Everywhere I look seems to blow by the statement that PIDs which are not fields have Krull dimension $1$. This relies on the fact:

A PID with Krull dimension $0$ is a field. (*)

It seems that the way to prove this is to show that $\left\{0\right\}$ is a maximal ideal. And to show this, one quotes the fact that maximal ideals in a PID are prime, so that no nonzero ideal can be maximal because the PID under consideration has Krull dimension $0$.

However, without evoking Zorn's Lemma to conclude that $\left\{0\right\}$ is contained in a maximal ideal, it seems that this argument cannot be used.

My question is: Is there a way to prove (*) without assuming Zorn's Lemma?



Given a UFD which is not a field, it has a prime element. It generates a non-zero prime ideal, so that we have a chain of prime ideals of length 1 and therefore the dimension is at least 1.

It remains to decide if PIDs are UFD. The usual proof uses the axiom of dependent choice, which is a weak form of the axiom of choice. In fact, it turns out that it is not provable in ZF. See MO/31507. In W. Hodges, Läuchli's algebraic closure of Q it is proven that it is consistent with ZF that there are PIDs without any maximal ideal, in particular these are no UFD. These have Krull dimension $0$, because in a PID every non-zero prime ideal is maximal.

  • $\begingroup$ I do not understand how this answers my question. I am trying to establish that $0$ is maximal under the hypothesis that there is only one prime ideal. $\endgroup$
    – Doug
    Nov 3 '14 at 21:29
  • $\begingroup$ And I believe then that what you are using is that a PID is a UFD, since a nonzero nonunit a fortiori gives the existence of a prime factor. $\endgroup$
    – Doug
    Nov 3 '14 at 22:13
  • $\begingroup$ Please see my edited answer. $\endgroup$ Nov 3 '14 at 22:26

Let $R$ be a PID and assume $a\in R$ is not invertible and $a\ne0$. If $a$ is irreducible, then we're done: if $bR$ is an ideal containing $aR$, then $a=bc$ so either $b$ is invertible or $c$ is invertible. In the former case $bR=R$, in the latter case $bR=aR$, so $aR$ is maximal.

If $a$ is not irreducible, we want to find it has an irreducible divisor. So let $a=a_0=a_1b_1$, with neither $a_1$ nor $b_1$ invertible. If one of them is irreducible, we're done. Otherwise, both are not irreducible and so $a_1=a_2b_2$. We have started a recursion; if it stops, we found an irreducible element. Otherwise we get $$ a_0R\subset a_1R\subset\dots\subset a_kR\subset\dotsb $$ and all of these are proper ideals. where $\subset$ denotes proper inclusion. We have a contradiction, because $$ I=\bigcup_{k\ge0} a_kR $$ is an ideal, so $I=cR$ and we conclude.

Note that this actually proves that a PID which is not a field has Krull dimension at least $1$. However it uses a principle that's not provable in ZF alone, because the existence of the whole sequence is needed in the “bad case” that we want to exclude.

The full Zorn's lemma (or some slightly weaker statement about existence of maximal ideals) is needed to prove that a domain which is not a field has Krull dimension at least $1$.

  • 1
    $\begingroup$ This uses the axiom of dependent choice DC, right? $\endgroup$ Nov 3 '14 at 22:25
  • $\begingroup$ @MartinBrandenburg Possibly, but that's something for Asaf. Surely it doesn't use the full axiom of choice, but rather noetherian induction. $\endgroup$
    – egreg
    Nov 3 '14 at 22:29
  • $\begingroup$ No noetherian induction. You define the $a_i$ recursively. This is where DC enters. $\endgroup$ Nov 3 '14 at 22:38
  • $\begingroup$ @MartinBrandenburg Yes, the link to MO was illuminating. $\endgroup$
    – egreg
    Nov 3 '14 at 22:39

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