Timeline for Any finite flat commutative group scheme of $p$-power order is etale if $p$ is invertible on the base
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| Jun 25, 2020 at 18:55 | comment | added | user267839 | Yes, I think I got it: $f_{p^N}^*\Omega_{G,e} \to \Omega_{G,e}$ is simultaneously in isomorphism ($P^N$ invertible) and zero map (because induced by zero map $f_{p^N}$). With the argument you remarked from Bhatt's notes then same must be true for $f_{p^N}^*\Omega_{G} \to \Omega_{G}$ as well. Consequently $\Omega_{G}=0$. Fini. Thank you a lot! | |
| Jun 25, 2020 at 17:31 | comment | added | R. van Dobben de Bruyn | I think you can apply it to the morphism of group schemes $G \to 1$, which is both the map $f_{p^N}$ (when viewing $1 \subseteq G$ as a closed subscheme) and the structure morphisms $G \to \operatorname{Spec} k$. | |
| Jun 25, 2020 at 12:05 | comment | added | user267839 | I understand. And how now we deduce from this observation that $G$ is unramified. As you said $f_{p^N}^*\Omega_{G,e} \to \Omega_{G,e}$ determine $f_{p^N}^*\Omega_{G} \to \Omega_{G}$. The first one was an isomorphism ($p^N$ invertible) therefore $f_{p^N}^*\Omega_{G} \to \Omega_{G}$ as well. Why this imply $G$ unramified? | |
| Jun 25, 2020 at 10:12 | history | edited | YCor |
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| Jun 24, 2020 at 22:06 | comment | added | R. van Dobben de Bruyn | Ok, you're right, but differentials on a group scheme are determined by what happens at the identity. See for example Prop 3.3 of Bhatt's notes on abelian varieties. To be precise, $\Omega_{G/S} = \pi^*e^*\Omega_{G/S}$ for $\pi \colon G \to S$ a group scheme with unit $e \colon S \to G$. Similarly for a morphism of group schemes $\phi \colon G \to H$, the map $\phi^*\Omega_{H/S} \to \Omega_{G/S}$ is determined by $e^*\Omega_{H/S} \to e^*\Omega_{G/S}$ under pullback to $G$. This is standard and used all the time. | |
| Jun 24, 2020 at 12:54 | comment | added | user267839 | @R.vanDobbendeBruyn: This is the important point: the map between tangent spaces $T_e G$ over $e$ induced by $f_{p^N}= \mu^{p^N}$ is as you correctly remarked the multiplication by $p^N$. Since $p^n$ is inverible in $k$ this map $T_e G \to T_e G$ is an isomorphism and dualizing the (1) in stacks.math.columbia.edu/tag/00UT implies only that $f_{p^N}$ is unramified at the stalk $\Omega_e$; that is at the unit point $e$ (!) . But Anonymous clamed that this should imply that $f_{p^N}$ is unramified; that means is unramified at every point of $G$. | |
| Jun 24, 2020 at 12:43 | comment | added | user267839 | @LSpice: you linked the correct comment | |
| Jun 24, 2020 at 3:58 | comment | added | LSpice | I finished the title (which ended mid-sentence), and added a link to the comment I think that you meant. If I got the wrong comment, please fix it or let me know. | |
| Jun 24, 2020 at 3:57 | history | edited | LSpice | CC BY-SA 4.0 |
Finished title; link to comment; proofreading
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| Jun 24, 2020 at 3:39 | comment | added | R. van Dobben de Bruyn | But if you insist on this argument, I follows since the derivative of $\mu \colon G \times G \to G$ is addition $T_e G \times T_e G \to T_e G$ (use $\mu \circ (\operatorname{id} \times e) = \operatorname{id} = \mu \circ (e \times \operatorname{id})$ where $e \colon \operatorname{Spec} k \to G$ is the unit). II is a near immediate consequence of the definitions (see e.g. Tags 00UT and 00RS). | |
| Jun 24, 2020 at 3:29 | comment | added | R. van Dobben de Bruyn | Nor do you need to assume it's commutative. Another reference is Waterhouse Introduction to affine group schemes, §11.4, §14.4. | |
| Jun 24, 2020 at 2:57 | comment | added | anon | You don't need Deligne's theorem to prove this. For the standard proof, see, for example, 11.31 of Milne, Algebraic Groups, CUP, 2017. | |
| Jun 24, 2020 at 2:39 | history | edited | user267839 | CC BY-SA 4.0 |
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| Jun 24, 2020 at 2:03 | history | edited | user267839 | CC BY-SA 4.0 |
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| Jun 24, 2020 at 1:33 | history | edited | user267839 | CC BY-SA 4.0 |
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| Jun 24, 2020 at 1:27 | history | asked | user267839 | CC BY-SA 4.0 |