Timeline for Can a compact object be a nontrivial self-retract?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jan 10, 2020 at 22:01 | comment | added | Ivan Di Liberti | Thanks for the comment, let me correct it: mathoverflow.net/questions/97495/a-categorical-nakayama-lemma/… | |
| Jan 10, 2020 at 21:34 | comment | added | Mare | @IvanDiLiberti Your first link leads to Gjergjis MO-profile but not to an answer. | |
| Jan 10, 2020 at 21:27 | answer | added | Martin Brandenburg | timeline score: 2 | |
| Nov 2, 2019 at 11:30 | comment | added | Ivan Di Liberti | @TimCampion, now I was studying how much you can generalize Nakayama lemma and I found this beautiful answer (mathoverflow.net/users/2384/gjergji-zaimi). Unfortunately this theory hasn't been exported to the semiabelian context (yet). Yet, some approximations do exist (rd.springer.com/article/10.1007/s41980-018-0020-2). | |
| Nov 2, 2019 at 11:27 | comment | added | Ivan Di Liberti | @TimCampion, I had a completely different proof in mind, which uses that a surjective endomorphism of a fin gen module is an isomorphism, but yours seems legit to me! | |
| Nov 2, 2019 at 11:18 | comment | added | Tim Campion | @IvanDiLiberti Nice, thanks! Let's see if I can flesh this out... If $M \cong M \oplus N$ as $R$-modules, then for any maximal ideal $m$, we have $M_m \cong M_m \oplus N_m$. Since $R_m$ is local, Nakayama's lemma implies that a minimal generating set for $M_m$ over $R_m$ is given by the union of minimal generating sets for $M_m,N_m$. So we have two minimal generating sets for $M_m$ which again by Nakayama have the same cardinality, which is finite. So the minimal generating set for $N_m$ has cardinality zero, so $N_m = 0$. Since this holds for all maximal ideals $m$, we have $N=0$. | |
| Oct 31, 2019 at 23:24 | comment | added | Ivan Di Liberti | I think that at least in the case of modules over a (commutative) ring (with 1) the result follows from the Nakayama lemma. | |
| Oct 31, 2019 at 22:49 | comment | added | YCor | About the group case, a closely related question is Sequence of proper retracting homomorphisms between finitely presented groups, which would be answered if one has a f.p. group isomorphic to a proper retract of itself. | |
| Oct 31, 2019 at 22:32 | history | edited | Tim Campion | CC BY-SA 4.0 |
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| Oct 31, 2019 at 22:23 | history | asked | Tim Campion | CC BY-SA 4.0 |