Let $R$ be a commutative ring, and $I$ be a principal prime ideal. Is it true that $I$ does not contain any non-zero prime ideal? (You may assume that $R$ is Noetherian)
The fact that the question says "You may assume that $R$ is Noetherian" makes this sound like a homework question; if it is, you should use the homework tag. Either way, you should let us know what you've tried so far. –  Michael Albanese Sep 25 '12 at 16:26
If $R$ is Noetherian, Krull's principal ideal theorem implies that $\operatorname{ht} I \le 1$, i.e., any chain of prime ideals $P_1 \subsetneq P_2 \subsetneq \ldots \subsetneq P_n = I$ has $n \le 1$. Taking a maximal chain of prime ideals, we know that $P_0$ is made up from zero-divisors, since the minimal prime ideals of a ring contain only 0 and a (possibly empty) subset of the zero-divisors. So, either $I$ is made up from zero-divisors, or it strictly contains only prime ideals made from zero-divisors.