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Tag 0519

Chapter 10: Commutative Algebra > Section 10.5: Finite modules and finitely presented modules

Lemma 10.5.3. Let $R$ be a ring. Let $$ 0 \to M_1 \to M_2 \to M_3 \to 0 $$ be a short exact sequence of $R$-modules.

  1. If $M_1$ and $M_3$ are finite $R$-modules, then $M_2$ is a finite $R$-module.
  2. If $M_1$ and $M_3$ are finitely presented $R$-modules, then $M_2$ is a finitely presented $R$-module.
  3. If $M_2$ is a finite $R$-module, then $M_3$ is a finite $R$-module.
  4. If $M_2$ is a finitely presented $R$-module and $M_1$ is a finite $R$-module, then $M_3$ is a finitely presented $R$-module.
  5. If $M_3$ is a finitely presented $R$-module and $M_2$ is a finite $R$-module, then $M_1$ is a finite $R$-module.

Proof. Proof of (1). If $x_1, \ldots, x_n$ are generators of $M_1$ and $y_1, \ldots, y_m \in M_2$ are elements whose images in $M_3$ are generators of $M_3$, then $x_1, \ldots, x_n, y_1, \ldots, y_m$ generate $M_2$.

Part (3) is immediate from the definition.

Proof of (5). Assume $M_3$ is finitely presented and $M_2$ finite. Choose a presentation $$ R^{\oplus m} \to R^{\oplus n} \to M_3 \to 0 $$ By Lemma 10.5.2 there exists a map $R^{\oplus n} \to M_2$ such that the solid diagram $$ \xymatrix{ & R^{\oplus m} \ar[r] \ar@{..>}[d] & R^{\oplus n} \ar[r] \ar[d] & M_3 \ar[r] \ar[d]^{\text{id}} & 0 \\ 0 \ar[r] & M_1 \ar[r] & M_2 \ar[r] & M_3 \ar[r] & 0 } $$ commutes. This produces the dotted arrow. By the snake lemma (Lemma 10.4.1) we see that we get an isomorphism $$ \text{Coker}(R^{\oplus m} \to M_1) \cong \text{Coker}(R^{\oplus n} \to M_2) $$ In particular we conclude that $\text{Coker}(R^{\oplus m} \to M_1)$ is a finite $R$-module. Since $\text{Im}(R^{\oplus m} \to M_1)$ is finite by (3), we see that $M_1$ is finite by part (1).

Proof of (4). Assume $M_2$ is finitely presented and $M_1$ is finite. Choose a presentation $R^{\oplus m} \to R^{\oplus n} \to M_2 \to 0$. Choose a surjection $R^{\oplus k} \to M_1$. By Lemma 10.5.2 there exists a factorization $R^{\oplus k} \to R^{\oplus n} \to M_2$ of the composition $R^{\oplus k} \to M_1 \to M_2$. Then $R^{\oplus k + m} \to R^{\oplus n} \to M_3 \to 0$ is a presentation.

Proof of (2). Assume that $M_1$ and $M_3$ are finitely presented. The argument in the proof of part (1) produces a commutative diagram $$ \xymatrix{ 0 \ar[r] & R^{\oplus n} \ar[d] \ar[r] & R^{\oplus n + m} \ar[d] \ar[r] & R^{\oplus m} \ar[d] \ar[r] & 0 \\ 0 \ar[r] & M_1 \ar[r] & M_2 \ar[r] & M_3 \ar[r] & 0 } $$ with surjective vertical arrows. By the snake lemma we obtain a short exact sequence $$ 0 \to \text{Ker}(R^{\oplus n} \to M_1) \to \text{Ker}(R^{\oplus n + m} \to M_2) \to \text{Ker}(R^{\oplus m} \to M_3) \to 0 $$ By part (5) we see that the outer two modules are finite. Hence the middle one is finite too. By (4) we see that $M_2$ is of finite presentation. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 381–400 (see updates for more information).

