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Let $\mathbb{K}$ be an arbitrary field with a subfield $\mathbb{F}$ of index 2. Let $a,b\in\mathbb{K}[X]$ be univariate non-vanishing polynomials over $\mathbb{K}$ of degree $\leq 3$ each. Edit: Due to how this problem arises, one may assume that $a,b$ have no common zeros and at least one has degree exactly 3. However, I am also interested in what happens without this extra assumption, but instead the assumption that $char\mathbb{K}\neq2$. The latter to avoid the counterexamples described in the comments.

Consider the set of points $T:=\{(a(x),b(x)) \mid x\in\mathbb{K}\}$, a subset of $\mathbb{K}^2$.

Prove (or give a counterexample) for the following:

Claim: Assume every point $T$ is projectively equivalent to a point in $\mathbb{F}\times\mathbb{F}$ (i.e. for every $x\in\mathbb{K}$ we have $a(x)=b(x)\cdot f_x$ for some $f_x\in\mathbb{F}$). Then either $\lvert\mathbb{K}\rvert=4$ or 9, or all points in $T$ are projectively equivalent (that is, $T$ is contained in a one-dimensional $\mathbb{K}$-subspace of $\mathbb{K}^2$); put another way, $a/b$ is a constant.

Since we assumed $a,b$ to have no common zeros, we can think of this in terms of projective coordinates. Then the question becomes: If all points on the curve $T$ are $\mathbb{F}$-rational, does this imply that $T$ consists of a single point?

For finite fields, this can be shown using a simple counting argument (had to check the field with 9 elements manually, and found an exception over the field with four elements). It is also not hard to see that the condition of the claim implies that $a,b$ must have coefficients in $\mathbb{F}$.

For $\mathbb{K}=\mathbb{C}$ and $\mathbb{F}=\mathbb{R}$ I have an argument involving the topology and metric of these fields. For the general case, I tried various approaches, and one of them might still work out (but all my ideas seem at some point to end up in heavy, uninsightful and simply ugly computations)...

But I keep wondering if this isn't a problem that somebody with a better background in number theory or algebraic geometry or something like that could quickly solve with "standard" methods.... so before I keep going on with my little knowledge, I though it best to ask here for any pointer or even solutions :).


Lastly, here is one thing I was trying, but couldn't quite complete. It's quite possibly a dead end, so you may not want to get yourself overly distracted by it :): Pick $\alpha\in\mathbb{K}\setminus\mathbb{F}$. For each $t\in\mathbb{K}$, define a polynomial $p_t(x):=a(x+t)b(\alpha+t)-a(\alpha+t) b(x+t)$. They all have $\alpha$ as a zero. And (up to some rescaling), the coefficients of $p_t$ are in $\mathbb{F}$ by hypothesis.

Now if any of the $p_t$ vanishes everywhere, then all do, and $a/b$ is constant. So assume the $p_t$ do not vanish. Then every $p_t$ is divisible by the minimal polynomial of $\alpha$, and so has degree 2 or 3. Indeed, looking at the coefficients, for at most three $t$ can $p_t$ have degree 2, so for almost all it has degree 3, and is divisible by the minimal polynomial of $\alpha$. This sounds quite improbable to me (but that proves nothing, only that I lack imagination ;). So we could now compare several of the $p_t$, and try to derive a contradiction, but this (at least in the naive ways I tried) quickly gets very messy, uninsightful and ugly ;).

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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. –  Qiaochu Yuan Jun 7 '11 at 13:09
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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 $f \in K$. –  Qiaochu Yuan Jun 7 '11 at 13:16
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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. –  Felipe Voloch Jun 7 '11 at 14:12
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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. –  Kevin Buzzard Jun 7 '11 at 14:15
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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. –  Kevin Buzzard Jun 7 '11 at 20:44

1 Answer 1

This isn't an answer but I think it's progress. It started off by thinking of restriction of scalars but I've translated it down to a rather more mundane point of view.

Let me call the fields $K$ and $F$ to save some typing.

