The Stacks project

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. $\square$

Proof. Combine Lemma 11.7.4 with Theorem 11.8.2. $\square$

Proof. By Theorem 11.8.2 there exists a finite central simple algebra $B$ in the Brauer class of $K$ such that $[B : k] = [k' : k]^2$. By Lemma 11.5.1 we see that $B = \text{Mat}(n \times n, K)$ for some $n$. Then $[k' : k]^2 = n^2d^2$ whence the result. $\square$

Proof. Since every Brauer class is represented by a finite central skew field over $k$, we see that the second statement follows from the first by Lemma 11.8.3.

To prove the first statement, suppose that we are given a separable subfield $k' \subset K$. Then the centralizer $K'$ of $k'$ in $K$ has center $k'$, and the problem reduces to finding a maximal subfield of $K'$ separable over $k'$. Thus it suffices to prove, if $k \not= K$, that we can find an element $x \in K$, $x \not\in k$ which is separable over $k$. This statement is clear in characteristic zero. Hence we may assume that $k$ has characteristic $p > 0$. If the ground field $k$ is finite then, the result is clear as well (because extensions of finite fields are always separable). Thus we may assume that $k$ is an infinite field of positive characteristic.

To get a contradiction assume no element of $K$ is separable over $k$. By the discussion in Fields, Section 9.28 this means the minimal polynomial of any $x \in K$ is of the form $T^ q - a$ where $q$ is a power of $p$ and $a \in k$. Since it is clear that every element of $K$ has a minimal polynomial of degree $\leq \dim _ k(K)$ we conclude that there exists a fixed $p$-power $q$ such that $x^ q \in k$ for all $x \in K$.

Consider the map

and write it out in terms of a $k$-basis $\{ a_1, \ldots , a_ n\} $ of $K$ with $a_1 = 1$. So

Since multiplication on $K$ is $k$-bilinear we see that each $f_ i$ is a polynomial in $x_1, \ldots , x_ n$ (details omitted). The choice of $q$ above and the fact that $k$ is infinite shows that $f_ i$ is identically zero for $i \geq 2$. Hence we see that it remains zero on extending $k$ to its algebraic closure $\overline{k}$. But the algebra $K \otimes _ k \overline{k}$ is a matrix algebra (for example by Lemmas 11.4.9 and 11.5.3), which implies there are some elements whose $q$th power is not central (e.g., $e_{11}$). This is the desired contradiction. $\square$

Proof. The equivalence of (1) and (2) is a consequence of the definitions, see Section 11.2. Assume (1). By Proposition 11.8.5 there exists a separable splitting field $k \subset k'$ for $A$. Of course, then a Galois closure of $k'/k$ is a splitting field also. Thus we see that (1) implies (5). It is clear that (5) $\Rightarrow $ (4) $\Rightarrow $ (3). Assume (3). Then $A \otimes _ k \overline{k}$ is a finite central simple $\overline{k}$-algebra for example by Lemma 11.4.5. This trivially implies that $A$ is a finite central simple $k$-algebra. Finally, the equivalence of (1) and (6) is Wedderburn's theorem, see Theorem 11.3.3. $\square$