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edited Nov 15 2011 at 13:48 |
The classical Bézout theorem works for curves in the projective space $\mathbb{P}^2$.
In the case of weighted homogeneous polynomials one needs a Bézout theorem in the weighted projective plane $\mathbb{P}^2(w_1, w_2, w_3)$. Such a result can be found, for instance, in the paper by Bartolo, Martin-Morales and Ortigas-Galindo Q-resolutions and intersection numbers, Section 5. It turns out that the intersection number of two curves of equation $f=0$ and $g=0$ in $\mathbb{P}^2(w_1, w_2, w_3)$ is given by
$$\frac{1}{w_1 w_2 w_3} \deg_{\omega}(f) \deg_{\omega}(g),$$
where $\deg_{\omega}$ denotes the weighted degree, see also auniket's comment below.
Since $\mathbb{P}^2(w_1, w_2, w_3)$ is a singular variety (with cyclic quotient singularities, hence $\mathbb{Q}$-factorial), this formula makes sense only as an intersection formula for $\mathbb{Q}$-divisors. In fact, if the zero locus of your polynomials intersect the singular locus of the weighted projective plane, it may happen that the corresponding Weil divisors are not Cartier. Consequently, one can obtain a rational intersection number, insted of an integer one.
For instance, let us consider $\mathbb{P}(1,1,2)$, which is isomorphic to a quadric cone in $\mathbb{P}^3$. If $x, y, z$ are the weighted homogeneous coordinates, the ruling of the cone is generated by $x=0$ and $y=0$; furthermore, the intersection numbers for any two curves $L_1$, $L_2 \subset \mathbb{P}(1,1,2)$ of equation $$\lambda_1x+\mu_1y=0, \quad \lambda_2x+\mu_2y=0 \quad (\lambda_i, \mu_i \in \mathbb{C})$$
is equal to $\frac{1}{ \frac{1 \cdot 1}{ 1 \cdot 1 \cdot 2} = \frac{1}{2}$.
This happens because a line $L$ in the ruling is not a Cartier divisor (since it passes through the vertex ) $[0:0:1]$) but $2L$ is Cartier, being linearly equivalent to a conic, i.e. to a hyperplane section of the cone. Now two hyperplane sections intersect in two points, so we have $(2L_1)(2L_2)=2$, that is $L_1L_2=\frac{1}{2}$.
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edited Nov 14 2011 at 23:44 |
The classical Bézout theorem works for curves in the projective space $\mathbb{P}^2$.
In your the case , you have of weighted homogeneous polynomials , so you need one needs a Bézout theorem in the weighted projective plane $\mathbb{P}^2(w_1, w_2, w_3)$. This actually exists, and Such a version of it result can be found, for instance, in the paper by Bartolo, Martin-Morales and Ortigas-Galindo Q-resolutions and intersection numbers, Section 5. In fact one obtain It turns out that the intersection number for of two curves of equation $f=0$ and $g=0$ in $\mathbb{P}^2(w_1, w_2, w_3)$ is given by
$$\frac{1}{w_1 w_2 w_3} \deg_{\omega}(f) \deg_{\omega}(g),$$
where $\deg_{\omega}$ denotes the weighted degree, see also auniket's comment below.
Since $\mathbb{P}^2(w_1, w_2, w_3)$ is a singular variety (with cyclic quotient singularities, hence $\mathbb{Q}$-factorial), this formula makes sense only as an intersection formula for $\mathbb{Q}$-divisors. In fact, if the zero locus of your polynomials intersect the singular locus of the weighted projective plane, it may happen that the corresponding Weil divisors are not Cartier. Consequently, one can obtain a rational intersection number, insted of an integer one.
For instance, let us consider $\mathbb{P}(1,1,2)$, which is isomorphic to a quadric cone in $\mathbb{P}^3$. If $x, y, z$ are the weighted homogeneous coordinates, the ruling of the cone is generated by $x=0$ and $y=0$; furthermore, the intersection numbers for any two curves $L_1$, $L_2 \subset \mathbb{P}(1,1,2)$ of equation $$\lambda_1x+\mu_1y=0, \quad \lambda_2x+\mu_2y=0 \quad (\lambda_i, \mu_i \in \mathbb{C})$$
is equal to $\frac{1}{ 1 \cdot 1 \cdot 2} = \frac{1}{2}$.
This happens because a line $L$ in the ruling is not a Cartier divisor (since it passes through the vertex) but $2L$ is Cartier, being linearly equivalent to a conic, i.e. to a hyperplane section of the cone. Since Now two hyperplane sections intersect in two points, so we have $(2L_1)(2L_2)=2$, so that is $L_1L_2=\frac{1}{2}$.
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edited Nov 14 2011 at 23:19 |
The classical Bézout theorem works for curves in the projective space $\mathbb{P}^2$.
