Order of finite fields is $p^n$












31












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Let $F$ be a finite field.
.How do I prove that the order of $F$ is always of order $p^n$ where $p$ is prime?










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$endgroup$








  • 4




    $begingroup$
    What is the cardinality of any finite-dimensional vector space over the field with $p$ elements?
    $endgroup$
    – NKS
    Oct 15 '11 at 17:50










  • $begingroup$
    What exactly is the scalar field and the multiplication law?
    $endgroup$
    – Mohan
    Oct 15 '11 at 18:14










  • $begingroup$
    The scalar field is the subfield consisting of the elements $0$, $1$, $1+1$, $1+1+1$, $1+1+1+1$ and so forth. (You first have to prove this is in fact a subfield, of course). Multiplication is whatever passes for multiplication in the finite field.
    $endgroup$
    – Henning Makholm
    Oct 15 '11 at 18:37












  • $begingroup$
    First prove that if $charF=p $ then $F_p$ is a subfield of $F$.
    $endgroup$
    – Frank Murphy
    Oct 15 '11 at 19:27










  • $begingroup$
    The answers to this question contain all the information that you need. IOW this is almost an exact duplicate.
    $endgroup$
    – Jyrki Lahtonen
    Oct 16 '11 at 6:17


















31












$begingroup$


Let $F$ be a finite field.
.How do I prove that the order of $F$ is always of order $p^n$ where $p$ is prime?










share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    What is the cardinality of any finite-dimensional vector space over the field with $p$ elements?
    $endgroup$
    – NKS
    Oct 15 '11 at 17:50










  • $begingroup$
    What exactly is the scalar field and the multiplication law?
    $endgroup$
    – Mohan
    Oct 15 '11 at 18:14










  • $begingroup$
    The scalar field is the subfield consisting of the elements $0$, $1$, $1+1$, $1+1+1$, $1+1+1+1$ and so forth. (You first have to prove this is in fact a subfield, of course). Multiplication is whatever passes for multiplication in the finite field.
    $endgroup$
    – Henning Makholm
    Oct 15 '11 at 18:37












  • $begingroup$
    First prove that if $charF=p $ then $F_p$ is a subfield of $F$.
    $endgroup$
    – Frank Murphy
    Oct 15 '11 at 19:27










  • $begingroup$
    The answers to this question contain all the information that you need. IOW this is almost an exact duplicate.
    $endgroup$
    – Jyrki Lahtonen
    Oct 16 '11 at 6:17
















31












31








31


23



$begingroup$


Let $F$ be a finite field.
.How do I prove that the order of $F$ is always of order $p^n$ where $p$ is prime?










share|cite|improve this question









$endgroup$




Let $F$ be a finite field.
.How do I prove that the order of $F$ is always of order $p^n$ where $p$ is prime?







finite-fields






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Oct 15 '11 at 17:45









MohanMohan

6,0271356102




6,0271356102








  • 4




    $begingroup$
    What is the cardinality of any finite-dimensional vector space over the field with $p$ elements?
    $endgroup$
    – NKS
    Oct 15 '11 at 17:50










  • $begingroup$
    What exactly is the scalar field and the multiplication law?
    $endgroup$
    – Mohan
    Oct 15 '11 at 18:14










  • $begingroup$
    The scalar field is the subfield consisting of the elements $0$, $1$, $1+1$, $1+1+1$, $1+1+1+1$ and so forth. (You first have to prove this is in fact a subfield, of course). Multiplication is whatever passes for multiplication in the finite field.
    $endgroup$
    – Henning Makholm
    Oct 15 '11 at 18:37












  • $begingroup$
    First prove that if $charF=p $ then $F_p$ is a subfield of $F$.
    $endgroup$
    – Frank Murphy
    Oct 15 '11 at 19:27










  • $begingroup$
    The answers to this question contain all the information that you need. IOW this is almost an exact duplicate.
    $endgroup$
    – Jyrki Lahtonen
    Oct 16 '11 at 6:17
















  • 4




    $begingroup$
    What is the cardinality of any finite-dimensional vector space over the field with $p$ elements?
    $endgroup$
    – NKS
    Oct 15 '11 at 17:50










  • $begingroup$
    What exactly is the scalar field and the multiplication law?
    $endgroup$
    – Mohan
    Oct 15 '11 at 18:14










  • $begingroup$
    The scalar field is the subfield consisting of the elements $0$, $1$, $1+1$, $1+1+1$, $1+1+1+1$ and so forth. (You first have to prove this is in fact a subfield, of course). Multiplication is whatever passes for multiplication in the finite field.
    $endgroup$
    – Henning Makholm
    Oct 15 '11 at 18:37












  • $begingroup$
    First prove that if $charF=p $ then $F_p$ is a subfield of $F$.
    $endgroup$
    – Frank Murphy
    Oct 15 '11 at 19:27










  • $begingroup$
    The answers to this question contain all the information that you need. IOW this is almost an exact duplicate.
    $endgroup$
    – Jyrki Lahtonen
    Oct 16 '11 at 6:17










4




4




$begingroup$
What is the cardinality of any finite-dimensional vector space over the field with $p$ elements?
$endgroup$
– NKS
Oct 15 '11 at 17:50




$begingroup$
What is the cardinality of any finite-dimensional vector space over the field with $p$ elements?
$endgroup$
– NKS
Oct 15 '11 at 17:50












$begingroup$
What exactly is the scalar field and the multiplication law?
$endgroup$
– Mohan
Oct 15 '11 at 18:14




$begingroup$
What exactly is the scalar field and the multiplication law?
$endgroup$
– Mohan
Oct 15 '11 at 18:14












$begingroup$
The scalar field is the subfield consisting of the elements $0$, $1$, $1+1$, $1+1+1$, $1+1+1+1$ and so forth. (You first have to prove this is in fact a subfield, of course). Multiplication is whatever passes for multiplication in the finite field.
$endgroup$
– Henning Makholm
Oct 15 '11 at 18:37






$begingroup$
The scalar field is the subfield consisting of the elements $0$, $1$, $1+1$, $1+1+1$, $1+1+1+1$ and so forth. (You first have to prove this is in fact a subfield, of course). Multiplication is whatever passes for multiplication in the finite field.
$endgroup$
– Henning Makholm
Oct 15 '11 at 18:37














$begingroup$
First prove that if $charF=p $ then $F_p$ is a subfield of $F$.
$endgroup$
– Frank Murphy
Oct 15 '11 at 19:27




$begingroup$
First prove that if $charF=p $ then $F_p$ is a subfield of $F$.
$endgroup$
– Frank Murphy
Oct 15 '11 at 19:27












$begingroup$
The answers to this question contain all the information that you need. IOW this is almost an exact duplicate.
$endgroup$
– Jyrki Lahtonen
Oct 16 '11 at 6:17






$begingroup$
The answers to this question contain all the information that you need. IOW this is almost an exact duplicate.
$endgroup$
– Jyrki Lahtonen
Oct 16 '11 at 6:17












6 Answers
6






active

oldest

votes


















42












$begingroup$


  1. Prove that the smallest multiple $m$ of 1 that gives zero has to be a prime. (Otherwise there are divisors of $m$ which are then divisors of zero.)


