An integral domain whose every prime ideal is principal is a PID

Does anyone has a simple proof of the following fact:

An integral domain whose every prime ideal is principal is a principal ideal domain (PID).

Here is a proof followed by conceptual elaboration, from my 2008/11/9 Ask an Algebraist post.

Let R be an integral domain. Let every prime ideal in R be principal. Prove that R is a principal ideal domain (PID)

Below I present a simpler way to view the proof, and some references. First let's recall one well-known proof, as presented by P.L. Clark (edited):

Proof$_{\,1}$ $\ $ Suppose not. Then the set of all nonprincipal ideals is nonempty. Let $\{I_i\}$ be a chain of nonprincipal ideals and put $\,I = \cup_i I_i.\,$ If $I = (x)$ then $x \in I_i$ for some $i,$ so $I = (x) \subset I_i$ implies $I = I_i$ is principal, contradiction. Thus by Zorn's Lemma there is an ideal $I$ which is maximal with respect to the property of not being principal. As is so often the case for ideals maximal with respect to some property or other, we can show that I must be prime. Indeed, suppose that $ab \in I$ but neither $a$ nor $b$ lies in I. Then the ideal $J = (I,a)$ is strictly larger than $I,$ so principal: say $J = (c).$ $I:a := \{r \in R\ :\ ra \in I\}$ is an ideal containing $I$ and $b,$ so strictly larger than $I$ and thus principal: say $I:a = (d).$ Let $i \in I,$ so $i = uc.$ Now $u(c) \subset I$ so $ua \in I$ so $u \in I:a.$ Thus we may write $u = vd$ and $i = vcd.$ This shows $I \subset (cd).\ $ Conversely, $d \in I:a$ implies $da \in I$ so $d(I,a) = dJ \subset I$ so $cd \in I.$ Therefore $I = (cd)$ is principal, contradiction. $\ \ $ QED

We show that the second part of the proof is just an ideal theoretic version of a well-known fact about integers. Namely suppose that the integer $i>1$ isn't prime. Then, by definition, there are integers $\,a,b\,$ such that $\,i\mid ab,\ \,i\nmid a,b.\,$ But this immediately yields a proper factorization of $\,i,\,$ namely $\,i = c\, (i\!:\!c),$ where $c = (i,a).\,$ Hence: $ $ not prime $\Rightarrow$ reducible (or: irreducible $\Rightarrow$ prime). $ $ A similar constructive proof works much more generally, namely

Lemma $ $ If ideal $I\ne 1$ satisfies: ideal $\,J \supset I \Rightarrow J\,|\,I\,$ then $I$ not prime $\Rightarrow I\,$ reducible (properly).

Proof $\ $ $I$ not prime $\Rightarrow$ exists $\,a,b \not\in I\,$ and $\,ab \in I.$ $\ A := (I,a)\supset I \Rightarrow A\mid I,\,$ say $\,I = AB;$ wlog we may assume $\,b \in B\,$ since $A(B,b) = AB\,$ via $Ab = (I,a)b \subset I = AB.$ The factors $A,B$ are proper: $A = (I,a),\, a \not\in I;\,\ B \supset (I,b),\, b \not\in I.\quad$ QED

Note that the contains $\Rightarrow$ divides hypothesis: $J\supset I \Rightarrow J\,|\,I\,$ is trivially true for principal ideals $J$ (hence proof$_{\,1}$), and also holds true for all ideals in a Dedekind domain. Generally such ideals J are called multiplication ideals. Rings whose ideals satisfy this property are known as multiplication rings and their study goes back to Krull.

The OP's problem is well-known: it is Exercise $1\!-\!1\!-\!10\ p.8$ in Kaplansky: Commutative Rings, namely:

10. (M. Isaacs) In a ring R let $I$ be maximal among non-principal ideals. Prove that $I$ is prime. (Hint: adapt the proof of Theorem 7. We have $(I,a) = (c).$ This time take $J =$ all $x$ with $xc \in I.$ Since $J \supset (I,b),\ J$ is principal. Argue that $I = Jc$ and so is principal.)

