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Kevin Buzzard
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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+3txy^2, ty^3+3x^2y$$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".

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 $x^3+3txy^2, ty^3+3x^2y$. 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".

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".

Source Link
Kevin Buzzard
  • 41.4k
  • 13
  • 166
  • 245

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 $x^3+3txy^2, ty^3+3x^2y$. 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".