-1

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
  • 1
  • 13
  • 62
  • 125
jpatrick
  • 884
  • 5
  • 16

1 Answers1

3

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
  • 1
  • 13
  • 62
  • 125
Bernard
  • 1
  • 9
  • 62
  • 162
  • 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
  • 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