Mixing
seeing the differential dx/y on an elliptic curve as an element of the sheaf of differentials
$begingroup$
$newcommand{CC}{mathbb{C}}$
$newcommand{Spec}{operatorname{Spec}}$
It's a well known fact that every elliptic curve (say, over a field $k$) has a global holomorphic nowhere vanishing differential. If the curve is given by $y^2 = x^3 + ax + b$, then it's computed in Silverman (p33, example 4.6), that the differential $dx/y is holomorphic and nonvanishing. Hence, the line bundle of holomorphic differentials should be trivial, corresponding to a free rank 1 sheaf of relative differentials.
On the other hand, let $E$ be some elliptic curve over $k = CC$, say given by the equation $y^2 = x^3 - 1$. Then the existence of a nowhere vanishing holomorphic differential should imply that the sheaf of relative differentials $Omega_{E/k}$ is free of rank 1. In particular, we can restrict the sheaf $Omega_{E/k}$ to the affine locus given by the ring
$$R := CC[x,y]/(y^2-x^3+1)$$
The restriction of $Omega_{E/k}$ to $U := Spec R$ is then just the sheaf associated to the $R$ module
$$Omega_{R/k} = (Rdxoplus Rdy)/(2ydy - 3x^2dx)$$
The global sections of $Omega_{E/k}|_U$ should then just be the elements of $Omega_{R/k}$. In particular, $dx/y$ should be an element of the module $Omega_{R/k}$.
My question is: how do you see $dx/y$ as an element of $Omega_{R/k}$? Am I missing some algebra trick?
algebraic-geometry elliptic-curves
edited Jan 29 at 9:47
Martin Sleziak
44.9k10122277
asked May 5 '14 at 20:56
oxeimonoxeimon
9,10811127
$endgroup$
add a comment |
$begingroup$
$newcommand{CC}{mathbb{C}}$
$newcommand{Spec}{operatorname{Spec}}$
It's a well known fact that every elliptic curve (say, over a field $k$) has a global holomorphic nowhere vanishing differential. If the curve is given by $y^2 = x^3 + ax + b$, then it's computed in Silverman (p33, example 4.6), that the differential $dx/y is holomorphic and nonvanishing. Hence, the line bundle of holomorphic differentials should be trivial, corresponding to a free rank 1 sheaf of relative differentials.
On the other hand, let $E$ be some elliptic curve over $k = CC$, say given by the equation $y^2 = x^3 - 1$. Then the existence of a nowhere vanishing holomorphic differential should imply that the sheaf of relative differentials $Omega_{E/k}$ is free of rank 1. In particular, we can restrict the sheaf $Omega_{E/k}$ to the affine locus given by the ring
$$R := CC[x,y]/(y^2-x^3+1)$$
The restriction of $Omega_{E/k}$ to $U := Spec R$ is then just the sheaf associated to the $R$ module
$$Omega_{R/k} = (Rdxoplus Rdy)/(2ydy - 3x^2dx)$$
The global sections of $Omega_{E/k}|_U$ should then just be the elements of $Omega_{R/k}$. In particular, $dx/y$ should be an element of the module $Omega_{R/k}$.
My question is: how do you see $dx/y$ as an element of $Omega_{R/k}$? Am I missing some algebra trick?
algebraic-geometry elliptic-curves
edited Jan 29 at 9:47
Martin Sleziak
44.9k10122277
asked May 5 '14 at 20:56
oxeimonoxeimon
9,10811127
$endgroup$
1
$begingroup$
The point of the global differential is that it changes forms when you change charts. Is that a chart where $frac{dx}{y}$ makes sense?
$endgroup$
– Alex Youcis
May 5 '14 at 21:32
1
$begingroup$
I don't know, but $Omega_{E/k}$ is quasi-coherent, so over $U$ it's just the sheaf associated to the $R$-module $Omega_{R/k}$, which means that global sections are literally just elements of the module... In fact I'd be happy to even find an $R$-basis for $Omega_{R/k}$.
$endgroup$
– oxeimon
May 5 '14 at 21:50
add a comment |
$begingroup$
$newcommand{CC}{mathbb{C}}$
$newcommand{Spec}{operatorname{Spec}}$
It's a well known fact that every elliptic curve (say, over a field $k$) has a global holomorphic nowhere vanishing differential. If the curve is given by $y^2 = x^3 + ax + b$, then it's computed in Silverman (p33, example 4.6), that the differential $dx/y is holomorphic and nonvanishing. Hence, the line bundle of holomorphic differentials should be trivial, corresponding to a free rank 1 sheaf of relative differentials.
