Timeline for Does "all points rational" imply "constant" for this "cubic" curve over an arbitrary field?
Current License: CC BY-SA 3.0
27 events
| when toggle format | what | by | license | comment | |
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| Jan 30, 2016 at 12:57 | vote | accept | Max Horn | ||
| Jun 10, 2011 at 20:47 | comment | added | Max Horn | Thanks, will take a look. Since I am just interested in the case where $F$ has index $2$ in $K$, then unless the characteristic is 2 I am in the separable case anyway, I think. | |
| Jun 10, 2011 at 19:54 | comment | added | Kevin Buzzard | Max: I'm just writing this comment because I know the system will alert you to it -- I just wanted to point out that currently there are two answers for my more general question, and both of these should resolve your question too, at least in the separable case when the base is infinite. | |
| Jun 9, 2011 at 10:04 | history | edited | Max Horn | CC BY-SA 3.0 |
Mention |K|=9 is an exception, too; add missing )
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| Jun 9, 2011 at 10:01 | comment | added | Max Horn | This requires that $q:=\lvert F \rvert$ is at least 8. For the remaining cases $q=2,3,4,5,7$, one can use other tools. E.g. I just fired up GAP and verified the claim there. For $q=2,3$ there are counterexamples. Oops, I previously claimed that there is no counterexample for $q=3$, but obviously there is: In a field with $q^2$ elements, of course $x+x^q$ and $x^{q+1}$ are in the subfield with $q$ elements, and so we get $a(x)=x^q+x$, $b(x)=1$ for $q=2,3$ as counterexamples. But no others. | |
| Jun 9, 2011 at 9:44 | comment | added | Max Horn | The idea is to exploit that $x^i$ and $x^j$ are linearly independent for $i\neq j$, and use that to describe the coefficients $a_i$ and $b_i$ of $a$ resp. $b$ in terms of elements of these: For all $x\in K$ (ignoring zeros of $b$ for the moment) we have by our hypothesis that $a(x)=b(x) c_x$, where $c_x\in F$. For $x\in F$, this gives a linear equation with $a_i,b_j$ as variables (!) and with the $x^i$ and $c_x$ as coefficients in $F$. Doing this for eight $x\in F$, we find the $a_i$, $b_j$ as the solution of a system of $F$-linear equations. | |
| Jun 8, 2011 at 23:16 | comment | added | Kevin Buzzard | Back to the point, can you explain the reasoning behind your assertion "It is also not hard to see that the condition of the claim implies that $a,b$ must have coefficients in $F$" in the original question? I asked a more general question, using a more geometric language, at mathoverflow.net/questions/67304/… . Yours is a special case of this. | |
| Jun 8, 2011 at 23:14 | comment | added | Kevin Buzzard | [it was your little 'logo', whatever they're called, that made me ask] | |
| Jun 8, 2011 at 8:45 | comment | added | Max Horn | Thanks for the explanation on the degree. Makes sense. And no, my full name is not Apple Max ;) (if you actually want to know it, you can find out on my homepage, linked from the user profile). | |
| Jun 7, 2011 at 21:50 | answer | added | Kevin Buzzard | timeline score: 3 | |
| Jun 7, 2011 at 21:08 | comment | added | Kevin Buzzard | [PS is your full name Apple Max?] | |
| Jun 7, 2011 at 20:44 | comment | added | Kevin Buzzard | If one has degree exactly 3 then you can make the other one have degree exactly three by e.g. taking the reciprocal if necessary, and adding 1 if necessary. Similarly if both have degree 3 you can lower the degree of one of them. The degree of $a/b$ as a rational function is just the max of the degrees. I don't know if it helps, but perhaps the theory of "thin sets" etc in Serre's lectures on the Mordell-Weil theorem would show that the map had to be constant in the case that $K$ and $F$ are number fields. | |
| Jun 7, 2011 at 20:09 | history | edited | Max Horn | CC BY-SA 3.0 |
Clarify assumptions a bit more
