every prime ideal of height 1in an UFD is principal

2017-12-03 03:03:13

if $R$ is a

UFD, then every prime ideal of height 1 in $R$ is principal.

I find it is a direct consequence of kaplansky's theorem.But the proof of kaplansky theorem is difficult for me.

So i wonder how to prove the statement directly.

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.

  • 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.

    2017-12-03 05:00:04