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)

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)

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)

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


 * 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)


 * 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)

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)