2004

Abstract. In these notes we give a criterion for a central simple algebra A to be split in terms of the essential dimension of the algebraic group SL_{1}(A). This criterion also provides an example of an algebraic group G with ed(G) = n which does not possess any non-trivial cohomological invariant.

2000 Mathematics Subject Classification: 11E72, 16K20. Keywords and Phrases: Essential dimension, central simple algebras, Galois cohomology, cohomological invariants.

1. Introduction

Let A be a central simple algebra over a field k. For every field extension K/k we will denote by A_{K} the algebra A ⊗_{k} K. We consider the algebraic group SL_{1}(A), defined over k, of elements of reduced norm equal to 1. This group is defined by the exact sequence of algebraic groups

\xymatrix{1 \ar[r] & SL_{1}(A) \ar[r] & GL_{1}(A) \ar[r]^{Nrd} & 𝔾_{m} \ar[r] & 1}  (1)

where GL_{1}(A) stands for the algebraic group defined by GL_{1}(A)(K) = A_{K}^{×} and Nrd : A_{K}^{×} → K^{×} is the reduced norm homomorphism.

Applying Galois cohomology to the exact sequence (1), one finds, for every field extension K/k, the exact sequence of pointed sets

\xymatrix{GL_{1}(A)(K) \ar[r]^{Nrd} & 𝔾_{m}(K) \ar[r]^{∂} & H^{1}(K, SL_{1}(A)) \ar[r] & H^{1}(K, GL_{1}(A))} .

Since H^{1}(K, GL_{1}(A)) = 1 for all K/k (see [4], pp. 392-393), one obtains a surjection

∂ : \xymatrix{𝔾_{m}(K) \ar@{->>}[r] & H^{1}(K, SL_{1}(A))}  (2)

and, much more precisely, an isomorphism

H^{1}(K, SL_{1}(A)) ≃ K^{×}/Nrd(A_{K}^{×})  (3)

which is functorial in K/k (see [4], Corollary 28.2, p.385 for details).

We will give a criterion for A to be split in terms of the triviality of the functor K ↦ H^{1}(K, SL_{1}(A)).

2. Essential dimension

Let k be a field. We denote by ℭ_{k} the category of field extensions of k, i.e. the category whose objects are field extensions K over k and whose morphisms are field homomorphisms which fix k. We will consider covariant functors F : ℭ_{k} → Sets from ℭ_{k} to the category of sets. For such a functor F and for a field extension K/k, we will write F(K) instead of F(K/k). We shall say that a morphism F ⟶ G between functors is a surjection if, for any field extension K/k, the corresponding map F(K) ⟶ G(K) is a surjection of sets.

Definition 2.1. Let F : ℭ_{k} → Sets be a covariant functor, K/k a field extension and a ∈ F(K). For n ∈ ℕ, we say that the essential dimension of a is ≤ n (and we write ed(a) ≤ n) if there exists a subextension E/k of K/k such that:

i) the transcendence degree of E/k is ≤ n,

ii) the element a is in the image of the map F(E) ⟶ F(K).

We say that ed(a) = n if ed(a) ≤ n and ed(a) ≰ n − 1. The essential dimension of F is the supremum of ed(a) for all a ∈ F(K) and for all K/k. The essential dimension of F will be denoted by ed_{k}(F).

Examples 2.2.

i) Consider the trivial functor ∗ : ℭ_{k} → Sets which sends each K/k to a one-element set ∗. Clearly one has ed_{k}(∗) = 0.

ii) The functor 𝔾_{m}, which assigns to each K the set K^{×}, satisfies

ed_{k}(𝔾_{m}) = 1. iii) Let \xymatrix{F \ar@{->>}[r] & G} be a surjection. Then ed_{k}(G) ≤ ed_{k}(F).

These facts are easy consequences of the definition. See [1] or [6] for example.

Definition 2.3. For an algebraic group G defined over k, the essential dimension of the Galois cohomology functor K ↦ H^{1}(K, G) is denoted by ed_{k}(G).

For an account on the notion of essential dimension of algebraic groups see for instance [1, 2, 6] or [7].

