# Existence of conjugation on a quaternion algebra given a separable subalgebra

In Vignéras' book Arithmétique des algèbres de quaternions, a quaternion algebra $$A$$ over a field $$K$$ is defined as a $$4$$-dim central algebra for which there exists a separable $$2$$-dim (necessarily commutative, so étale) algebra $$L/K$$ and $$u \in A^\times$$ such that conjugation by $$u$$ induces the nontrivial automorphism $$\sigma$$ of $$L$$: $$u l u^{-1} = \sigma(l) \quad \forall l \in L$$

With the data from the definition, the text works out the multiplication of two elements, and one sees from this computation that the nontrivial automorphism $$\sigma$$ of $$L$$ extends uniquely to an anti-automorphism of $$A$$ that sends $$u$$ to $$-u$$. We call it the conjugation on $$A$$.

Question. Is there a conceptual way to understand why $$\sigma$$ extends uniquely to an anti-automorphism that sends $$u$$ to $$-u$$?

## 1 Answer

Note that, by conjugating by $$u$$, it is equivalent to ask for an anti-automorphism of $$A$$ that leaves $$L$$ invariant and sends $$u$$ to $$-u$$. Writing $$\theta := u^2 \in K$$ (it lies in the center of $$A$$), this is very similar to the problem of constructing homomorphisms between algebraic field extensions:

Fact. When $$A/K$$ and $$B/K$$ are field extensions and $$u \in A$$ and $$v \in B$$ are generators with the same minimal polynomial over $$K$$, then there exist an isomorphism of $$K$$-algebras $$A \to B$$ that sends $$u$$ to $$v$$.

What we are led to here is

Fact. When $$A, B$$ are $$4$$-dim algebras over $$K$$, and $$L$$ is a $$2$$-dim separable $$K$$-algebra embedded in $$A$$ and $$B$$, and $$u \in A^\times$$ and $$v \in A^\times$$ are such that:

1. $$u$$ and $$v$$ generate $$A$$ resp. $$B$$ together with $$L$$
2. $$u^2 = v^2 = \theta \in K$$
3. conjugation by $$u$$ and $$v$$ leaves $$L$$ invariant, and induces the same $$K$$-automorphism of $$L$$

then there exists an $$L$$-linear isomorphism of $$K$$-algebras $$A \to B$$ that sends $$u$$ to $$v$$.

Proof. In the commutative case above, one uses the universal property of the polynomial ring $$K[X]$$ (free commutative algebra) to write $$A$$ and $$B$$ as quotients of $$K[X]$$.

In this associative case, we proceed similarly: Let $$\sigma$$ be that automorphism of $$L$$. Let $$K \langle L, X \rangle$$ be the free associative algebra on the elements of $$L$$ and another variable $$X$$, and let $$I$$ be the ideal generated by all algebraic relations in $$L$$, as well as $$Xl - \sigma(l) X$$ for all $$l \in L$$ and $$X^2 - \theta$$. Inspecting degrees, it is not hard to see that $$K \langle L, X \rangle / I$$ has dimension $$4$$, so that the natural maps $$K \langle L, X \rangle / I \to A, B$$ sending $$X$$ to $$u, v$$, are $$L$$-linear $$K$$-algebra isomorphisms. $$\square$$

Note how every step in the proof has its commutative counterpart.

Back to the problem, we can apply this to $$(A, u)$$ and $$(A^{\text{op}}, -u)$$ to obtain the required anti-automorphism.