Tensor product of simple finite-dimensional modules I am trying to understand the following one-liner that appears in http://arxiv.org/abs/0901.0827v5, Theorem 2.26 (and I'm afraid I must be missing something really basic):
Let $V$ and $W$ be irreducible finite dimensional representations of an algebra $A$.

By the density theorem, the maps $A\to End(V)$ and $B\to End(W)$ are surjective. Therefore, the map $A\otimes B \to End(V) \otimes End(W) = End(V\otimes W )$ is surjective. Thus, $V\otimes W$ is irreducible.

I only have a problem with the final statement.  Put differently: is it obvious that surjectivity of $C\to End(M)$ implies irreducibility of $M$?  Or am I completely off-track?
 A: $\newcommand{\End}{\operatorname{End}}$ Proposition. Let $k$ be a field (not necessarily algebraically closed). Let $C$ be a $k$-algebra. Let $M$ be a $C$-module such that the map $C \to \End M$ (which describes the action of $C$ on $M$ as a representation of $C$) is surjective (where $\End M$ means the endomorphism ring of the $k$-vector space $M$). Then, the $C$-module $M$ is simple.
Proof. Let $V$ be a nonzero $C$-submodule of $M$. We need to prove that $V = M$.
Indeed, assume the contrary. Thus, $V \neq M$. Hence, there exists a $w \in M$ satisfying $w \not\in V$. Consider such a $w$.
Also, $V \neq 0$, and thus there exists a nonzero $v \in V$. Consider such a $v$. There exists a $P \in \End M$ satisfying $P\left(v\right) = w$ (because we can extend the one-element family $\left(v\right)$ to a basis $\left(b_i\right)_{i \in I}$ of $M$, and extend the one-element family $\left(w\right)$ by zeroes to a family $\left(c_i\right)_{i \in I}$ of elements of $M$; then we can define $P$ to be the linear map $M \to M$ sending each $b_i$ to $c_i$). Consider such a $P$. Then, $P$ is the action of some element $c \in C$ on $M$ (since the map $C \to \End M$ is surjective). Consider this $c$. We have $c\underbrace{v}_{\in V} \in cV \subseteq V$ (since $V$ is a $C$-submodule of $M$). But the action of the element $c \in C$ on $M$ is $P$; thus, $cv = P\left(v\right) = w \not\in V$. This contradicts $cv \in V$. This contradiction shows that our assumption was wrong, and the Proposition is proven.