On the other hand, let $E$ be some elliptic curve over $k = CC$, say given by the equation $y^2 = x^3 - 1$. Then the existence of a nowhere vanishing holomorphic differential should imply that the sheaf of relative differentials $Omega_{E/k}$ is free of rank 1. In particular, we can restrict the sheaf $Omega_{E/k}$ to the affine locus given by the ring
$$R := CC[x,y]/(y^2-x^3+1)$$
The restriction of $Omega_{E/k}$ to $U := Spec R$ is then just the sheaf associated to the $R$ module
$$Omega_{R/k} = (Rdxoplus Rdy)/(2ydy - 3x^2dx)$$
The global sections of $Omega_{E/k}|_U$ should then just be the elements of $Omega_{R/k}$. In particular, $dx/y$ should be an element of the module $Omega_{R/k}$.
My question is: how do you see $dx/y$ as an element of $Omega_{R/k}$? Am I missing some algebra trick?
algebraic-geometry elliptic-curves
edited Jan 29 at 9:47
Martin Sleziak
44.9k10122277
asked May 5 '14 at 20:56
oxeimonoxeimon
9,10811127
$endgroup$
$newcommand{CC}{mathbb{C}}$
$newcommand{Spec}{operatorname{Spec}}$
It's a well known fact that every elliptic curve (say, over a field $k$) has a global holomorphic nowhere vanishing differential. If the curve is given by $y^2 = x^3 + ax + b$, then it's computed in Silverman (p33, example 4.6), that the differential $dx/y is holomorphic and nonvanishing. Hence, the line bundle of holomorphic differentials should be trivial, corresponding to a free rank 1 sheaf of relative differentials.
On the other hand, let $E$ be some elliptic curve over $k = CC$, say given by the equation $y^2 = x^3 - 1$. Then the existence of a nowhere vanishing holomorphic differential should imply that the sheaf of relative differentials $Omega_{E/k}$ is free of rank 1. In particular, we can restrict the sheaf $Omega_{E/k}$ to the affine locus given by the ring
$$R := CC[x,y]/(y^2-x^3+1)$$
The restriction of $Omega_{E/k}$ to $U := Spec R$ is then just the sheaf associated to the $R$ module
$$Omega_{R/k} = (Rdxoplus Rdy)/(2ydy - 3x^2dx)$$
The global sections of $Omega_{E/k}|_U$ should then just be the elements of $Omega_{R/k}$. In particular, $dx/y$ should be an element of the module $Omega_{R/k}$.
My question is: how do you see $dx/y$ as an element of $Omega_{R/k}$? Am I missing some algebra trick?
algebraic-geometry elliptic-curves
algebraic-geometry elliptic-curves
edited Jan 29 at 9:47
Martin Sleziak
44.9k10122277
asked May 5 '14 at 20:56
oxeimonoxeimon
9,10811127
edited Jan 29 at 9:47
Martin Sleziak
44.9k10122277
asked May 5 '14 at 20:56
oxeimonoxeimon
9,10811127
edited Jan 29 at 9:47
Martin Sleziak
44.9k10122277
edited Jan 29 at 9:47
Martin Sleziak
44.9k10122277
edited Jan 29 at 9:47
Martin Sleziak
44.9k10122277
44.9k10122277
asked May 5 '14 at 20:56
oxeimonoxeimon
9,10811127
asked May 5 '14 at 20:56
oxeimonoxeimon
9,10811127
asked May 5 '14 at 20:56
oxeimonoxeimon
9,10811127
9,10811127
1
$begingroup$
The point of the global differential is that it changes forms when you change charts. Is that a chart where $frac{dx}{y}$ makes sense?
$endgroup$
– Alex Youcis
May 5 '14 at 21:32
1
$begingroup$
I don't know, but $Omega_{E/k}$ is quasi-coherent, so over $U$ it's just the sheaf associated to the $R$-module $Omega_{R/k}$, which means that global sections are literally just elements of the module... In fact I'd be happy to even find an $R$-basis for $Omega_{R/k}$.
$endgroup$
– oxeimon
May 5 '14 at 21:50
add a comment |
1
$begingroup$
The point of the global differential is that it changes forms when you change charts. Is that a chart where $frac{dx}{y}$ makes sense?
$endgroup$
– Alex Youcis
May 5 '14 at 21:32
1
$begingroup$
I don't know, but $Omega_{E/k}$ is quasi-coherent, so over $U$ it's just the sheaf associated to the $R$-module $Omega_{R/k}$, which means that global sections are literally just elements of the module... In fact I'd be happy to even find an $R$-basis for $Omega_{R/k}$.
$endgroup$
– oxeimon
May 5 '14 at 21:50
1
1
$begingroup$
The point of the global differential is that it changes forms when you change charts. Is that a chart where $frac{dx}{y}$ makes sense?
