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5
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I think that the answer is no.
Here is a somewhat related problem. Every curve curve $C$ of genus >0, has meromorphic functions $x,y$ on it which are
not related by any equation of the form $f(x)=g(y)$.
I denote by $(x)=(x)^+-(x)^-$ the principal divisor of an element $x$,
zeros minus poles.
If $x$ and $y$ are two elemets of this filed, the field of meromorphic functions on $C$, related by $f(x)=g(y)$,
where $f,g$ are ratonal functions,
then the divisors of poles of $x$ and $y$ are related as follows:
$$m(x)^-\sim n(y)^-,$$
where $\sim$ means the usual equivalence of divisors.
(Two dividors $d$ and $e$ are equivalent if $d=e+(z)$). And $m,n$ are degrees of $f,g$.
Now the factor of the set of all divisors over this equivalent equation
is a torus of dimension $g$ ($g$ is the genus of $C$). We only need the fact that it is
uncountable for $g>0$. So
we can always find incommensurable divisors of the form $(x)^-$ and $(y)^-$.
These $x$ and $y$ are related by some polynomial relation $F(x,y)=0$, but cannot be related
by an equation of the form $f(x)=g(y)$.
This solution was explained me by Drinfeld in 1980 when I asked him more
general question: Can every algebraic relation $F(x,y)=0$ be obtained
from a chain $x=x_1,x_2,x_3,\ldots,x_n=y$ where $x_i$ and $x_{i+1}$
are related by $f_i(x_i)=f_{i+1}(x_{i+1})$, with some rational functions
$f_i$, by elimination of $x_2,...x_{n-1}$? The answer is no, for the same reason).
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4
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The answer is no. I consider everything over complex numbers
(the only numbers I know; let other people explain
what happens in other fleds).
EDIT:
Let $d$ be the maximal degree of $d$ and $g$. Then think that the equations $f(x)=g(y)$
depend on $4d+2$ parameters. The curve $(f(x)-f(y))/(x-y)$ answer is of degree $d-1$ and
genericaly non-singularno.Therefore its genus
Here is $g=(d-2)(d-3)/2$. But generic curves of genus $g$,
depend on $g$ parameters. This implies that for high genus, a generic
curve cannot be represented in this way. A somewhat related problem. Every curve curve $C$ of genus >0, has meromorphic functions $x,y$ on it which are
not related by any equation of the form $f(x)=g(y)$.
I denote by $(x)=(x)^+-(x)^-$ the principal divisor of an element $x$,
zeros minus poles.
If $x$ and $y$ are two elemets of this filed, related by $f(x)=g(y)$,
where $f,g$ are ratonal functions,
then the divisors of poles of $x$ and $y$ are related as follows:
$$m(x)^-\sim n(y)^-,$$
where $\sim$ means the usual equivalence of divisors.
(Two dividors $d$ and $e$ are equivalent if $d=e+(z)$). And $m,n$ are degrees of $f,g$.
Now the factor of the set of all divisors over this equivalent equation
is a torus of dimension $g$ ($g$ is the genus of $C$). We only need the fact that it is
uncountable for $g>0$. So
we can always find incommensurable divisors of the form $(x)^-$ and $(y)^-$.
These $x$ and $y$ are related by some polynomial relation $F(x,y)=0$, but cannot be related
by an equation of the form $f(x)=g(y)$.
This solution was explained me by Drinfeld in 1980 when I asked him more
general question: Can every algebraic relation $F(x,y)=0$ be obtained
from a chain $x=x_1,x_2,x_3,\ldots,x_n=y$ where $x_i$ and $x_{i+1}$
are related by $f_i(x_i)=f_{i+1}(x_{i+1})$, with some rational functions
$f_i$, by elimination of $x_2,...x_{n-1}$? The answer is no, for the same reason).
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3
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The answer is no. I consider everything over complex numbers
(the only numbers I know; let other people explain
what happens in other fleds).
EDIT:
Let $d$ be the maximal degree of $d$ and $g$. Then the equations $f(x)=g(y)$
depend on $4d+2$ parameters, while generic curves . The curve $(f(x)-f(y))/(x-y)$ is of degree $d$ have d-1$ and
genericaly non-singular. Therefore its genus is
$(d-1)(d-2)/2$,
thus g=(d-2)(d-3)/2$. But generic curves of genus $g$,
depend in on $3(d-1)(d-2)/2-3$ g$ parameters. This implies that for high genus, a generic
curve cannot be represented in this way. A somewhat related problem. Every curve curve $C$ of genus >0, has meromorphic functions $x,y$ on it which are
not related by any equation of the form $f(x)=g(y)$.
I denote by $(x)=(x)^+-(x)^-$ the principal divisor of an element $x$,
zeros minus poles.
