Finitely many zeros and poles for a function in a function field of a smooth curve
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Let $bar{K}$ be a perfect field and let $f in bar{K}(C)$ be a nonzero function in the function field of $C$, a smooth curve (projective variety of dimension 1). I'm trying to understand why (even though it feels obvious) any nonzero function $f$ has only finitely many zeros and poles. I only have a really minor knowledge on varieties - because I'm working through Silverman's The Arithmetic of Elliptic Curves. I've been trying to think in order to find a contradiction that occurs when $ord_P(f) > 0$ for all $P in C$ to no avail. Is it that the set of all local rings, $bar{K}[C]_P$ where $f notin bar{K}[C]_P$ is finite?
Some relevant definitions, $f$ has a zero at $P$ if $ord_P(f) > 0$ and a pole if $ord_P(f) < 0$.
algebraic-geometry elliptic-curves
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add a comment |
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Let $bar{K}$ be a perfect field and let $f in bar{K}(C)$ be a nonzero function in the function field of $C$, a smooth curve (projective variety of dimension 1). I'm trying to understand why (even though it feels obvious) any nonzero function $f$ has only finitely many zeros and poles. I only have a really minor knowledge on varieties - because I'm working through Silverman's The Arithmetic of Elliptic Curves. I've been trying to think in order to find a contradiction that occurs when $ord_P(f) > 0$ for all $P in C$ to no avail. Is it that the set of all local rings, $bar{K}[C]_P$ where $f notin bar{K}[C]_P$ is finite?
Some relevant definitions, $f$ has a zero at $P$ if $ord_P(f) > 0$ and a pole if $ord_P(f) < 0$.
algebraic-geometry elliptic-curves
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What about this : If $u,v in overline{K}(C)$ are non-constant then $overline{K}(u,v)/overline{K}(u)$ is a finite extension so $overline{K}(v)capoverline{K}(u)$ contains a non-constant element : there are some polynomials such that $f(u) =h(u) g(v)$ and if $u$ has finitely many zeros then $v$ has finitely many zeros.
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– reuns
Dec 9 '18 at 5:17
add a comment |
$begingroup$
Let $bar{K}$ be a perfect field and let $f in bar{K}(C)$ be a nonzero function in the function field of $C$, a smooth curve (projective variety of dimension 1). I'm trying to understand why (even though it feels obvious) any nonzero function $f$ has only finitely many zeros and poles. I only have a really minor knowledge on varieties - because I'm working through Silverman's The Arithmetic of Elliptic Curves. I've been trying to think in order to find a contradiction that occurs when $ord_P(f) > 0$ for all $P in C$ to no avail. Is it that the set of all local rings, $bar{K}[C]_P$ where $f notin bar{K}[C]_P$ is finite?
Some relevant definitions, $f$ has a zero at $P$ if $ord_P(f) > 0$ and a pole if $ord_P(f) < 0$.
algebraic-geometry elliptic-curves
$endgroup$
Let $bar{K}$ be a perfect field and let $f in bar{K}(C)$ be a nonzero function in the function field of $C$, a smooth curve (projective variety of dimension 1). I'm trying to understand why (even though it feels obvious) any nonzero function $f$ has only finitely many zeros and poles. I only have a really minor knowledge on varieties - because I'm working through Silverman's The Arithmetic of Elliptic Curves. I've been trying to think in order to find a contradiction that occurs when $ord_P(f) > 0$ for all $P in C$ to no avail. Is it that the set of all local rings, $bar{K}[C]_P$ where $f notin bar{K}[C]_P$ is finite?
Some relevant definitions, $f$ has a zero at $P$ if $ord_P(f) > 0$ and a pole if $ord_P(f) < 0$.
algebraic-geometry elliptic-curves
algebraic-geometry elliptic-curves
asked Dec 9 '18 at 4:31
agaveagave
385
385
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What about this : If $u,v in overline{K}(C)$ are non-constant then $overline{K}(u,v)/overline{K}(u)$ is a finite extension so $overline{K}(v)capoverline{K}(u)$ contains a non-constant element : there are some polynomials such that $f(u) =h(u) g(v)$ and if $u$ has finitely many zeros then $v$ has finitely many zeros.
$endgroup$
– reuns
Dec 9 '18 at 5:17
add a comment |
$begingroup$
What about this : If $u,v in overline{K}(C)$ are non-constant then $overline{K}(u,v)/overline{K}(u)$ is a finite extension so $overline{K}(v)capoverline{K}(u)$ contains a non-constant element : there are some polynomials such that $f(u) =h(u) g(v)$ and if $u$ has finitely many zeros then $v$ has finitely many zeros.
$endgroup$
– reuns
Dec 9 '18 at 5:17
$begingroup$
What about this : If $u,v in overline{K}(C)$ are non-constant then $overline{K}(u,v)/overline{K}(u)$ is a finite extension so $overline{K}(v)capoverline{K}(u)$ contains a non-constant element : there are some polynomials such that $f(u) =h(u) g(v)$ and if $u$ has finitely many zeros then $v$ has finitely many zeros.
$endgroup$
– reuns
Dec 9 '18 at 5:17
$begingroup$
What about this : If $u,v in overline{K}(C)$ are non-constant then $overline{K}(u,v)/overline{K}(u)$ is a finite extension so $overline{K}(v)capoverline{K}(u)$ contains a non-constant element : there are some polynomials such that $f(u) =h(u) g(v)$ and if $u$ has finitely many zeros then $v$ has finitely many zeros.
$endgroup$
– reuns
Dec 9 '18 at 5:17
add a comment |
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$begingroup$
What about this : If $u,v in overline{K}(C)$ are non-constant then $overline{K}(u,v)/overline{K}(u)$ is a finite extension so $overline{K}(v)capoverline{K}(u)$ contains a non-constant element : there are some polynomials such that $f(u) =h(u) g(v)$ and if $u$ has finitely many zeros then $v$ has finitely many zeros.
$endgroup$
– reuns
Dec 9 '18 at 5:17