ScalingStacks

Proof. [02V3]

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Proof.

Let p∈im⁡θΣp\in\operatorname{im}\theta_{\Sigma} be such that valK⁡(p)>a{\operatorname{val}}_{K}(p)>a, hence |t⁡(p)|<|ϖ|a|t(p)|<|\varpi|^{a}. By Lemma 5.57, this implies that red⁡(p)=q0{\operatorname{red}}(p)=q_{0}. In a neighbourhood of q0q_{0}, the divisor of the rational section sΨ⊗e​te​m0​ϖ−α0−e​m0​as_{\Psi}^{\otimes e}t^{em_{0}}\varpi^{-\alpha_{0}-em_{0}a} is zero, and so

‖sΨ⊗e​(p)​te​m0​(p)​ϖ−α0−e​m0​a‖=1.\|s_{\Psi}^{\otimes e}(p)t^{em_{0}}(p)\varpi^{-\alpha_{0}-em_{0}a}\|=1.

Set u=val⁡(p)u={\operatorname{val}}(p). Then,

ψ∥⋅∥(u)\displaystyle\psi_{\|\cdot\|}(u) =log⁡‖sΨ⊗e​(p)‖e​λK\displaystyle=\frac{\log\|s_{\Psi}^{\otimes e}(p)\|}{e\lambda_{K}}
=−e​m0​log⁡|t⁡(p)​|+(α0+e​m0​a)​log|​ϖ|−e​log⁡|ϖ|\displaystyle=\frac{-em_{0}\log|t(p)|+(\alpha_{0}+em_{0}a)\log|\varpi|}{-e\log|\varpi|}
=m0​(u−a)−α0e.\displaystyle=m_{0}(u-a)-\frac{\alpha_{0}}{e}.

The other cases are proved in a similar way. ∎

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