Theorem 11.7.1. Let $A$ be a finite central simple algebra over $k$, and let $B$ be a simple subalgebra of $A$. Then

the centralizer $C$ of $B$ in $A$ is simple,

$[A : k] = [B : k][C : k]$, and

the centralizer of $C$ in $A$ is $B$.

Theorem 11.7.1. Let $A$ be a finite central simple algebra over $k$, and let $B$ be a simple subalgebra of $A$. Then

the centralizer $C$ of $B$ in $A$ is simple,

$[A : k] = [B : k][C : k]$, and

the centralizer of $C$ in $A$ is $B$.

**Proof.**
Throughout this proof we use the results of Lemma 11.4.6 freely. Choose a simple $A$-module $M$. Set $L = \text{End}_ A(M)$. Then $L$ is a skew field with center $k$ which acts on the left on $M$ and $A = \text{End}_ L(M)$. Then $M$ is a right $B \otimes _ k L^{op}$-module and $C = \text{End}_{B \otimes _ k L^{op}}(M)$. Since the algebra $B \otimes _ k L^{op}$ is simple by Lemma 11.4.7 we see that $C$ is simple (by Lemma 11.4.6 again).

Write $B \otimes _ k L^{op} = \text{Mat}(m \times m, K)$ for some skew field $K$ finite over $k$. Then $C = \text{Mat}(n \times n, K^{op})$ if $M$ is isomorphic to a direct sum of $n$ copies of the simple $B \otimes _ k L^{op}$-module $K^{\oplus m}$ (the lemma again). Thus we have $\dim _ k(M) = nm [K : k]$, $[B : k] [L : k] = m^2 [K : k]$, $[C : k] = n^2 [K : k]$, and $[A : k] [L : k] = \dim _ k(M)^2$ (by the lemma again). We conclude that (2) holds.

Part (3) follows because of (2) applied to $C \subset A$ shows that $[B : k] = [C' : k]$ where $C'$ is the centralizer of $C$ in $A$ (and the obvious fact that $B \subset C')$. $\square$

Lemma 11.7.2. Let $A$ be a finite central simple algebra over $k$, and let $B$ be a simple subalgebra of $A$. If $B$ is a central $k$-algebra, then $A = B \otimes _ k C$ where $C$ is the (central simple) centralizer of $B$ in $A$.

**Proof.**
We have $\dim _ k(A) = \dim _ k(B \otimes _ k C)$ by Theorem 11.7.1. By Lemma 11.4.7 the tensor product is simple. Hence the natural map $B \otimes _ k C \to A$ is injective hence an isomorphism.
$\square$

Lemma 11.7.3. Let $A$ be a finite central simple algebra over $k$. If $K \subset A$ is a subfield, then the following are equivalent

$[A : k] = [K : k]^2$,

$K$ is its own centralizer, and

$K$ is a maximal commutative subring.

**Proof.**
Theorem 11.7.1 shows that (1) and (2) are equivalent. It is clear that (3) and (2) are equivalent.
$\square$

Lemma 11.7.4. Let $A$ be a finite central skew field over $k$. Then every maximal subfield $K \subset A$ satisfies $[A : k] = [K : k]^2$.

**Proof.**
Special case of Lemma 11.7.3.
$\square$

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 (0)