Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 20F (Number Fields)

For a nonzero ideal $I\subset O_K$, its norm isFor $d=[K:\mathbb{Q}]$ and embeddings ${\sigma }_1,\dots ,{\sigma }_d$, defineThe field discriminant is the discriminant of any integral basis of $O_K$. If ${\alpha }_1,\dots ,{\alpha }_d$ is a $\mathbb{Z}$-basis of $I$, then

Multiplication by $\alpha$ on an integral basis has integer matrix $M_{\alpha }$. Its image lattice is $\alpha O_K$, soThe determinant of this $\mathbb{Q}$-linear map is the field norm $N_{K/\mathbb{Q}}(\alpha )$, proving

If $Q\mid P{O}_L$, then $P{O}_L\subseteq Q$, so $P\subseteq Q\cap O_K$. The contraction is a nonzero prime ideal of $O_K$ containing the maximal ideal $P$, hence equals $P$. Conversely, if $P=Q\cap O_K$, every element of $P$ lies in $Q$, so the ideal it generates satisfies $P{O}_L\subseteq Q$, equivalently $Q\mid P{O}_L$.

The equality of residue-field sizes says that the residue degree $f(Q/P)$ is one. Since $Q=\alpha O_L$, the principal ideal factorization of $\alpha$ has valuation one at $Q$ and zero at every other prime. Taking the relative ideal norm givesThus $P=N_{L/K}(\alpha )O_K$.