Irreducible Representation of A_5 Knowing the fact that standard representation arising out of permutation representation of $A_5$ over $\mathbb{C}$ is irreducible and of degree $4$. What can we conclude about the irreducibility over general field, whose characteristics does not divide the order of $A_5$. Is it irreducible ? Can we use Clifford Theory here ? How ?
 A: Here is a fairly general answer for the characteristic zero question you ask, which I will mark as Community Wiki:
Let $G$ be any doubly transitive permutation group of degree $n$ on a set $\Omega$. Let $V$ be the underlying  permutation module, viewed as the $\mathbb{Q}G$-module $\mathbb{Q}\Omega$. Then $V \cong U \oplus W$, where $U$ is the trivial $\mathbb{Q}G$-module and $W$ is $n-1$-dimensional ($U$ may be realised as the space of $\mathbb{Q}$-linear combinations of elements of $\Omega$ with all coefficients equal, and $W$ is the set of all $\mathbb{Q}$-linear combinations of elements of $\Omega$ in which the sum of coefficients appearing is zero.
It is an easy exercise that ${\rm End}_{\mathbb{Q}G}(W) \cong \mathbb{Q}$ (this uses the double transitivity of the permutation action).  This implies that the representation afforded by $W$ is absolutely irreducible- that is, remains irreducible after any extension of scalars enlarging $\mathbb{Q}$ to any extension field. Since any field $\mathbb{F}$ of characteristic zero has prime subfield isomorphic to $\mathbb{Q}$, we see
that $W \otimes_{\mathbb{Q}} \mathbb{F}$ is an irreducible $\mathbb{F}G$-module.
( This is, of course, all standard theory)
Later edit: In view of some of the comments, let me point out that this argument does not always generalize to prime characteristic dividing the group order. Possibly The easiest case it fails is when the prime characteristic $p$ is a divisor of $n$, the degree of the permutation representation, a case mentioned by Derek Holt in comments.
For if $\mathbb{F}$ is a field of prime characteristic $p$ which divides
$n = |\Omega|$, then the permutation module $\mathbb{F}\Omega$ has two obvious submodules $U$ and $W$ defined as before, but note that $U \subseteq W$ in this case because the coefficients appearing in any element of  $U$ sum to zero. Hence $W$ is not irreducible (if $n >2$ to discount a trivial exception). We also find that $\mathbb{F}\Omega/W \cong U$. Hence the permutation module $\mathbb{F}\Omega$ has at least three composition factors, at least two of which are trivial. The $n-2$-dimensional module $W/U$ is sometimes called the Green heart (after J.A. Green). It is often irreducible, but I think examples exist when it is not.
