nLab
free module

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 $\mathrm{Alg}(C)$ the category of monoids in $C$ and for $A\in \mathrm{Alg}(C)$ let $A$Mod$(C)$ be the category of $A$-modules in $C$.

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

Definition

The left adjoint $C\to A\mathrm{Mod}(C)$ 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 $\mid S\mid$-fold direct sum of $R$ with itself

$$R^{(S)}\simeq {\oplus }_{s\in S}R.$$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

$$R\stackrel{{\lambda }_x}{\to }(x)\cong {\oplus }_{J}R\stackrel{\nabla }{\to }R$$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 ${\lambda }_r(s)=rs=1$ 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

$$0\ne fg\in R\cdot f\cap R\cdot g$$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.

Related concepts

• locally free module

• finitely generated module, finitely presented module?

• free module $\Rightarrow$ projective module $\Rightarrow$ flat module $\Rightarrow$ torsion-free module

• basis of a free module

• projective object, projective presentation, projective cover, projective resolution

• injective object, injective presentation, injective envelope, injective resolution

  • injective module

• free object, free resolution

• flat object, flat resolution

  • flat module

References

• Rotman Advanced Modern Algebra, pp. 650–651