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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).

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  • $\begingroup$ it is exactly that $\endgroup$ Commented Oct 20, 2015 at 20:10

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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.

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  • $\begingroup$ Fair point, thanks! I guess it feels somewhat unsatisfactory to me for some reason. Nevertheless I'm sure it is useful to have so many examples of free modules at your disposal. Hopefully this is useful to someone else as well! $\endgroup$
    – Ducky
    Commented Oct 20, 2015 at 20:15

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