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There are only finitely many algebraic numbers of bounded degree and bounded Absolute multiplicative Weil height. The finite set can be enumerated effectively from the bounded coefficients of their primitive minimal polynomials.

Let $K$ be a number field. Its places consist of its real embeddings, conjugate pairs of complex embeddings, and the finite places associated with nonzero prime ideals $\mathfrak{p}$ of its ring of integers of a number field. At a real or complex place use the usual absolute value. If $\mathfrak{p}$ lies over the prime number $p$ with ramification index $e_{\mathfrak{p}}$, normalize its absolute value byThese normalizations extend the standard absolute values on $\mathbb{Q}$.

Write $d_v=[K_v:{\mathbb{Q}}_v]$ for the local degree of a place. Thus $d_v$ is $1$ at a real place, $2$ at a complex place, and $e_v{f}_v$ at a finite place. The Absolute multiplicative Weil height isThe product formula shows that this is unchanged when $K$ is replaced by a larger number field containing $\alpha$.

Solved by gpt-5.6-sol high.

Choose a number field $L$ containing ${\alpha }_1,\dots ,{\alpha }_k$. At every place $v$ of $L$, putThe triangle inequality giveswhere $c_v=1$ at every non-Archimedean place because the coefficients are integers, while at an Archimedean place one may takeRaise these inequalities to the local weights and multiply over all places. The definition of the Absolute multiplicative Weil height and the product formula then giveThis is the height bound for a polynomial evaluation.

Solved by gpt-5.6-sol high.