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:
- (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
To prove the result I will use two facts which can be easily find in internet (if not, just tell me):
- Let $R$ be a integral domain. Then, $R$ is a UFD iff every non zero prime ideal contains a prime element (Kaplansky criterion).
- In an integral domain every prime ideal is principal iff it is a PID.
Now we are going to prove:
If $R$ is a UFD of Krull dimension one, then it is a PID.
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
Best Answer
The whole ring is not a prime ideal, by convention. It's the same way that $1$ is not a prime number. So the only prime ideal in a field is $\{0\}$.
Apart from that, the proof looks good (assuming you are allowed to use the theorem that nonzero prime ideals are maximal in PIDs).