Let \(R\) be a ring. An $R$-module \(M\) is free if it has a basis: a family \((\beta_i)_{i\in I}\) in \(M\) such that every \(m\in M\) can be written as \(m=\sum_{i\in I} r_i\beta_i\) with \(r_i\in R\), all but finitely many zero, and \(\sum_{i\in I} r_i\beta_i = 0\) implies \(r_i = 0\) for all \(i\).
Let \(R\) be a ring and \(M\) an \(R\)-module. The following are equivalent:
\(M\) is free, with basis \((\beta_i)_{i\in I}\);
there are a set \(\mathcal{B}\) and a map of sets \(\iota\colon\mathcal{B}\to M\) such that for every \(R\)-module \(N\) and every map of sets \(f\colon\mathcal{B}\to N\) there is a unique \(R\)-linear map \(\tilde f\colon M\to N\) with \(\tilde f\circ\iota = f\):
- there is a family \((\beta_i)_{i\in I}\) in \(M\) such that \(M = \bigoplus_{i\in I} R\beta_i\) is the internal direct sum of the cyclic submodules \(R\beta_i\), and \(r\mapsto r\beta_i\) is injective for each \(i\), so that \(R\beta_i\cong R\).
Given a basis \((\beta_i)_{i\in I}\), take \(\mathcal B=I\) and \(\iota(i)=\beta_i\); the unique extension of \(f\) is \(\tilde f(\sum_i r_i\beta_i)=\sum_i r_if(i)\), which is well defined because the coefficients \(r_i\) are unique. Conversely, if \((\mathcal B,\iota)\) has the universal property, compare \(M\) with the module \(R^{(\mathcal B)}\) of finitely supported functions \(\mathcal B\to R\) and its standard basis \((e_b)_{b\in\mathcal B}\): the linear maps \(M\to R^{(\mathcal B)}\) extending \(b\mapsto e_b\) and \(R^{(\mathcal B)}\to M\) extending \(e_b\mapsto\iota(b)\) compose in both orders to maps that restrict to the identity on the generators, so by uniqueness they are inverse isomorphisms, and \((\iota(b))_{b\in\mathcal B}\) is a basis of \(M\). Finally, the spanning condition says that \(M=\sum_i R\beta_i\), and linear independence says exactly that this sum is direct and each \(r\mapsto r\beta_i\) is injective.
Let \(R\) be a ring, \(M\) a left \(R\)-module, \(S\subseteq M\) a subset, and \(R^{(S)} \coloneqq \bigoplus_{s\in S} R\) with standard basis \((e_s)_{s\in S}\). There is a unique \(R\)-linear map \(\varphi\colon R^{(S)}\to M\) with \(\varphi(e_s) = s\) for all \(s\in S\).
\(\varphi\) is surjective if and only if \(S\) is a generating set of \(M\); in that case \(M\cong R^{(S)}/\ker\varphi\).
\(\varphi\) is an isomorphism if and only if \(S\) is a basis of \(M\). In particular, \(M\) is free if and only if \(M\cong R^{(T)}\) for some set \(T\) [@DF04].