$endgroup$
– Alex Youcis
May 5 '14 at 21:32
$begingroup$
The point of the global differential is that it changes forms when you change charts. Is that a chart where $frac{dx}{y}$ makes sense?
$endgroup$
– Alex Youcis
May 5 '14 at 21:32
1
1
$begingroup$
I don't know, but $Omega_{E/k}$ is quasi-coherent, so over $U$ it's just the sheaf associated to the $R$-module $Omega_{R/k}$, which means that global sections are literally just elements of the module... In fact I'd be happy to even find an $R$-basis for $Omega_{R/k}$.
$endgroup$
– oxeimon
May 5 '14 at 21:50
$begingroup$
I don't know, but $Omega_{E/k}$ is quasi-coherent, so over $U$ it's just the sheaf associated to the $R$-module $Omega_{R/k}$, which means that global sections are literally just elements of the module... In fact I'd be happy to even find an $R$-basis for $Omega_{R/k}$.
$endgroup$
– oxeimon
May 5 '14 at 21:50
add a comment |
1 Answer
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$begingroup$
$newcommand{Spec}{operatorname{Spec}}$So here's the answer. Since $dx/y$ is defined on $D(y)$, and over $D(xy)$, $dx/y = 2dy/3x^2$, which is defined on $D(x)$, and since $D(x),D(y)$ cover $Spec R$, they must glue to a global section, ie, an element of $Omega_{R/k}$.
Ie, we want to find an element (which we now know exists!) of $Omega_{R/k} = (Rdxoplus Rdy)/(2ydy-3x^2dx)$ which gets mapped to (under the localization map) to $dy/x$ on $D(y)$.
Ie, you want to find $f,gin R$ such that $fdx + gdy = dx/y$ in $Omega_{R/k}|_{D(y)}$, which is to say that there exists some $t = y^k$ such that
$$t(y(fdx+gdy) - dx) = 0qquadtext{in $Omega_{R/k}$}$$
Equivalently, $t((yf-1)dx + ygdy)$ is a multiple of $2ydy-3x^2dx$. Using the relation $y^2 = x^3-1$, we may choose $f = -y, g = frac{2}{3}x, t = 1$, and we see that
$$(-y^2-1)dx + frac{2}{3}xydy = -x^3dx + frac{2}{3}xydy = frac{x}{3}(2ydy - 3x^2dx) = 0$$
edited Jan 29 at 9:47
Martin Sleziak
44.9k10122277
answered May 6 '14 at 18:22
oxeimonoxeimon
9,10811127
$endgroup$
add a comment |
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$begingroup$
$newcommand{Spec}{operatorname{Spec}}$So here's the answer. Since $dx/y$ is defined on $D(y)$, and over $D(xy)$, $dx/y = 2dy/3x^2$, which is defined on $D(x)$, and since $D(x),D(y)$ cover $Spec R$, they must glue to a global section, ie, an element of $Omega_{R/k}$.
Ie, we want to find an element (which we now know exists!) of $Omega_{R/k} = (Rdxoplus Rdy)/(2ydy-3x^2dx)$ which gets mapped to (under the localization map) to $dy/x$ on $D(y)$.
Ie, you want to find $f,gin R$ such that $fdx + gdy = dx/y$ in $Omega_{R/k}|_{D(y)}$, which is to say that there exists some $t = y^k$ such that
$$t(y(fdx+gdy) - dx) = 0qquadtext{in $Omega_{R/k}$}$$
Equivalently, $t((yf-1)dx + ygdy)$ is a multiple of $2ydy-3x^2dx$. Using the relation $y^2 = x^3-1$, we may choose $f = -y, g = frac{2}{3}x, t = 1$, and we see that
$$(-y^2-1)dx + frac{2}{3}xydy = -x^3dx + frac{2}{3}xydy = frac{x}{3}(2ydy - 3x^2dx) = 0$$
edited Jan 29 at 9:47
Martin Sleziak
44.9k10122277
answered May 6 '14 at 18:22
oxeimonoxeimon
9,10811127
$endgroup$
add a comment |
$begingroup$
$newcommand{Spec}{operatorname{Spec}}$So here's the answer. Since $dx/y$ is defined on $D(y)$, and over $D(xy)$, $dx/y = 2dy/3x^2$, which is defined on $D(x)$, and since $D(x),D(y)$ cover $Spec R$, they must glue to a global section, ie, an element of $Omega_{R/k}$.
Ie, we want to find an element (which we now know exists!) of $Omega_{R/k} = (Rdxoplus Rdy)/(2ydy-3x^2dx)$ which gets mapped to (under the localization map) to $dy/x$ on $D(y)$.
