Timeline for Does a smooth relative curve $X/S$ embed into $\mathbb{P}^3_S$?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 5, 2022 at 18:20 | vote | accept | Somatic Custard | ||
| Sep 5, 2022 at 12:25 | answer | added | Daniel Loughran | timeline score: 4 | |
| Sep 5, 2022 at 11:09 | comment | added | Jason Starr | For every infinite field $k$ (not necessarily algebraically closed), every smooth projective $k$-variety $X$ of dimension $d$ embeds in $\mathbb{P}^n_k$ for every integer $n\geq 2d+1$. This follows from the usual proof (Whitney embedding theorem, roughly) together with Bertini's theorem for infinite fields. | |
| Sep 5, 2022 at 3:22 | comment | added | Somatic Custard | Ah ha, and here we have an clear and incontrovertible obstruction. Thank you, @NoamD.Elkies | |
| Sep 5, 2022 at 2:48 | comment | added | Noam D. Elkies | Consider the case that S is (the spectrum of) a finite field k. If a curve C/k embeds into P^n(k), then C has no more k-rational points than P^n(k). But there are smooth projective curves over k with arbitrarily many k-rational points. Thus for every n there are curves C/k that do not embed into P^n over k. | |
| Sep 5, 2022 at 2:41 | comment | added | Jef | Sorry, i was talking nonsense! The degree of the embedding in $\mathbb{P}^3$ could be large. Please ignore my first comment. | |
| Sep 5, 2022 at 2:25 | history | edited | Somatic Custard | CC BY-SA 4.0 |
honed question to account for comment
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| Sep 5, 2022 at 2:25 | comment | added | Somatic Custard | @Jef Thank you. I don't fully understand your comment, but I am happy to upgrade hypotheses so that $\pi$ has a section. Is there anything you can say about that situation? Also, what is the method for showing the result for question 2 (locally)? | |
| Sep 5, 2022 at 1:07 | comment | added | Jason Starr | @Jef Why does an embedding give an effective divisor of degree $3$? Quartic elliptic normal curves in $\mathbb{P}^3$ have an effective divisor of degree $3$, but they usually do not have an effective divisor of degree $3$. | |
| Sep 5, 2022 at 1:00 | comment | added | Jef | Your second question is true Zariski locally on $S$ (Dedekind or not), and in general $X$ can be embedded in a rank $2$ projective bundle over $S$. (Exactly by the argument you give) | |
| Sep 5, 2022 at 0:59 | comment | added | Jef | This is not even true for $S$ the spectrum of a non-algebraically closed field. If every curve $X/\mathbb{Q}$ would embed in $\mathbb{P}^3$, then every such curve would have a $\Q$-rational effective divisor of degree $3$. But not every curve has this property, e.g. take a genus $1$ curve that corresponds to an element of order $n$ in the Weil--Chatelet group of its Jacobian for $n>>0$ | |
| Sep 5, 2022 at 0:22 | history | asked | Somatic Custard | CC BY-SA 4.0 |