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Quadratic forms over $F_p$ in more than $2$ variables are isotropic

1

$begingroup$

I'm trying to find all anisotropic quadratic forms over $F_p=Z/pZ$. I have found that "If $F$ is a finite field and $(V, q)$ is a quadratic space of dimension at least three, then it is isotropic" (here https://en.wikipedia.org/wiki/Isotropic_quadratic_form). So I wonder if the fact is true and if it is, why. Thank you

geometry finite-fields quadratic-forms

share|cite|improve this question

edited Apr 2 at 12:22

Jose Brox

3,45711129

asked Apr 2 at 10:41

user653342

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you assuming $p$ odd?
  $endgroup$
  – Lord Shark the Unknown
  Apr 2 at 10:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  p is a prime number
  $endgroup$
  – user653342
  Apr 2 at 10:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you assuming $p$ is an odd prime number?
  $endgroup$
  – Lord Shark the Unknown
  Apr 2 at 10:58
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $p$ is an arbitrary prime number (except 2)
  $endgroup$
  – user653342
  Apr 2 at 11:01
  

  
  
  
  

add a comment | 

1

$begingroup$

I'm trying to find all anisotropic quadratic forms over $F_p=Z/pZ$. I have found that "If $F$ is a finite field and $(V, q)$ is a quadratic space of dimension at least three, then it is isotropic" (here https://en.wikipedia.org/wiki/Isotropic_quadratic_form). So I wonder if the fact is true and if it is, why. Thank you

geometry finite-fields quadratic-forms

share|cite|improve this question

edited Apr 2 at 12:22

Jose Brox

3,45711129

asked Apr 2 at 10:41

user653342

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you assuming $p$ odd?
  $endgroup$
  – Lord Shark the Unknown
  Apr 2 at 10:50
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  p is a prime number
  $endgroup$
  – user653342
  Apr 2 at 10:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you assuming $p$ is an odd prime number?
  $endgroup$
  – Lord Shark the Unknown
  Apr 2 at 10:58
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $p$ is an arbitrary prime number (except 2)
  $endgroup$
  – user653342
  Apr 2 at 11:01
  

  
  
  
  

add a comment | 

$begingroup$

I'm trying to find all anisotropic quadratic forms over $F_p=Z/pZ$. I have found that "If $F$ is a finite field and $(V, q)$ is a quadratic space of dimension at least three, then it is isotropic" (here https://en.wikipedia.org/wiki/Isotropic_quadratic_form). So I wonder if the fact is true and if it is, why. Thank you

geometry finite-fields quadratic-forms

share|cite|improve this question

edited Apr 2 at 12:22

Jose Brox

3,45711129

asked Apr 2 at 10:41

user653342

$endgroup$

share|cite|improve this question

edited Apr 2 at 12:22

Jose Brox

3,45711129

asked Apr 2 at 10:41

user653342

share|cite|improve this question

edited Apr 2 at 12:22

Jose Brox

3,45711129

asked Apr 2 at 10:41

user653342

edited Apr 2 at 12:22

Jose Brox

3,45711129

edited Apr 2 at 12:22

Jose Brox

3,45711129

1 Answer
1

active

oldest

votes

0

$begingroup$

First, an observation: if the number of variables of $q$ is less than the dimension, then we can get an isotropic vector by assigning $0$ to the appearing variables and $1$ to the non appearing ones. So we now assume the number of variables $n$ to be the dimension.

Then your result is an specific instance of the Chevalley-Warning theorem: If a set of polynomials $f_i$ over a finite field satisfies $n>sum_i d_i$, where $n$ is the number of variables and $d_i$ the degree of $f_i$, then the system of equations $f_i=0$ has a number of solutions which is a multiple of the characteristic of the finite field.

