Timeline for Questions about the algebraic space $\mathbb{A}^1/\mathbb{Z}$
Current License: CC BY-SA 3.0
21 events
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| Nov 10, 2017 at 11:47 | comment | added | Bernd | @nfdc23 But there is "a" relation, right? I mean, the analytification of the stack $[\mathbb{A}^1/\mathbb{Z}]$ is isomorphic to $\mathbb{C}^\ast$, no? This kind of perplexes me, because it means that quasi-compactness of the diagonal of an algebraic space can't be "seen" in its analytification... | |
| Nov 5, 2017 at 21:36 | comment | added | nfdc23 | @KevinCasto: There is no relationship since the quotient you're thinking of is of very complex-analytic nature and has no meaning in terms of algebraic geometry. (Note that the way the "Tate curve" uniformization is linked up with algebraic geometry goes via formal schemes and algebraization thereof, so in effect an "analytic" construction with no meaning if working only over a field rather than over a ring with some completeness properties for a suitable ideal-adic topology.) | |
| Nov 5, 2017 at 18:15 | comment | added | Kevin Casto | By the way, is there any relationship between $X$ and ${\mathbb{G}_m}_{/\mathbb{C}}$, given that the quotient 'wants to be' $\mathbb{C}^*$? | |
| Nov 5, 2017 at 13:51 | comment | added | nfdc23 | @JasonStarr: I agree about clarifying seemingly contradictory hypotheses in the literature. One reason it seems better not to impose quasi-separatedness in the definition of "locally finite type" (admittedly a non-issue when one only works with quasi-separated algebraic spaces) is that the notion is then etale-local on the source as it is for schemes, whereas otherwise it isn't since quasi-separatedness isn't etale-local on the source (which seems to lie at the core of what makes the quotients in the question posed so disorienting at first sight). | |
| Nov 5, 2017 at 13:44 | comment | added | nfdc23 | @JasonStarr: of course when working with qcqs algebraic spaces (as most people do, at least implicitly, when using algebraic spaces) we don't need to explicitly mention the automatic quasi-separatedness of maps and so one can say "every finite type map [between qcqs algebraic spaces!] factors as a closed immersion followed by a map of finite presentation" (Theorem 3.2.1 in arxiv.org/pdf/0910.5008.pdf), which doesn't hold for finite type maps that aren't q-s (though one could ask if it holds for q-s finite type maps to an arbitrary q-c alg. space; I doubt it's true but don't know). | |
| Nov 5, 2017 at 13:27 | comment | added | Jason Starr | @nfdc23. I do not hold any dogmatic views about what one "should" do. I try to help confused students navigate contradictory hypotheses in different parts of the literature. | |
| Nov 5, 2017 at 13:25 | comment | added | nfdc23 | @JasonStarr: the notions of "finite type" and "locally finite type" shouldn't require quasi-separatedness, but rather one should instead just be more attentive than is necessary for schemes to the distinction between "finite type" and "finite presentation" (and even the meaning of "noetherian") when considering general algebraic spaces (which need not be Zariski-locally quasi-separated). See Tag 03E9 in the Stacks Project for a nice discussion of this. Since even with schemes we don't "spread out" in the loc. finite type context, focusing on "finite type" vs. "finite presentation" seems best. | |
| Nov 5, 2017 at 13:11 | comment | added | Ariyan Javanpeykar | @Bernd Let $E$ be an elliptic curve, and let $p$ be a point of infinite order. Let $G$ be the subgroup generated by $p$. This acts freely on $E$. Thus, the sheaf $E/G$ is an algebraic space, and the morphism $E\to E/G$ is a $\mathbb{Z}$-torsor. This is another example. [Also, I don't think there are any "other" real examples, because the automorphism group of a curve is finite, in general.] | |
| Nov 5, 2017 at 13:08 | history | edited | Bernd | CC BY-SA 3.0 |
There is only one question left, so I edited the question
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| Nov 5, 2017 at 13:05 | comment | added | Bernd | @JasonStarr Ok. Got it. With the conventions of the Stacks Project, the algebraic space $\mathbb{A}^1_{\mathbb{C}}/\mathbb{Z}$ is of finite type over $\mathbb{C}$, but it is not of finite presentation; see stacks.math.columbia.edu/tag/01TP. So, the algebraic stack $[\mathbb{A}^1_{\mathbb{Z}}/\mathbb{Z}]$ is of finite type over $\mathbb{Z}$ but not of finite presentation. | |
| Nov 5, 2017 at 12:49 | comment | added | Jason Starr | @Bernd. I suspect the issue is that many authors (myself included) would not speak of a morphism being "finite type" or even "locally finite type" unless it is quasi-separated. Certainly Proposition B.3 seems to require this. | |
| Nov 5, 2017 at 11:04 | comment | added | Bernd | @LaurentMoret-Bailly Yes, I agree. But doesn't that contradict Prop. B.3 in arxiv.org/pdf/0904.0227.pdf ? It says that representability of the morphism $X\to Y$ does spread out (with $Y=S$) in the notation of loc. cit. | |
| Nov 5, 2017 at 10:40 | comment | added | Laurent Moret-Bailly | @Bernd: Clearly representability doesn't "spread out" here since you would have to remove all primes from $\mathrm{Spec}\,(\mathbb{Z})$. | |
| Nov 5, 2017 at 10:26 | comment | added | Bernd | @LaurentMoret-Bailly Yes, you are right. I thought this would contradict "spreading out". The stack $[\mathbb{A}^1_{\mathbb{Z}}/\mathbb{Z}]$ is a finite type algebraic stack which is generically an algebraic space. This ought to imply that it is an algebraic space over a dense open of Spec $\mathbb{Z}$. I guess "being an algebraic space" spreads out only for finite type algebraic stacks over $\mathbb{Z}$ with a finite type diagonal. Is that right? [Here I'm confused by Prop. B.3 in arxiv.org/pdf/0904.0227.pdf which says that "representability" spreads out.] | |
| Nov 5, 2017 at 8:35 | comment | added | Laurent Moret-Bailly | Question 2: no, because $\mathbb{Z}$ doesn't act freely on $\mathbb{A}^1_\mathbb{Z}$. A prime $p$ acts trivially on the subscheme $\mathbb{A}^1_{\mathbb{F}_p}$. | |
| Nov 4, 2017 at 8:29 | history | edited | Bernd | CC BY-SA 3.0 |
I removed one question that was answered in the comments, and removed another question.
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| Nov 4, 2017 at 8:27 | comment | added | Bernd | @AlexM. Ok. I will change the format of the question. Thank you. | |
| Nov 3, 2017 at 21:31 | comment | added | Jason Starr | The morphism from $X$ to $X\times X$ is not a finite type morphism for precisely the same reason that the morphism $\left(\sqcup_{n\in \mathbb{Z}} \text{Spec}\ k \right) \to \mathbb{A}^1_k$ is not finite type: the inverse images of quasi-compact opens in the target are not quasi-compact in the domain. | |
| Nov 3, 2017 at 19:51 | comment | added | Alex M. | The general rule here is: one question per post. I'm not flagging this for closure, but I'm writing this in case your question gets closed, and you might not know why. | |
| Nov 3, 2017 at 19:27 | review | First posts | |||
| Nov 3, 2017 at 19:51 | |||||
| Nov 3, 2017 at 19:24 | history | asked | Bernd | CC BY-SA 3.0 |