TITLE: Bound on nilpotency index of endomorphisms
QUESTION [20 upvotes]: Let $A$ be a Noetherian ring (commutative with $1$) and $M$ a finitely generated $A$-module. I want to show that there exists a bound $n$ such that for every nilpotent endomorphism $T : M \to M$ we have $T^n = 0$.

There are two examples in mind:

$M$ a module over a field $K$, that is $M = K^n$ a finite-dimensional vector space. In that case, $n$ is the required upper bound.
$A=K[x]/(x^d)$ as a module over itself. In that case the dimension is $1$ (a basis is $\{ 1 \}$) but the upper bound is $d$ (consider the endomorphism $y \mapsto x y$).

REPLY [5 votes]: Here is the quicker proof I alluded to. I write in a separated answer because the first one is already fairly long and becomes hard to edit and has its own interest I think. We also need the remark in that answer on the fact that any nilpotent endomorphism remains nilpotent after ring extensions (base change). 
Let $T$ be a nilpotent endomorphism of $M$. Let $\mathfrak p_1, \dots, \mathfrak p_r$ be the minimal prime ideals of $J:=\mathrm{Ann}M$. They represent the generic points of $V(J)=\mathrm{Supp}(M)$. So the localizations $M_{\mathfrak p_i}=M\otimes_A A_{\mathfrak p_i}$ are Artinian and noetherian over $A_{\mathfrak p_i}$ because $M=M/JM$, $M_{\mathfrak p_i}=(M/JM)_{\bar{\mathfrak p}_i}$ where $\bar{\mathfrak p}_i$ is the image of $\mathfrak p_i$ in $A/J$ and $(A/J)_{\bar{\mathfrak p}_i}$ is Artinian and noetherian. By YACP's anwer, $T_{\mathfrak p_i}^n=0$ where $n$ is the maximum of the lengths $\mathrm{length}(M_{\mathfrak p_i})$. 
Consider the canonical map 
$$ M\to \prod_{1\le i\le r} M_{\mathfrak p_i}$$ 
and its kernel $N$. If there exists $n'\ge 1$ such that all nilpotent endomorphisms of $N$ vanish at power $n'$, then the same is true for $M$ and the power $n'+n$. As $N_{\mathfrak p_i}=0$ for all $i$ by construction, the support of $N$ is strictly contained in that of $M$ (in fact $\mathrm{Supp}(N)$ has positive codimension in $\mathrm{Supp}(M)$). We then use noetherian induction on $\mathrm{Supp}(M)$ (the initial case is when the support is empty, so $M=0$ and the property is trivially satisfied). 
Edit On the noetherian induction. Rather than use induction on the noetherian topological space $\mathrm{Spec}(A)$, one can use the very definition of noetherian rings. Namely, consider the set of ideals $I$ such that there is a finitely generated $A$-module $M$ such that $I=\mathrm{Ann}(M)$ and $M$ doesn't satisfy the bounded nilpotence index property. This set of ideals, if non-empty, admits a maximal (for the inclusion relation) element $I$. Let $M$ be a bad $A$-module corresponding to $I$. Then the $N$ in the above proof has $\mathrm{Ann}(N)$ containing strictly $I$. So $N$ has the bounded nilpotence index property. We conclude as above that $M$ also has this property. Contradiction. So the set of bad ideals is empty.