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Let $R$ be Noetherian domain and $P\neq 0 $ is the unique prime ideal in $R$. Is it possible to prove $R$ is UFD ?

If $P=(p)$ is principal ideal then $R$ is regular local ring so it is UFD. But I can not show that.

More general, in Noetherian domain, is it true that prime elements always exist ?

user26857
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Anh_Rose 1210
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  • For the latter: Fields don't contain prime elements but one could still ask if for non-fields prime elements exist. –  May 12 '17 at 12:51
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    Try $R=k[[t^2,t^3]]$ which satisfies your requirement, but not a UFD and has no prime elements. – Mohan May 12 '17 at 12:52
  • @Mohan I'm not very good with UFD's, and even less so with formal power series in multiple variables! So I would like to understand better... It seems apparent that $t^2$ and $t^3$ are both irreducibles, and lead to distinct factorizations of $t^6$, say. I need help seeing why there are no irreducibles! I'm used to the normal power series ring over a field and a single variable being local... but this one gives me pause. Can you comment on the maximal ideals and/or units? Thanks! – rschwieb May 12 '17 at 13:06
  • @Mohan I think I get why this setup is an easy way to make $(t^2, t^3)$ the unique, nonprincipal maximal ideal. The argument for units being the things with nonzero constant term still seems to be the same. And the only candidates for primes are irreducibles, which seem to be only $t^2$ and $t^3$, but neither of these is prime. Is that about right? – rschwieb May 12 '17 at 13:25
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    @axlp apparently you mean unique nonzero prime ideal? Saying "$P\neq 0$ is the unique prime ideal" is not quite the same. If $P$ is principal, another easy way is: by Kaplansky's theorem for Noetherian rings, $R$ is a PID. So yeah, the maximal ideal cannot be principal. – rschwieb May 12 '17 at 13:36
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    Yes. I mean except from 0, $P$ is the only prime ideal in $R$. – Anh_Rose 1210 May 12 '17 at 13:57
  • @rschwieb Yes, what you say is correct. Another way of saying it is to see that the only prime ideals are $0, (t^2,t^3)$. If $f\neq 0$ is a prime, then it generates a prime ideal and then $fR=(t^2,t^3)$. But Nakayama will tell you that $(t^2,t^3)$ can not be a principal ideal. – Mohan May 12 '17 at 20:24

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In fact one can show the following in this situation:

If $R$ is a noetherian domain with a unique non-zero prime ideal $P$, then $R$ is a UFD if and only if $P$ is principal.

You have already shown the one direction, let me proof the other one, which is very easy: Since $R$ is a UFD, which is not a field, there is at least one irreducible element. This element generates a non-zero prime ideal, which of course coincides with $P$ by the uniqueness of $P$. Thus $P$ is principal.

With a little more work, it can be shown that any ideal of $R$ is principal in this case: If $I \neq 0$, we find a maximal $n$ with $I \subset P^n$. Take some $f \in I$ with $f \notin P^{n+1}$. Since $P^n/P^{n+1}$ is a one-dimensional vector space, we obtain that the residue class of $f$ generates it. Nakayama tells us that $f$ generates $P^n$, i.e. $I=P^n$ is indeed principal.

MooS
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