THE CHAIN LEMMA FOR BIQUATERNION ALGEBRAS

A.S. Sivatski

St.Petersburg Electrotechnical University 197376, St.Petersburg, Russia

Abstract. Any two decompositions of a biquaternion algebra over a field F into a sum of two quaternion algebras can be connected by a chain of decompositions such that any two neighboring decompositions are (a, b)+(c, d) and (ac, b)+(c, bd) for some a, b, c, d ∈ F^{∗}. A similar result is established for decompositions of a biquaternion algebra into a sum of three quaternions if F has no cubic extension.

Let A be a biquaternion algebra (i.e. a tensor product of two quaternion algebras) over a field F of characteristic different from 2. A decomposition of A into a tensor product of two quaternion algebras is not unique, and there is no canonical one. However, it turns out that any two decompositions of A can be connected by a chain of decompositions in which neighboring ones do not differ ”too much”. In fact in this note we prove an analogue of the chain lemma (see, for instance [L], where it is called ”Common Slot Theorem”) for a quaternion algebra.

So let A = D_{1} + D′_{1} = D_{2} + D′_{2} two decompositions of A into a sum of two quaternion algebras (the signs = and + will always mean equality and addition in the Brauer group of F). Dimension count shows that this means

A ≃ D_{1} ⊗_{F} D′_{1} ≃ D_{2} ⊗_{F} D′_{2}.

We call these decompositions equal if D_{1} = D_{2} and D′_{1} = D′_{2}, and simply-equivalent if there exist elements x, y, a, c ∈ F such that D_{1}_{F (\sqrt{a})} = D′_{1}_{F (\sqrt{c})} = 0 and

D_{2} = D_{1} + (a, x^{2} − acy^{2}), D′_{2} = D′_{1} + (c, x^{2} − acy^{2}). (∗)

Notice that, since (ac, x^{2} − acy^{2}) = 0, we have D_{1} + D′_{1} = D_{2} + D′_{2} as soon as the equalities (∗) hold. We say that two decompositions of A are equivalent if they can be connected by a chain of decompositions in such a way that every two neighboring decompositions in this chain are simply-equivalent. The following result justifies this definition.

Typeset by AMS-TEX

Key words and phrases. .

Proposition 1. Any two biquaternion decompositions of A are equivalent to one another, and can be connected by a chain of length 3. Moreover, this bound is strict, i.e. in general two decompositions of A cannot be connected by a chain of length 2.

Proof. Let A = (a_{1}, b_{1}) + (c_{1}, d_{1}) = (a_{2}, b_{2}) + (c_{2}, d_{2}) be two decompositions of A. Assume first that the algebras (a_{1}, b_{1}) and (a_{2}, b_{2}) have a common splitting quadratic extension. In this case we may suppose that a_{1} = a_{2}. Hence (c_{1}, d_{1}) + (c_{2}, d_{2}) = (a_{1}, b_{1}b_{2}), so (c_{1}, d_{1}) and (c_{2}, d_{2}) have a common splitting quadratic extension ([A]). Therefore, we may suppose that c_{1} = c_{2}. This implies that (a_{1}, b_{1}b_{2}) = (c_{1}, d_{1}d_{2}). Denote this algebra by Q. We have Q_{F (\sqrt{a}_{1})} = Q_{F (\sqrt{c}_{1})} = 0. It is easy to verify that Q ≃ (a_{1}, x^{2} − a_{1}c_{1}y^{2}) for some x, y ∈ F. Hence

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

and

(c_{2}, d_{2}) = (c_{1}, d_{1}) + (c_{1}, d_{1}d_{2}) = (c_{1}, d_{1}) + (a_{1}, x^{2} − a_{1}c_{1}y^{2}).

In particular, the decompositions (a_{1}, b_{1})+(c_{1}, d_{1}) and (a_{2}, b_{2})+(c_{2}, d_{2}) are simplyequivalent. This implies that in the general case it suffices to find x_{1}, y_{1}, x_{2}, y_{2} ∈ F such that the algebras (a_{1}, b_{1}(x_{1}^{2}−a_{1}c_{1}y_{1}^{2})) and (a_{2}, b_{2}(x_{2}^{2}−a_{2}c_{2}y_{2}^{2})) have a common quadratic splitting extension. This certainly will be the case if the form

⟨a_{1}, b_{1}(x_{1}^{2} − a_{1}c_{1}y_{1}^{2}), −a_{2}, −b_{2}(x_{2}^{2} − a_{2}c_{2}y_{2}^{2})⟩

is isotropic. Notice that we can modify c_{1} and c_{2} to any values of the forms ⟨c_{1}, d_{1}, −c_{1}d_{1}⟩ and ⟨c_{2}, d_{2}, −c_{2}d_{2}⟩ respectively. Thus it suffices to show that the form

⟨a_{1}, b_{1}⟩ ⊥ −a_{1}b_{1}⟨c_{1}, d_{1}, −c_{1}d_{1}⟩ ⊥ ⟨−a_{2}, −b_{2}⟩ ⊥ a_{2}b_{2}⟨c_{2}, d_{2}, −c_{2}d_{2}⟩

is isotropic. But the last form is 10-dimensional, belongs to I^{2}(F) and its Clifford invariant is equal to (a_{1}, b_{1}) + (c_{1}, d_{1}) + (a_{2}, b_{2}) + (c_{2}, d_{2}) = 0. In particular, this form belongs to I^{3}(F) ([P]). Since any 10-dimensional form from I^{3}(F) is isotropic ([P]), we are done.