In your case, you have just one polynomial $q$ of degree $2$ and $ngeq 3>2$, so the number of solutions is a multiple of $p$, hence can be $0,p,2pldots$. But we already know that $(0,ldots,0)$ is a solution, so there must be at least other $p-1$ nonzero solutions $v$ which give $q(v)=0$.

share|cite|improve this answer

answered Apr 2 at 12:17

Jose BroxJose Brox

3,45711129

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Kelly I'm going to add a link in the Wikipedia page to the Chevalley-Warning theorem. Thanks!
  $endgroup$
  – Jose Brox
  Apr 2 at 12:23
  

  
  
  
  

add a comment | 

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0

$begingroup$

First, an observation: if the number of variables of $q$ is less than the dimension, then we can get an isotropic vector by assigning $0$ to the appearing variables and $1$ to the non appearing ones. So we now assume the number of variables $n$ to be the dimension.

Then your result is an specific instance of the Chevalley-Warning theorem: If a set of polynomials $f_i$ over a finite field satisfies $n>sum_i d_i$, where $n$ is the number of variables and $d_i$ the degree of $f_i$, then the system of equations $f_i=0$ has a number of solutions which is a multiple of the characteristic of the finite field.

In your case, you have just one polynomial $q$ of degree $2$ and $ngeq 3>2$, so the number of solutions is a multiple of $p$, hence can be $0,p,2pldots$. But we already know that $(0,ldots,0)$ is a solution, so there must be at least other $p-1$ nonzero solutions $v$ which give $q(v)=0$.

share|cite|improve this answer

answered Apr 2 at 12:17

Jose BroxJose Brox

3,45711129

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Kelly I'm going to add a link in the Wikipedia page to the Chevalley-Warning theorem. Thanks!
  $endgroup$
  – Jose Brox
  Apr 2 at 12:23
  

  
  
  
  

add a comment | 

0

$begingroup$

First, an observation: if the number of variables of $q$ is less than the dimension, then we can get an isotropic vector by assigning $0$ to the appearing variables and $1$ to the non appearing ones. So we now assume the number of variables $n$ to be the dimension.

Then your result is an specific instance of the Chevalley-Warning theorem: If a set of polynomials $f_i$ over a finite field satisfies $n>sum_i d_i$, where $n$ is the number of variables and $d_i$ the degree of $f_i$, then the system of equations $f_i=0$ has a number of solutions which is a multiple of the characteristic of the finite field.

In your case, you have just one polynomial $q$ of degree $2$ and $ngeq 3>2$, so the number of solutions is a multiple of $p$, hence can be $0,p,2pldots$. But we already know that $(0,ldots,0)$ is a solution, so there must be at least other $p-1$ nonzero solutions $v$ which give $q(v)=0$.

share|cite|improve this answer

answered Apr 2 at 12:17

Jose BroxJose Brox

3,45711129

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Kelly I'm going to add a link in the Wikipedia page to the Chevalley-Warning theorem. Thanks!
  $endgroup$
  – Jose Brox
  Apr 2 at 12:23
  

  
  
  
  

add a comment | 

$begingroup$

First, an observation: if the number of variables of $q$ is less than the dimension, then we can get an isotropic vector by assigning $0$ to the appearing variables and $1$ to the non appearing ones. So we now assume the number of variables $n$ to be the dimension.

Then your result is an specific instance of the Chevalley-Warning theorem: If a set of polynomials $f_i$ over a finite field satisfies $n>sum_i d_i$, where $n$ is the number of variables and $d_i$ the degree of $f_i$, then the system of equations $f_i=0$ has a number of solutions which is a multiple of the characteristic of the finite field.

In your case, you have just one polynomial $q$ of degree $2$ and $ngeq 3>2$, so the number of solutions is a multiple of $p$, hence can be $0,p,2pldots$. But we already know that $(0,ldots,0)$ is a solution, so there must be at least other $p-1$ nonzero solutions $v$ which give $q(v)=0$.

share|cite|improve this answer

answered Apr 2 at 12:17

Jose BroxJose Brox

3,45711129

$endgroup$

share|cite|improve this answer

answered Apr 2 at 12:17

Jose BroxJose Brox

3,45711129

answered Apr 2 at 12:17

Jose BroxJose Brox

3,45711129

answered Apr 2 at 12:17

Jose BroxJose Brox

3,45711129