## Splitting fields of central simple algebras

Posted: October 2, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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Throughout this post, $k$ is a field and $A$ is a finite dimensional central simple $k$-algebra.

Definition. A field $L$ is called a splitting field of $A$ if $k \subseteq L$ and $A \otimes_k L \cong M_n(L),$ as $L$-algebras, for some integer $n \geq 1.$

Remark. By this lemma, the algebraic closure of $k$ is a splitting field of any finite dimensional central simple $k$-algebra. Also, if f $L$ is a splitting field of $A,$ then $\dim_k A = \dim_L A \otimes_k L = \dim_L M_n(L)=n^2$ and so $n = \deg A.$

Theorem 1. Let $A = M_n(D),$ where $D$ is some finite dimensional central division $k$-algebra and let $L$ be a field containing $k.$ Then $L$ is a splitting field of $A$ if and only if $L$ is a splitting field of $D.$

Proof. Let $\dim_k D = m^2.$ Then $\dim_k A = (mn)^2$ and so $\deg A = mn.$ If $L$ is any field containg $k,$ then

$A \otimes_k L \cong M_n(D) \otimes_k L \cong M_n(D \otimes_k L). \ \ \ \ \ \ \ \ \ (*)$

Now, suppose that $L$ is a splitting field of $A.$ Then $A \otimes_k L \cong M_{mn}(L).$ Also, since $D \otimes_k L$ is a finite dimensional central simple $L$-algebra, $D \otimes_k L \cong M_r(D')$ for some division ring $D'$ and some integer $r.$ Therefore, by $(*),$ we have $M_{mn}(L) \cong A \otimes_k L \cong M_{rn}(D')$ and thus $m=r$ and $D' \cong L.$ Hence $D \otimes_k L \cong M_m(L).$ Conversely, if $L$ is a splitting field of $D,$ then $D \otimes_k L \cong M_m(L)$ and $(*)$ gives us $A \otimes_k L \cong M_{mn}(L). \ \Box$

Corollary. There exists a finite Galois extension $L/k$ such that $L$ is a splitting field of $A.$

Proof. Trivial by Theorem 1 and the theorem in this post. $\Box$

Notation. Let $F/k$ and $E/k$ be field extensions and suppose that $\phi : F \longrightarrow E$ is a $k$-algebra homomorphism. Note that, since $F$ is a field, $\phi$ is injective. Now, given an integer $n \geq 1,$ we define the map $\phi_m: M_n(F) \longrightarrow M_n(E)$ by $\phi_m([u_{ij}]) = [\phi(u_{ij})] \in M_n(E).$ Clearly $\phi_m$ is a $k$-algebra injective homomorphisms.

Theorem 2. Let $F/k$ and $E/k$ be field extensions and suppose that $\phi : F \longrightarrow E$ is a $k$-algebra homomorphism. Suppose also that $F$ is a spiltting field of $A$ with an $F$-algebra isomorphism $f : A \otimes_k F \longrightarrow M_n(F).$ Then $E$ is a splitting field of $A$ and there exists an $E$-algebra isomorphism $g : A \otimes_k E \longrightarrow M_n(E)$ such that $g(a \otimes_k 1) = \phi_m(f(a \otimes_k 1))$ for all $a \in A.$

Proof. The map $\phi_m: M_n(F) \longrightarrow M_n(\phi(F))$ is an isomorphism and so

$A \otimes_k \phi(F) \cong A \otimes_k F \cong M_n(F) \cong M_n(\phi(F)).$

Therefore

$A \otimes_k E \cong A \otimes_k (\phi(F) \otimes_{\phi(F)} E) \cong (A \otimes_k \phi(F)) \otimes_{\phi(F)} E \cong M_n(\phi(F)) \otimes_{\phi(F)} E \cong$

$M_n(\phi(F) \otimes_{\phi(F)} E) \cong M_n(E).$

So we have an isomorphism $g: A \otimes_k E \longrightarrow M_n(E).$ It is easy to see that $g(a \otimes_k 1)=\phi_m(f(a \otimes_k 1))$ for all $a \in A. \ \Box$

For the definition of splitting fields of central simple algebras see here. Throughout, $D$ is a division algebra with the center $k$ and $\dim_k D < \infty.$ For a subalgebra $B$ of $D,$ we will denote the centralizer of $B$ in $D$ by $C_D(B).$

Theorem. Let $L$ be a subfield of $D$ and suppose that $k \subseteq L.$ Then $D \otimes_k L \cong M_n(C_D(L)),$ where $n = \dim_{C_D(L)} D = \dim_k L.$

Proof. Let $R = D \otimes_k L.$ As we saw in here, $D$ has a structure of a (right) $R$-module and

$End_R(D) \cong C_D(L). \ \ \ \ \ \ \ (1)$

Note that $R$ is a finite dimensional central simple $L$-algebra. In particular, it is primitive. We are now going to show that $D$ is a faithful simple $R$-module. First, $D$ is faithful because the annihilator of $D$ in $R$ is an ideal of $R$ and, since $R$ is a simple algebra, the annihilator is zero. Now, let $I$ be a non-zero $R$-submodule of $D.$ Let $x \in D$ and $y \in I.$ Then, since $I$ is an $R$-module, we have $yx = y(x \otimes_k 1) \in I.$ Thus $I$ is a right ideal of $D$ and so $I=D.$ Hence $D$ is a simple $R$-module. So, by $(1)$ and the structure theorem for primitive rings, we have $R \cong M_n(C_D(L)),$ where $n = \dim_{C_D(L)} D.$ To complete the proof of the theorem, we only need to show that $\dim_k L=n.$ Well, we have

