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

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.