Hartshorne exercise 1.1 (c)
$begingroup$
I want to know how can I find the conditions where $A(W)$, the affine coordinate ring of a variety given by an irreducible quadratic polynomial in $k[x,y]$, is isomorphic to $A(V)$ or to $A(Z)$ where $V$ is a parabola defined by $y=x^2$ and $Z$ is the hyperbola given by $xy=1$. Thanks in advance for your answers.
algebraic-geometry
asked Aug 31 '17 at 20:39
Billy BonesBilly Bones
408
$endgroup$
add a comment |
$begingroup$
I want to know how can I find the conditions where $A(W)$, the affine coordinate ring of a variety given by an irreducible quadratic polynomial in $k[x,y]$, is isomorphic to $A(V)$ or to $A(Z)$ where $V$ is a parabola defined by $y=x^2$ and $Z$ is the hyperbola given by $xy=1$. Thanks in advance for your answers.
algebraic-geometry
asked Aug 31 '17 at 20:39
Billy BonesBilly Bones
408
$endgroup$
$begingroup$
$A(V) cong k[t]$ and $A(Z) cong k[t,t^{-1}]$
$endgroup$
– user171326
Aug 31 '17 at 21:07
$begingroup$
Well... Yes, but how is this related with a general quadratic polynomial? I mean, when is isomorphic with one of those affine coordinate rings?
$endgroup$
– Billy Bones
Aug 31 '17 at 21:13
$begingroup$
If $k$ is algebraically closed, I believe all the time, as $W$ is the complement of a line in $Bbb P^2$ which can be a parabola (your conic is tangent to the line) or an hyperbola (your conic intersects the line twice).
$endgroup$
– user171326
Aug 31 '17 at 21:16
$begingroup$
It is supposed I can't use the projective varieties because there aren't defined yet. I understand the point, but I want to read a more constructive proof of this fact, using the affine coordinate rings, if there is no problem. Thanks in advance.
$endgroup$
– Billy Bones
Aug 31 '17 at 21:19
add a comment |
$begingroup$
I want to know how can I find the conditions where $A(W)$, the affine coordinate ring of a variety given by an irreducible quadratic polynomial in $k[x,y]$, is isomorphic to $A(V)$ or to $A(Z)$ where $V$ is a parabola defined by $y=x^2$ and $Z$ is the hyperbola given by $xy=1$. Thanks in advance for your answers.
algebraic-geometry
asked Aug 31 '17 at 20:39
Billy BonesBilly Bones
408
$endgroup$
I want to know how can I find the conditions where $A(W)$, the affine coordinate ring of a variety given by an irreducible quadratic polynomial in $k[x,y]$, is isomorphic to $A(V)$ or to $A(Z)$ where $V$ is a parabola defined by $y=x^2$ and $Z$ is the hyperbola given by $xy=1$. Thanks in advance for your answers.
algebraic-geometry
algebraic-geometry
asked Aug 31 '17 at 20:39
Billy BonesBilly Bones
408
asked Aug 31 '17 at 20:39
Billy BonesBilly Bones
408
asked Aug 31 '17 at 20:39
Billy BonesBilly Bones
408
asked Aug 31 '17 at 20:39
Billy BonesBilly Bones
408
asked Aug 31 '17 at 20:39
Billy BonesBilly Bones
408
408
$begingroup$
$A(V) cong k[t]$ and $A(Z) cong k[t,t^{-1}]$
$endgroup$
– user171326
Aug 31 '17 at 21:07
$begingroup$
Well... Yes, but how is this related with a general quadratic polynomial? I mean, when is isomorphic with one of those affine coordinate rings?
$endgroup$
– Billy Bones
Aug 31 '17 at 21:13
$begingroup$
If $k$ is algebraically closed, I believe all the time, as $W$ is the complement of a line in $Bbb P^2$ which can be a parabola (your conic is tangent to the line) or an hyperbola (your conic intersects the line twice).
$endgroup$
– user171326
Aug 31 '17 at 21:16
$begingroup$
It is supposed I can't use the projective varieties because there aren't defined yet. I understand the point, but I want to read a more constructive proof of this fact, using the affine coordinate rings, if there is no problem. Thanks in advance.
$endgroup$
– Billy Bones
Aug 31 '17 at 21:19
add a comment |
$begingroup$
$A(V) cong k[t]$ and $A(Z) cong k[t,t^{-1}]$
$endgroup$
– user171326
Aug 31 '17 at 21:07
$begingroup$
Well... Yes, but how is this related with a general quadratic polynomial? I mean, when is isomorphic with one of those affine coordinate rings?
