Noetherian, finitely generated R-module

**poincare4223** · April 29th 2009, 08:57 PM

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.

**NonCommAlg** · April 30th 2009, 04:30 AM

Originally Posted by poincare4223

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.)

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