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Nov 7, 2022 at 8:47 comment added Martin Brandenburg Notice that the property means that $\varphi : \mathrm{Spec}(B) \to \mathrm{Spec}(A)$ satisfies a topological property: for all closed subsets $T \subseteq \mathrm{Spec}(B)$ we have $$T = \varphi^{-1}(\overline{\varphi(T)}).$$
S Nov 7, 2022 at 1:11 review First questions
Nov 7, 2022 at 3:24
S Nov 7, 2022 at 1:11 history asked hello CC BY-SA 4.0