Timeline for Discriminant ideal in a member of Barsotti-Tate Group
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Jul 10, 2020 at 17:06 | comment | added | Anonymous | Yup, this is indeed what I had mind | |
| Jul 10, 2020 at 16:35 | comment | added | user267839 | As Remy pointed out with reference to Bhatt's notes it's indeed a fact that that a map on differentials of group schemes are completly determined on how the map act on the stalk $\Omega_{G,e}$. For sake of completness: was this exactly the argument you implicitly used in your proof in the comment to show that $f_{p^N}$ is unramified? | |
| Jul 10, 2020 at 16:34 | comment | added | user267839 | And this is an isomorphism (since $p^n \in k^*$) and therefore the dual map on differentials at stalk of $e$ $f_{p^N}^*\Omega_{G,e} \to \Omega_{G,e}$ is isomorphism as well, especially $f_{p^N}$ is unramified at $e$. The crucial point that confused me before was why in order to check that $f_{p^N}^*\Omega_{G} \to \Omega_{G}$ is unramified it was sufficient to check it on $e$-stalk $\Omega_{G,e}$. | |
| Jul 10, 2020 at 16:34 | comment | added | user267839 | @Anonymous: No problem, yes meanwhile I got an answer. But could you tell me: was this exactly the idea you had in mind? namely we want to deduce that the multiplication map $f_{p^n}: G \to G$ (which is in our case the zero map as we already know) is unramified. Therefore we have to analyse the induced map on differentials $f_{p^N}^*\Omega_{G} \to \Omega_{G}$. What we know that as you said the induced map $f_{p^N}^*: T_e G \to T_e G$ on the tangent space at the stalk on the neutral element $e \in G$ is liteally given by multiplication with $p^n$. | |
| Jun 30, 2020 at 16:52 | comment | added | Anonymous | Sorry, I missed this until now. Looks like you got an answer elsewhere | |
| Jun 24, 2020 at 0:18 | comment | added | user267839 | More precisely I understand that the induced map over tangent spaces over neutral element $e \in G$ is the multiplication by $p^N$-map. But over other points of $G$ I'm not sure why this is true. Could you give a reference for the proof you gave above. Meanwhile I found another proof of this claim but I would like to understand yours. | |
| Jun 24, 2020 at 0:07 | comment | added | user267839 | @Anonymous: Sorry for digging out this thread but there is still a detail in your proof of the claim that Any finite flat group scheme over $k$ of $p$-power order is \'etale if $p$ is invertible on the base. We By Deligne theorem the multiplication by $p^N$ on $G$ indeed kills $G$; in other words the map $p^N: G \to G$ factorize over $Spec(k)$. Then you claim that the map on tangent spaces of $G$ induced by $p^N$ is also given by multiplication by $p^N$? Question: why the induced maps on tangent spaces are given by multiplications by $p^N$? | |
| May 11, 2020 at 16:42 | comment | added | Anonymous | A finitely presented flat map is \'etale if it is fibrewise \'etale | |
| May 11, 2020 at 10:20 | comment | added | user267839 | yes, I see. Last remark: how do you perform the reduction of the problem from "over $R$" to "over a field"? | |
| May 11, 2020 at 4:34 | comment | added | Anonymous | Unramified finite type schemes over a field are \'etale and hence geometrically reduced. | |
| May 11, 2020 at 1:15 | comment | added | user267839 | The last part of your argument I not understand: Why the fact that multiplication by $p^N$ for $N$ big enough ($N$ bigger then the group order) acts as zero and induce simultaneously an unramified morphism, imply that $G$ is geometrically reduced? | |
| May 11, 2020 at 1:15 | comment | added | user267839 | Ok, so over a field the story follows from a decomposition into etale and connected part. Structure theorem tells about the connected part that it's of order coprime to $p$, so it's trivial by Lagrange. So over field it's fine. | |
| May 10, 2020 at 23:51 | comment | added | Anonymous | @MortyPB. The following is a standard fact (found in most introductory references on group schemes): Any finite flat group scheme of $p$-power order is \'etale if $p$ is invertible on the base. To prove this, reduce to checking this over a field, and then use that multiplication by $p^N$ is simultaneously $0$ for $N \gg 0$ (by assumption) and unramified (as the map on tangent spaces is given by $p^N$, which is invertible), so the group scheme is geometrically reduced (and thus \'etale). | |
| May 10, 2020 at 23:14 | comment | added | user267839 | In Tate's paper I mentioned above I found indeed that the discriminant ideal is generated by a power of $p$ (page 164, Prop. 2), but it's a non trivial result. SoI not see why the implication $p \in R^\times \Rightarrow G_n$ etale is "elementary". Could you sketch the proof or give a reference for the argument? | |
| May 10, 2020 at 23:10 | comment | added | user267839 | @DavidHansen: but why $p \in R^\times$ imply that $G_n$ is etale? | |
| May 10, 2020 at 23:02 | comment | added | David Hansen | If $p \in R^\times$ then the discriminant ideal is trivial because all $G_n$ are etale. | |
| May 10, 2020 at 22:47 | history | edited | user267839 | CC BY-SA 4.0 |
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| May 10, 2020 at 22:39 | history | edited | user267839 | CC BY-SA 4.0 |
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| May 10, 2020 at 22:12 | history | asked | user267839 | CC BY-SA 4.0 |