Timeline for Exact short sequences of vector spaces
Current License: CC BY-SA 2.5
8 events
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Dec 7, 2009 at 22:17 | comment | added | Kevin Buzzard | I only just noticed this comment (AFAIK I don't get notified if someone comments on a comment I made). What is "the Ext computation"? If C isn't free (i.e. we're in some zany model of ZF with vector spaces with no bases) then how are you going to prove Ext^1(C,-) is zero? i.e. "tell me what computation you're doing! Then I might be able to say where it fails." Hint: it will fail at the point where you use AC ;-) | |
Dec 2, 2009 at 23:31 | comment | added | Ilya Nikokoshev |
I have a vague idea, but still where does the Ext computation fail without AC?
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Dec 2, 2009 at 21:54 | comment | added | Kevin Buzzard | The obvious rule of thumb is: any construction where the standard argument involves Zorn or AC will almost certainly not work without AC (i.e. there will be models of ZF where it fails), becuase if it worked without Zorn then everyone would say "use Zorn, although there is a non-Zorn argument". Every vector space has a basis because of Zorn, and you've never heard anyone say "...and it can be done without Zorn", so you can be sure there will be some crazy model of ZF containing a vector space with no basis. | |
Dec 2, 2009 at 4:01 | comment | added | Ben Webster♦ | Fair enough. I'm blissfully unconcerned about axiom of choice issues. | |
Dec 1, 2009 at 19:15 | comment | added | Kevin Buzzard | A more explicit form of that comment is: Under AC, C is free so Hom(C,-) is exact. But if AC fails sufficiently badly, C may not have a basis, and now suddenly it's not so clear that Hom(C,-) is exact I guess. | |
Dec 1, 2009 at 18:07 | history | edited | Ben Webster♦ | CC BY-SA 2.5 |
added 3 characters in body
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Dec 1, 2009 at 18:03 | comment | added | Leonid Positselski | How do you know that it remains exact? | |
Dec 1, 2009 at 17:50 | history | answered | Ben Webster♦ | CC BY-SA 2.5 |