Past exam of the number theory course of the University of Oxford / 2011 / B / 6

Choose a nonzero integral ideal $B$ whose class is $[A{]}^{-1}$; one obtains such an ideal by multiplying the fractional ideal $A^{-1}$ by a suitable nonzero integer. The assumed Minkowski bound for ideal classes gives $0\ne \beta \in B$ withBecause $(\beta )\subseteq B$ and nonzero ideals of the ring of integers are invertible,It is an integral ideal in class $[B{]}^{-1}=[A]$. By multiplicativity of the ideal norm,The same argument applies if the starting $A$ is a fractional ideal, since its class is unchanged by clearing denominators.

Write $s=\sqrt{-33}$. The discriminant is $-132$, and the bound of part (a) isEvery ideal class therefore has an integral representative of norm at most $7$. The Kummer-Dedekind theorem givesModulo $5$, $x^2+33=x^2+3$ is irreducible, so there is no ideal of norm $5$. Modulo $7$ it has roots $\pm 3$, giving $(7)=R\bar{R}$, where $R=(7,s+3)$. The ideal of $3+s$ has norm $42$, and reduction in the three residue fields shows that it lies in $P,Q,R$. Equality of norms therefore givesIt follows that $[R]=[P][Q]$, because $[P{]}^2=[Q{]}^2=1$. The other prime over $7$ has the inverse class, which is the same class here.

By unique factorization of nonzero ideals, all norms at most $7$ now give only the classes $1,[P],[Q],[PQ]$: norm $4$ gives $P^2=(2)$, norm $6$ gives $PQ$, and norm $7$ was handled above. None of $P,Q,PQ$ is principal, since their norms are $2,3,6$ andAlso $[P]\ne [Q]$, since equality would make $PQ$ principal. Thus there are exactly four classes, all of exponent two:

For degree $3$ there are either three real embeddings or one real embedding and one complex pair. The respective Minkowski bound for ideal classes areUnder the given discriminant bound they are at most $\frac{2}{9}\sqrt{45}<2$ and $\frac{8}{9\pi }\sqrt{45}<2$. Hence every ideal class contains an integral ideal with positive integer norm strictly below $2$, so with norm $1$. The only such ideal is ${\mathcal{O}}_L$. Therefore

Part (a), applied to any ideal class, produces an integral ideal of positive integer norm at most $c_K$, so $c_K\ge 1$. Writing $n=[K:\mathbb{Q}]$ and using $s\le n/2$ givesFor the last inequality, write $n^n/n!={\prod }_{j=1}^n(n/j)$; the first $\lfloor n/2\rfloor$ factors are at least $2$, and the remaining factors are at least $1$. Since $\lfloor n/2\rfloor \ge (n-1)/2$, squaring yieldsAs $\pi /2>1$, the assumed upper bound $D$ forcesFor $D\ge 1$, the ceiling of the right-hand side is a valid integer $M$. If $D=0$, no number field satisfies the premise, so any positive $M$ works.