Skip to main content
19 events
when toggle format what by license comment
S Dec 30, 2020 at 22:00 history bounty ended CommunityBot
S Dec 30, 2020 at 22:00 history notice removed CommunityBot
Dec 28, 2020 at 21:55 history edited user267839 CC BY-SA 4.0
added 6 characters in body
Dec 22, 2020 at 21:04 history edited YCor CC BY-SA 4.0
formatting
Dec 22, 2020 at 20:39 history edited user267839 CC BY-SA 4.0
added 1 character in body
S Dec 22, 2020 at 20:15 history bounty started user267839
S Dec 22, 2020 at 20:15 history notice added user267839 Canonical answer required
Dec 22, 2020 at 20:14 history edited user267839 CC BY-SA 4.0
added 494 characters in body
S Feb 21, 2020 at 3:03 history bounty ended CommunityBot
S Feb 21, 2020 at 3:03 history notice removed CommunityBot
S Feb 13, 2020 at 1:34 history bounty started user267839
S Feb 13, 2020 at 1:34 history notice added user267839 Canonical answer required
Feb 13, 2020 at 1:34 comment added user267839 That is that $H^0$ and $\otimes$ commute. I'm facing two problems now: first one: to show that $m O_V$ is invertible $O_V$-module. (then it would flat and $H^0$ and $\otimes$ commute and we are done). The second problem is the initial induction step: how to show that $A/m \to H^0(Z,O_{Z})$ is surjection?
Feb 13, 2020 at 1:33 comment added user267839 I think I have (almost) the answer I was looking for: The problem is to show that $a$ surjective and by induction hypothesis $c_n: A/m^k \to H^0(Z,O_{kZ})$ is surjective for all $k < n+1$. Assume $A/m \to H^0(Z,O_{Z})$ is surjective (that's a seriuos problem: see below problem 2, but let at first assume we know it). We tensor it with $m^n$ and obtain the surjection $m^n/m^{n+1} \to m^n \otimes H^0(V, K_n)$. The goal would be to show $m^n \otimes H^0(Z,O_{Z})= H^0(V, K_n)$.
S Feb 10, 2020 at 6:41 history suggested AG learner CC BY-SA 4.0
typo correction
Feb 9, 2020 at 18:28 review Suggested edits
S Feb 10, 2020 at 6:41
Feb 9, 2020 at 3:00 comment added user267839 I think that it was just a typo in the paper. That is we consider indeed $A/{\mathfrak m}^n\to H^0(nZ,O_{nZ})$ induced canonically by $nZ \subset V \to \bar{V}=Spec(A)$ on the ring side ($A \to H^0(nZ,O_{nZ})$ factorizes through $A/{\mathfrak m}^n$)
Feb 8, 2020 at 11:24 comment added inkspot I don't know whether this will answer the question, but you write $A/{\mathfrak m}^n\to H^0(nZ,O_{nZ})$ while Artin's homomorphism goes in the opposite direction.
Feb 8, 2020 at 0:51 history asked user267839 CC BY-SA 4.0