Let $F=\mathbb{F}_q$ be the field of $q$ elements and $G$ be a finite group. I'm trying to show that for an irreducible $FG$-module $V$, we have $\mathrm{End}_{FG}(V)=F \cdot 1$, i.e. that $F$ is a splitting field for $FG$.
My attempts thus far at this problem have involved trying to show that $V$ must be absolutely irreducible, as I believe this is equivalent to the statement I am trying to prove. This is where I am at a loss. I have tried to show that $V \otimes_F \mathbb{F}_{p^{n!}}$ is irreducible for all $n$ (as $\mathbb{F}_{p^{\infty}}$ is the algebraic closure of $\mathbb{F}_q$), though I'm not too sure exactly how to do this. I'm feeling a little bit out of my depth here so any help/pointers would be greatly appreciated!
The claim is false as stated. For a small example let $G$ be the cyclic group of order three generated by the element $g$, and let $q=2, F=\mathbb{F}_2$. Let $$ V=\{a_01+a_1g+a_2g^2\in FG\mid a_0+a_1+a_2=0\}. $$ It is easy to see that $V$ is irreducible. Yet multiplications by fixed elements of $G$ obviously give distinct non-zero endomorphisms of $V$ so there are four endomorphisms of $V$.
You correctly judged that this has a lot to do with absolute irreducibility. This module $V$ splits into a sum of two 1-dimensional representations over $E=\mathbb{F}_4$. Let $\omega$ be a primitive cubic root of unity in $E$. Then as modules over $EG$ we get $$ E\otimes_FV=E(1+\omega g+\omega^2g^2)\oplus E(1+\omega^2g+\omega g^2). $$ It is also easy to see that $\operatorname{End}_{FG}(V)\cong E$ as $F$-algebras.