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We can easily construct free $R$-module when $R$ is unital by setting $$R[S] = \{ f\colon S\to R\,|\, f\ \text{finitely supported}\}$$ and defining operations pointwise. The key here is that we can include the set of generators $S$ into $R[S]$ via map $s\mapsto \delta_s$ where $\delta_s(x) = 1_R$ if $x = s$ and zero otherwise. This allows us to uniquely express any $f\in R[S]$ as $$f=\sum_{s\in S} f(s)\delta_s$$ and thus extend by linearity any function $\varphi\colon S\to M$ to linear map $\overline\varphi\colon R[S]\to M$. Thus, $R[S]$ is indeed free.

Can we construct free module for non-unital rings? If possible, please give construction or reference.

P.S. One needs adequate notion of module here, of course. It's just representation of non-unital ring over abelian group, i.e. we just drop axiom $1_R\cdot a = a$.

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    $\begingroup$ If $R$ is not unital then there is no canonical $\theta : S \to R^S$ for our universal property. $\endgroup$ – basket Dec 28 '15 at 19:22
  • $\begingroup$ @basket, yes, that is what's bothering me. But the underlying set doesn't necessarily have to be $R^S$, or is there an argument for that? $\endgroup$ – Ennar Dec 28 '15 at 19:29
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    $\begingroup$ When $R$ is not unital $R[S]= \{f:S\to \mathbb{Z}\ltimes R\,|\, f \text{ finitely supported}\}$ (where $\mathbb{Z}\ltimes R$ is the ring obtained by freely adding a unit to $R$). $\endgroup$ – Nex Dec 29 '15 at 8:16
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    $\begingroup$ @Nex, thank you for the comment. It's rather obvious when told the answer: $R$ modules are same as $\mathbb{Z}\ltimes R$ modules because adding a unit is left adjoint (perhaps more is needed to argue that $R$-linearity is same as $\mathbb{Z}\ltimes R$-linearity). $\endgroup$ – Ennar Dec 29 '15 at 12:18

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