Absolute multiplicative Weil height ($H(\alpha )$)

For an algebraic number $\alpha$ in a number field $L$, the absolute multiplicative Weil height isIt is independent of the field $L$ containing $\alpha$, satisfies $H({\alpha }^n)=H(\alpha {)}^{|n|}$, and obeys $H(\alpha +\beta )\le 2H(\alpha )H(\beta )$.

For coprime integers $p,q$ with $q>0$,This follows directly from the real and p-adic absolute values, or from the height-Mahler measure formula for the primitive polynomial $qX-p$.

The absolute logarithmic Weil height is $h(\alpha )=\log H(\alpha )$.

For a nonzero vector $\mathbf{a}=(a_0,\dots ,a_n)$ over a number field $K$, its absolute projective height isThe product formula makes this unchanged by multiplying all coordinates by the same nonzero scalar. The projective height of a linear form is the projective height of its coefficient vector.

The naive height of a polynomial is the largest absolute value of its coefficients. For a primitive polynomial in $\mathbb{Z}[X_1,\dots ,X_n]$, this equals the projective height of its coefficient vector.

If the primitive minimal polynomial of an algebraic number $\alpha$ has degree $d$, then

Let $P,Q\in \mathbb{Z}[X_1,\dots ,X_k]$ have degree at most $n_j$ in $X_j$. Thenwhenever the quotient is defined. At each non-Archimedean place, the integral coefficients contribute at most one; at the Archimedean places, the triangle inequality contributes the polynomial length. Multiplying these local estimates and using the product formula gives the claim.

If $\gamma \ne 0$ is algebraic of degree at most $D$, then every embedding satisfiesIndeed, $H({\gamma }^{-1})=H(\gamma )$ and the contribution of one Archimedean place to the height bounds $|\gamma {|}^{-1/D}$.

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.