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 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.Gille & Szamuely (2006) p. 40Lorenz (2008) p. 174

Proof

First suppose $B = \operatorname_n(k) = \operatorname_k(k^n)$. 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 $\operatorname_n(k) = B$. For the general case, $B \otimes_k B^$ is a matrix algebra and that $A \otimes_k B^$ is simple. By the first part applied to the maps $f \otimes 1, g \otimes1 : A \otimes_k B^ \to B \otimes_k B^$, there exists $b \in B \otimes_k B^$ such that
:$(f \otimes 1)(a \otimes z) = b (g \otimes 1)(a \otimes z) b^$
for all $a \in A$ and $z \in B^$. Taking $a = 1$, we find
:$1 \otimes z = b (1\otimes z) b^$
for all ''z''. That is to say, ''b'' is in $Z_(k \otimes B^) = B \otimes k$ and so we can write $b = b' \otimes 1$. Taking $z = 1$ this time we find
:$f(a)= b' g(a)$,
which is what was sought.

Notes

References

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* A discussion in Chapter IV of Milne, class field theor

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{{DEFAULTSORT:Skolem-Noether theorem
Theorems in ring theory