Automorphisms of central simple algebras (1)

Posted: January 26, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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Throughout this two-part note $k$ is a field and $A$ is a finite dimensional central simple $k$-algebra. That means, as we mentioned in Remark 1 in this post, $A=M_n(D)$ for some finite dimensional central $k$-division algebra $D,$ i.e. the center of $D$ is $k$ and $[D:k] < \infty.$ We are going now to prove a result due to Skolem and Noether: every automorphism of $A$ is inner. That means if $f: A \longrightarrow A$ is a $k$-algebra isomorphism, then there exists a unit $u \in A$ such that $f(a)= uau^{-1}$ for all $a \in A.$ Note that if $R$ is a simple $k$-algebra and $S$ is any $k$-algebra, then any nonzero $k$-algebra homomorphism $f: R \longrightarrow S$ is injective. The reason is that $\ker f$ is an ideal of $R$ and since $R$ is simple, we must either have $\ker f=(0)$ or $\ker f = R.$ Since $f$ is nonzero, $\ker f \neq R$ and so $\ker f = (0),$ i.e. $f$ is injective. In particular, if $R$ is a finite dimensional simple $k$-algebra, then ${\rm{End}}(R)={\rm{Aut}}(R).$

Remark 1. Let $R \cong M_r(D),$ where $D$ is a division ring. By the Wedderburn-Artin theorem, $R$ has only one simple $R$-module, say $M,$ up to isomorphism. This $M$ is nothing but $D^r,$ the set of all $r \times 1$ vectors with entries in $D.$ It is also true that $D \cong \text{End}_R(M).$ Since $R$ is semisimple, every $R$-module is a finite direct product of $M.$ Thus if $V$ is any $R$-module, then $V \cong M^n,$ for some integer $n.$

Remark 2. Let $S = M_n(k).$ Since $M_n(k) \cong End_k(k^n),$ we may look at an element $s \in S$ as both an $n \times n$ matrix and a $k$-linear transformations $s: k^n \longrightarrow k^n.$ We will use both of them in the proof of the following lemma.

Lemma. Let $S=M_n(k)$ and suppose that $R$ is any simple $k$-subalgebra of $R.$ If $f: R \longrightarrow S$ is a $k$-algebra homomorphism,  then there exists an invertible element $u \in S$ such that $f(r)=uru^{-1},$ for all $r \in R.$

Proof. Clearly $R$ is Artinian because it is finite dimensional over $k.$ Let $M$ be the unique simple $R$-module (see Remark 1). Let $V =k^n.$ Clearly $V$ is a left $S$-module, and hence a left $R$-module, where the multiplication of an element of $S$ by an element of $V$ is just the multiplication of an $n \times n$ matrix by an $n \times 1$ vector. So by Remark 1,

$V \cong M^p, \ \ \ \ \ \ \ \ \ \ \ \ (1),$

for some integer $p.$ We can also define the multiplication of an element of $r \in R$ by an element of $v \in V$ in this way:

$r \cdot v = f(r)v. \ \ \ \ \ \ \ \ \ \ \ \ (2)$

With this definition $V$ will have a second structure as an $R$-module because $f$ is a ring homomorphism. Let’s put $V_1=(V,+, \cdot).$ So, by Remark 1,

$V_1 \cong M^q, \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$

for some integer $q.$ Note that as $k$-modules, $V$ and $V_1$ have the same structure because if $r \in k,$ then $f(r)=r$ and hence by (2), $r \cdot v = rv.$ Thus $\dim_k V = \dim_k V_1.$ But by (1), $\dim_k V = p \dim_k M$ and by (3), $\dim_k V_1 = q \dim_k M.$ Hence $p=q$ and so, as $R$-modules, $V \cong V_1.$ Let $u: V \longrightarrow V_1$ be an $R$-module isomorphism. Then obviously $u$ is also an isomorphism between $k$-vector spaces and so $u$ is an invertible element of $S = M_n(k) \cong End_k(V).$ Finally, let $r \in R$ and $v \in V.$ Let $u(v)=v_1.$ Then

$uru^{-1}(v_1)=u(rv)=r \cdot u(v)=r \cdot v_1 = f(r)(v_1).$

So $uru^{-1}=f(r). \ \Box$