  2. Prove that a field is a vector space over a subfield.


  3. Count the elements of the field if the dimension of this vector space is $n$.







share|cite|improve this answer









$endgroup$





















    74












    $begingroup$

    Let $p$ be the characteristic of a finite field $F$. Then since $1$ has order $p$ in $(F,+)$, we know that $p$ divides $|F|$. Now let $qneq p$ be any other prime dividing $|F|$. Then by Cauchy's Theorem, there is an element $xin F$ whose order in $(F,+)$ is $q$.



    Then $qcdot x=0$. But we also have $pcdot x=0$. Now since $p$ and $q$ are relatively prime, we can find integers $a$ and $b$ such that $ap+bq=1$.



    Thus $(ap+bq)cdot x=x$. But $(ap+bq)cdot x=acdot(pcdot x)+bcdot(qcdot x)=0$, giving $x=0$, which is not possible since $x$ has positive order in $(F,+)$.



    So there is no prime other than $p$ which divides $|F|$.






    share|cite|improve this answer









    $endgroup$









    • 3




      $begingroup$
      I like this argument better than the vector space one. Surprising that it has so little attention.
      $endgroup$
      – R R
      May 7 '15 at 17:38










    • $begingroup$
      why $(ap+bq)cdot x = a cdot (p cdot x) + b cdot (q cdot x)?$ thanks!
      $endgroup$
      – scitamehtam
      Aug 27 '15 at 5:26






    • 3




      $begingroup$
      @scitamehtam An inductive argument should easily settle that $(m+n)cdot x=mcdot x+ncdot x$ for all integers $m$ and $n$ and any $xin F$. Further, an inductive argument can be used to settle $(mn)cdot x= mcdot(ncdot x)$ for all integers $m$ and $n$ and $xin F$. I guess the source of the confusion is probably the fact that here the '$cdot$' does not denote the multiplication in $F$. It is simply a notation that if $m$ is a positive integer, then $mcdot x$ is $x$ summed $m$ times. If $m$ is negative then $mcdot x:=(-m)cdot x$. If $m=0$ then $mcdot x=0$.
      $endgroup$
      – caffeinemachine
      Aug 27 '15 at 7:06






    • 8




      $begingroup$
      After the first paragraph, can’t you just argue: In the equation $x+…+x=0$ ($q$ $x$s) factor out $x$ to get $x(1+…+1)=0$. Since we’re in a field we know that $x^{-1}$ exists so multiply it to both sides to get $1+…+1=0$ ($q$ $1$s). But we know that the order of $1$ in $(F,+)$ is $p$, hence $p|q$.
      $endgroup$
      – Coffee_Table
      Nov 19 '17 at 0:19






    • 1




      $begingroup$
      Coffee Table combined with caffeinemachine gives best result :)
      $endgroup$
      – Silent
      Apr 5 at 12:08



















    11












    $begingroup$

    Let $F$ be a finite field. Then the underlying additive group of the field (let's denote this by $F^+$) has this interesting property:




    For every two non-identity (i.e. non-zero) elements $a$ and $bin F^+$, there is an automorphism $phi$ of the additive group such that $phi(a)=b$.




    This can be seen by examining the map $(xmapsto ba^{-1}x)$.



    This means the set of automorphisms of $F^+$ act transitively on $F^+$. Since automorphisms permute elements of the same order, we can conclude that every element in $F$ has the same order.



    But a finite group in which all non-identity elements have the same order is necessarily a $p$-group such that every element has prime order. This can be shown by Cauchy's Theorem.



    Suppose the order of $F^+$ had two distinct prime factors $p$ and $q$. Then $F^+$ would contain an element of order $p$ and another element of order $q$ by Cauchy's Theorem. This contradicts that every element has the same order. So the order of $F$ is indeed a prime-power. Cauchy's Theorem implies that $F$ has an element of order $p$, thus all elements have order $p$ by the hypothesis.



    So, $F$ must be of prime-power order $p^n$, and we have that $px=0$ for all $xin F$.






    share|cite|improve this answer









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      2












      $begingroup$

      First note that if $R$ is a commutative ring with identity then there exists a ring homomorphism from $Bbb Z$ to $R$ given by $1 mapsto 1_R$ having kernel $n Bbb Z$ for $n ge 0$. If $R$ is a field then the kernel would be ${0 }$ or $p Bbb Z$ for $p$ a prime. Hence the characteristic of a field is either $0$ or $p$ for $p$ a prime. Now let $R=F$ be a finite field. Consider a map $f : Bbb Z longrightarrow F$ as above. Then $f$ is a homomorphism. Since $F$ is finite $f$ cannot be one-one; for otherwise there will be a copy of $Bbb Z$ sitting inside $F$. But then $F$ will be of infinite order, a contradiction. Hence $Ker (f) ne {0 }$. Since $F$ is a field $Ker (f) = p Bbb Z$ for some prime number $p$. Then $Bbb Z/ p Bbb Z$ is embedded in $F$. So $F$ can be thought of as a vector space over $Bbb Z / p Bbb Z$. Since $F$ is finite, then dimension of $F$ as a vector space over $Bbb Z / p Bbb Z$ is finite. Lets say the dimension to be $n$. Then $|F| = p^n$. This proves that the order of any field is some power $n$ of a prime say $p$.



      QED






      share|cite|improve this answer









      $endgroup$





















        1












        $begingroup$

        A slight variation on caffeinmachine's answer that I prefer, because I think it shows more of the structure of what's going on:



        Let $F$ be a finite field (and thus has characteristic $p$, a prime).




        1. Every element of $F$ has order $p$ in the additive group $(F,+)$. So $(F,+)$ is a $p$-group.

        2. A group is a $p$-group iff it has order $p^n$ for some positive integer $n$.


        The first claim is immediate, by the distributive property of the field. Let $x in F, x neq 0_F$. We have



        begin{align}
        p cdot x &= p cdot (1_{F} x) = (p cdot 1_{F}) x
        \
        & = 0
        end{align}



        This is the smallest positive integer for which this occurs, by the definition of the characteristic of a field. So $x$ has order $p$.



        The part that we need of the second claim is a well-known corollary of Cauchy's theorem (the reverse direction is just an application of Lagrange's theorem).






        share|cite|improve this answer









        $endgroup$





















          1












          $begingroup$

          Let $n,m$ be positive integers and $F$ be a finite field. Define the operations.



          begin{align}
          (n cdot mathbb{1}_F) (m cdot mathbb{1}_F) &= (nm cdot mathbb{1_F})\
          (-n)cdot mathbb{1}_F &= -(ncdot mathbb{1}_F)\
          0 cdot mathbb{1}_F &= 0\
          end{align}



          This then suggests there is a natural homomorphism



          begin{align}
          mathbb{Z} &stackrel{phi}to F\
          n &to n cdot mathbb{1}_F
          end{align}



          Now since $F$ is finite, it has prime characteristic $p$ and we may see that $ker phi = pmathbb{Z}$. Subsequently, we also see



          $$mathbb{Z}/kerphi = mathbb{Z}/pmathbb{Z} stackrel{phi}to text{img} phi approx F_p subset F$$



          and in fact $F$ is a finite extension of $F_p$ (say $[F: F_p] = n$), that is, $F_p leq F$ or in other words, $F$ is a vector space over $F_p$. So any element $z in F$ can be written as a linear combination of basis elements $(z_1, dots, z_n) subset F$ and $a_i in F_p$



          $$z = a_1z_1 + dots a_nz_n$$



          Now by counting, each $a_i$ has $p$ choices. So going through $n$ basis elements, there are exactly $p^n$ total choices and this counts all the $z$s in $F$. Therefore $|F| = p^n$






          share|cite|improve this answer









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            6 Answers
            6






            active

            oldest

            votes








            6 Answers
            6






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            42












            $begingroup$


            1. Prove that the smallest multiple $m$ of 1 that gives zero has to be a prime. (Otherwise there are divisors of $m$ which are then divisors of zero.)


            2. Prove that a field is a vector space over a subfield.


            3. Count the elements of the field if the dimension of this vector space is $n$.







            share|cite|improve this answer









            $endgroup$


















              42












              $begingroup$


              1. Prove that the smallest multiple $m$ of 1 that gives zero has to be a prime. (Otherwise there are divisors of $m$ which are then divisors of zero.)


              2. Prove that a field is a vector space over a subfield.


              3. Count the elements of the field if the dimension of this vector space is $n$.







              share|cite|improve this answer









              $endgroup$
















                42












                42








                42





                $begingroup$


                1. Prove that the smallest multiple $m$ of 1 that gives zero has to be a prime. (Otherwise there are divisors of $m$ which are then divisors of zero.)


                2. Prove that a field is a vector space over a subfield.


                3. Count the elements of the field if the dimension of this vector space is $n$.







                share|cite|improve this answer









                $endgroup$




                1. Prove that the smallest multiple $m$ of 1 that gives zero has to be a prime. (Otherwise there are divisors of $m$ which are then divisors of zero.)


                2. Prove that a field is a vector space over a subfield.


                3. Count the elements of the field if the dimension of this vector space is $n$.








                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Oct 15 '11 at 17:52









                PhiraPhira

                18.6k245100




                18.6k245100























                    74












                    $begingroup$

                    Let $p$ be the characteristic of a finite field $F$. Then since $1$ has order $p$ in $(F,+)$, we know that $p$ divides $|F|$. Now let $qneq p$ be any other prime dividing $|F|$. Then by Cauchy's Theorem, there is an element $xin F$ whose order in $(F,+)$ is $q$.



                    Then $qcdot x=0$. But we also have $pcdot x=0$. Now since $p$ and $q$ are relatively prime, we can find integers $a$ and $b$ such that $ap+bq=1$.



                    Thus $(ap+bq)cdot x=x$. But $(ap+bq)cdot x=acdot(pcdot x)+bcdot(qcdot x)=0$, giving $x=0$, which is not possible since $x$ has positive order in $(F,+)$.



                    So there is no prime other than $p$ which divides $|F|$.






                    share|cite|improve this answer









                    $endgroup$









                    • 3




                      $begingroup$
                      I like this argument better than the vector space one. Surprising that it has so little attention.
                      $endgroup$
                      – R R
                      May 7 '15 at 17:38










                    • $begingroup$
                      why $(ap+bq)cdot x = a cdot (p cdot x) + b cdot (q cdot x)?$ thanks!
                      $endgroup$
                      – scitamehtam
                      Aug 27 '15 at 5:26






                    • 3




                      $begingroup$
                      @scitamehtam An inductive argument should easily settle that $(m+n)cdot x=mcdot x+ncdot x$ for all integers $m$ and $n$ and any $xin F$. Further, an inductive argument can be used to settle $(mn)cdot x= mcdot(ncdot x)$ for all integers $m$ and $n$ and $xin F$. I guess the source of the confusion is probably the fact that here the '$cdot$' does not denote the multiplication in $F$. It is simply a notation that if $m$ is a positive integer, then $mcdot x$ is $x$ summed $m$ times. If $m$ is negative then $mcdot x:=(-m)cdot x$. If $m=0$ then $mcdot x=0$.
                      $endgroup$
                      – caffeinemachine
                      Aug 27 '15 at 7:06






                    • 8




                      $begingroup$
                      After the first paragraph, can’t you just argue: In the equation $x+…+x=0$ ($q$ $x$s) factor out $x$ to get $x(1+…+1)=0$. Since we’re in a field we know that $x^{-1}$ exists so multiply it to both sides to get $1+…+1=0$ ($q$ $1$s). But we know that the order of $1$ in $(F,+)$ is $p$, hence $p|q$.
                      $endgroup$
                      – Coffee_Table
                      Nov 19 '17 at 0:19






                    • 1




                      $begingroup$
                      Coffee Table combined with caffeinemachine gives best result :)
                      $endgroup$
                      – Silent
                      Apr 5 at 12:08
















                    74












                    $begingroup$

                    Let $p$ be the characteristic of a finite field $F$. Then since $1$ has order $p$ in $(F,+)$, we know that $p$ divides $|F|$. Now let $qneq p$ be any other prime dividing $|F|$. Then by Cauchy's Theorem, there is an element $xin F$ whose order in $(F,+)$ is $q$.



                    Then $qcdot x=0$. But we also have $pcdot x=0$. Now since $p$ and $q$ are relatively prime, we can find integers $a$ and $b$ such that $ap+bq=1$.



                    Thus $(ap+bq)cdot x=x$. But $(ap+bq)cdot x=acdot(pcdot x)+bcdot(qcdot x)=0$, giving $x=0$, which is not possible since $x$ has positive order in $(F,+)$.



                    So there is no prime other than $p$ which divides $|F|$.






                    share|cite|improve this answer









                    $endgroup$









                    • 3




                      $begingroup$
                      I like this argument better than the vector space one. Surprising that it has so little attention.
                      $endgroup$
                      – R R
                      May 7 '15 at 17:38










                    • $begingroup$
                      why $(ap+bq)cdot x = a cdot (p cdot x) + b cdot (q cdot x)?$ thanks!
                      $endgroup$
                      – scitamehtam
                      Aug 27 '15 at 5:26






                    • 3




                      $begingroup$
                      @scitamehtam An inductive argument should easily settle that $(m+n)cdot x=mcdot x+ncdot x$ for all integers $m$ and $n$ and any $xin F$. Further, an inductive argument can be used to settle $(mn)cdot x= mcdot(ncdot x)$ for all integers $m$ and $n$ and $xin F$. I guess the source of the confusion is probably the fact that here the '$cdot$' does not denote the multiplication in $F$. It is simply a notation that if $m$ is a positive integer, then $mcdot x$ is $x$ summed $m$ times. If $m$ is negative then $mcdot x:=(-m)cdot x$. If $m=0$ then $mcdot x=0$.
                      $endgroup$
                      – caffeinemachine
                      Aug 27 '15 at 7:06






                    • 8




                      $begingroup$
                      After the first paragraph, can’t you just argue: In the equation $x+…+x=0$ ($q$ $x$s) factor out $x$ to get $x(1+…+1)=0$. Since we’re in a field we know that $x^{-1}$ exists so multiply it to both sides to get $1+…+1=0$ ($q$ $1$s). But we know that the order of $1$ in $(F,+)$ is $p$, hence $p|q$.
                      $endgroup$
                      – Coffee_Table
                      Nov 19 '17 at 0:19






                    • 1




                      $begingroup$
                      Coffee Table combined with caffeinemachine gives best result :)
                      $endgroup$
                      – Silent
                      Apr 5 at 12:08














                    74












                    74








                    74





                    $begingroup$

                    Let $p$ be the characteristic of a finite field $F$. Then since $1$ has order $p$ in $(F,+)$, we know that $p$ divides $|F|$. Now let $qneq p$ be any other prime dividing $|F|$. Then by Cauchy's Theorem, there is an element $xin F$ whose order in $(F,+)$ is $q$.



                    Then $qcdot x=0$. But we also have $pcdot x=0$. Now since $p$ and $q$ are relatively prime, we can find integers $a$ and $b$ such that $ap+bq=1$.



                    Thus $(ap+bq)cdot x=x$. But $(ap+bq)cdot x=acdot(pcdot x)+bcdot(qcdot x)=0$, giving $x=0$, which is not possible since $x$ has positive order in $(F,+)$.



                    So there is no prime other than $p$ which divides $|F|$.






                    share|cite|improve this answer









                    $endgroup$



                    Let $p$ be the characteristic of a finite field $F$. Then since $1$ has order $p$ in $(F,+)$, we know that $p$ divides $|F|$. Now let $qneq p$ be any other prime dividing $|F|$. Then by Cauchy's Theorem, there is an element $xin F$ whose order in $(F,+)$ is $q$.



                    Then $qcdot x=0$. But we also have $pcdot x=0$. Now since $p$ and $q$ are relatively prime, we can find integers $a$ and $b$ such that $ap+bq=1$.



                    Thus $(ap+bq)cdot x=x$. But $(ap+bq)cdot x=acdot(pcdot x)+bcdot(qcdot x)=0$, giving $x=0$, which is not possible since $x$ has positive order in $(F,+)$.



                    So there is no prime other than $p$ which divides $|F|$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Apr 11 '15 at 14:32









                    caffeinemachinecaffeinemachine

                    6,75121458




                    6,75121458








                    • 3




                      $begingroup$
                      I like this argument better than the vector space one. Surprising that it has so little attention.
                      $endgroup$
                      – R R
                      May 7 '15 at 17:38










                    • $begingroup$
                      why $(ap+bq)cdot x = a cdot (p cdot x) + b cdot (q cdot x)?$ thanks!
                      $endgroup$
                      – scitamehtam
                      Aug 27 '15 at 5:26






                    • 3




                      $begingroup$
                      @scitamehtam An inductive argument should easily settle that $(m+n)cdot x=mcdot x+ncdot x$ for all integers $m$ and $n$ and any $xin F$. Further, an inductive argument can be used to settle $(mn)cdot x= mcdot(ncdot x)$ for all integers $m$ and $n$ and $xin F$. I guess the source of the confusion is probably the fact that here the '$cdot$' does not denote the multiplication in $F$. It is simply a notation that if $m$ is a positive integer, then $mcdot x$ is $x$ summed $m$ times. If $m$ is negative then $mcdot x:=(-m)cdot x$. If $m=0$ then $mcdot x=0$.
                      $endgroup$
                      – caffeinemachine
                      Aug 27 '15 at 7:06






                    • 8




                      $begingroup$
                      After the first paragraph, can’t you just argue: In the equation $x+…+x=0$ ($q$ $x$s) factor out $x$ to get $x(1+…+1)=0$. Since we’re in a field we know that $x^{-1}$ exists so multiply it to both sides to get $1+…+1=0$ ($q$ $1$s). But we know that the order of $1$ in $(F,+)$ is $p$, hence $p|q$.
                      $endgroup$
                      – Coffee_Table
                      Nov 19 '17 at 0:19






                    • 1




                      $begingroup$
                      Coffee Table combined with caffeinemachine gives best result :)
                      $endgroup$
                      – Silent
                      Apr 5 at 12:08














                    • 3




                      $begingroup$
                      I like this argument better than the vector space one. Surprising that it has so little attention.
                      $endgroup$
                      – R R
                      May 7 '15 at 17:38










                    • $begingroup$
                      why $(ap+bq)cdot x = a cdot (p cdot x) + b cdot (q cdot x)?$ thanks!
                      $endgroup$
                      – scitamehtam
                      Aug 27 '15 at 5:26






                    • 3




                      $begingroup$
                      @scitamehtam An inductive argument should easily settle that $(m+n)cdot x=mcdot x+ncdot x$ for all integers $m$ and $n$ and any $xin F$. Further, an inductive argument can be used to settle $(mn)cdot x= mcdot(ncdot x)$ for all integers $m$ and $n$ and $xin F$. I guess the source of the confusion is probably the fact that here the '$cdot$' does not denote the multiplication in $F$. It is simply a notation that if $m$ is a positive integer, then $mcdot x$ is $x$ summed $m$ times. If $m$ is negative then $mcdot x:=(-m)cdot x$. If $m=0$ then $mcdot x=0$.
                      $endgroup$
                      – caffeinemachine
                      Aug 27 '15 at 7:06






                    • 8




                      $begingroup$
                      After the first paragraph, can’t you just argue: In the equation $x+…+x=0$ ($q$ $x$s) factor out $x$ to get $x(1+…+1)=0$. Since we’re in a field we know that $x^{-1}$ exists so multiply it to both sides to get $1+…+1=0$ ($q$ $1$s). But we know that the order of $1$ in $(F,+)$ is $p$, hence $p|q$.
                      $endgroup$
                      – Coffee_Table
                      Nov 19 '17 at 0:19






                    • 1




                      $begingroup$
                      Coffee Table combined with caffeinemachine gives best result :)
                      $endgroup$
                      – Silent
                      Apr 5 at 12:08








                    3




                    3




                    $begingroup$
                    I like this argument better than the vector space one. Surprising that it has so little attention.
                    $endgroup$
                    – R R
                    May 7 '15 at 17:38




                    $begingroup$
                    I like this argument better than the vector space one. Surprising that it has so little attention.
                    $endgroup$
                    – R R
                    May 7 '15 at 17:38












                    $begingroup$
                    why $(ap+bq)cdot x = a cdot (p cdot x) + b cdot (q cdot x)?$ thanks!
                    $endgroup$
                    – scitamehtam
                    Aug 27 '15 at 5:26




                    $begingroup$
                    why $(ap+bq)cdot x = a cdot (p cdot x) + b cdot (q cdot x)?$ thanks!
                    $endgroup$
                    – scitamehtam
                    Aug 27 '15 at 5:26




                    3




                    3




                    $begingroup$
                    @scitamehtam An inductive argument should easily settle that $(m+n)cdot x=mcdot x+ncdot x$ for all integers $m$ and $n$ and any $xin F$. Further, an inductive argument can be used to settle $(mn)cdot x= mcdot(ncdot x)$ for all integers $m$ and $n$ and $xin F$. I guess the source of the confusion is probably the fact that here the '$cdot$' does not denote the multiplication in $F$. It is simply a notation that if $m$ is a positive integer, then $mcdot x$ is $x$ summed $m$ times. If $m$ is negative then $mcdot x:=(-m)cdot x$. If $m=0$ then $mcdot x=0$.
                    $endgroup$
                    – caffeinemachine
                    Aug 27 '15 at 7:06




                    $begingroup$
                    @scitamehtam An inductive argument should easily settle that $(m+n)cdot x=mcdot x+ncdot x$ for all integers $m$ and $n$ and any $xin F$. Further, an inductive argument can be used to settle $(mn)cdot x= mcdot(ncdot x)$ for all integers $m$ and $n$ and $xin F$. I guess the source of the confusion is probably the fact that here the '$cdot$' does not denote the multiplication in $F$. It is simply a notation that if $m$ is a positive integer, then $mcdot x$ is $x$ summed $m$ times. If $m$ is negative then $mcdot x:=(-m)cdot x$. If $m=0$ then $mcdot x=0$.
                    $endgroup$
                    – caffeinemachine
                    Aug 27 '15 at 7:06




                    8




                    8




                    $begingroup$
                    After the first paragraph, can’t you just argue: In the equation $x+…+x=0$ ($q$ $x$s) factor out $x$ to get $x(1+…+1)=0$. Since we’re in a field we know that $x^{-1}$ exists so multiply it to both sides to get $1+…+1=0$ ($q$ $1$s). But we know that the order of $1$ in $(F,+)$ is $p$, hence $p|q$.
                    $endgroup$
                    – Coffee_Table
                    Nov 19 '17 at 0:19




                    $begingroup$
                    After the first paragraph, can’t you just argue: In the equation $x+…+x=0$ ($q$ $x$s) factor out $x$ to get $x(1+…+1)=0$. Since we’re in a field we know that $x^{-1}$ exists so multiply it to both sides to get $1+…+1=0$ ($q$ $1$s). But we know that the order of $1$ in $(F,+)$ is $p$, hence $p|q$.
                    $endgroup$
                    – Coffee_Table
                    Nov 19 '17 at 0:19




                    1




                    1




                    $begingroup$
                    Coffee Table combined with caffeinemachine gives best result :)
                    $endgroup$
                    – Silent
                    Apr 5 at 12:08




                    $begingroup$
                    Coffee Table combined with caffeinemachine gives best result :)
                    $endgroup$
                    – Silent
                    Apr 5 at 12:08











                    11












                    $begingroup$

                    Let $F$ be a finite field. Then the underlying additive group of the field (let's denote this by $F^+$) has this interesting property:




                    For every two non-identity (i.e. non-zero) elements $a$ and $bin F^+$, there is an automorphism $phi$ of the additive group such that $phi(a)=b$.




                    This can be seen by examining the map $(xmapsto ba^{-1}x)$.



                    This means the set of automorphisms of $F^+$ act transitively on $F^+$. Since automorphisms permute elements of the same order, we can conclude that every element in $F$ has the same order.



                    But a finite group in which all non-identity elements have the same order is necessarily a $p$-group such that every element has prime order. This can be shown by Cauchy's Theorem.



                    Suppose the order of $F^+$ had two distinct prime factors $p$ and $q$. Then $F^+$ would contain an element of order $p$ and another element of order $q$ by Cauchy's Theorem. This contradicts that every element has the same order. So the order of $F$ is indeed a prime-power. Cauchy's Theorem implies that $F$ has an element of order $p$, thus all elements have order $p$ by the hypothesis.



                    So, $F$ must be of prime-power order $p^n$, and we have that $px=0$ for all $xin F$.






                    share|cite|improve this answer









                    $endgroup$


















                      11












                      $begingroup$

                      Let $F$ be a finite field. Then the underlying additive group of the field (let's denote this by $F^+$) has this interesting property:




                      For every two non-identity (i.e. non-zero) elements $a$ and $bin F^+$, there is an automorphism $phi$ of the additive group such that $phi(a)=b$.




                      This can be seen by examining the map $(xmapsto ba^{-1}x)$.



                      This means the set of automorphisms of $F^+$ act transitively on $F^+$. Since automorphisms permute elements of the same order, we can conclude that every element in $F$ has the same order.



                      But a finite group in which all non-identity elements have the same order is necessarily a $p$-group such that every element has prime order. This can be shown by Cauchy's Theorem.



                      Suppose the order of $F^+$ had two distinct prime factors $p$ and $q$. Then $F^+$ would contain an element of order $p$ and another element of order $q$ by Cauchy's Theorem. This contradicts that every element has the same order. So the order of $F$ is indeed a prime-power. Cauchy's Theorem implies that $F$ has an element of order $p$, thus all elements have order $p$ by the hypothesis.



                      So, $F$ must be of prime-power order $p^n$, and we have that $px=0$ for all $xin F$.






                      share|cite|improve this answer









                      $endgroup$
















                        11












                        11








                        11





                        $begingroup$

                        Let $F$ be a finite field. Then the underlying additive group of the field (let's denote this by $F^+$) has this interesting property:




                        For every two non-identity (i.e. non-zero) elements $a$ and $bin F^+$, there is an automorphism $phi$ of the additive group such that $phi(a)=b$.




                        This can be seen by examining the map $(xmapsto ba^{-1}x)$.



                        This means the set of automorphisms of $F^+$ act transitively on $F^+$. Since automorphisms permute elements of the same order, we can conclude that every element in $F$ has the same order.



                        But a finite group in which all non-identity elements have the same order is necessarily a $p$-group such that every element has prime order. This can be shown by Cauchy's Theorem.



                        Suppose the order of $F^+$ had two distinct prime factors $p$ and $q$. Then $F^+$ would contain an element of order $p$ and another element of order $q$ by Cauchy's Theorem. This contradicts that every element has the same order. So the order of $F$ is indeed a prime-power. Cauchy's Theorem implies that $F$ has an element of order $p$, thus all elements have order $p$ by the hypothesis.



                        So, $F$ must be of prime-power order $p^n$, and we have that $px=0$ for all $xin F$.






                        share|cite|improve this answer









                        $endgroup$



                        Let $F$ be a finite field. Then the underlying additive group of the field (let's denote this by $F^+$) has this interesting property:




                        For every two non-identity (i.e. non-zero) elements $a$ and $bin F^+$, there is an automorphism $phi$ of the additive group such that $phi(a)=b$.




                        This can be seen by examining the map $(xmapsto ba^{-1}x)$.



                        This means the set of automorphisms of $F^+$ act transitively on $F^+$. Since automorphisms permute elements of the same order, we can conclude that every element in $F$ has the same order.



                        But a finite group in which all non-identity elements have the same order is necessarily a $p$-group such that every element has prime order. This can be shown by Cauchy's Theorem.



                        Suppose the order of $F^+$ had two distinct prime factors $p$ and $q$. Then $F^+$ would contain an element of order $p$ and another element of order $q$ by Cauchy's Theorem. This contradicts that every element has the same order. So the order of $F$ is indeed a prime-power. Cauchy's Theorem implies that $F$ has an element of order $p$, thus all elements have order $p$ by the hypothesis.



                        So, $F$ must be of prime-power order $p^n$, and we have that $px=0$ for all $xin F$.







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Aug 1 '14 at 19:29









                        Robert WolfeRobert Wolfe

                        6,01922763




                        6,01922763























                            2












                            $begingroup$

                            First note that if $R$ is a commutative ring with identity then there exists a ring homomorphism from $Bbb Z$ to $R$ given by $1 mapsto 1_R$ having kernel $n Bbb Z$ for $n ge 0$. If $R$ is a field then the kernel would be ${0 }$ or $p Bbb Z$ for $p$ a prime. Hence the characteristic of a field is either $0$ or $p$ for $p$ a prime. Now let $R=F$ be a finite field. Consider a map $f : Bbb Z longrightarrow F$ as above. Then $f$ is a homomorphism. Since $F$ is finite $f$ cannot be one-one; for otherwise there will be a copy of $Bbb Z$ sitting inside $F$. But then $F$ will be of infinite order, a contradiction. Hence $Ker (f) ne {0 }$. Since $F$ is a field $Ker (f) = p Bbb Z$ for some prime number $p$. Then $Bbb Z/ p Bbb Z$ is embedded in $F$. So $F$ can be thought of as a vector space over $Bbb Z / p Bbb Z$. Since $F$ is finite, then dimension of $F$ as a vector space over $Bbb Z / p Bbb Z$ is finite. Lets say the dimension to be $n$. Then $|F| = p^n$. This proves that the order of any field is some power $n$ of a prime say $p$.



                            QED






                            share|cite|improve this answer









                            $endgroup$


















                              2












                              $begingroup$

                              First note that if $R$ is a commutative ring with identity then there exists a ring homomorphism from $Bbb Z$ to $R$ given by $1 mapsto 1_R$ having kernel $n Bbb Z$ for $n ge 0$. If $R$ is a field then the kernel would be ${0 }$ or $p Bbb Z$ for $p$ a prime. Hence the characteristic of a field is either $0$ or $p$ for $p$ a prime. Now let $R=F$ be a finite field. Consider a map $f : Bbb Z longrightarrow F$ as above. Then $f$ is a homomorphism. Since $F$ is finite $f$ cannot be one-one; for otherwise there will be a copy of $Bbb Z$ sitting inside $F$. But then $F$ will be of infinite order, a contradiction. Hence $Ker (f) ne {0 }$. Since $F$ is a field $Ker (f) = p Bbb Z$ for some prime number $p$. Then $Bbb Z/ p Bbb Z$ is embedded in $F$. So $F$ can be thought of as a vector space over $Bbb Z / p Bbb Z$. Since $F$ is finite, then dimension of $F$ as a vector space over $Bbb Z / p Bbb Z$ is finite. Lets say the dimension to be $n$. Then $|F| = p^n$. This proves that the order of any field is some power $n$ of a prime say $p$.



                              QED






                              share|cite|improve this answer









                              $endgroup$
















                                2












                                2








                                2





                                $begingroup$

                                First note that if $R$ is a commutative ring with identity then there exists a ring homomorphism from $Bbb Z$ to $R$ given by $1 mapsto 1_R$ having kernel $n Bbb Z$ for $n ge 0$. If $R$ is a field then the kernel would be ${0 }$ or $p Bbb Z$ for $p$ a prime. Hence the characteristic of a field is either $0$ or $p$ for $p$ a prime. Now let $R=F$ be a finite field. Consider a map $f : Bbb Z longrightarrow F$ as above. Then $f$ is a homomorphism. Since $F$ is finite $f$ cannot be one-one; for otherwise there will be a copy of $Bbb Z$ sitting inside $F$. But then $F$ will be of infinite order, a contradiction. Hence $Ker (f) ne {0 }$. Since $F$ is a field $Ker (f) = p Bbb Z$ for some prime number $p$. Then $Bbb Z/ p Bbb Z$ is embedded in $F$. So $F$ can be thought of as a vector space over $Bbb Z / p Bbb Z$. Since $F$ is finite, then dimension of $F$ as a vector space over $Bbb Z / p Bbb Z$ is finite. Lets say the dimension to be $n$. Then $|F| = p^n$. This proves that the order of any field is some power $n$ of a prime say $p$.



                                QED






                                share|cite|improve this answer









                                $endgroup$



                                First note that if $R$ is a commutative ring with identity then there exists a ring homomorphism from $Bbb Z$ to $R$ given by $1 mapsto 1_R$ having kernel $n Bbb Z$ for $n ge 0$. If $R$ is a field then the kernel would be ${0 }$ or $p Bbb Z$ for $p$ a prime. Hence the characteristic of a field is either $0$ or $p$ for $p$ a prime. Now let $R=F$ be a finite field. Consider a map $f : Bbb Z longrightarrow F$ as above. Then $f$ is a homomorphism. Since $F$ is finite $f$ cannot be one-one; for otherwise there will be a copy of $Bbb Z$ sitting inside $F$. But then $F$ will be of infinite order, a contradiction. Hence $Ker (f) ne {0 }$. Since $F$ is a field $Ker (f) = p Bbb Z$ for some prime number $p$. Then $Bbb Z/ p Bbb Z$ is embedded in $F$. So $F$ can be thought of as a vector space over $Bbb Z / p Bbb Z$. Since $F$ is finite, then dimension of $F$ as a vector space over $Bbb Z / p Bbb Z$ is finite. Lets say the dimension to be $n$. Then $|F| = p^n$. This proves that the order of any field is some power $n$ of a prime say $p$.



                                QED







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Jan 9 at 6:32









                                Dbchatto67Dbchatto67

                                3,185625




                                3,185625























                                    1












                                    $begingroup$

                                    A slight variation on caffeinmachine's answer that I prefer, because I think it shows more of the structure of what's going on:



                                    Let $F$ be a finite field (and thus has characteristic $p$, a prime).




                                    1. Every element of $F$ has order $p$ in the additive group $(F,+)$. So $(F,+)$ is a $p$-group.

                                    2. A group is a $p$-group iff it has order $p^n$ for some positive integer $n$.


                                    The first claim is immediate, by the distributive property of the field. Let $x in F, x neq 0_F$. We have



                                    begin{align}
                                    p cdot x &= p cdot (1_{F} x) = (p cdot 1_{F}) x
                                    \
                                    & = 0
                                    end{align}



                                    This is the smallest positive integer for which this occurs, by the definition of the characteristic of a field. So $x$ has order $p$.



                                    The part that we need of the second claim is a well-known corollary of Cauchy's theorem (the reverse direction is just an application of Lagrange's theorem).






                                    share|cite|improve this answer









                                    $endgroup$


















                                      1












                                      $begingroup$

                                      A slight variation on caffeinmachine's answer that I prefer, because I think it shows more of the structure of what's going on:



                                      Let $F$ be a finite field (and thus has characteristic $p$, a prime).




                                      1. Every element of $F$ has order $p$ in the additive group $(F,+)$. So $(F,+)$ is a $p$-group.

                                      2. A group is a $p$-group iff it has order $p^n$ for some positive integer $n$.


                                      The first claim is immediate, by the distributive property of the field. Let $x in F, x neq 0_F$. We have



                                      begin{align}
                                      p cdot x &= p cdot (1_{F} x) = (p cdot 1_{F}) x
                                      \
                                      & = 0
                                      end{align}



                                      This is the smallest positive integer for which this occurs, by the definition of the characteristic of a field. So $x$ has order $p$.



                                      The part that we need of the second claim is a well-known corollary of Cauchy's theorem (the reverse direction is just an application of Lagrange's theorem).






                                      share|cite|improve this answer









                                      $endgroup$
















                                        1












                                        1








                                        1





                                        $begingroup$

                                        A slight variation on caffeinmachine's answer that I prefer, because I think it shows more of the structure of what's going on:



                                        Let $F$ be a finite field (and thus has characteristic $p$, a prime).




                                        1. Every element of $F$ has order $p$ in the additive group $(F,+)$. So $(F,+)$ is a $p$-group.

                                        2. A group is a $p$-group iff it has order $p^n$ for some positive integer $n$.


                                        The first claim is immediate, by the distributive property of the field. Let $x in F, x neq 0_F$. We have



                                        begin{align}
                                        p cdot x &= p cdot (1_{F} x) = (p cdot 1_{F}) x
                                        \
                                        & = 0
                                        end{align}



                                        This is the smallest positive integer for which this occurs, by the definition of the characteristic of a field. So $x$ has order $p$.



                                        The part that we need of the second claim is a well-known corollary of Cauchy's theorem (the reverse direction is just an application of Lagrange's theorem).






                                        share|cite|improve this answer









                                        $endgroup$



                                        A slight variation on caffeinmachine's answer that I prefer, because I think it shows more of the structure of what's going on:



                                        Let $F$ be a finite field (and thus has characteristic $p$, a prime).




                                        1. Every element of $F$ has order $p$ in the additive group $(F,+)$. So $(F,+)$ is a $p$-group.

                                        2. A group is a $p$-group iff it has order $p^n$ for some positive integer $n$.


                                        The first claim is immediate, by the distributive property of the field. Let $x in F, x neq 0_F$. We have



                                        begin{align}
                                        p cdot x &= p cdot (1_{F} x) = (p cdot 1_{F}) x
                                        \
                                        & = 0
                                        end{align}



                                        This is the smallest positive integer for which this occurs, by the definition of the characteristic of a field. So $x$ has order $p$.



                                        The part that we need of the second claim is a well-known corollary of Cauchy's theorem (the reverse direction is just an application of Lagrange's theorem).







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered Jul 9 '18 at 10:13









                                        Jonathan RaynerJonathan Rayner

                                        164112




                                        164112























                                            1












                                            $begingroup$

                                            Let $n,m$ be positive integers and $F$ be a finite field. Define the operations.



                                            begin{align}
                                            (n cdot mathbb{1}_F) (m cdot mathbb{1}_F) &= (nm cdot mathbb{1_F})\
                                            (-n)cdot mathbb{1}_F &= -(ncdot mathbb{1}_F)\
                                            0 cdot mathbb{1}_F &= 0\
                                            end{align}



                                            This then suggests there is a natural homomorphism



                                            begin{align}
                                            mathbb{Z} &stackrel{phi}to F\
                                            n &to n cdot mathbb{1}_F
                                            end{align}



                                            Now since $F$ is finite, it has prime characteristic $p$ and we may see that $ker phi = pmathbb{Z}$. Subsequently, we also see



                                            $$mathbb{Z}/kerphi = mathbb{Z}/pmathbb{Z} stackrel{phi}to text{img} phi approx F_p subset F$$



                                            and in fact $F$ is a finite extension of $F_p$ (say $[F: F_p] = n$), that is, $F_p leq F$ or in other words, $F$ is a vector space over $F_p$. So any element $z in F$ can be written as a linear combination of basis elements $(z_1, dots, z_n) subset F$ and $a_i in F_p$



                                            $$z = a_1z_1 + dots a_nz_n$$



                                            Now by counting, each $a_i$ has $p$ choices. So going through $n$ basis elements, there are exactly $p^n$ total choices and this counts all the $z$s in $F$. Therefore $|F| = p^n$






                                            share|cite|improve this answer









                                            $endgroup$


















                                              1












                                              $begingroup$

                                              Let $n,m$ be positive integers and $F$ be a finite field. Define the operations.



                                              begin{align}
                                              (n cdot mathbb{1}_F) (m cdot mathbb{1}_F) &= (nm cdot mathbb{1_F})\
                                              (-n)cdot mathbb{1}_F &= -(ncdot mathbb{1}_F)\
                                              0 cdot mathbb{1}_F &= 0\
                                              end{align}



                                              This then suggests there is a natural homomorphism



                                              begin{align}
                                              mathbb{Z} &stackrel{phi}to F\
                                              n &to n cdot mathbb{1}_F
                                              end{align}



                                              Now since $F$ is finite, it has prime characteristic $p$ and we may see that $ker phi = pmathbb{Z}$. Subsequently, we also see



                                              $$mathbb{Z}/kerphi = mathbb{Z}/pmathbb{Z} stackrel{phi}to text{img} phi approx F_p subset F$$



                                              and in fact $F$ is a finite extension of $F_p$ (say $[F: F_p] = n$), that is, $F_p leq F$ or in other words, $F$ is a vector space over $F_p$. So any element $z in F$ can be written as a linear combination of basis elements $(z_1, dots, z_n) subset F$ and $a_i in F_p$



                                              $$z = a_1z_1 + dots a_nz_n$$



                                              Now by counting, each $a_i$ has $p$ choices. So going through $n$ basis elements, there are exactly $p^n$ total choices and this counts all the $z$s in $F$. Therefore $|F| = p^n$






                                              share|cite|improve this answer









                                              $endgroup$
















                                                1












                                                1








                                                1





                                                $begingroup$

                                                Let $n,m$ be positive integers and $F$ be a finite field. Define the operations.



                                                begin{align}
                                                (n cdot mathbb{1}_F) (m cdot mathbb{1}_F) &= (nm cdot mathbb{1_F})\
                                                (-n)cdot mathbb{1}_F &= -(ncdot mathbb{1}_F)\
                                                0 cdot mathbb{1}_F &= 0\
                                                end{align}



                                                This then suggests there is a natural homomorphism



                                                begin{align}
                                                mathbb{Z} &stackrel{phi}to F\
                                                n &to n cdot mathbb{1}_F
                                                end{align}



                                                Now since $F$ is finite, it has prime characteristic $p$ and we may see that $ker phi = pmathbb{Z}$. Subsequently, we also see



                                                $$mathbb{Z}/kerphi = mathbb{Z}/pmathbb{Z} stackrel{phi}to text{img} phi approx F_p subset F$$



                                                and in fact $F$ is a finite extension of $F_p$ (say $[F: F_p] = n$), that is, $F_p leq F$ or in other words, $F$ is a vector space over $F_p$. So any element $z in F$ can be written as a linear combination of basis elements $(z_1, dots, z_n) subset F$ and $a_i in F_p$



                                                $$z = a_1z_1 + dots a_nz_n$$



                                                Now by counting, each $a_i$ has $p$ choices. So going through $n$ basis elements, there are exactly $p^n$ total choices and this counts all the $z$s in $F$. Therefore $|F| = p^n$






                                                share|cite|improve this answer









                                                $endgroup$



                                                Let $n,m$ be positive integers and $F$ be a finite field. Define the operations.



                                                begin{align}
                                                (n cdot mathbb{1}_F) (m cdot mathbb{1}_F) &= (nm cdot mathbb{1_F})\
                                                (-n)cdot mathbb{1}_F &= -(ncdot mathbb{1}_F)\
                                                0 cdot mathbb{1}_F &= 0\
                                                end{align}



                                                This then suggests there is a natural homomorphism



                                                begin{align}
                                                mathbb{Z} &stackrel{phi}to F\
                                                n &to n cdot mathbb{1}_F
                                                end{align}



                                                Now since $F$ is finite, it has prime characteristic $p$ and we may see that $ker phi = pmathbb{Z}$. Subsequently, we also see



                                                $$mathbb{Z}/kerphi = mathbb{Z}/pmathbb{Z} stackrel{phi}to text{img} phi approx F_p subset F$$



                                                and in fact $F$ is a finite extension of $F_p$ (say $[F: F_p] = n$), that is, $F_p leq F$ or in other words, $F$ is a vector space over $F_p$. So any element $z in F$ can be written as a linear combination of basis elements $(z_1, dots, z_n) subset F$ and $a_i in F_p$



                                                $$z = a_1z_1 + dots a_nz_n$$



                                                Now by counting, each $a_i$ has $p$ choices. So going through $n$ basis elements, there are exactly $p^n$ total choices and this counts all the $z$s in $F$. Therefore $|F| = p^n$







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                                                answered Feb 17 at 6:03









                                                IAmNoOneIAmNoOne

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