Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 2 / 20G (Number Fields)

Put $s=\sqrt{30}$. Since $30\equiv 2(\mathrm{mod}4)$, the ring of integers is $\mathbb{Z}[s]$ and its field discriminant is $120$. All three primes ramify:For example ${\mathfrak{p}}_p^2\subseteq (p)$ and both ideals have norm $p^2$, proving equality. The Minkowski bound for ideal classes is $\sqrt{120}/2=\sqrt{30}<6$. Thus the ideal class group is generated by the classes of these prime ideals; an ideal of norm four contributes only the already principal ${\mathfrak{p}}_2^2$.

The elements $6+s$ and $5+s$ have norms $6$ and $-5$. Their ideal factorizations give ${\mathfrak{p}}_2{\mathfrak{p}}_3=(6+s)$ and ${\mathfrak{p}}_5=(5+s)$. This also explains the printed hint: $N((6+s)/2)=3/2$, a fraction, not the integer $32$. Therefore $[{\mathfrak{p}}_3]=[{\mathfrak{p}}_2]$ and $[{\mathfrak{p}}_5]=1$, leaving at most two ideal classes. The class of ${\mathfrak{p}}_2$ is nontrivial: a principal generator would have norm $\pm 2$, but $x^2-30y^2=\pm 2$ would require $x^2\equiv 2$ or $3(\mathrm{mod}5)$, impossible. Hence

Units have norm $\pm 1$. Norm $-1$ is impossible modulo three. The least positive unit greater than one isIndeed its norm is one. A smaller positive norm-one unit would have $0<y=({\epsilon }^{\prime }-1/{\epsilon }^{\prime })/(2\sqrt{30})<2$, forcing $y=1$, but then $x^2=31$. Multiplying an arbitrary positive unit by powers of $\epsilon$ reduces it to $[1,\epsilon )$, so minimality proves that all units are $\pm {\epsilon }^n$.

The equation with right side $+5$ is impossible modulo three, since it would require $x^2\equiv 2(\mathrm{mod}3)$. For norm $-5$, let $\beta =5+\sqrt{30}$, so ${\beta }^2=5\epsilon$ and $\beta /\epsilon =\sqrt{30}-5$. For unit reduction for quadratic norm equations, change the sign of $x+y\sqrt{30}$ if necessary to make it positive, then multiply by a power of the fundamental unit to place it in $[\beta /\epsilon ,\beta )$. Its conjugate is $-5$ divided by itself, soThe maximum on this interval occurs at its endpoints. Thus $y=1$, and $x^2=25$; only $x=-5$ lies in the half-open interval. Reversing the reduction gives all solutions:Every displayed element has norm $-5$, and the reduction proves completeness, including both signs of $x$ and $y$.