It follows from surjection (2) and from Examples 2.2 ii) and iii) above that ed_{k}(SL_{1}(A)) ≤ 1. The question is to know when this number is equal to 0 or 1. If A is split, say A ≃ M_{n}(k), then SL_{1}(A) ≃ SL_{n} and the reduced norm homomorphism is clearly surjective. Hence the functor K ↦ K^{×}/Nrd(A_{K}^{×}) is trivial and thus ed_{k}(SL_{1}(A)) = 0. We will see that the converse is true.

3. Statement of the result

Let t be an indeterminate and denote by [t] the class of t in k(t)^{×}/Nrd(A_{k(t)}^{×}).

Theorem 3.1. Let A be a central simple algebra over k. The following conditions are equivalent:

1) A is split,

2) the functor K ↦ K^{×}/Nrd(A_{K}^{×}) is trivial,
3) ed_{k}(SL_{1}(A)) = 0,
4) ed([t]) = 0.

Proof. The implications 1) ⇒ 2) ⇒ 3) ⇒ 4) are trivial and follow easily from the definitions. It remains to show that 4) ⇒ 1). It will be a consequence of the following fact. □

Proposition 3.2. Let A be a central simple algebra over k. Denote by ind(A) its index and consider the group homomorphism

θ_{A} : \xymatrix{(A ⊗_{k} k(t))^{×} \ar[r]^{Nrd} & k(t)^{×} \ar[r]^{deg} & ℤ}

where deg stands for the degree in t. Then im(θ_{A}) = ind(A) ℤ.

To prove this fact, we use the following lemma.

Lemma 3.3. Let K be a commutative field and A and B two simple central algebras over K. If A and B are Brauer equivalent, then Nrd(A^{×}) = Nrd(B^{×}).

Proof. Let D be a central division algebra over K such that A ≃ M_{r}(D) and B ≃ M_{s}(D). We show that Nrd(A^{×}) = Nrd(D^{×}). We have that A^{×} ≃ GL_{r}(D). Now for invertible matrices over a division algebra there is the notion of Dieudonné determinant det_{D} : GL_{r}(D) → D^{×}/[D^{×}, D^{×}] (see [3], Definition 3, p. 135). One has the following commutative diagram

\xymatrix{GL_{r}(D) \ar[d]_{_{det_{D}}} \ar[r]^{Nrd} & K^{×} \\ D^{×}/[D^{×}, D^{×}] \ar[ur]_{Nrd}}

which shows that Nrd(A^{×}) = Nrd(D^{×}) since det_{D} is surjective (see [3], Theorem 1, p. 146). Similarly Nrd(B^{×}) = Nrd(D^{×}) which finishes the proof. □

Remark 3.4. It follows from the above lemma and the functor isomorphism (3) that, if A and B are two simple central algebras over k which are Brauer equivalent, then the functors K → H^{1}(K, SL_{1}(A)) and K → H^{1}(K, SL_{1}(B)) are isomorphic.

Proof of Proposition 3.2. By Lemma 3.3, we can suppose that A is a division algebra. In this case, the index of A and the degree of A are equal.

Consider now the element 1⊗t ∈ (A⊗_{k}k(t))^{×}. We have that Nrd(1⊗t) = t^{ind(A)} and thus ind(A) ∈ im(θ_{A}). It follows that ind(A) ℤ ⊆ im(θ_{A}). To show the converse inclusion take a ∈ (A ⊗_{k} k(t))^{×} and write it a = \frac{b}{c} where c ∈ k[t] and b ∈ A⊗_{k} k[t]. Since c ∈ k[t] we have that Nrd(c) = c^{ind(A)} and thus it suffices to show that ind(A) divides the degree of Nrd(b). We then write

b = b_{0} ⊗ 1 + b_{1} ⊗ t + ⋯ + b_{n} ⊗ t^{n}

where b_{i} ∈ A and b_{n} ≠ 0. An easy computation shows that

Nrd(b) = Nrd(b_{n}) t^{n·ind(A)} + ⋯

Since A is a division algebra Nrd(b_{n}) ≠ 0 whence the result. □

Proposition 3.2 enables us to establish the implication 4) ⇒ 1) of Theorem 3.1.

