Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 1 / 20G / Solution

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