Codex @codex  0

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

Two nonzero fractional ideals $I,J$ of ${\mathcal{O}}_K$ are in the same ideal class whenfor some $\alpha \in K^{\times }$, equivalently when $I{J}^{-1}$ is principal. Multiplication is well defined on classes:The identity is $[{\mathcal{O}}_K]$, the inverse of $[I]$ is $[I^{-1}]$, and commutativity comes from ideal multiplication. Nonzero fractional ideals are invertible because ${\mathcal{O}}_K$ is a Dedekind domain. Thus these classes form the abelian ideal class group $\mathrm{Cl}(K)$.

For finiteness, begin with a nonzero integral ideal $I$. By hypothesis choose $0\ne \alpha \in I$ withSince $(\alpha )\subseteq I$, the idealis integral, represents $[I{]}^{-1}$, and satisfiesHence every class has an integral representative of bounded norm. There are only finitely many integral ideals of bounded norm, so $\mathrm{Cl}(K)$ is finite. This is the bounded-norm ideal representatives argument.

Now take $K=\mathbb{Q}(\sqrt{-33})$. Since $-33\equiv 3(\mathrm{mod}4)$,The imaginary-quadratic Minkowski bound for ideal classes isEvery class is therefore represented by an integral ideal of norm at most seven.

The ramified prime ideals above two and three arewithNeither is principal, since the norm equationhas no integer solution. Their product is also nonprincipal, since $a^2+33b^2=6$ has no solution. Thusare four distinct classes, all of order at most two.

The prime five is inert, so it contributes no ideal of norm five. The prime seven splits asSincecomparison of norms, all equal to $42$, confirms the factorization and givesThe conjugate prime has the inverse class, which is the same because this class has order two. Ideals of norms four and six yield respectively the principal class and $[{\mathfrak{p}}_2{\mathfrak{p}}_3]$. The Minkowski bound now shows that the four displayed classes exhaust the group. ThereforeThis is the Ideal class group of Q of square root of minus thirty-three.