# 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 centre of B. Notice that 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. Suppose further that the dimension of B over k is finite, i.e. that B is a central simple algebra of finite dimension. Then given k-algebra homomorphisms

f, g : AB,

there exists a unit b in B such that for all a in A[1][2]

g(a) = b · f(a) · b−1.

In particular, every automorphism of a central simple k-algebra is an inner automorphism.[3][4]

## Proof

First suppose ${\displaystyle B=\operatorname {M} _{n}(k)=\operatorname {End} _{k}(k^{n})}$. 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. Any 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 ${\displaystyle \operatorname {M} _{n}(k)=B}$. For the general case, note that ${\displaystyle B\otimes B^{\text{op}}}$ is a matrix algebra and thus by the first part this algebra has an element b 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 ${\displaystyle b=b'\otimes 1}$. Taking ${\displaystyle z=1}$ this time we find

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

which is what was sought.

## Notes

1. ^ Lorenz (2008) p.173
2. ^ Farb, Benson; Dennis, R. Keith (1993). Noncommutative Algebra. Springer. ISBN 9780387940571.
3. ^ Gille & Szamuely (2006) p.40
4. ^ Lorenz (2008) p.174