The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

11.7 The centralizer theorem

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

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

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

  2. $K$ is its own centralizer, and

  3. $K$ 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. $\square$

slogan

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. $\square$


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