Talk:Wedderburn's little theorem

Name

Does anybody know why it's called the "little theorem"? I assume he had two theorems? Or was this a way to belittle him? Some explanation would be great in the article. — Preceding unsigned comment added by 2003:C3:2BE0:4B00:E8B5:C68:EF57:B63E (talk) 17:47, 3 February 2018 (UTC)[reply]

Jacobson theorem

It is good to mention, that Jacobson theorem is generalization. Example of link: https://www-fourier.ujf-grenoble.fr/~marin/une_autre_crypto/articles_et_extraits_livres/Herstein-Wedderburn.pdf My English is not good to do it alone.Meproun (talk) 13:33, 17 April 2015 (UTC)[reply]

Noether-Skolem

"Alternatively, the theorem is a consequence of the Noether-Skolem theorem." Does anyone understand this? First, I thought this would follow by applying Skolem-Noether to the Frobenius, but it isn't an algebra homomorphism for noncommutative algebras. Ringspectrum (talk) 18:09, 15 March 2009 (UTC)[reply]

I also don't see that it helps to restrict the Frobenius to the F_q = Z(B)-algebra generated by some element of B. Ringspectrum (talk) 18:19, 15 March 2009 (UTC)[reply]

If B is the finite skew-field, and K is its center, then to use N-S in B, the homomorphism from A to B must be a K-algebra homomorphism. In particular, it acts as the identity on the center, or any central subalgebra.

The proof they had in mind was probably for splitting fields. Every splitting field has the same degree, and since finite fields are determined by their degree, there is some K-isomorphism from F1 to F2, for each pair of splitting fields F1,F2 ≤ B. This isomorphism is inner by N-S, so F1 and F2 are conjugate in B. Indeed, their multiplicative groups are conjugate in the multiplicative group of B. Since every nonzero element of B is contained in the multiplicative group of a splitting field, the multiplicative group G of B is the union of the G-conjugates of the multiplicative group of F1. Since a finite group is not the union of the conjugates of its proper subgroups, G = F1* and B=F1 is a field. JackSchmidt (talk) 19:07, 15 March 2009 (UTC)[reply]

Thanks! That each minimal separable splitting field has the same degree ind(A) can be found in Gille, Szamuely, Proposition 4.5.4/Corollary 4.5.9. The proof works over any quasi-finite field (or field with Galois group Z^) since, more generally, a group is not the union of the conjugates of a proper subgroup of finite index (take the union of the unit groups of the splitting fields with {0}).

The cohomological proof is easier: $H^{2}({\hat {\mathbf {Z} }},{\bar {K}}^{sep})=0$ since the latter group is divisible. Ringspectrum (talk) 09:03, 16 March 2009 (UTC)[reply]

The proof here assumes there is an identity. This actually has to be proved. Let a be an element of A. Then the set of products ab for b in A spans A, as does the set of products ba for b in A. In particular, there is an e such that ae=a. If b is another element of A, then there is a b* such that b= b*a. But be = b*ae =b*a =b, so e is a right identity. Similarly, there is a left identity, 1. But 1e=e since 1 is a left identity, and 1e=1 since e is a right identity. So e is a two-sided identity. Syd Henderson (talk) 22:30, 14 April 2011 (UTC)[reply]