Skolem–Noether theorem

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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 Bx = xB 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, 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[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 B = M n ⁡ ( k ) = End k ⁡ ( k n ) {\displaystyle B=\operatorname {M} _{n}(k)=\operatorname {End} _{k}(k^{n})} {\displaystyle B=\operatorname {M} _{n}(k)=\operatorname {End} _{k}(k^{n})}. Then f and g define the actions of A on k n {\displaystyle k^{n}} {\displaystyle k^{n}}; let V f , V g {\displaystyle V_{f},V_{g}} {\displaystyle V_{f},V_{g}} denote the A-modules thus obtained. Since f ( 1 ) = 1 ≠ 0 {\displaystyle f(1)=1\neq 0} {\displaystyle 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 {\displaystyle V_{f},V_{g}} {\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 b : V g → V f {\displaystyle b:V_{g}\to V_{f}} {\displaystyle b:V_{g}\to V_{f}}. But such b must be an element of M n ⁡ ( k ) = B {\displaystyle \operatorname {M} _{n}(k)=B} {\displaystyle \operatorname {M} _{n}(k)=B}. For the general case, B ⊗ k B op {\displaystyle B\otimes _{k}B^{\text{op}}} {\displaystyle B\otimes _{k}B^{\text{op}}} is a matrix algebra and that A ⊗ k B op {\displaystyle A\otimes _{k}B^{\text{op}}} {\displaystyle A\otimes _{k}B^{\text{op}}} is simple. By the first part applied to the maps f ⊗ 1 , g ⊗ 1 : A ⊗ k B op → B ⊗ k B op {\displaystyle f\otimes 1,g\otimes 1:A\otimes _{k}B^{\text{op}}\to B\otimes _{k}B^{\text{op}}} {\displaystyle f\otimes 1,g\otimes 1:A\otimes _{k}B^{\text{op}}\to B\otimes _{k}B^{\text{op}}}, there exists b ∈ B ⊗ k B op {\displaystyle b\in B\otimes _{k}B^{\text{op}}} {\displaystyle b\in B\otimes _{k}B^{\text{op}}} such that

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

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

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

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

f ( a ) = b ′ g ( a ) b ′ − 1 {\displaystyle f(a)=b'g(a){b'^{-1}}} {\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

References