Timeline for Are all exotic affine spaces count equivalent to affine space?
Current License: CC BY-SA 4.0
12 events
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| May 28, 2018 at 2:40 | history | edited | Sean Lawton | CC BY-SA 4.0 |
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| May 24, 2018 at 13:23 | vote | accept | Sean Lawton | ||
| May 24, 2018 at 13:22 | comment | added | Sean Lawton | @WillSawin Thanks for clarifying. Yes, I had in mind explicit varieties defined over $\mathbb{Q}$ and the counting polynomial holding for all but a finite number of primes. | |
| May 24, 2018 at 13:13 | comment | added | Will Sawin | @Qfwfq It's easy to see that there are. For instance once can take $\operatorname{Spec} \mathbb Z[x_1,\dots,x_n, 1/2]$, which has no $\mathbb F_2$-points, or, more interestingly, $\mathbb Z[x_1,x_2, p x_1 - x_2 (x_2-1) ]$ which has $2p^2$ $\mathbb F_p$-points. | |
| May 24, 2018 at 13:11 | comment | added | Will Sawin | @SeanLawton The issue is not with the model of affine space but the model of your affine variety. It all depends on the definition of "polynomial count", which you don't specify. The definition that is correct for most purposes is that the polynomial holds for all but finitely many primes. This is a perfectly well-defined concept for varieties over $\mathbb Q$. On the other hand, if you define "polynomial count" at every prime, then the answer is false, already in the case when $V_\mathbb C$ is algebraically isomorphic to $\mathbb A^n_{\mathbb C}$. | |
| May 24, 2018 at 13:10 | comment | added | Sean Lawton | @Qfwfq Yup, that's the model I am using for affine space. | |
| May 24, 2018 at 12:58 | comment | added | Qfwfq | Maybe R. van Dobben de Bruyn meant something in the sense of mathoverflow.net/questions/18747/… So the "model" of $\mathbb{A}^n_{\mathbb{C}}$ you're using to define what are its $\mathbb{F}_q$-points is the usual scheme $\mathbb{A}^n_{\mathbb{Z}}$, which is not a variety but okay... (Btw, I have no idea if there are other models of $\mathbb{A}^n_{\mathbb{C}}$ over $\mathbb{Z}$ that have a different number of $\mathbb{F}_q$-points) | |
| May 24, 2018 at 12:47 | comment | added | Sean Lawton | @R.vanDobbendeBruyn I thought the context made it pretty clear what I meant by "affine space". But I guess if you are confused, others might be too, so let me spell it out: I mean the usual one. | |
| May 24, 2018 at 12:17 | comment | added | R. van Dobben de Bruyn | Don't you have to choose a model for this question to make sense? I'm pretty sure there are models of $\mathbb A^n$ whose point counts over finite fields are not always $q^n$... | |
| May 24, 2018 at 12:16 | answer | added | Will Sawin | timeline score: 11 | |
| May 24, 2018 at 12:14 | history | edited | Sean Lawton | CC BY-SA 4.0 |
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| May 24, 2018 at 12:02 | history | asked | Sean Lawton | CC BY-SA 4.0 |