An example of two decompositions which cannot be connected by a chain of length 2 is as follows. Let k be a field, a, b, c ∈ k^{∗}, ⟨⟨a, b, c⟩⟩ ≠ 0, (a, b)_{k(\sqrt{c})} ≠ 0, F = k((t)), A = (a, b) + (c, t) = (c, t) + (a, b). Suppose that these decompositions are connected by a chain of length at most 2. Then the index of (a, b) + (c, t) + (c′, x^{2}−a′c′y^{2}) is at most 2 for some x, y ∈ F, a′ ∈ D(⟨a, b, −ab⟩), c′ ∈ D(⟨c, t, −ct⟩), where, as usual, by D(φ) we denote the set of nonzero values of the quadratic form φ. Obviously, we may assume that c′ equals either c, or t, or −ct. We will consider these cases one by one.

i) Assume c′ = c. The condition (a, b)_{k(\sqrt{c})} ≠ 0 is equivalent to that the form ⟨a, b, −ab, −c⟩ is anisotropic. Suppose x, y ∈ F, and either x ≠ 0, or y ≠ 0. Then x^{2} − a′cy^{2} ∈ k^{∗}F^{∗}^{2}, hence (a, b) + (c, t) + (c, x^{2} − a′cy^{2}) = (a, b) + (c, et) for some e ∈ k^{∗}. Since (a, b)_{k(\sqrt{c})} ≠ 0, and c ∉ k^{∗}^{2} (for ⟨⟨a, b, c⟩⟩ ≠ 0), we get by Prop.2.4. in [T] that ind(a, b) ⊗ (c, et) = 4, a contradiction.

ii) Assume c′ = t. Obviously, x^{2} − a′ty^{2} ∈ F^{∗}^{2} ∪ −a′tF^{∗}^{2}, hence (a, b) + (c, t) + (t, x^{2} − a′ty^{2}) equals either (a, b) + (c, t), or (a, b) + (a′c, t). Since the index of the last algebra is 2, then again by Prop. 2.4 of [T] either a′c ∈ D(⟨a, b, −ab⟩), or a′c ∈ k^{∗}^{2}, which implies that c ∈ D(⟨⟨a, b⟩⟩), a contradiction in view of the hypothesis ⟨⟨a, b, c⟩⟩ ≠ 0.

iii) The case c′ = −ct is quite similar to case ii). The algebra (a, b) + (c, t) + (−ct, x^{2} + a′cty^{2}) equals either (a, b) + (c, t), or (a, b) + (c, t) + (−ct, a′) = (a, b) + (−c, a′) + (a′c, t). If the index of the last algebra is 2, then as in case ii) a′c ∈ D(⟨a, b, −ab⟩), or a′c ∈ k^{∗}^{2}, which is impossible. □

Remark. Notice that in the above example cd_{2} F ≥ 4. We do not know, however, whether any two decompositions can be connected by a chain of length 2 if cd_{2} F ≤ 3.

Now we introduce another equivalence between biquaternion decompositions. Namely, we say that two decompositions A = D_{1} + D′_{1} = D_{2} + D′_{2} are strongly simply-equivalent if there exist elements x, a, c ∈ F such that D_{1}_{F (\sqrt{a})} = D′_{1}_{F (\sqrt{c})} = 0 and D_{2} = D_{1}+(a, c), D′_{2} = D′_{1}+(a, c). Clearly, if the decompositions are strongly simply equivalent, they are simply equivalent (one can put x = 0, y = 1). As earlier we call two decompositions strongly equivalent if they can be connected by a chain of decompositions where any two neighboring decompositions are strongly simply-equivalent. The following statement shows that in fact there is no difference between equivalence and strong equivalence.

Proposition 2. Any two biquaternion decompositions of A are strongly equivalent to one another, and can be connected by a chain of length 6.

Proof. The proof is clear in view of the chain of length 2, connecting two simplyequivalent decompositions

(a, b)+(c, d) ∼ (ac, b)+(c, bd) = (ac, b(x^{2}−acy^{2}))+(c, bd) ∼ (a, b(x^{2}−acy^{2}))+(c, d(x^{2}−acy^{2})),

where each of the two steps is a strongly simply-equivalence. □

Recall that any biquaternion decomposition of A determines the corresponding Albert quadratic form, namely, the Albert form of the decomposition A ≃ (a, b) ⊗ (c, d) is ⟨a, b, −ab, −c, −d, cd⟩. As a consequence of Proposition 2 we obtain a strengthening of a well-known theorem of Jacobson on similarity of any two Albert forms of a biquaternion algebra ([J],[L]).

