Let $A$ be a ring (not necessarily commutative). If it helps, one can assume that $A$ is right hereditary (i.e. every submodule of a projective right $A$-module is projective). Let $M$ be a finitely generated module and $B=\text{End}(M)$. Is it true that $B$ is Noetherian?
I think this is true for commutative $A$. Then $B$ would be a Noetherian $A$-module, and since any ideal in $B$ could be considered an $A$-module, the fact that $B$ is a Noetherian $A$-module would imply that $B$ is a Noetherian $B$-module. That is, a Noetherian ring.
But is this true in the non-commutative case?