Section 10.5 (0517): Finite modules and finitely presented modules

10.5 Finite modules and finitely presented modules

Just some basic notation and lemmas.

Definition 10.5.1. Let $R$ be a ring. Let $M$ be an $R$ -module.

We say $M$ is a finite $R$ -module, or a finitely generated $R$ -module if there exist $n \in \mathbf{N}$ and $x_1, \ldots , x_ n \in M$ such that every element of $M$ is an $R$ -linear combination of the $x_ i$ . Equivalently, this means there exists a surjection $R^{\oplus n} \to M$ for some $n \in \mathbf{N}$ .
We say $M$ is a finitely presented $R$ -module or an $R$ -module of finite presentation if there exist integers $n, m \in \mathbf{N}$ and an exact sequence

$R^{\oplus m} \longrightarrow R^{\oplus n} \longrightarrow M \longrightarrow 0$

Informally, $M$ is a finitely presented $R$ -module if and only if it is finitely generated and the module of relations among these generators is finitely generated as well. A choice of an exact sequence as in the definition is called a presentation of $M$ .

Lemma 10.5.2. Let $R$ be a ring. Let $\alpha : R^{\oplus n} \to M$ and $\beta : N \to M$ be module maps. If $\mathop{\mathrm{Im}}(\alpha ) \subset \mathop{\mathrm{Im}}(\beta )$ , then there exists an $R$ -module map $\gamma : R^{\oplus n} \to N$ such that $\alpha = \beta \circ \gamma$ .

Proof. Let $e_ i = (0, \ldots , 0, 1, 0, \ldots , 0)$ be the $i$ th basis vector of $R^{\oplus n}$ . Let $x_ i \in N$ be an element with $\alpha (e_ i) = \beta (x_ i)$ which exists by assumption. Set $\gamma (a_1, \ldots , a_ n) = \sum a_ i x_ i$ . By construction $\alpha = \beta \circ \gamma$ . $\square$

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.

If $M_1$ and $M_3$ are finite $R$ -modules, then $M_2$ is a finite $R$ -module.
If $M_1$ and $M_3$ are finitely presented $R$ -modules, then $M_2$ is a finitely presented $R$ -module.
If $M_2$ is a finite $R$ -module, then $M_3$ is a finite $R$ -module.
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.
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

$\mathop{\mathrm{Coker}}(R^{\oplus m} \to M_1) \cong \mathop{\mathrm{Coker}}(R^{\oplus n} \to M_2)$

In particular we conclude that $\mathop{\mathrm{Coker}}(R^{\oplus m} \to M_1)$ is a finite $R$ -module. Since $\mathop{\mathrm{Im}}(R^{\oplus m} \to M_1)$ 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

$0 \to \mathop{\mathrm{Ker}}(R^{\oplus n} \to M_1) \to \mathop{\mathrm{Ker}}(R^{\oplus n + m} \to M_2) \to \mathop{\mathrm{Ker}}(R^{\oplus m} \to M_3) \to 0$

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. $\square$

Lemma 10.5.4.slogan Let $R$ be a ring, and let $M$ be a finite $R$ -module. There exists a filtration by finite $R$ -submodules

$0 = M_0 \subset M_1 \subset \ldots \subset M_ n = M$

such that each quotient $M_ i/M_{i - 1}$ is isomorphic to $R/I_ i$ for some ideal $I_ i$ of $R$ .

Proof. By induction on the number of generators of $M$ . Let $x_1, \ldots , x_ r \in M$ be generators. Let $M' = Rx_1 \subset M$ . Then $M/M'$ has $r - 1$ generators and the induction hypothesis applies. And clearly $M' \cong R/I_1$ with $I_1 = \{ f \in R \mid fx_1 = 0\}$ . $\square$

Lemma 10.5.5. Let $R \to S$ be a ring map. Let $M$ be an $S$ -module. If $M$ is finite as an $R$ -module, then $M$ is finite as an $S$ -module.

Proof. In fact, any $R$ -generating set of $M$ is also an $S$ -generating set of $M$ , since the $R$ -module structure is induced by the image of $R$ in $S$ . $\square$

Comments (11)

Comment #269 by Keenan Kidwell on August 03, 2013 at 19:10

In 055Z, would it be convenient to have the extra generality of allowing $R^n$ to be replaced by any finite $R$ -module? It doesn't change the proof at all, since all that is used of $R^n$ is that it is finitely generated.

Comment #270 by Johan on August 04, 2013 at 14:29

It could, but that result is contained in Lemma 10.5.3. What must have happened is that somebody (me) thought it was necessary to first prove the result of Lemma 115.4.1 for free modules and then conclude it for general ones. Also, the proof of Lemma 10.5.3 should refer to Lemma 115.4.1.

Just overall bad writing! I'll fix it up a bit.

Comment #271 by Johan on August 04, 2013 at 15:21

OK, I tried to improve it a bit. Note that Lemma 115.4.1 got moved to the obsolete chapter but it still exists (it isn't wrong). Changeset is here.

Comment #272 by Keenan Kidwell on August 05, 2013 at 02:13

I think the changes you made improve it a lot. The only thing I find slightly confusing is the invocation of part (4) at the end of the proof of part (2). You've concluded that the surjection $R^{n+m}\rightarrow M_2$ is finitely generated, so $M_2$ is finitely presented by definition, and there is no need to invoke (4), because the module playing the role of $M_2$ in (4) is $R^{n+m}$ , not an arbitrary finitely presented module.

Comment #274 by Johan on August 08, 2013 at 00:34

@#272: We use (4) because the definition of finitely presented modules specifies that the module is a cokernel between a map of free modules and $\text{Ker}(R^{\oplus n + m} \to M_2)$ won't be free in general.

Comment #4338 by ExcitedAlgebraicGeometer on August 09, 2019 at 13:36

A potential lemma of interest here: if a module is finitely presented, then any generating set has a finite presentation. The

Comment #4488 by Johan on September 02, 2019 at 12:19

Dear ExcitedAlgebraicGeometer, this follows from Lemma 10.5.3 part (5).

Comment #4647 by Mathcal on November 06, 2019 at 05:44

In the proof of Lemma:10.5.3 , 4) you don't need the fact that $M_1$ is finite I think and in the proof of 5) you didn't use the fact that $M_2$ is finite ?

Comment #4793 by Johan on December 10, 2019 at 09:31

@#4647: Try reading it again. It seems to me that in both cases we do use the finiteness of $M_1$ respectively $M_2$ . Also, without those assumptions the conclusions would be wrong.

Comment #9499 by Jack Gallahan on July 10, 2024 at 12:56

Is there an implied 0 of the left in the exact sequence that appears in the definition of a finitely presented module?

Comment #9500 by Stacks project on July 10, 2024 at 13:01

no

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10.5 Finite modules and finitely presented modules

Comments (11)

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