Timeline for For finite group G and field k of char=p, if P,P′ are projective k[G]-modules with [P]=[P′], is it true that P=P′ ?
Current License: CC BY-SA 3.0
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| Dec 11, 2012 at 20:11 | comment | added | George | Thanks, Jim, Geoff. Unfortunately, I am not allowed to accept both answers (or I don't know how), so accepting from Jim. | |
| Dec 11, 2012 at 20:09 | vote | accept | George | ||
| Dec 11, 2012 at 18:28 | comment | added | Jim Humphreys | I think what I wrote is accurate: your original question gets answered only by Corollary 2 of Theorem 35 in 16.1. (Section 14 is more elementary and preliminary.) Section 16.1 is very concise, so you have to follow up with details of Serre's proofs later on. His formalism is efficient but it takes real work to identify the main points in the proofs and their logical dependence. You might find the concrete Brauer style in the older Curtis-Reiner book to be more attractive. But the theorems are not easy in any case. Good luck. | |
| Dec 11, 2012 at 4:00 | comment | added | George | Jim - yes, 16.1, my bad. I still don't get the arugment. Corr. 2 of Prop. 42 in 14.4 says: P=Q if [P]=[Q] in $$ P_A[G] $$, not in $$ R_A[G] $$. That is: if P and Q have same indecoposable projective factors, then they are isomorphic (which is already stated in the preceding Corr 1.). From this I can't deduct that if P and Q have same irreducible factors, then they are isomophic. As for using the cde-triangle: Serre uses 16.1 as a building step to prove cde properties in Ch. 17, so I am not sure that using it here wouldn't create a cycled reference. Thanks for the ref to Curtis-Reiner. | |
| Dec 10, 2012 at 13:18 | history | answered | Jim Humphreys | CC BY-SA 3.0 |