Show that this linear projection with center is an isomorphism
$begingroup$
Let $n in mathbb{N}$ and $f,g in k[x_1,x_2]$ homogenous and coprime polynomials where $deg (f)=n$ and $deg(g)=n-1$. Let $C$ be the variety $V(f-x_0g) subset mathbb{P}^2$. Show that the linear projection $phi : mathbb{P}^2 to mathbb{P}^1$ with center $z = (1:0:0)$ is an isomorphism between $C backslash { z }$ and an open subset $U subset mathbb{P}^1$.
I am trying to solve this exercise right now. I followed a hint that said I should prove that $C$ is irreducible first, which I have done by proving that $f-x_0g$ is irreducible (by reduction). Now I am stuck and would appreciate some help. I am not that familiar with morphisms of varieties and am not sure how to tackle the problem.
Thx in advance!
Edit: I am not so sure anymore why $C$ is irreducible. Can't I just say that the polynomial $f-x_0g in k[x_1,x_2][x_0]$ has degree $1$ and is therefore irreducible?
algebraic-geometry projective-space projection
asked Jan 20 at 19:02
Johny HunterJohny Hunter
10510
$endgroup$
add a comment |
$begingroup$
Let $n in mathbb{N}$ and $f,g in k[x_1,x_2]$ homogenous and coprime polynomials where $deg (f)=n$ and $deg(g)=n-1$. Let $C$ be the variety $V(f-x_0g) subset mathbb{P}^2$. Show that the linear projection $phi : mathbb{P}^2 to mathbb{P}^1$ with center $z = (1:0:0)$ is an isomorphism between $C backslash { z }$ and an open subset $U subset mathbb{P}^1$.
I am trying to solve this exercise right now. I followed a hint that said I should prove that $C$ is irreducible first, which I have done by proving that $f-x_0g$ is irreducible (by reduction). Now I am stuck and would appreciate some help. I am not that familiar with morphisms of varieties and am not sure how to tackle the problem.
Thx in advance!
Edit: I am not so sure anymore why $C$ is irreducible. Can't I just say that the polynomial $f-x_0g in k[x_1,x_2][x_0]$ has degree $1$ and is therefore irreducible?
algebraic-geometry projective-space projection
asked Jan 20 at 19:02
Johny HunterJohny Hunter
10510
$endgroup$
add a comment |
$begingroup$
Let $n in mathbb{N}$ and $f,g in k[x_1,x_2]$ homogenous and coprime polynomials where $deg (f)=n$ and $deg(g)=n-1$. Let $C$ be the variety $V(f-x_0g) subset mathbb{P}^2$. Show that the linear projection $phi : mathbb{P}^2 to mathbb{P}^1$ with center $z = (1:0:0)$ is an isomorphism between $C backslash { z }$ and an open subset $U subset mathbb{P}^1$.
I am trying to solve this exercise right now. I followed a hint that said I should prove that $C$ is irreducible first, which I have done by proving that $f-x_0g$ is irreducible (by reduction). Now I am stuck and would appreciate some help. I am not that familiar with morphisms of varieties and am not sure how to tackle the problem.
Thx in advance!
Edit: I am not so sure anymore why $C$ is irreducible. Can't I just say that the polynomial $f-x_0g in k[x_1,x_2][x_0]$ has degree $1$ and is therefore irreducible?
algebraic-geometry projective-space projection
asked Jan 20 at 19:02
Johny HunterJohny Hunter
10510
$endgroup$
Let $n in mathbb{N}$ and $f,g in k[x_1,x_2]$ homogenous and coprime polynomials where $deg (f)=n$ and $deg(g)=n-1$. Let $C$ be the variety $V(f-x_0g) subset mathbb{P}^2$. Show that the linear projection $phi : mathbb{P}^2 to mathbb{P}^1$ with center $z = (1:0:0)$ is an isomorphism between $C backslash { z }$ and an open subset $U subset mathbb{P}^1$.
I am trying to solve this exercise right now. I followed a hint that said I should prove that $C$ is irreducible first, which I have done by proving that $f-x_0g$ is irreducible (by reduction). Now I am stuck and would appreciate some help. I am not that familiar with morphisms of varieties and am not sure how to tackle the problem.
Thx in advance!
