THE CHAIN LEMMA FOR KUMMER ELEMENTS OF DEGREE 3

MARKUS ROST

Abstract. Let A be a skew field of degree 3 over a field containing the 3^{rd} roots of unity. We prove a sort of chain equivalence for Kummer elements in A. As a consequence one obtains a common slot lemma for presentations of A as a cyclic algebra.

Chaînes d’éléments de Kummer en degré 3

Résumé. Soit k un corps contenant les racines cubiques de l’unité, et soit A un corps gauche de centre k, avec [A: k] = 9. Nous montrons que deux éléments de Kummer de A peuvent être joints par une chaîne de longueur 4.

Version française abrégée

Soit k un corps contenant une racine primitive n-ème de l’unité ζ, et soit A une k-algèbre centrale simple de degré n. Un élément de Kummer de A est un élément dont le polynôme caractéristique est de la forme t^{n} −a, avec a ∈ k^{∗}. Par une ζ-paire on entend un couple (X, Y) d’éléments de Kummer de A tels que Y X = ζXY. Une telle paire donne une présentation de A comme produit croisé cyclique :

A = ⟨ X, Y | X^{n} = a, Y^{n} = b, Y X = ζXY ⟩, avec a, b ∈ k^{∗}.

Soient X, Y deux éléments de Kummer de A, et soit m un entier ≥ 1. Une chaîne de longueur m joignant X à Y est une suite de m + 1 éléments de Kummer :

X = Z_{0}, Z_{1}, …, Z_{m} = Y

tels que (Z_{i−1}, Z_{i}) soit une ζ-paire pour i = 1, …, m.

Supposons que A soit un corps gauche. Si n = 2 (i.e. si A est un corps de quaternions), il est facile de voir que tout couple d’éléments de Kummer peut être joint par une chaîne de longueur 2. Si n = 3, J.-P. Tignol a donné des exemples de couples (X, Y) d’éléments de Kummer tels qu’il n’existe aucune chaîne de longueur 2 joignant X à Y (ni même à un conjugué de Y, cf. Appendice) ; dans ce qui suit, nous montrons qu’un tel couple peut être joint par une chaîne de longueur 4. La démonstration s’inspire de celle donnée par Petersson-Racine [1] pour un résultat analogue dans les algèbres de Jordan exceptionnelles. Comme conséquence, on obtient un “common slot lemma” pour les algèbres de degré 3.

Date: February 19, 1999.

The text appeared as: C. R. Acad. Sci. Paris Sér. I Math. 328 (1999), no. 3, 185–190.

Introduction

The well known common slot lemma for quaternion algebras asserts that if (a, b) is split over k(\sqrt{c}), then (a, b) ≃ (a, e) ≃ (c, e) for some e.

Till a few years ago not much has been known about similar statements for algebras of degree > 2. Tignol has given an example (cf. Appendix) which shows that a common slot lemma with just one additional “slot” does not hold in general for algebras of degree 3. The first positive result was obtained by Petersson and Racine [1] who proved, taking up a suggestion of J-P. Serre, a common slot lemma for exceptional Jordan algebras over quadratically closed fields.

The major purpose of this Note is to present the Petersson-Racine arguments in the much simpler case of central simple algebras of degree 3. They yield a sort of chain equivalence for Kummer elements. As a consequence one obtains a common slot lemma for such algebras.

I am indebted to Jean-Pierre Tignol for leaving his text on the counterexample as an appendix to this Note.

1. Kummer elements

Let n ≥ 2 and let k be a field containing a primitive n^{th} root of unity ζ. For a, b ∈ k^{∗} we denote by (a, b) the k-algebra defined by the presentation

⟨ X, Y | X^{n} = a, Y^{n} = b, Y X = ζXY ⟩. (∗)

Let A be a central simple algebra of degree n over k. A Kummer element in A is an element X ∈ A whose characteristic polynomial P_{X} is of the form P_{X}(t) = t^{n} −a for some a ∈ k^{∗}.

Lemma 1.1. Let X ∈ A be a Kummer element and let

E(X, ζ) = { Z ∈ A | ZX = ζXZ }.

(i) L = k[X] is the centralizer of X in A. (ii) There exists Y ∈ A^{∗} such that Y XY^{−1} = ζX. (iii) For Y as in (ii) one has E(X, ζ) = Y L = LY.

