**Proof plan.** Translate the prime ideal condition and the [integral](/page/Integral) domain condition via the quotient ring $R/I$. **Step 1: ($\Rightarrow$) Prime ideal implies $R/I$ is an integral domain.** [claim: Prime Implies Integral Domain] If $I$ is prime, then $R/I$ is an integral domain. [/claim] [proof] $R/I \neq 0$ since $I \neq R$. Let $(a + I)(b + I) = 0_{R/I}$, i.e. $ab + I = I$, so $ab \in I$. Since $I$ is prime, $a \in I$ or $b \in I$, i.e. $a + I = 0_{R/I}$ or $b + I = 0_{R/I}$. So $R/I$ has no zero divisors, hence is an integral domain. [/proof] **Step 2: ($\Leftarrow$) $R/I$ an integral domain implies $I$ is prime.** [claim: Integral Domain Implies Prime] If $R/I$ is an integral domain, then $I$ is prime. [/claim] [proof] $I \neq R$ since $R/I \neq 0$. Let $ab \in I$. Then $(a+I)(b+I) = ab + I = 0_{R/I}$. Since $R/I$ is an integral domain, $a + I = 0$ or $b + I = 0$, i.e. $a \in I$ or $b \in I$. So $I$ is prime. [/proof] These two claims give the equivalence. $\square$