# Free R-module as a direct sum of copies of R

I am currently taking a class on module theory, and we recently defined a free $R$-module (over some fixed ring $R$) as follows:

An $R$-module $F$ is free if there exists a set $S\subseteq F$ such that $S$ is both linearly independent and a generating set for $F$. We call such an $S$ a basis for $F$.

We proved shortly after that, given a set $S$, we can construct a free $R$-module $F(S)$ with basis elements that are in one-to-one correspondence with the elements of $S$ as such: $$F(S)=\bigoplus_{s\in S}R_s$$

where $R_s$ is the ring $R$ considered as a left $R$-module over itself.

The $R_s$ notation confuses me - does it mean that we are taking as many copies of $R$ as there are elements in $S$, and direct summing them? As a simple-minded example, then, if I want a free $R$-module with basis $S=\lbrace s_1,...,s_n \rbrace$, then I should let $F(S)=R\oplus R\oplus ...\oplus R$, where there are $n$ factors of $R$?

This seems like $F(S)$ will be fairly complicated if $S$ is a sufficiently nasty set (although that is more of an observation than a strike against the result).

• it is exactly that Oct 20, 2015 at 20:10

What you say is correct. About your last sentence, notice that what really matters for $F(S)$ is just the cardinality of $S$, not how $S$ is defined concretely.