Template:Short description 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.

StatementEdit

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 Template:Nowrap 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 : AB,

there exists a unit b in B such that for all a in A<ref>Lorenz (2008) p.173</ref><ref>Template:Cite book</ref>

g(a) = b · f(a) · b−1.

In particular, every automorphism of a central simple k-algebra is an inner automorphism.<ref name=GS40>Gille & Szamuely (2006) p. 40</ref><ref name=Lor174>Lorenz (2008) p. 174</ref>

ProofEdit

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.

NotesEdit

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ReferencesEdit