$A$ is a ring, $M$ a left noetherian $A$-module, $f:M \to M$ is a left homomorphism of modules.

Part A -- Suppose $Ker f^n=Ker f^{2n}$, then prove $Ker f^n \cap Im f^n = \{0\}$.

Suppose that $a \in Ker f^n$ and $a\ne0$. Then $f^n(a)= 0$. But since $Ker f^n = Ker f^{2n}$ then $f^{2n}(a)=f^n(f^n(a))=f^n(0)=0$, so $0\in ker f^n$ a contradiction. So, if $a \in Ker f^n$, then $a = 0$. Also\, $Im f^n$ must be a module, so must obey the criteria of an abelian group under addition, so $0\in Im f^n$.

Since $Ker f^n = {0}$, and $0\in Im f^n$, then $Ker f^n \cap Im f^n = \{0\}$.

This seems like a good proof to me, however it doesn't seem to use any properties of noetherian modules. To me it seems to be just a property of homomorphic maps in general.

Part B -- Suppose $f$ is surjective and $M$ is noetherian, then show $f$ is an isomorphism.

The only thing we need to show is that $f$ is injective. So, let's supposed that $f$ is not injective, that is, for $a,b \in M$, $f(a) = f(b)$. Its seems clear here that then if $M$ is finite, then the cardinality of $f(M)$ must be strictly less that $M$, so $f$ could not be surjective, a contradiction. The only remaining possibility is that $M$ is infinite.

However, here I have a problem, I am not finding the contradiction here. Again, I am not really seeing how to use the fact that $M$ is noetherian, ie. every chain of submodules terminates. I see an example of an infinite set that is a noetherian module ( the integers ).

  • 1
    $\begingroup$ In Part A, you say that $0 \in Ker(f^n)$ is a contradiction. What exactly do you think does this contradict? Also, you claim that this would imply $Ker(f^n) = \{0\}$ and I do not see how this follows. In fact, this is not true in general (take a nontrivial idempotent endomorphism of some vector space). To show that $Ker(f^n) \cap Im(f^n) = \{0 \}$, take an arbitrary element $ a \in Ker(f^n) \cap Im(f^n)$ and use the given assumptions to show that $a = 0$. You will not need $M$ to be noetherian for this. For the second part, you need it. Consider $Ker(f) \subseteq Ker(f^2) \subseteq \dots$ $\endgroup$ – Matthias Klupsch May 15 '20 at 13:23
  • 1
    $\begingroup$ Additional thing to link for the first question that I can't link as a dupe since people chose to answer in the comments: math.stackexchange.com/q/428760/29335 $\endgroup$ – rschwieb May 15 '20 at 13:35
  • $\begingroup$ @MatthiasKlupsch thanks for your help. I see me my mistake. I was trying to supposed that $0 \notin Ker f^n$, and the contradiction still follows from this, showing that $0 \in Ker f^n$, however the conclusion that $Ker f^n = \{0\}$ does not follow direction from that. $\endgroup$ – jeffery_the_wind May 15 '20 at 13:45