2
$\begingroup$

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 .

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.