Proof. (i) follows from dim_{k} L = deg A, (ii) from the Skolem-Noether theorem, and (iii) from (i) and (ii).

By a ζ-pair we understand a pair (X, Y) of invertible elements X, Y ∈ A such that Y X = ζXY.

Lemma 1.2. Let (X, Y) be a ζ-pair.

(i) X and Y are Kummer elements. (ii) If A = M_{n}(k) and X^{n} = Y^{n} = 1, then the pair (X, Y) is conjugate to the

pair (X_{0}, Y_{0}), where X_{0} is the diagonal matrix diag(1, ζ, ζ^{2}, …, ζ^{n−1}) and where Y_{0} is the permutation matrix e_{i} ↦ e_{i−1} with i taken mod n.

(iii) The algebra A has the presentation (∗).

Proof. Since Y XY^{−1} = ζX, the n different powers of ζ are roots of P_{X}, whence P_{X}(t) = t^{n} − a for some a ∈ k. Further, X is invertible and therefore a ≠ 0. Similarly one sees P_{Y} (t) = t^{n} − b for some b ∈ k^{∗}. This proves (i). For (ii) note that any matrix X with P_{X}(t) = t^{n} − 1 is conjugate to X_{0} and we may therefore assume X = X_{0}. Then necessarily Y = UY_{0} where U is in the centralizer L = k[X] of X. One has N_{L/k}(U) = Y^{n} = 1. Therefore there exist V ∈ L^{∗} such that U = V Y_{0}V^{−1}Y_{0}^{−1}. It follows that V^{−1}Y V = Y_{0}, which proves the claim. For (iii) one may assume that k is algebraically closed and that A = M_{n}(k). The claim follows from (ii) after replacing X by X/\sqrt[n]{a} and Y by Y/\sqrt[n]{b}.

2. Chains

Let X, Y ∈ A be Kummer elements. By a chain from X to Y of length m we understand a sequence X = Z_{0}, Z_{1}, …, Z_{m} = Y of Kummer elements in A such that (Z_{i−1}, Z_{i}) is a ζ-pair for i = 1, …, m.

Let Z_{0}, …, Z_{m} be a chain of Kummer elements in A and let a_{i} = Z_{i}^{n}. Then

A ≃ (a_{i−1}, a_{i})

for i = 1, …, m. This shows that a chain of Kummer elements gives rise to a sequence of presentations (∗) with “common slots”.

If there exists a chain from X to Y of length m, then there exists also a chain from X to Y of length m′ for any m′ ≥ m (if X, Y is a chain of length 1, then X, Y X, Y is a chain of length 2).

Given Kummer elements X and Y, does there exist a chain from X to Y?

Let us consider the case n = 2. Then A is a quaternion algebra and X ∈ A is a Kummer element if and only if X is invertible and trace(X) = 0. Given Kummer elements X and Y, let Z = XY − Y X. If Z is invertible, then X, Z, Y is chain from X to Y. If Z = 0, then X and Y are scalar multiples of each other and any Kummer element Z′ anti-commuting with X gives rise to a chain X, Z′, Y. It follows that for quaternion skew fields there exist always chains from X to Y of length 2. In the case A = M_{2}(k) is not difficult to see that there exist always chains of length 3 and to give examples of Kummer elements X, Y for which there does not exist a chain of length 2.

We now assume n = 3.

Proposition 2.1. Let A a skew field of degree 3 over a field containing a primitive 3^{rd} root of unity ζ. Then for any two Kummer elements X, Y ∈ A there exists a chain of length 4 from X to Y.

As an immediate consequence of the proposition one obtains:

Corollary 2.2. Suppose that (a, b) is split over k(\sqrt[3]{c}). Then there exist e, f, g ∈ k^{∗} such that

(a, b) ≃ (a, e) ≃ (f, e) ≃ (f, g) ≃ (c, g).

Proof. Let A = (a, b). If A is split, one takes e = f = g = 1. Assume that A is a skew field and choose Kummer elements X, Y ∈ A with X^{3} = a and Y^{3} = c. By Proposition 2.1 there exists a chain X, Z_{1}, Z_{2}, Z_{3}, Y. It suffices to take e = Z_{1}^{3}, f = Z_{2}^{−3}, and g = Z_{3}^{−3}.

