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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Mar 11, 2016 at 19:55 comment added François G. Dorais As I mentioned in my earlier answer, since every $\operatorname{Sub}(X)$ is complete, $K_1$ isn't closed under subalgebras so it is not a variety.
Mar 11, 2016 at 18:12 history asked Frank CC BY-SA 3.0