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I'm trying to solve this question:

I didn't understand why the hint is true and how to apply it. I really need help, because it's my first question on this subject and my experience on this field is zero.

I need some help.

Thanks a lot

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  • $\begingroup$ Already proved here. $\endgroup$ – user89712 Nov 17 '13 at 21:36
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For any $m\in M$, if you have $u(m)=0$, then you have $$(u\circ u)(m)=u(u(m))=u(0)=0.$$ Therefore $\ker(u)\subseteq\ker(u\circ u)$. By the same argument, you have $$\ker(u)\subseteq\ker(u\circ u)\subseteq\ker(u\circ u\circ u)\subseteq\cdots$$ which should look familiar if you have just read a chapter on Noetherian modules.

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