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Thread: Noetherian, finitely generated R-module

  1. #1
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    Noetherian, finitely generated R-module

    I need help proving the following:

    1. Let $\displaystyle R$ be a Noetherian ring, $\displaystyle I$ an ideal, and $\displaystyle N \subseteq M$ be $\displaystyle R$-modules. Let $\displaystyle \frac{M}{N}$ be a finitely generated $\displaystyle R$-module. Prove that $\displaystyle \frac{IM}{IN}$ is a finitely generated $\displaystyle R$-module.

    2. Let $\displaystyle R$ be a Noetherian ring, $\displaystyle I$ an ideal of R. Prove that $\displaystyle M$ is a finitely generated $\displaystyle R$-module iff $\displaystyle \frac{M}{\text{Nil}(R)M}$ is a finitely generated $\displaystyle \frac{R}{\text{Nil}(R)}$-module.

    The first problem seems to be straightforward. Can I just take the generators and multiply them by $\displaystyle I$ to show this is finitely generated? Thanks for any suggestions with this one.

    As for the second one, I am not seeing how to use the first part in this problem. I am guessing that I just let $\displaystyle I=\text{Nil}(R)$ for one direction; but the other direction is confusing me.
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  2. #2
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    Quote Originally Posted by poincare4223 View Post
    I need help proving the following:

    1. Let $\displaystyle R$ be a Noetherian ring, $\displaystyle I$ an ideal, and $\displaystyle N \subseteq M$ be $\displaystyle R$-modules. Let $\displaystyle \frac{M}{N}$ be a finitely generated $\displaystyle R$-module. Prove that $\displaystyle \frac{IM}{IN}$ is a finitely generated $\displaystyle R$-module.
    since $\displaystyle R$ is Noetherian, $\displaystyle I$ is finitely generated. let $\displaystyle I=\sum_{i=1}^r Ra_i$ and $\displaystyle \frac{M}{N}=\sum_{i=1}^s R(m_i + N).$ it's easy to see that $\displaystyle \frac{IM}{IN}=\sum_{i,j}R(a_im_j + IN).$


    2. Let $\displaystyle R$ be a Noetherian ring and let $\displaystyle \color{red}I={\text{Nil}(R)}.$ Prove that $\displaystyle M$ is a finitely generated $\displaystyle R$-module iff $\displaystyle \frac{M}{IM}$ is a finitely generated $\displaystyle \frac{R}{I}$-module.
    if $\displaystyle M$ is generated by $\displaystyle m_1, \cdots , m_k$ as $\displaystyle R$ module, then $\displaystyle \frac{M}{IM}$ is obviously generated by $\displaystyle m_1 + IM, \cdots , m_k + IM$ as $\displaystyle \frac{R}{I}$ module. this is true for any ideal $\displaystyle I$ of any ring $\displaystyle R.$

    conversely: since $\displaystyle R$ is Noetherian, $\displaystyle I={\text{Nil}(R)}$ is nilpotent, i.e. $\displaystyle I^n=(0),$ for some $\displaystyle n.$ now $\displaystyle \frac{M}{IM}$ is finitely generated $\displaystyle R$ module because it's finitely generated $\displaystyle \frac{R}{I}$ module.

    applying part 1) of the problem gives us that $\displaystyle \frac{I^{n-2}M}{I^{n-1}M}$ and $\displaystyle \frac{I^{n-1}M}{I^nM} \cong I^{n-1}M$ are both finitely generated (and hence Noetherian) $\displaystyle R$ module. thus $\displaystyle I^{n-2}M$ is also Noetherian.

    again $\displaystyle \frac{I^{n-3}M}{I^{n-2}M}$ and $\displaystyle I^{n-2}M$ are Noetherian and so $\displaystyle I^{n-3}M$ is Noetherian too. continuing this we'll get that $\displaystyle M=I^{n-n}M$ is Noetherian (and so finitely generated) $\displaystyle R$ module.


    (recall that a finitely generated module over a Noetherian ring is Noetherian and a module $\displaystyle A$ is Noetherian iff for any submodule $\displaystyle B$ of $\displaystyle A,$ both $\displaystyle B$ and $\displaystyle \frac{A}{B}$ are Noetherian.)
    Last edited by NonCommAlg; Apr 30th 2009 at 04:40 AM.
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