Criterion for Simpleness of Modules over an Algebra Let $k$ be a (if necessary algebraically closed) field. Let $A$ be a finite-dimensional $k$-algebra and $M$ be an $A$-left-module. Consider $M$ to be a right-module over $B = \mathrm{End}_A(M)$. Assume that $N$ is a simple $B$-left-module. Is it true that $M\otimes_B N$ is a simple $A$-left-module?
In particular, is it true, that if $\mathrm{End}_A(M) \cong k$, we have $M$ is a simple $A$-module?
 A: No. Let $A$ be the three dimensional path algebra of the quiver $1 \to 2$, then the projective cover $M$ of the first simple module is not simple but has $\operatorname{End}_A(M) \cong k$.
More explicitly, let $$
A = \left\{ \begin{bmatrix} a & b \\ 0 & c \end{bmatrix} : a,b,c \in k \right\}, \qquad 
M = \left\{ \begin{bmatrix} x \\ y \end{bmatrix} : x,y \in k \right\}, \qquad 
N = \left\{ \begin{bmatrix} x \\ 0 \end{bmatrix} : y \in k \right\}$$
Then $M$ has exactly three submodules, $0$, $N$, and $M$, so $M$ is indecomposable. The general element of $A$ acts as multiplication by $a$ on $N$, and as multiplication by $c$ on $M/N$, so $N$ and $M/N$ are not isomorphic. Hence every endomorphism is 0 or an isomorphism, and so $B$ is a division ring. While it is not too hard to see $B\cong k$ already (especially if $k$ is algebraically closed), one can also give a direct calculation:
Consider $\operatorname{End}_k(M) = \left\{ \begin{bmatrix} d & e \\ f & g \end{bmatrix} : d,e,f,g \in k \right\}$ and its subring $B = \operatorname{End}_A(M) = \left\{ F \in \operatorname{End}_k(M) : FG=GF \text{ for every } G \in A \right\}$.  Finding $B$ is then just a matter of solving the equations $FG=GF$ for $d,e,f,g$. One gets $e=f=0$ by taking $a=1$ and $b=c=0$, and one gets $d=g$ by taking $a=c=0$ and $b=1$. Hence $$B = \left\{ \begin{bmatrix} d & 0 \\ 0 & d \end{bmatrix} : d \in k \right\} \cong k$$
