In the last Algebraic Number Theory lecture we were proving that a Noetherian domain $A$ is a DVR iff it is intergrally closed and it has a unique non-zero prime ideal. At some point the lecturer used an argument that if $A$ is Noetherian, then $\textrm{End}_A(\mathfrak{p})$ is Noetherian as an $A$-module (where $\mathfrak{p}$ is the unique non-zero prime ideal).
- Why is what I described above true? (If it is.)
- Is a more general statement true: if we exchange $\mathfrak{p}$ with any Noetherian module $M$, will $\textrm{End}_A(M)$ be Noetherian as an $A$-module?
More generally, if $A$ is a commutative ring and $M$ and $N$ are $A$-modules with $M$ finitely generated and $N$ Noetherian, then $\operatorname{Hom}_A(M,N)$ is a Noetherian $A$-module. Indeed, if $M$ is generated by $x_1,\dots,x_n$, then $\operatorname{Hom}_A(M,N)$ is isomorphic to a submodule of $N^n$, by mapping a homomorphism $f$ to $(f(x_1),\dots,f(x_n))$. Since $N$ is Noetherian, so is any submodule of $N^n$, so $\operatorname{Hom}_A(M,N)$ is Noetherian.