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Any affine conic is isomorphic to $V(y-x^2)$ or $V(xy-1)$

1

$begingroup$

I'm trying to solve the above exercise in Ghatmann's notes on algebraic geometry.

I tried to consider the real case and I suspect that X1 should correspond to a plane that intersects with exactly one of the cones, whereas X2 is the case in which both cones are intersected (not at the origin).

Next i tried to look at the general case of a quadratic polynomial in two variables to see what happens if I act on it with an affine change of coordinates(i.e. linear transformation followed by translation), this quickly got very complicated.

My guess is that I need to figure out how the irreducibility condition affects the general equation, and maybe extend the intuition I gained for the real case to arbitrary fields(of characteristic $ne$ 2).

However, right now I'm kind of stuck, so any help would be appreciated. Thanks.

Note that the field $K$ is assumed to be algebraically closed.

abstract-algebra algebraic-geometry commutative-algebra conic-sections

share|cite|improve this question

edited Mar 24 at 20:19

user26857

39.6k124284

asked Mar 12 at 17:50

Gauss57Gauss57

174

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I assume $K$ is algebraically closed?
  $endgroup$
  – jgon
  Mar 12 at 18:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, it's closed. I will edit my question to reflect this.
  $endgroup$
  – Gauss57
  Mar 12 at 18:53
  

  
  
  
  

add a comment | 

1

$begingroup$

I'm trying to solve the above exercise in Ghatmann's notes on algebraic geometry.

I tried to consider the real case and I suspect that X1 should correspond to a plane that intersects with exactly one of the cones, whereas X2 is the case in which both cones are intersected (not at the origin).

Next i tried to look at the general case of a quadratic polynomial in two variables to see what happens if I act on it with an affine change of coordinates(i.e. linear transformation followed by translation), this quickly got very complicated.

My guess is that I need to figure out how the irreducibility condition affects the general equation, and maybe extend the intuition I gained for the real case to arbitrary fields(of characteristic $ne$ 2).

However, right now I'm kind of stuck, so any help would be appreciated. Thanks.

Note that the field $K$ is assumed to be algebraically closed.

abstract-algebra algebraic-geometry commutative-algebra conic-sections

share|cite|improve this question

edited Mar 24 at 20:19

user26857

39.6k124284

asked Mar 12 at 17:50

Gauss57Gauss57

174

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I assume $K$ is algebraically closed?
  $endgroup$
  – jgon
  Mar 12 at 18:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah, it's closed. I will edit my question to reflect this.
  $endgroup$
  – Gauss57
  Mar 12 at 18:53
  

  
  
  
  

add a comment | 

$begingroup$

I'm trying to solve the above exercise in Ghatmann's notes on algebraic geometry.

I tried to consider the real case and I suspect that X1 should correspond to a plane that intersects with exactly one of the cones, whereas X2 is the case in which both cones are intersected (not at the origin).

Next i tried to look at the general case of a quadratic polynomial in two variables to see what happens if I act on it with an affine change of coordinates(i.e. linear transformation followed by translation), this quickly got very complicated.

My guess is that I need to figure out how the irreducibility condition affects the general equation, and maybe extend the intuition I gained for the real case to arbitrary fields(of characteristic $ne$ 2).

However, right now I'm kind of stuck, so any help would be appreciated. Thanks.

Note that the field $K$ is assumed to be algebraically closed.

abstract-algebra algebraic-geometry commutative-algebra conic-sections

share|cite|improve this question

edited Mar 24 at 20:19

user26857

39.6k124284

asked Mar 12 at 17:50

Gauss57Gauss57

174

$endgroup$

share|cite|improve this question

edited Mar 24 at 20:19

user26857

39.6k124284

asked Mar 12 at 17:50

Gauss57Gauss57

174

share|cite|improve this question

edited Mar 24 at 20:19

user26857

39.6k124284

asked Mar 12 at 17:50

Gauss57Gauss57

174

edited Mar 24 at 20:19

user26857

39.6k124284

edited Mar 24 at 20:19

user26857

39.6k124284

asked Mar 12 at 17:50

Gauss57Gauss57

174

asked Mar 12 at 17:50

Gauss57Gauss57

174

1 Answer
1

active

oldest

votes

0

$begingroup$

Assume $K$ is algebraically closed, with characteristic not $2$.

Suppose your irreducible conic is given by $ax^2+bxy+cy^2+dx+ey+f=0$ (up to a constant factor).

Consider the quadratic form $ax^2+bxy+cy^2$. Since $K$ is algebraically closed, this factors into a product of homogeneous linear forms, $aell_1ell_2$, with $ell_1$ and $ell_2$ monic in $x$. If $ell_1=ell_2$, then let $dx+ey=dell_3$. If $ell_3=ell_1$ as well, then we could rewrite our equation as $aell_3^2 + dell_3 + f=0$, which is reducible, since $K$ is algebraically closed. Otherwise, $ell_3 ne ell_1$, so let $x'=sqrt{-a}ell_1$, $y'=dell_3 + f$ to get that our original conic is given by $y' - x'^2$ in our new variables.

Otherwise, if $ell_1ne ell_2$, take $x_1=aell_1$, $y_1=ell_2$, and rewrite our conic in terms of $x_1$ and $y_1$, so that it is of the form
$$x_1y_1+Ax_1 + By_1+C =0.$$
Let $x_2=x_1+B$, $y_2=y_1+A$, so that the conic can be rewritten as
$$x_2y_2 + (C-AB)=0.$$
If $C-AB=0$, then the conic is reducible, so finally
let $x'=x_2$, $y'=frac{-1}{C-AB}y_2$, so that
$$x'y' -1=0,$$
as desired.

share|cite|improve this answer

edited Mar 24 at 19:54

Community♦

1

answered Mar 12 at 19:16

jgonjgon

16.7k32244

$endgroup$

add a comment | 

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0

$begingroup$

Assume $K$ is algebraically closed, with characteristic not $2$.

