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Oct 17, 2018 at 16:11 comment added LSpice Is it OK to speak purely algebraically of an infinite- (even uncountable-) fold tensor product? I guess that one has to construct it "all at once" by an obvious analogue of the 2-fold construction, satisfying an obvious analogue of the 2-fold universal property, since one can't sneak up on it by tensoring only 2 modules at a time.
May 11, 2011 at 4:58 vote accept Amritanshu Prasad
Oct 4, 2010 at 8:13 vote accept Amritanshu Prasad
Oct 4, 2010 at 8:14
Oct 1, 2010 at 3:52 vote accept Amritanshu Prasad
Oct 1, 2010 at 3:52
Oct 1, 2010 at 3:51 vote accept Amritanshu Prasad
Oct 1, 2010 at 3:51
Sep 30, 2010 at 12:46 comment added Martin Brandenburg Note that this description is also valid if $S$ is infinite.
Sep 30, 2010 at 12:18 history answered Martin Brandenburg CC BY-SA 2.5