Automorphisms of central simple algebras (1)

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

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


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s