Let me first deal with the finite field case. My understanding of the question as it currently stands is that we have a morphism $t:\mathbf{P}^1_K\to\mathbf{P}^1_K$ of degree exactly 3, with the property that the image of $\mathbf{P}^1(K)$ is contained within $\mathbf{P}^1(F)$ and we want to show that $F$ has size 2.

The case $F$ finite is easy to deal with. The pre-image of an $F$-point has size at most 3, so if $q$ is the size of $F$ then $q^2+1\leq 3(q+1)$ and we quickly deduce $q\leq 3$ and we deal with the case $q=3$ by hand.

Now for the case $F$ infinite. My understanding is that we can assume that the characteristic isn't 2. So we can write $K=F(\sqrt{d})$ for some $d\in F$, not a square. Let me now think of $K$ as a 2-dimensional vector space over $F$ with basis $[1,\sqrt{d}]$ and let's translate the question into a messy algebra one.

We see $a(x+y\sqrt{d})=L+M\sqrt{d}$, where $L$ and $M$ are $F$-linear combinations of the six polynomials $x$, $y$ [real and imag parts of $x+y\sqrt{d}$], $x^2+dy^2, 2xy$ [this comes from $(x+y\sqrt{d})^2$ and $x^3+3dxy^2, dy^3+3x^2y$. Similarly $b(x+y\sqrt{d})=N+P\sqrt{d}$.

We are given that for all $x,y\in F$ we have $L(x,y)+M(x,y)\sqrt{d}=f(x,y)(N(x,y)+P(x,y)\sqrt{d})$ with $f(x,y)\in F$ (forget the finitely many points where $f$ has a pole), and we deduce that $L(x,y)P(x,y)=M(x,y)N(x,y)$ for all $x,y\in F$. But $F$ is infinite and this implies that, as polynomials in $x$ and $y$, we have $LP=MN$ identically. This is a piece of information that wasn't clear before.

This means that we can base change our entire situation to the algebraic closure, and replace $F$ with $\overline{F}$ and $K$ with $\overline{F}\oplus\overline{F}$, and (calling these new rings $F$ and $K$) we now have maps $\mathbf{P}^1_K\to\mathbf{P}^1_K$ which are defined over $K$ and such that the image of $\mathbf{P}^1(K)$ is in $\mathbf{P}^1(F)$. [Note: it's at this point that I'm assuming $K/F$ separable.]. Now I want to say "and now we should be done because of alg geom" but in fact what I mean is "and now someone else will have to take over because I have to clear up the kitchen".

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Over the algebraic closure it seems like I have something like a rational map from affine 2-space to affine 2-space, of the form $(s,t)\mapsto (g(s),h(t))$ (this is the statement that the map is defined over $K$), and whose image is contained in a diagonal affine 1-space. This means each horizontal line and each vertical line must be mapped to a point, and this means that the image of all of affine 2-space must be a point. Am I now done? –  Kevin Buzzard Jun 7 '11 at 23:50
    
In your answer, everything up to the final paragraph seems clear. But why exactly does $LP=MN$ imply you can change the setting as described? You somehow seem to arrive at the conclusion that the curve $(a(x,y),b(x,y))$ in the old $K$ (where multiplication "mixes" real and imaginary parts) can be ``untangled'' to $(g(x),h(y))$ and we can thus switch to $\overline F\oplus\overline F$. Since you mention it so casually, it's probably something terribly basic. But I don't see how it follows. Can you give me a hint? –  Max Horn Jun 8 '11 at 9:21
    
Accepting this as a black box for the moment, the rest seems clear, assuming new-$F$ maps diagonally into new-$K$. Indeed, can't we then immediately conclude $g=h$? –  Max Horn Jun 8 '11 at 9:22
    
Max -- I just raced down some comments. I'm not saying I've answered the question -- I hoped someone could either take over or point out a problem. If $K=F(\sqrt{d})$ then the moment I adjoin another $\sqrt{d}$ to the situation I can unravel everything. $LP=MN$ guarantees that I don't lose control at this point. That's the proof plan anyway. –  Kevin Buzzard Jun 8 '11 at 20:04

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