Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 2 / 16H (Number Fields)

Use the customary squarefree convention for the quadratic radicand $d$. For $d\equiv 2,3(\mathrm{mod}4)$, the ring of integers of a number field is $\mathbb{Z}[\sqrt{d}]$, soThe ideal $(p)$ is prime exactly when this quotient is an integral domain. A polynomial quotient of a field is an integral domain exactly when the defining polynomial is irreducible. This proves the assertion, including the possibility of ramified primes where the reduction has a repeated root.

Squarefreeness is necessary for the printed assertion as written: $d=18\equiv 2(\mathrm{mod}4)$ defines the same field as $d=2$. The prime $3$ is inert in $\mathbb{Q}(\sqrt{2})$ because $X^2-2$ is irreducible over ${\mathbb{F}}_3$, although $X^2-18=X^2$ is reducible. Thus the result requires a squarefree quadratic radicand.

Put $s=\sqrt{-14}$. Then ${\mathcal{O}}_K=\mathbb{Z}[s]$ and the field discriminant is $-56$. Reduction of $X^2+14$ modulo $2$ and $3$ givesTheir ideal norms are $2,3,3$, respectively. The Minkowski bound for ideal classes is $(2/\pi )\sqrt{56}<5$, so every ideal class has an integral representative of norm at most $4$. The only possible representatives are the unit ideal, ${\mathfrak{p}}_2$, ${\mathfrak{p}}_3$, ${\overline{\mathfrak{p}}}_3$, and the norm-$4$ ideal ${\mathfrak{p}}_2^2=(2)$. Thus there are at most four classes.

The class of ${\mathfrak{p}}_2$ is nontrivial, since a principal ideal of norm $2$ would require $x^2+14y^2=2$. Also ${\mathfrak{p}}_3^2$ is not principal: norm $9$ forces $x+ys=\pm 3$, whose ideal is ${\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$, not ${\mathfrak{p}}_3^2$.

Now $N(2+s)=18$. It is divisible by ${\mathfrak{p}}_2$ and ${\mathfrak{p}}_3$, but not by ${\overline{\mathfrak{p}}}_3$, as substitution of $s=-1$ modulo $3$ gives $1$. Unique ideal factorization and norms therefore give $(2+s)={\mathfrak{p}}_2{\mathfrak{p}}_3^2$. Consequently $[{\mathfrak{p}}_3{]}^2=[{\mathfrak{p}}_2{]}^{-1}=[{\mathfrak{p}}_2]$, a nontrivial class of order $2$. The class of ${\mathfrak{p}}_3$ has order $4$, and the four-class bound is attained: