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 $F(X)\to G$ 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 $\mathbf{\text{Grp}}\to \mathbf{\text{Set}}$. 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 $U:R\mathbf{\text{-Mod}}\to \mathbf{\text{Set}}$ be the forgetful functor from $R$-modules to sets. We will show there is a functor $F:\mathbf{\text{Set}}\to R\mathbf{\text{-Mod}}$ 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

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 ${\varphi }_{X,M}$ 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.

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 ${\displaystyle \sum _{x\in X}{r}_{x}x}$, where $r_x=0_R$ for all but finitely many $x\in X$.

We then define a binary operation on $F(X)$ by "combining coefficients." In other words, if $m_1={\displaystyle \sum _{x\in X}{r}_{x}x}$ and $m_2={\displaystyle \sum _{x\in X}{s}_{x}x}$ are two elements of $F(X)$, then we define $m_1+m_2={\displaystyle \sum _{x\in X}(r_x+s_x)x}$. 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: $s\cdot {\displaystyle \sum _{x\in X}{r}_{x}x=\sum _{x\in X}(s{r}_x)x}$. It is then straightforward to verify this gives $F(X)$ the structure of an $R$-module.

In summary:

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 $x_0\in X$ with the $R$-linear combination that has coefficient $1_R$ for $x_0$, and coefficient $0_R$ for all other $x\in X$. 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 $f:F(X)\to M$ 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 $e_{x_0}\in F(X)$ be the "standard basis element" corresponding to $x_0$, i.e., the $R$-linear combination that has coefficient $1_R$ for $x_0$, and coefficient $0_R$ for all other $x\in X$. We can then send $x_0$ to the image of $e_{x_0}$ under the given morphism $f$. This is how we construct our set map

We can verify this set map is bijective by constructing the inverse set map. Suppose $g:X\to U(M)$ 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, suppose $v={\displaystyle \sum _{x\in X}{r}_{x}x}$ is an element in the free module $F(X)$. Define a map ${\psi }_{X,M}(g):F(X)\to M$ that sends $v$ to the element in $M$ given by ${\displaystyle \sum _{x\in X}{r}_{x}g(x)}$.

Free modules in general

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 $m\in M$ can be written as ${\displaystyle m=\sum _{x\in X}{r}_{x}x}$ 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 $M$:

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

Suggested next note

Examples of free modules