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Isomorphism between $mathbbP^1$ and $V : X^2 + Y^2 = pZ^2$

2

2

$begingroup$

My question is about an exercise from "arithmetic of elliptic curves":

Let $$ V : X^2 + Y^2 = pZ^2$$ be a projective vareity in $mathbbP^2$ and $p$ be a prime number.

prove that $V$ is isomorphic to $mathbbP^1$ iff $ p equiv 1$(mod $4$).

and for $p equiv 3$(mod $4$) no two of such vareities are isomorphic.

---

I was trying to solve this exercise but I have stuck in defining the morphism. And as it is the case that $p equiv 1$(mod $4$) I think we should use an integer solution of $a^2 + b^2 =p$. but most of my efforts didn't turn out to be a morphism. So if anyone could help with this, it would be great.

algebraic-geometry elliptic-curves

share|cite|improve this question

asked Mar 27 at 7:29

AmirhosseinAmirhossein

867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When $pequiv1pmod 4$ the conic has a rational point. You can then use the usual tricks to transform it into standard form.
  $endgroup$
  – Lord Shark the Unknown
  Mar 27 at 7:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LordSharktheUnknown could you explain a little bit more about the process?
  $endgroup$
  – Amirhossein
  Mar 27 at 8:07
  

  
  
  
  

add a comment | 

2

2

$begingroup$

My question is about an exercise from "arithmetic of elliptic curves":

Let $$ V : X^2 + Y^2 = pZ^2$$ be a projective vareity in $mathbbP^2$ and $p$ be a prime number.

prove that $V$ is isomorphic to $mathbbP^1$ iff $ p equiv 1$(mod $4$).

and for $p equiv 3$(mod $4$) no two of such vareities are isomorphic.

---

I was trying to solve this exercise but I have stuck in defining the morphism. And as it is the case that $p equiv 1$(mod $4$) I think we should use an integer solution of $a^2 + b^2 =p$. but most of my efforts didn't turn out to be a morphism. So if anyone could help with this, it would be great.

algebraic-geometry elliptic-curves

share|cite|improve this question

asked Mar 27 at 7:29

AmirhosseinAmirhossein

867

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When $pequiv1pmod 4$ the conic has a rational point. You can then use the usual tricks to transform it into standard form.
  $endgroup$
  – Lord Shark the Unknown
  Mar 27 at 7:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @LordSharktheUnknown could you explain a little bit more about the process?
  $endgroup$
  – Amirhossein
  Mar 27 at 8:07
  

  
  
  
  

add a comment | 

$begingroup$

My question is about an exercise from "arithmetic of elliptic curves":

Let $$ V : X^2 + Y^2 = pZ^2$$ be a projective vareity in $mathbbP^2$ and $p$ be a prime number.

prove that $V$ is isomorphic to $mathbbP^1$ iff $ p equiv 1$(mod $4$).

and for $p equiv 3$(mod $4$) no two of such vareities are isomorphic.

---

I was trying to solve this exercise but I have stuck in defining the morphism. And as it is the case that $p equiv 1$(mod $4$) I think we should use an integer solution of $a^2 + b^2 =p$. but most of my efforts didn't turn out to be a morphism. So if anyone could help with this, it would be great.

algebraic-geometry elliptic-curves

share|cite|improve this question

asked Mar 27 at 7:29

AmirhosseinAmirhossein

867

$endgroup$

share|cite|improve this question

asked Mar 27 at 7:29

AmirhosseinAmirhossein

867

share|cite|improve this question

asked Mar 27 at 7:29

AmirhosseinAmirhossein

867

asked Mar 27 at 7:29

AmirhosseinAmirhossein

867

asked Mar 27 at 7:29

AmirhosseinAmirhossein

867