| 29 Jan 2024 |
 gm_z | i kept trying to see what factorisations of g in k(α)[X] had to look like | 02:57:59 |
| gm_z | so the full argument is, notice that k(β) ⊆ k(β,α) has min poly divide into g. since deg g | [k(α, β) : k(β)][k(β) : k], deg g | [k(α, β) : k(β)], so we have equality | 03:01:25 |
| gm_z | switch the α, β around whoops | 03:03:39 |
| gm_z | * switch all the α, β around whoops | 03:10:17 |
| 31 Jan 2024 |
|  beetapoornam joined the room. | 01:25:18 |
| 1 Feb 2024 |
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| 2 Feb 2024 |
|  Timo ⚡️ changed their display name from Timo ⚡️ to Timo ⚡️ (Sat&Sun at FOSDEM, DM me :). | 08:15:33 |
| 4 Feb 2024 |
| gm_z | if M is an R-module that is presented by finitely presented modules; i.e. if there exists an exact sequence
F \to G \to M \to 0
where F,G are finitely presented, must M be finitely presented as well?
| 04:52:48 |
| gm_z | 
Download image.png | 08:49:28 |
| gm_z | the highlighted part looks something like splitting lemma, and has to rely on the cartesian (pullback) property | 08:49:59 |
| gm_z | but i'm not quite seeing it | 08:50:03 |
| gm_z | tbh i'm not quite sure where homological algebra belongs lol | 08:50:33 |
| gm_z | * tbh idk where homological algebra belongs lol | 08:50:40 |
 matipau | In reply to @gm_z:matrix.org
the highlighted part looks something like splitting lemma, and has to rely on the cartesian (pullback) property It's just the universal property of w being the fiber product of k and h: You have the two morphisms j: Ker(k) -> x and 0: Ker(k) -> y, and they satisfy k ° j = h ° 0 (because both compositions are zero), so by the universal property there exists t: Ker(k) -> w with the claimed properties. | 12:06:00 |
| matipau | In reply to @gm_z:matrix.org
if M is an R-module that is presented by finitely presented modules; i.e. if there exists an exact sequence
F \to G \to M \to 0
where F,G are finitely presented, must M be finitely presented as well?
Yes, and it suffices for F to be finitely generated. Proof sketch: You have a surjection A^n -> G with finitely generated kernel. Now the kernel of the induced surjection A^n -> M can be generated by the kernel of A^n -> G together with lifts of generators for the kernel of G -> M, which can be taken to be the images under F -> G of the finitely many generators for F. | 12:28:54 |
| gm_z | In reply to @matipau:matrix.org
Yes, and it suffices for F to be finitely generated. Proof sketch: You have a surjection A^n -> G with finitely generated kernel. Now the kernel of the induced surjection A^n -> M can be generated by the kernel of A^n -> G together with lifts of generators for the kernel of G -> M, which can be taken to be the images under F -> G of the finitely many generators for F. (without explicitly writing in the kernels) is it something like this? | 23:15:26 |
| gm_z | 
Download image.png | 23:15:28 |
| gm_z | In reply to @matipau:matrix.org
It's just the universal property of w being the fiber product of k and h: You have the two morphisms j: Ker(k) -> x and 0: Ker(k) -> y, and they satisfy k ° j = h ° 0 (because both compositions are zero), so by the universal property there exists t: Ker(k) -> w with the claimed properties. ah right! thank you | 23:15:36 |
| gm_z | wlog F is free right | 23:16:27 |
| gm_z | well it doesn't matter | 23:43:34 |
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| 5 Feb 2024 |
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| 14 Feb 2024 |
| gm_z invited  @-kasper:matrix.org. | 08:02:26 |
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