Timeline for Cotangent spaces of finite flat group schemes in short exact sequences
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 28, 2021 at 1:48 | comment | added | Bjorn Poonen | When I wrote "that rule is not exact", I meant specifically that the functor sending $G$ to $\operatorname{Map}(G^*,\mathbf{G}_m)$ is not exact, so certainly the functor sending $G$ to the Mazur-Roberts sequence $0 \to G \to A \to B \to 0$ is not exact. | |
| Jul 28, 2021 at 1:37 | comment | added | Bjorn Poonen | I think that that rule is not exact. It sends $1 \to \mu_2 \to \mu_6 \to \mu_3 \to 1$ to $1 \to \mathbf{G}_m^2 \to \mathbf{G}_m^6 \to \mathbf{G}_m^3 \to 1$, and $2 + 3 \ne 6$. Anyway, the pushout argument is not hard: Choose $G' \hookrightarrow A'$ and $G \hookrightarrow \mathcal{A}$; then $0 \to G' \to G \to G/G' \to 0$ embeds in $0 \to A' \to (A' \times \mathcal{A})/G' \to \mathcal{A}/G' \to 0$ (with $G'$ in the quotient in the middle being the antidiagonal copy), and then one can take cokernels to get $0 \to B' \to B \to B'' \to 0$. | |
| Jul 27, 2021 at 21:56 | comment | added | Jay | The rule sending G to this particular s.e.s. [0 to G to A to B to 0] is concrete enough that it should be possible to verify directly that it is exact in G, no pushouts neded? The remaining argument on differentials then makes sense (to me). | |
| Jul 27, 2021 at 21:56 | comment | added | Jay | Thanks! To summarize what's behind that Mazur–Roberts reference, for the sake of anyone reading this post later on: Writing G* for the Cartier dual of G, one embeds G = Hom(G*,G_m) into A = Map(G*,G_m) (maps of S-schemes), and takes B to be the quotient. (The Map in question is representable and that A,B have the stated properties.) | |
| Jul 27, 2021 at 21:47 | vote | accept | Jay | ||
| Jul 24, 2021 at 5:16 | history | answered | Bjorn Poonen | CC BY-SA 4.0 |