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

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...p_r and b = vq₁...q_s for unique irreducible elements p_i and q_j, and units u and v ∈ R. By unique factorisation, p = one of the p_i's, or p = one of the q_j'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 = ∪ J_i. J is an ideal of R because the subset was totally ordered. If J = (x) for some x ∈ R, then x ∈ J_i for some i, and hence (x) = J_i. 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.

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.