Prime ideals in $A$ and prime ideals in $S^{-1}A$

Question

Let $A$ be a ring and $S$ be a multiplicative closed subset. Then there is a 1 to 1 correspondence between the prime ideals in $A$ (intersect $S$ is empty) and prime ideals in $S^{-1}A$.
My question is in what sense they are corresponded?
I know we can do the canonical map $A \rightarrow S^{-1}A$, by $a \rightarrow a/1$, but assume $I$ is an ideal not meet $S$, I don't think $f(I) = S^{-1}(I)$, and I'm not sure if it's true that $f^{-1}(S^{-1}(I)) = I$?

Answer 1

They correspond as follows:

1. If $\mathfrak p$ is a prime ideal disjoint from $S$, then $\mathfrak p \mapsto S^{-1}\mathfrak p$.
2. If $\widetilde {\mathfrak p}$ is a prime ideal in $S^{-1}A$, then $\widetilde{ \mathfrak p} \mapsto f^{-1}(\widetilde{\mathfrak p})$ where $f : A \to S^{-1}A$ is defined by $a \mapsto a/1$. and $f^{-1}(\widetilde{\mathfrak p})$ is a prime ideal disjoint from $S$.

and then notice that

3. $S^{-1}\big(f^{-1}(\widetilde{\mathfrak p})\big) = \widetilde{\mathfrak p}$ and
4. $f^{-1}\big(S^{-1}\mathfrak p\big) = \mathfrak p$.

shows the maps are bijective.

Proof: For (1) let $\mathfrak p$ be a prime ideal disjoint from $S$, let $\displaystyle \frac{x}{m} \cdot \frac{y}{n} = \frac{xy}{mn} \in S^{-1}\mathfrak p$. There exists $z \in \mathfrak p$, $s \in S$ such that $\displaystyle \frac{xy}{mn} = \frac{z}{s}$ which means there exists $t \in S$ such that $(xys - mnz)t = 0$ or equivalently $xyst = zmnt \in \mathfrak p$. Notice that $st \in S$ and $S \cap \mathfrak p = \emptyset$ so $xy \in \mathfrak p$ must hold since $\mathfrak p$ is prime. Again since $\mathfrak p$ is prime, either $x \in \mathfrak p$, in which case $\displaystyle \frac{x}{m} \in S^{-1}\mathfrak p$, or $y \in \mathfrak p$, in which case $\displaystyle \frac{y}{n} \in S^{-1} \mathfrak p$. Conclude that $S^{-1}\mathfrak p$ is a prime ideal in $S^{-1}A$.

For (2) let $\widetilde{\mathfrak p}$ be a prime ideal in $S^{-1}A$. Consider $\mathfrak p = f^{-1}(\widetilde{\mathfrak p}) = \{a \in A : a/1 \in \widetilde{\mathfrak p}\}$ which is prime since the inverse image of a prime ideal is prime (via a homomorphism). If, by way of contradiction, $\mathfrak p$ intersects with $S$, then it would imply that $\widetilde{\mathfrak p} = S^{-1}A$, a contradiction to it being prime by hypothesis (and hence proper).

For (3) we look at two containments:

1. ($\subseteq$) Let $\displaystyle \frac{x}{m} \in S^{-1}\big( f^{-1}(\widetilde{\mathfrak p}) \big)$ which means $m \in S$ and $x \in f^{-1}(\widetilde{\mathfrak p})$ or equivalently $\displaystyle \frac{x}{1} = f(x) \in \widetilde{\mathfrak p}$. Hence $\displaystyle \frac{x}{m} = \frac{x}{1} \cdot \frac{1}{m} \in \widetilde{\mathfrak p}$ since ideals absorb.
2. ($\supseteq$) Let $\displaystyle \frac{x}{m} \in \widetilde{\mathfrak p}$ with $x \in A$ and $m \in S$, then $\displaystyle f(x) = \frac{x}{1} = \frac{x}{1} \cdot \frac{m}{m} = \frac{x}{m} \cdot \frac{m}{1} \in \widetilde{\mathfrak p}$ since ideals absorb. Hence $x \in f^{-1}(\widetilde{\mathfrak p})$ which means $\displaystyle \frac{x}{m} \in S^{-1}\big(f^{-1}(\widetilde{\mathfrak p})\big)$ as desired.

For (4) we look at two containments:

1. ($\subseteq$) Let $x \in f^{-1}\big(S^{-1}\mathfrak p\big)$, then $\displaystyle f(x) = \frac{x}{1} \in S^{-1}\mathfrak p$. Then $\displaystyle \frac{x}{1} = \frac{y}{n}$ for some $y \in \mathfrak p$ and $n \in S$. This means that there exists $s \in S$ such that $(xn - y)s = 0$ or equivalently $xns = ys \in \mathfrak p$. But notice that $ns \in S$ and hence $ns \notin \mathfrak p$ since $\mathfrak p \cap S = \emptyset$ by hypothesis. Thus $x \in \mathfrak p$ since $\mathfrak p$ is prime.
2. ($\supseteq$) Let $x \in \mathfrak p$, then $\displaystyle f(x) = \frac{x}{1} \in S^{-1}\mathfrak p$ so $x \in f^{-1}\big(S^{-1}\mathfrak p\big)$ as desired.