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Apr 13, 2017 at 12:19 history edited CommunityBot
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
Sep 23, 2015 at 23:48 vote accept Aldo Guzmán Sáenz
Sep 23, 2015 at 17:47 comment added Eric Wofsey It follows easily from the universal coefficient theorem that this is equivalent to the existence of a group $A$ such that $\operatorname{Ext}(A,\mathbb{Z})\cong\mathbb{Q}$ (as $\mathbb{Q}$ is indecomposable, and $\operatorname{Hom}(A,\mathbb{Z})\cong\mathbb{Q}$ is impossible).
Sep 23, 2015 at 17:31 answer added Oscar Randal-Williams timeline score: 15
Sep 23, 2015 at 16:03 answer added Chris Gerig timeline score: 2
Sep 23, 2015 at 15:31 history asked Aldo Guzmán Sáenz CC BY-SA 3.0