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Prime ideal in $k[x,y]$, which is not maximal, is generated by one element. ($k$ is a field)

Why it is true? I know that dimension of $k[x,y]$ is equal 2 and also that $k[x,y]$ is UFD. I have problems to connect these two facts. I would like any hint.

user26857
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jpatrick
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1 Answers1

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A non-zero prime ideal in a dimension $2\,$ U.F.D. which is not maximal has height $1$. Furthermore, every non-zero prime ideal $\mathfrak p$ in a U.F.D. contains a prime element $p$, by a theorem of Kaplansky. So $(p)\subset\mathfrak p$ and both have height $1$: necessarily $(p)=\mathfrak p$.

user26857
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Bernard
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  • Your first sentence does not make sense in general. – MooS Nov 17 '17 at 15:38
  • @MooS: You're right. I had in mind the case of $k[X,Y]$. I've completed my sentence. Thanks for pointing the problem! – Bernard Nov 17 '17 at 15:55
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    I think I've already seen this claim: every non-zero prime ideal $\mathfrak p$ in a U.F.D. contains a prime element $p$ attributed to Kaplansky. This is a trivial fact for UFDs. It occurs in Kaplansky's criterion for UFDs, but ... – user26857 Nov 17 '17 at 21:33