ScalingStacks

Remark 1.13 . [025X]

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Remark 1.13.

We assume that |.|′|\raisebox{1.72218pt}{.}|^{\prime} is discrete and

|a′|′=exp⁡(−α​ord𝔬k′⁡(a′))(a′∈k′)|a^{\prime}|^{\prime}=\exp(-\alpha\operatorname{ord}_{\mathfrak{o}_{k^{\prime}}}(a^{\prime}))\qquad(a^{\prime}\in k^{\prime})

for α∈ℝ>0\alpha\in\mathbb{R}_{>0}. If

α∉⋃v,v′∈V∖{0}ℚ⁡(log⁡‖v‖−log⁡‖v′‖),\alpha\not\in\bigcup_{v,v^{\prime}\in V\setminus\{0\}}\mathbb{Q}(\log\|v\|-\log\|v^{\prime}\|),

then the assumption (ii) holds. Indeed, we suppose that |a′|′=‖v‖/‖v′‖|a^{\prime}|^{\prime}=\|v\|/\|v^{\prime}\| for some a′∈k′×a^{\prime}\in{k^{\prime}}^{\times} and v,v′∈V∖{0}v,v^{\prime}\in V\setminus\{0\}. Then

−α​ord𝔬k′⁡(a′)=log⁡‖v‖−log⁡‖v′‖,-\alpha\operatorname{ord}_{\mathfrak{o}_{k^{\prime}}}(a^{\prime})=\log\|v\|-\log\|v^{\prime}\|,

so that ord𝔬k′⁡(a′)=0\operatorname{ord}_{\mathfrak{o}_{k^{\prime}}}(a^{\prime})=0, and hence ‖v‖=‖v′‖\|v\|=\|v^{\prime}\|, as required.

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