$endgroup$
– Billy Bones
Aug 31 '17 at 21:13
$begingroup$
If $k$ is algebraically closed, I believe all the time, as $W$ is the complement of a line in $Bbb P^2$ which can be a parabola (your conic is tangent to the line) or an hyperbola (your conic intersects the line twice).
$endgroup$
– user171326
Aug 31 '17 at 21:16
$begingroup$
It is supposed I can't use the projective varieties because there aren't defined yet. I understand the point, but I want to read a more constructive proof of this fact, using the affine coordinate rings, if there is no problem. Thanks in advance.
$endgroup$
– Billy Bones
Aug 31 '17 at 21:19
$begingroup$
$A(V) cong k[t]$ and $A(Z) cong k[t,t^{-1}]$
$endgroup$
– user171326
Aug 31 '17 at 21:07
$begingroup$
$A(V) cong k[t]$ and $A(Z) cong k[t,t^{-1}]$
$endgroup$
– user171326
Aug 31 '17 at 21:07
$begingroup$
Well... Yes, but how is this related with a general quadratic polynomial? I mean, when is isomorphic with one of those affine coordinate rings?
$endgroup$
– Billy Bones
Aug 31 '17 at 21:13
$begingroup$
Well... Yes, but how is this related with a general quadratic polynomial? I mean, when is isomorphic with one of those affine coordinate rings?
$endgroup$
– Billy Bones
Aug 31 '17 at 21:13
$begingroup$
If $k$ is algebraically closed, I believe all the time, as $W$ is the complement of a line in $Bbb P^2$ which can be a parabola (your conic is tangent to the line) or an hyperbola (your conic intersects the line twice).
$endgroup$
– user171326
Aug 31 '17 at 21:16
$begingroup$
If $k$ is algebraically closed, I believe all the time, as $W$ is the complement of a line in $Bbb P^2$ which can be a parabola (your conic is tangent to the line) or an hyperbola (your conic intersects the line twice).
$endgroup$
– user171326
Aug 31 '17 at 21:16
$begingroup$
It is supposed I can't use the projective varieties because there aren't defined yet. I understand the point, but I want to read a more constructive proof of this fact, using the affine coordinate rings, if there is no problem. Thanks in advance.
$endgroup$
– Billy Bones
Aug 31 '17 at 21:19
$begingroup$
It is supposed I can't use the projective varieties because there aren't defined yet. I understand the point, but I want to read a more constructive proof of this fact, using the affine coordinate rings, if there is no problem. Thanks in advance.
$endgroup$
– Billy Bones
Aug 31 '17 at 21:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Do you know the classification of quadrics using completing the square? For convenience let me assume that characteristic is not 2 and $k$ is algebraically closed. Then, a quadratic equation can be assumed to look as $x^2+a(y)x+b(y)$ and by completing squares (changing variables), you can assume it looks like $x^2+b(y)$ with $deg b(y)leq 2$. So, the polynomial looks like $x^2+ay^2+by+c$. Both $a,b$ can not be zero, since the polynomial is irreducible. Then $aY^2+by+c$ can be written as $-y^2+1$ or $y$ depending on whether $aneq 0$ or $a=0, bneq 0$, after changing the variable.So, the polynomial looks like $x^2-y^2=1$ or $x^2=y$. Use $x^2-y^2=(x-y)(x+y)$.
answered Sep 1 '17 at 0:27
MohanMohan
11.9k1817
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add a comment |
$begingroup$
Here's a solution that'll work for all characteristics - Factorize the degree $2$ homogeneous part into linear factors (can do this because algebraically closed). Now, if the linear factors are linearly dependent, w.l.o.g. change coordinates to make this linear factor the new $X$. The equation now becomes $X^2 + aX + bY + c$. By now making $aX+bY+c$ the new $Y$ ($b neq 0$), we get $X^2+Y$.
If the factors are linearly independent, change coordinates to make one of them $X$ and the other $Y$. The equation now becomes $XY + aX + bY + c$ which we can write as $(X+b)(Y+a)+d$. We can now change coordinates to make it $XY+d$ and so, we're done.
