Timeline for How many curves in a family possess a rational point?
Current License: CC BY-SA 3.0
25 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 10, 2013 at 16:52 | comment | added | duje | For "explicit variant" of your question you may consult the paper by Andrew Granville "Rational and integral points on quadratic twists of a given hyperelliptic curve" IMRN (2007), in particular Corollary 1 (ii) and Conjecture 1 (ii). I am wondering what is known (or conjectured) concering the same question for degree 4 (the mentioned paper deals with integer points for degree 4, but not with rational points). | |
| Feb 9, 2013 at 12:56 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
deleted 513 characters in body
|
| Feb 9, 2013 at 10:53 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
added 320 characters in body
|
| Feb 9, 2013 at 8:55 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
deleted 219 characters in body
|
| Feb 9, 2013 at 8:44 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
added 4 characters in body
|
| Feb 9, 2013 at 8:39 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
added 385 characters in body; added 13 characters in body; added 3 characters in body
|
| Feb 8, 2013 at 9:40 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
deleted 45 characters in body; added 9 characters in body; Post Made Community Wiki
|
| Feb 8, 2013 at 9:14 | comment | added | Vesselin Dimitrov | Indeed... Sorry about this. I have edited again. | |
| Feb 8, 2013 at 9:14 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
added 47 characters in body
|
| Feb 8, 2013 at 7:15 | comment | added | user30035 | Although the meaning of this question has been clear from the start (and I haven't got a clue what the answer is), there are still issues with the precise formulation. If $B$ is a variety over $\mathbf{Q}$ then $B(\mathbf{Z})$ is empty. | |
| Feb 7, 2013 at 23:20 | comment | added | Felipe Voloch | "would it indeed be true that the K-rank of a random elliptic curve over ℚ is strictly >1/2?" Sorry, I don't know the answer to that and I don't see a reason for it to be true or false. I have somehow missed the earlier question. I'll have a look. | |
| Feb 7, 2013 at 23:14 | comment | added | Vesselin Dimitrov | @ Felipe: Thanks! Let me ask you the question which came up in the above link. Given an appropriate number field $K$ (e.g. a polyquadratic field), would it indeed be true that the $K$-rank of a random elliptic curve over $\mathbb{Q}$ is strictly $> 1/2$? | |
| Feb 7, 2013 at 23:08 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
added 297 characters in body
|
| Feb 7, 2013 at 22:56 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
added 37 characters in body
|
| Feb 7, 2013 at 22:55 | comment | added | Vesselin Dimitrov | It is better, in view of both comments, to say that $B$ is a quasi-projective variety over $\mathbb{Q}$, rather than a finite-type scheme over $\mathbb{Z}$. I will edit to adjust this. | |
| Feb 7, 2013 at 22:54 | comment | added | Felipe Voloch | That's expected but has not been proved. There has been some recent progress for the universal family of hyperelliptic curves by Bharghava, Gross, Poonen and Stoll. | |
| Feb 7, 2013 at 22:49 | comment | added | Vesselin Dimitrov | It means the following. Let $h$ be a height function on $B$ given by some projective embeding (assume, to fix ideas, that $B$ is quasi-projective). Then the proportion is the lim, or the lim sup, as $H \to \infty$ of the ratio $|b \in Z \mid h(b) \leq H| / |b \in B(\mathbb{Z}) \mid h(b) \leq H|$, where $Z \subset B(\mathbb{Z})$ is the set of $b$ for which $X_b$ does have a rational point. | |
| Feb 7, 2013 at 22:36 | comment | added | user30035 | What does "proportion" mean in this general context? | |
| Feb 7, 2013 at 22:01 | comment | added | Vesselin Dimitrov | Edited: what I really wanted to say is that $X_{\mathbb{Q}} \to B_{\mathbb{Q}}$ was a smooth proper family of curves. So I have replaced "smooth" by "generically smooth." | |
| Feb 7, 2013 at 21:57 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
added 12 characters in body; added 12 characters in body; deleted 42 characters in body
|
| Feb 7, 2013 at 21:56 | comment | added | R.P. | I guess the answer is trivially yes, because the hypotheses are never fulfilled. If $B(\mathbb{Z})$ contains an element $b$, then $X_b$ must be a proper smooth curve over $\operatorname{Spec}(\mathbb{Z})$ of genus $>1$, and such curves don't exist. | |
| Feb 7, 2013 at 21:00 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
deleted 12 characters in body
|
| Feb 7, 2013 at 19:44 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
deleted 7 characters in body
|
| Feb 7, 2013 at 19:37 | history | edited | Vesselin Dimitrov | CC BY-SA 3.0 |
added 7 characters in body; added 1 characters in body
|
| Feb 7, 2013 at 18:22 | history | asked | Vesselin Dimitrov | CC BY-SA 3.0 |