# $A$ noetherian, $A$-endomorphism not injective for all invariant submodules is nilpotent

Let $A$ be a noetherian ring, $M$ a finitely generated $A$-module, $T: M \to M$ an endomorphism. Assume that for all $T$-invariant proper submodules $N$ of $M$, the induced endomorphism $\overline T : M/N \to M/N$ is not injective. Show that there is $n \in \mathbb{N}$ such that $T^n=0$.

My approaches:

The condition on being $T$-invariant immediately suggests using Nakayama's lemma: $T(N) \subseteq N \subseteq A \cdot N$, so T satisfies a monic polynomial with coefficients in $A$. (By proposition 2.4 in Atiyah-Macdonald.) However, in this phrasing we don't get to use the different choices of N or the noetherian assumption.

Another natural approach is to look at the ascending chain of ideals $\ker(T^i)$ which must stabilize. Assuming by contrary that $T^n$ is always nonzero, these are all proper ideals and hence the stabilized ideal is contained in a maximal ideal $m$ which contains all of the chain. This too seems not to use the different choices of N.

The condition on $N$, stated explicitly, says that $\exists m \not \in N: T(m) \in N$.

M is finitely generated, so it is enough to show that for each generator $m$ of $M$, $\exists n \in \mathbb{N}: T^n(m) =0$.

How do we 'connect the dots'?

$M$ is a finitely generated $A[X]$-module via $T$, and $T$-invariant submodules correspond to the $A[X]$-submodules of $M$. Then the multiplication by $X$ on $M/N$ is not injective for every proper $A[X]$-submodule $N$ of $M$. You want to show that $X^nM=0$ for some $n\ge1$.
Let $M$ be a noetherian $R$-module, and $x\in R$. If $M/N\stackrel{x\cdot}\to M/N$ is not injective for every proper submodule $N$ of $M$, then there is $n\ge1$ such that $x^nM=0$.
Since $M$ is noetherian and $\ker(x\cdot)\subseteq\ker(x^2\cdot)\subseteq\cdots$, there is $n\ge1$ such that $\ker(x^n\cdot)=\ker(x^{n+1}\cdot)=\cdots$. If $N:=\ker(x^n\cdot)\ne M$, then the map $M/N\stackrel{x\cdot}\to M/N$ is not injective, so there is $m\in M$, $m\notin\ker(x^n\cdot)$ such that $xm\in\ker(x^n\cdot)\Rightarrow x^{n+1}m=0$ and thus $m\in\ker(x^{n+1}\cdot)=\ker(x^n\cdot)$, a contradiction.
• $x^k\cdot$ stands for the multiplication by $x^k$. – user26857 Jan 18 '16 at 13:00
$0$ is an invariant module so there exists $x$ such that $T(x)=0$, construct the sequence of invariant submodule $N_m$ such that $N_1$ is the module generated by $x$ suppose defined $N_m$, since the map induced by $T$ on $M/N_m$ is not injective, if $N_m\neq M$ there exists $y_m\in M$ such that $y_m\neq N_m$ and $T(y_m)\in N_m$. We denote by $N_{m+1}$ the submodule generated by $N_m$ and $y_m$, it is invariant by $T$. Since $A$ is Noetherian and $M$ is finitely generated there exists $m_0$ such that $N_m=N_{m+1}=M$ for $m\geq m_0$. This implies that $T^{m_0+1}=0$ since the image of $T^i$ is contained in $N_{m_0-i}$.