Prove the class number of $\mathbb{Z}[\sqrt{-5}]$ is $2$. 
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?
 A: 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. 
A: 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 $2$.
