Splitting fields of symmetric groups

Question

Is it true that $k$ is a splitting field of $S_n$ if and only if the characteristic $p$ of $k$ is zero or larger than $n$? The fact that the character table (over $\mathbb C$) has only integer entries smaller or equal to $n$, seems to imply this, or am I mistaken? If the statement is true, could someone give a citation?

Edit: $k$ is a splitting field of $S_n$ if the $k$-algebra $kS_n$ splits over $k$, i.e. if for every simple ($=$ irreducible) $kS_n$-left-module $M$, we have $\mathrm{End}_{kS_n}(M) \cong k$.

Answer 1

The irreducible $S_n$-modules are all realizable over the integers. Specifically, there is a family of $\mathbb{Z} S_n$-modules $S^\lambda$ indexed by partitions $\lambda$ of $n$, called Specht modules, together with symmetric $\mathbb{Z}$-bilinear forms, such that over a field $k$, the quotient by the radical of the form is either zero or irreducible, and the set of non-zero irreducibles obtained this way is a complete set of representatives for the isoclasses of irreducibles. What is true is that the group algebra is semisimple exactly if the characteristic $p$ is bigger than $n$. In general, the blocks are in bijection with the set of $p$-cores of partitions of $n$.

You can read about this in chapter 4 of James' book "The representation theory of the symmetric group".

Answer 2

I don't see why you need $p$ larger than $n$ if non-zero -more precisiely, you don't. In that case, an absolutely irreducible module for a finite group is realizable over the field of its character, and for the symmetric group, this is always the prime subfield. This boils down to the fact that Schur indices are trivial over finite fields, which in turn is a consequence of the fact that finite division algebras are fields. It is explained in many standard texts on representation theory, eg Curtis and Reiner.