1
$\begingroup$

For a finite group G and a finite field $\mathbb{F}_p$ of characteristic $p$, J($\mathbb{F}_{p^k} \otimes_{\mathbb{F}_p }\mathbb{F}_p G ) = J(\mathbb{F}_{p^k}G)$? where $J(\mathbb{F}_{p^k}G)$ is the Jacobson radical of the group algebra $\mathbb{F}_{p^k}G$ and $\mathbb{F}_{p^k}$ is an extension field of $\mathbb{F}_p$.

$\endgroup$
3
  • $\begingroup$ The two algebras you're taking the Jacobson radical of are naturally isomorphic, so... $\endgroup$ Nov 10, 2016 at 6:41
  • 3
    $\begingroup$ I'd guess it'y a typo and the question is whether $J(F_{p^k}\otimes F_pG)=F_{p^k}\otimes J(F_p G)$? If so the question should be edited (fixing along the way typos in the title and in the text) $\endgroup$
    – YCor
    Nov 10, 2016 at 8:13
  • $\begingroup$ Ah, I was going to fix the error in the title, but after YCor's comment I leave it there... $\endgroup$
    – Vincent
    Nov 10, 2016 at 11:03

2 Answers 2

3
$\begingroup$

Assuming that the question is the one of YCor: $J(F_{p^k}\otimes F_p G) = F_{p^k}\otimes J(F_pG)$, the answer is yes. The reason is the following: For a finite dimensional algebra $A$ over a field $K$ one can describe $J(A)$ as the minimal ideal such that $A/J(A)$ is a semisimple algebra. It is also the maximal nilpotent ideal of $A$. Let us show that if we have a finite separable extension of fields $L/K$, then $J(L\otimes_K A) = L\otimes J(A)$. On the one hand, $L\otimes_K J(A)$ is a nilpotent ideal, and is therefore contained in $J(L\otimes A)$. On the other hand, $A/J(A)$ is semisimple and can therefore be written as a direct sum of matrix algebras over division algebras. But then $$(L\otimes_K A) / (L\otimes_K J(A)) \cong L\otimes_K (A/J(A)).$$ The last algebra is semi-simple due to the fact that if $D$ is a division algebra over $K$, then $L\otimes_K D$ is a semi-simple $L$-algebra (here we use in an essential way the fact that $L/K$ is finite and separable). This implies the inclusion in the other direction, and the two ideals are therefore equal.

$\endgroup$
3
$\begingroup$

More generally, let $k$ be any field, let $K/k$ be any separable algebraic extension, and let $R$ be any $k$-algebra. Then $J(K\otimes_k R)=K\otimes_k J(R)$. [See Theorem 5.17 in Lam's "First Course in Noncommutative Rings," for instance.]

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.