$P$ is a prime ideal, and $ R/P$ has no nilpotent elements. Then $R/P$ is a domain.

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Let $P$ be a prime ideal. Suppose that $R/P$ has no nonzero nilpotent elements. Show that $R/P$ is a domain.

What I did :

WTS : $(a+P)(b+P)=ab+P=0+P$ implies $a+P=0+P$ or $b+P=0+P$.

but it means that $ab \in P$ implies $a \in P$ or $ b \in P$. The nilpotent condition means that $a \notin R $ implies $a^n \notin R$ for all $n \in \mathbb{N}$. Since P is a prime, $AB \subseteq P $ implies $A \subseteq P$ or $B \subseteq P$ for ideals $A,B$ in a ring $R$. To use this, I made ideals $(a),(b),(ab)$.

It suffices show that $(a)(b) \subseteq P$ but I failed to show this.

Thanks in advance.