Timeline for Decide two indices of Ext functor
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Apr 7, 2015 at 13:37 | comment | added | Strongart | I see, if b≠t, the Ext functor with the index b is 0 by prop11.33, thanks. | |
| Apr 7, 2015 at 13:33 | vote | accept | Strongart | ||
| Apr 6, 2015 at 7:39 | comment | added | user 1 | no we (I, at least) cant get the isomorphism for a and b, a+b=i+t. see the 1st paragraph of proof. | |
| Apr 6, 2015 at 6:48 | comment | added | Strongart | If we set T=$Ext^{b}_{R}(S,\omega))$, 0≤b≤t+i, then do the same discussing, can we get the isomorphism for a and b, a+b=i+t? | |
| Apr 5, 2015 at 6:15 | comment | added | user 1 | set $T=Ext^t_R(S, ω)$. we want to have $Ext^i_{S}(l,T)$. for this we should write "an injective resolution of the S-module, $T$", apply $Hom(l,-)$ and calculate homology module at the i-th point. Now as you see in the answer (which is from book) "$Hom_R(S, I^•)$ is a finite injective resolution of the S-module T" | |
| Apr 5, 2015 at 5:43 | comment | added | Strongart | Yes, I know here t is special, but it seems that the step from the Hom to Ext maybe is a little jump. | |
| Apr 4, 2015 at 17:56 | history | edited | user 1 | CC BY-SA 3.0 |
deleted 3 characters in body
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| Apr 4, 2015 at 17:47 | history | answered | user 1 | CC BY-SA 3.0 |