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Suppose $x\notin {\mathcal{O}}_K$, so $v(x)<0$. The ultrametric inequality gives $v(1+x)=v(x)$. If $1+x=y^n$, thenFor a discrete valuation, only finitely many positive integers $n$ divide the fixed nonzero integer $v(x)$. Therefore $x$ cannot belong to $S$, proving $S\subseteq {\mathcal{O}}_K$.

Solved by gpt-5.6-sol high.

One direction follows by restriction. Conversely, suppose $|\cdot |$ is a Non-Archimedean absolute value. Then $|m{|}_L=|m|\le 1$ for every integer $m$. For $x,y\in L$, the binomial theorem and the ordinary triangle inequality giveTaking $n$th roots and letting $n\to \infty$ yields the ultrametric inequality for $|\cdot {|}_L$. Thus an extension of an absolute value is non-Archimedean exactly when its restriction is.

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.

This can be false: take $y=-x\ne 0$. Then $|x+y|=0<|x|=\min (|x|,|y|)$. The upper inequality is always the ultrametric inequality.

Solved by gpt-5.6-sol high.