# Quaternion algebras Brauer equivalent iff isomorphic

Let $$k$$ be a field, and for any $$a,b\in k^\times$$ let $$(a,b)$$ be the quaternion algebra over $$k$$ with parameters $$a,b$$. Is the following correct? I haven't been able to find a place where this is spelled out, so I have doubts about its validity.

For any $$a,b,c,d\in k^\times$$, the quaternion algebras $$(a,b),(c,d)$$ are Brauer equivalent (i.e. $$(a,b)\otimes M_n(k)\simeq(c,d)\otimes M_n(k)$$ for some $$n$$) if and only if $$(a,b)\simeq(c,d)$$.

Proof. Each Brauer class contains only one division algebra. This is because by Wedderburn's theorem, any finite-dimensional simple algebra over a field $$k$$ is isomorphic to $$M_n(D)$$ for some division algebra $$D$$, with $$n$$ unique and $$D$$ unique up to isomorphism. Moreover, for any division algebra $$D$$ over $$k$$, $$D\otimes M_n(k)\simeq M_n(D)$$.

We conclude by using the fact that each quaternion algebra over $$k$$ is either split (isomorphic to $$M_2(k)$$) or a division algebra.

In fact the following is true if general:

Claim. Two central simple $$k$$-algebras of same degree are Brauer equivalent if and only if they are isomorphic.

Note that Brauer equivalence for two arbitrary central simple algebras (not necessarily of same degree) is

$$A,B$$ are Brauer equivalent if there exists $$r,s$$ such that $$M_r(k)\otimes_k A\simeq M_s(k)\otimes_k B$$.

Note that the isomorphism $$M_r(k)\otimes A\simeq M_r(A)$$ is true for any $$k$$-algebra.

Now it is useful to rephrase Brauer equivalence in terms of division algebras: $$A$$ and $$B$$ are Brauer equivalent if and only if they correspond to the same division algebra (up to isomorphism) via Wedderburn's theorem.

In other words: write $$A\simeq M_p(D), B\simeq M_q(D')$$, where $$D,D'$$ are central division algebras. Then $$A$$ and $$B$$ are Brauer equivalent if and only if $$D\simeq D'$$.

Proof. If $$D\simeq D'$$, then $$M_q(k)\otimes A\simeq M_q(A)\simeq M_q(M_p(D))\simeq M_{qp}(D)\simeq M_{qp}(D')\simeq M_p(M_q(D'))\simeq M_p(B)\simeq M_p(k)\otimes B$$.

Conversely, if $$M_r(k)\otimes_k A\simeq M_s(k)\otimes_k B$$, then similar computations show that $$M_{qr}(D)\simeq M_{ps}(D')$$. By Wedderburn, $$D\simeq D'$$.

The proof of the claim is then clear: if $$A,B$$ are Brauer equivalent of same degree, then $$A\simeq M_p(D), B\simeq M_q(D')$$. Since they are Brauer equivalent, $$D\simeq D'$$. In particular, $$\deg(D)=\deg(D')$$. Since $$\deg(A)=\deg(B),$$ we get $$p\deg(D)=q\deg(D')=q\deg(D)$$, hence $$p=q$$, ans thus $$A\simeq M_p(D)\simeq B$$.

Remark. I think you can safely drop the assumption of being central everywhere