Definition 11.8.1. Let $A$ be a finite central simple $k$-algebra. We say a field extension $k'/k$ *splits* $A$, or $k'$ is a *splitting field* for $A$ if $A \otimes _ k k'$ is a matrix algebra over $k'$.

## 11.8 Splitting fields

Another way to say this is that the class of $A$ maps to zero under the map $\text{Br}(k) \to \text{Br}(k')$.

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. $\square$

Lemma 11.8.3. A maximal subfield of a finite central skew field $K$ over $k$ is a splitting field for $K$.

Lemma 11.8.4. Consider a finite central skew field $K$ over $k$. Let $d^2 = [K : k]$. For any finite splitting field $k'$ for $K$ the degree $[k' : k]$ is divisible by $d$.

**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.
$\square$

Proposition 11.8.5. Consider a finite central skew field $K$ over $k$. There exists a maximal subfield $k \subset k' \subset K$ which is separable over $k$. In particular, every Brauer class has a finite separable spitting field.

**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 $A$ 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 $A \otimes _ k \overline{k}$ is a matrix algebra, which implies there are some elements whose $q$th power is not central (e.g., $e_{11}$). This is the desired contradiction. $\square$

The results above allow us to characterize finite central simple algebras as follows.

Lemma 11.8.6. Let $k$ be a field. For a $k$-algebra $A$ the following are equivalent

$A$ is finite central simple $k$-algebra,

$A$ is a finite dimensional $k$-vector space, $k$ is the center of $A$, and $A$ has no nontrivial two-sided ideal,

there exists $d \geq 1$ such that $A \otimes _ k \bar k \cong \text{Mat}(d \times d, \bar k)$,

there exists $d \geq 1$ such that $A \otimes _ k k^{sep} \cong \text{Mat}(d \times d, k^{sep})$,

there exist $d \geq 1$ and a finite Galois extension $k'/k$ such that $A \otimes _ k k' \cong \text{Mat}(d \times d, k')$,

there exist $n \geq 1$ and a finite central skew field $K$ over $k$ such that $A \cong \text{Mat}(n \times n, K)$.

The integer $d$ is called the *degree* of $A$.

**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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #7742 by Tao on

Comment #7989 by Stacks Project on