Let $R = k[x]$ be a polynomial ring in one indeterminant and $S \supset R$ a finite free $R$-module of rank $n$ over $R$ and of Krull-dimension one.
Consider $M = \operatorname{Hom}_R(S,R)$ as an $S$-module via $(s\cdot\phi)(x) = \phi(sx)$ for all $s,x \in S$ and $\phi \in M$.
My question is: $\quad$ Is $M$ as an $S$-module torsionless?
That is, is the canonical map $M \to M^{**}$ from $M$ to its double dual (dual as an $S$-module!) injective? Or equivalently, does for every non-zero $\phi \in M$ exist some $f \in M^*$ such that $f(\phi) \neq 0$?
I know that the dual of a module is torsionless, that is $M$ considered as an $R$-module is torsionless.
The first ansatz that comes to mind might be the following: For given $R$-linear $\phi: S \to R$ we need to find an assignment that produces an element in $S$ and is $S$-linear. If $\phi \neq 0$, there is some $x \in S$ with $\phi(x) \neq 0$. Hence we could try the assignment $f$ that maps $\psi \mapsto \psi(x)$ for every $\psi \in M$. Then we would have $f(\phi) \neq 0$. But $f$ is not $S$-linear since $\phi$ need not be $S$-linear: $$f(s \cdot \phi) = (s \cdot \phi)(x) = \phi(sx) \neq s\ \phi(x) = s\ f(\phi).$$