Examples of free modules

#module_theory

For each finite set $X$ , the structure of the free module $F (X)$ is entirely determined by the cardinality of $X$ . Let's consider some specific examples.

The free module on the empty set

What is the free module on the empty set? According to our construction the set of elements of the $R$ -module $F (\emptyset)$ consists of all formal finite $R$ -linear combinations of elements $\emptyset$ . But what the heck is the set of combinations of nothing, you might ask? Let's look at the universal property $F (\emptyset)$ should enjoy, namely that there is a natural bijection

ϕ_{\emptyset, M} : {Hom}_{R} (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 ${Hom}_{R} (F (\emptyset), M)$ 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, $F (\emptyset) = 0$ , the zero module.

This is why you sometimes see books/people declare (usually by fiat) that the "empty combination" is the zero element.

Summary

The free module on the empty set is the zero module.

The free module on a singleton set

Now let's consider a singleton set $X = {x}$ . By our construction, the module $F ({x})$ consists of all $R$ -multiples $r x$ with $r \in R$ . We therefore have a set bijection $F ({x}) ≃ R$ , which you can quickly verify is actually an $R$ -module isomorphism. We could have already predicted this, since we've seen that ${Hom}_{R} (R, M) ≃ M$ as an $R$ -module (at least for commutative rings $R$ ), and the universal property for $F ({x})$ is that we have a natural set bijection

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

Summary

For any singleton set $X$ , the free module on $X$ is isomorphic to $R$ as an $R$ -module.

The free module on a finite set

Next let's consider an arbitrary finite set $X = {x_{1}, x_{2}, \dots, x_{n}}$ . By our construction, the module $F (X)$ consists of all formal $R$ -linear combinations of the form $r_{1} x_{1} + r_{2} x_{2} + \dots + r_{n} x_{n}$ , which his evidently isomorphic to the $R$ -module $R \oplus R \oplus \dots \oplus R$ (the direct sum of $R$ with itself $n$ times). This $R$ -module is usually denoted $R^{n}$ .

Is this bad notation?

The notation $R^{n}$ is usually shorthand for the direct product $R \times R \times \dots \times R$ , not the direct sum. Fortunately, for finite families in the category $R - Mod$ the direct product and direct sum are isomorphic $R$ -modules.

Suggested next note

Generators for modules and submodules