How to prove that $Hom_A(V,W)$ and $End_A(V)$ are noetherian Consider the following problem from  module theory assignment of mine:
Let A be a commutative ring.
(a) Let V be a finite A-module and W a noetherian  A-module. Then $Hom_A(V,W)$ is also noetherian.
Attempt: There exists a finite basis for V and W a noetherian A-module which means that any ascending chain of submodules terminates. If f  belongs to $Hom_A(V,W)$ , then f is a A-module homomorphism from $V\to W$ .
(b) Let V be an A-module which is noetherian. Then prove that $End_A V$ is a noetherian A-module.
I have to show that both $Hom_A(V,W)$ and $End_A V$ are finitely generated.
Attempt for (b) : I thought if {$x_1,...,x_n$} is a generating set  for V and $f\in End_A V$ then if $End_A V$ is a basis for V ? But it is not as f is not an isomorphism. Even if it were it is a generating set for V not $End_A V$.

Question : Can you please tell how should I construct a generating set for $End_A V$? I Would Like  to do (a) by myself after it.

Thanks!
 A: $\DeclareMathOperator\End{End}\DeclareMathOperator\Hom{Hom}$ I am not sure how to show that $\End_{A} V$ is finitely generated without first showing that it is Noetherian, which then implies it is finitely generated anyways. In any case, since you need to prove $\End_{A} V$ is a Noetherian $A$-module, it is not enough to show it is finitely generated (there are finitely generated modules which are not Noetherian).
So, I will sketch a proof that for a Noetherian $A$-module $V,$ $\End_{A} V$ is Noetherian. First, since $V$ is Noetherian, it is finitely generated, hence we can let $v_{1}, v_{2}, \ldots, v_{k}$ be a finite generating set for $V$ over $A$. For any $A$-module endomorphism $\varphi \colon V \to V,$ note that $\varphi$ is uniquely determined by the values it takes on the generating set: if you know the values $\varphi(v_{1}), \varphi(v_{2}), \ldots, \varphi(v_{k}),$ you can calculate $\varphi(v)$ for any $v \in V$.
Using this fact, we define a map $\Phi \colon \End_{A}(V) \to V^{k}$, which sends $\varphi \in \End_{A}(V)$ to $$(\varphi(v_{1}), \varphi(v_{2}), \ldots, \varphi(v_{k})) \in V^{k}.$$ You should check that $\Phi$ is a homomorphism of $A$-modules. The fact that each $\varphi \in \End_{A} V$ is uniquely determined by the values it takes on the generating set tells us that $\Phi$ is injective.
So, we have an injective $A$-module homomorphism $\Phi$ from $\End_{A}(V)$ to $V^{k}$, hence we can identify $\End_{A}(V)$ with a submodule of $V^{k}.$ Now, since $V^{k}$ is a direct sum of finitely many Noetherian modules, $V^{k}$ is Noetherian. So, $\End_{A}(V)$ is a submodule of a Noetherian module, hence itself is Noetherian. (Of course, since every submodule of a Noetherian module is finitely generated, $\End_{A}(V)$ is finitely generated. But we have shown something stronger.)
The proof for $\Hom_{A}(V, W)$ when $V$ is finite and $W$ is Noetherian is very similar, so I will leave that for you.