In your case, you have weighted homogeneous polynomials, so you need a Bézout theorem in the weighted projective plane $\mathbb{P}^2(w_1, w_2, w_3)$. This actually exists, and a version of it can be found in the paper by Bartolo, Martin-Morales and Ortigas-Galindo Q-resolutions and intersection numbers, Section 5. In fact one obtain that the intersection number for two curves of equation $f=0$ and $g=0$ is given by
$$\frac{1}{w_1 w_2 w_3} \deg_{\omega}(f) \deg_{\omega}(g),$$
where $\deg_{\omega}$ denotes the weighted degree, see also auniket's comment below.
Since $\mathbb{P}^2(w_1, w_2, w_3)$ is a singular variety (with cyclic quotient singularities, hence $\mathbb{Q}$-factorial), this formula makes sense only as an intersection formula for $\mathbb{Q}$-divisors. In fact, if the zero locus of your polynomials intersect the singular locus of the weighted projective plane, it may happen that the corresponding Weil divisors are not Cartier. Consequently, one can obtain a rational intersection number, insted of an integer one.
For instance, let us consider $\mathbb{P}(1,1,2)$, which is isomorphic to a quadric cone in $\mathbb{P}^3$. If $x, y, z$ are the weighted homogeneous coordinates, the ruling of the cone is generated by $x=0$ and $y=0$; therefore furthermore, the intersection numbers for any two curves $L_1$, $L_2 \subset \mathbb{P}(1,1,2)$ of equation $$\lambda_1x+\mu_1y=0, \quad \lambda_2x+\mu_2y=0 \quad (\lambda_i, \mu_i \in \mathbb{C})$$
is equal to $1/2$.\frac{1}{ 1 \cdot 1 \cdot 2} = \frac{1}{2}$.
This happens because a line $L$ in the ruling is not a Cartier divisor (since it passes through the vertex) but $2L$ is Cartier, being linearly equivalent to a conic, i.e. to a hyperplane section of the cone. Since two hyperplane sections intersect in two points, we have $(2L_1)(2L_2)=2$, so $L_1L_2=1/2$.L_1L_2=\frac{1}{2}$.
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edited Nov 14 2011 at 23:13 |
The classical Bézout theorem works for curves in the projective space $\mathbb{P}^2$.
In your case, you have weighted homogeneous polynomials, so you need a Bézout theorem in the weighted projective plane $\mathbb{P}^2(w_1, w_2, w_3)$. This actually exists, and a version of it can be found in the paper by Bartolo, Martin-Morales and Ortigas-Galindo Q-resolutions and intersection numbers, Section 5. In fact one obtain that the intersection number for two curves of equation $f=0$ and $g=0$ is given by
$$\frac{1}{w_1 w_2 w_3} \deg_{\omega}(f) \deg_{\omega}(g),$$
where $\deg_{\omega}$ denotes the weighted degree, see also auniket's comment below.
Since $\mathbb{P}^2(w_1, w_2, w_3)$ is a singular variety (with cyclic quotient singularities, hence $\mathbb{Q}$-factorial), this formula makes sense only as an intersection formula for $\mathbb{Q}$-divisors. In fact, if the zero locus of your polynomials intersect the singular locus of the weighted projective plane, it may happen that the corresponding Weil divisors are not Cartier. Consequently, one can obtain a rational intersection number, insted of an integer one.
For instance, let us consider $\mathbb{P}(1,1,2)$, which is isomorphic to a quadric cone in $\mathbb{P}^3$. If $x, y, z$ are the weighted homogeneous coordinates, the ruling of the cone is generated by $x=0$ and $y=0$; therefore the intersection numbers for any two curves $L_1$, $L_2 \subset \mathbb{P}(1,1,2)$ of equation $$\lambda_1x+\mu_1y=0, \quad \lambda_2x+\mu_2y=0 \quad (\lambda_i, \mu_i \in \mathbb{C})$$
is the rational number equal to $1/2$.
This happens because a line $L$ in the ruling is not a Cartier divisor (since it passes through the vertex) but $2L$ is Cartier, being linearly equivalent to a conic, i.e. to a hyperplane section of the cone. Since two hyperplane sections intersect in two points, we have $(2L_1)(2L_2)=2$, so $L_1L_2=1/2$.
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edited Nov 14 2011 at 23:07 |
The classical Bézout theorem works for curves in the projective space $\mathbb{P}^2$.
In your case, you have weighted homogeneous polynomials, so you need a Bézout theorem in the weighted projective plane $\mathbb{P}^2(w_1, w_2, w_3)$. This actually exists, and a version of it can be found in the paper by Bartolo, Martin-Morales and Ortigas-Galindo Q-resolutions and intersection numbers, Section 5. In fact one obtain that the intersection number for two curves of equation $f=0$ and $g=0$ is given by
$$\frac{1}{w_1 w_2 w_3} \deg_{\omega}(f) \deg_{\omega}(g),$$
where $\deg_{\omega}$ denotes the weighted degree, see also auniket's comment below.
