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The following question is probably straightforward for those who know. However, I am used to working either over splitting fields or in characteristic zero.

Question. Let $G$ be a finite group and $k$ a field of characteristic $p>0$. The quotient $kG/rad(kG)$ decomposes as a direct product of matrix rings over division algebras, finite dimensional over $k$. What can be said about these division algebras? If $D$ is one of these division algebras, is it true that $p$ does not divide the dimension of $D$ over its center?

I sort of convinced myself that in fact these division algebras must be fields in characteristic $p$ (and hence what I want is true), but since I couldn't find this stated anywhere, I suspect there is a flaw in my logic. Anyway, here is my "proof". Please let me know where it goes astray.

Let $F_p$ be the field with $p$ elements. Then $F_pG/rad(F_pG)$ is a separable algebra because $F_p$ is perfect.

Claim. $kG/rad(KG)\cong k\otimes_{F_p}(F_pG/rad(F_pG))$.

Pf. Note that $k\otimes_{F_p}(F_pG/rad(F_pG))= kG/(k\otimes_{F_p}rad(F_pG))$. But since $F_pG/rad(F_pG)$ is separable, we have that $k\otimes_{F_p}(F_pG/rad(F_pG))$ is semisimple. As $k\otimes_{F_p}rad(F_pG)$ is a nilpotent ideal, we conclude that it is $rad(kG)$.

Since finite division rings are fields and $F_p$ is perfect, we have that $F_pG/rad(F_pG)\cong \prod_{i=1}^m M_{n_i}(L_i)$ where the $L_i$ are finite separable extensions of $F_p$. Thus by the claim $kG/rad(kG) = k\otimes_{F_p}(F_pG/rad(F_pG))=\prod_{i=1}^m M_{n_i}(k\otimes_{F_p}L_i)$. But since the $L_i$ are finite separable extensions, each $k\otimes_{F_p}L_i$ is a finite product of separable extensions of $k$.

I guess if there is a mistake here, it is in the claim.

Olexandr Konovalov
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Benjamin Steinberg
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    My understanding of this topic is not perfect, but I think that your proof is correct, and that the result is essentially Exercise (9.7) Page 157 of Isaacs' Character Theory of Finite Groups. – Derek Holt Jun 03 '14 at 19:53
  • @DerekHolt, I don't have Isaacs book, which was where I wanted to look. Thanks. – Benjamin Steinberg Jun 03 '14 at 19:58
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    Although this may not technically be a research level problem, in practice there would probably be many more people who read mathoverflow who would have the necessary expertise to answer this properly. – Derek Holt Jun 04 '14 at 10:45
  • Hi I have a question to the last line of the proof: If one takes a finite and hence separable field $L_i$ and an arbitrary (not necessary separable, but (semi-)simple) field K why is the tensor product $K\otimes L_i$ like you described it? I know that a tensor product of a separable and semisimple algebra is again semisimple (e.g. in the book of Curtis/Reiner). Thanks for commend as this proof is related to the question when a group algebra is a reduced algebra:-) – Sven Wirsing Aug 27 '14 at 07:02
  • If $L_i$ is a finite separable extension of $F$ and $K$ is any extension of $F$, then by the primitive element theorem $L_i\cong F[x]/(p(x))$ where $p(x)$ factors into linear factors over the algebraic closure of $F$. Then $K\otimes L_i\cong K[x]/(p(x))$ and $p(x)$ still has distinct roots in the algebraic closure of $K$ (which contains an algebraic closure of $F$ which will have the roots of $p(x)$). The general case follows because a separable extension is a direct limit of finite separable extensions. – Benjamin Steinberg Aug 27 '14 at 14:40
  • I have another question to your Claim that $KG/rad(KG)\cong K\otimes_{F_p} (F_pG/rad(F_pG))$. The left side is finite-dimensionl the right side could be infinite dimensional if $K$ is choosen to be an infinite field of $char(K)=p$ with prime-field $F_p$. – Sven Wirsing Aug 29 '14 at 07:19
  • Tensoring a fd vector space with an extension field K results in a K-vector space of the same dimension as a K-vector space as the original had over the ground field. – Benjamin Steinberg Aug 29 '14 at 07:31
  • Yes, indeed, so in the claim the isom. is viewed as $K$-algebras. From this one can deduce the $K$-dimension of $KG/rad(KG)$ is precisely the $F_p$-dimension of $F_pG/rad(F_pG)$ and in other words for every fields $K,L$ of characteristic $p$ the $K$-dimension of $KG/rad(KG)$ and $L$-dim. of $LG/rad(LG)$ are equal. Thats very nice. – Sven Wirsing Aug 29 '14 at 10:48
  • Just a comment: This theorem is proven at the beginning of Finite groups II by Huppert and Blackburn. – Sven Wirsing Sep 29 '16 at 13:44
  • It is true that an infinite field can be replaced by a finite subfield in Benjamin's question. Hence the result follows since a finite division ring is a field. See Section 7 of my joint paper MR1380609 with Laci Kovács. – Glasby Feb 06 '18 at 02:11
  • Under what circumstances would $k \otimes_{F_p} L_i$ be Galois over $k$? – John Toh Apr 28 '20 at 13:48

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