# Localization at a multiplicative set is a localization at a prime ideal if local

Let $$R$$ be an integral domain. Assume $$S$$ is a multiplicative set in $$R$$ such that the localization $$S^{-1}R$$ is a local ring. Does there exist a prime ideal $$\mathfrak{p}\subset R$$ such that $$S^{-1}R\simeq R_{\mathfrak{p}}$$ as $$R$$-algebras (if not, at least as unital rings)?

Edit: In the following, $$S$$ is assumed to be saturated i.e. $$\forall a,b \in R: ab \in S \Rightarrow a \in S \wedge b \in S$$. A counterexample for when $$S$$ is not saturated is $$\mathbb{R} [[x]]$$ with $$S = \{1\}$$. Of course there still exists a prime ideal $$p \subseteq \mathbb{R}[[x]]$$ s.t. $$R \simeq R_p$$, i.e. $$p = \mathbb{R}[[x]] \setminus (\mathbb{R}[[x]])^{\times}$$.

If $$S$$ is saturated, then via contra positiv we have $$\forall a,b \in R: a \in R \setminus S \vee b \in R \setminus S \Rightarrow ab \in R \setminus S$$ which then gives closure under left multiplication and prime.

Yes, that is the case assuming $$0 \not\in S$$. Given that that $$S$$ is saturated we have $$p = R \setminus S$$. For the general case see reuns answer.

We want to show that there exists some $$p \in \text{Spec}(R)$$ s.t.

$$\{\frac{a}{b} \vert a \in R, b \in S\} = S^{-1}R = R_p = (R \setminus p)^{-1} R = \{\frac{a}{b} \vert a \in R, b \in R \setminus p\}$$

Hence we want show that $$R \setminus S$$ is a prime ideal.

Write $$A := S^{-1}R$$. Since $$A$$ is local it has a unique maximal ideal given by $$A \setminus A^{\times}$$. Then for any $$a \in R, s \in S$$ we have

$$\frac{a}{s} \in A \setminus A^{\times} \Leftrightarrow \frac{s}{a} \not\in A \Leftrightarrow a \not\in S \Leftrightarrow a \in R \setminus S$$

Now just check that $$\forall a, b \in R \setminus S$$, $$r \in R$$

• $$0 \in R \setminus S$$ since $$0 \not\in S$$
• $$-a \in R \setminus S$$ since $$A \setminus A^{\times}$$ an ideal i.e. given that $$\frac{a}{1} \in A \setminus A^{\times}$$ also $$\frac{-a}{1} \in A \setminus A^{\times}$$.
• $$a+b \in R \setminus S$$ since $$A \setminus A^{\times}$$ an ideal and $$S$$ saturated i.e. $$a,b \in R \setminus S$$ we have $$\frac{a}{1}, \frac{b}{1} \in A \setminus A^{\times}$$ and thus $$\frac{a+b}{1} \in A \setminus A^{\times}$$
• $$rb \in R \setminus S$$ since $$A \setminus A^{\times}$$ an ideal and $$S$$ saturated as above. $$\color{red}{\text{This can fail with x \in R^{\times} and r = x, b = \frac{1}{x} if S is not saturated}}$$

• prime: given $$ab \in R \setminus S$$ since $$A \setminus A^{\times}$$ an ideal and $$S$$ saturated as above. $$\color{red}{\text{This can fail with x \in R^{\times} and \frac{ab}{1} = \frac{a}{x} \frac{b}{1/x}}}$$

Hence $$R \setminus S$$ is a prime ideal in $$R$$.

$$R$$ is an integral domain and $$A=S^{-1} R$$. That it is a local domain means it has a unique maximal ideal $$m\subset A - A^\times$$.

For every non-zero $$a \in A - A^\times$$ then $$(a)$$ is a proper ideal of $$A$$ so it is contained in a maximal ideal which must be $$m$$, thus $$a \in m$$ and $$m = A-A^\times$$.

Let $$P = m \cap R= (A-A^\times) \cap R = R - (A^\times \cap R)$$ which is a prime ideal of $$R$$.

Whence $$A = (A^\times \cap R)^{-1}R=(R-P)^{-1} R=R_P$$