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Nov 18, 2016 at 5:11 comment added Tyler Lawson Are you asking if taking units commutes with p-completion? If so, then the answer is no because it is also false for ordinary rings: $(R^\wedge_p)^\times \neq (R^\times)^\wedge_p$ as you can see from the example of $\Bbb Z$. However, the issue is almost entirely a $\pi_0$-issue.
Nov 18, 2016 at 1:30 comment added Qiaochu Yuan I don't understand the relationship between the second paragraph and the first.
Nov 17, 2016 at 19:06 history asked Nicolas Boerger CC BY-SA 3.0