# nLab free module

### Context

#### Algebra

higher algebra

universal algebra

## Theorems

#### Homological algebra

homological algebra

and

nonabelian homological algebra

diagram chasing

# Contents

## Idea

A free module over some ring $R$ is freely generated on a set of basis elements.

Under the interpretation of modules as generalized vector bundles a free module corresponds to a trivial bundle.

## Definition

### General

Let $C$ be a monoidal category, and $Alg(C)$ the category of monoids in $C$ and for $A \in Alg(C)$ let $A$Mod$(C)$ be the category of $A$-modules in $C$.

There is the evident forgetful functor $U : A Mod(C) \to C$ that sends each module $(N,\rho)$ to its underlying object $N \in C$.

###### Definition

The left adjoint $C \to A Mod(C)$ is the corresponding free construction. The modules in the image of this functor are free modules.

### Over rings

Let $R$ be a ring. We discuss free modules over $R$.

###### Proposition

For $R \in$ Ring a ring and $S \in$ Set, the free $R$-module on $S$ is isomorphic to the ${\vert S\vert}$-fold direct sum of $R$ with itself

$R^{(S)}\simeq \oplus_{s \in S} R \,.$

## Properties

### Submodules of free modules

Let $R$ be a commutative ring.

###### Proposition

Assuming the axiom of choice, the following are equivalent

1. every submodule of a free $R$-module is itself free;

2. every ideal in $R$ is a free $R$-module;

3. $R$ is a principal ideal domain.

###### Proof

(See also Rotman, pages 650-651.) Condition 1. immediately implies condition 2., since ideals of $R$ are the same as submodules of $R$ seen as an $R$-module. Now assume condition 2. holds, and suppose $x \in R$ is any nonzero element. Let $\lambda_x$ denote multiplication by $x$ (as an $R$-module map). We have a sequence of surjective $R$-module maps

$R \stackrel{\lambda_x}{\to} (x) \cong \oplus_J R \stackrel{\nabla}{\to} R$

(where $\nabla$ is the codiagonal map); by the Yoneda lemma, the composite map $R \to R$ is of the form $\lambda_r$, where $r \in R$ is the value of the composite at $1 \in R$. Since $\lambda_r$ is surjective, we have $\lambda_r(s) = r s = 1$ for some $s$, so that $r$ is invertible. Hence $\lambda_r$ is invertible, and this implies $\lambda_x$ is monic. Therefore $R$ is a domain. From that, we infer that if $f$ and $g$ belong to a basis of an ideal $I$, then

$0 \neq f g \in R\cdot f \cap R \cdot g$

whence $f$ and $g$ are not linearly independent, so $f = g$ and $I$ as an $R$-module is generated by a single element, i.e., $R$ is a principal ideal domain.

That condition 3. implies condition 1. is proved here.

###### Corollary

Assuming the axiom of choice, over a ring $R$ which is a principal ideal domain, every module has a projective resolution of length 1.

See at projective resolution – Resolutions of length 1 for more.

### Over a field: vector spaces

Assuming the axiom of choice, if $R = k$ is a field then every $R$-module is free: it is $k$-vector space and every such has a basis.

## References

• Rotman Advanced Modern Algebra, pp. 650–651

Revised on February 8, 2013 12:03:06 by Urs Schreiber (89.204.138.214)