Proof of Theorem 3.1. Suppose that ed([t]) = 0. This means that there exist an element λ ∈ k^{×} and an element x ∈ (A ⊗_{k} k(t))^{×} such that t = λ Nrd(x). But this implies that 1 ∈ im(θ_{A}). By the preceeding Proposition, this means that ind(A) = 1, saying that A is split. □

Actually Theorem 3.1 can be sharpened in order to give a more precise result:

Corollary 3.5. Let A be a central simple algebra over k. Let r ∈ ℕ, then

i) one has ed([t^{r}]) = 0 if and only if ind(A) divides r;

ii) one has ind(A) = r if and only if ed([t^{r}]) = 0 and ed([t^{m}]) ≠ 0 for all m ≠ r with m | r.

Proof. By Lemma 3.3 one can suppose A to be a division algebra.

i) Suppose that ind(A) divides r, say r = ind(A)m. One has t^{r} = Nrd(1 ⊗ t^{m}) saying that ed([t^{r}]) = 0. Conversely, if ed([t^{r}]) = 0 then there exists u ∈ k and x ∈ (A ⊗_{k} k(t))^{×} such that t^{r} = uNrd(x). This shows that r ∈ im(θ_{A}). By Proposition 3.2, it follows that ind(A) divides r. ii) It follows easily from i). □

4. A slight generalization

For two algebraic groups G and H very little is known about the behaviour of ed(G × H) (see [1, 2] or [7] for partial results). Even when G = H the computation of the essential dimension is not well understood. The preceeding discussion and the interpretation of the essential dimension of SL_{1}(A) in terms of the algebra A allows to give a precise description in this case. Our aim is to prove the following result for a product of copies of SL_{1}(A).

Theorem 4.1. Let A be a central simple algebra defined over a field k and let SL_{1}(A) × ⋯ × SL_{1}(A) the product of n copies of SL_{1}(A). Then

ed_{k}(SL_{1}(A) × ⋯ × SL_{1}(A)) = n ed_{k}(SL_{1}(A)).

Proof. It is enough to prove that if A is a non-split algebra then we have ed_{k}(SL_{1}(A)×⋯×SL_{1}(A)) = n. Notice that, since there is no non-split central simple algebra over a finite field, we may suppose that the ground field k is infinite. For simplicity we will denote the functor H^{1}(_{−}, SL_{1}(A)×⋯×SL_{1}(A)), with n copies of SL_{1}(A), by F_{n}.

Let t_{1}, …, t_{n} be algebraically independent variables over k. In the sequel we will denote by K_{n} the field k(t_{1}, …, t_{n}). Consider the element a = (t̄_{1}, …, t̄_{n}) ∈ F_{n}(K_{n}) where t̄_{i} denotes the class of t_{i} in the quotient K_{n}^{×}/Nrd(A_{K_{n}}^{×}). We will show, by induction on n, that ed(a) = n.

Suppose that n ≥ 2 and that there is a subextension L/k ⊂ K_{n}/k and an element b ∈ F_{n}(L) such that b maps to a under the map

F_{n}(L) → F_{n}(k(t_{1}, …, t_{n})).

Suppose moreover that trdeg(L : k) < n. We write b = (b̄_{1}, …, b̄_{n}) for some elements b_{i} ∈ L^{×}. Saying that b maps to a means that there exist elements x_{i} ∈ (A ⊗_{k} K_{n})^{×} such that

t_{1} = b_{1} Nrd(x_{1})
t_{2} = b_{2} Nrd(x_{2})
⋮
t_{n} = b_{n} Nrd(x_{n})

Recall that for a field extension K/k a k-valuation on K is a valuation v on K which is trivial on k, that is v(x) = 0 for all x ∈ k. We denote by O_{v} = {x ∈ K | v(x) ≥ 0} the ring of the valuation, by 𝔪_{v} = {x ∈ K | v(x) > 0} its unique maximal ideal and by κ_{v} the residue field, that is the quotient O_{v}/𝔪_{v}. Since v is trivial on k it follows that O_{v} is a k-algebra and that κ_{v} is again a field extension of k. It is well known that

trdeg(κ_{v} : k) ≤ trdeg(K : k)

and that the equality holds if and only if v is trivial on K.

