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Thread: Noetherian ring, finitely generated module

  1. #1
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    Noetherian ring, finitely generated module

    I need help with these two questions:

    1. Let $\displaystyle R$ be a Noetherian ring, $\displaystyle I$ an ideal, and $\displaystyle N \subseteq M$ be $\displaystyle R$-modules. Suppose $\displaystyle R$ is reduced and $\displaystyle P_1, \ldots, P_n$ are the minimal primes of $\displaystyle R$. Prove that $\displaystyle M$ is a finitely generated $\displaystyle R$-module iff $\displaystyle \frac{M}{P_iM}$ is a finitely generated $\displaystyle \frac{R}{P_i}$-module for each $\displaystyle i=1, \ldots, n$.

    2. Now suppose $\displaystyle R$ is Artinian (need not be reduced). Prove that $\displaystyle M$ is Noetherian iff $\displaystyle M$ is Artinian.

    I know how to do #1 $\displaystyle (\Rightarrow)$. But, I don't see how to do #1 $\displaystyle (\Leftarrow)$. Also, for #2, I am stuck on both implications right now. Thanks in advance.
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  2. #2
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    Quote Originally Posted by xianghu324 View Post
    I need help with these two questions:

    1. Let $\displaystyle R$ be a Noetherian ring, $\displaystyle I$ an ideal, and $\displaystyle N \subseteq M$ be $\displaystyle R$-modules. Suppose $\displaystyle R$ is reduced and $\displaystyle P_1, \ldots, P_n$ are the minimal primes of $\displaystyle R$. Prove that $\displaystyle M$ is a finitely generated $\displaystyle R$-module iff $\displaystyle \frac{M}{P_iM}$ is a finitely generated $\displaystyle \frac{R}{P_i}$-module for each $\displaystyle i=1, \ldots, n$.
    since $\displaystyle R$ is reduced, the nilradical of $\displaystyle R$ is (0) and thus $\displaystyle P_1 P_2 \cdots P_n=(0).$ now see my solution to part 2) of the problem in this thread: http://www.mathhelpforum.com/math-he...-r-module.html


    2. Now suppose $\displaystyle R$ is Artinian (need not be reduced). Prove that $\displaystyle M$ is Noetherian iff $\displaystyle M$ is Artinian.
    it'd help if instead of just posting your problem, you'd also tell us what you know! for example, do you know that every Artinian ring is Noetherian? (Hopkins-Levitzki theorem) or do you know

    about semisimple rings or composition series?
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    since $\displaystyle R$ is reduced, the nilradical of $\displaystyle R$ is (0) and thus $\displaystyle P_1 P_2 \cdots P_n=(0).$ now see my solution to part 2) of the problem in this thread: http://www.mathhelpforum.com/math-he...-r-module.html



    it'd help if instead of just posting your problem, you'd also tell us what you know! for example, do you know that every Artinian ring is Noetherian? (Hopkins-Levitzki theorem) or do you know

    about semisimple rings or composition series?
    Hi NonCommAlg,

    We have not covered semi-simple rings yet. However, I do know that:

    $\displaystyle M$ has a comp series $\displaystyle \Leftrightarrow$ $\displaystyle M$ is Artinian and Noetherian.

    $\displaystyle R$ is Artin ring $\displaystyle \Leftrightarrow$ $\displaystyle R$ is Noetherian and $\displaystyle \text{dim} (R)=0$.

    $\displaystyle R$ is Artin ring $\displaystyle \Leftrightarrow$ $\displaystyle R$ is Noetherian and each prime ideal is maximal.


    Using comp series seems the best way to go. But I am not seeing how to use a comp series on $\displaystyle M$, as we have much info on M right now.
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  4. #4
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    Quote Originally Posted by xianghu324 View Post

    2. Now suppose $\displaystyle R$ is Artinian (need not be reduced). Prove that $\displaystyle M$ is Noetherian iff $\displaystyle M$ is Artinian.
    if $\displaystyle M$ is Noetherian, then it's finitely generated and we know that a finitely generated module over an Artinian ring is Artinian. conversely, suppose $\displaystyle M$ is Artinian. let $\displaystyle \overline{R}=\frac{R}{\text{Nil}(R)}.$

    then $\displaystyle \overline{R}$ is a reduced Noetherian ring (because every Artinian ring is Noetherian). we know that $\displaystyle R$ has finitely many primes and every prime is maximal (so all primes are minimal).

    let $\displaystyle P_1, \cdots , P_n$ be the prime ideals of $\displaystyle R.$ let $\displaystyle \overline{P_j}=\frac{P_j}{\text{Nil}(R)}.$ then $\displaystyle \overline{P_j}$ are the (minimal) primes of $\displaystyle \overline{R}.$ let $\displaystyle \overline{M}=\frac{M}{\text{Nil}(R)M}.$ then $\displaystyle \overline{M}$ is an $\displaystyle \overline{R}$ module and $\displaystyle \overline{M_j}=\frac{\overline{M}}{\overline{P_j} \ \overline{M}}$ is a Noetherian

    $\displaystyle R_j=\frac{\overline{R}}{\overline{P_j}}$ module because $\displaystyle R_j$ is a field and $\displaystyle \overline{M_j}$ is an Artinian $\displaystyle R_j$ module and we know that over a field "Artinian" and "Noetherian" are equivalent. thus $\displaystyle \overline{M_j}$ is finitely generated $\displaystyle R_j$

    module and so by part 1) of your problem, $\displaystyle \overline{M}$ is finitely generated $\displaystyle \overline{R}$ module. so by the link i already gave you, $\displaystyle M$ is finitely generated $\displaystyle R$ module and hence Noetherian, since $\displaystyle R$

    is Noetherian.
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