Skolem–Noether theorem

In ring theory, a branch of mathematics, the Skolem–Noether theorem characterizes the automorphisms of simple rings. It is a fundamental result in the theory of central simple algebras.

The theorem was first published by Thoralf Skolem in 1927 in his paper Zur Theorie der assoziativen Zahlensysteme (German: On the theory of associative number systems) and later rediscovered by Emmy Noether.

Statement
In a general formulation, let A and B be simple unitary rings, and let k be the center of B. The center k is a field since given x nonzero in k, the simplicity of B implies that the nonzero two-sided ideal BxB = (x) is the whole of B, and hence that x is a unit. If the dimension of B over k is finite, i.e. if B is a central simple algebra of finite dimension, and A is also a k-algebra, then given k-algebra homomorphisms


 * f, g : A → B,

there exists a unit b in B such that for all a in A


 * g(a) = b · f(a) · b−1.

In particular, every automorphism of a central simple k-algebra is an inner automorphism.

Proof
First suppose $$B = \operatorname{M}_n(k) = \operatorname{End}_k(k^n)$$. Then f and g define the actions of A on $$k^n$$; let $$V_f, V_g$$ denote the A-modules thus obtained. Since $$f(1) = 1 \neq 0 $$ the map f is injective by simplicity of A, so A is also finite-dimensional. Hence two simple A-modules are isomorphic and $$V_f, V_g$$ are finite direct sums of simple A-modules. Since they have the same dimension, it follows that there is an isomorphism of A-modules $$b: V_g \to V_f$$. But such b must be an element of $$\operatorname{M}_n(k) = B$$. For the general case, $$B \otimes_k B^{\text{op}}$$ is a matrix algebra and that $$A \otimes_k B^{\text{op}}$$ is simple. By the first part applied to the maps $$f \otimes 1, g \otimes1 : A \otimes_k B^{\text{op}} \to B \otimes_k B^{\text{op}}$$, there exists $$b \in B \otimes_k B^{\text{op}}$$ such that
 * $$(f \otimes 1)(a \otimes z) = b (g \otimes 1)(a \otimes z) b^{-1}$$

for all $$a \in A$$ and $$z \in B^{\text{op}}$$. Taking $$a = 1$$, we find
 * $$1 \otimes z = b (1\otimes z) b^{-1}$$

for all z. That is to say, b is in $$Z_{B \otimes B^{\text{op}}}(k \otimes B^{\text{op}}) = B \otimes k$$ and so we can write $$b = b' \otimes 1$$. Taking $$z = 1$$ this time we find
 * $$f(a)= b' g(a) {b'^{-1}}$$,

which is what was sought.