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. :-)

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

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.

  • $\begingroup$ 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
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    $\begingroup$ (By the way, finite gneration plays no role here) $\endgroup$ – Mariano Suárez-Álvarez Apr 23 '14 at 15:11

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