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

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.

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