Prove that any divisor of order 0 on non-singular projective curve of genus $g$ is equivalent to other
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Could you please check whether the solution below is ok?
There is an exercise from Shafarevich's Basic Algebraic Geometry, vol. 1, ex. 7.7.21.
Let $o$ be a point of an smooth algebraic curve $X$ of genus $g$. Using the Riemann–Roch theorem, prove that any divisor $D$ with $deg D=0$ is equivalent to a divisor of the form $D_0-go$, where $D_0 > 0, deg D_0=g$.
It's equivalent to show that the space $mathcal{L}(go+D)$ has dimension $geqslant 1$. Suppose it has dimension $0$. By Riemann-Roch theorem, we have
$ell(go+D)-ell(K-go-D) = 1$. Therefore, $ell(K-go-D)=-1$. Since $K-go+D$ and $K-go$ are of the same degree, we have
$ell(K-go) leqslant ell(K-go+D)$. Now apply Riemann-Roch to $go$:
$$ell(go)-ell(K-go) = 1$$
But the space $mathcal{L}(go)$ is the space of all rational functions having pole of order $leqslant g$ at $o$, so of dimension $g+1$,
and $ell(K-go)$ is of dimension $g$ - contradiction.
I don't like the last argument and that we didn't use the structure of a canonical divisor.
Thanks in advance.
proof-verification algebraic-geometry divisors-algebraic-geometry
asked Dec 6 '18 at 19:29
Nicholas SNicholas S
299110
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add a comment |
$begingroup$
Could you please check whether the solution below is ok?
There is an exercise from Shafarevich's Basic Algebraic Geometry, vol. 1, ex. 7.7.21.
Let $o$ be a point of an smooth algebraic curve $X$ of genus $g$. Using the Riemann–Roch theorem, prove that any divisor $D$ with $deg D=0$ is equivalent to a divisor of the form $D_0-go$, where $D_0 > 0, deg D_0=g$.
It's equivalent to show that the space $mathcal{L}(go+D)$ has dimension $geqslant 1$. Suppose it has dimension $0$. By Riemann-Roch theorem, we have
$ell(go+D)-ell(K-go-D) = 1$. Therefore, $ell(K-go-D)=-1$. Since $K-go+D$ and $K-go$ are of the same degree, we have
$ell(K-go) leqslant ell(K-go+D)$. Now apply Riemann-Roch to $go$:
$$ell(go)-ell(K-go) = 1$$
But the space $mathcal{L}(go)$ is the space of all rational functions having pole of order $leqslant g$ at $o$, so of dimension $g+1$,
and $ell(K-go)$ is of dimension $g$ - contradiction.
I don't like the last argument and that we didn't use the structure of a canonical divisor.
Thanks in advance.
proof-verification algebraic-geometry divisors-algebraic-geometry
asked Dec 6 '18 at 19:29
Nicholas SNicholas S
299110
$endgroup$
add a comment |
$begingroup$
Could you please check whether the solution below is ok?
There is an exercise from Shafarevich's Basic Algebraic Geometry, vol. 1, ex. 7.7.21.
Let $o$ be a point of an smooth algebraic curve $X$ of genus $g$. Using the Riemann–Roch theorem, prove that any divisor $D$ with $deg D=0$ is equivalent to a divisor of the form $D_0-go$, where $D_0 > 0, deg D_0=g$.
It's equivalent to show that the space $mathcal{L}(go+D)$ has dimension $geqslant 1$. Suppose it has dimension $0$. By Riemann-Roch theorem, we have
$ell(go+D)-ell(K-go-D) = 1$. Therefore, $ell(K-go-D)=-1$. Since $K-go+D$ and $K-go$ are of the same degree, we have
$ell(K-go) leqslant ell(K-go+D)$. Now apply Riemann-Roch to $go$:
$$ell(go)-ell(K-go) = 1$$
But the space $mathcal{L}(go)$ is the space of all rational functions having pole of order $leqslant g$ at $o$, so of dimension $g+1$,
and $ell(K-go)$ is of dimension $g$ - contradiction.
I don't like the last argument and that we didn't use the structure of a canonical divisor.
Thanks in advance.
proof-verification algebraic-geometry divisors-algebraic-geometry
asked Dec 6 '18 at 19:29
Nicholas SNicholas S
299110
$endgroup$
Could you please check whether the solution below is ok?
There is an exercise from Shafarevich's Basic Algebraic Geometry, vol. 1, ex. 7.7.21.
Let $o$ be a point of an smooth algebraic curve $X$ of genus $g$. Using the Riemann–Roch theorem, prove that any divisor $D$ with $deg D=0$ is equivalent to a divisor of the form $D_0-go$, where $D_0 > 0, deg D_0=g$.
