Let $R$ denote a commutative ring, $S\subset R$ a subset that is closed under multiplication and $\pi$ the canonical ring homomorphism, i.e., $\pi: R\rightarrow S^{-1}R, r\mapsto \frac{r}{1}$.
(i) Prove that for every prime ideal $J\subset S^{-1}R$, $\pi^{-1}(J)\subset R$ is a prime ideal with $S \ \cap \pi^{-1}(J)=\emptyset$.
(ii) Let $P\subset R$ be a prime ideal. Then prove that $S=R\ \backslash P$ is closed under multiplication and the fraction of rings $S^{-1}R$ has exactly one maximal ideal (i.e., is local).
$\textbf{Own ideas:}$
ad (i). (Note that $\pi^{-1}(J) := \{ r \mid \pi(r)\in J \}$) I have already proven that $\pi^{-1}(J)$ is a prime ideal, but am currently stuck at why $S \cap \pi^{-1}(J)=\emptyset$. Let $x\in \pi^{-1}(J)\Rightarrow \pi(x)\in J$. Suppose $x\in S$ holds as well. Where is the contradiction? I don't see one, yet ...
ad (ii). Showing that $S$ is closed under multiplication isn't difficult. In a previous exercise sheet, we showed that $R \ \text{local}\Leftrightarrow \text{set of non-units of $S^{-1}R$ forms an ideal.}$ Let $\text{NU}(S^{-1}R)$ denote this set. I am not quite managing it yet to show that this set is really an ideal. I always check that $\left( \text{NU}(S^{-1}R), +\right)\subset $(S^{-1}R, +)$ is an abelian subgroup:
- $0\in \text{NU}(S^{-1}R)$,
- $\underbrace{\frac{a}{b}}_{\in NU}+\underbrace{\frac{e}{f}}_{\in NU} = \frac{af+eb}{bf}$,
- $\dots$
A bit help would be appreciated concerning that $NU(S^{-1}R)$ is supposed to be an ideal.
Kind regards, MathIsFun