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| Jun 7, 2011 at 20:04 | comment | added | Max Horn | With "we may assume that they have no common zero" I meant that "due to the way these polynomials turn up, we can arrange things so that they have no common zero". I'll try to clarify the question accordingly. In addition, I can arrange it so that the degree of $a$ (or $b$) is exactly 3; not sure whether I can force this simultaneously, need to think about that. However, this requires lengthy technical arguments. So, I am also still interested in the original question (degree of a and b $\leq 3$), under the additional assumption that $char K\neq 2$, if that is not much harder to handle. | |
| Jun 7, 2011 at 19:50 | comment | added | Kevin Buzzard | If you assume that $a$ has degree exactly three then the statement "we may assume that they have no common zero" becomes false. Is that still an assumption, or can I just say $a(x)=x^3$ and $b(x)=x$ and same counterexample? Your question is the following: you have a morphism $f:P^1\to P^1$ defined over $K$, either constant or of degree at most 3, and it maps $P^1(K)$ into $P^1(F)$, and you want to prove it's constant. I am wondering now whether you're furthermore demanding that the morphism have degree exactly 3. | |
| Jun 7, 2011 at 15:05 | comment | added | Max Horn | By the way: I can probably also add the assumption that $a$ has degree exactly $3$. That rules out the counterexamples by Felipe and Kevin, but I wonder if there are others then (in characteristic 2) ? | |
| Jun 7, 2011 at 14:24 | comment | added | Felipe Voloch | @Kevin: Great minds think alike and fools seldom differ... | |
| Jun 7, 2011 at 14:20 | history | edited | Max Horn | CC BY-SA 3.0 |
Trying to fix weird formatting glitch ?!
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| Jun 7, 2011 at 14:18 | comment | added | Kevin Buzzard | Darn Felipe, curse my slow typing :-) | |
| Jun 7, 2011 at 14:18 | comment | added | Max Horn | Thanks, I should have expected that characteristic 2 will produce counterexamples, it always does for me ;). However, I have to exclude characteristic 2 for other reasons anyway, so I am not overly concerned. It is very good to have concrete counterexamples for this, though. Thanks again! | |
| Jun 7, 2011 at 14:16 | comment | added | Max Horn | @Qiaochu: So what you are saying is that $f:=a/b$ as function from $K\to K$ is either constant, or ``almost'' surjective if $K$ is algebraically closed. That is, its image is dense in $K$ in the Zariski topology -- i.e. $f$ is a dominant rational mapping. True enough, that's a very nice and simple argument to solve the algebraically closed case. Thanks :). Now, can this be extended to the general case somehow? | |
| Jun 7, 2011 at 14:15 | comment | added | Kevin Buzzard | You need a separability assumption or else there are counterexamples in characteristic 2. Let $a(x)=x^2$, $b(x)=1$, and consider an inseparable quadratic extension e.g. $k(t)/k(t^2)$ with char(k)=2. | |
| Jun 7, 2011 at 14:12 | comment | added | Felipe Voloch | If $\mathbb{K}$ is a non-perfect field of characteristic two, $\mathbb{F} = \mathbb{K}^2$, then $a(x)=1,b(x)=x^2$ is a counterexample. | |
| Jun 7, 2011 at 14:10 | comment | added | Max Horn | Here's a sketch of my metric argument for $\mathbb{C,R}$: Consider $T$ as subset of the vectorspace $K^2$. The hypothesis now is that $T$ is contained in the union $S$ of all 1-dimensional subspaces / rays with ``real slope''. Pick $x$ such that $p_1:=p(x):=(a(x),b(x))$ is far from 0. Then $p_1$ and $p_2:=p(x+\epsilon)$ are connected by a path in $T$ which doesn't contain 0. But in $S$, all paths between points are either in a ray, or they pass through 0 and are in two rays. Since the curve crosses 0 at most 3 times, it is in the union of finitely many $S$-rays. Then it is in only one ray. | |
| Jun 7, 2011 at 13:16 | comment | added | Qiaochu Yuan | More generally, for $K$ algebraically closed it's obvious that $a(x) = b(x) f$ has a root in $K$ for all but at most one $f \in K$. | |
| Jun 7, 2011 at 13:09 | comment | added | Qiaochu Yuan | For $\mathbb{C}, \mathbb{R}$ try to show that no non-constant function $f : \mathbb{C} \setminus \{ p_1, ... p_n \} \to \mathbb{R}$ can be holomorphic. | |
| Jun 7, 2011 at 12:53 | history | asked | Max Horn | CC BY-SA 3.0 |