Timeline for Why is, for a group scheme of finite type, "smooth" (resp. irreducible) equivalent to "geometrically reduced" (resp. geometrically irreducible)?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 1, 2019 at 7:24 | comment | added | Alex Youcis | Just to comment to @Karl_Peter if $X/k$ is any finite type scheme and $X(k)\ne\varnothing$ then $X$ is connected if and only if it's geometrically connected (e.g. see this: math.stackexchange.com/a/3168956/16497) | |
| Mar 9, 2019 at 19:20 | comment | added | anon | @Karl_Peter See Section b of Chapter 1 of Milne: Algebraic Groups, The theory of group schemes of finite type over a field, Cambridge UP, 2017. | |
| Mar 9, 2019 at 14:19 | comment | added | user267839 | @PiotrAchinger: Sorry again for the annoying asking again but I'm not sure if I understood your argument correctly about how concretely the action on $|G|$ works. As you explained there is given a canonical action of ${\rm Gal}(\bar K/K)$ on $\bar{G}(\bar{K})$ by conjugating. According to your last comment that suffice. Do you mean this in the sense that this action on $\bar{G}(\bar{K})$ automatically extends/induces to an action on $|G|$? Since $\bar{G}(\bar{K})$ is dense in in $|G|$? Or did you mean that in another way? | |
| Mar 9, 2019 at 14:06 | comment | added | user267839 | @anon: do you know a recommendable reference where the case with positive characteristic is explained? | |
| Mar 9, 2019 at 7:01 | comment | added | anon | The argument in the answer that the identity component of $G$ over the algebraic closure of $K$ is defined over $K$ applies only in characteristic zero. Otherwise the proof is more difficult. | |
| Mar 7, 2019 at 23:36 | comment | added | user267839 | But what is concretely the "canonical" action of ${\rm Gal}(\bar K/K)$ on the underlying topological space $|\bar G|$? | |
| Mar 7, 2019 at 22:56 | comment | added | Piotr Achinger | @Karl_Peter I think the latter action induces the former on $\bar K$-points. Here it is enough to say that the profinite group ${\rm Gal}(\bar K/K)$ acts continuously on the underlying topological space $|\bar G|$ of $\bar G$ (or just the subspace of closed points $\bar G(\bar K)$) and the quotient space is identified with $|G|$. | |
| Mar 7, 2019 at 21:06 | comment | added | user267839 | Thank you for your answer. One penible question: When you talk about the ${\rm Gal}(\bar K/K)$-action on $\bar G = G \otimes \bar{K}$ do you implicitely mean the action on "points" $\bar{G}(\bar{K})= Hom(\bar{K}, \bar{G})$ via composing $\phi \mapsto \phi \circ g$ for a $g \in {\rm Gal}(\bar K/K)$ or the action via "base change" namely that $g$ induces automorphism on $\bar{G}$ via $id \otimes g$? | |
| Mar 7, 2019 at 19:40 | vote | accept | user267839 | ||
| Mar 7, 2019 at 19:18 | history | answered | Piotr Achinger | CC BY-SA 4.0 |