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Is there a general criterion which tells me whether $\mathbb{Z}[\sqrt{d}]$, $d \in \mathbb{Z}$ is a unique factorization domain?

$\mathbb{Z}[\sqrt{-5}]$ is a frequent example for non-unique factorization domains because 6 has two different factorizations. $\mathbb{Z}[\sqrt{-1}]$ on the other hand is a Euclidean domain. But I'm not even sure about simple examples like $\mathbb{Z}[\sqrt{2}]$.

Marc
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    I believe it's not even known whether there are infinitely many. At least when $d$ is squarefree and $d\equiv 2,3\bmod 4$, we have $$\mathbb{Z}[\sqrt{d}]=\text{ring of integers of }\mathbb{Q}(\sqrt{d})$$ and then it's a question of what the class number of $\mathbb{Q}(\sqrt{d})$ is (Wikipedia link). – Zev Chonoles Jun 09 '15 at 21:33
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    See also this question. – Dietrich Burde Jun 09 '15 at 21:43
  • Interesting. So it is known for which negative integers it is true but not for which positive integers? – Marc Jun 09 '15 at 21:51
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    Yes. Note, though, that it's more usual to consider the ring of integers of $\mathbb{Q}(\sqrt{d})$ rather than $\mathbb{Z}[\sqrt{d}]$ -- the former is the integral closure of the latter inside $\mathbb{Q}(\sqrt{d}$). Assuming $d$ is squarefree, this agrees when $d$ isn't $1$ mod $4$, but for $d\equiv 1 \pmod{4}$ the former is slightly larger. In this case, $\mathbb{Z}[\sqrt{d}]$ cannot possibly be a UFD, as it isn't integrally closed in its fraction field. Which could be taken as one reason why we look at the ring of integers instead, which could be. – Harry Altman Jun 09 '15 at 22:15
  • A ring of algebraic integers is a UFD if an only if it is principal. Moreover $\mathbf Z[\sqrt d]$ is a ring of algebraic integers, for a square-free $d$ if and only id $d\not\equiv 1\mod 4$. If $d<0$, there are only $9$ values for which its a PID. – Bernard Jun 09 '15 at 22:15
  • @Harry: I'm not familiar with the concept of the integral closure. Could you make an easy example where I can see your point? – Marc Jun 10 '15 at 08:25
  • I'd just look up what the other commenters have already written (see in particular here). But basically, to be brief, to say that a ring $A$ is integrally closed in its fraction field $K$ means that for any monic polynomial over $A$, all its roots in $K$ in fact lie in $A$. That this is true for the integers is known as the Rational Root Theorem; the same proof works for UFDs in general. – Harry Altman Jun 10 '15 at 17:56
  • This link may be of some help: http://oeis.org/A061574 –  Jun 30 '15 at 16:35

2 Answers2

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The general criterion is that no number can be found with more than one valid, distinct factorization. This might sound like I'm merely rephrasing the question, but it's actually a reframing of the question.

Plenty of numbers (infinitely many, to be precise) in $\mathbb{Z}[\sqrt{-5}]$ have more than one factorization. $6$ is just the easiest to find. To oversimplify matters, your main concern is with the "natural" primes from 2 to $p < 4|d|$ or $p \leq d$ as needed.

Now, the case of $\mathbb{Z}[\sqrt{2}]$ is actually more complicated than you might realize. Part of the complication is that $\sqrt{2}$ is a real number and so $\mathbb{Z}[\sqrt{2}]$ has infinitely many units. This sets up the trap of infinitely many factorizations that are not distinct because they involve multiplication by units, e.g., $$7 = (3 - \sqrt{2})(3 + \sqrt{2}) = (-1)(1 - 2\sqrt{2})(1 + 2\sqrt{2}) = (5 - 3\sqrt{2})(5 + 3\sqrt{2}) = \ldots$$

But $7$ really does have only one distinct factorization in $\mathbb{Z}[\sqrt{2}]$, as you can see by dividing these numbers by $1 + \sqrt{2}$, and $\mathbb{Z}[\sqrt{2}]$ really is a UFD. But the full explanation may require me to make several assumptions about what you know.