Tignol’s example in the appendix shows that there exist an algebra A of degree 3 and Kummer elements X, Y ∈ A for which there is no chain of length 2 from X to any conjugate of Y. The question for chains of length 3 is more delicate: it turns out that for generic X, Y there exist exactly 2 chains of length 3 which however might be defined only over a quadratic extension of the ground field. We hope to provide details for this at another occasion.

3. Proof of Proposition 2.1

Let k be a field with char k ≠ 3 containing a primitive 3^{rd} root of unity ζ. Moreover let A be a skew field of degree 3 and let X, Y ∈ A be Kummer elements. Let L = k[X] ⊂ A be the subfield generated by X. Then

A = L ⊕ E(X, ζ) ⊕ E(X, ζ^{2}). (∗∗)

We show that there exist invertible elements Z_{1}, Z_{2}, Z_{3} ∈ A such that:

(1) Z_{1} ∈ E(X, ζ),

(2) Z_{2}Z_{1} = ζZ_{1}Z_{2},

(3) Z_{3}Z_{2} = ζZ_{2}Z_{3},

(4) Z_{3} ∈ E(Y, ζ^{2}),

(5) Z_{2} ∈ X^{2}k ⊕ E(X, ζ^{2}),

(6) Z_{3} ∈ E(X, ζ) ⊕ E(X, ζ^{2}).

Conditions (1)–(4) mean that X, Z_{1}, Z_{2}, Z_{3}, Y is a chain. The additional conditions (5) and (6) are taken from [1]. Their significance lies in the fact that for generic X, Y the system of equations (1)–(6) has a solution (Z_{1}, Z_{2}, Z_{3}), Z_{i} ≠ 0 which is unique up to scalar factors of the Z_{i}. It would be interesting to understand more about the geometry of the system (1)–(6). In the following we merely present a solution.

Lemma 3.1. There exist Z_{3} ≠ 0 satisfying (4) and (6).

Proof. One has dim_{k} E(Y, ζ^{2}) = 3 and dim_{k}(E(X, ζ) ⊕ E(X, ζ^{2})) = 6. Both vector spaces lie in the 8-dimensional vector subspace of A of trace zero elements. Hence they have a nontrivial intersection.

We choose Z_{3} as in Lemma 3.1. It remains to find Z_{1}, Z_{2} ∈ A^{∗} satisfying (1), (2), (3), and (5).

Let Z ∈ E(X, ζ), Z ≠ 0. Then E(X, ζ) = ZL and E(X, ζ^{2}) = LZ^{−1}. Write

Z_{3} = Zμ′ + μ′′Z^{−1}

with μ′, μ′′ ∈ L.

If μ′ = 0, then Z_{3} ∈ E(X, ζ^{2}) and Z_{1} = Z_{3}^{−1}, Z_{2} = X^{2} do the job.

If μ′′ = 0, then Z_{3} ∈ E(X, ζ) and Z_{1} = Z_{3}X, Z_{2} = Z_{3}^{2}X do the job.

Assume that μ′ ≠ 0 and μ′′ ≠ 0. After replacing Z by Zμ′′ we have Z_{3} = Zμ + Z^{−1} for some nonzero μ ∈ L.

Lemma 3.2. Let (X, Z) be a ζ-pair, let μ = m_{0} + m_{1}X + m_{2}X^{2}, m_{i} ∈ k, and let T = Zμ + Z^{−1}. Let further c_{2} be the second coefficient of the characteristic polynomial of T. Then c_{2} = −3m_{0}.

Proof. One has trace(T) = 0 and trace(T^{2}) = 2 trace(μ) = 6m_{0}. Since 2c_{2} = trace(T)^{2} − trace(T^{2}), it follows that 2c_{2} = −6m_{0}. This proves the claim for char k ≠ 2. For char k = 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 (X, Z) in M_{3}(ℤ[ζ]). This follows from the characteristic 0 case.

For the Kummer element T = Z_{3} one has c_{2} = 0 and Lemma 3.2 shows that

μ = m_{1}X + m_{2}X^{2}, Z_{3} = Z(m_{1}X + m_{2}X^{2}) + Z^{−1}

for some m_{1}, m_{2} ∈ k.

