Prove the class number of $\mathbb{Z}[\sqrt{-5}]$ is $2$ .
This argument Prove that the class number of $\mathbb{Z}[\zeta_3]$ is $1$ doesn't work since $\mathbb{Z}[\sqrt{-5}]$ is not a unique factorization domain. How do we prove this?
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user167857
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1What have you done so far? Have you already calculated this ring's Minkowski Bound, or its discriminant, in order to know what prime ideals to take into account? – Timbuc Jun 05 '15 at 08:16
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@Timbuc Is the Minkowski Bound $(2/ \pi) \sqrt{ \operatorname{disc}(\mathbb{Z}[\sqrt{-5} \hspace{1mm} ] \hspace{1mm} )}$? – user167857 Jun 05 '15 at 08:21
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3Take a look. – Jyrki Lahtonen Jun 05 '15 at 08:28
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@JyrkiLahtonen Very helpful, thank you – user167857 Jun 05 '15 at 08:30
2 Answers
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The ideal $J = (2, 1 + \sqrt{-5})$ is not principal (can be shown by considering the norm of a potential generator); hence the class group is not trivial and the class numer is greater than $1$.
If $I \subsetneq \mathbb Z[\sqrt{-5}]$ is any non-principal ideal, let $\alpha \in I$ have minimal nonzero norm, let $\beta \in I\setminus \langle \alpha \rangle$ have minimal norm. Then $\beta/\alpha$ (which without loss of generality has positive imaginary part) must lie in a region bounded by the unit circle, the lines $x = \frac12$, $x = -\frac12$, and $y = \frac12\sqrt 5$. For any $\gamma \in I$ with $\gamma \ne \beta$ and $\gamma/\alpha$ in the same region, $\beta - \gamma$ is closer to one of $0, \pm \alpha$ than the minimality of $\alpha$ allows. We conclude that $I = \langle \alpha, \beta \rangle$. Now show that $IJ$ is principal (e.g., again by considering elements of small norm).
We conclude that the class group has precisely two elements.
Robert Soupe
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Hagen von Eitzen
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3A shot in the dark, since this is an old post, but is it possible to be more explicit about showing that $IJ$ is principal? We're trying to show that if $I$ is not principal, then $IJ$ is principal, where $J = (2,1+ \sqrt{-5})$, and all we actually know is that $I$ is generated by two elements. It's hard for me to imagine how to use the fact that $I$ is generated by two elements (but not one) in any kind of argument... – tcamps Nov 29 '18 at 22:26
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First note that $(1+\sqrt{-5})(1-\sqrt{-5})=3\cdot 2=6$, and so the field does not have unique factorization, and so the class number cannot equal $1$. Now, the Minkowski bound on the class number for a number field $K$ is $$\frac{n!}{n^n}\left(\frac{4}{\pi}\right)^{r_2}\sqrt{|\text{disc}(K)|},$$ where $r_2$ is the number of complex places. In your this bounds the class number above by $$\frac{2}{4}\frac{4}{\pi}\sqrt{20}<3,$$ so we conclude that it equals $2$.
Eric Naslund
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6The Minkowski bound says you can find an integral ideal representing every class with norm less than or equal to the bound. It does not directly bound the class number (though it can certainly help). – vacant Feb 18 '20 at 02:47