Let $R$ be a UFD and $P$ be a nonzero prime ideal of $R$. Suppose that $P$ does not contain any nonzero prime ideal other than $P$. What can you say about $P$?
I think $P$ should be a principal ideal. Now to prove this, I tried this idea. Say $P$ is not principal ideal and say $x$ and $y$ both are in the generating set of $P$. Then $\langle x\rangle$ must be a prime ideal in $P$ and is not the whole of $P$. Hence contradiction arrives. My problem is we are not using UFD criteria anywhere.
Any comments on this proof.
Why is $(x)$ a prime ideal in what you've written ? The ideal $(x)$ is prime if and only if $x$ is zero or prime. Also, what exactly do you call the generating set of $P$ ?
A proof of the fact that $P$ must be principal is the following: Since $P$ is a non-zero proper ideal of $R$, it contains a non-zero non-unit element $x$. As $R$ is a UFD, there exist prime elements $p_{1}, \dotsc, p_{r}$ of $R$ such that $x = \prod\limits_{i = 1}^{r} p_{i}$. Since $P$ is prime, there exists $k \in \lbrace 1, \dotsc, r \rbrace$ such that $p_{k} \in P$. The ideal $\left( p_{k} \right)$ is a non-zero prime ideal contained in $P$, and hence $P = \left( p_{k} \right)$ is principal.
Edit: My initial answer was incorrect.