Why $ S^{-1}R $ is local if and only if the saturation of $ S $ (set of divisors of elements of $ S $) is the complement of a prime ideal?

Question

I was trying to understand $ S^{-1}R $ is local if and only if the saturation of $ S $ (set of divisors of elements of $ S $) is the complement of a prime ideal. Is that true? How to prove it?

Answer 1

$R_P$ is a local ring, i.e. it has a unique maximal ideal of the form $$m_P=\Bigl\{\frac{p}{s}\,;\, p\in P, s\in S=R-P \Bigr\}$$ note that an element of $R_P$ not in $m_P$ is of the form $\frac{s'}{s}$ with $s'\in S$, and so has an inverse $\frac{s}{s'}$, so is a unit.

Answer 2

Hint: First show that if $S$ is a multiplicative subset of a ring $R$, with saturation $S'$, then $R\setminus S'$ is the union of the prime ideal $\mathfrak p$ such that $\mathfrak p\cap S = \emptyset$.

Then, note that the maximal elements of the set $\{\mathfrak p\subset R\mid \text{prime & }\mathfrak p\cap S = \emptyset\}$ correspond to the maximal ideals of $S^{-1}R$. Hence, $S^{-1}R$ has a unique maximal ideal if and only if $\{\mathfrak p\subset R\mid \text{prime & }\mathfrak p\cap S = \emptyset\}$ has a unique maximal element. What does this mean for $S'$?