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Math Help - Noetherian, finitely generated R-module

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    Noetherian, finitely generated R-module

    I need help proving the following:

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

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

    The first problem seems to be straightforward. Can I just take the generators and multiply them by 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 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 R be a Noetherian ring, I an ideal, and N \subseteq M be R-modules. Let \frac{M}{N} be a finitely generated R-module. Prove that \frac{IM}{IN} is a finitely generated R-module.
    since R is Noetherian, I is finitely generated. let I=\sum_{i=1}^r Ra_i and \frac{M}{N}=\sum_{i=1}^s R(m_i + N). it's easy to see that \frac{IM}{IN}=\sum_{i,j}R(a_im_j + IN).


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

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

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

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


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