I am currently stuck on this problem:
"Let $R$ be an integral domain, let $0 \notin S \subseteq R$ be a multiplicative group of $R$. If $S^{-1}R$ is finitely generated as a $R$-module, then every element of $S$ is invertible (in $R$)"
I just started doing Algebra last month and I have some difficulties grasping these abstract concepts. So let's assume, that $S^{-1}R$ is finally generated by $\frac{1}{s_1}, … , \frac{1}{s_n}$ with $s_1, … , s_n \in S$. What can I do now? My first guess, that $S$ has to be finite as well, because if I understood it right, the localization $S^{-1}R$ describes a Ring, that is a little bit bigger than $R$ that also contains the multiplicative inverses of all Elements of $S$. So for example, if $R = \mathbb{Z}$ and $S = \mathbb{N}$, then $S^{-1}R = \mathbb{Q}$.
Is that so far correct? But in this case, $\mathbb{Q}$ is not a finitely generated $\mathbb{Z}$-module, so it's not a very good example, I tried it on $Z_{11}$, and as a field it obviously contains multiplicative inverses, so every multiplicative group is a unit. Also, it looks like if all such Rings $R$ would already satisfy $R = S^{-1}R$, is that true?
Nevertheless, I still don't know how to formally prove it, can someone maybe give me a small tip, because right now I don't even know where to start.
Best Answer
Hint for a direct argument: suppose that $S^{-1}R$ is generated as an $R$-module by finitely many $r_i/s_i$. Let $s = \prod s_i$. Rewrite $r_i / s_i = r_i'/ s$ where $r_i' = r_i \prod_{j \not= i} s_j$. Now for any $t \in S$, consider the element $1 /ts \in S^{-1}R$. Write $1/ts$ as an $R$-linear combination of the $r_i'/s$ and mess around with the expression to deduce that $t$ is a unit of $R$.