Timeline for Counting algebraic points of bounded height
Current License: CC BY-SA 3.0
20 events
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| Nov 14, 2017 at 10:06 | comment | added | var | Actually I want this upper bound is uniform for all $X$ with fixed dimension and degree. | |
| Nov 14, 2017 at 9:59 | comment | added | var | Oh, I don't think your argument about "constant $C$ doesn't depend on $X$ " is complete. Could you offer me more details about it? | |
| Nov 13, 2017 at 0:58 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
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| Nov 13, 2017 at 0:52 | vote | accept | var | ||
| Nov 14, 2017 at 10:56 | |||||
| Nov 12, 2017 at 23:57 | history | undeleted | Vesselin Dimitrov | ||
| Nov 12, 2017 at 23:56 | history | deleted | Vesselin Dimitrov | via Vote | |
| Nov 12, 2017 at 23:54 | history | undeleted | Vesselin Dimitrov | ||
| Nov 12, 2017 at 23:51 | history | deleted | Vesselin Dimitrov | via Vote | |
| Nov 12, 2017 at 23:50 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
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| Nov 12, 2017 at 21:36 | comment | added | Vesselin Dimitrov | @var: That is actually obvious: the bijection is just $Q \leftrightarrow g'^{-1} \cdot Q$. | |
| Nov 12, 2017 at 17:13 | comment | added | var | Do you mean that the last two sets are isomorphism due to "Masser-Vaaler generalization of Northcott's asymptotic count"? Could you point out the refered theorem explicitly? Thank you. | |
| Nov 12, 2017 at 9:40 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
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| Nov 12, 2017 at 9:34 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
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| Nov 12, 2017 at 9:28 | comment | added | Vesselin Dimitrov | @var: I edited the above. In my comment, I was trying to make the point that it is not really a problem that $Z$ can't be taken independently of $X$, due to the asymptotic lemma I formulated. | |
| Nov 12, 2017 at 9:26 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
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| Nov 12, 2017 at 8:32 | comment | added | var | Yes, the constant $C$ depends only on the center $Z$ of the projection, and the choosing of $Z$ is essentially the choosing of a rational point in $Gr(d+1, n)$ outside the Chow variety (or called Chow form) of $X$. For a fixed degree $X$, the height of such a rational point is at lest $\delta+1$, and this estimate is almost optimal. So I don't think the linear projection method works for this problem, since by this argument, the height of $Z$ is larger if $\delta$ is larger. For your add information, I don't know whether it works, could you provide me more information? Thank you. | |
| Nov 12, 2017 at 4:34 | comment | added | Vesselin Dimitrov | Sorry, I wasn't very clear. The constant $C$ depends only on the center $Z$ of the linear projection: a linear subspace of dimension $n-d-1$ disjoint from $X$. In general, there needn't be such a fixed $Z$ if $\delta$ is unbounded. But choose any $Z$ (depending on $X$) and take $g$ a linear automorphism (over $K$) moving $Z$ to a fixed $\mathbb{P}_K^{n-d-1}$. Then use that $\# \{ H_K(\xi) \leq B \}$ and $\# \{ H_K(g \cdot \xi) \leq B \}$ have the same asymptotic count (where $\xi$ ranges over the points of $\mathbb{P}_K^d \subset \mathbb{P}_K^n$ with $[K(\xi):K] \leq D$). Does it make sense? | |
| Nov 12, 2017 at 3:24 | comment | added | var | Do you know any method which can bound the constant as in the requiremen? Thank you very much. | |
| Nov 12, 2017 at 3:15 | comment | added | var | I don't think your argument is complete. The key problem is the constant $C$ in your answer. I can only bound $C$ as $O(\delta^{n-d+1})$, which is far from the requirement. For the rational points case, I consider the corresponding affine cone, and count integral points by induction. For the hypersurfaces' case, I consider the polynomial directly. | |
| Nov 11, 2017 at 20:41 | history | answered | Vesselin Dimitrov | CC BY-SA 3.0 |