Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

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 $u(1-u)$.

share|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.