Equivalently, an $R$-module $M$ is projective iff it has a dual basis$\{(x_i,x_i^*)\in M\times Hom_k(M,R)\}_{i\in I}$ in the following sense. There is a free $R$-module $F = \oplus_{i\in I} R$, an epimorphism of $R$-modules $F\to M$, $F=\oplus_i R \ni \sum_i r_i\mapsto \sum_i r_i x_i$ which is split, i.e. has a right inverse, where this inverse, by the universal property of the direct sum, must be of the form $x\mapsto \sum_i x_i^*(x)\in \oplus_i R$ where $x_i^*\in Hom_k(M,R)$. The right inverse condition translates to $\forall x \in M : x=\sum_i x_i^*(x)x_i$. In particular for every $x\in M$$x_i^*(x)\neq 0$ for only finitely many $i\in I$.

Relation to dual vector spaces

This terminology is related to but a bit different than in the case of $k$-vector spaces (cf. at dual vector space). If $V$ has a vector space basis $(x_j)_{j\in I}$ is a linear basis of $V$ then $x_i^*$defined by$x_i^*(x_j) = \delta_i^j$ is not necessarily a basis of $V^* = Hom_k(V,k)$; it is if $V$ is finite dimensional. In the case of projective module $M$, $x_i$ do not form a basis in a free sense, but only a set of generators and with the property that there exist another set $x_i^*$ in $M^*$ such that they together form a “dual basis”. (Still, one sometimes says that $x_i^*$ form a basis dual to $x_i$.)