I have been looking at extensions of irreducible representation over fields of positive characteristic. Specifically at the moment, to get the hang of things, I'm looking at $S_3$ over $\mathbb F_3$, and considering the modules corresponding to the trivial representation and the sign representation, looking for $Ext^1(Triv, Sign)$ and also looking at $Ext^1(Triv, Triv)$. This is the first time I've met Ext groups and had to compute them, so I don't feel very confident.
So first I looked for a projective resolution of the 'Triv' module, $A$. I came up with the following, where $P_1 = R(e_1) = R( -s-1)$, s is a transposition:
$... P_1 \oplus P_1 \rightarrow P_1 \rightarrow A \rightarrow 0$ where I keep taking direct sums to preserve the image and kernel dimensions as required.
So then I dropped A and took $Hom(-,B), B$ the sign module.
Then I claimed that the first two entries in the complex were zero, and thus, so was the ext group I was looking for. But I read somewhere that my answer should be $\mathbb F_3$, so I'm not happy.
So I'll outline what I thought and hopefully someone can point out where I'm wrong:
For $Hom(P_1, B)$:
If something, $T$ existed here, we would have, as matrices:
$TP_1(g) = S(g)T$ for all $g$ in $G$, where $S(g)$ is the sign representation. First let $g$ = $(123)$. Then we get that the three columns in $T$ are equal. Then letting $g$ = $(12)$, we get the third column = 0, thus $T$ is 0. This reasoning was basically identical in the case of the second object in the complex. I know in the ordinary case where we work over $\mathbb C$ that these are 0, because we have different decompositions into irreducibles, but I did think it was different here (or maybe I was wrong, which would be nice). I guess my main concern is that I haven't used the fact that I have characteristic three, and I feel I need to do that (apart from identifying idempotents I suppose).
I then looked at the $Ext(A,A)$ case, and had similar uncertainties. I use the same projective resolution. I then get:
$0 \rightarrow Hom(P_1,B) \rightarrow Hom($P_1 $\bigoplus$ P_1$ ,B) ...$
So in this case, we have a 1-dimensional space of homomorphisms (first object), where each basis vector gets sent to the basis vector of the trivial space. For second, we can either have zero or this homomorphism in each summand. So moving on from here I'm having a little trouble. The kernel for my first map is zero? As the original map I described from $P_1 \oplus P_1$ to $P_1$ gets sent to a map where we're just sending everything to the trivial basis vector, so must have kernel zero? I'm really not sure. I'm even less sure about how to finish this homology calculation.
If someone could help me and say whats wrong so far (if anything) I'd really appreciate it, thanks.