Timeline for Disjoint images of polynomials
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Nov 15, 2014 at 0:11 | comment | added | Bjorn Poonen | If $f$ and $g$ are of degree $5$, with leading coefficients $\alpha$ and $\beta$, then $f(x)=g(y)$ and the line $\alpha^{1/5} x = \beta^{1/5} y$ intersect in either the whole line or in four points with multiplicity (the fifth being at infinity in the projective closure); in either case, one finds solutions in $\mathbb{Q}^{\operatorname{sol}}$, by solving a quartic equation if necessary. | |
| Nov 14, 2014 at 7:05 | comment | added | Pablo | @BjornPoonen another way to see this is that you have used the fact that $(\mathbb{Q}^{\text{ab}})^{\times}$ is not divisible, which obviously does not generalize to the field of radicals. Also note that curves which arise in this manner (i.e $f(x) - g(y)$) are of very specific type (no?) so it well may be that for these curves the existence of a rational point can be proved. I am even ready to restrict my attention to $f,g$ of degree $5$. What do you think? | |
| Nov 14, 2014 at 3:02 | comment | added | Bjorn Poonen | @Pablo: I don't know how to do it for $\mathbb{Q}^{\operatorname{sol}}$ instead of $\mathbb{Q}^{\operatorname{ab}}$. The trick behind my example above was to choose $f$ and $g$ so that $f(x)=g(y)$ has no geometrically irreducible component over $\mathbb{Q}^{\operatorname{ab}}$, but that idea doesn't seem to be implementable over $\mathbb{Q}^{\operatorname{sol}}$. Whether every geometrically irreducible curve over $\mathbb{Q}^{\operatorname{sol}}$ has a rational point is an open question, as far as I know. | |
| Nov 14, 2014 at 2:49 | history | edited | Bjorn Poonen | CC BY-SA 3.0 |
added 1 character in body
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| Nov 14, 2014 at 2:48 | comment | added | Bjorn Poonen | @ACL: Oops, yes. I'll fix this. | |
| Nov 13, 2014 at 23:26 | comment | added | Pablo | And what if $\mathbb{Q}^{\text{sol}}$ is taken instead of $\mathbb{Q}^{\text{ab}}$? | |
| Nov 13, 2014 at 23:26 | vote | accept | Pablo | ||
| Nov 13, 2014 at 23:20 | vote | accept | Pablo | ||
| Nov 13, 2014 at 23:26 | |||||
| Nov 13, 2014 at 23:13 | comment | added | ACL | So doesn't it mean that the answer is yes? | |
| Nov 13, 2014 at 23:10 | comment | added | Joe Silverman | You beat me to it. I was about to post "A silly observation is you can take $f(x)=1$ and $g(x)=2$. So presumably you want $f$ and $g$ non-constant," and then follow that with the observation that the OP is looking at points on curves defined over $\mathbb{Q}^{\text{ab}}$, for which it shouldn't be hard to produce examples with no points (as you just did). | |
| Nov 13, 2014 at 23:08 | history | answered | Bjorn Poonen | CC BY-SA 3.0 |