Timeline for Surjective morphisms dominated by some fpqc covering
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 12, 2020 at 12:41 | comment | added | Daebeom Choi | Thanks for all of you! Those comments really helped me. Now I feel comfortable with the `right’ statement. | |
| Apr 11, 2020 at 1:21 | comment | added | Will Sawin | Bhargav Bhatt told me that the statement the author may have meant that this is true if $G \to Q$ induces a surjection on sheaves in the fpqc topology - i.e. if any map from a scheme $Y$ to $Q$ arises fpqc-locally on $Y$ from a map to $G$. This implication is clear on taking $Y = Q$. The idea is that at this point in the argument the sheaf of $Q$ is already known as the quotient sheaf and so this condition is automatic. | |
| Apr 10, 2020 at 16:47 | comment | added | Laurent Moret-Bailly | The author refers to Raynaud's paper in the Driebergen conference on local fields. Raynaud's argument there makes clever use of the equivalence relation defined by the action; this essential part is missing in Shatz's paper. | |
| Apr 10, 2020 at 15:19 | comment | added | Daebeom Choi | However, considering your counterexample, I think it will be very hard to find a reasonable setup where this statement is true.. | |
| Apr 10, 2020 at 15:19 | comment | added | Daebeom Choi | The author used this statement to prove the following theorem: If $G$ is a finite flat group scheme over $S$ ad $H$ is a (finite flat) subgroup scheme of $G$, then $G/H$ exists and it is flat over $S$. The statement is used to prove $G/H$ is flat. We assume every scheme is Noetherian, but there are no further assumptions up to this point. Since I found another proof which does not use the statement, so I can just skip this proof. However, I am curious whether this statement is a totally wrong statement or right in some good circumstances. | |
| Apr 10, 2020 at 13:32 | comment | added | Will Sawin | Presumably there is no way to dominate $\mathbb A^1 \setminus \{0\} \cup \{0\} \to \mathbb A^1$ by an fpqc covering. This is maybe the nicest possible counterexample - everything smooth, finite type. Is there something in the context of the book that would rule out this kind of counterexample? | |
| Apr 10, 2020 at 10:15 | history | asked | Daebeom Choi | CC BY-SA 4.0 |