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If $I$ is a finitely generated ideal of a commutative ring $R$ with $1$ such that $I^n = \{0\}$ and $R/I^{n-1}$ is noetherian, then $R$ is also noetherian.

I don't know what I should do. If I can prove for example that $I^{n-1}$ is finitely generated as $R/I^{n-1}$ module then I can deduce that it is noetherian as $R$-module and then I am done. But I don't know if that is true and how to prove it.

Thanks.

Added: also isn't it possible to prove that $I^{n-1}$ is finitely generated without requiring $I^n = \{0\}$? (just realized that)

I mean I can prove that if $I$ and $J$ are finitely generated ideals then $IJ$ is also finitely generated so by induction if $I_1,...,I_k$ are finitely generated then $I_1...I_k$ is so. So $I^{n-1}$ is finitely generated as $R/I^{n-1}$ module so it is noetherian as $R$-module so $R$ is then noetherian. Is there something wrong with this?

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    $\begingroup$ Yes, you can prove that $I^{n-1}$ is finitely generated without knowing $I^n=0$ (actually you don't need this at all!): in fact, every power of a finitely generated ideal is finitely generated. But you have a flaw in the last reasoning: this doesn't entail that $I^{n-1}$ is an $R/I^{n-1}$-module. For this you need $I^n=0$. $\endgroup$
    – user26857
    Commented May 15, 2016 at 10:19

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Since $I^n=0$ the ideal $I^{n-1}$ is a finitely generated $R/I^{n-1}$-module: it is finitely generated (any power of a finitely generated ideal is finitely generated) and $I^{n-1}\cdot I^{n-1}=0$; see also here. Now use the exact sequence of $R$-modules $$0\to I^{n-1}\to R\to R/I^{n-1}\to 0.$$

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  • $\begingroup$ sorry but I am little bit confused. I want to prove that $I^{n-1}$ is a finitely generated $R/I^{n-1}$ module to conclude that it is noetherian. I don't understand the bit with exact sequences (newbie here) $\endgroup$
    – user290300
    Commented May 15, 2016 at 10:09
  • $\begingroup$ Can you finish the proof knowing that $I^{n-1}$ is a finitely generated $R/I^{n-1}$-module (and without using exact sequences)? $\endgroup$
    – user26857
    Commented May 15, 2016 at 10:15
  • $\begingroup$ yes if I know that $I^{n-1}$ is a finitely generated $R/I^{n-1}$ module then it is a noetherian $R/I^{n-1}$ module so because $I^{n-1}$ annihilates itself I can say that $I^{n-1}$ is a noetherian $R$ module so finally from "if $N$ is a submodule of $M$ then $M$ is noetherian if and only if $N$ and $M/N$ are noetherian" I get that $R$ is a noetherian ring $\endgroup$
    – user290300
    Commented May 15, 2016 at 10:20
  • $\begingroup$ thanks a lot. things are much more clear now! $\endgroup$
    – user290300
    Commented May 15, 2016 at 10:27

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