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Sep 30, 2013 at 16:52 answer added Samuel Chamberlin timeline score: 1
Jul 24, 2011 at 4:51 comment added Torsten Ekedahl Minor quibble: I think you want $\mathcal B$ to be something like free as $\mathbb Z$-module. Otherwise $\mathcal B$ could be for instance a $\mathbb Q$-vector space (note that for a $\mathbb Q$-algebra $\mathcal B$ we have $\mathcal B\bigotimes_{\mathbb Z}\mathbb C=\mathcal B\bigotimes_{\mathbb Q}\mathbb C$.
Jul 24, 2011 at 1:59 answer added Mariano Suárez-Álvarez timeline score: 4
Jul 24, 2011 at 1:54 answer added Binai timeline score: 3
Apr 14, 2011 at 21:03 history edited Charles Matthews CC BY-SA 3.0
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Apr 14, 2011 at 20:34 history asked Najdorf CC BY-SA 3.0