I've been trying to calculate projective modules in an effort to eventually classify extensions of modules of $S_3$ over a field of three elements.
So I know that projective modules can be found by multiplying the ring by an idempotent. So I got two, $ e_1 = (-s-1)$ and $ e_2 = (s-1)$, where $s$ is a transposition.
As I'm working over the field of characteristic three, the sum of these is $1$, so then $R(e_1+e_2) = R $, and each of $Re_i$ must be three dimensional.
However, when I take the actual product (taking for example, the transposition $ (12) $, I get something two-dimensional. Apparently though I should be getting representation of trivial/sign on $S_3$ over $S_2$. Which kind of makes sense to me because I get the same coefficient on the first two basis elements? But I still have the wrong dimensions, so I'm not sure what I've managed to do. Should I be going over all transpositions?
Just starting doing things like this, so there may be a really silly error in here, but any help would be appreciated, thanks.
Take $s = (12)$ then right multiplication by $e_1$ corresponds to the map:
$e \to -(12)-e =:v_1$
$(12) \to -e-(12)= v_1$
$(23) \to -(132)-(23) =: v_2$
$(132) \to -(23)-(132) = v_2$
$(13) \to -(123)-(13) =: v_3$
$(123) \to -(13)-(123) = v_3$
So indeed we see the image of this map is three dimensional, spanned by $v_1$, $v_2$, and $v_3$, and $S_3$ acts by left multiplication. Okay, but what is this representation? It is not too hard to see that it's just the standard representation of $S_3$ on $k^3$ permuting the coordinates, the map is given by $v_1 \to (0,0,1)$, $v_2 \to (0,1,0)$, $v_3 \to (1,0,0)$.
If you do the same thing for $e_2$ you get the same thing but twisted by the sign representation. These are the two projective modules for $S_3$ in characteristic $3$. They both have length $3$ composition series with factors triv-sign-triv and sign-triv-sign respectively.