The difficulty non-ramp
Feb. 21st, 2009 08:28 am...However, it would appear that there's nothing about sheaves or presheaves on next week's homework.
There is, however, one of the most useless hints ever. "Hint: requires some work." Nice. In essence "Hint: this is hard." I mean, I guess it's a helpful note, but I don't think "hint" is really the right word.
There is also the Grothendieck group of finitely generated A-modules, which I think is just funny. Take all isomorphism classes of finitely generated A-modules [yes this is set-sized, if you think about it, though Nori doesn't bother to note this in his definition], take the free abelian group with them as basis, then mod out by the relation M'+M''=M whenever there's a short exact sequence 0→M'→M→M''→0. Wow. You just made (finitely generated) A-modules into an abelian group, even if the modules themselves aren't the elements (the image of M in this group is denoted cl(M)). You know, I feel like it would somehow be more appropriate if you used the free A-module rather than the free abelian group, so as to make an A-module of A-modules, but it doesn't seem like you'd actually gain any additional information that way, and the fact that that isn't what's done suggests that my suspicion is right.
-Harry
There is, however, one of the most useless hints ever. "Hint: requires some work." Nice. In essence "Hint: this is hard." I mean, I guess it's a helpful note, but I don't think "hint" is really the right word.
There is also the Grothendieck group of finitely generated A-modules, which I think is just funny. Take all isomorphism classes of finitely generated A-modules [yes this is set-sized, if you think about it, though Nori doesn't bother to note this in his definition], take the free abelian group with them as basis, then mod out by the relation M'+M''=M whenever there's a short exact sequence 0→M'→M→M''→0. Wow. You just made (finitely generated) A-modules into an abelian group, even if the modules themselves aren't the elements (the image of M in this group is denoted cl(M)). You know, I feel like it would somehow be more appropriate if you used the free A-module rather than the free abelian group, so as to make an A-module of A-modules, but it doesn't seem like you'd actually gain any additional information that way, and the fact that that isn't what's done suggests that my suspicion is right.
-Harry
no subject
Date: 2009-02-21 04:56 pm (UTC)Also, the "non-hint" might actually be one, if there is an obvious non-answer that doesn't require much work.
no subject
Date: 2009-02-21 04:57 pm (UTC)no subject
Date: 2009-02-21 07:04 pm (UTC)no subject
Date: 2009-02-25 05:32 am (UTC)(Josh Z)
no subject
Date: 2009-02-25 05:57 am (UTC)no subject
Date: 2009-02-22 10:35 pm (UTC)