Over Silverman’s differential forms on curves
2
I’m reading Silverman’s Arithmetic of Ellipctic curves. In II.4 he gives an definition of differential form as the set of $dx$ for $x$ in $overline{K}(C)$. Taking the complex circle $x^2+y^2=1$ in $mathbb{P}^2$. By 4.2(a), $xdy$ is a differential form. By definition there exists $tinoverline{K}(C)$ such that $dt=xdy$.
If we integrate $sqrt{1-y^2}dy$, it yields an $arcsin$, which suggests that such a function $t$ wouldn’t be algebraic...
What I’m I missing here? What is the function $t$ such that $dt=xdy$?
algebraic-geometry elliptic-curves
asked Nov 22 '18 at 22:23
BUI Quang-TuBUI Quang-Tu
1,211718
|
show 2 more comments
2
I’m reading Silverman’s Arithmetic of Ellipctic curves. In II.4 he gives an definition of differential form as the set of $dx$ for $x$ in $overline{K}(C)$. Taking the complex circle $x^2+y^2=1$ in $mathbb{P}^2$. By 4.2(a), $xdy$ is a differential form. By definition there exists $tinoverline{K}(C)$ such that $dt=xdy$.
If we integrate $sqrt{1-y^2}dy$, it yields an $arcsin$, which suggests that such a function $t$ wouldn’t be algebraic...
What I’m I missing here? What is the function $t$ such that $dt=xdy$?
algebraic-geometry elliptic-curves
asked Nov 22 '18 at 22:23
BUI Quang-TuBUI Quang-Tu
1,211718
2
"By definition"? What definition is that?
– Lord Shark the Unknown
Nov 22 '18 at 22:29
1
@LordSharktheUnknown I think they mean the definition in the first sentence: Silverman defines differential forms to be things of the form $dt$ where $tinoverline{K}(C)$. So if $xdy$ is a differential form then it should have this form
– Alex Mathers
Nov 22 '18 at 22:32
4
Silverman says the $overline{K}$-vector space generated by the symbols $dx, x in overline{K}(C)$. He meant the $overline{K}(C)$-vector space ? ($dx$ is an exact meromorphic differential form for $x in overline{K}(C)$ and $x_2 dx$ is a (closed) meromorphic differential form for $x,x_2 in overline{K}(C)$) @LordSharktheUnknown
– reuns
Nov 22 '18 at 23:45
2
I agree with reuns: I think it's a typo and it should be the $overline{K}(C)$-vector space generated by $dx$ for $x in overline{K}(C)$. Indeed, without an action of $overline{K}(C)$, how can one even define $x dy$? Or state the Leibniz rule: $d(xy) = x dy + y dx$?
– André 3000
Nov 23 '18 at 0:16
1
Agree with reuns and André. The space of differential forms is a 1-dimensional space over $overline{K}(C)$. Observe that differentiating both sides of $x^2+y^2=1$ yields $x,dx+y,dy=0$. In other words, $$dy=-frac xy,dx.$$ All in line with $dx$ and $dy$ being linearly dependent over $K(C)$.
– Jyrki Lahtonen
Nov 26 '18 at 6:47
|
show 2 more comments
2
2
2
I’m reading Silverman’s Arithmetic of Ellipctic curves. In II.4 he gives an definition of differential form as the set of $dx$ for $x$ in $overline{K}(C)$. Taking the complex circle $x^2+y^2=1$ in $mathbb{P}^2$. By 4.2(a), $xdy$ is a differential form. By definition there exists $tinoverline{K}(C)$ such that $dt=xdy$.
If we integrate $sqrt{1-y^2}dy$, it yields an $arcsin$, which suggests that such a function $t$ wouldn’t be algebraic...
What I’m I missing here? What is the function $t$ such that $dt=xdy$?
algebraic-geometry elliptic-curves
asked Nov 22 '18 at 22:23
BUI Quang-TuBUI Quang-Tu
1,211718
I’m reading Silverman’s Arithmetic of Ellipctic curves. In II.4 he gives an definition of differential form as the set of $dx$ for $x$ in $overline{K}(C)$. Taking the complex circle $x^2+y^2=1$ in $mathbb{P}^2$. By 4.2(a), $xdy$ is a differential form. By definition there exists $tinoverline{K}(C)$ such that $dt=xdy$.
