Existence of conjugation on a quaternion algebra given a separable subalgebra

Question

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 $-u$?

Answer 1

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.