# How does one prove that $\mathbb{Z}\left[\frac{1 + \sqrt{-43}}{2}\right]$ is a unique factorization domain?

By extending the Euclidean algorithm one can show that $\mathbb{Z}[i]$ has unique factorization.

This logic extends to show $\mathbb{Z}\left[\frac{1 + \sqrt{-3}}{2}\right]$, $\mathbb{Z}\left[\frac{1 + \sqrt{-3}}{7}\right]$ and $\mathbb{Z}\left[\frac{1 + \sqrt{-11}}{2}\right]$ have unique factorization.

However, there are principal ideal domains which are not Euclidean. How do we check that $\mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right]$ and $\mathbb{Z}\left[\frac{1 + \sqrt{-43}}{2}\right]$ have unique factorization?

There are two other interesting rings: $\mathbb{Z}\left[\frac{1 + \sqrt{-41}}{2}\right]$ and $\mathbb{Z}[\sqrt{-41}]$.

• See this answer. – Bill Dubuque May 27 '15 at 1:43
• @johnmangual Presumably he meant $\mathbb{Z}[\sqrt{-41}]$, since $\mathbb{Z}\left[\frac{1 + \sqrt{-41}}{2}\right]$ contains algebraic numbers that are not algebraic integers, e.g., $$\left(\frac{1 - \sqrt{-41}}{2}\right) \left(\frac{1 + \sqrt{-41}}{2}\right) = \frac{21}{2}.$$ – user155234 May 27 '15 at 20:42
• @John I meant $\mathbb{Z}[\sqrt{-41}]$. But I did not compute the class number, I had Mathematica do it with NumberFieldClassNumber[Sqrt[-41]]. There's also a table in Alaca & Williams and a few other books with these values. – user153918 May 28 '15 at 15:57

## 2 Answers

The first thing to keep in mind is that almost all complex quadratic integer rings are non-UFDs. Put another way, if a random $d$ is negative, then you can be almost certain that the ring of algebraic integers of $\mathbb{Q}(\sqrt{d})$ (often denoted $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$) has class number greater than $1$. If $d < -7$ is odd, then $2$ is irreducible but $d + 1 = (1 - \sqrt{d})(1 + \sqrt{d})$. For example, in $\textbf{Z}[\sqrt{-21}]$, we have $22 = 2 \times 11 = (1 - \sqrt{-21})(1 + \sqrt{-21})$, yet neither of the last two factors is divisible by $2$ or $11$.

In domains like $\mathcal{O}_{\mathbb{Q}(\sqrt{-19})}$ and a very few cases you've probably already seen in your textbooks (e.g., Heegner numbers), things get a lot more interesting. It is true that $$20 = 2^2 \times 5 = 2^2 \left(\frac{1 - \sqrt{-19}}{2}\right) \left(\frac{1 + \sqrt{-19}}{2}\right),$$ but it turns out that $$\left(\frac{1 - \sqrt{-19}}{2}\right) \left(\frac{1 + \sqrt{-19}}{2}\right) = 5.$$ This means that $2^2 \times 5$ is an incomplete factorization of $20$ in this ring.

And as you already know, this ring is not Euclidean, so that shortcut for proving unique factorization is not available. It is worth reviewing the fact that every principal ideal domain is a unique factorization domain (Theorem $2.3$ in Peric & Vukovic). If you can prove that the ring at hand (no pun intended) is a principal ideal domain, then you have also proven that it is UFD. There is a paper where they do just that, but I can't remember at the moment how I found it (it's on the Internet, I can tell you that much).

Peric & Vukovic take a different approach: they prove that $\mathcal{O}_{\mathbb{Q}(\sqrt{-19})}$ is an "almost Euclidean domain" (Definition $2.3$, Theorem $3.3$) and that all such domains are also UFDs (Theorem $2.2$).

A full and proper answer to your question would require about a dozen pages, which is about the length of Siqi Wei's paper "An example of a PID that is not a Euclidean domain." Here I'm just going to breeze through the relevant concepts in the hopes that it points you in the right direction towards a fuller understanding, but it's still going to be rather long for not being a proper answer.

All Yorkshire terriers are terriers, and all terriers are dogs. But not all dogs are terriers. Similarly, in algebraic number theory, we see all norm-Euclidean domains are Euclidean domains are principal ideal domains, and all principal ideal domains are unique factorization domains. But not all UFDs are PIDs, not all PIDs are Euclidean, and not all Euclidean domains are norm-Euclidean.

So you have verified that $\mathbb{Z}\left[\frac{1}{2} + \frac{\sqrt{-11}}{2}\right]$ (or $\mathcal{O}_{\mathbb{Q}(\sqrt{-11})}$ if you prefer) is norm-Euclidean, from which it follows that it is UFD. You have also been told that domains like $\mathcal{O}_{\mathbb{Q}(\sqrt{-19})}$ and $\mathcal{O}_{\mathbb{Q}(\sqrt{-43})}$ are PIDs but not Euclidean (that's true, and there's exactly only two others like that among the imaginary quadratic integer rings).

What Siqi Wei does is use the Dedekind-Hasse criterion to demonstrate that $\mathcal{O}_{\mathbb{Q}(\sqrt{-19})}$ is a PID. Then he uses the concept of the universal side divisor to prove that it is not Euclidean (I myself don't fully understand universal side divisors yet, but it's not all that relevant to your question here).

I haven't yet read the Perić and Vuković paper that Albert mentions, so I can't really speak to their approach with almost Euclidean domains. But it does suggest yet another approach to me: the classic proof that $\mathbb{Z}$ is UFD does without reference to PIDs. Niven & Zuckerman adapt that proof to show $\mathbb{Z}[i]$ is also UFD. Maybe this can also be done for $\mathcal{O}_{\mathbb{Q}(\sqrt{-19})}$, but I haven't tried to yet.