If we integrate $sqrt{1-y^2}dy$, it yields an $arcsin$, which suggests that such a function $t$ wouldn’t be algebraic...
What I’m I missing here? What is the function $t$ such that $dt=xdy$?
algebraic-geometry elliptic-curves
algebraic-geometry elliptic-curves
asked Nov 22 '18 at 22:23
BUI Quang-TuBUI Quang-Tu
1,211718
asked Nov 22 '18 at 22:23
BUI Quang-TuBUI Quang-Tu
1,211718
asked Nov 22 '18 at 22:23
BUI Quang-TuBUI Quang-Tu
1,211718
asked Nov 22 '18 at 22:23
BUI Quang-TuBUI Quang-Tu
1,211718
asked Nov 22 '18 at 22:23
BUI Quang-TuBUI Quang-Tu
1,211718
1,211718
2
"By definition"? What definition is that?
– Lord Shark the Unknown
Nov 22 '18 at 22:29
1
@LordSharktheUnknown I think they mean the definition in the first sentence: Silverman defines differential forms to be things of the form $dt$ where $tinoverline{K}(C)$. So if $xdy$ is a differential form then it should have this form
– Alex Mathers
Nov 22 '18 at 22:32
4
Silverman says the $overline{K}$-vector space generated by the symbols $dx, x in overline{K}(C)$. He meant the $overline{K}(C)$-vector space ? ($dx$ is an exact meromorphic differential form for $x in overline{K}(C)$ and $x_2 dx$ is a (closed) meromorphic differential form for $x,x_2 in overline{K}(C)$) @LordSharktheUnknown
– reuns
Nov 22 '18 at 23:45
2
I agree with reuns: I think it's a typo and it should be the $overline{K}(C)$-vector space generated by $dx$ for $x in overline{K}(C)$. Indeed, without an action of $overline{K}(C)$, how can one even define $x dy$? Or state the Leibniz rule: $d(xy) = x dy + y dx$?
– André 3000
Nov 23 '18 at 0:16
1
Agree with reuns and André. The space of differential forms is a 1-dimensional space over $overline{K}(C)$. Observe that differentiating both sides of $x^2+y^2=1$ yields $x,dx+y,dy=0$. In other words, $$dy=-frac xy,dx.$$ All in line with $dx$ and $dy$ being linearly dependent over $K(C)$.
– Jyrki Lahtonen
Nov 26 '18 at 6:47
|
show 2 more comments
2
"By definition"? What definition is that?
– Lord Shark the Unknown
Nov 22 '18 at 22:29
1
@LordSharktheUnknown I think they mean the definition in the first sentence: Silverman defines differential forms to be things of the form $dt$ where $tinoverline{K}(C)$. So if $xdy$ is a differential form then it should have this form
– Alex Mathers
Nov 22 '18 at 22:32
4
Silverman says the $overline{K}$-vector space generated by the symbols $dx, x in overline{K}(C)$. He meant the $overline{K}(C)$-vector space ? ($dx$ is an exact meromorphic differential form for $x in overline{K}(C)$ and $x_2 dx$ is a (closed) meromorphic differential form for $x,x_2 in overline{K}(C)$) @LordSharktheUnknown
– reuns
Nov 22 '18 at 23:45
2
I agree with reuns: I think it's a typo and it should be the $overline{K}(C)$-vector space generated by $dx$ for $x in overline{K}(C)$. Indeed, without an action of $overline{K}(C)$, how can one even define $x dy$? Or state the Leibniz rule: $d(xy) = x dy + y dx$?
– André 3000
Nov 23 '18 at 0:16
1
Agree with reuns and André. The space of differential forms is a 1-dimensional space over $overline{K}(C)$. Observe that differentiating both sides of $x^2+y^2=1$ yields $x,dx+y,dy=0$. In other words, $$dy=-frac xy,dx.$$ All in line with $dx$ and $dy$ being linearly dependent over $K(C)$.
– Jyrki Lahtonen
Nov 26 '18 at 6:47
2
2
"By definition"? What definition is that?
– Lord Shark the Unknown
Nov 22 '18 at 22:29
"By definition"? What definition is that?
