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Let $R$ be commutative ring with the identity $1\neq 0$. The ideal $P$ is a prime ideal in $R$ ring iff $R/P$ is an integral domain.

Proof. It is important to use the definition of a prime ideal in this proof.

“$\Rightarrow$“: Assume that $P$ is a prime ideal of $R$. Then $P \neq R$ and when $ab \in P$, then either $a \in P$ or $b \in P$. Take the quotient ring of $R$ by $P$, that is, $R/P$. Since $P$ is a prime ideal, we know firstly that $R/P \neq \overline{0}$.

Let $\overline{r} = r + P \in R/P$. Then $\overline{r} = 0$ in $R/P$ if and only if $r \in P$. Now take the elements $\overline{a} = a + P$ and $\overline{b} = b + P$ in $R/P$. Take by assumption that $\overline{a}\overline{b} = 0$. Then either $\overline{a} = \overline{0}$ or $\overline{b} = \overline{0}$ by the definition of a prime ideal. This means that $R/P$ has no zero divisors, and we saw earlier that $R/P \neq \overline{0}$, which is the definition of an integral domain.

“$\Leftarrow$“: repeat the proof above in the opposite direction.