Why an absolutely irreducible representation is irreducible under all field extensions? A representation $G \to\mathrm{GL}_n(F)$ over a field $F$ is called absolutely irreducible if it is irreducible over the algebraic closure of $F$. Why is this equivalent to saying that the representation is irreducible under any field extension of $F$?
I know that the representation is 1-dimensional over the closure, but why is it irreducible over any extension? In particular, why is it irreducible over the ground field?
 A: Let $V$ be a representation over a field $k$. If $W \subset V$ is a subrepresentation, then $W \otimes_k L \subset V \otimes_k L$ is a subrepresentation for the base change to $L$.  So we see that if $V$ is reducible over one field it is reducible over all field extensions. Therefore being irreducible over $\bar{F}$ implies irreducibility over all algebraic extensions of $F$.
Really I'd say that first part is the core part of the theorem, but to be complete we can address non-algebraic extensions of $F$. If $F \subset L$ is an arbitrary extension then we can take an algebraic closure so that $\bar{F} \subset \bar{L}$, and by the above reasoning it suffices to show irreducibility over $\bar{L}$.
Now since $\bar{F}$ is algebraically closed, the Jacobson density theorem tells us that irreducibility implies that $\bar{F}[G] \to End_\bar{F}(V)$ is surjective.  Base changing preserves surjectivity of linear maps, so $\bar{L}[G] \to End_\bar{L}(V)$ is also surjective, which means that $V$ is irreducible over $\bar{L}$.
