Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 1 / 20H (Number Fields)

Write $\alpha =\sqrt{-26}$. Since the squarefree radicand is not congruent to one modulo four, the ring of integers of a quadratic field is ${\mathcal{O}}_K=\mathbb{Z}[\alpha ]$. The discriminant of the basis $(1,\alpha )$ is $4(-26)$, giving

The ideal ${\mathfrak{p}}_2=(2,\alpha )$ has ideal norm two. Its square is contained in $(2)$ and has norm four, so $(2)={\mathfrak{p}}_2^2$. A generator would have field norm $u^2+26v^2=2$, impossible for integers $u,v$. Thus its ideal class has order two.

Modulo three, $X^2+26=(X-1)(X+1)$, giving $(3)={\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$ with ${\mathfrak{p}}_3=(3,1-\alpha )$. A generator of ${\mathfrak{p}}_3$ would have norm three, again impossible. If ${\mathfrak{p}}_3^2$ were principal, its generator would have norm nine. The only possibilities are $\pm 3$, whose ideal is ${\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$, not ${\mathfrak{p}}_3^2$.

Finally $N(1-\alpha )=27$. This element lies in ${\mathfrak{p}}_3$ but not ${\overline{\mathfrak{p}}}_3$, because $\alpha \equiv -1$ in the latter quotient. Unique unique factorization of ideals in a number field and norms therefore give

Modulo five, $X^2+26=(X-2)(X+2)$. Choose ${\mathfrak{p}}_5=(5,2+\alpha )$; then $(5)={\mathfrak{p}}_5{\overline{\mathfrak{p}}}_5$, and $N(2+\alpha )=4+26=30$. The element $2+\alpha$ belongs to ${\mathfrak{p}}_2$, ${\mathfrak{p}}_3$ and ${\mathfrak{p}}_5$. Their product has norm $2\cdot 3\cdot 5=30$, soThe imaginary-quadratic Minkowski bound for ideal classes is $(2/\pi )\sqrt{104}<7$. Every ideal class has an integral representative of norm at most six, whose prime ideal factors lie over two, three or five. Thus these prime ideals generate the ideal class group. Conjugate classes are inverses, and the displayed relation eliminates the class over five. The remaining generators have orders two and three, so their product has order six and generates the whole group: