# Prove that a UFD is a PID if and only if every nonzero prime ideal is maximal

Prove that a UFD is a PID if and only if every nonzero prime ideal is maximal.

The forward direction is standard, and the reverse direction is giving me trouble. In particular, I can prove that if every nonzero prime ideal is maximal then every maximal ideal is principal. From here, I know every ideal $I$ is contained in a maximal ideal, and that maximal ideal is principal, but I can't quite conclude that $I$ must also be principal.

In particular, this is not true for sub-ideals of principal ideals in general rings, so I was looking at the set of all maximal ideals containing I, and trying to argue something about factorization but I can't quite get there. Any hints?

• See this thread in Ask-an-Algebraist. You'll see some familiar names. – Arturo Magidin Nov 2 '11 at 3:49
• (I am only adding this aa an answer because i done have enough reputation to comment. Sorry!) For the forward direction, don't you need to assume your UFD is Noetherian ? The statement i have seen was for Dedekind domains, so the proof i have in mind works. EDIT: when i say forward, i mean assume the ring is UFD and every non-zero prime ideal is maximal. To prove: The ring is a PID – Manas Mar 24 '18 at 0:37

Ok, so it seems to me what you're saying is that you've proven that under the condition that every prime ideal is maximal that every maximal ideal is principal. The problem then follows from a nice condition you might want to remember:

Theorem: Let $R$ be an integral domain. Then, $R$ is a PID if and only if every element of $\text{Spec}(R)$ is principal.

I'll outline the proof here for you. Suppose that every element of $\text{Spec}(R)$ is principal and suppose the set $\mathscr{N}$ of non-prinicpal ideals was non-empty. It's pretty easy to show then that every ascending chain in $\mathscr{N}$ has an upper bound from where Zorn's lemma gives you that $\mathscr{N}$ has a maximal element, call it $\mathfrak{a}$ which is a non-prinicpal ideal. We claim that $\mathfrak{a}$ is actually not prime (towards a contradiction). We see then that there exists $a,b\notin \mathfrak{a}$ with $ab\in \mathfrak{a}$. Note then that $\mathfrak{a}_a=(\mathfrak{a},a)$ and $\mathfrak{a}_b=(\mathfrak{a},b)$ are proper supersets of $\mathfrak{a}$ and so must be principal, say equal to $(\alpha)=\mathfrak{a}_a$ (we only really care about the one). Define then $\mathfrak{b}$ to be the ideal quotient of $\mathfrak{a}$ in $\mathfrak{a}_a$. Prove that $\mathfrak{b}$ contains $\mathfrak{a}_b$ and so is principal, say $\mathfrak{b}=(\beta)$. Show then that $\mathfrak{a}=\mathfrak{a}_a\mathfrak{b}$ so that $\mathfrak{a}=(\alpha\beta)$ contradictory to assumption.

This contradiction allows us to conclude that $\mathscr{N}$ must have been empty.

• So this is all fine and dandy (though there seems to be a small typo), but I believe the point of the problem is to use the fact that every ideal is contained in a maximal ideal to get to the end, and not use this more general result. Is there another way to do it? – JeremyKun Nov 1 '11 at 23:52

This thread (including comments from both Jeremy and Alex) was very useful to me in PhD Quals preparation.

I would like to give a detailed proof using Jeremy's and Alex's outline, but without mention of Spec(R), in case the reader is less algebraic-geometry-inclined.

Corrections welcome. Thanks.

Proposition. Let R be a UFD in which every nonzero prime ideal is maximal. Then R is a PID.

Lemma 1. If p is a nonzero irreducible element of R, then (p) is a prime ideal of R.

Proof. Let a and b be nonzero elements of R such that ab ∈ (p), i.e. ab = pf for some nonzero element f ∈ R. If a and b are both non-units, then we can write their unique factorisations a = up1...pr and b = vq1...qs for unique irreducible elements pi and qj, and units u and v ∈ R. By unique factorisation, p = one of the pi's, or p = one of the qj's, or both. Hence either a ∈ (p) or b ∈ (p) or both. [Similar reasoning works if either a or b is a unit of R.] Thus, (p) is a prime ideal.

Lemma 2. If I is a nonzero prime ideal of R, then I is principal.

Proof. Given a nonzero, proper prime ideal, $I$, with some element $r\in I$, write $r= p_1\cdots p_n$ as a product of primes. By the primality of $I$ we know some $p_i\in I$. Since $p_i$ is prime, by assumption $(p_i)$ is maximal, so $(p_i)\subseteq I$ implies that $I=(p_i)$.

Proof of Proposition.

Following Alex's thread, let N be the set of non-principal ideals in R. Suppose that N is non-empty. N is a partially ordered set (with inclusion of ideals being the ordering relation). Any totally ordered subset of ideals J1 ⊆ J2 ⊆ ... in N has an upper bound, namely J = ∪ Ji. J is an ideal of R because the subset was totally ordered. If J = (x) for some x ∈ R, then x ∈ Ji for some i, and hence (x) = Ji. Contradiction. Thus, J ∈ N and by Zorn's Lemma, N has a maximal element. Call it n.

By Lemma 2, we see that n is not a prime ideal of R. Thus, there exist a, b ∈ R, such that a, b ∉ n, but ab ∈ n. But then n + (a) and n + (b) are ideals in R which are strictly bigger than n, and so must be principal ideals, i.e. n + (a) = (u) and n + (b) = (v) for u, v ∈ R. Then we have

n = n + (ab) = (n + (a)) (n + (b)) = (u)(v) = (uv). Contradiction.

Thus, N must be the empty set, i.e. all ideals in R are principal.

• @ Conam Wong Why is 'n' inside (n+(a))(n+(b)) ? This is not always true. Please clarify. – user100006 Oct 10 '13 at 10:38
• "I would like to give a detailed proof using Bean's and Alex's outline, but without mention of Spec(R), in case the reader is less algebraic-geometry-inclined." +1 vote :) – Vincent Dec 12 '13 at 14:50
• @Conam Wong: can you explain why $n + (ab) = (n + (a)) (n + (b))$? Thank you very much – Vincent Dec 12 '13 at 14:51
• For a correct proof that if all prime ideals are principal (Lemma 2 is true) then all ideals are principal, see math.stackexchange.com/questions/168082/… – KCd Aug 20 '14 at 23:06
• @KCd Can't we prove without Zorn's Lemma that UFD with every prime ideal maximal is PID – Sushil Feb 19 '15 at 12:07