# 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.

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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 $u(1-u)$.

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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). –  paul garrett Mar 3 '13 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 $$2^2 \left(\frac{1}{2} - \frac{\sqrt{-19}}{2}\right)\left(\frac{1}{2} + \frac{\sqrt{-19}}{2}\right) = 20.$$

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I will read this answer later and hopefully upvote. Thanks. –  daniel Oct 28 '14 at 4:37
Take your time. These numbers ain't going anywhere. –  Robert Soupe Oct 28 '14 at 12:21
Yes this seems clear and I am adding my vote. But is it really different from mercio's response? –  daniel Oct 30 '14 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. –  Robert Soupe Oct 30 '14 at 12:29