Ie, you want to find $f,gin R$ such that $fdx + gdy = dx/y$ in $Omega_{R/k}|_{D(y)}$, which is to say that there exists some $t = y^k$ such that
$$t(y(fdx+gdy) - dx) = 0qquadtext{in $Omega_{R/k}$}$$
Equivalently, $t((yf-1)dx + ygdy)$ is a multiple of $2ydy-3x^2dx$. Using the relation $y^2 = x^3-1$, we may choose $f = -y, g = frac{2}{3}x, t = 1$, and we see that
$$(-y^2-1)dx + frac{2}{3}xydy = -x^3dx + frac{2}{3}xydy = frac{x}{3}(2ydy - 3x^2dx) = 0$$
edited Jan 29 at 9:47
Martin Sleziak
44.9k10122277
answered May 6 '14 at 18:22
oxeimonoxeimon
9,10811127
$endgroup$
add a comment |
$begingroup$
$newcommand{Spec}{operatorname{Spec}}$So here's the answer. Since $dx/y$ is defined on $D(y)$, and over $D(xy)$, $dx/y = 2dy/3x^2$, which is defined on $D(x)$, and since $D(x),D(y)$ cover $Spec R$, they must glue to a global section, ie, an element of $Omega_{R/k}$.
Ie, we want to find an element (which we now know exists!) of $Omega_{R/k} = (Rdxoplus Rdy)/(2ydy-3x^2dx)$ which gets mapped to (under the localization map) to $dy/x$ on $D(y)$.
Ie, you want to find $f,gin R$ such that $fdx + gdy = dx/y$ in $Omega_{R/k}|_{D(y)}$, which is to say that there exists some $t = y^k$ such that
$$t(y(fdx+gdy) - dx) = 0qquadtext{in $Omega_{R/k}$}$$
Equivalently, $t((yf-1)dx + ygdy)$ is a multiple of $2ydy-3x^2dx$. Using the relation $y^2 = x^3-1$, we may choose $f = -y, g = frac{2}{3}x, t = 1$, and we see that
$$(-y^2-1)dx + frac{2}{3}xydy = -x^3dx + frac{2}{3}xydy = frac{x}{3}(2ydy - 3x^2dx) = 0$$
edited Jan 29 at 9:47
Martin Sleziak
44.9k10122277
answered May 6 '14 at 18:22
oxeimonoxeimon
9,10811127
$endgroup$
$newcommand{Spec}{operatorname{Spec}}$So here's the answer. Since $dx/y$ is defined on $D(y)$, and over $D(xy)$, $dx/y = 2dy/3x^2$, which is defined on $D(x)$, and since $D(x),D(y)$ cover $Spec R$, they must glue to a global section, ie, an element of $Omega_{R/k}$.
Ie, we want to find an element (which we now know exists!) of $Omega_{R/k} = (Rdxoplus Rdy)/(2ydy-3x^2dx)$ which gets mapped to (under the localization map) to $dy/x$ on $D(y)$.
Ie, you want to find $f,gin R$ such that $fdx + gdy = dx/y$ in $Omega_{R/k}|_{D(y)}$, which is to say that there exists some $t = y^k$ such that
$$t(y(fdx+gdy) - dx) = 0qquadtext{in $Omega_{R/k}$}$$
Equivalently, $t((yf-1)dx + ygdy)$ is a multiple of $2ydy-3x^2dx$. Using the relation $y^2 = x^3-1$, we may choose $f = -y, g = frac{2}{3}x, t = 1$, and we see that
$$(-y^2-1)dx + frac{2}{3}xydy = -x^3dx + frac{2}{3}xydy = frac{x}{3}(2ydy - 3x^2dx) = 0$$
edited Jan 29 at 9:47
Martin Sleziak
44.9k10122277
answered May 6 '14 at 18:22
oxeimonoxeimon
9,10811127
edited Jan 29 at 9:47
Martin Sleziak
44.9k10122277
edited Jan 29 at 9:47
Martin Sleziak
44.9k10122277
edited Jan 29 at 9:47
Martin Sleziak
44.9k10122277
44.9k10122277
answered May 6 '14 at 18:22
oxeimonoxeimon
9,10811127
answered May 6 '14 at 18:22
oxeimonoxeimon
9,10811127
answered May 6 '14 at 18:22
oxeimonoxeimon
9,10811127
9,10811127
add a comment |
add a comment |
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$begingroup$
The point of the global differential is that it changes forms when you change charts. Is that a chart where $frac{dx}{y}$ makes sense?
$endgroup$
– Alex Youcis
May 5 '14 at 21:32
1
$begingroup$
I don't know, but $Omega_{E/k}$ is quasi-coherent, so over $U$ it's just the sheaf associated to the $R$-module $Omega_{R/k}$, which means that global sections are literally just elements of the module... In fact I'd be happy to even find an $R$-basis for $Omega_{R/k}$.
$endgroup$
– oxeimon
May 5 '14 at 21:50