1-form on variety
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Let $k$ be an algebraically closed field and consider the affine curve $Z(-y^5+x^4-1) subset mathbb{A}^2(k)$. Define on it the rational 1-form $w=frac{dx}{5y^4}=frac{dy}{4x^3}$. Is there a (relatively straightforward) way to see that $w$ has no poles on the curve $textit{without}$ calculating its order at all points?
algebraic-geometry
asked Nov 24 at 17:46
Karl
366
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Let $k$ be an algebraically closed field and consider the affine curve $Z(-y^5+x^4-1) subset mathbb{A}^2(k)$. Define on it the rational 1-form $w=frac{dx}{5y^4}=frac{dy}{4x^3}$. Is there a (relatively straightforward) way to see that $w$ has no poles on the curve $textit{without}$ calculating its order at all points?
algebraic-geometry
asked Nov 24 at 17:46
Karl
366
2
Clearly this form has no poles when $x neq 0$ and when $y neq 0$. Moreover, $(0,0)$ is not on your curve so you can conclude that there is no poles.
– Nicolas Hemelsoet
Nov 24 at 19:24
Thanks for your answer Nicolas. What makes you say without any calculation that surely when say x is not zero, it doesn't have a pole?
– Karl
Nov 24 at 21:08
2
Because if $z$ is some coordinate on a curve, the poles of $frac{f}{g}dz$ are by definition the zeroes of $g$. In your case, since $omega$ can be written as $frac{dy}{4x^3}$ this shows that if $x neq 0$ there is no poles.
– Nicolas Hemelsoet
Nov 24 at 21:12
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $k$ be an algebraically closed field and consider the affine curve $Z(-y^5+x^4-1) subset mathbb{A}^2(k)$. Define on it the rational 1-form $w=frac{dx}{5y^4}=frac{dy}{4x^3}$. Is there a (relatively straightforward) way to see that $w$ has no poles on the curve $textit{without}$ calculating its order at all points?
algebraic-geometry
asked Nov 24 at 17:46
Karl
366
Let $k$ be an algebraically closed field and consider the affine curve $Z(-y^5+x^4-1) subset mathbb{A}^2(k)$. Define on it the rational 1-form $w=frac{dx}{5y^4}=frac{dy}{4x^3}$. Is there a (relatively straightforward) way to see that $w$ has no poles on the curve $textit{without}$ calculating its order at all points?
algebraic-geometry
algebraic-geometry
asked Nov 24 at 17:46
Karl
366
asked Nov 24 at 17:46
Karl
366
asked Nov 24 at 17:46
Karl
366
asked Nov 24 at 17:46
Karl
366
asked Nov 24 at 17:46
Karl
366
366
2
Clearly this form has no poles when $x neq 0$ and when $y neq 0$. Moreover, $(0,0)$ is not on your curve so you can conclude that there is no poles.
– Nicolas Hemelsoet
Nov 24 at 19:24
Thanks for your answer Nicolas. What makes you say without any calculation that surely when say x is not zero, it doesn't have a pole?
– Karl
Nov 24 at 21:08
2
Because if $z$ is some coordinate on a curve, the poles of $frac{f}{g}dz$ are by definition the zeroes of $g$. In your case, since $omega$ can be written as $frac{dy}{4x^3}$ this shows that if $x neq 0$ there is no poles.
– Nicolas Hemelsoet
Nov 24 at 21:12
add a comment |
2
Clearly this form has no poles when $x neq 0$ and when $y neq 0$. Moreover, $(0,0)$ is not on your curve so you can conclude that there is no poles.
– Nicolas Hemelsoet
Nov 24 at 19:24
Thanks for your answer Nicolas. What makes you say without any calculation that surely when say x is not zero, it doesn't have a pole?
– Karl
Nov 24 at 21:08
2
Because if $z$ is some coordinate on a curve, the poles of $frac{f}{g}dz$ are by definition the zeroes of $g$. In your case, since $omega$ can be written as $frac{dy}{4x^3}$ this shows that if $x neq 0$ there is no poles.
– Nicolas Hemelsoet
Nov 24 at 21:12
2
2
Clearly this form has no poles when $x neq 0$ and when $y neq 0$. Moreover, $(0,0)$ is not on your curve so you can conclude that there is no poles.
– Nicolas Hemelsoet
Nov 24 at 19:24
Clearly this form has no poles when $x neq 0$ and when $y neq 0$. Moreover, $(0,0)$ is not on your curve so you can conclude that there is no poles.
– Nicolas Hemelsoet
Nov 24 at 19:24
Thanks for your answer Nicolas. What makes you say without any calculation that surely when say x is not zero, it doesn't have a pole?
– Karl
Nov 24 at 21:08
Thanks for your answer Nicolas. What makes you say without any calculation that surely when say x is not zero, it doesn't have a pole?
– Karl
Nov 24 at 21:08
2
2
Because if $z$ is some coordinate on a curve, the poles of $frac{f}{g}dz$ are by definition the zeroes of $g$. In your case, since $omega$ can be written as $frac{dy}{4x^3}$ this shows that if $x neq 0$ there is no poles.
– Nicolas Hemelsoet
Nov 24 at 21:12
Because if $z$ is some coordinate on a curve, the poles of $frac{f}{g}dz$ are by definition the zeroes of $g$. In your case, since $omega$ can be written as $frac{dy}{4x^3}$ this shows that if $x neq 0$ there is no poles.
– Nicolas Hemelsoet
Nov 24 at 21:12
add a comment |
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2
Clearly this form has no poles when $x neq 0$ and when $y neq 0$. Moreover, $(0,0)$ is not on your curve so you can conclude that there is no poles.
– Nicolas Hemelsoet
Nov 24 at 19:24
Thanks for your answer Nicolas. What makes you say without any calculation that surely when say x is not zero, it doesn't have a pole?
– Karl
Nov 24 at 21:08
2
Because if $z$ is some coordinate on a curve, the poles of $frac{f}{g}dz$ are by definition the zeroes of $g$. In your case, since $omega$ can be written as $frac{dy}{4x^3}$ this shows that if $x neq 0$ there is no poles.
– Nicolas Hemelsoet
Nov 24 at 21:12