## Automorphisms of central simple algebras (2)

Posted: January 26, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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Corollary 1. If $k$ is a field, then every automorphism of $M_n(k)$ is inner.

Proof. Let $f : M_n(k) \longrightarrow M_n(k)$ be an automorphism. Now let $R=S=M_n(k)$ and apply the lemma we proved in part (1)$\Box$

We now extend the lemma in part (1) to any finite dimensional central simple $k$-algebra $S.$

Lemma. Let $A$ be a finite dimensional central simple $k$-algebra and let $B$ be a simple $k$-subalgebra of $A.$ If $g: B \longrightarrow A$ is a $k$-algebra homomorphism, then there exists an invertible element $v \in A$ such that $g(b)=vbv^{-1}$ for all $b \in B.$

Proof. We proved in here that $A \otimes_k A^{op} \cong M_n(k),$ for some integer $n.$ Let $S = A \otimes_k A^{op}$ and let $R=B \otimes_k A^{op}.$ By this corollary,  $R$ is a simple $k$-subalgebra of $S.$ Now let

$f = g \otimes_k \text{Id}_{A^{op}}.$

So $f : R \longrightarrow S$ is an $k$-algebra homomorphism and thus, by the lemma in part (1), there exists an invertible element $u \in S$ such that $f(r)=uru^{-1},$ for all $r \in R.$ Let $a \in A^{op}$ and choose $r = 1 \otimes_k a.$ Then

$r = 1 \otimes_k a = g(1) \otimes_k \text{Id}_{A^{op}}(a) =f(1 \otimes_k a)=f(r)=uru^{-1}.$

So $ru=ur.$ Thus $u$ is in the centralizer of $1 \otimes_k A^{op}$ and hence $u \in A \otimes_k 1$ by the second part of this lemma.  Therefore

$u =v \otimes_k 1,$

for some $v \in A.$ Note that since $u$ is invertible, $v$ is invertible too and $u^{-1}=v^{-1} \otimes_k 1.$ Therefore for every $b \in B$ we have

$g(b) \otimes_k 1 = f(b \otimes_k 1)=u(b \otimes_k 1)u^{-1}=vbv^{-1} \otimes_k 1.$

Thus $g(b)=vbv^{-1}. \ \Box$

Corollary 2. (Skolem-Noether, 1929) Let $A$ be a finite dimensional central simple $k$-algebra. Every automorphism of $A$ is inner. Also, if $B$ is a simple $k$-subalgebra of $A$ and $f,g : B \longrightarrow A$ are $k$-algebra homomorphisms, then there exists an invertible element $v \in A$ such that $f(b)=vg(b)v^{-1}$ for all $b \in B.$

Proof. The first part is clear from the lemma. For the second part, we know from the lemma that there exist invertible elements $v_1,v_2 \in A$ such that $f(b)=v_1bv_1^{-1}$ and $g(b)=v_2bv_2^{-1}$ for all $b \in B.$ So if we let $v=v_1v_2^{-1},$ then $f(b)=vg(b)v^{-1}$ for all $b \in B. \ \Box$