When the localization of a ring is a finitely generated module over the ring via the canonical map from the ring to the localization

Question

Let $R$ be a commutative ring with unity and let $S$ be a multiplicative closed subset of $R$ ; let $f:R \to S^{-1}R$ be the natural map $f(r)=r/1 , \forall r \in R$ ; then $f$ is a ring homomorphism sending unity to unity , so $f(R)$ is a subring of $S^{-1}R$ with the same unity as that of $S^{-1}R$ ; so we can naturally consider $S^{-1}R$ as an $f(R)$-module ; now my question is :

If this $S^{-1}R$ as an $f(R)$-module is finitely generated , then is $f$ surjective ?

I can only show that if $S^{-1}R$ is cyclic as an $f(R)$-module then I can show that $f$ is surjective .

Answer 1

If $S^{-1}R$ is finitely generated then it is automatically cyclic. Indeed, let $\frac{r_1}{s_1},\dots,\frac{r_n}{s_n}$ be a finite collection of generators, with $r_i\in R$ and $s_i\in S$ for all $i$. Note that all of these generators are in the submodule generated by $\frac{1}{s_1s_2\dots s_n}$: for instance, $\frac{r_1}{s_1}=(r_1s_2s_3\dots s_n)\cdot \frac{1}{s_1s_2\dots s_n}$. So actually, $S^{-1}R$ is generated by $\frac{1}{s_1s_2\dots s_n}$ and hence is cyclic.