For generalizations of such Kaplansky-style Zorn Lemma arguments see the papers referenced in my post here.

Below is an interesting reference on multiplication rings.

Mott, Joe Leonard. Equivalent conditions for a ring to be a multiplication ring.
Canad. J. Math. 16 1964 429--434. MR 29:119 13.20 (16.00)

If "ring" is taken to mean a commutative ring with identity and a multiplication ring is a "ring" in which, when A and B are ideals with A $\subset$ B, there is an ideal C such that A = BC , then it is shown that the following statements are equivalent.

• (I) R is a multiplication ring;
• (II) if P is a prime ideal of R containing the ideal A, then there is an ideal C such that A = PC;
• (III) R is a ring in which the following three conditions are valid:
  $\qquad$ (a) every ideal is equal to the intersection of its isolated primary components;
  $\qquad$ (b) every primary ideal is a power of its radical;
  $\qquad$ (c) if P is a minimal prime of B and n is the least positive integer such that $\rm P^n$ is an isolated primary component of B , and if $\rm P^n \ne P^{n+1},$ then P does not contain the intersection of the remaining isolated primary components of B . (Here an isolated P-primary component of A is the intersection of all P-primary ideals that contain A .)

Reviewed by H. T. Muhly

We do this in a few steps. Let $R$ be our integral domain, and we are supposing that every prime ideal in $R$ is principal.

1. Let $S$ be the set of non-principal ideals of $R$, and suppose that $S$ is nonempty. Throw an inclusion partial order on $S$ (or note that it has one, but I like to be physical with my math sometimes). WRT this partial order, let $\{C_i\}$ be a chain in $S$, and let $C = \cup C_i$. Then we know that $C$ is an ideal.

2. Prove that $C$ is not principal:
   Suppose that $C$ is principal, so that $C = \langle c \rangle$. Then $c \in C_j$ for some $j$, so that $C \subseteq C_j$. Then $C_j$ is principal, contradicting our hypothesis.

3. Appeal to Zorn to get an inclusion-maximal nonprincipal ideal:
   As $C$ isn't principal, the chain $\{C_i\}$ has an upper bound in $S$. By Zorn's Lemma, $S$ contains a maximal element. Let $M$ be one such element.

4. Let $a,b \in R$ s.t. $ab \in M$ while $a,b \not \in M$, which exist because $M \in S$ and thus is not prime. Since $a \not \in M$, we have that $M \subsetneq (M,a)$, $M \subsetneq (M,b)$. As $M$ is maximal in terms of non-principality, we know that both $(M,a)$ and $(M,b)$ are principal. Suppose that $(M,a) = \langle \alpha \rangle, (M,b) = \langle \beta \rangle$.

   Let $N = \{ r \in R\;|\;ra \in M\}$, which is also larger than $M$ and thus principal. Note that $(M,a)(M,b) = (M^2, Ma, Mb, ab) \subseteq M$, so that $(M,b) \subseteq N$. And further, $(M,a)N \subseteq M$, and $(M,a)N$ is a product of two principal ideals, and is thus principal.

5. If $x \in M$, we'll show that $x = s\alpha$ for some $s \in N$, and conclude that $M = (M,a)N$ is principal - a contradiction:
   Let $x \in M$. Since $M \subset (M,a)$, we know that $x = s\alpha$ for some $s \in R$. Note that in fact $s(M,a) = s\langle \alpha \rangle \subseteq M$, so $s \in N$. Thus $M \subseteq (M,a)N$, and more importantly $M$ is principal.

This is a contradiction on the non-principality of $M$, so we are wrong to assume that $S$ is nonempty.

Thus $R$ has no nonprincipal ideals and is a principal ideal domain.