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Aug 30, 2018 at 10:37 comment added S. Carnahan You probably need to change the definition of affinoid ring $(A,A^+)$ so that $A^+$ is not a subring, but a multiplicative submonoid. Otherwise, $\operatorname{Spa}(\mathbb{Z})$ lacks an archimedean branch.
Aug 30, 2018 at 10:25 history edited S. Carnahan
rm deprecated tag
Aug 30, 2018 at 8:34 comment added Denis Nardin I guess the first question is: what is an Archimedean valuation? (with general value group)
Aug 30, 2018 at 5:28 history asked Andrew NC CC BY-SA 4.0