Localization is finitely generated implies, that multiplicative group only contains invertible elements

Question

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.

Answer 1

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 $R$.

Answer 2

I apologise for the somewhat misleading comments I've been posting above, for I have only now realised you include in the problem context the hypothesis that your original ring is an integral domain.

The problem statement remains true in a slightly more general context, which I would like to present. Assume that $A$ is a commutative ring and $S \leqslant_{\mathrm{Mon}}A$ (the notation $\leqslant_{\mathrm{Mon}}$ means submonoid of) is a multiplicative system consisting only of cancellable elements, which is equivalent to claiming that the localisation map $\iota \colon A \to S^{-1}A$ is injective. Then if $S^{-1}A$ is finitely generated as an $A$-module than necessarily $S \subseteq \mathrm{U}(A)$. For arbitrary $t \in A$ and $x \in S^{-1}A$ I will use the notation $t.x\colon=\iota(t)x$.

Let $Y \subseteq S^{-1}(A)$ be a finite generating system for $S^{-1}(A)$ as an $A$-module. By virtue of this finiteness, considering the product of all the (finitely many) denominators occurring in some fixed representations of the elements of $Y$ as fractions, one easily infers the existence of $a \in S$ such that $a.Y \subseteq \mathrm{Im}\iota$ and subsequently the existence of $X \subseteq A$ such that $a.Y=\iota[X]$ (we can argue that a finite such $X$ exists, but it is not relevant for the rest of the argument).

Since $A$ and its localisation are both commutative, the map of multiplication by $\iota(a)$ is an endomorphism of $S^{-1}(A)$ as a module over itself and hence in particular as an $A$-module. As this map is surjective (it is actually an automorphism, since $\iota(a)$ is invertible as $a \in S$), it follows that it maps the generating system $Y$ onto yet another generating system: $$S^{-1}(A)=\iota(a)S^{-1}(A)=\iota(a)\langle Y \rangle_A=\langle \iota(a)Y \rangle_A=\langle a.Y \rangle_A=\langle \iota[X] \rangle_A=\iota[(X)],$$ where we remark that the last equality makes use of the fact that $\iota$ is in particular $A$-linear.

At this point we have ascertained that $\iota$ is surjective ($S^{-1}A=\iota[(X)]\subseteq \iota[A] \subseteq S^{-1}(A)$) and we know by hypothesis that it is also injective. As I remarked in one of the comments above, it is an elementary fact of localisation theory that the localisation map is bijective if and only if the multiplicative system consists only of units, which is in particular the case here.