    \begin{lemma}
    \label{lemma-extension}
    Let $R$ be a ring.
    Let
    $$
    0 \to M_1 \to M_2 \to M_3 \to 0
    $$
    be a short exact sequence of $R$-modules.
    \begin{enumerate}
    \item If $M_1$ and $M_3$ are finite $R$-modules, then $M_2$ is a finite
    $R$-module.
    \item If $M_1$ and $M_3$ are finitely presented $R$-modules, then $M_2$
    is a finitely presented $R$-module.
    \item If $M_2$ is a finite $R$-module, then $M_3$ is a finite $R$-module.
    \item If $M_2$ is a finitely presented $R$-module and $M_1$ is a
    finite $R$-module, then $M_3$ is a finitely presented $R$-module.
    \item If $M_3$ is a finitely presented $R$-module and $M_2$ is a finite
    $R$-module, then $M_1$ is a finite $R$-module.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Proof of (1). If $x_1, \ldots, x_n$ are generators of $M_1$ and
    $y_1, \ldots, y_m \in M_2$ are elements whose images in $M_3$ are
    generators of $M_3$, then $x_1, \ldots, x_n, y_1, \ldots, y_m$
    generate $M_2$.
    
    \medskip\noindent
    Part (3) is immediate from the definition.
    
    \medskip\noindent
    Proof of (5). Assume $M_3$ is finitely presented and $M_2$ finite.
    Choose a presentation
    $$
    R^{\oplus m} \to R^{\oplus n} \to M_3 \to 0
    $$
    By Lemma \ref{lemma-lift-map} there exists a map
    $R^{\oplus n} \to M_2$ such that
    the solid diagram
    $$
    \xymatrix{
    & R^{\oplus m} \ar[r] \ar@{..>}[d] & R^{\oplus n} \ar[r] \ar[d] &
    M_3 \ar[r] \ar[d]^{\text{id}} & 0 \\
    0 \ar[r] & M_1 \ar[r] & M_2 \ar[r] & M_3 \ar[r] & 0
    }
    $$
    commutes. This produces the dotted arrow. By the snake lemma
    (Lemma \ref{lemma-snake}) we see that we get an isomorphism
    $$
    \Coker(R^{\oplus m} \to M_1)
    \cong
    \Coker(R^{\oplus n} \to M_2)
    $$
    In particular we conclude that $\Coker(R^{\oplus m} \to M_1)$
    is a finite $R$-module. Since $\Im(R^{\oplus m} \to M_1)$
    is finite by (3), we see that $M_1$ is finite by part (1).
    
    \medskip\noindent
    Proof of (4). Assume $M_2$ is finitely presented and $M_1$ is finite.
    Choose a presentation $R^{\oplus m} \to R^{\oplus n} \to M_2 \to 0$.
    Choose a surjection $R^{\oplus k} \to M_1$. By Lemma \ref{lemma-lift-map}
    there exists a factorization $R^{\oplus k} \to R^{\oplus n} \to M_2$
    of the composition $R^{\oplus k} \to M_1 \to M_2$. Then
    $R^{\oplus k + m} \to R^{\oplus n} \to M_3 \to 0$
    is a presentation.
    
    \medskip\noindent
    Proof of (2). Assume that $M_1$ and $M_3$ are finitely presented.
    The argument in the proof of part (1) produces a commutative diagram
    $$
    \xymatrix{
    0 \ar[r] & R^{\oplus n} \ar[d] \ar[r] & R^{\oplus n + m} \ar[d] \ar[r] &
    R^{\oplus m} \ar[d] \ar[r] & 0 \\
    0 \ar[r] & M_1 \ar[r] & M_2 \ar[r] & M_3 \ar[r] & 0
    }
    $$
    with surjective vertical arrows. By the snake lemma we obtain a short
    exact sequence
    $$
    0 \to \Ker(R^{\oplus n} \to M_1) \to
    \Ker(R^{\oplus n + m} \to M_2) \to
    \Ker(R^{\oplus m} \to M_3) \to 0
    $$
    By part (5) we see that the outer two modules are finite. Hence the
    middle one is finite too. By (4) we see that $M_2$ is of finite presentation.
    \end{proof}

    Comments (1)

    Comment #691 by Anfang Zhou on June 14, 2014 a 11:40 pm UTC

    Typo. In the proof of (4), it should be " $R^{\oplus k} \to R^{\oplus n} \to M_2$" , not" $R^{\oplus k} \to R^{\oplus n} \to M_1$".

    There are also 5 comments on Section 10.5: Commutative Algebra.

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