If $R$ is a UFD, then every prime ideal of height 1 in $R$ is principal.
I find this is a direct consequence of Kaplansky's theorem. But the proof of Kaplansky's theorem is difficult for me. So I wonder how to prove the statement directly.
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3Let $P \subset R$ prime ideal and $a \in P$ element not zero. Take an $q$ prime that occurs in decomposition of $a$ and $q \in P$ ($P$ is prime). Then, $P = (q)$, by consideration of height. – numberwat Dec 03 '17 at 01:45
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Suppose $R$ is a UFD and $P\subset R$ is a height $1$ prime. Then $P\neq 0$, so there is some nonzero element $x\in P$. Then $x$ can be written as a product of irreducible elements of $R$. Since $P$ is a prime ideal, one of those irreducible elements must be in $P$; say $p\in P$ for some irreducible element $p$. Since $R$ is a UFD, the ideal $(p)$ is prime. But then $(p)$ is a nonzero prime ideal contained in $P$. Since $P$ has height $1$, this means $(p)=P$, so $P$ is principal.
Eric Wofsey
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1so (0) is considered as a prime ideal since Ufd rings are domains – User not found Aug 01 '22 at 12:06
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