Fulton, "Algebraic Curves," Exercise 1.39(a):

Let $R$ be a UFD, and $P = (t)$ a principal, proper, prime ideal. Show there is no prime ideal $Q$ with $0 \subset Q \subset P$.

After being stumped for some time, I came up with the following proof while attending a concert (my ears are still ringing an hour later):

Suppose there is such an ideal. Take some $0 \neq q \in Q$. Since $Q \subset P$, $q = rt$ for some $r \in R$. Since $Q$ is prime, and $t \not \in Q$ by assumption, we must have $r \in Q$. Applying the preceding to $r$ instead of $q$, we have $r = r't$ for some $r' \in Q$. So $q = rt = (r't)t = r't^2$. Proceeding along these lines, $q$ is divisible by $t^n$ for all $n \geq 0$. Notice that $t$ is irreducible, since $P = (t)$ is prime. Now, what possible factorization into irreducibles could $q$ have?

So far as I can tell, this seems to work, but it also seems exceedingly silly. What's the "right" way to prove this?

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    $\begingroup$ Your proof is very clever not silly. Finally, because $R$ is UFD, $q$ can be only divided by finitely many $t$. $\endgroup$ – wxu May 28 '12 at 8:58
  • $\begingroup$ Clever can still be silly, though, when there's a more straightforward way of doing things but you don't know what it is. (The question in the last sentence of my proof was rhetorical, by the way -- I'm well aware that that's how the proof concludes. This style of writing is probably more common in popular expository work than in research papers, though.) $\endgroup$ – Daniel McLaury May 28 '12 at 9:04

Your idea is not stupid as noted in a comment above. Recall the definition of a UFD:

A domain $R$ is a UFD if

  1. $R$ is a factorisation domain, that is every non-zero non-unit can be factored into a finite product of irreducibles

  2. The factorisation into irreducibles is unique upto order and associates.

You could provide an alternative ("less stupid" if you want to call it) proof like this. Suppose that $Q \subsetneqq (t)$. Now since $Q$ is a prime ideal and $R$ is a UFD by a theorem of Kaplansky there is a prime element $q \in Q$. Since $Q \subsetneqq (t)$ we can write $q = rt$ for some $r \in R$. Now because $Q$ is a prime ideal and $t \notin Q$ we are forced to conclude that $r \in Q$. However $q$ being a prime element is irreducible and this forces $r$ to be a unit. It follows that $Q$ contains a unit and hence $1 \in Q$, contradicting $Q$ being prime.

  • $\begingroup$ @DanielMcLaury TBH this the first time I have heard of the book you mentioned above. I don't think I would know what the intedended solution is! :D :P $\endgroup$ – user38268 May 28 '12 at 9:21
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    $\begingroup$ Note that this employs only the simple (trivial) direction of Kaplansky's theorem, namely, that a prime ideal $\ne 0$ contains a prime $\ne 0$. Indeed, it is trivial to prove that prime ideals in a UFD may be generated by primes, which immediately yields the sought result - see my answer. $\endgroup$ – Bill Dubuque May 28 '12 at 21:08
  • $\begingroup$ @BillDubuque Thanks for your heads up, I learned a lot from your post below! $\endgroup$ – user38268 May 29 '12 at 6:56

Yes, your proof is fine. Here is another way to view it. In a UFD every prime ideal $\rm\:P\:$ may be generated by primes. Indeed, $\rm\:0\ne p_1\cdots p_n\in P\:\Rightarrow\:$ some $\rm\: p_i\in P,\:$ so we may replace each generator by one of its prime factors. Hence if prime $\rm\:(p)\supseteq Q = (p_1,p_2,\ldots)\ne 0\ne p_i\:$ then $\rm\:(p)\supseteq (p_i)\:$ $\Rightarrow$ $\rm\:(p) = (p_i)\:$ $\Rightarrow$ $\rm\: Q = (p).\,$ QED $\, $ A converse is a famous theorem of Kaplansky.

Theorem $\ $ TFAE for an integral domain D

$\rm(1)\ \ \:D\:$ is a UFD (Unique Factorization Domain)
$\rm(2)\ \ $ In $\rm\:D\:$ every prime ideal is generated by primes.
$\rm(3)\ \ $ In $\rm\:D\:$ every prime ideal $\ne 0$ contains a prime $\ne 0.$

Proof $\ (1 \Rightarrow 2)\ $ Proved above. $\rm\ (2\Rightarrow 3)\ $ Clear.
$(3 \Rightarrow 1)\ $ The set $\rm\:S\subseteq D\:$ of products of units and nonzero primes forms a saturated monoid, i.e. $\rm\:S\:$ is closed under products (clear) and under divisors, since the only nonunit divisors of a prime product are subproducts (up to associates), due to uniqueness of factorization of prime products. Since $\rm\:S\:$ is a saturated monoid, its complement $\rm\:\bar S\:$ is a union of prime ideals. So $\rm\:\bar S = \{0\}\:$ (else it contains some prime ideal $\rm\:P\ne 0\:$ which contains a prime $\rm\:0\ne p\in P\subseteq \bar S,\:$ contra $\rm\:p\in S).\:$ Hence every $\rm\:0\ne d\in D\:$ lies in $\rm S,\:$ i.e. $\rm\:d\:$ is a unit or prime product. Thus $\rm\:D\:$ is a UFD. $\ $ QED

Remark $\ $ The essence of the proof is clearer when one learns localization. Then ones sees from general principles that prime ideals in $\rm\bar S\:$ correspond to maximal ideals of the localization $\rm\:S^{-1} D.\:$

In fact one can view this as a special case of how UFDs behave under localization. Generally the localization of a UFD remains a UFD. Indeed, such localizations are characterized by the sets of primes that survive (don't become units) in the localizations.

The converse is also true for atomic domains, i.e. domains where nonzero nonunits factor into atoms (irreducibles). Namely, if $\rm\:D\:$ is an atomic domain and $\rm\:S\:$ is a saturated submonoid of $\rm\:D^*$ generated by primes, then $\rm\: D_S$ UFD $\rm\:\Rightarrow\:D$ UFD $\:\!$ (popularized by Nagata). This yields a slick proof of $\rm\:D$ UFD $\rm\Rightarrow D[x]$ UFD, viz. $\rm\:S = D^*\:$ is generated by primes, so localizing yields the UFD $\rm\:F[x],\:$ $\rm\:F =\:$ fraction field of $\rm\:D.\:$ Therefore $\rm\:D[x]\:$ is a UFD, by Nagata. This yields a more conceptual, more structural view of the essence of the matter (vs. traditional argument by Gauss' Lemma).

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    $\begingroup$ This seems like a bit much, given that we only need the $(\Rightarrow)$ direction. But thanks -- I'm sure this will be very helpful in understanding the structure of UFDs! $\endgroup$ – Daniel McLaury May 28 '12 at 20:18
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    $\begingroup$ @Daniel Since Benjamin mentioned Kaplansky's theorem only generally, my point was to emphasize that this proof uses only the simple direction (and to lend further insight on related matters). I edited to clarify the simple direction. $\endgroup$ – Bill Dubuque May 28 '12 at 21:04

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