Noetherian submodules and isomorphisms

Question

Suppose $M$ is an $A$-Module, and $N$ is a submodule of $M$. Let $f:N\to M$ be an $A$-module epimorphism. How could I show that if $N$ is noetherian, then $f$ is an isomorphism?

Thanks

Answer 1

A partial answer for the case M=N:

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Let $x\in \ker f$. Consider $\ker f \subseteq \ker f^2 \subseteq \dotsb$ . Since $N$ is noetherian, for some $t$ we have $\ker f^t = \ker f^{t+1}$. Since $f^t$ is surjective (because f is surjective), there exists $y\in M$ such that $f^t(y)=x$. So $f^{t+1}(y)=f(x)=0$. Hence $y\in \ker f^{t+1}=\ker f^t$. So $x=f^t(y)=0$.

Answer 2

For the general case, where $N$ and $M$ are not necessarily equal, suppose $f$ is not injective, and let $K_0=\ker(f)$ and $L_0=N$. Then we have inclusions of modules $$0<K_0<L_0=N\leq M,$$ with $L_0/K_0\cong M$.

Submodules of $L_0$ containing $K_0$ correspond to submodules of $L_0/K_0\cong M$. More precisely, there is an inclusion-preserving bijection from the set of submodules of $L_0$ containing $K_0$ to the set of submodules of $M$ given by $V\mapsto f(V)$.

Let $K_1$ and $L_1$ be those corresponding to $K_0$ and $L_0$: i.e., $K_1=f^{-1}(K_0)$ and $L_1=f^{-1}(L_0)$. Then we have $$0<K_0<K_1<L_1\leq L_0=N\leq M,$$ with $L_1/K_1\cong M$.

Repeating, we get an infinite ascending chain $$0<K_0<K_1<\dots$$ of submodules of $N$, contradicting the assumption that $N$ is Noetherian.