For a commutative ring $A$, $S$ multiplicative subset, if $p \in A$ is prime, then $p/s$ is prime in $S^{-1}A$.

Question

For a commutative ring $A$, $S$ multiplicative subset, if $p \in A$ is prime with $(p) \cap S = \emptyset$, then $p/s$ is prime in $S^{-1}A$.

How do I prove this? Any suggestion or hint?

Answer 1

Hint: Suppose $(p/s)|ab$, for $a,b\in S^{-1}A$. Then prove that $(p/s)|a$ or $(p/s)|b$, using the structure of $a,b$ as elements of $S^{-1}A$.

Answer 2

Suppose $$ \frac{p}{s} \mid \frac{a}{c}\frac{b}{d} $$ Then, $\exists m/n \in S^{-1}A$ such that $$ \frac{mp}{sn} = \frac{ab}{cd} $$ So $$ mpcd = absn \text{ in } A $$ Does this help?