The Stacks project

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

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

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

  3. 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 $B \subset C')$. $\square$

Comments (0)

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Your email address will not be published.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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 074T. Beware of the difference between the letter 'O' and the digit '0'.