$\dim_k D = \dim_L R = \dim_L M_n(C_D(L))=n^2 \dim_L C_D(L)$

and thus

$(\dim_k L)^2(\dim_k D)=n^2 (\dim_k C_D(L))(\dim_k L)=n^2 \dim_k D. \ \ \ \ \ \ \ (2)$

The last equality in $(2)$ is true by the lemma in this post. Therefore $\dim_k L = n. \ \Box$

Corollary 1. If $L$ is a maximal subfield of $D,$ then $L$ is a splitting subfield of $D.$ In fact, $D \otimes_k L \cong M_n(L),$ where $n = \deg D = \dim_L D=\dim_k L.$

Proof. If $L$ is a maximal subfield, then $k \subset L$ and $C_D(L)=L. \ \Box$

Corollary 2. If $L$ is a splitting subfield of $D,$ then $L$ is a maximal subfield of $D.$

Proof. We only need to show that $C_D(L)=L.$ By Theorem 2, $D \otimes_k L \cong M_n(C_D(L)),$ where $n = \dim_k L.$ Since $L$ is a splitting subfield of $D,$ we also have $D \otimes_k L \cong M_m(L),$ where $m = \deg D.$ Thus $M_n(C_D(L)) \cong M_m(L).$ So $m = n$ and $C_D(L) \cong L.$ Thus $\dim_k C_D(L)=\dim_k L$ and hence $C_D(L)=L$ because $L \subseteq C_D(L). \ \Box$

Corollary 3. Let $L$ be a subfield of $D.$ Then the following statements are equivalent.

1) $L$ is a maximal subfield of $D.$

2) $L$ is a splitting subfield of $D.$

3) $k \subset L$ and $\dim_k L = \deg D.$

4) $C_D(L)=L.$

Proof. Straightforward! $\Box$

## The double centralizer theorem (2)

Posted: January 31, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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As in part (1), $k$ is a field, $A$ is a finite dimensional central simple $k$-algebra and $B$ is a simple $k$-subalgebra of $A.$ We will also be using notations and the result in the lemma in part (1), i.e. $R = A \otimes_k B^{op},$ $M$ is the unique simple $R$-module,  $A \cong M^n,$ $D = \text{End}_R(M)$ and $C_A(B) \cong M_n(D).$ Finally, as usual, we will denote the center of any algebra $S$ by $Z(S).$ We now prove a nice relationship between dimensions.

Lemma. $\dim_k C_A(B) \cdot \dim_k B = \dim_k A.$

Proof. We have $R \cong M_m(D)$ and $M \cong D^m,$ for some integer $m.$ Thus $A \cong D^{mn},$ as $k$-modules, and hence

$\dim_k A = mn \dim_k D. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

We also have

$\dim_k A \cdot \dim_k B = \dim_k R = \dim_k M_m(D)=m^2 \dim_k D \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

and

$\dim_k C_A(B)=\dim_k M_n(D)=n^2 \dim_k D. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$

Eliminating $\dim_k D$ in these three identities will give us the result. $\Box$

Theorem. If $B$ is a central simple $k$-algebra, then $C_A(B)$ is a central simple $k$-algebra too and we have $A =BC_A(B) \cong B \otimes_k C_A(B).$

Proof. By the lemma in part (1), $C_A(B)$ is simple and thus, since $B$ is central simple, $B \otimes_k C_A(B)$ is a simple algebra. Thus the map $\phi : B \otimes_k C_A(B) \longrightarrow BC_A(B)$ defined by $\phi(b \otimes_k c)=bc$ is a $k$-algebra isomorphism. Hence $B \otimes_k C_A(B) \cong BC_A(B)$ and so

$\dim_k BC_A(B)=\dim_k B \otimes_k C_A(B)=\dim_k A,$

by the above lemma. Therefore $BC_A(B)=A.$ Clearly if $c$ is in the center of $C_A(B),$ then $c$ commutes with every element of both $B$ and $C_A(B).$ Hence  $c$ commutes with every element of $A$ and thus $c \in k.$ So $Z(C_A(B))=k,$ i.e. $C_A(B)$ is a central simple $k$-algebra. $\Box$

The Double Centralizer Theorem. $C_A(C_A(B))=B.$

Proof. Obviously $B \subseteq C_A(C_A(B)).$ So, to prove that $B=C_A(C_A(B)),$ we only need to show that $\dim_k B = \dim_k C_A(C_A(B)).$  Applying the above lemma to $C_A(B)$ gives us

$\dim_k C_A(C_A(B)) \cdot \dim_k C_A(B) = \dim_k A. \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$

Plugging $\dim_k C_A(B)=\frac{\dim_k A}{\dim_k B},$ which is true by the above lemma, into $(*)$ finishes the proof. $\Box$

Corollary. If $B$ is a subfield of $A,$ then $\deg A = (\deg C_A(B))(\dim_k B).$

Proof. We first need to notice a couple of things. First, since $B$ is commutative, $B \subseteq C_A(B).$ Since $B$ is commutative, $C_A(C_A(B))=Z(C_A(B)).$ Thus, by the above theorem, $Z(C_A(B))=B$ and so $C_A(B)$ is a central simple $B$-algebra. Now, by the lemma

$(\deg A)^2=\dim_k A = (\dim_B C_A(B))(\dim_k B)^2=(\deg C_A(B))^2 (\dim_k B)^2. \ \Box$