Automorphisms of central simple algebras (2)

Posted: January 26, 2011 in Noncommutative Ring Theory Notes, Simple Rings
Tags: , ,

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

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