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 the notions of "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 $F (X) \to G$ are in natural bijection with set maps $X \to U (G)$ , where $U (G)$ is the underlying set of elements of $G$ ; i.e., where $U$ is the forgetful functor $Grp \to Set$ . 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 $F (X)$ 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 goal

Let $R$ be a fixed ring and $U : R - Mod \to Set$ be the forgetful functor from $R$ -modules to sets. We will show there is a functor $F : Set \to R - Mod$ 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 $F (A) \to M$ there will correspond a unique set map $A \to U (M)$ , and conversely. In other words, for every $R$ -module $M$ and set $X$ there will be a natural bijection of sets

ϕ_{X, M} : {Hom}_{R - Mod} (F (X), M) \tilde{\to} {Hom}_{Set} (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 recall what it means for $ϕ_{X, M}$ to be natural in $X$ and $M$ . Naturality "in $M$ " will mean that for every $R$ -module morphism $f : M \to N$ we have a commutative diagram

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

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

Anticipating the construction

Suppose for the moment there exists a functor $F : S e t \to R - Mod$ left adjoint to the forgetful functor $U : R - Mod \to S e t$ . Let's consider some special sets in $S e t$ , namely the initial object (i.e., the empty set) and the terminal object (i.e., the singleton set).

Let's first look at the universal property $F (\emptyset)$ should enjoy, namely that there is a natural bijection

ϕ_{\emptyset, M} : {Hom}_{R - Mod} (F (\emptyset), M) \to {Hom}_{Set} (\emptyset, U (M)) .

The empty set is the initial object in $Set$ , so there is a unique set map (the empty map) from it to any other set. In other words, the set ${Hom}_{Set} (\emptyset, U (M))$ is a singleton set. Our bijection above then implies ${Hom}_{R - Mod} (F (\emptyset), M)$ is a singleton set, for every $R$ -module $M$ . This exactly says that $F (\emptyset)$ is the initial object in the category $R -Mod$ , which we've already seen is the zero module. Thus, if the functor $F$ exists, then it must satisfy $F (\emptyset) ≃ 0$ .

Now let's consider a singleton set $X = {x}$ . The universal property for $F ({x})$ is that we have a natural set bijection

ϕ_{{x}, M} : {Hom}_{R - Mod} (F ({x}), M) \to {Hom}_{Set} ({x}, U (M)) ≃ U (M)

This type of property might look familiar. We've seen that ${Hom}_{R - Mod} (R, M) ≃ M$ as an $R$ -module (at least for commutative rings; for general rings there's at least an isomorphism of abelian groups). This is strong evidence that we likely must have $F ({x}) ≃ R$ .

Finally, suppose $X$ is a general set. We will eventually see that left adjoints commute with colimits; e.g., disjoint unions in $S e t$ , direct sums in $R - Mod$ , etc. Once we know this, we can immediately deduce that for any set we must have

F (X) ≃ F (⨆_{x \in X} {x}) ≃ ⨁_{x \in X} F ({x}) ≃ ⨁_{x \in X} R

Based on this quick analysis, we can now predict with decent confidence that the only possible definition of $F (X)$ must be $⨁_{x \in X} R$ . So let's do that and make sure it all works out.

The construction of $F (X)$

Inspired by the above analysis, for each set $X$ let's define $F (X) = ⨁_{x \in X} R$ , which we will call the free $R$ -module on $X$ .

Following our previous notes on direct sums of modules, this means that the elements of $F (X)$ are the set maps with the property that for all but finitely many $x \in X$ . Of course, as with all direct sums, it is common to instead think of $F (X)$ as "formal finite -linear combinations of elements of $X$ "; i.e., expressions of the form $\sum_{x \in X} r_{x} x$ , where $r_{x} = 0_{R}$ for all but finitely many $x \in X$ .

Note that the $R$ -module structure on $F (X)$ is as described for general direct sums. When we think of the elements of $F (X)$ as certain set maps $f : X \to U (R)$ , the additive operation is "addition of outputs" and the $R$ -action is "scaling outputs"; when we think of the elements of $F (X)$ as formal sums, the additive operation is "addition of coefficients" and the $R$ -action is "scaling coefficients."

