Theorem 11.8.2. Let $A$ be a finite central simple $k$-algebra. Let $k'/k$ be a finite field extension. The following are equivalent
$k'$ splits $A$, and
there exists a finite central simple algebra $B$ similar to $A$ such that $k' \subset B$ and $[B : k] = [k' : k]^2$.
Proof. Assume (2). It suffices to show that $B \otimes _ k k'$ is a matrix algebra. We know that $B \otimes _ k B^{op} \cong \text{End}_ k(B)$. Since $k'$ is the centralizer of $k'$ in $B^{op}$ by Lemma 11.7.3 we see that $B \otimes _ k k'$ is the centralizer of $k \otimes k'$ in $B \otimes _ k B^{op} = \text{End}_ k(B)$. Of course this centralizer is just $\text{End}_{k'}(B)$ where we view $B$ as a $k'$ vector space via the embedding $k' \to B$. Thus the result.
Assume (1). This means that we have an isomorphism $A \otimes _ k k' \cong \text{End}_{k'}(V)$ for some $k'$-vector space $V$. Let $B$ be the commutant of $A$ in $\text{End}_ k(V)$. Note that $k'$ sits in $B$. By Lemma 11.7.2 the classes of $A$ and $B$ add up to zero in $\text{Br}(k)$. From the dimension formula in Theorem 11.7.1 we see that
Hence $[B : k] = [k' : k]^2$. Thus we have proved the result for the opposite to the Brauer class of $A$. However, $k'$ splits the Brauer class of $A$ if and only if it splits the Brauer class of the opposite algebra, so we win anyway. $\square$
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