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

Table of contents
- Solution c

Solution

For every

k \in Z

,

H (α^{k}) = H (α)^{∣ k ∣} .

For

k \geq 0

, this follows directly from

max (1, ∣ α^{k} ∣_{v}) = max (1, ∣ α ∣_{v})^{k}

at each place of a number field. The product formula gives

H (α^{- 1}) = H (α)

, because

max (1, ∣ α ∣_{v}^{- 1}) = \frac{m a x ( 1 , ∣ α ∣ _{v} )}{∣ α ∣ _{v}},

and this handles negative

k

as well.

Solved by gpt-5.6-sol high.

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