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For $K=\mathbb{Q}(\sqrt{-163})$, the fundamental discriminant is $-163$ and the Minkowski bound for ideal classes isThus every ideal class contains an integral ideal of norm at most eight.

The ring of integers of a quadratic field is $\mathbb{Z}[\theta ]$ for $\theta =(1+\sqrt{-163})/2$, whose minimal polynomial is $X^2-X+41$. Its discriminant $-163$ is a quadratic nonresidue modulo each of $3,5,$ and $7$; directly modulo two, $X^2+X+1$ has no root. Hence these primes are inert by the splitting of rational primes in a quadratic field. The only prime ideal over one of them with norm at most eight is $(2)$, which has norm four and is principal. Unique prime ideal factorization now makes every ideal of norm at most eight principal. The Minkowski bound therefore provesThis is the Ideal class group of Q of square root of minus one hundred sixty-three.

The Minkowski convex body theorem states that if $\Lambda \subset {\mathbb{R}}^n$ is a full lattice and $C$ is a measurable, convex, centrally symmetric set withthen $C$ contains a nonzero point of $\Lambda$.

Here $d\not\equiv 3(\mathrm{mod}4)$, so the ring of integers of a quadratic field isUnder its complex embedding it has real-lattice covolume $\sqrt{d}$. An integral ideal $I$ has index $N(I)$ in ${\mathcal{O}}_K$, so its covolume is $\sqrt{d}N(I)$. Apply Minkowski's theorem to the disk $|z|\le R$, whose area is $\pi R^2$. Taking $R^2$ down to $4\sqrt{d}N(I)/\pi$ gives a nonzero $\alpha \in I$ withSince the field norm is $N_{K/\mathbb{Q}}(\alpha )=|\alpha {|}^2$ in this imaginary quadratic field, this proves the imaginary quadratic Minkowski disk bound

Because $(\alpha )\subseteq I$, the idealis integral, represents the inverse class of $I$, and hasThus every ideal class has a bounded-norm representative. Only finitely many integral ideals have bounded norm, proving the finiteness of the ideal class group.

Now take $K=\mathbb{Q}(\sqrt{-22})$. Its ring of integers is $\mathbb{Z}[\sqrt{-22}]$, its discriminant is $-88$, andEvery class therefore has an integral representative of norm at most five. The polynomial $X^2+22$ is irreducible modulo three and modulo five, so both primes are inert and there are no ideals of norm three or five. At two it reduces to $X^2$, and the splitting of rational primes in a quadratic field givesAn ideal of norm four is ${\mathfrak{p}}_2^2=(2)$ and is principal. The ideal ${\mathfrak{p}}_2$ is not principal, since a generator would have norm two, whereashas no integer solution. Consequently the Ideal class group of Q of square root of minus twenty-two is

Suppose finally that $y^3=x^2+22$. If $x$ were even, the right-hand side would be $2$ modulo four, which is not a cube modulo four. Hence $x$ and $y$ are odd. In ${\mathcal{O}}_K$,The two ideals on the left are coprime. Indeed, a common prime ideal would divide both $2x$ and $2\sqrt{-22}$. It cannot lie above two because their norms are odd. It would therefore lie above eleven and force $11\mid x$; but then $x^2+22$ has $11$-adic valuation exactly one, impossible for the cube $y^3$.

By unique factorization of ideals in a number field, coprimality impliesfor some ideal $\mathfrak{a}$. The class group has order two, so it has no element of order three; $[\mathfrak{a}{]}^3=1$ therefore forces $\mathfrak{a}$ to be principal. The only units are $\pm 1$, both of which can be absorbed into a cube, and hencefor integers $a,b$. Comparing the coefficient of $\sqrt{-22}$ givesThus $b=\pm 1$. For $b=1$ this requires $3a^2=23$, and for $b=-1$ it requires $3a^2=21$; neither is possible. Therefore the result on no integral solutions to y cubed equals x squared plus twenty-two is

The Dedekind factorization theorem says that, when a rational prime $p$ does not divide the index of an order $\mathbb{Z}[\theta ]\subseteq {\mathcal{O}}_K$, a factorizationof the minimal polynomial modulo $p$ gives

Let $K=\mathbb{Q}(\sqrt{d})$ with square-free $d$, and let $p$ be odd. The resulting splitting of rational primes in a quadratic field is:

Finally suppose $d>0$ and
$\alpha =A/C+(B/C)\sqrt{d}$ has norm $-1$, with integers
$A,B,C$ chosen coprime. ThenIf an odd $p\equiv 3(\mathrm{mod}4)$ ramified, then $p\mid d$. Reduction modulo $p$ gives
$A^2\equiv -C^2(\mathrm{mod}p)$. Since $-1$ is not a square modulo such a prime, $p\mid A,C$. Dividing the equation's divisibility shows $p\mid B$ as well (use that square-free $d$ has $p$-adic valuation one), contradicting coprimality. Hence no prime $p\equiv 3(\mathrm{mod}4)$ ramifies.

For each rational prime $p$, defineThese are the three cases in splitting of rational primes in a quadratic field.

Set $x=p^{-s}$. In the split case the local factor of the Dedekind zeta function is $(1-x{)}^{-2}$; in the inert case it is $(1-x^2{)}^{-1}$ because the prime ideal has norm $p^2$; and in the ramified case it is $(1-x{)}^{-1}$. On the other hand, the product of the local factors of ${\zeta }_{\mathbb{Q}}(s)$ and $L(\chi ,s)$ iswhich gives exactly those three expressions when $\chi (p)=1,-1,0$. Multiplying over all rational primes proves the quadratic Dedekind zeta factorizationformally.