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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 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 ;).

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 ;).

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. We may assume that they have no common zero. Edit: As was pointed out in the comments, there are counterexamples in characteristic 2; so assume $char\mathbb{K}\neq2$.

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 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 ;).

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 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 ;).

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. We may assume that they have no common zero.

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 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 ;).

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. We may assume that they have no common zero. Edit: As was pointed out in the comments, there are counterexamples in characteristic 2; so assume $char\mathbb{K}\neq2$.

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 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 ;).