"Dual function" for an element of an $R$-module

Question

Let $M$ be a finitely generated projective right module over a not-necessarily commutative ring $R$. For any non-zero $e \in M$ will there always exist a right $R$-module map $f:M \to R$ such that $f(e) \neq 0$?

Answer 1

Since $M$ is projective, it is isomorphic to a direct summand of a direct sum of copies of $R$ (in your case finite, but it's irrelevant): let $$ \alpha\colon M\to R^{(\Lambda)} \qquad \beta\colon R^{(\Lambda)}\to M $$ be the corresponding injection and projection, with $\beta\alpha$ the identity on $M$.

If $x\in M$, $x\ne0$, we have $$ \alpha(x)=\sum_{\lambda\in\Lambda}r_\lambda e_\lambda $$ (where $\{e_\lambda:\lambda\in\Lambda\}$ is the standard basis of the free module). Suppose $r_{\lambda_0}\ne0$. Compose with the projection on the $\lambda_0$-coordinate: under this map, we have $x\mapsto r_{\lambda_0}\ne0$.