The contraction of prime ideal is prime

Question

Let $R$ be a commutative ring with identity. Let $S$ be an extension ring of $R$ and $I \neq S$ be a prime ideal of $S$. The author says that $J=I \cap R$ is a prime ideal of $R$. I fail to understand that.

Let $ab \in J$. I have to show that $a \in J$ or $b \in J$. Since $I$ is prime ideal, $a \in I$ or $b \in I$, let $a\in I$. How can we derive that $a \in R$?

In the other way, we know that $S/J$ is an integral domain. It suffices to show that $R/R\cap I$ is an integral domain. But I fail to find an isomorphism.

I need your help. Thanks in advance.

Answer 1

$R \to S$ induces an embedding $R/(I \cap R) \to S/I$. Subrings of integral domains are integral domains. $\square$

Answer 2

Let $a,b \in R$ such that $ab \in J \implies ab \in I \cap R \implies ab \in I \implies a\in I \text{ or } b \in I $$\implies a\in I \cap R \text{ or } b \in I \cap R \implies a \in J \text{ or } b \in J$