Timeline for Classical $k$-prevarieties vs reduced $k$-schemes of finite type. What happens when $k$ is not algebraically closed?
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24 events
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| May 5 at 17:07 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 6 at 17:00 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Sep 8, 2023 at 14:06 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Aug 9, 2023 at 12:24 | answer | added | Elías Guisado Villalgordo | timeline score: 1 | |
| Aug 9, 2023 at 12:06 | history | edited | Elías Guisado Villalgordo | CC BY-SA 4.0 |
Added further facts
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| Aug 9, 2023 at 11:59 | history | edited | Elías Guisado Villalgordo | CC BY-SA 4.0 |
Added links to MSE
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| Jul 12, 2023 at 16:42 | comment | added | Giulio Bresciani | The action of the Galois group is given by simply composing a point $\operatorname{Spec}\bar{k} \to X$ with a Galois automorphism $\operatorname{Spec}\bar{k} \to \operatorname{Spec}\bar{k}$. Unfortunately, I do not have references for this approach. However, here is an instructive exercise: show that a point $x\in X(\bar{k})$ comes from a $k$-rational point if and only if it is Galois invariant. | |
| Jul 12, 2023 at 16:39 | comment | added | Giulio Bresciani | Learning scheme theory for the first time can be grueling indeed; if you want some help you can find my email on my website. One main motivation for studying schemes is in fact doing algebraic geometry over non-algebraically closed fields, since as you have seen the classical formalism just doesn't work in this case. A good thing to imagine when hearing "let $X$ be a scheme" is what RvDB was suggesting if $k$ is perfect: you think of $X(\bar{k})$, which is a classical variety over $\bar{k}$, together with an action of the Galois group. | |
| Jul 12, 2023 at 9:33 | comment | added | Elías Guisado Villalgordo | @GiulioBresciani My reason is: trying to find motivation to study schemes, or at least understanding what's different with the more naive-geometric case of class. varieties. After one year and a half of studying modern algebraic geometry (that felt quite grueling),I still don't know what I am supposed to imagine when hearing "let $X$ be a scheme." I guess motivations and intuitions have already been explained in literature such as Eisenbud-Harris, The Geometry of Schemes, and maybe other places. However, I haven't had the time yet to study from any of these sources (I eventually plan to do). | |
| Jul 11, 2023 at 18:15 | comment | added | Giulio Bresciani | May I ask why are you interested in this? If k is algebraically closed, classical k-varieties work smoothly and are easier than schemes. If k is not algebraically closed, trying to use them feels like shooting yourself in the foot. You are going to end up with extremely limiting assumptions on k and V, and still the theory will not work smoothly. So why? | |
| Jul 11, 2023 at 15:26 | comment | added | Elías Guisado Villalgordo | @GiulioBresciani Okay, yeah, I have just realized that asking for density of $X(k)$ makes this work over arbitrary $k$. This comes with the somewhat ugly fact that we rule out $\operatorname{Spec}(k[x_1,\dots,x_n])$ when $k$ is finite (as I explain on the last paragraph of the post). So I guess it might feel a little restrictive? | |
| Jul 11, 2023 at 13:32 | comment | added | Giulio Bresciani | Here is (probably) the missing point of the puzzle. Assume $X$ affine, let $F$ be the sheaf of regular functions on $X(k)$. The sheaf homomorphism $\mathcal{O}_{X}|_{X(k)}\to F$ is surjective, the kernel are the sections vanishing on each rational point. In particular, if $X$ is reduced and $X(k)$ is dense, this is an isomorphism. This allows you to glue the sheaves of regular functions. To prove the surjectivity: locally around a point $x\in X(k)\subset k^n$, a regular function is defined by a rational function, which in turn gives a local function on the scheme $X\subset\mathbb{A}^{n}$. | |
| S Jul 11, 2023 at 13:15 | history | suggested | ARA | CC BY-SA 4.0 |
fixed grammar
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| Jul 11, 2023 at 13:01 | review | Suggested edits | |||
| S Jul 11, 2023 at 13:15 | |||||
| Jul 11, 2023 at 12:51 | history | edited | Elías Guisado Villalgordo | CC BY-SA 4.0 |
Added better motivation
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| Jul 11, 2023 at 12:45 | comment | added | Elías Guisado Villalgordo | @GiulioBresciani The thing is that restricting $\mathcal{O}_X$ sometimes work, for instance, for $\mathbb{A}_k^n$ or $\mathbb{P}_k^n$. However, I'm unsure whether there is some condition we can impose on $X$ to guarantee this to work (in the particular case $k=\mathbb{R}$ or $k$ a real-closed field, this condition seems to be denseness of $X(k)$). On the other hand, if we cover $X$ by open affines $U_i$ and we endow the algebraic set $U_i(k)\subset k^{n_i}$ with the sheaf of regular functions, I don't see now how one could glue these to obtain a sheaf on $\bigcup_i U_i(k)=X(k)$. | |
| Jul 11, 2023 at 9:10 | comment | added | Giulio Bresciani | Once you have constructed the algebraic set $X(k) \subset k^n$, just take the sheaf of regular functions on $X(k)$. Restricting $\mathcal{O}_{X}$ doesn't work, as RvDB pointed out. | |
| Jul 9, 2023 at 9:59 | history | edited | Elías Guisado Villalgordo | CC BY-SA 4.0 |
Added second reference
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| Jul 8, 2023 at 10:29 | history | edited | Elías Guisado Villalgordo | CC BY-SA 4.0 |
grammatical error
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| Jul 5, 2023 at 10:11 | comment | added | Elías Guisado Villalgordo | @R.vanDobbendeBruyn Thank you for your comments 😀. Yes, what you propose in your first comment is a counterexample. I just edited the post adding a new hypothesis not present in your example (denseness of $X(k)$). Regarding your second comment: do you know any literature for the idea you explain? | |
| Jul 5, 2023 at 10:10 | history | edited | Elías Guisado Villalgordo | CC BY-SA 4.0 |
one question answered
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| Jul 4, 2023 at 20:43 | comment | added | R. van Dobben de Bruyn | As I'm sort of hinting in my comment, in the case of a perfect field $k$, there is a useful way to compare $k$-schemes to classical $\bar k$-varieties together with the natural $\operatorname{Gal}(\bar k/k)$-action on $\bar k^n$ given by $\sigma(x_1,\ldots,x_n) = (\sigma(x_1),\ldots,\sigma(x_n))$. | |
| Jul 4, 2023 at 19:57 | comment | added | R. van Dobben de Bruyn | I think that $I = (x^2+y^2) \subseteq \mathbf R[x,y]$ is a counterexample to what you're trying to do. It is a $1$-dimensional scheme consisting of two lines meeting at the origin, but the action of $\operatorname{Gal}(\mathbf C/\mathbf R)$ swaps the two lines so they are not defined over $\mathbf R$. But their point of intersection is, and $X(\mathbf R) = (0,0)$. If I understand your definitions correctly, then $\mathcal O_{X(k)}$ is the localisation of $\mathbf R[x,y]/(x^2+y^2)$ at the prime $(x,y)$, but $\mathcal O_{V(I)}$ is just $\mathbf R$. | |
| Jul 4, 2023 at 17:10 | history | asked | Elías Guisado Villalgordo | CC BY-SA 4.0 |