Past exam of the number theory course of the University of Oxford / 2007 / B / 7 / Solution

Minkowski convex body theorem states that a centrally symmetric convex measurable subset of ${\mathbb{R}}^n$ with volume greater than $2^n$ times the covolume of a full Euclidean lattice contains a nonzero lattice point. The equivalent closed-boundary form follows by compactness for compact convex bodies.

The set $S_t$ is centrally symmetric. It is convex because for $0\le \lambda \le 1$, the triangle inequality gives $\Vert \lambda u+(1-\lambda )v{\Vert }_1\le \lambda \Vert u{\Vert }_1+(1-\lambda )\Vert v{\Vert }_1\le t$. It is a diamond of area $2t^2$. For the embedded nonzero integral ideal $I$, the given covolume is $2N(I)\sqrt{d}$. Hence Minkowski convex body theorem applies whenFor the resulting nonzero $\alpha \in I$, write $x={\sigma }_1(\alpha )$, $y={\sigma }_2(\alpha )$. ThenDecrease $t$ to $2\sqrt{N(I)\sqrt{d}}$. A fixed slightly larger diamond contains only finitely many lattice points, so one nonzero $\alpha$ recurs along a sequence of such $t$ values. Passing to the limit provesGiven a target ideal class $C$, choose an integral ideal $I$ in $C^{-1}$. Since $\alpha \in I$, the ideal $J=(\alpha )I^{-1}$ is integral, belongs to $C$, and has ideal normFor $K=\mathbb{Q}(\sqrt{14})$, every ideal class thus has an integral representative of norm at most $\sqrt{14}<4$. The only possibilities are norms one, two or three. The prime two ramifies, giving the unique ideal of norm two $P_2=(2,\sqrt{14})$. The element $4+\sqrt{14}$ has field norm $16-14=2$, so $(4+\sqrt{14})=P_2$, making it principal. The prime three is inert because $X^2-14\equiv X^2+1(\mathrm{mod}3)$ is an irreducible polynomial; its prime ideal has norm nine, so there is no ideal of norm three. Every possible small representative is principal, proving class number one for Q square root of fourteen. Therefore