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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
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
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