As an interesting aside, this indicates as expected that the general equation is a hyperbola since it corresponds to the linearly independent case here.
answered Dec 18 '18 at 12:51
Sridhar VenkateshSridhar Venkatesh
163
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add a comment |
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2 Answers
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2 Answers
2
active
oldest
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votes
$begingroup$
Do you know the classification of quadrics using completing the square? For convenience let me assume that characteristic is not 2 and $k$ is algebraically closed. Then, a quadratic equation can be assumed to look as $x^2+a(y)x+b(y)$ and by completing squares (changing variables), you can assume it looks like $x^2+b(y)$ with $deg b(y)leq 2$. So, the polynomial looks like $x^2+ay^2+by+c$. Both $a,b$ can not be zero, since the polynomial is irreducible. Then $aY^2+by+c$ can be written as $-y^2+1$ or $y$ depending on whether $aneq 0$ or $a=0, bneq 0$, after changing the variable.So, the polynomial looks like $x^2-y^2=1$ or $x^2=y$. Use $x^2-y^2=(x-y)(x+y)$.
answered Sep 1 '17 at 0:27
MohanMohan
11.9k1817
$endgroup$
add a comment |
$begingroup$
Do you know the classification of quadrics using completing the square? For convenience let me assume that characteristic is not 2 and $k$ is algebraically closed. Then, a quadratic equation can be assumed to look as $x^2+a(y)x+b(y)$ and by completing squares (changing variables), you can assume it looks like $x^2+b(y)$ with $deg b(y)leq 2$. So, the polynomial looks like $x^2+ay^2+by+c$. Both $a,b$ can not be zero, since the polynomial is irreducible. Then $aY^2+by+c$ can be written as $-y^2+1$ or $y$ depending on whether $aneq 0$ or $a=0, bneq 0$, after changing the variable.So, the polynomial looks like $x^2-y^2=1$ or $x^2=y$. Use $x^2-y^2=(x-y)(x+y)$.
answered Sep 1 '17 at 0:27
MohanMohan
11.9k1817
$endgroup$
add a comment |
$begingroup$
Do you know the classification of quadrics using completing the square? For convenience let me assume that characteristic is not 2 and $k$ is algebraically closed. Then, a quadratic equation can be assumed to look as $x^2+a(y)x+b(y)$ and by completing squares (changing variables), you can assume it looks like $x^2+b(y)$ with $deg b(y)leq 2$. So, the polynomial looks like $x^2+ay^2+by+c$. Both $a,b$ can not be zero, since the polynomial is irreducible. Then $aY^2+by+c$ can be written as $-y^2+1$ or $y$ depending on whether $aneq 0$ or $a=0, bneq 0$, after changing the variable.So, the polynomial looks like $x^2-y^2=1$ or $x^2=y$. Use $x^2-y^2=(x-y)(x+y)$.
answered Sep 1 '17 at 0:27
MohanMohan
11.9k1817
$endgroup$
Do you know the classification of quadrics using completing the square? For convenience let me assume that characteristic is not 2 and $k$ is algebraically closed. Then, a quadratic equation can be assumed to look as $x^2+a(y)x+b(y)$ and by completing squares (changing variables), you can assume it looks like $x^2+b(y)$ with $deg b(y)leq 2$. So, the polynomial looks like $x^2+ay^2+by+c$. Both $a,b$ can not be zero, since the polynomial is irreducible. Then $aY^2+by+c$ can be written as $-y^2+1$ or $y$ depending on whether $aneq 0$ or $a=0, bneq 0$, after changing the variable.So, the polynomial looks like $x^2-y^2=1$ or $x^2=y$. Use $x^2-y^2=(x-y)(x+y)$.
answered Sep 1 '17 at 0:27
MohanMohan
11.9k1817
answered Sep 1 '17 at 0:27
MohanMohan
11.9k1817
answered Sep 1 '17 at 0:27
MohanMohan
11.9k1817
answered Sep 1 '17 at 0:27
MohanMohan
11.9k1817
11.9k1817
add a comment |
add a comment |
$begingroup$
Here's a solution that'll work for all characteristics - Factorize the degree $2$ homogeneous part into linear factors (can do this because algebraically closed). Now, if the linear factors are linearly dependent, w.l.o.g. change coordinates to make this linear factor the new $X$. The equation now becomes $X^2 + aX + bY + c$. By now making $aX+bY+c$ the new $Y$ ($b neq 0$), we get $X^2+Y$.
