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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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1 Answer 1

up vote 6 down vote accepted

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

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