A module over a noetherian ring and its dual

Question

Let $R$ be a noetherian ring, $M$ an $R$-module and $M^*$ be its dual. Is it true that if $M$ is noetherian then $f(x) = 0$ for all $f \in M^*$ implies $x =0$? How about the other way round?

Ok, so the above is clearly not true.

How about if I change the condition $M$ is noetherian into $M$ is a submodule of $R^n$ for some $n$?

Answer 1

Suppose there is $0\to M\stackrel{\varphi}\to R^n$. Let $\pi_i:R^n\to R$ be the canonical projections. Then $\pi_i\circ\varphi\in M^*$, so $(\pi_i\circ\varphi)(x)=0$. This shows that $\varphi(x)=0$ and since $\varphi$ is injective we get $x=0$.

(A module which is isomorphic to a submodule of a direct product of copies of $R$ is torsion-less, that is, the canonical map $M\to M^{**}$ is injective. The converse also holds.)

Answer 2

This is not true, suppose $R=Z$ and $M=Z/n$. The dual of $Z/n$ is trivial.