What is the arrow function of $F$ ? For each set map $g : X \to Y$ , the corresponding module morphism $F (g) : F (X) \to F (Y)$ is defined as follows. If we identify the elements of the free modules as finite formal sums, then $F (g)$ is given by

\sum_{x \in X} r_{x} x \mapsto \sum_{y \in Y} (\sum_{x \in g^{- 1} (y)} r_{x}) y .

In other words, we use the set map $g$ to send each $x$ to an element in $y$ , and then combine coefficients of "like terms".

If we view the elements of $F (X)$ as certain set maps $f : X \to U (R)$ , then $F (g)$ maps each such set map $f$ to the set map $\tilde{f} : Y \to U (R)$ defined by

\tilde{f} (y) = \sum_{x \in g^{- 1} (y)} r_{x} .

We leave it to the diligent reader to verify that we have now indeed defined a bona fide functor $F : S e t \to R - Mod$ .

One more observation (to be elaborated upon later): there is a "copy"^[2] of the set $X$ inside of the free module $F (X)$ , obtained by identifying each element $x_{0} \in X$ with the $R$ -linear combination that has coefficient $1_{R}$ for $x_{0}$ , and coefficient $0_{R}$ for all other $x \in X$ . (If we view the elements of $F (X)$ as certain set maps $f : X \to U (R)$ , then we are mapping each element $x_{0} \in X$ to the "characteristic function $f_{x_{0}}$ defined by $f_{x_{0}} (x_{0}) = 1_{R}$ and $f_{x_{0}} (x) = 0_{R}$ for all $x \neq x_{0}$ .) These specific linear combinations are the analogue in module theory of the standard basis vectors in linear algebra.

Verifying adjointness

If we're being careful, we should verify that our functor is indeed left adjoint to the forgetful functor, and that means explicitly describing the required natural bijections. Let's content ourselves with simply outlining how that goes.

First suppose $X$ is any set, $M$ is any $R$ -module and $g : F (X) \to M$ is a module morphism. By using the identification above of a copy of $X$ in $F (X)$ , we then also have a set map $X \to U (M)$ . More specifically, for each $x_{0} \in X$ let $f_{x_{0}} \in F (X)$ be the "standard basis element" corresponding to $x_{0}$ . We can then send $x_{0}$ to the image of $f_{x_{0}}$ under the given morphism $g$ . This is how we construct our set map

ϕ_{X, M} : {Hom}_{R} (F (X), M) \to {Hom}_{Set} (X, U (M)) .

We can verify this set map is bijective by constructing the inverse set map. Suppose now $g : X \to U (M)$ 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, in the formal sum notation, suppose $v = \sum_{x \in X} r_{x} x$ is an element in the free module $F (X)$ . Define a map $ψ_{X, M} (g) : F (X) \to M$ by sending $v$ to the element in $M$ given by the (actual) sum $\sum_{x \in X} r_{x} g (x)$ .

Challenge

Can you verify our two set maps $ϕ_{X, M}$ and $ψ_{X, M}$ are mutual inverses, and that $ϕ_{X, M}$ 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 it 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 set of elements of $X$ generates $M$ and are $R$ -linearly independent in $M$ ; the first condition guarantees that every $m \in M$ can be written as $m = \sum_{x \in X} r_{x} x$ for some $r_{x} \in X$ , 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 $M$ :

A subset $X \subseteq U (M)$ generates $M$ exactly when the corresponding module morphism $π : F (X) \to M$ is surjective; and
A subset $X \subseteq U (M)$ is "linearly independent in $M$ " exactly when the corresponding module morphism $π : F (X) \to M$ 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 $\ker (π) \subseteq F (X)$ .

Suggested next note

Examples of free modules

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). ↩︎
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 $X \to U (F (X))$ . This is the unit of the adjunction. ↩︎