Class group of Q(sqrt 10)

By [F1], ${\mathcal{O}}_K=\mathbb{Z}[\delta ]$ with $d_K=40$. Every field embedding $K\to \mathbb{C}$ fixing $\mathbb{Q}$ sends $\delta$ to a root of $X^2-10$, that is, to $\pm \delta$, and both of these are real; so $(r_1,r_2)=(2,0)$ with $n=2$ by [F2].

The map $\phi (a+b\delta )=a<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2$ is a surjective ring homomorphism ${\mathcal{O}}_K\to {\mathbb{F}}_2$: it is additive, and $(a+b\delta )(a^{\mathrm{\prime }}+b^{\mathrm{\prime }}\delta )=(a{a}^{\mathrm{\prime }}+10b{b}^{\mathrm{\prime }})+(a{b}^{\mathrm{\prime }}+a^{\mathrm{\prime }}b)\delta$ maps to $a{a}^{\mathrm{\prime }}+10b{b}^{\mathrm{\prime }}\equiv a{a}^{\mathrm{\prime }}=\phi (a+b\delta )\phi (a^{\mathrm{\prime }}+b^{\mathrm{\prime }}\delta )(\mathrm{m}\mathrm{o}\mathrm{d}2)$. Its kernel is $\{a+b\delta :a\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}2)\}$, which equals the ideal ${\mathfrak{p}}_2:=(2,\delta )$: the products $2x+\delta y$ have even coefficient of $1$, and conversely $a+b\delta =2\cdot \frac{a}{2}+b\delta$. Hence ${\mathcal{O}}_K\mathrm{/}{\mathfrak{p}}_2\cong {\mathbb{F}}_2$ and $N{\mathfrak{p}}_2=2$ by [F4].

Similarly $\psi (a+b\delta )=a-b<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>3$ is a surjective ring homomorphism ${\mathcal{O}}_K\to {\mathbb{F}}_3$: since $10\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}3)$, it sends $(a+b\delta )(a^{\mathrm{\prime }}+b^{\mathrm{\prime }}\delta )$ to $a{a}^{\mathrm{\prime }}+b{b}^{\mathrm{\prime }}-a{b}^{\mathrm{\prime }}-a^{\mathrm{\prime }}b=(a-b)(a^{\mathrm{\prime }}-b^{\mathrm{\prime }})$ modulo $3$. Its kernel is $\{a+b\delta :a\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}3)\}$, which equals ${\mathfrak{p}}_3:=(3,1+\delta )$ because $a+b\delta =3x+b(1+\delta )$ when $a=b+3x$ and conversely $3x+y(1+\delta )=(3x+y)+y\delta$ has congruent coefficients. So $N{\mathfrak{p}}_3=3$. Likewise ${\psi }^{\mathrm{\prime }}(a+b\delta )=a+b<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>3$ is a surjective ring homomorphism with kernel $\{a+b\delta :a\equiv -b(\mathrm{m}\mathrm{o}\mathrm{d}3)\}={\mathfrak{p}}_3^{\mathrm{\prime }}:=(3,1-\delta )$, so $N{\mathfrak{p}}_3^{\mathrm{\prime }}=3$.

${\mathfrak{p}}_2^2=(2)$: by [F7] the square is generated by the products of the generators $2,\delta$, namely $4$, $2\delta$ and ${\delta }^2=10$, so ${\mathfrak{p}}_2^2=(4,2\delta ,10)$. All three generators are multiples of $2$, giving ${\mathfrak{p}}_2^2\subseteq (2)$; conversely $2=10-2\cdot 4\in {\mathfrak{p}}_2^2$, so $(2)\subseteq {\mathfrak{p}}_2^2$. Hence ${\mathfrak{p}}_2^2=(2)$.

${\mathfrak{p}}_3{\mathfrak{p}}_3^{\mathrm{\prime }}=(3)$: by [F7] the product is generated by $9$, $3(1-\delta )$, $3(1+\delta )$ and $(1+\delta )(1-\delta )=1-10=-9$, so ${\mathfrak{p}}_3{\mathfrak{p}}_3^{\mathrm{\prime }}=3\cdot (3,1-\delta ,1+\delta )$. That second ideal contains $(1-\delta )+(1+\delta )=2$ and $3$, hence contains $3-2=1$, so it is ${\mathcal{O}}_K$ and ${\mathfrak{p}}_3{\mathfrak{p}}_3^{\mathrm{\prime }}=(3)$.

Minkowski constant: by step 1.1 and [F1], $M_K=(4\mathrm{/}\pi {)}^0\frac{2!}{2^2}\sqrt{40}=\frac{1}{2}\cdot 2\sqrt{10}=\sqrt{10}<4$, because $10<16$.

