Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 20G (Number Fields)

If $L$ has degree $n=r_1+2r_2$ and discriminant $d_L$, then every class in the ideal class group contains a nonzero integral ideal $\mathfrak{a}$ satisfying the Minkowski bound for ideal classes

The element $\omega$ satisfiesModulo both $2$ and $3$, this polynomial factors as $X(X-1)$. Henceandwhere the respective ideal norms are $2,2,3,3$.

Now $N(\omega )=12$, and $\omega$ lies in ${\mathfrak{p}}_2$ and ${\mathfrak{p}}_3$ but not in their conjugate primes. Similarly $N(2+\omega )=18$, while $2+\omega$ lies in ${\mathfrak{p}}_2$ and ${\mathfrak{q}}_3$. Comparing norms gives

Here $n=2$, $r_2=1$, and $d_K=-47$, so the Minkowski bound for ideal classes isEvery ideal class therefore has a representative of norm at most four. Such classes are generated by the prime ideals above $2$ and $3$. Putin additive notation. Since conjugate prime ideals represent inverse classes, the two principal-ideal factorizations from part (b) giveThus $y=-2x$ and $5x=0$, so the class group is generated by $x$ and has order dividing five.

It is not trivial. If ${\mathfrak{p}}_2$ were principal, there would be an element of norm two, butcannot equal two for integers $a,b$. Therefore $x\ne 0$. Since five is prime,