Endomorphism ring of an irreducible module is irreducible Let $M$ be an $A$-module and define $B:=End_A (M)$. Prove that if $M_A$ is irreducible (that is, the only decomposition of $M_A$ in direct sumands is the trivial decomposition) then $B_{B}$ is irreducible.
I have no idea on how to tackle this. I tried this.
Consider $N\leq B_B$ a direct summand of $B_B$, then there exists a projection $\pi:B_B\rightarrow B_B$ such that $\pi^2=\pi$ and $\pi (B_B)=N$. I want to prove that $\pi=0$ or $\pi=1_{B_B}$. 

You’re on the right track.
From where you left off, $\pi(M)\oplus (1-\pi)(M)=M$ is a direct decomposition of $M$.  Since one must be zero, you have $\pi=1$ or $\pi=0$.
