Minkowski Theory and Number Field Class Groups — Examples

By [F1], ${\mathcal{O}}_K=\mathbb{Z}[i]$ and $d_K=-4$. Every field embedding $K\to \mathbb{C}$ fixing $\mathbb{Q}$ sends $i$ to a root of $X^2+1$, that is, to $\pm i$, and both of these are nonreal; so $(r_1,r_2)=(0,1)$ with $n=2$ by [F2].

Minkowski constant: by step 1.1 and [F1], $M_K=(4\mathrm{/}\pi {)}^1\frac{2!}{2^2}\sqrt{4}=\frac{4}{\pi }\cdot \frac{1}{2}\cdot 2=\frac{4}{\pi }$; by [F5] $\pi >8\mathrm{/}3>2$, so $M_K=4\mathrm{/}\pi <2$.

Every class of $\mathrm{Cl}({\mathcal{O}}_K)$ contains an integral ideal $\mathfrak{b}$ with $N\mathfrak{b}\le M_K<2$ by [F3] and step 2.1. By [F4] the norm $N\mathfrak{b}$ is a positive integer, so $N\mathfrak{b}=1$ and therefore $\mathfrak{b}={\mathcal{O}}_K$, which is principal; hence every class is the principal class and $\mathrm{Cl}(\mathbb{Z}[i])$ is trivial.

By [F6] the Gaussian integers $\mathbb{Z}[i]={\mathcal{O}}_K$ form a Dedekind domain, so the criterion [F7] applies and the triviality of the class group from step 3.1 makes $\mathbb{Z}[i]$ a principal ideal domain.

In summary, $K=\mathbb{Q}(i)$ has $M_K=4\mathrm{/}\pi <2$, $\mathrm{Cl}(\mathbb{Z}[i])$ is trivial, and $\mathbb{Z}[i]$ is a principal ideal domain.

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 $\mathfrak{p}=(2,1+\sqrt{-5})$.

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 ${\mathfrak{p}}_2=(2,\sqrt{10})$.

Reduction modulo 3: $\bar{f}=X^5+2X+2$ has values $2,2,2$ at $0,1,2$, so it has no linear factor; division by the three monic irreducible quadratics leaves remainders $2$ for $X^2+1$ (where $X^2\equiv -1$), $X+2$ for $X^2+X+2$ (where $X^2\equiv -X-2$), and $X+2$ for $X^2+2X+2$ (where $X^2\equiv -2X-2$); hence $\bar{f}$ has no factor of degree at most $2$, and a degree-$5$ reducible polynomial would have one, so $\bar{f}$ is irreducible over ${\mathbb{F}}_3$.

Discriminant: in $\mathbb{C}$ write $f={\prod }_{i=1}^5(t-r_i)$; by [F2] and [F3], $\mathrm{Disc}(f)=\mathrm{Res}(f,f^{\mathrm{\prime }})={\prod }_i(5r_i^4-1)$ and ${\prod }_i{r}_i=-a_5=1$. Since $f(r_i)=0$ and $r_i\mathrm{\ne }0$ (as $f(0)=-1$), we have $r_i^4=(r_i+1)\mathrm{/}{r}_i$ and $5r_i^4-1=(4r_i+5)\mathrm{/}{r}_i$; moreover ${\prod }_i(4r_i+5)=4^5{\prod }_i(r_i+5\mathrm{/}4)=-4^{5}f(-5\mathrm{/}4)=-1024(-\frac{3125}{1024}+\frac{5}{4}-1)=2869$. Hence $\mathrm{Disc}(f)=2869$.

By [F1] with $p=3$ the polynomial $f$ is irreducible over $\mathbb{Q}$, so it is the minimal polynomial of $\alpha$, $K=\mathbb{Q}(\alpha )$ has degree $5$, and $\alpha \in {\mathcal{O}}_K$ by [F6].

No ideal of norm $2$ or $3$: if $N\mathfrak{a}=p\in \{2,3\}$, then ${\mathcal{O}}_K\mathrm{/}\mathfrak{a}$ is a commutative ring with $p$ elements, hence isomorphic to ${\mathbb{F}}_p$; the composite $\mathbb{Z}[X]\to {\mathcal{O}}_K\to {\mathbb{F}}_p$ with $X\mapsto \bar{\alpha }$ kills $f$, so $f$ has a root mod $p$ by [F7]; but $f<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>3$ has values $2,2,2$ and $f<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2$ has values $1,1$ at all elements of their prime fields, a contradiction.

The factorisation $2869=19\cdot 151$ consists of distinct primes, so $\mathrm{Disc}(f)$ is squarefree; by [F2] and [F4] the power-basis discriminant of $\alpha$ is $2869$, so [F5] gives ${\mathcal{O}}_K=\mathbb{Z}[\alpha ]$ and $d_K=2869$.

Signature: $f^{\mathrm{\prime }}=5X^4-1$ vanishes exactly at $\pm c$ with $c=5^{-1\mathrm{/}4}$; since $c^4=1\mathrm{/}5$, $f(-c)=-c^5+c-1=\frac{4}{5}c-1<0$ and $f(c)=c^5-c-1=-\frac{4}{5}c-1<0$, using $0<c<1$. The derivative is positive on $(-\mathrm{\infty },-c)$, negative on $(-c,c)$, and positive on $(c,\mathrm{\infty })$, so the local maximum and local minimum are both negative. Since $f(x)\to -\mathrm{\infty }$ as $x\to -\mathrm{\infty }$ and $f(x)\to +\mathrm{\infty }$ as $x\to +\mathrm{\infty }$, there is exactly one real root and two conjugate pairs of nonreal roots, that is $(r_1,r_2)=(1,2)$.

Minkowski constant: $M_K=(4\mathrm{/}\pi {)}^2\frac{5!}{5^5}\sqrt{2869}=\frac{120}{3125}(\frac{4}{\pi }{)}^2\sqrt{2869}<\frac{120}{3125}\cdot \frac{16}{9}\cdot 54=\frac{11520}{3125}<4$, using $\pi >3$ of [F9] and $\sqrt{2869}<54$.

Every class of $\mathrm{Cl}({\mathcal{O}}_K)$ has an integral representative $\mathfrak{b}$ with $N\mathfrak{b}\le M_K<4$ by [F8] and step 4.1; the norm is a positive integer, so $N\mathfrak{b}\in \{1,2,3\}$, and step 2.2 rules out $2$ and $3$, leaving $N\mathfrak{b}=1$, i.e. $\mathfrak{b}={\mathcal{O}}_K$; hence every class is principal and $\mathrm{Cl}({\mathcal{O}}_K)$ is trivial.

Therefore ${\mathcal{O}}_K=\mathbb{Z}[\alpha ]$, $d_K=2869=19\cdot 151$, $M_K<4$, and $\mathrm{Cl}({\mathcal{O}}_K)$ is trivial.