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Given that $R$ is a UFD which is not a PID, I want to show that $R$ must have a nonprincipal maximal ideal.

I tried several methods, including Zorn's lemma but didn't get anywhere. Any suggestions would be appreciated.

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By Zorn, every proper ideal is contained in a maximal one. I'd try using that fact. – Mike B May 7 '12 at 5:39
up vote 10 down vote accepted

Let $R$ be a UFD in which each maximal ideal is principal, and let $I$ be a nonzero ideal of $R$. Let $(x)$ be the greatest common divisor of all the nonzero elements of $I$, so $I = (x)J$, where $J$ is an ideal which is contained in no proper principal ideal. But then by our assumption $J$ is contained in no maximal ideal, so $J = R$ and $I = (x)$.

Added: Let me address why gcds of arbitrary sets of elements exist in a UFD. Let $R$ be a UFD and let $\mathcal{P}$ be the set of nonzero principal prime ideals of $R$, so that for all nonzero $x \in R$, $x = \prod_{p \in \mathcal{P}} p^{n_p(x)}$, where each $n_p(x) \in \mathbb{N}$ and all but finitely many are zero. For any subset $S = (x_s)$ of nonzero elements of $R$, the greatest common divisor of the elements of $S$ is $\prod_{p \in \mathcal{P}} p^{\min_{s \in S} n_p(x_s)}$: this exists! (In fact there is a finite subset $T \subset S$ such that $\operatorname{gcd}(S) = \operatorname{gcd}(T)$.)

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Is it possible to find the gcd of infinitely many elements? – Ehsan M. Kermani May 7 '12 at 8:12
@ehsanmo: in the present case $I$ is contained in only finitely many principal prime ideals, because $R$ is a UFD. Thus the gcd exists. – Hagen Knaf May 7 '12 at 10:35

Case 1. Let $R$ be a non-Noetherian UFD where all of its maximal ideals are principal, then it is a PID.

Assume the contrary and let $I$ be an ideal which is maximal w.r.t being non-principal, then $I$ is a prime ideal (Look at the Theorem 1.4. for more detail.) By assumption, $I$ cannot be maximal, thus there is a maximal ideal $m=(x)$ containing $I$ properly (note that $x$ is an irreducible element.) Now, $m \subseteq I+m \subseteq R.$

If $m=I+m,$ based on assumption, any prime ideal contains a prime element (See this). Let $i \in I$ be a prime element, then $xt=i+xr$ for some $t,r \in R,$ then $i=x(t-r).$ Obviously, $i$ cannot divide $x$ so $t-r=is$ for some $s \in R.$ Therefore, $i=x(t-r)=xis$ implying that $x$ is a unit which a contradiction.

If $I+m=R$ i.e. $is+xr=1$ for $i \in I,$ and $s,r \in R.$ Now, $i \in (x),$ or $i=xt$ for some $t \in R.$ Thus, $x(ts+r)=1$ implying that $x$ is a unit element, which is again a contradiction. Hence such an $I$ does not exist and we're done.

Case 2. Let $R$ be a Noetherian UFD where all of its maximal ideals are principal, then it is a PID.

Assume not, and let $I$ be the prime ideal above contained in a principal maximal ideal $m=(x)$ Since $R$ is a Noetherian UFD then any height one prime ideal is principal, therefore, $ht(I) \geq 2$ implying that $ht(m) \geq 3,$ hence is a contradiction, by Krull's principal ideal theorem.


You can see another short proof below by Pete L. Clark. In fact, I was not aware of the fact that Kaplansky's theorem (as mentioned, 15.1 here) holds for all domains.

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I think you have a very simple proof slightly hidden in here. Namely, suppose $R$ is a UFD with all maximal ideals principal. If $R$ is not a PID there is a nonprincipal prime $\mathfrak{p}$ contained in a principal maximal ideal $(x)$. By Kaplansky's Theorem, $\mathfrak{p}$ contains a nonzero prime element $y$. Thus $x \mid y$ but $x$ and $y$ are prime, so $(x) = (y)$ and $\mathfrak{p} = \mathfrak{m}$ is principal, contradiction. – Pete L. Clark May 7 '12 at 13:01
@PeteL.Clark: Yes, indeed, for the case 1. – Ehsan M. Kermani May 7 '12 at 17:47
@ehsanmmo: But your Case 1 argument doesn't use that $R$ is not just doesn't require that $R$ is Noetherian. So there is no need for Case 2. – Pete L. Clark May 7 '12 at 19:03
@ehsanmno: Ah, I see -- you are going by the wikipedia article. But in fact Kaplansky's theorem holds for all domains: see e.g. $\S 15.1$ of (Or see Kaplansky's book!) – Pete L. Clark May 7 '12 at 21:13
$@$ehsanmo: I put this result in my commutative algebra notes (Section 16). The proof that emerged from my comments on your answer is the one given in the text. The one in my answer is given as an exercise. – Pete L. Clark May 13 '12 at 17:39

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