Class group of Q(sqrt -5)

By [F1], ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{-5}]$, $d_K=-20$, and the signature is $(r_1,r_2)=(0,1)$ with $n=2$.

The rule $\phi (a+b\sqrt{-5})=a+b(\mathrm{m}\mathrm{o}\mathrm{d}2)$ is additive and multiplicative (the only nontrivial check is $(a+b\sqrt{-5})(a^{\mathrm{\prime }}+b^{\mathrm{\prime }}\sqrt{-5})=(a{a}^{\mathrm{\prime }}-5b{b}^{\mathrm{\prime }})+(a{b}^{\mathrm{\prime }}+a^{\mathrm{\prime }}b)\sqrt{-5}$ mapping to $a{a}^{\mathrm{\prime }}-5b{b}^{\mathrm{\prime }}+a{b}^{\mathrm{\prime }}+a^{\mathrm{\prime }}b\equiv a{a}^{\mathrm{\prime }}+b{b}^{\mathrm{\prime }}+a{b}^{\mathrm{\prime }}+a^{\mathrm{\prime }}b=(a+b)(a^{\mathrm{\prime }}+b^{\mathrm{\prime }})$ in ${\mathbb{F}}_2$), is surjective, and its kernel consists of the $a+b\sqrt{-5}$ with $a+b$ even, which are exactly the elements of the ideal $(2,1+\sqrt{-5})$: the kernel contains $2$ and $1+\sqrt{-5}$, and conversely $a+b\sqrt{-5}=b(1+\sqrt{-5})+(a-b)$ with $a-b$ even when $a+b$ is even. Hence ${\mathcal{O}}_K\mathrm{/}\mathfrak{p}\cong {\mathbb{F}}_2$ and $N\mathfrak{p}=2$.

Minkowski constant: $M_K=(4\mathrm{/}\pi {)}^{r_2}(n!\mathrm{/}{n}^n)\sqrt{\mathrm{\mid }{d}_K\mathrm{\mid }}=(4\mathrm{/}\pi )\cdot (1\mathrm{/}2)\cdot \sqrt{20}=\frac{2}{\pi }\sqrt{20}<\frac{2}{3}\cdot \frac{9}{2}=3$, since $\pi >3$ by [F2] and $\sqrt{20}<9\mathrm{/}2$.

$p^2=(2)$: the products of the generators $2$ and $1+\sqrt{-5}$ are $4$, $2(1+\sqrt{-5})=2+2\sqrt{-5}$ and $(1+\sqrt{-5}{)}^2=-4+2\sqrt{-5}$, all multiples of $2$, so ${\mathfrak{p}}^2\subseteq (2)$; by [F4] the norms are $N({\mathfrak{p}}^2)=(N\mathfrak{p}{)}^2=4$ and $N((2))=\mathrm{\mid }{N}_{K\mathrm{/}\mathbb{Q}}(2)\mathrm{\mid }=4$, so [F6] gives ${\mathfrak{p}}^2=(2)$.

Uniqueness of the norm-$2$ ideal: let $\mathfrak{b}$ be an integral ideal with $N\mathfrak{b}=2$. Then ${\mathcal{O}}_K\mathrm{/}\mathfrak{b}$ has two elements, so $2\in \mathfrak{b}$; the composite $\mathbb{Z}[X]\to {\mathcal{O}}_K\to {\mathcal{O}}_K\mathrm{/}\mathfrak{b}$ sends $X^2+5$ to $0$ and $X$ to an element $u$ of the two-element ring ${\mathbb{F}}_2$ with $u^2=-5\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}\mathfrak{b})$, so $u=1$ (as $0^2\mathrm{\ne }1$); hence $X-1$ and $2$ lie in the kernel, the image of $(2,X-1)$ is $(2,\sqrt{-5}-1)=(2,1+\sqrt{-5})=\mathfrak{p}$, and $\mathfrak{p}\subseteq \mathfrak{b}$; with $N\mathfrak{p}=N\mathfrak{b}=2$, [F6] yields $\mathfrak{b}=\mathfrak{p}$.

$\mathfrak{p}$ is not principal: if $\mathfrak{p}=(\beta )$, then $\beta \in \mathfrak{p}\subseteq {\mathcal{O}}_K$ and by [F4] and [F5] the equation $a^2+5b^2=2$ would hold for $\beta =a+b\sqrt{-5}$. But $b=0$ gives $a^2=2$, impossible, and $b\mathrm{\ne }0$ gives $a^2+5b^2\ge 5>2$; so no such $\beta$ exists.

In the class group, $[\mathfrak{p}{]}^2=[{\mathfrak{p}}^2]=[(2)]=1$ is the identity class, while $[\mathfrak{p}]\mathrm{\ne }1$ by step 2.4; hence $[\mathfrak{p}]$ has order exactly $2$.

Every class has an integral representative $\mathfrak{b}$ with $N\mathfrak{b}\le M_K<3$ by [F3] and step 2.1, so its norm is $1$ or $2$: norm $1$ forces $\mathfrak{b}={\mathcal{O}}_K$, and norm $2$ forces $\mathfrak{b}=\mathfrak{p}$ by step 2.3.

Hence every class is either the principal class or $[\mathfrak{p}]$, so $\mathrm{Cl}({\mathcal{O}}_K)=\{1,[\mathfrak{p}]\}\cong \mathbb{Z}\mathrm{/}2\mathbb{Z}$ is generated by the class of $\mathfrak{p}=(2,1+\sqrt{-5})$.