0
$\begingroup$

I am self studying some non-commutative algebra, and I want to make sure I don't confuse myself. Here is what I am thinking:

Let A and B be finitely generated $k[G]$-algebras, for $G$ a finite group and k a field, and $f : A \to B$ a surjective $k[G]$-algebra homomorphism.

Then the sequence $0 \to \ker f \to A \to B \to 0$ splits as an exact sequence of $k$-modules. (By restricting the representations of the ring $k[G]$ into $End_{Ab}(A)$ and $End_{Ab}(B)$ to $k$, and then constructing a $k$-basis from one of $Z$ - these should be finite dimensional vector spaces, since they are finitely generated over the group algebra of a finite group.)

Is it necessarily the case that this sequence also splits in the category of $k[G]$-modules? It seems like it should, since $\ker f$ and its complement (in $k$-mod) are both $G$ invariant. (But the complement is not necessarily $k[G]$-invariant - edit: this obstruction is obviously false.)

This would be a useful lemma for a problem I am working on (showing that $Q[S_3] \cong Q \times Q \times M_2[Q]$ as rings), but I don't want to waste time trying to prove something false or convince myself of something that is not true. (I would let $c = 1/6 \Sigma_{g \in S_3} g$, so that $Q[S_3] \cong Q[S_3]c \times Q[S_3](1 - c)$. The former is isomorphic to Q, and the latter surjects on $M_2[Q]$ by extending a faithful representation of $S_3$ into $GL_2(Q)$, and thus has a 1 dimensional kernel. I would like to conclude that $Q[S_3](1 -c) \cong Q \times M_2[Q]$ as $k[G]$-algebras from the lemma I described above.)

Does one need more conditions to show that this is true? I will be able to provide a proof if that is the case, but my algebraic intuition is not accurate enough to believe in any results I come up with on my own. :-)

$\endgroup$
1
  • $\begingroup$ Beging both $k$ and $G$-invariant, and being $k[G]$-invariant, are equivalent things. $\endgroup$ – Mariano Suárez-Álvarez Apr 23 '14 at 15:02
2
$\begingroup$

If the field is of characteristc zero or, more generally, does not divide the order of the group, then yes, it always splits: this is the content of Maschke's theorem.

If the characteristic is positive and divides th order of the group, then in general no. I suggest you find examples of this: pick a field $k$ of characteristic two, like the field with two elements, and consider the group $G$ which is cyclic of order two.

$\endgroup$
3
  • $\begingroup$ So I can conclude that $Q[S_3] \cong Q \times Q \times M_2[Q]$ in the way that I describe - since (as you point out), $\ker f$ and its complement are $k[G]$ invariant? But is that enough to split $Q[S_3]$ as rings, and not just modules? I would think that in general I need to describe an idempotent element... or rather, an basis of orthogonal idempotents for the kernel. $\endgroup$ – Lorenzo Najt Apr 23 '14 at 15:08
  • $\begingroup$ I did not say that «its complement is $k[G]$» invariant: for one thing, the kernel does not have one complement as vector spaces, but many, and only some of them are $k[G]$-invariant. I am not saying that your argument is correct, only answering the question of your title. $\endgroup$ – Mariano Suárez-Álvarez Apr 23 '14 at 15:11
  • 1
    $\begingroup$ (By the way, finite gneration plays no role here) $\endgroup$ – Mariano Suárez-Álvarez Apr 23 '14 at 15:11

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.