Corollary 3. Any two Albert forms of the same biquaternion algebra A are similar, and the coefficient of similarity can be chosen as a product of some u_{i} ∈ F^{∗} (1 ≤ i ≤ 6), where F (\sqrt{u}_{i}) is a quadratic subalgebra of A.

Proof. It is easy to find a similarity coefficient for two Albert forms corresponding to strongly simply-equivalent decompositions. Namely, let A = (a, b) + (c, d) = (a, bc) + (c, ad). Then

⟨a, bc, −abc, −c, −ad, acd⟩ ≃ −ac⟨a, b, −ab, −c, −d, cd⟩.

Since F (\sqrt{ac}) is a quadratic subalgebra of A, the corollary follows from Proposition

2. □

So far we have considered decompositions of biquaternion algebras into a sum of two quaternion algebras. Now we consider a similar problem, but this time decompositions into a sum of three quaternions. Let A be a central simple algebra of index 1, 2 or 4 over a field F of characteristic different from 2. Let further A = D_{1} +D′_{1} +D′′_{1} = D_{2} +D′_{2} +D′′_{2} be two decompositions of A into a sum of three quaternion algebras. We call these decompositions equal if D_{1} = D_{2}, D′_{1} = D′_{2}, and D′′_{1} = D′′_{2}, and simply-equivalent if there are 1 ≤ i < j ≤ 3 such that the sums of the ith and jth summands in both decompositions are simply equivalent in the previous sense (in particular, the remaining summands are equal). As earlier we say that these decompositions are equivalent if they can be connected by a chain of decompositions in such a way that every two neighboring decompositions in this chain are simply-equivalent. Unfortunately, we are not able to prove that any two decompositions are equivalent in full generality, restricting ourselves to the following

Proposition 4. 1) If ind A ≤ 2, then any two decompositions of A into a sum of three quaternions are equivalent.

2) If ind A = 4 and F has no cubic extension, then any two decompositions of A into a sum of three quaternions are equivalent.

Proof. 1) Let D_{1} +D′_{1} +D′′_{1} = Q, where Q is a quaternion algebra. Then D_{1} +D′_{1} = Q + D′′_{1}. By Proposition 1 we have D_{1} + D′_{1} ∼ Q + D′′_{1}. Hence

D_{1} + D′_{1} + D′′_{1} ∼ Q + D′′_{1} + D′′_{1} ∼ Q + 0 + 0,

which shows that any two decompositions of Q are equivalent.

2) Since the elements D_{1} + D′_{1} and A differ by a quaternion, there exists a field extension L/F of degree 4 such that (D_{1} + D′_{1})_{L} = A_{L} = 0 ([R]). Since F has no cubic extension, there exists an intermediate quadratic extension F (\sqrt{a}) between F and L. In particular, ind(D_{1} + D′_{1})_{F (\sqrt{a})} ≤ 2 and ind A_{F (\sqrt{a})} ≤ 2. Hence

D_{1} + D′_{1} = (a, b) + (c, d)

for some b, c, d ∈ F^{∗}. Since ((c, d) + D′′_{1})_{F (\sqrt{a})} = A_{F (\sqrt{a})} is a quaternion algebra over F (\sqrt{a}), we get (c, d) + D′′_{1} = (a, u) + (v, w) for some u, v, w ∈ F^{∗}. Therefore,

D_{1} + D′_{1} + D′′_{1} ∼ (a, b) + (c, d) + D′′_{1} ∼ (a, b) + (a, u) + (v, w) ∼ 0 + (a, bu) + (v, w).

Since all decompositions of A into a sum of two quaternions are equivalent, we are done. □

Open questions. Are lengths shorter than 6 generally impossible in Proposition 2 ? Can one drop the hypothesis on the absence of cubic extensions of F in part 2) of Proposition 4 ? What is the situation in the case ind A = 8 ?

Acknowledgements. I express my deep gratitude to Professor A.S. Merkurjev for very useful discussions and a careful check of this article.

REFERENCES

[A] Albert A.A., Tensor products of quaternion algebras, Proc.Amer.Math.Soc. 35 (1972), 65-66.

[J] Jacobson N., Some applications of Jordan norms to involutorial simple associative algebras, Adv.Math. 48 (1983), 149-165.

[L] Lam T.Y., Introduction to quadratic forms over fields, Graduate studies in Mathematics 67 (2005).

[P] Pfister A., Quadratische Formen in beliebigen Körpern, Invent. Math. 1 (1966), 116-132.

[R] Risman L.J., Zero divisors in tensor products of division algebras, Proceedings of the AMS 51 (1975), 35-36.

[T] Tignol J.P., Algebres indecomposables d’exposant premier, Advances in Math. 65 (1987), no. 3, 205-228.

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