7
$\begingroup$

I want to prove that if in a UFD every maximal ideal is principal then it is a PID.

My line of attack is: If it is a field i.e. it has no non-zero proper ideal, then we are done. Otherwise consider a non-zero proper ideal. Then, by Zorn's lemma, it is contained in some maximal ideal, hence contained in a principal ideal which is not the whole ring. Now we know that if in a UFD every proper ideal is contained a proper principal ideal, then it is a PID, hence we are done.

Is there any other solution without using Zorn's lemma?

$\endgroup$
  • $\begingroup$ @SaunDev See also this answer and and the papers cited there. $\endgroup$ – Bill Dubuque May 3 '15 at 16:41
  • $\begingroup$ @BillDubuque : thanks , I know that result of cohen-kaplansky . I will definitely look into the papers . The main problem is in a UFD every prime ideal need not necessarily be maximal , indeed if it so happens then it is either a field or a PID .... $\endgroup$ – user228168 May 4 '15 at 4:03
2
$\begingroup$

To prove the result I will use two facts which can be easily find in internet (if not, just tell me):

  1. Let $R$ be a integral domain. Then, $R$ is a UFD iff every non zero prime ideal contains a prime element (Kaplansky criterion).
  2. In an integral domain every prime ideal is principal iff it is a PID.

Now we are going to prove:

If $R$ is a UFD of Krull dimension one, then it is a PID.

Let $ \mathfrak{p} $ be a non-trivial prime ideal of $ R $ (the ideal $ \langle 0\rangle $ is principal and prime since we are working over integral domains). Since $ R $ is a UFD, we know that there exists some prime element $ p\in\mathfrak{p} $ (Kaplansky). But then $ \langle p\rangle\subseteq\mathfrak{p} $ which implies that $ \mathfrak{p}=\langle p\rangle$, as we assumed every prime ideal to be maximal. Since every non trivial prime ideal is now principal, we conclude that $ R $ is a PID.

Actually you can prove that the converse is also true, using the fact that every PID is a UFD. Then take a prime ideal which we know to be principal and suppose that it is not maximal, which leads you directly to a contradiction

$\endgroup$
  • $\begingroup$ Please read again the question, especially the last sentence. $\endgroup$ – user26857 Mar 15 '16 at 14:46
  • 1
    $\begingroup$ I am sorry! I forgot that one of the two results I use indeed require Zorns lemma. Nevertheless I think that this proof is the almost most elegant way to prove this result, but yes, that was not the question... $\endgroup$ – Patricio Mar 15 '16 at 15:47
1
$\begingroup$

The question reduces easily to Prove that a UFD is a PID if and only if every nonzero prime ideal is maximal.

Let $\mathfrak p$ be a non-zero prime ideal. Since the ring is a UFD there is a prime element $p\in\mathfrak p$. If $\mathfrak p$ is not maximal, then there exists a maximal ideal $\mathfrak m$ such that $\mathfrak p\subsetneq\mathfrak m$. By hypothesis $\mathfrak m$ is principal, so $\mathfrak m=(q)$ where $q$ is also prime. From $(p)\subsetneq (q)$ we get $q\mid p$ hence $q$ and $p$ are associates. This means $(p)=(q)$, a contradiction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy