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3

2

$begingroup$

Let $mathbbF_q$ be the finite field of order $q$, where $q$ is odd. Now, Let $S$ be the set of elements $x$ in $mathbbF_q^n^*$, such that $x$ doesnot belong to any smaller subfield(containing $mathbbF_q$) of $mathbbF_q^n$. Let $S^2= x^2 $. I want to find out $|Scap S^2|$.

I have made some calculations and found out the answer, but I am not sure whether it is perfectly correct, or whether there is easier way to see this!

Here is my idea: $|S|=sum_nmu(d)q^n/d$. Here $mu$ denotes the Mobius Function. Now I divide this in two case.

Case 1: Suppose $n$ is odd.

Then observe that $ain S implies a^2in S$. This is because there is no intermediate extension of even order. Now, since exactly two element of $S$ gives the same squared element, we have $|Scap S^2|=frac2= frac12sum_nmu(d)q^n/d$.

Case 2: Suppose $n$ is even

In this case: $ain S implies texteither a^2in S text or a^2in mathbbF_q^n/2^*$. Now, there are exactly $fracq^n/2-12$ non-squares in $mathbbF_q^n/2^*$. These non-squares are squares of elements in $S$. So, in this way there are $q^n/2-1$ elements in $S$, whose squares lie in $mathbbF_q^n/2^*$. Hence there are $|S|- (q^n/2-1)$ number of elements in $S$ whose squares are in $S$. Hence $|Scap S^2|=frac12[|S|-(q^n/2-1)]= frac12Big[sum_nmu(d)q^n/d - (q^n/2-1)Big]$.

This completes the solution. Please let me know whether this is a fine solution, or am I missing something! Let me know, if one can approach this problem any easier way.

Thank you!

abstract-algebra combinatorics field-theory galois-theory finite-fields

share|cite|improve this question

edited Mar 29 at 15:38

hardmath

29.3k953101

asked Mar 26 at 14:51

RijuRiju

2,294414

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  When you say "smaller subfield", do you mean "smaller field extension of $mathbbF_q$" or literally "smaller subfield"? Because that makes a difference.
  $endgroup$
  – darij grinberg
  Mar 27 at 17:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  no literally smaller subfield!! Will smaller subfield extension make sense in the question? Because not all finite fields are contained in $mathbbF_q^n$. $mathbbF_q^d$ is contained in $mathbbF_q^n$ iff $dmid n$.
  $endgroup$
  – Riju
  Mar 28 at 6:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you mean literally "smaller subfield", then your computation of $left|Sright|$ doesn't work already (because your Möbius inversion only accounts for the subfields that contain $mathbbF_q$).
  $endgroup$
  – darij grinberg
  Mar 28 at 6:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok I know sww what you are saying! I am taking subfields containing $mathbbF_q$.
  $endgroup$
  – Riju
  Mar 28 at 7:21
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Care to edit your question to correct it?
  $endgroup$
  – darij grinberg
  Mar 29 at 3:51
  

  
  
  
  

 | 
show 3 more comments

3

2

$begingroup$

Let $mathbbF_q$ be the finite field of order $q$, where $q$ is odd. Now, Let $S$ be the set of elements $x$ in $mathbbF_q^n^*$, such that $x$ doesnot belong to any smaller subfield(containing $mathbbF_q$) of $mathbbF_q^n$. Let $S^2= x^2 $. I want to find out $|Scap S^2|$.

I have made some calculations and found out the answer, but I am not sure whether it is perfectly correct, or whether there is easier way to see this!

Here is my idea: $|S|=sum_nmu(d)q^n/d$. Here $mu$ denotes the Mobius Function. Now I divide this in two case.

Case 1: Suppose $n$ is odd.

Then observe that $ain S implies a^2in S$. This is because there is no intermediate extension of even order. Now, since exactly two element of $S$ gives the same squared element, we have $|Scap S^2|=frac2= frac12sum_nmu(d)q^n/d$.

Case 2: Suppose $n$ is even

In this case: $ain S implies texteither a^2in S text or a^2in mathbbF_q^n/2^*$. Now, there are exactly $fracq^n/2-12$ non-squares in $mathbbF_q^n/2^*$. These non-squares are squares of elements in $S$. So, in this way there are $q^n/2-1$ elements in $S$, whose squares lie in $mathbbF_q^n/2^*$. Hence there are $|S|- (q^n/2-1)$ number of elements in $S$ whose squares are in $S$. Hence $|Scap S^2|=frac12[|S|-(q^n/2-1)]= frac12Big[sum_nmu(d)q^n/d - (q^n/2-1)Big]$.

This completes the solution. Please let me know whether this is a fine solution, or am I missing something! Let me know, if one can approach this problem any easier way.

Thank you!

abstract-algebra combinatorics field-theory galois-theory finite-fields

share|cite|improve this question

edited Mar 29 at 15:38

hardmath

29.3k953101

asked Mar 26 at 14:51

RijuRiju

2,294414

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  When you say "smaller subfield", do you mean "smaller field extension of $mathbbF_q$" or literally "smaller subfield"? Because that makes a difference.
  $endgroup$
  – darij grinberg
  Mar 27 at 17:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  no literally smaller subfield!! Will smaller subfield extension make sense in the question? Because not all finite fields are contained in $mathbbF_q^n$. $mathbbF_q^d$ is contained in $mathbbF_q^n$ iff $dmid n$.
  $endgroup$
  – Riju
  Mar 28 at 6:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you mean literally "smaller subfield", then your computation of $left|Sright|$ doesn't work already (because your Möbius inversion only accounts for the subfields that contain $mathbbF_q$).
  $endgroup$
  – darij grinberg
  Mar 28 at 6:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok I know sww what you are saying! I am taking subfields containing $mathbbF_q$.
  $endgroup$
  – Riju
  Mar 28 at 7:21
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Care to edit your question to correct it?
  $endgroup$
  – darij grinberg
  Mar 29 at 3:51
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

Let $mathbbF_q$ be the finite field of order $q$, where $q$ is odd. Now, Let $S$ be the set of elements $x$ in $mathbbF_q^n^*$, such that $x$ doesnot belong to any smaller subfield(containing $mathbbF_q$) of $mathbbF_q^n$. Let $S^2= x^2 $. I want to find out $|Scap S^2|$.

I have made some calculations and found out the answer, but I am not sure whether it is perfectly correct, or whether there is easier way to see this!

Here is my idea: $|S|=sum_nmu(d)q^n/d$. Here $mu$ denotes the Mobius Function. Now I divide this in two case.

Case 1: Suppose $n$ is odd.

Then observe that $ain S implies a^2in S$. This is because there is no intermediate extension of even order. Now, since exactly two element of $S$ gives the same squared element, we have $|Scap S^2|=frac2= frac12sum_nmu(d)q^n/d$.

Case 2: Suppose $n$ is even

In this case: $ain S implies texteither a^2in S text or a^2in mathbbF_q^n/2^*$. Now, there are exactly $fracq^n/2-12$ non-squares in $mathbbF_q^n/2^*$. These non-squares are squares of elements in $S$. So, in this way there are $q^n/2-1$ elements in $S$, whose squares lie in $mathbbF_q^n/2^*$. Hence there are $|S|- (q^n/2-1)$ number of elements in $S$ whose squares are in $S$. Hence $|Scap S^2|=frac12[|S|-(q^n/2-1)]= frac12Big[sum_nmu(d)q^n/d - (q^n/2-1)Big]$.

This completes the solution. Please let me know whether this is a fine solution, or am I missing something! Let me know, if one can approach this problem any easier way.

Thank you!

abstract-algebra combinatorics field-theory galois-theory finite-fields

share|cite|improve this question

edited Mar 29 at 15:38

hardmath

29.3k953101

asked Mar 26 at 14:51

RijuRiju

2,294414

$endgroup$

share|cite|improve this question

edited Mar 29 at 15:38

hardmath

29.3k953101

asked Mar 26 at 14:51

RijuRiju

2,294414

share|cite|improve this question

edited Mar 29 at 15:38

hardmath

29.3k953101

asked Mar 26 at 14:51

RijuRiju

2,294414

edited Mar 29 at 15:38

hardmath

29.3k953101

edited Mar 29 at 15:38

hardmath

29.3k953101

asked Mar 26 at 14:51

RijuRiju

2,294414

asked Mar 26 at 14:51

RijuRiju

2,294414