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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.

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.

The product formula says that every $x\in K^{\times }$ satisfiesFirst suppose that $x$ is an algebraic integer. Its principal ideal has the prime ideal factorizationTaking the ideal norm givesOn the other hand, the field norm is the product over embeddings, soEquating these expressions proves the formula for algebraic integers. Every nonzero element of $K$ is a quotient of two algebraic integers, and multiplicativity completes the proof.

Solved by gpt-5.6-sol high.

For every $k\in \mathbb{Z}$,For $k\ge 0$, this follows directly fromat each place of a number field. The product formula gives $H({\alpha }^{-1})=H(\alpha )$, becauseand this handles negative $k$ as well.

Solved by gpt-5.6-sol high.

Write $\mathcal{L}(R)$ for the polynomial length. The height bound for a polynomial evaluation isprovided the denominator is nonzero. At non-Archimedean places the integral coefficients and ultrametric inequality give the local estimate without an extra constant; at Archimedean places the triangle inequality gives the polynomial length. Multiplication over every place of a number field and the product formula produce the displayed bound.

Taking $P(X,Y)=X+Y$ and $Q=1$ givesTaking $P(X)=X+1$ gives

Solved by gpt-5.6-sol high.

For a linear form $L=a_1{X}_1+\cdots +a_N{X}_N$ over a number field $K$, define its height to be the projective height of its coefficient vector:The product formula makes this independent of multiplying $L$ by a nonzero scalar.

The Siegel lemma says the following. Let $D=[K:\mathbb{Q}]$, let $L_1,\dots ,L_M$ be linear forms in $N$ variables with $DM<N$, and suppose $H_p(L_i)\le \mathcal{H}$ for every $i$, where $\mathcal{H}\ge 1$. Then there is a nonzero $\mathbf{x}\in {\mathbb{Z}}^N$ annihilated by all the forms and satisfying

Solved by gpt-5.6-sol high.

One Archimedean form of the Schmidt subspace theorem is as follows. Let $L_1,\dots ,L_n$ be linearly independent linear forms in $n$ variables with algebraic coefficients. For every $\epsilon >0$, all nonzero $\mathbf{x}\in {\mathbb{Z}}^n$ satisfyingbelong to a finite union of proper rational linear subspaces of ${\mathbb{Q}}^n$.

We will also use its finite-place form: if $S$ is a finite set of places containing the Archimedean ones and, for each $v\in S$, the forms $L_{1,v},\dots ,L_{n,v}$ are independent, then the integer solutions oflie in finitely many proper rational subspaces. The absolute values are normalized so that the product formula holds.

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.

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.