Finitely generated R-module is a field iff R is a field?

Uncover the spoilers for solutions completing the hints below:

• Suppose $R$ is a field, and let $s \in S$ be a nonzero element. Then multiplication by $s$ is an $R$-linear endomorphism of $S$, which is injective since $s$ is nonzero and $S$ is a domain.

  Since $S$ is a finite-dimensional $R$-vector space, it follows that multiplication by $s$ is also surjective, and so $1$ is in the image of this map.

• Suppose $S$ is a field. This solution I have in mind for this direction is a bit trickier. Let $r \in R$ be nonzero, and let $s$ be the inverse to $r$ in $S$. As before, consider the $R$-linear map $\varphi_{s} \colon S \to S$ corresponding to multiplication by $s$. Since $S$ is a finitely generated $R$-module, $\varphi_{s}$ satisfies a monic polynomial relation with coefficients in $R$ by Cayley-Hamilton. That is, there exist $r_{1}, \ldots, r_{n} \in R$ such that multiplication by $s^{n}+r_{1}s^{n-1} + \cdots +r_{n}$ is the zero element of $\mathrm{End}_{R}(S)$.

  Since $S$ is a faithful $R$-module ($R$ is a subring of $S$,and so contains $1$), this implies that $s^{n}+r_{1}s^{n-1} + \cdots +r_{n} = 0$. Now multiply both sides by $r^{n-1}$ to conclude that $s \in R$.

Hints:

$\Rightarrow$: If $R$ is a field, let $s\in S$, and consider multiplication by $s$ in $S$. Check this is an injective $R$-linear map. What can you conclude, knowing $S$ is a finite dimensional $R$-vector space?

$\Leftarrow$: If $S$ is a field, consider $r\in R$; you know $r^{-1}\in S$, hence it is a root of a monic polynomial in $R[X]$. Deduce from this polynomial equation that $r^{-1}$ is a polynomial in $r$, hence it belongs to $R$.