Timeline for Dual of $End_A(M)$
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 10, 2020 at 18:36 | vote | accept | CommunityBot | ||
| Apr 10, 2020 at 7:34 | answer | added | metalspringpro | timeline score: 2 | |
| Dec 26, 2019 at 15:51 | comment | added | Mohan | If $M$ is not reflexive, this is false. As an example, take $M=A\oplus I$ where $A=k[x,y]$, $I=(x,y)$. | |
| Dec 22, 2019 at 10:12 | comment | added | Andrea Marino | Eheheh as you guess, I found a flaw while I was writing the proof down. Hope I will manage to correct it! | |
| Dec 17, 2019 at 22:27 | comment | added | Kapil | THe previous comment by me is missing a term. The first term of the complex should include $\{g\in\mathrm{Hom}(F_1,F_1)| d\circ g=9\}$. So it does not lead to a proof. | |
| Dec 17, 2019 at 16:31 | comment | added | Kapil | Using a two-term free resolution $F_1\to F_0\to M$, we can write $\mathrm{End}_A(M)$ as the (middle) cohomology of a complex $\mathrm{Hom}(F_0,F_1)\to\mathrm{Hom}(F_0,F_0)\oplus\mathrm{Hom}(F_1,F_1)\to\mathrm{Hom}(F_1,F_0)$. This should lead to a proof since we know what the dual of this complex is. | |
| Dec 16, 2019 at 16:00 | comment | added | user149914 | oh that would be wonderful...i think we have to define some sort of Trace map and show that it is non-degenerate | |
| Dec 16, 2019 at 7:48 | comment | added | Andrea Marino | I think I found that generally finitely generated torsion free modules over a domain are self dual - which would imply the statement here. I'll post a proof later! | |
| Dec 15, 2019 at 16:28 | comment | added | user149914 | I see ..so you are saying that every sheaf which is not locally free sheaf should work as a counterexample. | |
| Dec 15, 2019 at 14:44 | comment | added | Andrea Marino | I found that if $M$ is dualisable, which turns to be equivalent to be finitely presented and projective, then you have that $Hom(M,N) \simeq M^* \otimes N$. this case, the dual of $End(M)\simeq M^* \otimes M $ is $M^{**} \otimes M \simeq M \otimes M^* \simeq End(M)$, still because of dualisability. In case of gorentein rings, in particular they are noetherian, so that if I remember correctly finitely generate implies finitely presented. So I think that to find a counterexample you should search for a fgen which is torsion free but not projective (not flat is enough). | |
| Dec 14, 2019 at 21:13 | comment | added | user149914 | yes i meant "$\cong$" | |
| Dec 14, 2019 at 21:12 | history | edited | user149914 | CC BY-SA 4.0 |
deleted 4 characters in body; edited title
|
| Dec 14, 2019 at 20:44 | history | edited | user149914 | CC BY-SA 4.0 |
added 5 characters in body
|
| Dec 14, 2019 at 20:30 | review | Close votes | |||
| Dec 19, 2019 at 3:05 | |||||
| Dec 14, 2019 at 20:16 | comment | added | Jeremy Rickard | You clearly don't mean "=", since the two sides of the equality are sets of functions with different domains and codomains. Could you clarify what you do mean? | |
| Dec 14, 2019 at 20:15 | comment | added | user149914 | If M is free then this is true ..so i was wondering under what condition this is true for other modules? | |
| Dec 14, 2019 at 20:11 | comment | added | Andrea Marino | What's the intuition behind this you suggest? What makes you suspect this? | |
| Dec 14, 2019 at 20:10 | review | First posts | |||
| Dec 14, 2019 at 20:31 | |||||
| Dec 14, 2019 at 20:05 | history | asked | user149914 | CC BY-SA 4.0 |