If $x$ and $y$ are two elemets of this filed, related by $f(x)=g(y)$,
where $f,g$ are ratonal functions,
then the divisors of poles of $x$ and $y$ are related as follows:
$$m(x)^-\sim n(y)^-,$$
where $\sim$ means the usual equivalence of divisors.
(Two dividors $d$ and $e$ are equivalent if $d=e+(z)$). And $m,n$ are degrees of $f,g$.
Now the factor of the set of all divisors over this equivalent equation
is a torus of dimension $g$ ($g$ is the genus of $C$). We only need the fact that it is
uncountable for $g>0$. So
we can always find incommensurable divisors of the form $(x)^-$ and $(y)^-$.
These $x$ and $y$ are related by some polynomial relation $F(x,y)=0$, but cannot be related
by an equation of the form $f(x)=g(y)$.
This solution was explained me by Drinfeld in 1980 when I asked him more
general question: Can every algebraic relation $F(x,y)=0$ be obtained
from a chain $x=x_1,x_2,x_3,\ldots,x_n=y$ where $x_i$ and $x_{i+1}$
are related by $f_i(x_i)=f_{i+1}(x_{i+1})$, with some rational functions
$f_i$, by elimination of $x_2,...x_{n-1}$? The answer is no, for the same reason).
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2
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The answer is no. I consider everything over complex numbers
(the only numbers I know; let other people explain
what happens in other fleds).
Take
EDIT:
Let $d$ be the maximal degree of $d$ and $g$. Then the equations $f(x)=g(y)$
depend on $4d+2$ parameters, while generic curves of degree $d$ have genus $(d-1)(d-2)/2$,
thus depend in $3(d-1)(d-2)/2-3$ parameters. This implies that for high genus, a generic
curve cannot be represented in this way. A somewhat related problem. Every curve curve $C$ of genus >0, and consider its has meromorphic function filed.
functions $x,y$ on it which are
not related by any equation of the form $f(x)=g(y)$.
I denote by $(x)=(x)^+-(x)^-$ the principal divisor of an element $x$,
zeros minus poles.
If $x$ and $y$ are two elements elemets of this filed, related by $f(x)=g(y)$,
where $f,g$ are ratonal functions,
then the divisors of poles of $x$ and $y$ are related as follows:
$$m(x)^-\sim n(y)^-,$$
where $\sim$ means the usual equivalence of divisors.
(Two dividors $d$ and $e$ are equivalent if $d=e+(z)$). And $m,n$ are degrees of $f,g$.
Now the factor of the set of all divisors over this equivalent equation
is a torus of dimension $g$ ($g$ is the genus of $C$). We only need the fact that it is
uncountable for $g>0$. So
we can always find incommensurable divisors of the form $(x)^-$ and $(y)^-$.
These $x$ and $y$ are related by some polynomial relation $F(x,y)=0$, but cannot be related
by an equation of the form $f(x)=g(y)$.
This solution was explained me by Drinfeld in 1980 when I asked him more
general question: Can every algebraic relation $F(x,y)=0$ be obtained
from a chain $x=x_1,x_2,x_3,\ldots,x_n=y$ where $x_i$ and $x_{i+1}$
are related by $f_i(x_i)=f_{i+1}(x_{i+1})$, with some rational functions
$f_i$, by elimination of $x_2,...x_{n-1}$? The answer is no, for the same reason).
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1
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The answer is no. I consider everything over complex numbers
(the only numbers I know; let other people explain
what happens in other fleds).
Take a curve $C$ of genus >0, and consider its meromorphic function filed.
I denote by $(x)=(x)^+-(x)^-$ the principal divisor of an element $x$,
zeros minus poles.
If $x$ and $y$ are two elements of this filed, related by $f(x)=g(y)$,
where $f,g$ are ratonal functions,
then the divisors of poles of $x$ and $y$ are related as follows:
$$m(x)^-\sim n(y)^-,$$
where $\sim$ means the usual equivalence of divisors.
(Two dividors $d$ and $e$ are equivalent if $d=e+(z)$). And $m,n$ are degrees of $f,g$.
Now the factor of the set of all divisors over this equivalent equation
is a torus of dimension $g$ ($g$ is the genus of $C$). We only need the fact that it is
uncountable for $g>0$. So
we can always find incommensurable divisors of the form $(x)^-$ and $(y)^-$.
These $x$ and $y$ are related by some polynomial relation $F(x,y)=0$, but cannot be related
by an equation of the form $f(x)=g(y)$.
This solution was explained me by Drinfeld in 1980 when I asked him more
general question: Can every algebraic relation $F(x,y)=0$ be obtained
from a chain $x=x_1,x_2,x_3,\ldots,x_n=y$ where $x_i$ and $x_{i+1}$
are related by $f_i(x_i)=f_{i+1}(x_{i+1})$, with some rational functions
$f_i$, by elimination of $x_2,...x_{n-1}$? The answer is no, for the same reason).
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