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.
Every PID is a UFD. Not every UFD is a PID.
Example: A ring $R$ is a unique factorization domain if and only if the polynomial ring $R[X]$ is one. But $R[X]$ is a principal ideal domain if and only if $R$ is a field. So, $\mathbb{Z}[X]$ is an example of a unique factorization domain which is not a principal ideal domain.
The statement "In a PID every non-zero, non-unit element can be written as product of irreducibles" is true, but it is not the definition of a principal ideal domain. Nor is it the definition of a unique factorization domain: as you pointed out, it does not account for uniqueness.
Note that in an integral domain, every prime element is irreducible, but the converse is false.
Here are two equivalent definitions of unique factorization domain:
1 . An integral domain $R$ in which every nonzero nonunit is a product of prime elements,
2 . An integral domain $R$ in which every nonzero nonunit is a product of irreducible elements, and this product is unique up to associates.
In the proof you were reading, I guess what they are trying to say is the following: in a principal ideal domain $R$, it is true that irreducible is the same thing as prime. So once you show that every nonzero nonunit can be written as a product of irreducibles, it follows that $R$ is a unique factorization domain.
Best Answer
To prove the result I will use two facts which can be easily find in internet (if not, just tell me):
Now we are going to prove:
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