Uniqueness: let $\mathfrak{b}$ be an integral ideal with $N\mathfrak{b}=p\in \{2,3\}$. Then ${\mathcal{O}}_K\mathrm{/}\mathfrak{b}$ has $p$ elements, so its additive group is generated by $1$ and it is isomorphic to ${\mathbb{F}}_p$; the composite $\mathbb{Z}[X]\to {\mathcal{O}}_K\to {\mathbb{F}}_p$ sends $X$ to an element $u$ with $u^2=10$. For $p=2$ one has $u^2=0$, so $u=0$, hence $2$ and $X$ lie in the kernel and the image of $(2,X)$ is $(2,\delta )={\mathfrak{p}}_2\subseteq \mathfrak{b}$; with $N{\mathfrak{p}}_2=2=N\mathfrak{b}$, step 1.2 and [F6] give $\mathfrak{b}={\mathfrak{p}}_2$. For $p=3$ one has $u^2=1$, so $u=\pm 1$: if $u=1$ then $(3,\delta -1)=(3,1-\delta )={\mathfrak{p}}_3^{\mathrm{\prime }}\subseteq \mathfrak{b}$ and if $u=-1$ then $(3,\delta +1)={\mathfrak{p}}_3\subseteq \mathfrak{b}$, so by step 1.3 and [F6] $\mathfrak{b}$ is ${\mathfrak{p}}_3^{\mathrm{\prime }}$ or ${\mathfrak{p}}_3$.

By [F7] the product ${\mathfrak{p}}_2{\mathfrak{p}}_3$ is generated by the products $6$, $2(1+\delta )$, $3\delta$ and $\delta (1+\delta )=10+\delta$; the identity $4+\delta =(10+\delta )-6$ exhibits $4+\delta$ in the product, so $(4+\delta )\subseteq {\mathfrak{p}}_2{\mathfrak{p}}_3$.

${\mathfrak{p}}_2$ is not principal: if ${\mathfrak{p}}_2=(\beta )$ with $\beta =a+b\delta$, then by [F4] and [F5], $2=N{\mathfrak{p}}_2=N((\beta ))=\mathrm{\mid }{N}_{K\mathrm{/}\mathbb{Q}}(\beta )\mathrm{\mid }=\mathrm{\mid }{a}^2-10b^2\mathrm{\mid }$, so $a^2-10b^2=\pm 2$. Reducing modulo $5$ gives $a^2\equiv \pm 2(\mathrm{m}\mathrm{o}\mathrm{d}5)$, but the squares modulo $5$ are $0,1,4$ and neither $2$ nor $3$ occurs; this contradiction shows no such $\beta$ exists.

Inclusion and equal norms force equality: by [F5] and [F4], $N((4+\delta ))=\mathrm{\mid }16-10\mathrm{\mid }=6$, while $N({\mathfrak{p}}_2{\mathfrak{p}}_3)=2\cdot 3=6$ by steps 1.2 and 1.3; with $(4+\delta )\subseteq {\mathfrak{p}}_2{\mathfrak{p}}_3$ from step 2.3, [F6] gives ${\mathfrak{p}}_2{\mathfrak{p}}_3=(4+\delta )$.

Principal products are the identity class: step 1.4 gives $[{\mathfrak{p}}_2{]}^2=[{\mathfrak{p}}_2^2]=[(2)]=1$, and step 3.1 gives $[{\mathfrak{p}}_2][{\mathfrak{p}}_3]=[(4+\delta )]=1$, so $[{\mathfrak{p}}_3]=[{\mathfrak{p}}_2{]}^{-1}=[{\mathfrak{p}}_2]$; step 1.5 gives $[{\mathfrak{p}}_3][{\mathfrak{p}}_3^{\mathrm{\prime }}]=[(3)]=1$, so $[{\mathfrak{p}}_3^{\mathrm{\prime }}]=[{\mathfrak{p}}_3{]}^{-1}=[{\mathfrak{p}}_2]$.

Hence $[{\mathfrak{p}}_2{]}^2=1$ by step 4.1 while $[{\mathfrak{p}}_2]\mathrm{\ne }1$ by step 2.4, so $[{\mathfrak{p}}_2]$ has order exactly $2$.

By [F3] and step 2.1 every class of $\mathrm{Cl}({\mathcal{O}}_K)$ contains an integral ideal $\mathfrak{b}$ with $N\mathfrak{b}\le \sqrt{10}<4$, and $N\mathfrak{b}$ is a positive integer by [F4], so $N\mathfrak{b}\in \{1,2,3\}$. If $N\mathfrak{b}=1$ then ${\mathcal{O}}_K\mathrm{/}\mathfrak{b}$ is trivial, that is $\mathfrak{b}={\mathcal{O}}_K$; if $N\mathfrak{b}=2$ then $\mathfrak{b}={\mathfrak{p}}_2$; and if $N\mathfrak{b}=3$ then $\mathfrak{b}$ is ${\mathfrak{p}}_3$ or ${\mathfrak{p}}_3^{\mathrm{\prime }}$, by step 2.2. By step 4.1 all of these ideals represent either the identity class or $[{\mathfrak{p}}_2]$.

Therefore every class of $\mathrm{Cl}({\mathcal{O}}_K)$ is $1$ or $[{\mathfrak{p}}_2]$, so $\mathrm{Cl}({\mathcal{O}}_K)=\{1,[{\mathfrak{p}}_2]\}\cong \mathbb{Z}\mathrm{/}2\mathbb{Z}$ is generated by the class of ${\mathfrak{p}}_2=(2,\sqrt{10})$.