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I'm learning modular representation theory from the 3rd part of Serre's book. Through the process called "reduction mod. $\mathfrak{m}$", we obtain representations in positive characteristic from representations in characteristic $0$. Howerer, the reduction of irreducible representations may not be irreducible any longer.

I want to test myself, so I try to compute the representation of the symmetric group $\mathfrak{S}_p$ in $\mathbb{F}_p$ (or its algebraic closure). By Brauer--Nesbitt theorem, the number of irreducible $\mathbb{F}_p$-representations is equal to the number of $p$-regular conjugacy classes in $\mathfrak{S}_p$, so there is exactly one representation "missing" since the only $p$-singular conjugacy class is the one containing $(12\cdots p)$. Yet I still have no ideas about how to find them exactly.

Serre gives some examples in his book, say $\mathfrak{S}_4$ and $\mathfrak{A}_5$, but his computations are done by case-by-case analysis. Generally, we have many approaches to construct the representations of a symmetric group over $\mathbb{C}$ or $\mathbb{Q}$, say the Young symmetrizer and the Specht module. Can we generalize some constructions to positive characteristic?

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    $\begingroup$ Finding the irreducible representations of $\mathfrak{S}_n$ over $\mathbb{F}_p$, or even just their dimension, is a major problem in representation theory, and we know extremely little about them. So it's possible that you will be disappointed in what is known on that topic. $\endgroup$ Commented May 5, 2022 at 7:30
  • $\begingroup$ Thank you. I know the general case is complicated, Kleshchev wrote a book on this topic. But what about this special case? $\endgroup$
    – Estwald
    Commented May 5, 2022 at 8:13

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Here is one way to answer your question. I'll begin with the general story, then explain how it simplifies in your example. For any $n$ and $p$, the reduction modulo $p$ of the Specht modules $S^\lambda$ carries a symmetric bilinear form with the property that if it is non-zero, the quotient by its radical is irreducible; in addition, the non-zero quotients that arise this way are a complete set of non-isomorphic simple $F_p S_n$-modules. Moreover, the set of partitions for which the form is non-zero is the set of $p$-restricted partitions of $n$: those for which the difference between consecutive parts is at most $p-1$. Additionally, the irreducibles corresponding to $p$-restricted partitions belong to the same block of $F_p S_n$ if and only if the $p$-cores of the partitions are equal.

In your case $n=p$ this implies that the corresponding quotients of the Specht modules for all partitions of $n$ other than the trivial partition $(n)$ give a complete set of non-isomorphic irreducible representations, and that moreover there is a unique non-semisimple block in the category of $F_p S_p$-modules, containing the hooks. Every other Specht module is actually simple upon reduction modulo $p$. I believe it should be true that moreover there is an exact sequence

$$0 \to S^{(1^n)} \to S^{(2,1^{n-2})} \to \cdots \to S^{(n-2,1,1)} \to S^{(n-1,1)} \to S^{(n)} \to 0$$

with the property that the images are the desired unique irreducible quotients (and whose existence is closely related to the fact that we are in the case here of a block of defect 1). You can work out the dimensions of the interesting irreducible $F_p S_p$ modules from this.

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  • $\begingroup$ Thank you, sir. But I have a question: What do you mean by "the corresponding quotients of the Specht modules for all partitions of $n$ other than the trivial partition $(n)$"? I think the trivial partition corresponds to the trivial representation, which is irreducible under reduction. $\endgroup$
    – Estwald
    Commented May 6, 2022 at 2:24
  • $\begingroup$ @Estwald I mean the quotients of those Specht modules whose bilinear form is non-zero---depending on the conventions you use, it often happens for the trivial representation that its bilinear form is zero (as you point out it is nonetheless irreducible---the point is that it may occur as a quotient of another Specht module in certain cases such as yours, and we use the partition indexing that other Specht module instead of $(n)$ as part of our indexing set for the isoclasses of irreps mod p). $\endgroup$
    – Stephen
    Commented May 6, 2022 at 14:18
  • $\begingroup$ Understand now. Thank you! $\endgroup$
    – Estwald
    Commented May 7, 2022 at 0:11

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