Do prime ideals stay prime under localization?

Question

Let $A$ be a ring and $S$ a multiplicatively closed subset of $A$. Let $\mathfrak{p}$ be a prime ideal of $A$ with $\mathfrak{p} \cap S = \emptyset$. Is $S^{-1}\mathfrak{p}$ a prime ideal in $S^{-1}A$?

Answer 1

If $A$ is commutative, we have an order preserving bijection from the prime ideals of $A$ having empty intersection with $S$ to the prime ideals of $S^{-1} A$ via $P \mapsto S^{-1} P$. Its inverse is $Q \mapsto Q \cap A$. See for instance the commutative algebra books by Atiyah and Macdonald or Matsumura.

Answer 2

An ideal $I$ is a a prime ideal of a commutative ring $R$ iff $\forall a, b \in R$, $ab \in I \implies (a \in I \lor b \in I)$.

Let $P$ be a prime ideal of $R$. Let's localize the ring $R$ at $S$. So we're consider the ring $T \equiv S^{-1}R$, and the ideal $S^{-1} I$. Let $(r_1, s_1), (r_2, s_2) \in S^{-1}R$ such that $(r_1, s_1) \times (r_2, s_2) = (r_1 r_2, s_1 s_2) \in S^{-1} P$. That is, there exists $s' \in S, p' \in P$ such that $(r_1 r_2, s_1 s_2) = (p', s')$.

Recall that $P$ was a prime ideal in $R$, so since $(r_1 r_2 = p)$, $(r_1 r_2 \in P)$. Hence, $r_1 \in P \lor r_2 \in P$. Let $i$ be such that $r_i \in P$. Hence, $(r_i, s_i) \in S^{-1}P$.

We've shown that $(r_1, s_1) \times (r_2, s_2) \in S^{-1}P \implies (r_1, s_1) \in S^{-1} P \lor (r_2, s_2) \in S^{-1}P$, so $S^{-1}P$ is indeed prime if $P$ is prime.