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.


1 Answer 1


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


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