Let $k$ be a field of prime characteristic $p$ and $\Sigma\_n$ be the symmetric group.
If I have a concrete $k[\Sigma\_n]$-module $M,$ how to compute the direct summand corresponding to the principal block of the group algebra $k[\Sigma\_n]$? 


Is there a formula for the central idempotent $e$? If I had such a formula, I just need to compute $Me.$ 



Explicit description of the principal block of the symmetric group
If you just want to find the summand which lies in the principal block, that's easy: it's the submodule where the semi-simple part of every central element acts with the same eigenvalue as on the trivial module. That is, let $x=\sum a\_g g$ be a central element of $k\Sigma\_n$ (so $a\_g$ is constant on conjugacy classes); this acts on the trivial module with eigenvalue $a=\sum a\_g$, so a module is in the principal block if $x-a$ acts nilpotently on it.


