The Stacks Project


Tag 00IK

Chapter 10: Commutative Algebra > Section 10.50: More Noetherian rings

Lemma 10.50.1. Let $R$ be a Noetherian ring. Any finite $R$-module is of finite presentation. Any submodule of a finite $R$-module is finite. The ascending chain condition holds for $R$-submodules of a finite $R$-module.

Proof. We first show that any submodule $N$ of a finite $R$-module $M$ is finite. We do this by induction on the number of generators of $M$. If this number is $1$, then $N = J/I \subset M = R/I$ for some ideals $I \subset J \subset R$. Thus the definition of Noetherian implies the result. If the number of generators of $M$ is greater than $1$, then we can find a short exact sequence $0 \to M' \to M \to M'' \to 0$ where $M'$ and $M''$ have fewer generators. Note that setting $N' = M' \cap N$ and $N'' = \mathop{\mathrm{Im}}(N \to M'')$ gives a similar short exact sequence for $N$. Hence the result follows from the induction hypothesis since the number of generators of $N$ is at most the number of generators of $N'$ plus the number of generators of $N''$.

To show that $M$ is finitely presented just apply the previous result to the kernel of a presentation $R^n \to M$.

It is well known and easy to prove that the ascending chain condition for $R$-submodules of $M$ is equivalent to the condition that every submodule of $M$ is a finite $R$-module. We omit the proof. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 11561–11568 (see updates for more information).

    \begin{lemma}
    \label{lemma-Noetherian-basic}
    Let $R$ be a Noetherian ring.
    Any finite $R$-module is of finite presentation.
    Any submodule of a finite $R$-module is finite.
    The ascending chain condition holds for $R$-submodules
    of a finite $R$-module.
    \end{lemma}
    
    \begin{proof}
    We first show that any submodule $N$ of a finite $R$-module
    $M$ is finite. We do this by induction on the number of
    generators of $M$. If this number is $1$, then $N = J/I \subset
    M = R/I$ for some ideals $I \subset J \subset R$. Thus the definition
    of Noetherian implies the result. If the number of generators of
    $M$ is greater than $1$, then we can find a short exact sequence
    $0 \to M' \to M \to M'' \to 0$ where $M'$ and $M''$ have fewer
    generators. Note that setting $N' = M' \cap N$ and $N'' = \Im(N \to
    M'')$ gives a similar short exact sequence for $N$. Hence the result
    follows from the induction hypothesis
    since the number of generators of $N$ is at most the number of
    generators of $N'$ plus the number of generators of $N''$.
    
    \medskip\noindent
    To show that $M$ is finitely presented just apply the previous result
    to the kernel of a presentation $R^n \to M$.
    
    \medskip\noindent
    It is well known and easy to prove that the ascending chain condition for
    $R$-submodules of $M$ is equivalent to the condition that every submodule
    of $M$ is a finite $R$-module. We omit the proof.
    \end{proof}

    Comments (2)

    Comment #3028 by Brian Lawrence on December 13, 2017 a 3:01 am UTC

    Suggested slogan: Over a Noetherian ring: Any finite module is of finite presentation, any submodule of a finite module is finite, and the ascending chain condition holds for any finite module.

    Comment #3142 by Johan (site) on February 1, 2018 a 7:00 pm UTC

    Slogan ignored.

    Add a comment on tag 00IK

    Your email address will not be published. Required fields are marked.

    In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

    All contributions are licensed under the GNU Free Documentation License.




    In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?