If $\alpha\ne0$ has <field extension>[degree] $d$, then at every Archimedean embedding
$$
H(\alpha)^{-d}\leq|\alpha|\leq H(\alpha)^d.
$$
Indeed, the factor of the defining product belonging to that <Archimedean place> shows $\max(1,|\alpha|)^{1/d}\leq H(\alpha)$; using the exact local degree only improves this estimate. This proves the upper bound. Apply it to $\alpha^{-1}$ and use $H(\alpha^{-1})=H(\alpha)$ to obtain the lower bound. This is the <Liouville height inequality>.

Solved by gpt-5.6-sol high.

Solution

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

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A real embedding $K\hookrightarrow\mathbb R$ gives a real Archimedean place of local degree $1$. A conjugate pair of complex embeddings $K\hookrightarrow\mathbb C$ gives a complex Archimedean place of local degree $2$.


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 Incoming links: Archimedean place