It's equivalent to show that the space $mathcal{L}(go+D)$ has dimension $geqslant 1$. Suppose it has dimension $0$. By Riemann-Roch theorem, we have
$ell(go+D)-ell(K-go-D) = 1$. Therefore, $ell(K-go-D)=-1$. Since $K-go+D$ and $K-go$ are of the same degree, we have
$ell(K-go) leqslant ell(K-go+D)$. Now apply Riemann-Roch to $go$:
$$ell(go)-ell(K-go) = 1$$
But the space $mathcal{L}(go)$ is the space of all rational functions having pole of order $leqslant g$ at $o$, so of dimension $g+1$,
and $ell(K-go)$ is of dimension $g$ - contradiction.
I don't like the last argument and that we didn't use the structure of a canonical divisor.
Thanks in advance.
proof-verification algebraic-geometry divisors-algebraic-geometry
proof-verification algebraic-geometry divisors-algebraic-geometry
asked Dec 6 '18 at 19:29
Nicholas SNicholas S
299110
asked Dec 6 '18 at 19:29
Nicholas SNicholas S
299110
edited Dec 6 '18 at 19:55
Nicholas S
asked Dec 6 '18 at 19:29
Nicholas SNicholas S
299110
asked Dec 6 '18 at 19:29
Nicholas SNicholas S
299110
asked Dec 6 '18 at 19:29
Nicholas SNicholas S
299110
299110
add a comment |
add a comment |
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I don't understand the unnecessary proof by contradiction. You arrive, by Riemann-Roch, at
$$ell(go+D)-ell(K-go-D)=1.$$
Since $ell(E)ge 0$ for any divisor $E$, it follows that $ell(go+D)ge 1$, which is what you wanted to show. (And, no, you don't need the specific form of Serre duality here.)
answered Dec 6 '18 at 19:45
Ted ShifrinTed Shifrin
63.2k44489
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1
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Thank you! I need more sleep :)
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– Nicholas S
Dec 6 '18 at 19:51
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I don't understand the unnecessary proof by contradiction. You arrive, by Riemann-Roch, at
$$ell(go+D)-ell(K-go-D)=1.$$
Since $ell(E)ge 0$ for any divisor $E$, it follows that $ell(go+D)ge 1$, which is what you wanted to show. (And, no, you don't need the specific form of Serre duality here.)
answered Dec 6 '18 at 19:45
Ted ShifrinTed Shifrin
63.2k44489
$endgroup$
1
$begingroup$
Thank you! I need more sleep :)
$endgroup$
– Nicholas S
Dec 6 '18 at 19:51
add a comment |
$begingroup$
I don't understand the unnecessary proof by contradiction. You arrive, by Riemann-Roch, at
$$ell(go+D)-ell(K-go-D)=1.$$
Since $ell(E)ge 0$ for any divisor $E$, it follows that $ell(go+D)ge 1$, which is what you wanted to show. (And, no, you don't need the specific form of Serre duality here.)
answered Dec 6 '18 at 19:45
Ted ShifrinTed Shifrin
63.2k44489
$endgroup$
1
$begingroup$
Thank you! I need more sleep :)
$endgroup$
– Nicholas S
Dec 6 '18 at 19:51
add a comment |
$begingroup$
I don't understand the unnecessary proof by contradiction. You arrive, by Riemann-Roch, at
$$ell(go+D)-ell(K-go-D)=1.$$
Since $ell(E)ge 0$ for any divisor $E$, it follows that $ell(go+D)ge 1$, which is what you wanted to show. (And, no, you don't need the specific form of Serre duality here.)
answered Dec 6 '18 at 19:45
Ted ShifrinTed Shifrin
63.2k44489
$endgroup$
I don't understand the unnecessary proof by contradiction. You arrive, by Riemann-Roch, at
$$ell(go+D)-ell(K-go-D)=1.$$
Since $ell(E)ge 0$ for any divisor $E$, it follows that $ell(go+D)ge 1$, which is what you wanted to show. (And, no, you don't need the specific form of Serre duality here.)
answered Dec 6 '18 at 19:45
Ted ShifrinTed Shifrin
63.2k44489
edited Dec 6 '18 at 19:50
answered Dec 6 '18 at 19:45
Ted ShifrinTed Shifrin
63.2k44489
answered Dec 6 '18 at 19:45
Ted ShifrinTed Shifrin
63.2k44489
answered Dec 6 '18 at 19:45
Ted ShifrinTed Shifrin
63.2k44489
63.2k44489
1
$begingroup$
Thank you! I need more sleep :)
$endgroup$
– Nicholas S
Dec 6 '18 at 19:51
add a comment |
1
$begingroup$
Thank you! I need more sleep :)
$endgroup$
– Nicholas S
Dec 6 '18 at 19:51
1
1
$begingroup$
Thank you! I need more sleep :)
$endgroup$
– Nicholas S
Dec 6 '18 at 19:51
$begingroup$
Thank you! I need more sleep :)
$endgroup$
– Nicholas S
Dec 6 '18 at 19:51
add a comment |
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