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All rings in this post are commutative and with $1$.

Everyone knows the definition of a factorial ring, a. k. a. unique factorization domain (UFD). I have been wondering about some variations regarding this notion.

(a) A ring $R$ is called pre-pre-Schreier (this is my nomenclature) if and only if for any four elements $a$, $b$, $c$, $d$ of $R$ satisfying $ab = cd$, we can find four elements $x$, $y$, $z$, $w$ in $R$ such that $a = xy$, $b = zw$, $c = xz$, $d = yw$.

(b) A ring $R$ is called pre-Schreier if it is pre-pre-Schreier and an integral domain. (This is not my nomenclature.)

It is easy to see that a Noetherian ring is a UFD if and only if it is pre-Schreier; on the other hand, the condition on a ring to be pre-Schreier is a first-order logic formula (if I'm right; I'm not an expert in logic). This was actually my motivation to consider pre-Schreier rings: to first-orderize the UFD condition. (Is there a first-order logic formula that is equivalent to UFD always, no matter whether the ring is Noetherian or not?) As for pre-pre-Schreier rings, I was just trying to see what happens if we leave out the domain condition.

According to this paper (Remark 4.6. (1)), the polynomial ring $R\left[X\right]$ over a pre-Schreier ring $R$ doesn't have to be pre-Schreier. My questions are now:

(1) If a Noetherian ring $R$ is pre-pre-Schreier, then what can be said about $R\left[X\right]$ ?

(2) Can a pre-pre-Schreier ring contain nilpotents $\neq 0$ ? I used to think I have proven that it can't if it is Noetherian, but now I see flaws in my argument.

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    $\begingroup$ Just a side remark: the Schreier condition can be found in Euclid, book VII: look up the definition of four numbers in proportion. $\endgroup$
    – Franz Lemmermeyer
    Feb 7, 2010 at 21:30
  • $\begingroup$ Where exactly do my x, y, z, w appear in Euclid? $\endgroup$
    – darij grinberg
    Feb 7, 2010 at 22:07
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    $\begingroup$ +1. In response to one of your questions: I strongly suspect that being a UFD is not a first-order property, not even in the weaker sense of being preserved by elementary equivalence. In general, chain conditions do not fare well upon passage to ultrapowers, and UFD implies ACCP: the ascending chain condition on principal ideals. If I can see how to establish this for sure, I'll leave an answer. $\endgroup$
    – Pete L. Clark
    Feb 7, 2010 at 22:34
  • $\begingroup$ @DG: Regarding (2), why don't you show us your proof in the Noetherian case? $\endgroup$
    – Pete L. Clark
    Feb 8, 2010 at 8:05
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    $\begingroup$ As for Euclid's "Numbers are proportional when the first is the same multiple, or the same part, or the same parts, of the second that the third is of the fourth": The first number $a$ is parts of the second number $b$ if $a = \frac mn b$ for numbers $m, n$ such that $n \mid b$. Thus $b = nx$ and $a = mx$. Similarly, $c = \frac mn d$ with $d = ny$ and $c = my$. $\endgroup$
    – Franz Lemmermeyer
    Feb 8, 2010 at 13:38

1 Answer 1

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This is an answer to your parenthetical question:

Is there a first-order logic formula that is equivalent to UFD always, no matter whether the ring is Noetherian or not?

No. In fact, if T is any extension of the first-order theory of integral domains such that every model of T is a UFD, then every model of T is a field.

I think this is well known, but I couldn't think of a reference. I'll sketch a quick proof. Let φn(x) be the standard first-order formula which says that x is a product of n irreducible elements (and φ0(x) says that x is a unit). We argue that for some fixed n,

T ⊦ ∀x (x ≠ 0 → φ0(x) ∨ ... ∨ φn(x)).

If not, by the Compactness Theorem, T has a model R with a distinguished nonzero element x such that R ⊧ ¬φn(x) for each n; so x does not have a factorization into irreducibles. To conclude, show that every UFD which satisfies ∀x (x ≠ 0 → φ0(x) ∨ ... ∨ φn(x)) must be a field.

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  • $\begingroup$ +1: This is what I had, albeit only vaguely, in mind when I made my comment above. $\endgroup$
    – Pete L. Clark
    Feb 7, 2010 at 23:40
  • $\begingroup$ +1 from me, too. Nice argument. $\endgroup$
    – darij grinberg
    Feb 8, 2010 at 11:22

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