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The theorem is that if we have a finite length module M (Noetherian and Artinian), and a map f is endomorphism. Then, we can decompose M = Ker($f^n$) $\oplus$ Im($f^n$).

I understand the proof is that we use the idea: finite length property of M makes ascending chain of Ker($f^i$) stabilized, and the descending chain of Im($f^i$) stabilized.

I got a stupid question, why not use the first isomorphism thm directly, since f is R-linear, a module homomorphism, so M/Ker($f^n$)=Im($f^n$), so we can get M = Ker($f^n$)$\oplus$Im($f^n$). I know there must be some thing wrong......please help.

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    $\begingroup$ Note that $M/N\cong K$ doesn't imply $M\cong N\oplus K$ in general (eg let $M$ be $k[x]/(x^2)$ as a module over itself and $N$ be the unique 1 dimensional ideal). Also if $A$ and $B$ are submodules of $M$, then saying $M=A\oplus B$ (an "internal" direct sum) is a bit stronger than $M\cong A\oplus B$ (an "external" direct sum). $\endgroup$ – stewbasic Sep 23 '16 at 2:00
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Hint: Suppose $f$ is nilpotent $f^2=0, f\neq 0, Im(f)\subset Ker f$, so the result is not true for every n,but for $n$ for which $Ker f^{n+1}=Ker f^n$ and $Im f^{n+1}=Imf^n$ which existence is a consequence of the hypothesis.

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  • $\begingroup$ you mean im(f) is in ker(f^2)? but it is not in ker(f), so i am thinking the 1st isomorphism theorem still holds for M/ker(f)=im(f) and M/ker(f^2)=im(f^2). so we still have M = Ker($f^i$)⊕Im($f^i$). thanks for your reply $\endgroup$ – Zhang Rita Sep 23 '16 at 0:12
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    $\begingroup$ $f^2(x)=f(f(x))=0$, so $f^2=0$ implies $f(x)\in Ker f$. $\endgroup$ – Tsemo Aristide Sep 23 '16 at 0:13
  • $\begingroup$ I see!! thanks!! $\endgroup$ – Zhang Rita Sep 23 '16 at 2:39

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