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The functions by monomials $x_1^k_1x_2^k_2…x_n^k_n$ for $k_1,k_2,…,k_n

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-1

$begingroup$

Prove , that if a field K contains $q$ elements, the functions by monomyals $k_1^k_1x_2^k_2...x_n^k_n$ for $k_1,k_2,...,k_n<q$ form a basis in the space of all $K$-valued functions on $K^n$.

My solution: There are $q^q^n$ polynomials with $deg(F)<q$, since there are $q^n$ monomyals plus we can choose a coefficient $q$ ways. But the different polynomials of degree $<q$ correspond to different functions. Then there is a bijection between polynomials of degree $<q$ and functions $K^n to K$.

Is it correct? Could someone check it, please?

linear-algebra abstract-algebra

share|cite|improve this question

edited Mar 17 at 18:34

Metso

asked Mar 17 at 8:44

MetsoMetso

15510

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is not a proof. It lacks all justification. Also the statement it purports to prove is false. Consider $BbbF_2$, and the functions on the plane. There are only 3 monomials of degree less than 2, $1,x_1,x_2$, but the space of functions is clearly 4-dimensional. The correct statement should be $k_i<q$ for all $i$. This is what the "proof" seems to be assuming anyway.
  $endgroup$
  – jgon
  Mar 17 at 15:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know what it means for a subset of a vector space to be a basis?
  $endgroup$
  – jgon
  Mar 17 at 15:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @jgon: A subset of a vector space with n elements is a basis if and only if it is linearly independent. What do mean: Also the statement it purports to prove is false
  $endgroup$
  – Metso
  Mar 17 at 17:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $n$ in your definition? A better definition is a basis for a vector space $V$ is a subset $B$ that is linearly independent and spans $V$. My next question is how does your proof seek to prove that the monomials are linearly independent? Also again, the statement you are trying to prove is false as written.
  $endgroup$
  – jgon
  Mar 17 at 17:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, regardless of where it's from, I provided a counterexample to the claim as written in my first comment. Also, try using the definition of basis I gave in my previous comment. Although, actually, you may get more mileage from the following hint. Prove that the monomials with $k_i<q$ span the space of functions, and then prove that the dimension of the space of functions is equal to the number of monomials. This also would prove that the monomials form a basis, and is significantly easier than proving that the monomials are linearly independent.
  $endgroup$
  – jgon
  Mar 17 at 17:36
  

  
  
  
  

 | 
show 1 more comment

-1

$begingroup$

Prove , that if a field K contains $q$ elements, the functions by monomyals $k_1^k_1x_2^k_2...x_n^k_n$ for $k_1,k_2,...,k_n<q$ form a basis in the space of all $K$-valued functions on $K^n$.

My solution: There are $q^q^n$ polynomials with $deg(F)<q$, since there are $q^n$ monomyals plus we can choose a coefficient $q$ ways. But the different polynomials of degree $<q$ correspond to different functions. Then there is a bijection between polynomials of degree $<q$ and functions $K^n to K$.

Is it correct? Could someone check it, please?

linear-algebra abstract-algebra

share|cite|improve this question

edited Mar 17 at 18:34

Metso

asked Mar 17 at 8:44

MetsoMetso

15510

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is not a proof. It lacks all justification. Also the statement it purports to prove is false. Consider $BbbF_2$, and the functions on the plane. There are only 3 monomials of degree less than 2, $1,x_1,x_2$, but the space of functions is clearly 4-dimensional. The correct statement should be $k_i<q$ for all $i$. This is what the "proof" seems to be assuming anyway.
  $endgroup$
  – jgon
  Mar 17 at 15:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you know what it means for a subset of a vector space to be a basis?
  $endgroup$
  – jgon
  Mar 17 at 15:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @jgon: A subset of a vector space with n elements is a basis if and only if it is linearly independent. What do mean: Also the statement it purports to prove is false
  $endgroup$
  – Metso
  Mar 17 at 17:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is $n$ in your definition? A better definition is a basis for a vector space $V$ is a subset $B$ that is linearly independent and spans $V$. My next question is how does your proof seek to prove that the monomials are linearly independent? Also again, the statement you are trying to prove is false as written.
  $endgroup$
  – jgon
  Mar 17 at 17:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, regardless of where it's from, I provided a counterexample to the claim as written in my first comment. Also, try using the definition of basis I gave in my previous comment. Although, actually, you may get more mileage from the following hint. Prove that the monomials with $k_i<q$ span the space of functions, and then prove that the dimension of the space of functions is equal to the number of monomials. This also would prove that the monomials form a basis, and is significantly easier than proving that the monomials are linearly independent.
  $endgroup$
  – jgon
  Mar 17 at 17:36
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Prove , that if a field K contains $q$ elements, the functions by monomyals $k_1^k_1x_2^k_2...x_n^k_n$ for $k_1,k_2,...,k_n<q$ form a basis in the space of all $K$-valued functions on $K^n$.

My solution: There are $q^q^n$ polynomials with $deg(F)<q$, since there are $q^n$ monomyals plus we can choose a coefficient $q$ ways. But the different polynomials of degree $<q$ correspond to different functions. Then there is a bijection between polynomials of degree $<q$ and functions $K^n to K$.

Is it correct? Could someone check it, please?

linear-algebra abstract-algebra

share|cite|improve this question

edited Mar 17 at 18:34

Metso

asked Mar 17 at 8:44

MetsoMetso

15510

$endgroup$

share|cite|improve this question

edited Mar 17 at 18:34

Metso

asked Mar 17 at 8:44

MetsoMetso

15510

share|cite|improve this question

edited Mar 17 at 18:34

Metso

asked Mar 17 at 8:44

MetsoMetso

15510

asked Mar 17 at 8:44

MetsoMetso

15510

asked Mar 17 at 8:44

MetsoMetso

15510