We consider now on K_{n} the (t_{n} − λ)-valuation where λ ∈ k. We denote it by v and its restriction to L by v′. Since the field k is infinite, we can find a λ such that the b_{i}’s, the x_{i}’s and their inverses y_{i} are unramified at v. That means that b_{i} ∈ O_{v} and that x_{i}, y_{i} ∈ A ⊗_{k} O_{v}. Thus we may specialise the above equations at t_{n} = λ and find

t_{1} = b_{1}(λ) Nrd(x_{1}(λ))
t_{2} = b_{2}(λ) Nrd(x_{2}(λ))
⋮
λ = b_{n}(λ) Nrd(x_{n}(λ))

where b_{i}(λ) denotes the image of b_{i} in the residue field κ_{v′} and x_{i}(λ) denote the image of x_{i} in (A ⊗_{k} κ_{v})^{×} = (A ⊗_{k} K_{n−1})^{×}. Hence b′ = (b̄_{1}, …, b̄_{n−1}) ∈ F_{n−1}(κ_{v′}) maps to (t̄_{1}, …, t̄_{n−1}) ∈ F_{n−1}(K_{n−1}) under the induced map F_{n−1}(κ_{v′}) → F_{n−1}(κ_{v}). Since by induction hypothesis ed((t̄_{1}, …, t̄_{n−1})) = n − 1 and since trdeg(L : k) < n it follows that the valuation v′ has to be trivial over L. Consequently b_{n}(λ) = b_{n} ∈ L and the equation λ = b_{n}(λ) Nrd(x_{n}(λ)) actually shows that b_{n} ∈ K_{n−1}. Coming back to the equation t_{n} = b_{n} Nrd(x_{n}) this contradicts Theorem 3.1 for the central simple algebra A′ = A ⊗_{k} K_{n−1}. □

5. A side remark

Let F : ℭ_{k} → Sets be a covariant functor and let M be any torsion Γ_{k}- module. A degree d cohomological invariant for F with values in M is, by definition, a morphism of functors η : F → H^{d}(_{−}, M). An invariant η is said to be non-trivial if, for every field extension K/k, there exists a field extension L/K and an element a ∈ F(L) such that η_{L}(a) ≠ 0 ∈ H^{d}(L, M). It is shown in [1, 6, 7] that if η is a non-trivial degree d cohomological invariant then ed_{k}(F) ≥ d. Now, since the algebra A splits over some field extension K/k, it follows that for any L/K the cohomology set H^{1}(L, SL_{1}(A)) is reduced to one element. Thus one sees that H^{1}(_{−}, SL_{1}(A)) cannot have any non-trivial cohomological invariant of degree ≥ 1. Hence, when A is non-split, SL_{1}(A)×⋯×SL_{1}(A) gives an easy example of a group without any non-trivial cohomological invariant which has essential dimension n.

6. Description of H^{1}(k(t), SL_{1}(A))

For completeness we give a detailed study of H^{1}(k(t), SL_{1}(A)) inspired by [8]. Let X denote the set of irreducible monic polynomials in k[t].

Proposition 6.1. Let A be a central simple algebra over k and x ∈ X. Consider the group homomorphism

η_{x} : \xymatrix{(A ⊗_{k} k(t))^{×} \ar[r]^{Nrd} & k(t)^{×} \ar[r]^{v_{x}} & ℤ}

where v_{x} denotes the x-adic valuation on k(t). Then im(η_{x}) = ind(A)ℤ.