If the factors are linearly independent, change coordinates to make one of them $X$ and the other $Y$. The equation now becomes $XY + aX + bY + c$ which we can write as $(X+b)(Y+a)+d$. We can now change coordinates to make it $XY+d$ and so, we're done.
As an interesting aside, this indicates as expected that the general equation is a hyperbola since it corresponds to the linearly independent case here.
answered Dec 18 '18 at 12:51
Sridhar VenkateshSridhar Venkatesh
163
$endgroup$
add a comment |
$begingroup$
Here's a solution that'll work for all characteristics - Factorize the degree $2$ homogeneous part into linear factors (can do this because algebraically closed). Now, if the linear factors are linearly dependent, w.l.o.g. change coordinates to make this linear factor the new $X$. The equation now becomes $X^2 + aX + bY + c$. By now making $aX+bY+c$ the new $Y$ ($b neq 0$), we get $X^2+Y$.
If the factors are linearly independent, change coordinates to make one of them $X$ and the other $Y$. The equation now becomes $XY + aX + bY + c$ which we can write as $(X+b)(Y+a)+d$. We can now change coordinates to make it $XY+d$ and so, we're done.
As an interesting aside, this indicates as expected that the general equation is a hyperbola since it corresponds to the linearly independent case here.
answered Dec 18 '18 at 12:51
Sridhar VenkateshSridhar Venkatesh
163
$endgroup$
add a comment |
$begingroup$
Here's a solution that'll work for all characteristics - Factorize the degree $2$ homogeneous part into linear factors (can do this because algebraically closed). Now, if the linear factors are linearly dependent, w.l.o.g. change coordinates to make this linear factor the new $X$. The equation now becomes $X^2 + aX + bY + c$. By now making $aX+bY+c$ the new $Y$ ($b neq 0$), we get $X^2+Y$.
If the factors are linearly independent, change coordinates to make one of them $X$ and the other $Y$. The equation now becomes $XY + aX + bY + c$ which we can write as $(X+b)(Y+a)+d$. We can now change coordinates to make it $XY+d$ and so, we're done.
As an interesting aside, this indicates as expected that the general equation is a hyperbola since it corresponds to the linearly independent case here.
answered Dec 18 '18 at 12:51
Sridhar VenkateshSridhar Venkatesh
163
$endgroup$
Here's a solution that'll work for all characteristics - Factorize the degree $2$ homogeneous part into linear factors (can do this because algebraically closed). Now, if the linear factors are linearly dependent, w.l.o.g. change coordinates to make this linear factor the new $X$. The equation now becomes $X^2 + aX + bY + c$. By now making $aX+bY+c$ the new $Y$ ($b neq 0$), we get $X^2+Y$.
If the factors are linearly independent, change coordinates to make one of them $X$ and the other $Y$. The equation now becomes $XY + aX + bY + c$ which we can write as $(X+b)(Y+a)+d$. We can now change coordinates to make it $XY+d$ and so, we're done.
As an interesting aside, this indicates as expected that the general equation is a hyperbola since it corresponds to the linearly independent case here.
answered Dec 18 '18 at 12:51
Sridhar VenkateshSridhar Venkatesh
163
answered Dec 18 '18 at 12:51
Sridhar VenkateshSridhar Venkatesh
163
answered Dec 18 '18 at 12:51
Sridhar VenkateshSridhar Venkatesh
163
answered Dec 18 '18 at 12:51
Sridhar VenkateshSridhar Venkatesh
163
163
add a comment |
add a comment |
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$begingroup$
$A(V) cong k[t]$ and $A(Z) cong k[t,t^{-1}]$
$endgroup$
– user171326
Aug 31 '17 at 21:07
$begingroup$
Well... Yes, but how is this related with a general quadratic polynomial? I mean, when is isomorphic with one of those affine coordinate rings?
$endgroup$
– Billy Bones
Aug 31 '17 at 21:13
$begingroup$
If $k$ is algebraically closed, I believe all the time, as $W$ is the complement of a line in $Bbb P^2$ which can be a parabola (your conic is tangent to the line) or an hyperbola (your conic intersects the line twice).
$endgroup$
– user171326
Aug 31 '17 at 21:16
$begingroup$
It is supposed I can't use the projective varieties because there aren't defined yet. I understand the point, but I want to read a more constructive proof of this fact, using the affine coordinate rings, if there is no problem. Thanks in advance.
$endgroup$
– Billy Bones
Aug 31 '17 at 21:19