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Free module

## Summary

In mathematics, a free module is a module that has a basis – that is, a generating set consisting of linearly independent elements. Every vector space is a free module,[1] but, if the ring of the coefficients is not a division ring (not a field in the commutative case), then there exist non-free modules.

Given any set S and ring R, there is a free R-module with basis S, which is called the free module on S or module of formal R-linear combinations of the elements of S.

A free abelian group is precisely a free module over the ring Z of integers.

## Definition

For a ring ${\displaystyle R}$  and an ${\displaystyle R}$ -module ${\displaystyle M}$ , the set ${\displaystyle E\subseteq M}$  is a basis for ${\displaystyle M}$  if:

• ${\displaystyle E}$  is a generating set for ${\displaystyle M}$ ; that is to say, every element of ${\displaystyle M}$  is a finite sum of elements of ${\displaystyle E}$  multiplied by coefficients in ${\displaystyle R}$ ; and
• ${\displaystyle E}$  is linearly independent, that is, for every subset ${\displaystyle \{e_{1},e_{2},\ldots ,e_{n}\}}$  of distinct elements of ${\displaystyle E}$ , ${\displaystyle r_{1}e_{1}+r_{2}e_{2}+\cdots +r_{n}e_{n}=0_{M}}$  implies that ${\displaystyle r_{1}=r_{2}=\cdots =r_{n}=0_{R}}$  (where ${\displaystyle 0_{M}}$  is the zero element of ${\displaystyle M}$  and ${\displaystyle 0_{R}}$  is the zero element of ${\displaystyle R}$ ).

A free module is a module with a basis.[2]

An immediate consequence of the second half of the definition is that the coefficients in the first half are unique for each element of M.

If ${\displaystyle R}$  has invariant basis number, then by definition any two bases have the same cardinality. For example, nonzero commutative rings have invariant basis number. The cardinality of any (and therefore every) basis is called the rank of the free module ${\displaystyle M}$ . If this cardinality is finite, the free module is said to be free of finite rank, or free of rank n if the rank is known to be n.

## Examples

Let R be a ring.

• R is a free module of rank one over itself (either as a left or right module); any unit element is a basis.
• More generally, If R is commutative, a nonzero ideal I of R is free if and only if it is a principal ideal generated by a nonzerodivisor, with a generator being a basis.[3]
• If R is commutative, the polynomial ring ${\displaystyle R[X]}$  in indeterminate X is a free module with a possible basis 1, X, X2, ....
• Let ${\displaystyle A[t]}$  be a polynomial ring over a commutative ring A, f a monic polynomial of degree d there, ${\displaystyle B=A[t]/(f)}$  and ${\displaystyle \xi }$  the image of t in B. Then B contains A as a subring and is free as an A-module with a basis ${\displaystyle 1,\xi ,\dots ,\xi ^{d-1}}$ .
• For any non-negative integer n, ${\displaystyle R^{n}=R\times \cdots \times R}$ , the cartesian product of n copies of R as a left R-module, is free. If R has invariant basis number, then its rank is n.
• A direct sum of free modules is free, while an infinite cartesian product of free modules is generally not free (cf. the Baer–Specker group.)
• Kaplansky's theorem states a projective module over a local ring is free.

## Formal linear combinations

Given a set E and ring R, there is a free R-module that has E as a basis: namely, the direct sum of copies of R indexed by E

${\displaystyle R^{(E)}=\bigoplus _{e\in E}R}$ .

Explicitly, it is the submodule of the Cartesian product ${\textstyle \prod _{E}R}$  (R is viewed as say a left module) that consists of the elements that have only finitely many nonzero components. One can embed E into R(E) as a subset by identifying an element e with that of R(E) whose e-th component is 1 (the unity of R) and all the other components are zero. Then each element of R(E) can be written uniquely as

${\displaystyle \sum _{e\in E}c_{e}e,}$

where only finitely many ${\displaystyle c_{e}}$  are nonzero. It is called a formal linear combination of elements of E.

A similar argument shows that every free left (resp. right) R-module is isomorphic to a direct sum of copies of R as left (resp. right) module.

### Another construction

The free module R(E) may also be constructed in the following equivalent way.