– Lord Shark the Unknown
Nov 22 '18 at 22:29
1
1
@LordSharktheUnknown I think they mean the definition in the first sentence: Silverman defines differential forms to be things of the form $dt$ where $tinoverline{K}(C)$. So if $xdy$ is a differential form then it should have this form
– Alex Mathers
Nov 22 '18 at 22:32
@LordSharktheUnknown I think they mean the definition in the first sentence: Silverman defines differential forms to be things of the form $dt$ where $tinoverline{K}(C)$. So if $xdy$ is a differential form then it should have this form
– Alex Mathers
Nov 22 '18 at 22:32
4
4
Silverman says the $overline{K}$-vector space generated by the symbols $dx, x in overline{K}(C)$. He meant the $overline{K}(C)$-vector space ? ($dx$ is an exact meromorphic differential form for $x in overline{K}(C)$ and $x_2 dx$ is a (closed) meromorphic differential form for $x,x_2 in overline{K}(C)$) @LordSharktheUnknown
– reuns
Nov 22 '18 at 23:45
Silverman says the $overline{K}$-vector space generated by the symbols $dx, x in overline{K}(C)$. He meant the $overline{K}(C)$-vector space ? ($dx$ is an exact meromorphic differential form for $x in overline{K}(C)$ and $x_2 dx$ is a (closed) meromorphic differential form for $x,x_2 in overline{K}(C)$) @LordSharktheUnknown
– reuns
Nov 22 '18 at 23:45
2
2
I agree with reuns: I think it's a typo and it should be the $overline{K}(C)$-vector space generated by $dx$ for $x in overline{K}(C)$. Indeed, without an action of $overline{K}(C)$, how can one even define $x dy$? Or state the Leibniz rule: $d(xy) = x dy + y dx$?
– André 3000
Nov 23 '18 at 0:16
I agree with reuns: I think it's a typo and it should be the $overline{K}(C)$-vector space generated by $dx$ for $x in overline{K}(C)$. Indeed, without an action of $overline{K}(C)$, how can one even define $x dy$? Or state the Leibniz rule: $d(xy) = x dy + y dx$?
– André 3000
Nov 23 '18 at 0:16
1
1
Agree with reuns and André. The space of differential forms is a 1-dimensional space over $overline{K}(C)$. Observe that differentiating both sides of $x^2+y^2=1$ yields $x,dx+y,dy=0$. In other words, $$dy=-frac xy,dx.$$ All in line with $dx$ and $dy$ being linearly dependent over $K(C)$.
– Jyrki Lahtonen
Nov 26 '18 at 6:47
Agree with reuns and André. The space of differential forms is a 1-dimensional space over $overline{K}(C)$. Observe that differentiating both sides of $x^2+y^2=1$ yields $x,dx+y,dy=0$. In other words, $$dy=-frac xy,dx.$$ All in line with $dx$ and $dy$ being linearly dependent over $K(C)$.
– Jyrki Lahtonen
Nov 26 '18 at 6:47
|
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2
"By definition"? What definition is that?
– Lord Shark the Unknown
Nov 22 '18 at 22:29
1
@LordSharktheUnknown I think they mean the definition in the first sentence: Silverman defines differential forms to be things of the form $dt$ where $tinoverline{K}(C)$. So if $xdy$ is a differential form then it should have this form
– Alex Mathers
Nov 22 '18 at 22:32
4
Silverman says the $overline{K}$-vector space generated by the symbols $dx, x in overline{K}(C)$. He meant the $overline{K}(C)$-vector space ? ($dx$ is an exact meromorphic differential form for $x in overline{K}(C)$ and $x_2 dx$ is a (closed) meromorphic differential form for $x,x_2 in overline{K}(C)$) @LordSharktheUnknown
– reuns
Nov 22 '18 at 23:45
2
I agree with reuns: I think it's a typo and it should be the $overline{K}(C)$-vector space generated by $dx$ for $x in overline{K}(C)$. Indeed, without an action of $overline{K}(C)$, how can one even define $x dy$? Or state the Leibniz rule: $d(xy) = x dy + y dx$?
– André 3000
Nov 23 '18 at 0:16
1
Agree with reuns and André. The space of differential forms is a 1-dimensional space over $overline{K}(C)$. Observe that differentiating both sides of $x^2+y^2=1$ yields $x,dx+y,dy=0$. In other words, $$dy=-frac xy,dx.$$ All in line with $dx$ and $dy$ being linearly dependent over $K(C)$.
– Jyrki Lahtonen
Nov 26 '18 at 6:47