Since $\mathbb{P}^2(w_1, w_2, w_3)$ is a singular variety (with cyclic quotient singularities, hence $\mathbb{Q}$-factorial), this formula makes sense only as an intersection formula for $\mathbb{Q}$-divisors. In fact, if the zero locus of your polynomials intersect the singular locus of the weighted projective plane, it may happen that the corresponding Weil divisors are not Cartier. Consequently, one can obtain a rational intersection number, insted of an integer one.
For instance, let us consider $\mathbb{P}(1,1,2)$ \mathbb{P}(1,1,2)$, which is isomorphic to a quadric cone . in $\mathbb{P}^3$. If $x, y, z$ are the weighted homogeneous coordinates, the ruling of the cone is generated by $x=0$ and $y=0$; therefore one checks that the correct intersection numbers for any two curves in $\mathbb{P}(1,1,2)$ L_1$, $L_2 \subset \mathbb{P}(1,1,2)$ of equation $$\lambda_1x+\mu_1y=0, \quad \lambda_2x+\mu_2y=0 \quad (\lambda_i, \mu_i \in \mathbb{C})$$
is the rational number $1/2$.
This happens because a line $L$ in the ruling is not a Cartier divisor (since it passes through the vertex) but $2L$ is Cartier, being linearly equivalent to a conic(i.e, , i.e. to a hyperplane section)section of the cone. Since two conics in the cone hyperplane sections intersect in two points, we have $(2L_1)(2L_2)=2$, so $L_1L_2=1/2$.
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edited Nov 14 2011 at 22:21 |
The classical Bézout theorem works for curves in the projective space $\mathbb{P}^2$.
In your case, you have weighted homogeneous polynomials, so you need a Bézout theorem in the weighted projective plane $\mathbb{P}^2(w_1, w_2, w_3)$. This actually exists, and a version of it can be found in the paper by Bartolo, Martin-Morales and Ortigas-Galindo Q-resolutions and intersection numbers, Section 5.
Since $\mathbb{P}^2(w_1, w_2, w_3)$ is a singular variety (with cyclic quotient singularities, hence $\mathbb{Q}$-factorial), this formula makes sense only as an intersection formula for $\mathbb{Q}$-divisors. In fact, if the zero locus of your polynomials intersect the singular locus of the weighted projective plane, it may happen that the corresponding Weil divisors are not Cartier. Consequently, one can obtain a rational intersection number, insted of an integer one.
For instance, $\mathbb{P}(1,1,2)$ is isomorphic to a quadric cone. If $x, y, z$ are the weighted homogeneous coordinates, the ruling of the cone is generated by $x=0$ and $y=0$; therefore one checks that the correct intersection of numbers for any two lines curves in $\mathbb{P}(1,1,2)$ of equation $$\lambda_1x+\mu_1y=0, \quad \lambda_2x+\mu_2y=0 \quad (\lambda_i, \mu_i \in \mathbb{C})$$
is the ruling equals rational number $1/2$.
This happens because the a line $L$ in the ruling is not a Cartier divisor but $2L$ is Cartier, being linearly equivalent to a conic (i.e, to a hyperplane section), and . Since two conics in the cone intersect in two points., we have $(2L_1)(2L_2)=2$, so $L_1L_2=1/2$.
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answered Nov 14 2011 at 22:09 |
The classical Bézout theorem works for curves in the projective space $\mathbb{P}^2$.
In your case, you have weighted homogeneous polynomials, so you need a Bézout theorem in the weighted projective plane $\mathbb{P}^2(w_1, w_2, w_3)$. This actually exists, and a version of it can be found in the paper by Bartolo, Martin-Morales and Ortigas-Galindo Q-resolutions and intersection numbers, Section 5.
Since $\mathbb{P}^2(w_1, w_2, w_3)$ is a singular variety (with cyclic quotient singularities, hence $\mathbb{Q}$-factorial), this formula makes sense only as an intersection formula for $\mathbb{Q}$-divisors. In fact, if the zero locus of your polynomials intersect the singular locus of the weighted projective plane, it may happen that the corresponding Weil divisors are not Cartier. Consequently, one can obtain a rational intersection number, insted of an integer one.
For instance, $\mathbb{P}(1,1,2)$ is a quadric cone, and the intersection of two lines in the ruling equals $1/2$.
This happens because the a line $L$ is not a Cartier divisor but $2L$ is Cartier, being linearly equivalent to a conic (i.e, a hyperplane section), and two conics intersect in two points.
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