# Unique factorization in $\mathbb Z(\sqrt{-19})$

An elementary confusion about class number:

In $\mathbb Z(\sqrt{-19})$ we have $N(1+\sqrt{-19}) = (1+\sqrt{-19})(1-\sqrt{-19}) = 2^2\cdot 5.$

I see that 2 and 5 are irreducible, 4 is not.

In a UFD a non-zero, non-unit element can be factored uniquely (up to associates) as a finite product of irreducibles. What is it about the two factorizations of 20 above that prevents them from being non-trivial distinct factorizations into a finite product of irreducibles?

Thank you.

• @John-Luke Unless your bounty it's a joke, I can't see any "exemplary" answer here. Both are saying the same (more or less trivial) thing: $\mathbb Z[\sqrt{-19}]$ is not a UFD since it's not integrally closed. Commented Apr 8, 2015 at 7:58
• @user26857 Maybe he meant to click on a different bounty type. Or maybe—and is this really so hard to believe?—he really does think one of these answers is "exemplary and worthy of an additional bounty." But I suppose either way my answer will be ineligible for the bounty, since under this type the bounty must be awarded to an existing answer, right?
– user153918
Commented Apr 8, 2015 at 13:40
• @user26857 That's fine if it isn't. The important thing is for Daniel to find an answer that is correct and addresses all his confusions.
– user153918
Commented Apr 8, 2015 at 14:28
• I do like Alonzo's answer better, but I guess I'm at least honor-bound (if not technically bound) to award the bounty to an answer from before the bounty was offered. Commented Apr 13, 2015 at 20:53

Let $u = \frac{1+ \sqrt {-19}}2$. $u$ is an algebraic integer (because $u^2 = u-5$). And so the ring of integers of $\Bbb Q(\sqrt {-19})$ is $\Bbb Z[u]$ and not $\Bbb Z[\sqrt{ -19}]$.

$\Bbb Z[u]$ has unique factorisation. For example, $u$ and $1-u$ are irreducible, and $5$ factors as $u(1-u)$.

• And it might be noted that "failing" to take the full ring of algebraic integers will definitely prevent the resulting ring from being a principal ideal domain, because principal ideal domains are integrally closed (in their fraction fields). Commented Mar 3, 2013 at 18:05

That's because $\mathbb{Z}[\sqrt{-19}]$ is not integrally closed. Since $-19 \equiv 1 \pmod 4$, numbers of the form $\frac{a}{2} + \frac{b \sqrt{-19}}{2}$ (with $a$ and $b$ of the same parity) are also algebraic integers. Observe that $$\left(\frac{1}{2} - \frac{\sqrt{-19}}{2}\right)\left(\frac{1}{2} + \frac{\sqrt{-19}}{2}\right) = 5.$$ The former factor is an algebraic integer with minimal polynomial $x^2 - x + 5$, and the latter factor has the same minimal polynomial. This means that in $\mathcal{O}_{\mathbb{Q}(\sqrt{-19})}$, 5 is actually "composite"! It should also be clear that $1 + \sqrt{-19}$ is also reducible; to claim it as a prime factor of 20 in $\mathcal{O}_{\mathbb{Q}(\sqrt{-19})}$ would be just as erroneous as saying 10 is a prime factor of 20 in $\mathbb{Z}$. The complete factorization of 20 in $\mathcal{O}_{\mathbb{Q}(\sqrt{-19})}$ is then $$2^2 \left(\frac{1}{2} - \frac{\sqrt{-19}}{2}\right)\left(\frac{1}{2} + \frac{\sqrt{-19}}{2}\right) = 20.$$

• I will read this answer later and hopefully upvote. Thanks. Commented Oct 28, 2014 at 4:37
• Take your time. These numbers ain't going anywhere. Commented Oct 28, 2014 at 12:21
• Yes this seems clear and I am adding my vote. But is it really different from mercio's response? Commented Oct 30, 2014 at 5:49
• Maybe not, but hopefully it does offer those new to the topic more basic facts they can verify for themselves rather than accept on faith. Commented Oct 30, 2014 at 12:29

I'm going to try to take your question more literally than the others who've answered so far. I'm going to assume that by "$\mathbb{Z}(\sqrt{-19})$" you mean only the numbers of the form $a + b \sqrt{-19}$, where $a$ and $b \in \mathbb{Z}$. (I can't remember if it's in Stewart & Tall's book on the Fermat-Wiles theorem or maybe Watkins that there's a similar exercise with numbers of the form $a + b \sqrt{5}$, but the notation used for the ring is definitely not "$\mathbb{Z}(\sqrt{5})$").

In "$\mathbb{Z}(\sqrt{-19})$", the $\mathbb{Z}$-primes $2, 3, 5, 7, 11, 13, 17$ must be inert, $19$ ramifies (as $(-1)(\sqrt{-19})^2$, $23$ splits as $(2 - \sqrt{-19})(2 + \sqrt{-19})$ and after that the others might either split or stay inert. The factorization of $20$ from $\mathbb{Z}$ carries over to "$\mathbb{Z}(\sqrt{-19})$": $20 = 2^2 \times 5$, with three prime factors.

But as you noticed, $20$ also splits as $(1 - \sqrt{-19})(1 + \sqrt{-19})$. These two factors are irreducible in "$\mathbb{Z}(\sqrt{-19})$". I leave this to you to check, but I will tell you that they are not associates of either $2$ or $5$: $\frac{1 - \sqrt{-19}}{2}$ is outside of "$\mathbb{Z}(\sqrt{-19})$", as is $\frac{1 - \sqrt{-19}}{5}$.

The inescapable conclusion here is that this "$\mathbb{Z}(\sqrt{-19})$" is not a unique factorization domain and its class number can't be $1$ (or $2$ for that matter). Understandably confusion will arise when you look up $-19$ in a table of class numbers for quadratic rings and see a $1$ rather than $3$ or higher.

That's because they're not talking about "$\mathbb{Z}(\sqrt{-19})$". They're talking about all numbers of the form $n = a + b \sqrt{-19}$, where $a$ and $b \in \mathbb{Q}$ (not just $\mathbb{Z}$) such that $n$ is an algebraic integer. This concept of algebraic integers is very important in algebraic number theory.

The ring of algebraic integers within $\mathbb{Q}(\sqrt{-19})$ may be notated a few different ways: $\mathcal{O}_{\mathbb{Q}(\sqrt{-19})}$, $\textbf{Z}\left[ \frac{1 - \sqrt{-19}}{2} \right]$, $\mathbb{Z}\left[ \frac{1}{2} + \frac{\sqrt{-19}}{2} \right]$, etc. Some may first define $K = \mathbb{Q}(\sqrt{-19})$ and then refer to $\mathcal{O}_K$. Others may define $u = \frac{1 + \sqrt{-19}}{2}$ and then refer to $\mathbb{Z}[u]$. The important point is to show that you're aware of numbers of the form $\frac{a}{2} + \frac{b \sqrt{-19}}{2}$.

• Appreciate the late contribution. I will look at it as time allows. @mercio's answer did completely address my confusion, which was due to forgetting a definition. Commented Apr 8, 2015 at 18:12