Timeline for Closure of the locus of polynomials vanishing to a given order at two points
Current License: CC BY-SA 4.0
11 events
when toggle format | what | by | license | comment | |
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May 31, 2021 at 19:39 | comment | added | Joel Kamnitzer | Ok, I agree, this makes sense, thanks! | |
May 31, 2021 at 19:39 | vote | accept | Joel Kamnitzer | ||
May 27, 2021 at 19:45 | comment | added | R. van Dobben de Bruyn | I actually confused myself about this for a bit. But the point is that $k[s]$-points of $P_{n,k[s]}$ are sections of $P_{n,k[s]} \to \mathbf A^1$, not just points of $P_{n,k[s]}$ in any classical way. Unlike geometry over algebraically closed fields, in this relative setting you don't just want to look at sections over $R$, but over any $R \to S$ (the functor of points point of view). Maybe my main observation is that $P_n \times \mathbf A^1$ really wants to be a relative gadget over $\mathbf A^1$. | |
May 27, 2021 at 18:24 | comment | added | Joel Kamnitzer | Thanks for this proof! I followed the individual steps, but I am somehow still confused. It seems that you have replaced my space $ P_n \times \mathbb A^1 $, whose $ k $ pts were pairs of a polynomial and a number, with $ P_{n, k[s]} $ whose $ k[s] $ pts are polynomials with coefficients in $ k[s] $. I understand that $ P_n \times_{Spec k} Spec R = P_{n,R}$, but somehow I am still confused. | |
May 27, 2021 at 15:55 | comment | added | R. van Dobben de Bruyn | @NeilStrickland thanks, fixed! | |
May 27, 2021 at 15:55 | history | edited | R. van Dobben de Bruyn | CC BY-SA 4.0 |
Correction
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May 27, 2021 at 8:47 | comment | added | Neil Strickland | I don't think that $\phi$ is a closed immersion if $R=\mathbb{Z}$ and $g=2$. If $g$ is monic then $\phi$ is a closed immersion. | |
May 27, 2021 at 6:43 | history | edited | R. van Dobben de Bruyn | CC BY-SA 4.0 |
Included definition
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May 27, 2021 at 6:42 | history | undeleted | R. van Dobben de Bruyn | ||
May 27, 2021 at 5:48 | history | deleted | R. van Dobben de Bruyn | via Vote | |
May 27, 2021 at 4:24 | history | answered | R. van Dobben de Bruyn | CC BY-SA 4.0 |