Let's look at a "simpler" domain, $\mathbb{Z}[\sqrt{10}]$, though it certainly has some of the same traps as $\mathbb{Z}[\sqrt{2}]$: $$31 = (-1)(3 - 2\sqrt{10})(3 + 2\sqrt{10}) = (11 - 3\sqrt{10})(11 + 3\sqrt{10}) = (-1)(63 - 20\sqrt{10})(63 \ldots$$

You have to look at numbers that are already composite in $\mathbb{Z}$ to begin with. And if $d = pq$, where $p$ and $q$ are distinct primes, the choice of where to look first is obvious: $$10 = 2 \times 5 = (\sqrt{10})^2.$$

Verify that $$\frac{\sqrt{10}}{2} \not\in \mathbb{Z}[\sqrt{10}], \frac{\sqrt{10}}{5} \not\in \mathbb{Z}[\sqrt{10}], \frac{2}{\sqrt{10}} \not\in \mathbb{Z}[\sqrt{10}], \frac{5}{\sqrt{10}} \not\in \mathbb{Z}[\sqrt{10}].$$ This means that $\mathbb{Z}[\sqrt{10}]$ is not UFD and we didn't need to compute any logarithms or sines to come to this conclusion. (You're starting to see why integral closure matters in making these determinations, right?)

Contrast $\mathbb{Z}[\sqrt{6}]$: $$6 = (2 - \sqrt{6})(2 + \sqrt{6})(3 - \sqrt{6})(3 + \sqrt{6}) = (\sqrt{6})^2$$ but $$\frac{\sqrt{6}}{2 + \sqrt{6}} = 3 - \sqrt{6}$$ and so on and so forth. This means that $6 = (\sqrt{6})^2$ is an incomplete factorization, just as, say, $81 = 9^2$ is in $\mathbb{Z}$. But this is not enough to prove that $\mathbb{Z}[\sqrt{6}]$ is or is not UFD.

As it turns out, both $\mathbb{Z}[\sqrt{2}]$ and $\mathbb{Z}[\sqrt{6}]$ are UFDs, and what is probably the simplest, most common way of proving this requires a full understanding of ideals. Adapting the proof that $\mathbb{Z}$ is a UFD to these domains can be done, but that has its own pitfalls.

Robert Soupe
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Yes, there is. But first there is the matter of integral closure (I prefer to think of it as "completeness") to attend to. If $d \equiv 1 \pmod 4$, then you need $\frac{1 + \sqrt{d}}{2}$ instead of $d$. For example, with $d = 13$, $1 + \sqrt{13}$ is an algebraic integer, but so is $\frac{1 + \sqrt{13}}{2}$. For fun, try this in your calculator: $$\left(\frac{1 - \sqrt{13}}{2}\right)\left(\frac{1 + \sqrt{13}}{2}\right) = ?$$

The criterion is $h(4d) = 1$ if $d \not \equiv 1 \pmod 4$, $h(d) = 1$ if so. $h(d)$ is a function that tells you the class number of $\mathcal{O}_{\textbf{Q}(\sqrt{d})}$. I now quote the formula for $h(d)$ from Mathworld: if $d$ is positive, $$h(d) = -\frac{1}{2 \log \eta(d)} \sum_{r = 1}^{d - 1} \left(\frac{d}{r}\right) \log \sin \left(\frac{\pi r}{d}\right),$$ see http://mathworld.wolfram.com/ClassNumber.html for a full explanation of this, make sure to substitute $4d$ as needed. That also gives the formula for $d$ negative, but since there are only nine such values...

David R.
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    I would hate to try to compute this by hand, even for $\sqrt{10}$, i.e. $d=40$. – Lubin Jul 01 '15 at 02:54
  • That's a good point. In my haste to crank out the answer I didn't have to time to get a sense of how often preliminary calculations would clearly show $h > 1$, making it unnecessary to complete the calculation. – David R. Jul 02 '15 at 21:26