Proof. As in Lemma 3.3 we can suppose that A is a division algebra. One has Nrd(1 ⊗ x) = x^{ind(A)}, thus ind(A)ℤ ⊆ im(η_{x}). Conversely, let a ∈ (A ⊗_{k} k(t))^{×} and write it a = \frac{b}{c} where c ∈ k[t] and b ∈ A ⊗_{k} k[t]. Since Nrd(c) = c^{ind(A)} it suffices to prove that ind(A) | v_{x}(Nrd(b)). To do this, write

b = ∑_{r≤i≤s} b_{i} ⊗ x^{i}f_{i}

where deg(f_{i}) < deg(x) and b_{r} ≠ 0. An easy computation shows then that

Nrd(b) = Nrd(b_{r})x^{r·ind(A)}f_{r}^{ind(A)} + x^{r·ind(A)+1}g

where g ∈ k[t]. Since Nrd(b_{r}) ≠ 0 this shows that v_{x}(Nrd(b)) = r · ind(A). Hence im(η_{x}) ⊆ ind(A)ℤ. This concludes the proof. □

Definition 6.2. For each x ∈ X, let

∂_{x} : H^{1}(k(t), SL_{1}(A)) → ℤ/ind(A)ℤ

be the group homomorphism induced by the valuation v_{x}.

Theorem 6.3. Let A be a central simple algebra over k. Then there is a split exact sequence

1 → H^{1}(k, SL_{1}(A)) → H^{1}(k(t), SL_{1}(A)) → ⨁_{x∈X} ℤ/ind(A)ℤ → 0

where the first map is induced by k → k(t) and the second is ⊕∂_{x}.

Proof. Though it is well known that the natural map

ι : H^{1}(k, G) → H^{1}(k(t), G)

is injective (see [5]), we show injectivity in this particular case. As above we may suppose that A is a division algebra. So let λ ∈ k^{×} and x ∈ (A ⊗ k(t))^{×} such that λ = Nrd(x). Write x as x = (1/p(t))y, where p(t) is a unitary polynomial and y ∈ A ⊗ k[t]. One has

Nrd(x) = (1/p(t)^{deg(A⊗k(t))})Nrd(y),

and hence λp(t)^{ind(A)} = Nrd(y). Write y = a_{r} ⊗ t^{r} + ⋯ + a_{0} ⊗ 1, where a_{i} ∈ A and a_{r} ≠ 0. Then Nrd(y) is a polynomial of degree ind(A)r and its leading coefficient is Nrd(a_{r}). Comparing the leading terms in the equality λp(t)^{ind(A)} = Nrd(y), one gets λ = Nrd(a_{r}). This proves the injectivity.

The composition

k^{×}/Nrd(A) → k(t)^{×}/Nrd(A_{k(t)}^{×}) → ⨁_{x∈X} ℤ/ind(A)ℤ

is clearly zero. We show that ker(⊕δ_{x}) ⊆ im(ι). Let f ∈ k(t)^{×} and write it f = λ ∏_{x∈X} x^{v_{x}(f)} with λ ∈ k^{×}. Since ∂_{x}(f) = 0 one has that ind(A) divides

v_{x}(f) for all x ∈ X. Hence ∏_{x∈X} x^{v_{x}(f)} ∈ Nrd(A_{k(t)}^{×}). Thus f ∈ im(ι).

To end the proof, we give a section to ⊕∂_{x}. This will show surjectivity as well. We let s : ⨁_{x∈X} ℤ/ind(A)ℤ ⟶ H^{1}(k(t), SL_{1}(A)) defined by sending the element

([n_{x}])_{x∈X} to the class of ∏_{x∈X} x^{n_{x}} in k(t)^{×}/Nrd(A_{k(t)}^{×}). This clearly gives the

desired section. □

Corollary 6.4. The following conditions are equivalent:

i) the algebra A is split, ii) the group H^{1}(k(t), SL_{1}(A)) is trivial, iii) the group H^{1}(k(t), SL_{1}(A)) is finite.

Acknowledgements

The authors would warmly like to thank Grégory Berhuy and Mohammad Mahmoudi for helpful remarks.

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Emmanuel Lequeu École Polytéchnique de Lausanne CH-1015 Lausanne, Switzerland

emmanuel.lequeu@epfl.ch

Giordano Favi University of Basel CH-4051 Basel, Switzerland

                               giordano.favi@unibas.ch