Let $R$ be a unique factorization domain and $P \neq R$ a prime ideal of $R$.
How to show:
The ideal $P$ is a principal ideal $\iff$ For all prime ideals $P'$ of $R$ with $P' \subset P$ it's $P'=${$0$} or $P'=P$
For $\Rightarrow$ I got:
Since $P$ is a principal ideal and a prime ideal: $\forall a,b \in R$ with $ab \in P$ it's $a \in P$ or $b \in P$.
Now $P'$ is a subset of $P$. So it's $P' \subset P \subset R$.
This implies that $P'$ is a maximal ideal, so $P'=P$ or $P=R$.
Is this way correct? And how to conclude that $P'=${$0$}?
For $\Leftarrow$ I got:
$P'=${$0$} or $P'=P$.
If $P'=P$ then $P$ is a principal ideal since prime ideals als principal ideals in a unique factorization domain.
Is this argumentation right or is there another way to show it?
This is false. Let $R = \mathbb{Z}$ and $P = \langle 2 \rangle$, then $P$ is both principal and prime. Let $P' = \langle 4 \rangle$, then $P'$ is a subset of $P$, but $P'$ is neither $\{0\}$ nor $P$.