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 and then define the "free object on " as the "smallest" or "simplest" object one can create (of the desired type) beginning from only the set . But how do we formalize "smallest" or "simplest"? We might reasonably suggest that the definition be such that has a desirable universal property. For example, the free group should be the group such that group morphisms are in natural bijection with set maps , where is the underlying set of elements of ; i.e., where is the forgetful functor . In this way, the set would act like a "basis" for the group , in that maps from are entirely determined by the images of the "basis" elements . The "free" adjective could then be interpreted as the fact that there are no conditions on the maps from to ; they are simply set maps. You can map the elements of "freely" to any elements you'd like in , and to every such choice there is a unique corresponding group morphism from the free object to . So it's not so much that is "free" from "relations," but rather that the basis elements in 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 be a fixed ring and be the forgetful functor from -modules to sets. We will show there is a functor that associates to each set a "minimal" -module with a "free" property as described above. Specifically, to every -module morphism there will correspond a unique set map , and conversely. In other words, for every -module and set there will be a natural bijection of sets
Put more simply still, our functor will be a left adjoint to the forgetful functor .
Before we construct the functor , let's take a minute to explain what it means for to be natural in and . Naturality "in " will mean that for every -module morphism we have a commutative diagram
Similarly, naturality "in " will mean that for every set map we have a commutative diagram[1]
This naturality condition will have many consequences for our construction, which we'll investigate later.
The construction of
Let be a fixed set. We first define the set to be the collection of formal finite -linear combinations of elements of .[2] In other words, consists of expressions of the form , where for all but finitely many .
We then define a binary operation on by "combining coefficients." In other words, if and are two elements of , then we define . 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 the structure of an abelian group[3].
Finally, we let act on the left of by left multiplication of the coefficients: . It is then straightforward to verify this gives the structure of an -module.
In summary:
The construction/definition of free modules
Given a set , the free -module on 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 is addition of coefficients, and the -action is by left multiplication of coefficients.
One more observation (to be elaborated upon later): there is a "copy"[4] of the set inside of the free module , obtained by identifying each element with the -linear combination that has coefficient for , and coefficient for all other . These specific linear combinations are the analogue in module theory of the standard basis vectors in linear algebra.
Verifying the natural bijections
First suppose is any set, is any -module, and is a module morphism. By using the identification above of a copy of in , we then also have a set map . More specifically, for each let be the "standard basis element" corresponding to , i.e., the -linear combination that has coefficient for , and coefficient for all other . We can then send to the image of under the given morphism . This is how we construct our set map
We can verify this set map is bijective by constructing the inverse set map. Suppose is any set map. We can then define a map on by applying to the "basis" elements and "using linearity." More precisely, suppose is an element in the free module . Define a map that sends to the element in given by .
Challenge
Can you verify our two set maps and are mutual inverses, and that satisfies the naturality condition outlined above?
Free modules in general
Definition of free
Suppose is an -module. We say is free if is isomorphic (as an -module) to for some finite set . In that case we can also say is free on .
In terms of elements, this is just like a basis for a vector space. If is a subset of the elements of , then is free on exactly when the elements of generate and are -linearly independent in ; the first condition guarantees that every can be written as for some , while the second condition guarantees that such an expression is unique.
For a more categorical approach, observe the following about subsets of elements of :
A subset generates exactly when the corresponding module morphism is surjective; and
A subset is "linearly independent in " exactly when the corresponding module morphism is injective. We can take this as the definition of linear independence, and declare the relations on to be the elements of the submodule .
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). âŠī¸
Equivalently, we can define to be the collection of all set maps with the property that for all but finitely many âŠī¸
This is precisely the construction of the free abelian group on the set . Do you see how one could convert any finite formal sum of the given form into such a set map, and conversely? âŠī¸
More precisely, there is an injective set map from to the underlying set of elements of , i.e., there is a set map . This is the unit of the adjunction. âŠī¸