Bound on nilpotency index of endomorphisms

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:

1. $M$ 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.
2. $A=K[x]/(x^d)$ 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 $y \mapsto x y$).

Solution 1:

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.

Solution 2:

Let $A$ be a (commutative) noetherian ring. Let $M$ be a finitely generated $A$-module. Then there exists $n\ge 1$ (depending on $M$) such that for any nilpotent endomorphism $T$ of $M$, we have $T^n=0$.

For the proof we proceed by noetherian induction. A trivial but important fact is that for any $A$-algebra $B$, $T$ induces canonically a nilpotent endomorphism of $M\otimes_A B$. We will use this fact mainly for $B$ a quotient or a localization of $A$.

We first make some observations.

(A) The statement is true if $A$ is reduced and $M$ is flat over $A$.

Let $n$ be the smallest number of generators of $M$, let $\mathfrak p_1, \dots, \mathfrak p_s$ be the minimal prime ideals of $A$ and let $K_i$ be the field $A_{\mathfrak p_i}=\mathrm{Frac}(A/\mathfrak p_i)$. Then $M\otimes_A K_i$ is a vector space over $K_i$ of dimension $\le n$. Let $T_i$ be the endomorphism of $M\otimes_A K_i$ induced by $T$. Then $T_i^n=0$. As $M$ is flat over $A$, the canonical injection $$ A\to \prod_{i} A/\mathfrak p_i\to \prod_i K_i $$ induces an injection $$ M\to \prod_i M\otimes_A K_i.$$ This implies that $T^n=0$.

(B) Let $I=\sqrt{0_A}$ be the nilradical of $A$. If the statement is true for the $A$-module $M/IM$, then it is true for $M$.

Let $n_1$ be such that the $n_1$-th power of any nilpotent endomorphism of $M/IM$ vanishes. Let $r\ge 1$ be such that $I^r=0$. Let $T$ be a nilpotent endomorphism of $M$. Then $T^{n_1}(M)\subseteq IM.$ As $T$ is $A$-linear, for any positive integer $m$, we have $(T^{n_1})^m(M)\subseteq I^mM$. This implies that $T^{n_1r}=0$.

(C) Let $f\in A$. Then there exists $m\ge 1$ such that the kernel of $M\to M_f$ is anihilated by $f^m$.

Consider the increasing sequence of sub-modules $M_k:=\{ x\in M \mid f^kx=0\}$ of $M$. Note that their union is the kernel of $M\to M_f$. As $M$ is noetherian, this sequence is stationnary at some $m$. So $f^m(\cup_{k\ge 1} M_k)=0$.

Now we prove the main statement. We consider the set $S$ of the radical ideals $I$ (i.e. $I=\sqrt{I}$) of $A$ such that the statement is true for the $A$-module $M/IM$. This set contains the unit ideal so is non-empty. We are going to show $\sqrt{0_A}\in S$. This will imply the main statement by Observation (B) above.

Let $I$ be a radical ideal of $A$ such that any radical ideal containing properly $I$ belongs to $S$. We are going to show that $I\in S$. By the noetherian induction principle on $\mathrm{Spec}(A)$, this will imply that $\sqrt{0_A}\in S$.

Let $B=A/I$ and let $N=M/IM$. By the generic flatness theorem (see the reference to EGA for the case of reduced rings) applied to $N$ on $B$, there exists a non-zero $f\in B$ such that $N_f$ is flat over $B_f$. Let $n$ be the smallest number of generators of $M$. Then $N_f$ is generated by $n$ elements. By (A), the $n$-th power any nilpotent endomorphism of $N_f$ is zero. Let $T_1$ be a nilpotent endomorphism of $N$. Then $T_1^n(N)$ is contained in the kernel of $N\to N_f$. By (C), there exists $m\ge 1$ (depending only on $f$ and $N=M/IM$) such that $$f^m(T_1^n(N))=0.$$ As $V(f)$ is a proper subset of $V(I)$, by noetherian induction hypothesis, there exists $r\ge 1$ such that for all $T_1$, $T_1^r(N)\subseteq \sqrt{fB}N$. As in (B), we can increase $r$ so that for all $T$, we have $T_1^r(N)\subseteq fN$. Thus $$T_1^{mnr}(N)\subseteq f^m T_1^{n}(N)=0.$$ So $I\in S$ and we are done.

Solution 3:

This is not a complete answer, but covers both examples where the modules have finite length ($n$, respectively $d$).

Consider $M$ of finite length $n$. Then, by Fitting's Lemma, for every $T\in\operatorname{End}_A(M)$ we have $M=\operatorname{Im}T^n\oplus\ker T^n$. If $T$ is nilpotent, then there exists $m\ge 1$ such that $T^m=0$. If $m\le n$, then $T^n=0$. If $m>n$, then we also have $M=\operatorname{Im}T^m\oplus\ker T^m$. Since $T^m=0$ it follows that $M=\ker T^m=\ker T^n$, thus $\operatorname{Im}T^n=0$, i.e. $T^n=0$.

Remark. The proof of Fitting's Lemma goes like this: Assume that $\ker T\neq 0$. Then $\ker T\subseteq\ker T^2\subseteq\cdots$. Suppose the following inclusions $\ker T\subset \cdots\subset\ker T^i$ are strict, and $\ker T^i=\ker T^{i+1}$. Now we can deduce two things: the length of $\ker T^i$ is at least $i$, so $i\le n$, and $\ker T^{i+1}=\ker T^{i+2}=\cdots$. On the other side, $\operatorname{Im}T\supseteq\operatorname{Im}T^2\supseteq\cdots$. Suppose the following inclusions $\operatorname{Im}T\supset\cdots\supset\operatorname{Im}T^j$ are strict, and $\operatorname{Im}T^j=\operatorname{Im}T^{j+1}$. As before we deduce that $j\le n$ and $\operatorname{Im}T^j=\operatorname{Im}T^{j+1}=\cdots$. Now for any $k\ge\max(i,j)$ we get $M=\operatorname{Im}T^k\oplus\ker T^k$.

(I've added this proof since the usual proofs of Fitting's Lemma do not say it explicitely that the equality holds for any $k\ge n$, where $n$ is the length of $M$.)