If m_{1} = 0, then Z_{1} = Z and Z_{2} = (ZX)^{−1} do the job.

Otherwise let

b = Z^{−3}, c = ζm_{1}b/N_{L/k}(μ), λ = cμX,
Z_{1} = Zλ, Z_{2} = X^{2}(1 + (Zλ)^{−1}).

With these settings, (1), (2), and (5) are obvious. It remains to verify (3):

(Zμ + Z^{−1})X^{2}(1 + (Zλ)^{−1}) = ζX^{2}(1 + (Zλ)^{−1})(Zμ + Z^{−1}).

To check this, one considers the components with respect to the decomposition (∗∗). For the first component one gets ZμX^{2}λ^{−1}Z^{−1} = ζX^{2}λ^{−1}Z^{−1}Zμ, which follows from μX^{2}λ^{−1} = Xc^{−1} and ZX = ζXZ. For the third component one gets Z^{−1}X^{2} = ζX^{2}Z^{−1}, which is immediate from ZX = ζXZ. For the second component one gets

ZμX^{2} + Z^{−1}X^{2}λ^{−1}Z^{−1} = ζX^{2}Zμ + ζX^{2}λ^{−1}Z^{−1}Z^{−1}.

This is equivalent to both of the following equations:

X^{2}μ + Z^{−2}X^{2}λ^{−1}Z^{−1} = ζ^{2}X^{2}μ + ζ^{2}X^{2}Z^{−1}λ^{−1}Z^{−2},
(1 − ζ^{2})X^{2}μ = ζ^{2}X^{2}Z^{−1}λ^{−1}Z^{−2} − Z^{−2}X^{2}λ^{−1}Z^{−1}.

For the right hand side of the last equation one computes

r. h. s. = ζ^{2}X^{2}Z^{−1}(cμX)^{−1}Z^{−2} − ζ^{2}X^{2}Z^{−2}(cμX)^{−1}Z^{−1}
= c^{−1}XZ^{−1}μ^{−1}Z^{−2} − c^{−1}ζXZ^{−2}μ^{−1}Z^{−1}
= bc^{−1}X(Z^{−1}μZ)^{−1} − bc^{−1}ζX(Z^{−2}μZ^{2})^{−1}.

We multiply both sides with the conjugates Z^{−1}μZ and Z^{−2}μZ^{2} of μ. Then our equation reads as

(1 − ζ^{2})X^{2}N_{L/k}(μ) = bc^{−1}X(Z^{−2}μZ^{2}) − bc^{−1}ζX(Z^{−1}μZ)
= bc^{−1}X(m_{1}ζX + m_{2}ζ^{2}X^{2}) − bc^{−1}ζX(m_{1}ζ^{2}X + m_{2}ζX^{2})
= bc^{−1}m_{1}ζX^{2}(1 − ζ^{2}).

The equality is now clear.

Appendix

With the kind permission of Jean-Pierre Tignol we reproduce here his text on

A “common slot” counterexample in degree 3

Notation: For a, b nonzero elements in a field F containing a primitive cube root of unity ω, the symbol (a, b) denotes the element of the Brauer group of F represented by the F -algebra generated by elements α, β subject to

α^{3} = a,  β^{3} = b,  βα = ωαβ.

Let a_{1}, b_{1}, a_{2} ∈ F^{×}. If there exist x, y ∈ F^{×} such that

(a_{1}, b_{1}) = (a_{1}, x) + (a_{1}, y), (a_{1}, x) = −(a_{2}, x), (a_{1}, y) = (a_{2}, y), (∗)

then the additivity of symbols yields (a_{1}, b_{1}) = (a_{2}, x^{−1}y). However, the next example shows that when (a_{1}, b_{1}) is split by F (\sqrt[3]{a_{2}}), there need not exist elements x, y satisfying (∗).

Example: A global field F containing a primitive cube root of unity and elements a_{1}, b_{1}, a_{2}, b_{2} such that (a_{1}, b_{1}) = (a_{2}, b_{2}), but no couple of elements x, y satisfying (∗). In particular (taking x = 1), the field F does not contain any element y such that

(a_{1}, b_{1}) = (a_{1}, y) = (a_{2}, y) = (a_{2}, b_{2}).

