Explicit description of the principal block of the symmetric group 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 $Me.$ 
 A: Of little practical use in this instance, but perhaps of some theoretical interest, is that if we let $G_{p^{\prime}}^{+}$ denote the sum of the $p$-regular ( ie order prime to $p$) elements (in the group algebra) in any finite group $G$, then multiplication by $G_{p^{\prime}}^{+}$ annihilates any indecomposable module which does not lie in the principal $p$-block, but annihilates no indecomposable module in the principal $p$-block. Over a field of characteristic $p$, some power of $G_{p^{\prime}}^{+}$
is the primitive central idempotent of the center of the group algebra corresponding to the principal $p$-block (and there is an analogous though not identical statement when working over a complete dvr $R$ with $R/J(R)$ of characteristic $p$), so the multiplication can't annihilate anything in the principal $p$-block. On the other hand, by block orthogonality relations due to R. Brauer (which are also one way to see that the previous claim is true), it follows that multiplication by $G_{p^{\prime}}^{+}$ annihilates any indecomposable module outside the principal $p$-block.
In view of Sergei's comment, let me expand. By block orthogonality relations due to R. Brauer (which can be found in most reasonably modern texts on representation theory of finite groups), it follows that $\sum_{g \in G_{p^{\prime}}} \chi(g)\mu(g^{-1}) = 0$ whenever $\chi$ and $\mu$ are complex irreducible characters of $G$ in different $p$-blocks. Letting $\chi$ be the trivial character, we see that $\sum_{g \in G_{p^{\prime}}} \mu(g) = 0$ for each irreducible character $\mu$ which does not lie in the principal $p$-block(note that the set of $p$-regular elements is closed under inversion).
Now in the complex group algebra $\mathbb{C}G$, any central element $X$ may be written as $\sum_{\chi \in {\rm Irr}(G)} \omega_{\chi}(X) e_{\chi},$ where $\omega_{\chi}: Z(\mathbb{C}G) \to \mathbb{C}$ is the algebra homomorphism which sends the conjugacy class of $y$ to $\frac{[G:C_{G}(y)]\chi(y)}{\chi(1)}.$ It follows that when $G_{p^{\prime}}^{+}$ is so expressed, it is a linear combination of primitive idempotents $e_{\chi}$ for which $\chi$ is in the principal block, since (from above) $\omega_{\mu}$ annihilates $G_{p^{\prime}}^{+}$ whenever $\mu$ lies outside the principal $p$-block.
This implies that $1_{B} G_{p^{\prime}}^{+} =  G_{p^{\prime}}^{+}$, where $1_{B}$ is the identity of the principal $p$-block $B$ (which may be written in the complex group algebra as $\sum_{ \chi \in {\rm Irr}(B)} e_{\chi}$). It follows that $G_{p^{\prime}}^{+}$ annihilates any indecomposable module with no indecomposable summand in the principal $p$-block (even over a field $F$ of characteristic $p$ or over a dvr $R$ with $R/J(R)$ of characteristic $p$). (Recall that each indecomposable module is in one and only one $p$-block in these last cases).
On the other hand, if $P$ is a Sylow $p$-subgroup of $G$, then $P$ acts by conjugation on the $p$-regular elements of $G$, and $P$ fixes only the normal complement $M$ to $Z(P)$ in $C_{G}(P)$. Hence $|G_{p^{\prime}}| \equiv |M| \not \equiv 0$ (mod $p$). ( I think this observation may first have been used in block theory by Osima in a proof of Brauer's Third Main Theorem).
For a complete dvr $R$ as above, letting $\chi$ denote the trivial character, we know that for each irreducible character $\mu$ in the principal $p$-block $B$, we have 
$\omega_{\mu}(z) \equiv \omega_{\chi}(z)$ (mod $J(R)$) for each $z \in Z(RG)$.
It follows that $\omega_{\mu}(G_{p^{\prime}}^{+}) \equiv \omega_{\chi}(G_{p^{\prime}}^{+}) \equiv |G_{p^{\prime}}| \not \equiv 0$ (mod $J(R)$).
Now we  know that $\omega_{\mu}(G_{p^{\prime}}^{+}) \equiv |G_{p^{\prime}}| \omega_{\mu}(1_{B})$ (mod $J(R)$) for each irreducible character $\mu$ of $G$, so that 
$G_{p^{\prime}}^{+} \equiv |G_{p^{\prime}}|1_{B}$ (mod $J(Z(RG))$). From this it follows that working over $F = R/J(R)$, some power of $G_{p^{\prime}}^{+}$ is $1_{B}$.
For this purpose, it may be better to normalize $G_{p^{\prime}}^{+}$ by replacing it with $G_{p^{\prime}}^{+}/|G_{p^{\prime}}|$).
The power needed may be very high, though it is possible to give explicit bounds in terms of $|P|$, where $P \in {\rm Syl}_{p}(G)$.
Direct ring theoretic proofs of this type of results have been given by B. K\"ulshammer in a series of papers (some in German) called "Remarks on the Group Algebra as Symmetric Algebra". A very clean result of his is that (over a field $F$ of characteristic $p$),
we have $G_{p^{\prime}}^{+}G_{p}^{+} = c1_{B}$ (+ a linear combination of $p$-singular elements), where $c \neq 0 \in F$, and $G_{p}$ denotes the set of elements of $G$ whose order is a power of $p$.
A: 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.
