Timeline for Variety acquiring rational point over any quadratic extension
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Oct 17, 2019 at 11:36 | comment | added | Emil Jeřábek | That probably works. In fact, I think that the simpler polynomial $P^2+(x_{n+1}^2-a)^2+y^2$ is also irreducible. | |
| Oct 17, 2019 at 10:33 | comment | added | joro | @LaurentMoret-Bailly Is irreducibility really an issue? Add nonnegative summand on new variables with a rational point over Q. Isn't $P^2+(x^2-a)^2+(y(y-1))^4$ absolutely irreducible? | |
| Oct 17, 2019 at 10:32 | comment | added | joro | @EmilJeřábek Is irreducibility really an issue? Add nonnegative summand on new variables with a rational point over Q. Isn't $P^2+(x^2-a)^2+(y(y-1))^4$ absolutely irreducible? | |
| Oct 17, 2019 at 10:29 | comment | added | joro | @EmilJeřábek Many thanks for editing the question! Also many thanks for your numerous "bug reports" in stuff I have written! (some of your bug reports were unfixable) | |
| Oct 17, 2019 at 8:21 | comment | added | Laurent Moret-Bailly | @EmilJeřábek Indeed, in these questions "k-variety" usually just means "k-scheme of finite type". | |
| Oct 17, 2019 at 8:13 | comment | added | Emil Jeřábek | One thing is bothering me, though. In my book, varieties are required to be absolutely irreducible. The algebraic set in this answer is not (the polynomial factors over the algebraic closure as $(P+i(x_{n+1}^2-a))(P-i(x_{n+1}^2-a)))$. | |
| Oct 17, 2019 at 7:46 | history | edited | Emil Jeřábek | CC BY-SA 4.0 |
Actually, it's better to link directly to the specific answer, in case more will come in the future. Also, give credit where credit is due.
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| Oct 17, 2019 at 7:01 | history | edited | Emil Jeřábek | CC BY-SA 4.0 |
fix one more integer, and fix link so that it does not go to an irrelevant comment
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| Oct 17, 2019 at 6:55 | history | edited | YCor | CC BY-SA 4.0 |
fixed irrelevant issue by removing reference to integrality
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| Oct 16, 2019 at 19:54 | comment | added | Emil Jeřábek | By the way, the result of Dittmann mentioned in the linked answer suggests that the following generalization is also true: for any $n$, there exists a variety over $\mathbb Q$ that has no rational points over $\mathbb Q$, but it has rational points over every proper extension of $\mathbb Q$ of degree at most $n$. | |
| Oct 16, 2019 at 19:45 | comment | added | Emil Jeřábek | @LaurentMoret-Bailly Yes, I know, YCor already suggested it above. I’m just pointing out that the answer needs corrections. | |
| Oct 16, 2019 at 17:06 | comment | added | joro | @EmilJeřábek Thanks, I need to fix this (probably by the previous comment). | |
| Oct 16, 2019 at 16:46 | comment | added | Laurent Moret-Bailly | @EmilJeřábek: Integers are irrelevant here. Just replace "integer non-squares" by "rational non-squares" in the answer and everything is OK. | |
| Oct 16, 2019 at 15:02 | comment | added | Emil Jeřábek | The linked answer does not give a polynomial that has a rational solution iff $a$ is integer non-square. It gives a polynomial that has a rational solution iff $a$ is not an integer square. That is, the polynomial always has a rational solution when $a$ is a non-integer rational. Whether the set of integer non-squares is diophantine over the rationals is a hard open problem, as it easily implies that the set of integers itself is diophantine over the rationals. | |
| Oct 16, 2019 at 10:54 | comment | added | joro | @Ycor Thanks. I edited adding equality. | |
| Oct 16, 2019 at 10:52 | history | edited | joro | CC BY-SA 4.0 |
added equality
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| Oct 16, 2019 at 10:14 | comment | added | YCor | I guess you mean the variety "$P^2+(x_{n+1}^2-a)^2=0$", viewed as hypersurface of the $(n+2)$-dimensional affine space with coordinates $(a,x_1,\dots,x_{n+1})$. Also it seems you want, for $a\in\mathbf{Q}$, that $P(a,x_1,\dots,x_n)=0$ have a rational solution iff $a$ is not a rational square (which is covered by the linked post, actually easier than the half-integral version). | |
| Oct 16, 2019 at 9:44 | history | edited | joro | CC BY-SA 4.0 |
added 19 characters in body
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| Oct 16, 2019 at 9:30 | history | answered | joro | CC BY-SA 4.0 |