# nLab finitely presented module

Contents

### Context

#### Algebra

higher algebra

universal algebra

# Contents

## Definition

Given a (not necessarily commutative) unital ring $R$, an $R$-module is finitely presented (or of finite presentation) if there exists an exact sequence $R^q\to R^p\to M\to 0$ where $p,q$ are natural numbers.

## Equivalent characterisations

###### Proposition

For a module $M$ over a ring $R$, the following are equivalent :

1. $M$ is finitely presented ;
2. $M$ is a compact object of the category of $R$-modules.

###### Proposition

For a module $M$ over a ring $R$, the following are equivalent :

1. $M$ is finitely presented ;
2. for every family $\{Q_\alpha\}_{\alpha \in A}$ of $R$-modules, the canonical map
$M \otimes_R \prod_\alpha Q_\alpha \rightarrow \prod_\alpha M \otimes_R Q_\alpha$

is an isomophism ;

3. for every $R$-module $Q$ and every set $A$, the canonical map
$M \otimes_R R^A \rightarrow M^A$

is an isomorphism.

###### Proof

$1 \Rightarrow 2$. Using a finite presentation $R^q \to R^p \to M$ of $M$ one can write the commutative diagram

and deduce the isomorphism.

$2 \Rightarrow 3$. Obvious

$3 \Rightarrow 1$. The map $M \otimes R^M \to M^M$ is surjective, so there exists a finite number of elements $\{m_i\}_{i \in I}$ and a finite number of functions $\{f_i\}_{i \in I} : M \to R$ such that the sum $\sum_i m_i f^i$ be the identity of $M$. As a consequence $M$ must be finitely generated.

Consider a short exact sequence $0 \to K \to R^p \to M \to 0$. Then using the commutative diagram

one gets that $K \otimes R^K \to K^K$ is surjective and thus that $K$ is finitely generated.

Last revised on July 25, 2023 at 16:38:55. See the history of this page for a list of all contributions to it.