5
$\begingroup$

Let $k$ be a field and $G$ is finite group. I want to prove that a $kG$ module $P$ is projective iff it's injective. I proved that if module is projective then it's injective.

1) $kG$ is injective because $Hom_{kG}(M,kG)=Hom_{k}(M,k)$. So every free $kG$ module is injective.

2) Every projective module is a free summand of free, so it's injective.

But I don't know how can I prove that injective modules are projective

$\endgroup$
  • $\begingroup$ We have to additionally observe that $kG$is Artinian to draw your first conclusion. What you wrote proves that free modules of finite rank are injective, but knowing the ring is Noetherian allows us to also say infinite direct sums of the regular module are injective too. $\endgroup$ – rschwieb Jul 8 '14 at 10:34
1
$\begingroup$

If $M$ is any $kG$-module, then $kG \bigotimes_{k} M$ is a free $kG$-module, and the kG-module homomorphism

$$ m \mapsto \sum_{g \in G} g \otimes g^{-1} m$$

is injective. If $M$ is injective, then since free $kG$-modules are injective, $M$ is now a direct summand of a free $kG$-module and is therefore projective.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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