Timeline for What is the difference between Grothendieck groups K_0(X) vs K^0(X) on schemes?
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| when toggle format | what | by | license | comment | |
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| Jul 22, 2010 at 17:50 | vote | accept | Matt | ||
| Apr 22, 2010 at 4:26 | comment | added | Hailong Dao | @hilbertthm90: That's right, one takes alternating sum. We probably can't avoid regular, because finite projective dimension of the residue field characterize regular local rings. | |
| Apr 22, 2010 at 4:14 | comment | added | Matt | Oh, of course. The homological definition of regular of dimension $d$ is that $Ext^{d+1}$ vanishes which implies projective dimension less than or equal to $d+1$. I'm just learning this stuff, so I wanted a resolution of length 2 because from that it was clear where to send the sheaf. I'll need to think what the map is in this case. Probably just an alternating sum of the vector bundle associated to the free sheaf in each part of the resolution? I still feel like regular is strange. | |
| Apr 22, 2010 at 4:08 | comment | added | Hailong Dao | To finish and show that k can not have even have finite projective resolution, we can use the resolution in the last sentence to compute $Tor_i^A(k,k)$ (the resolution is periodic, so one only needs to do a few steps) and see that they are all non-zero. But if k has finite projective dimension one must have $Tor_i(k,k)=0$ for $i>>0$. | |
| Apr 22, 2010 at 1:57 | history | answered | David E Speyer | CC BY-SA 2.5 |