Skolem–Noether theorem

Proof

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

${\displaystyle (f\otimes 1)(a\otimes z)=b(g\otimes 1)(a\otimes z)b^{-1}}$

for all ${\displaystyle a\in A}$ and ${\displaystyle z\in B^{\text{op}}}$. Taking ${\displaystyle a=1}$, we find

${\displaystyle 1\otimes z=b(1\otimes z)b^{-1}}$

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

${\displaystyle f(a)=b^{\prime }g(a)b^{\prime -1}}$,

which is what was sought.