Suppose your irreducible conic is given by $ax^2+bxy+cy^2+dx+ey+f=0$ (up to a constant factor).

Consider the quadratic form $ax^2+bxy+cy^2$. Since $K$ is algebraically closed, this factors into a product of homogeneous linear forms, $aell_1ell_2$, with $ell_1$ and $ell_2$ monic in $x$. If $ell_1=ell_2$, then let $dx+ey=dell_3$. If $ell_3=ell_1$ as well, then we could rewrite our equation as $aell_3^2 + dell_3 + f=0$, which is reducible, since $K$ is algebraically closed. Otherwise, $ell_3 ne ell_1$, so let $x'=sqrt{-a}ell_1$, $y'=dell_3 + f$ to get that our original conic is given by $y' - x'^2$ in our new variables.

Otherwise, if $ell_1ne ell_2$, take $x_1=aell_1$, $y_1=ell_2$, and rewrite our conic in terms of $x_1$ and $y_1$, so that it is of the form
$$x_1y_1+Ax_1 + By_1+C =0.$$
Let $x_2=x_1+B$, $y_2=y_1+A$, so that the conic can be rewritten as
$$x_2y_2 + (C-AB)=0.$$
If $C-AB=0$, then the conic is reducible, so finally
let $x'=x_2$, $y'=frac{-1}{C-AB}y_2$, so that
$$x'y' -1=0,$$
as desired.

share|cite|improve this answer

edited Mar 24 at 19:54

Community♦

1

answered Mar 12 at 19:16

jgonjgon

16.7k32244

$endgroup$

add a comment | 

0

$begingroup$

Assume $K$ is algebraically closed, with characteristic not $2$.

Suppose your irreducible conic is given by $ax^2+bxy+cy^2+dx+ey+f=0$ (up to a constant factor).

Consider the quadratic form $ax^2+bxy+cy^2$. Since $K$ is algebraically closed, this factors into a product of homogeneous linear forms, $aell_1ell_2$, with $ell_1$ and $ell_2$ monic in $x$. If $ell_1=ell_2$, then let $dx+ey=dell_3$. If $ell_3=ell_1$ as well, then we could rewrite our equation as $aell_3^2 + dell_3 + f=0$, which is reducible, since $K$ is algebraically closed. Otherwise, $ell_3 ne ell_1$, so let $x'=sqrt{-a}ell_1$, $y'=dell_3 + f$ to get that our original conic is given by $y' - x'^2$ in our new variables.

Otherwise, if $ell_1ne ell_2$, take $x_1=aell_1$, $y_1=ell_2$, and rewrite our conic in terms of $x_1$ and $y_1$, so that it is of the form
$$x_1y_1+Ax_1 + By_1+C =0.$$
Let $x_2=x_1+B$, $y_2=y_1+A$, so that the conic can be rewritten as
$$x_2y_2 + (C-AB)=0.$$
If $C-AB=0$, then the conic is reducible, so finally
let $x'=x_2$, $y'=frac{-1}{C-AB}y_2$, so that
$$x'y' -1=0,$$
as desired.

share|cite|improve this answer

edited Mar 24 at 19:54

Community♦

1

answered Mar 12 at 19:16

jgonjgon

16.7k32244

$endgroup$

add a comment | 

$begingroup$

Assume $K$ is algebraically closed, with characteristic not $2$.

Suppose your irreducible conic is given by $ax^2+bxy+cy^2+dx+ey+f=0$ (up to a constant factor).

Consider the quadratic form $ax^2+bxy+cy^2$. Since $K$ is algebraically closed, this factors into a product of homogeneous linear forms, $aell_1ell_2$, with $ell_1$ and $ell_2$ monic in $x$. If $ell_1=ell_2$, then let $dx+ey=dell_3$. If $ell_3=ell_1$ as well, then we could rewrite our equation as $aell_3^2 + dell_3 + f=0$, which is reducible, since $K$ is algebraically closed. Otherwise, $ell_3 ne ell_1$, so let $x'=sqrt{-a}ell_1$, $y'=dell_3 + f$ to get that our original conic is given by $y' - x'^2$ in our new variables.

Otherwise, if $ell_1ne ell_2$, take $x_1=aell_1$, $y_1=ell_2$, and rewrite our conic in terms of $x_1$ and $y_1$, so that it is of the form
$$x_1y_1+Ax_1 + By_1+C =0.$$
Let $x_2=x_1+B$, $y_2=y_1+A$, so that the conic can be rewritten as
$$x_2y_2 + (C-AB)=0.$$
If $C-AB=0$, then the conic is reducible, so finally
let $x'=x_2$, $y'=frac{-1}{C-AB}y_2$, so that
$$x'y' -1=0,$$
as desired.

share|cite|improve this answer

edited Mar 24 at 19:54

Community♦

1

answered Mar 12 at 19:16

jgonjgon

16.7k32244

$endgroup$

share|cite|improve this answer

edited Mar 24 at 19:54

Community♦

1

answered Mar 12 at 19:16

jgonjgon

16.7k32244

edited Mar 24 at 19:54

Community♦

1

edited Mar 24 at 19:54

Community♦

1

answered Mar 12 at 19:16

jgonjgon

16.7k32244

answered Mar 12 at 19:16

jgonjgon

16.7k32244