Free modules

Motivation

You often hear it said that a "free object" is an object with no other "relations" beyond those required of every object of that type, e.g., a "free group" is a group with no relations beyond those required of every group. But what does that actually mean, and how do you formalize it?

As a first attempt, we could start with a set X and then define the "free object on X" as the "smallest" or "simplest" object F(X) one can create (of the desired type) beginning from only the set X. But how do we formalize "smallest" or "simplest"? We might reasonably suggest that the definition be such that F(X) has a desirable universal property. For example, the free group F(X) should be the group such that group morphisms F(X)→G are in natural bijection with set maps X→U(G), where U(G) is the underlying set of elements of G; i.e., where U is the forgetful functor Grp→Set. In this way, the set X would act like a "basis" for the group F(X), in that maps from F(X) are entirely determined by the images of the "basis" elements X. The "free" adjective could then be interpreted as the fact that there are no conditions on the maps from X to U(G); they are simply set maps. You can map the elements of X "freely" to any elements you'd like in U(G), and to every such choice there is a unique corresponding group morphism from the free object F(X) to G. So it's not so much that F(X) is "free" from "relations," but rather that the basis elements in X can be freely mapped to any elements of your chosen target group.

Let's try following this idea for modules, but this time filling in all of the details.

The free module functor

The goal

Let R be a fixed ring and U:R-Mod→Set be the forgetful functor from R-modules to sets. We will show there is a functor F:Set→R-Mod that associates to each set X a "minimal" R-module F(X) with a "free" property as described above. Specifically, to every R-module morphism F(A)→M there will correspond a unique set map A→U(M), and conversely. In other words, for every R-module M and set X there will be a natural bijection of sets

Ī•X,M:HomR(F(X),M)→âˆŧHomSet(X,U(M)).

Put more simply still, our functor F will be a left adjoint to the forgetful functor U.

Before we construct the functor F, let's take a minute to explain what it means for Ī•X,M to be natural in X and M. Naturality "in M" will mean that for every R-module morphism f:M→N we have a commutative diagram

Similarly, naturality "in X" will mean that for every set map g:X→Y we have a commutative diagram[1]

This naturality condition will have many consequences for our construction, which we'll investigate later.

The construction of F(X)

Let X be a fixed set. We first define the set F(X) to be the collection of formal finite R-linear combinations of elements of X.[2] In other words, F(X) consists of expressions of the form ∑x∈Xrxx, where rx=0R for all but finitely many x∈X.

We then define a binary operation on F(X) by "combining coefficients." In other words, if m1=∑x∈Xrxx and m2=∑x∈Xsxx are two elements of F(X), then we define m1+m2=∑x∈X(rx+sx)x. You can then verify that this operation is commutative and associative, that there is an identity element (the trivial combination), and that every element has a inverse (obtained by taking the combination with the additive inverse coefficients). In other words, this gives the set F(X) the structure of an abelian group[3].

Finally, we let R act on the left of F(X) by left multiplication of the coefficients: s⋅∑x∈Xrxx=∑x∈X(srx)x. It is then straightforward to verify this gives F(X) the structure of an R-module.

In summary:

The construction/definition of free modules

Given a set X, the free R-module on X is the set $$F(X)=\left{\sum_{x\in X}r_x x\mid r_x\in R,, r_x=0_R\text{ for all but finitely many }x\right}.$$
The group operation in F(X) is addition of coefficients, and the R-action is by left multiplication of coefficients.

One more observation (to be elaborated upon later): there is a "copy"[4] of the set X inside of the free module F(X), obtained by identifying each element x0∈X with the R-linear combination that has coefficient 1R for x0, and coefficient 0R for all other x∈X. These specific linear combinations are the analogue in module theory of the standard basis vectors in linear algebra.

Verifying the natural bijections

First suppose X is any set, M is any R-module, and f:F(X)→M is a module morphism. By using the identification above of a copy of X in F(X), we then also have a set map X→U(M). More specifically, for each x0∈X let ex0∈F(X) be the "standard basis element" corresponding to x0, i.e., the R-linear combination that has coefficient 1R for x0, and coefficient 0R for all other x∈X. We can then send x0 to the image of ex0 under the given morphism f. This is how we construct our set map

Ī•X,M:HomR(F(X),M)→HomSet(X,U(M)).

We can verify this set map is bijective by constructing the inverse set map. Suppose g:X→U(M) is any set map. We can then define a map on F(X) by applying g to the "basis" elements and "using linearity." More precisely, suppose v=∑x∈Xrxx is an element in the free module F(X). Define a map ĪˆX,M(g):F(X)→M that sends v to the element in M given by ∑x∈Xrxg(x).

Challenge

Can you verify our two set maps Ī•X,M and ĪˆX,M are mutual inverses, and that Ī•X,M satisfies the naturality condition outlined above?

Free modules in general

Definition of free

Suppose M is an R-module. We say M is free if M is isomorphic (as an R-module) to F(X) for some finite set X. In that case we can also say M is free on X.

In terms of elements, this is just like a basis for a vector space. If X is a subset of the elements of M, then M is free on X exactly when the elements of X generate M and are R-linearly independent in M; the first condition guarantees that every m∈M can be written as m=∑x∈Xrxx for some rx∈X, while the second condition guarantees that such an expression is unique.

For a more categorical approach, observe the following about subsets X of elements of M:

  1. A subset X⊆M generates M exactly when the corresponding module morphism Ī€:F(X)→M is surjective; and
  2. A subset X⊆M is "linearly independent in M" exactly when the corresponding module morphism Ī€:F(X)→M is injective. We can take this as the definition of linear independence, and declare the relations on X to be the elements of the submodule ker⁥(Ī€)⊆F(X).

Suggested next note

Examples of free modules


  1. If you're wondering why the vertical arrows in the second diagram are flipped, it's because a certain functor is "contravariant" (or equivalently, we need to use an "opposite" category as part of the formal formulation of what's going on). ↩ī¸Ž

  2. Equivalently, we can define F(X) to be the collection of all set maps f:X→R with the property that f(x)=0R for all but finitely many x∈X ↩ī¸Ž

  3. This is precisely the construction of the free abelian group on the set X. Do you see how one could convert any finite formal sum of the given form into such a set map, and conversely? ↩ī¸Ž

  4. More precisely, there is an injective set map from X to the underlying set of elements of F(X), i.e., there is a set map X→U(F(X)). This is the unit of the adjunction. ↩ī¸Ž