Dual, Projective Modules, and Isomorphisms

Question

Given a finitely generated right $R$-module $M$ with a generating set $\{m_i\}$, can one define a map $M$ to its dual left-module of right module functions by $m_i \mapsto m_i^*$ in analogy with the vector space case? I'm guessing one needs to assume projectivity here, but I can't see how to use it.

Answer 1

I doubt this is possible because the generators need not be independent. I don't think $m_i^*$ is even well-defined for this reason. Given a free module $M$ with basis $e_1, \ldots, e_n$, recall that we define the dual basis $e_1^*, \ldots, e_n^*$ by $$ e_i^*(e_j) = \begin{cases} 1 & i = j\\ 0 & i \neq j \end{cases} $$ and then extending $R$-linearly by the universal property of free modules. But if the module isn't free, why should this map exist?

For example, take $R = \mathbb{Z} \oplus \mathbb{Z}$, $M = \mathbb{Z} \oplus 0$, (so $M$ is a direct summand of $R$, hence is projective) and take $m_1 = (1,0)$ as the generator of $M$. Then what is $m_1^*(2,0)$? On one hand, $(2,0) = (2,0) m_1$ so $m_1^*((2,0)m_1) = (2,0)$, but we could also consider $(2,0) = (2,5)m_1$, so $m_1^*((2,5)m_1) = (2,5)$. Thus there is no way to give a well-defined map $m_1^*$.

The key here is that for a free module $M$ with basis $e_1, \ldots, e_n$, for any $m \in M$ there exist unique $r_1, \ldots, r_n \in R$ such that $m = r_1 e_1 + \cdots + r_n e_n$. It is this uniqueness that allows to give a well-defined definition for the dual basis in this case, and the failure of uniqueness that makes it impossible otherwise.