Let F = 𝔽_{7}(t), where t is an indeterminate, a_{1} = t and a_{2} = t(1 − t). Note that (a_{1}, a_{2}) = 0. Therefore, for all places v of F, the local invariant (a_{1}, a_{2})_{v} is trivial. It follows that in the completion F_{v} of F at v we have either a_{1} ∈ F_{v}^{×3} or a_{1} ≡ a_{2} mod F_{v}^{×3} or a_{1} ≡ a_{2}^{2} mod F_{v}^{×3} or a_{2} ∈ F_{v}^{×3}, since the (generalized) Hilbert symbol ( , )_{v}: (F_{v}^{×}/F_{v}^{×3}) × (F_{v}^{×}/F_{v}^{×3}) → \frac{1}{3}ℤ/ℤ is a nondegenerate alternating pairing.

Consider in particular v_{1} the t-adic place and v_{2} the (t+3)-adic place. Since a_{1}, a_{2} are uniformizing parameters at v_{1}, we have a_{1}, a_{2} ∉ F_{v_{1}}^{×3}; but a_{1} ≡ a_{2} mod F_{v_{1}}^{×3}. On the other hand, a_{1} and a_{2} have non-cube residues at v_{2}, hence a_{1}, a_{2} ∉ F_{v_{2}}^{×3} but a_{1} ≡ a_{2}^{−1} mod F_{v_{2}}^{×3}.

Let now A be the central simple F -algebra with local invariants 1/3 at v_{1}, 2/3 at v_{2} and 0 everywhere else. If v is a place of F where a_{1} ∈ F_{v}^{×3}, then v ≠ v_{1}, v_{2} hence [A]_{v} = 0. It follows that A is split by F (\sqrt[3]{a_{1}}), hence we may find b_{1} ∈ F^{×} such that [A] = (a_{1}, b_{1}) in the Brauer group of F. Similarly, A is split by F (\sqrt[3]{a_{2}}) hence we may find b_{2} ∈ F^{×} such that [A] = (a_{2}, b_{2}); thus,

(a_{1}, b_{1}) = (a_{2}, b_{2}).

Suppose now x, y ∈ F^{×} satisfy (∗). Since a_{1} ≡ a_{2} mod F_{v_{1}}^{×3}, the relation (a_{1}, x)_{v_{1}} = −(a_{2}, x)_{v_{1}} implies (a_{1}, x)_{v_{1}} = 0. On the other hand, since a_{1} ≡ a_{2}^{−1} mod F_{v_{2}}^{×3}, it follows from (a_{1}, y)_{v_{2}} = (a_{2}, y)_{v_{2}} that (a_{1}, y)_{v_{2}} = 0, hence (a_{1}, x)_{v_{2}} = (a_{1}, b_{1})_{v_{2}} = 2/3.

For v ≠ v_{1}, v_{2}, we consider four cases, according to the relation between a_{1} and a_{2} in the group of cube classes:

• if a_{1} ∈ F_{v}^{×3}, then clearly (a_{1}, x)_{v} = 0.

• if a_{1} ≡ a_{2} mod F_{v}^{×3}, then (a_{1}, x)_{v} = 0 as for v = v_{1} above.

• if a_{1} ≡ a_{2}^{−1} mod F_{v}^{×3}, then (a_{1}, x)_{v} = (a_{1}, b_{1})_{v} as for v = v_{2} above, hence (a_{1}, x)_{v} = 0.

• if a_{2} ∈ F_{v}^{×3}, then (a_{1}, x)_{v} = 0 follows from (a_{1}, x) = (a_{2}, x^{−1}).

Thus, the invariants of (a_{1}, x) are:

(a_{1}, x)_{v_{2}} = 2/3, and (a_{1}, x)_{v} = 0 for v ≠ v_{2},

a contradiction to the reciprocity law.

Jean-Pierre Tignol, June 1996.

References

[1] H. P. Petersson and M. L. Racine, An elementary approach to the Serre-Rost invariant of Albert algebras, Indag. Mathem., N.S. 7 (1996), no. 3, 343–365.

NWF I - Mathematik, Universität Regensburg, D-93040 Regensburg, Germany

   E-mail address: markus.rost@mathematik.uni-regensburg.de
   URL: http://www.physik.uni-regensburg.de/~rom03516