Representation of an algebra is absolutely irreducible if and only the representation map is surjective This should be well known but I can't seem to locate a reference:
Let $k$ be a field, $V$ a $n$-dimensional vector space over $k$ with an action of a $k$-algebra $A$. We say that $V$ is an absolutely irreducible $A$-module if for any field extension $L/k$, $V\otimes_k L$ is irreducible as an $A\otimes_{k}L$ module.
In this situation, we also have a representation map $r: A \to M_n(k) = End_k(V)$. Why is it true that $V$ is absolutely irreducible if and only if $r$ is surjective?
It is at least easy to show that if $r$ is surjective, then $V$ is absolutely irreducible. This is simply equivalent to showing that $M_n(k)$ (even $GL_n(k)$) acts transitively on non zero vectors in $k^n$ (plus the fact that tensor products preserve surjectivity).
What about the other direction?
 A: Let $L$ be an algebraic closure of $k$.  Note that $r$ is surjective iff $r\otimes L:A\otimes_k L \to M_n(L)$ is surjective since $L$ is faithfully flat over $k$.  So, we may as well assume $k=L$ is algebraically closed.  We may further replace $A$ with the image $r(A)$ to suppose $A$ is a subalgebra of $M_n(k)$.  
So, suppose $k$ is algebraically closed and $A$ is a subalgebra of $M_n(k)$ such that $V=k^n$ is irreducible over $A$.  Note that we can consider $M_n(k)$ as an $A$-module as well and that $M_n(k)$ is a direct sum of $n$ copies of $V$ as an $A$-module (corresponding to the columns of the matrix).  In particular, since $V$ is a simple $A$-module, $M_n(k)$ is a semisimple $A$-module.  But $A$ itself (as a left $A$-module) is a submodule of $M_n(k)$, so $A$ is semisimple as a module over itself and hence a semisimple ring.
The conclusion that $A$ is all of $M_n(k)$ now follows from Artin-Wedderburn.  For instance, part of Artin-Wedderburn says that if $V$ is a simple $A$-module, $\operatorname{End}_A(V)$ is a division ring $D$ and the image of the action of $A$ on $V$ is the full endomorphism ring $\operatorname{End}_D(V)$ (since this full endomorphism ring is one of the factors in the product decomposition of $A$).  Since $k$ is contained in the center of $A$, we have $k\subseteq D\subseteq \operatorname{End}_k(V)$ and in particular every element of $D$ is algebraic over $k$ since $V$ is finite dimensional over $k$.  Since $k$ is algebraically closed, this means $D=k$.
Alternatively, this follows from the Jacobson density theorem (which is indeed often used to prove the relevant part of Artin-Wedderburn).  As above, we must have $\operatorname{End}_A(V)=k$ since $k$ is algebraically closed, and since $V$ is finite dimensional over $k$ the Jacobson density theorem says immediately that every $k$-endomorphism of $V$ is realized by an element of $A$.