Given a ring R and a set E, first as a set we let

${\displaystyle R^{(E)}=\{f:E\to R\mid f(x)=0{\text{ for all but finitely many }}x\in E\}.}$

We equip it with a structure of a left module such that the addition is defined by: for x in E,

${\displaystyle (f+g)(x)=f(x)+g(x)}$

and the scalar multiplication by: for r in R and x in E,

${\displaystyle (rf)(x)=r(f(x))}$

Now, as an R-valued function on E, each f in ${\displaystyle R^{(E)}}$  can be written uniquely as

${\displaystyle f=\sum _{e\in E}c_{e}\delta _{e}}$

where ${\displaystyle c_{e}}$  are in R and only finitely many of them are nonzero and ${\displaystyle \delta _{e}}$  is given as

${\displaystyle \delta _{e}(x)={\begin{cases}1_{R}\quad {\mbox{if }}x=e\\0_{R}\quad {\mbox{if }}x\neq e\end{cases}}}$

(this is a variant of the Kronecker delta.) The above means that the subset ${\displaystyle \{\delta _{e}\mid e\in E\}}$  of ${\displaystyle R^{(E)}}$  is a basis of ${\displaystyle R^{(E)}}$ . The mapping ${\displaystyle e\mapsto \delta _{e}}$  is a bijection between E and this basis. Through this bijection, ${\displaystyle R^{(E)}}$  is a free module with the basis E.

## Universal property

The inclusion mapping ${\displaystyle \iota :E\to R^{(E)}}$  defined above is universal in the following sense. Given an arbitrary function ${\displaystyle f:E\to N}$  from a set E to a left R-module N, there exists a unique module homomorphism ${\displaystyle {\overline {f}}:R^{(E)}\to N}$  such that ${\displaystyle f={\overline {f}}\circ \iota }$ ; namely, ${\displaystyle {\overline {f}}}$  is defined by the formula:

${\displaystyle {\overline {f}}\left(\sum _{e\in E}r_{e}e\right)=\sum _{e\in E}r_{e}f(e)}$

and ${\displaystyle {\overline {f}}}$  is said to be obtained by extending ${\displaystyle f}$  by linearity. The uniqueness means that each R-linear map ${\displaystyle R^{(E)}\to N}$  is uniquely determined by its restriction to E.

As usual for universal properties, this defines R(E) up to a canonical isomorphism. Also the formation of ${\displaystyle \iota :E\to R^{(E)}}$  for each set E determines a functor

${\displaystyle R^{(-)}:{\textbf {Set}}\to R-{\mathsf {Mod}},\,E\mapsto R^{(E)}}$ ,

from the category of sets to the category of left R-modules. It is called the free functor and satisfies a natural relation: for each set E and a left module N,

${\displaystyle \operatorname {Hom} _{\textbf {Set}}(E,U(N))\simeq \operatorname {Hom} _{R}(R^{(E)},N),\,f\mapsto {\overline {f}}}$

where ${\displaystyle U:R-{\mathsf {Mod}}\to {\textbf {Set}}}$  is the forgetful functor, meaning ${\displaystyle R^{(-)}}$  is a left adjoint of the forgetful functor.

## Generalizations

Many statements about free modules, which are wrong for general modules over rings, are still true for certain generalisations of free modules. Projective modules are direct summands of free modules, so one can choose an injection into a free module and use the basis of this one to prove something for the projective module. Even weaker generalisations are flat modules, which still have the property that tensoring with them preserves exact sequences, and torsion-free modules. If the ring has special properties, this hierarchy may collapse, e.g., for any perfect local Dedekind ring, every torsion-free module is flat, projective and free as well. A finitely generated torsion-free module of a commutative PID is free. A finitely generated Z-module is free if and only if it is flat.

## Notes

1. ^ Keown (1975). An Introduction to Group Representation Theory. p. 24.
2. ^ Hazewinkel (1989). Encyclopaedia of Mathematics, Volume 4. p. 110.
3. ^ Proof: Suppose ${\displaystyle I}$  is free with a basis ${\displaystyle \{x_{j}|j\}}$ . For ${\displaystyle j\neq k}$ , ${\displaystyle x_{j}x_{k}}$  must have the unique linear combination in terms of ${\displaystyle x_{j}}$  and ${\displaystyle x_{k}}$ , which is not true. Thus, since ${\displaystyle I\neq 0}$ , there is only one basis element which must be a nonzerodivisor. The converse is clear.${\displaystyle \square }$