0
$\begingroup$

In $\mathbb{Z}$, the rules are fairly well established, a few minor quibbles notwithstanding. But in, say, $\mathbb{Z}[\sqrt{7}]$, there are, as far as I can tell, no established rules. What I've seen in some places is that the factorization roughly follows what the canonical factorization would be in $\mathbb{Z}$. This seems inadequate to me whether or not we're in a UFD.

So I propose the following rules:

  • First, if applicable, a unit other than 1 (e.g., $-1$, $i$). Negative rational primes (like $-3$ or $-47$) are not used.
  • Then, primes/irreducibles in order by norm (lowest to highest), with exponents for any factor present more than once. Where there is a choice between $-a \pm b \theta$ and $a \pm b \theta$, the latter shall be preferred.
  • If two or more factors have the same norm, they are sorted by signs thus: $-a - b \theta$, $-a + b \theta$, $a - b \theta$, $a + b \theta$. (In some places I've seen the opposite of this).

If these rules are incomplete, then there is some domain of algebraic integers in which there are numbers with factorizations for which these rules would fail to provide a definitive way of resolving questions of order.

For example, factorize $-405$ in $\mathbb{Z}[\sqrt{-14}]$. This is not a UFD and one of the distinct factorizations is the same as in $\mathbb{Z}$: $(-1) \times 3^4 \times 5$. It can also be factorized as $(-1) \times 3^3 \times (1 - \sqrt{-14})(1 - \sqrt{-14})$ or as $(-1) \times 5 \times (5 - 2\sqrt{-14})(5 + 2\sqrt{-14})$ (these are distinct, but if I'm wrong you'll loudly let me know they're not distinct). I suppose in the former there is no conflict between my proposed rules and a desire to parallel the canonical factorization in $\mathbb{Z}$, but in the latter, my proposed rules dictate 5, with its norm of 25, precede those numbers with a norm of 81, even though they kinda correspond to 3 in the $\mathbb{Z}$ factorization.

Whether or not these rules are complete, I will appreciate comments as to whether you like them on an aesthetic level.

EDIT: By $\theta$, I mean that algebraic number that is being adjoined to $\mathbb{Z}$. So for example, if we're talking about $\mathbb{Z}[\sqrt{7}]$, then $\theta = \sqrt{7}$. Thanks to anon for pointing out an important detail I left out.

$\endgroup$
  • 2
    $\begingroup$ They might be complete for quadratic rings, but I doubt they are for cubic, quartic, etc. $\endgroup$ – user153918 Sep 22 '14 at 14:54
  • 1
    $\begingroup$ Suppose a factor of something I'm trying to factor is $1+\sqrt{-10}$, up to units. If we set $\theta=\sqrt{-10}$, then then the factor is $1+\theta$, and your rules say to keep it as-is. However if $\theta=2+\sqrt{-10}$ then your rules say to change $1+\sqrt{-10}=-1+\theta$ to $1-\theta$, which is $-1-\sqrt{-10}$. So your rules contradict themselves. The problem is that since you haven't defined $\theta$, you can write an algebraic number as $a+b\theta$ in different ways by varying the $\theta$. $\endgroup$ – anon Sep 27 '14 at 19:11
  • 1
    $\begingroup$ Good point. I've amended the question. If we're talking about $\mathbb{Z}[\sqrt{-10}]$ then I mean for $\theta$ to be $\sqrt{-10}$. $\endgroup$ – Robert Soupe Sep 27 '14 at 19:24
  • 1
    $\begingroup$ I don't understand what this is about. Are you proposing your favorite ordering of prime factors in prime factorizations as a canonical one? What possible purpose does fixing an order have? $\endgroup$ – Mariano Suárez-Álvarez Sep 27 '14 at 20:06
  • 3
    $\begingroup$ @Mariano Agreeing on a canonical order would be a big help in the study of factorizations in non-UFDs. We take order for granted in $\mathbb{Z}$. Some people might suggest we make descending order canonical, but no one would seriously suggest something like $60 = 3 \times 2 \times 5 \times 2$ as canonical. $\endgroup$ – Bill Thomas Sep 27 '14 at 21:14
0
+100
$\begingroup$

No, these rules are incomplete. Consider for example, $\mathbb{Z}[\sqrt{2} + \sqrt{3}]$, a quartic domain in which the integers have the form $a + b\sqrt{2} + c\sqrt{3} + d\frac{\sqrt{2} + \sqrt{3}}{2}$, with $\{a, b, c, d\} \in \mathbb{Z}$. I don't know what the norms are for numbers in this domain, but humor me for just a minute and assume for the sake of argument that $4 + 3\sqrt{2} - 2\sqrt{3} + \frac{\sqrt{2} - \sqrt{3}}{2}$ and $4 + 3\sqrt{2} - 2\sqrt{3} - \frac{\sqrt{2} + \sqrt{3}}{2}$ have the same norm (they probably don't, but humor me). How do you resolve which one goes first? For these rules to be complete, it is necessary to have a full understanding of norm in any arbitrary domain. Also, if $\theta = \sqrt{2} + \sqrt{3}$, the whole thing becomes very awkward before we even think about norms.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ But, as far as I can tell, these rules are complete for quadratic integer rings, though I could be wrong. For example, in $\mathbb{Z}[\omega]$, we have $7 = (2 - \omega)(2 - \omega^2)$. But the way you've phrased the rules, $2 - \omega^2$ needs to be rewritten. Since $\omega^2 = -1 - \omega$, it follows that $2 - \omega^2 = 3 + \omega$. So we have $7 = (2 - \omega)(3 + \omega)$. Though I don't know, I guess I would like for the rules to allow the use of $\omega^2$. $\endgroup$ – Bill Thomas Oct 7 '14 at 21:28
0
$\begingroup$

The rules are actually incomplete not only for cubic rings and beyond, but also for quadratic rings. Your rules address the ordering of factors, but the problem I'm talking about is a non-problem for factorization in $\mathbb{Z}$, and that's the ordering of "nomials" within factors. For example, consider this factorization of $4$ in $\mathbb{Z}[\sqrt{65}]$: $$\left(-\frac{7}{2} + \frac{\sqrt{65}}{2}\right)\left(\frac{7}{2} + \frac{\sqrt{65}}{2}\right)$$ This factorization came up in a question recently asked on this site, and the answerer expressed it something like this: $$\left(\frac{\sqrt{65} - 7}{2}\right)\left(\frac{\sqrt{65} + 7}{2}\right)$$ Now, I believe you prefer the former to the latter, but I see nothing in your proposed rules to resolve this. Furthermore, these two ways of expressing this factorization should suggest other different but potentially canonical expressions.

This problem of course becomes more acute in cubic rings, to say nothing of quartics.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.