For every $k\in\mathbb Z$,
$$
H(\alpha^k)=H(\alpha)^{|k|}.
$$
For $k\geq0$, this follows directly from
$$
\max(1,|\alpha^k|_v)=\max(1,|\alpha|_v)^k
$$
at each <place of a number field>. The <product formula> gives $H(\alpha^{-1})=H(\alpha)$, because
$$
\max(1,|\alpha|_v^{-1})
=\frac{\max(1,|\alpha|_v)}{|\alpha|_v},
$$
and this handles negative $k$ as well.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 166 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-166/2/c/solution