Mixing
Canonical Divisors On Projective Curves
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I am working on an expository paper and need help with a problem from Algebraic Geometry: A Problem Solving Approach by Thomas Garrity (and many more authors). Suppose $x$ and $y$ are both local coordinates at a point $p$ on a smooth curve $C = V(P)$ for some homogeneous polynomial $P(x, y, z)$. I want to show that the divisor of a differential form $w$ does not depend on the choice of local coordinates. This will lead me into the definition of the canonical divisor. The book walks me through a proof that considers 3 cases: if $x$ and $y$ are in the same affine patch of $mathbbP^2$, if they're in two different affine patches, and then a third case if they're in overlapping affine patches. If I could get help with the first case, when they're in the same affine patch, I'm hopeful I can finish the other two cases.
Assume $x$ and $y$ are in the same affine patch. We can describe $C$ near $p$ by $P(x, y) = 0$. Prove the divisors of $dx$ and $dy$ both have order zero at $p$. And we're given a hint that since $P(x,y) = 0$, we have $0 = dP = P_xdx + P_ydy$. Since these are local coordinates in the same affine patch, simplification yields $dy = -dfracP_xP_ydx$, and similar for $dx$. Not sure where to go from here. Any help is much appreciated, hints or a proof of this part. Probably missing something small.
algebraic-geometry divisors-algebraic-geometry
asked Jul 30 at 6:04
Travis62
112
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up vote
0
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I am working on an expository paper and need help with a problem from Algebraic Geometry: A Problem Solving Approach by Thomas Garrity (and many more authors). Suppose $x$ and $y$ are both local coordinates at a point $p$ on a smooth curve $C = V(P)$ for some homogeneous polynomial $P(x, y, z)$. I want to show that the divisor of a differential form $w$ does not depend on the choice of local coordinates. This will lead me into the definition of the canonical divisor. The book walks me through a proof that considers 3 cases: if $x$ and $y$ are in the same affine patch of $mathbbP^2$, if they're in two different affine patches, and then a third case if they're in overlapping affine patches. If I could get help with the first case, when they're in the same affine patch, I'm hopeful I can finish the other two cases.
Assume $x$ and $y$ are in the same affine patch. We can describe $C$ near $p$ by $P(x, y) = 0$. Prove the divisors of $dx$ and $dy$ both have order zero at $p$. And we're given a hint that since $P(x,y) = 0$, we have $0 = dP = P_xdx + P_ydy$. Since these are local coordinates in the same affine patch, simplification yields $dy = -dfracP_xP_ydx$, and similar for $dx$. Not sure where to go from here. Any help is much appreciated, hints or a proof of this part. Probably missing something small.
algebraic-geometry divisors-algebraic-geometry
asked Jul 30 at 6:04
Travis62
112
You've shown $dy$ and $dx$ differ by a scalar, and you're in projective space. Can you conclude from here?
– Samir Canning
Jul 30 at 15:48
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am working on an expository paper and need help with a problem from Algebraic Geometry: A Problem Solving Approach by Thomas Garrity (and many more authors). Suppose $x$ and $y$ are both local coordinates at a point $p$ on a smooth curve $C = V(P)$ for some homogeneous polynomial $P(x, y, z)$. I want to show that the divisor of a differential form $w$ does not depend on the choice of local coordinates. This will lead me into the definition of the canonical divisor. The book walks me through a proof that considers 3 cases: if $x$ and $y$ are in the same affine patch of $mathbbP^2$, if they're in two different affine patches, and then a third case if they're in overlapping affine patches. If I could get help with the first case, when they're in the same affine patch, I'm hopeful I can finish the other two cases.
Assume $x$ and $y$ are in the same affine patch. We can describe $C$ near $p$ by $P(x, y) = 0$. Prove the divisors of $dx$ and $dy$ both have order zero at $p$. And we're given a hint that since $P(x,y) = 0$, we have $0 = dP = P_xdx + P_ydy$. Since these are local coordinates in the same affine patch, simplification yields $dy = -dfracP_xP_ydx$, and similar for $dx$. Not sure where to go from here. Any help is much appreciated, hints or a proof of this part. Probably missing something small.
algebraic-geometry divisors-algebraic-geometry
asked Jul 30 at 6:04
Travis62
112
I am working on an expository paper and need help with a problem from Algebraic Geometry: A Problem Solving Approach by Thomas Garrity (and many more authors). Suppose $x$ and $y$ are both local coordinates at a point $p$ on a smooth curve $C = V(P)$ for some homogeneous polynomial $P(x, y, z)$. I want to show that the divisor of a differential form $w$ does not depend on the choice of local coordinates. This will lead me into the definition of the canonical divisor. The book walks me through a proof that considers 3 cases: if $x$ and $y$ are in the same affine patch of $mathbbP^2$, if they're in two different affine patches, and then a third case if they're in overlapping affine patches. If I could get help with the first case, when they're in the same affine patch, I'm hopeful I can finish the other two cases.
Assume $x$ and $y$ are in the same affine patch. We can describe $C$ near $p$ by $P(x, y) = 0$. Prove the divisors of $dx$ and $dy$ both have order zero at $p$. And we're given a hint that since $P(x,y) = 0$, we have $0 = dP = P_xdx + P_ydy$. Since these are local coordinates in the same affine patch, simplification yields $dy = -dfracP_xP_ydx$, and similar for $dx$. Not sure where to go from here. Any help is much appreciated, hints or a proof of this part. Probably missing something small.
algebraic-geometry divisors-algebraic-geometry
asked Jul 30 at 6:04
Travis62
112
asked Jul 30 at 6:04
Travis62
112
asked Jul 30 at 6:04
Travis62
112
asked Jul 30 at 6:04
Travis62
112
112
You've shown $dy$ and $dx$ differ by a scalar, and you're in projective space. Can you conclude from here?
– Samir Canning
Jul 30 at 15:48
add a comment |Â
You've shown $dy$ and $dx$ differ by a scalar, and you're in projective space. Can you conclude from here?
– Samir Canning
Jul 30 at 15:48
You've shown $dy$ and $dx$ differ by a scalar, and you're in projective space. Can you conclude from here?
– Samir Canning
Jul 30 at 15:48
You've shown $dy$ and $dx$ differ by a scalar, and you're in projective space. Can you conclude from here?
– Samir Canning
Jul 30 at 15:48
add a comment |Â
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You've shown $dy$ and $dx$ differ by a scalar, and you're in projective space. Can you conclude from here?
– Samir Canning
Jul 30 at 15:48