Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 2 / 20H (Number Fields)

Choose a nonzero integral ideal $I$ in the inverse of any desired ideal class. For the nonzero element $x\in I$ supplied by the norm bound, $J=(x)I^{-1}$ is integral, belongs to the desired class and satisfiesThere are only finitely many integral ideals of bounded norm: identify ${\mathcal{O}}_K$ with ${\mathbb{Z}}^n$ as an additive group, so ideals of norm $m$ are among its subgroups of index $m$, of which there are finitely many. Since every ideal class has one of these bounded-norm representatives, the ideal class group is finite.

For the ideal class group of the imaginary quadratic field of discriminant minus eighty-four, write $s=\sqrt{-21}$. Here ${\mathcal{O}}_K=\mathbb{Z}[s]$, $D_K=-84$, and the Minkowski bound for ideal classes is $(2/\pi )\sqrt{84}<6$. Thus every class has a representative of norm at most five. The relevant prime ideals areThe first two are ramified: ${\mathfrak{p}}_2^2=(2)$ and ${\mathfrak{p}}_3^2=(3)$. Also ${\mathfrak{p}}_5^2=(2-s)$, as both sides have norm 25 and $s\equiv 2(\mathrm{mod}{\mathfrak{p}}_5)$. The element $3+s$ has norm 30 andThere are no elements of norms 2, 3, 5 or 6, since the element norm is $a^2+21b^2$. Hence the classes of ${\mathfrak{p}}_2$, ${\mathfrak{p}}_3$ and their product are all nontrivial and distinct. The relation makes the last product equal to the class of ${\mathfrak{p}}_5$. Ideals of norm four are principal, and the two primes above five have the same class of order two. The norm bound therefore gives no further classes. Thus

Suppose $y^2+21=x^3$. Its positive left side gives $x>0$. Reduction modulo four shows that $x$ is odd and $y$ even. Moreover $3\mid y$ would make the valuation of $y^2+21$ at 3 equal to one, impossible for a cube. The same argument excludes $7\mid y$.

In $\mathbb{Z}[s]$, factor $(y+s)(y-s)=(x{)}^3$. Any common prime ideal divisor divides both $2y$ and $2s$, so must lie above 2, 3 or 7. The odd norm excludes 2, and the previous divisibility arguments exclude 3 and 7. The two ideals are therefore coprime. Unique factorization of ideals gives $(y+s)=I^3$. The class group has exponent two, so $[I{]}^3=1$ implies $[I]=1$. Thus $I=(a+bs)$, and the only units, $\pm 1$, can be absorbed into a cube. It follows thatComparing the coefficient of $s$ gives $1=3b(a^2-7b^2)$, an impossibility over the integers. Consequently there are no integer solutions: this is the no integer solutions of the Mordell equation with constant minus twenty-one result. The class-group argument is essential: element factorization was not assumed to be unique.