## 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.
$\square$

slogan
Lemma 10.5.4. 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 $S$.
$\square$

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