Edit: I am not so sure anymore why $C$ is irreducible. Can't I just say that the polynomial $f-x_0g in k[x_1,x_2][x_0]$ has degree $1$ and is therefore irreducible?
algebraic-geometry projective-space projection
algebraic-geometry projective-space projection
asked Jan 20 at 19:02
Johny HunterJohny Hunter
10510
asked Jan 20 at 19:02
Johny HunterJohny Hunter
10510
edited Jan 20 at 20:14
Johny Hunter
asked Jan 20 at 19:02
Johny HunterJohny Hunter
10510
asked Jan 20 at 19:02
Johny HunterJohny Hunter
10510
asked Jan 20 at 19:02
Johny HunterJohny Hunter
10510
10510
add a comment |
add a comment |
1 Answer
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$begingroup$
This linear projection is the map $phi : mathbb P^2 setminus z to mathbb P^1$ sending $[x_0 : x_1 : x_2] mapsto [x_1, x_2]$. So the image $phi(C setminus {z})$ consists of all $[x_1 : x_2 ] in mathbb P^1$ for which there exists some $x_0 in k$ such that $f(x_1, x_2) = x_0 g(x_1, g_2)$. In other words, $phi(C setminus { z })$ consists of all points in $mathbb P^1$ except for points where $g$ vanishes but $f$ doesn't vanish. But since $g$ and $f$ are coprime, there are no points where $g$ and $f$ both vanish. So $phi(C setminus { z })$ is simply the set of points in $mathbb P^1$ where $g$ doesn't vanish, i.e. $phi(C setminus { z }) = mathbb P^1 setminus V(g)$. This is an open set.
To show that $phi$ is an isomorphism, we need to exhibit an inverse morphism from $mathbb P^1 setminus V(g)$ to $C setminus {z }$. This inverse is given by $[x_1 : x_2 ] mapsto [tfrac{f(x_1, x_2) }{ g(x_1, x_2)} : x_1, : x_2]$.
As for irreducibility, the fact that $f - x_0 g in k[x_1, x_2][x_0]$ has degree $1$ means that the only way that $f - x_0 g$ can be written as a product is for there to exist $p, q, rin k[x_1, x_2]$ such that $f - x_0 g = (p + x_0 q)r$. But then, $r$ would be a common factor of $f$ and $g$, so since $f$ and $g$ are coprime, $r$ must be a constant. I'm not sure why irreducibility is relevant to this exercise though.
answered Jan 21 at 9:21
Kenny WongKenny Wong
19k21440
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$begingroup$
This linear projection is the map $phi : mathbb P^2 setminus z to mathbb P^1$ sending $[x_0 : x_1 : x_2] mapsto [x_1, x_2]$. So the image $phi(C setminus {z})$ consists of all $[x_1 : x_2 ] in mathbb P^1$ for which there exists some $x_0 in k$ such that $f(x_1, x_2) = x_0 g(x_1, g_2)$. In other words, $phi(C setminus { z })$ consists of all points in $mathbb P^1$ except for points where $g$ vanishes but $f$ doesn't vanish. But since $g$ and $f$ are coprime, there are no points where $g$ and $f$ both vanish. So $phi(C setminus { z })$ is simply the set of points in $mathbb P^1$ where $g$ doesn't vanish, i.e. $phi(C setminus { z }) = mathbb P^1 setminus V(g)$. This is an open set.
To show that $phi$ is an isomorphism, we need to exhibit an inverse morphism from $mathbb P^1 setminus V(g)$ to $C setminus {z }$. This inverse is given by $[x_1 : x_2 ] mapsto [tfrac{f(x_1, x_2) }{ g(x_1, x_2)} : x_1, : x_2]$.
As for irreducibility, the fact that $f - x_0 g in k[x_1, x_2][x_0]$ has degree $1$ means that the only way that $f - x_0 g$ can be written as a product is for there to exist $p, q, rin k[x_1, x_2]$ such that $f - x_0 g = (p + x_0 q)r$. But then, $r$ would be a common factor of $f$ and $g$, so since $f$ and $g$ are coprime, $r$ must be a constant. I'm not sure why irreducibility is relevant to this exercise though.
answered Jan 21 at 9:21
Kenny WongKenny Wong
19k21440
$endgroup$
add a comment |
$begingroup$
This linear projection is the map $phi : mathbb P^2 setminus z to mathbb P^1$ sending $[x_0 : x_1 : x_2] mapsto [x_1, x_2]$. So the image $phi(C setminus {z})$ consists of all $[x_1 : x_2 ] in mathbb P^1$ for which there exists some $x_0 in k$ such that $f(x_1, x_2) = x_0 g(x_1, g_2)$. In other words, $phi(C setminus { z })$ consists of all points in $mathbb P^1$ except for points where $g$ vanishes but $f$ doesn't vanish. But since $g$ and $f$ are coprime, there are no points where $g$ and $f$ both vanish. So $phi(C setminus { z })$ is simply the set of points in $mathbb P^1$ where $g$ doesn't vanish, i.e. $phi(C setminus { z }) = mathbb P^1 setminus V(g)$. This is an open set.
