# $A$ is a ring, $M$ a left noetherian $A$-module, $f:M \to M$ is a left homomorphism of modules. 2 part question. [duplicate]

$$A$$ 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 ).

• 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 $Ker(f) \subseteq Ker(f^2) \subseteq \dots$ – Matthias Klupsch May 15 '20 at 13:23
• 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 – rschwieb May 15 '20 at 13:35
• @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. – jeffery_the_wind May 15 '20 at 13:45