Timeline for Trace of finitely generated projective module
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Jan 31, 2014 at 11:03 | comment | added | user43326 | Trace is defined for endomorphisms, and we have the additivity, so Tr(hozizontal) + Tr(vertical for Q) = Tr(vertical ofr Q') + Tr(horizontal) | |
| Jan 31, 2014 at 11:00 | vote | accept | CommunityBot | ||
| Jan 31, 2014 at 11:00 | comment | added | Jeremy Rickard | By the way, $[A,A]$ is not naturally a left $A$-module, so $T(A)$ is just a $k$-module, not an $A$-module. | |
| Jan 31, 2014 at 10:58 | answer | added | Jeremy Rickard | timeline score: 10 | |
| Jan 31, 2014 at 10:57 | comment | added | user1688 | Ok, fine, but the horizontal maps are not isomorphisms. | |
| Jan 31, 2014 at 10:52 | comment | added | user43326 | For two horizontal arrows, both zeros are zero map from $Q$ to $Q'$, so they are same, aren't they? | |
| Jan 31, 2014 at 10:44 | comment | added | user1688 | There are two different zeros around. If you don't take the horizontal arrow to be the same, it won't prove the independence. | |
| Jan 31, 2014 at 10:41 | comment | added | user43326 | with $u\oplus 0$ for vertical arrows, $id_p\oplus 0$ for horizontal arrows. | |
| Jan 31, 2014 at 10:03 | comment | added | user1688 | I still don't see how to get this square? | |
| Jan 31, 2014 at 9:52 | comment | added | user43326 | Actually it is much simpler, we don't really have to bother with the projectivity of $Q$. You get $P\oplus Q$ on the left $P\oplus Q'$ on the right, vertical maps are $u\oplus 0_Q$ and $u\oplus 0_{Q'}$. I guess the additivity of the trace suffices to conclude that the trace doesn't depend on the choice of $Q$. | |
| Jan 31, 2014 at 9:46 | comment | added | user1688 | What does the square look like? | |
| Jan 31, 2014 at 9:34 | history | edited | Ricardo Andrade |
added top level tag
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| Jan 31, 2014 at 9:22 | comment | added | user43326 | Since the complement of $P$ in $A^n$ is also projective, one can construct endomorphisms of $A^n$ which fit in a commutative square involving $u\oplus 0_Q$ and $u\oplus 0_{Q ^\prime}$. Does this suffice to show that the trace is unique? | |
| Jan 31, 2014 at 9:14 | comment | added | user1688 | Oh sorry, should have been $End_A ( P )$. I changed it. | |
| Jan 31, 2014 at 9:13 | history | edited | user1688 | CC BY-SA 3.0 |
added 8 characters in body
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| Jan 31, 2014 at 9:11 | comment | added | user43326 | Is the source of $Tr _P$ $End _P$? | |
| Jan 31, 2014 at 8:51 | history | asked | user1688 | CC BY-SA 3.0 |