To show that $phi$ is an isomorphism, we need to exhibit an inverse morphism from $mathbb P^1 setminus V(g)$ to $C setminus {z }$. This inverse is given by $[x_1 : x_2 ] mapsto [tfrac{f(x_1, x_2) }{ g(x_1, x_2)} : x_1, : x_2]$.
As for irreducibility, the fact that $f - x_0 g in k[x_1, x_2][x_0]$ has degree $1$ means that the only way that $f - x_0 g$ can be written as a product is for there to exist $p, q, rin k[x_1, x_2]$ such that $f - x_0 g = (p + x_0 q)r$. But then, $r$ would be a common factor of $f$ and $g$, so since $f$ and $g$ are coprime, $r$ must be a constant. I'm not sure why irreducibility is relevant to this exercise though.
answered Jan 21 at 9:21
Kenny WongKenny Wong
19k21440
$endgroup$
add a comment |
$begingroup$
This linear projection is the map $phi : mathbb P^2 setminus z to mathbb P^1$ sending $[x_0 : x_1 : x_2] mapsto [x_1, x_2]$. So the image $phi(C setminus {z})$ consists of all $[x_1 : x_2 ] in mathbb P^1$ for which there exists some $x_0 in k$ such that $f(x_1, x_2) = x_0 g(x_1, g_2)$. In other words, $phi(C setminus { z })$ consists of all points in $mathbb P^1$ except for points where $g$ vanishes but $f$ doesn't vanish. But since $g$ and $f$ are coprime, there are no points where $g$ and $f$ both vanish. So $phi(C setminus { z })$ is simply the set of points in $mathbb P^1$ where $g$ doesn't vanish, i.e. $phi(C setminus { z }) = mathbb P^1 setminus V(g)$. This is an open set.
To show that $phi$ is an isomorphism, we need to exhibit an inverse morphism from $mathbb P^1 setminus V(g)$ to $C setminus {z }$. This inverse is given by $[x_1 : x_2 ] mapsto [tfrac{f(x_1, x_2) }{ g(x_1, x_2)} : x_1, : x_2]$.
As for irreducibility, the fact that $f - x_0 g in k[x_1, x_2][x_0]$ has degree $1$ means that the only way that $f - x_0 g$ can be written as a product is for there to exist $p, q, rin k[x_1, x_2]$ such that $f - x_0 g = (p + x_0 q)r$. But then, $r$ would be a common factor of $f$ and $g$, so since $f$ and $g$ are coprime, $r$ must be a constant. I'm not sure why irreducibility is relevant to this exercise though.
answered Jan 21 at 9:21
Kenny WongKenny Wong
19k21440
$endgroup$
This linear projection is the map $phi : mathbb P^2 setminus z to mathbb P^1$ sending $[x_0 : x_1 : x_2] mapsto [x_1, x_2]$. So the image $phi(C setminus {z})$ consists of all $[x_1 : x_2 ] in mathbb P^1$ for which there exists some $x_0 in k$ such that $f(x_1, x_2) = x_0 g(x_1, g_2)$. In other words, $phi(C setminus { z })$ consists of all points in $mathbb P^1$ except for points where $g$ vanishes but $f$ doesn't vanish. But since $g$ and $f$ are coprime, there are no points where $g$ and $f$ both vanish. So $phi(C setminus { z })$ is simply the set of points in $mathbb P^1$ where $g$ doesn't vanish, i.e. $phi(C setminus { z }) = mathbb P^1 setminus V(g)$. This is an open set.
To show that $phi$ is an isomorphism, we need to exhibit an inverse morphism from $mathbb P^1 setminus V(g)$ to $C setminus {z }$. This inverse is given by $[x_1 : x_2 ] mapsto [tfrac{f(x_1, x_2) }{ g(x_1, x_2)} : x_1, : x_2]$.
As for irreducibility, the fact that $f - x_0 g in k[x_1, x_2][x_0]$ has degree $1$ means that the only way that $f - x_0 g$ can be written as a product is for there to exist $p, q, rin k[x_1, x_2]$ such that $f - x_0 g = (p + x_0 q)r$. But then, $r$ would be a common factor of $f$ and $g$, so since $f$ and $g$ are coprime, $r$ must be a constant. I'm not sure why irreducibility is relevant to this exercise though.
answered Jan 21 at 9:21
Kenny WongKenny Wong
19k21440
answered Jan 21 at 9:21
Kenny WongKenny Wong
19k21440
answered Jan 21 at 9:21
Kenny WongKenny Wong
19k21440
answered Jan 21 at 9:21
Kenny WongKenny Wong
19k21440
19k21440
add a comment |
add a comment |
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