Editing Skolem–Noether theorem (section)

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