What does it mean that field $mathbb{F}_{p^n}$ “contains” the prime field $mathbb{Z}_p$?












4












$begingroup$


I have read in few books (example Computational Number Theory, page 77) that any extension field $mathbb{F}_{p^n}$ "contains" as a subfield the prime field $mathbb{Z}_p$?



What exactly does "contains" mean? Since the representations of the two fields are not same (we have to use polynomials or some other structure to represent $mathbb{F}_{p^n}$, while we use modular arithmetic to represent $mathbb{Z}_p$), how can we say that one contains the other? Does it actually mean that it contains a subfield that is isomorphic to $mathbb{Z}_p$?



For instance, we can represent $mathbb{F}_{3^2}$ using arbitrary symbols ${0, 1, a, b, c, d, e, f, g}$ or by pairs ${(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)}$. How does this "contain" $mathbb{Z}_3$?



This is more related to the computational perspective, so for example if I am working in $mathbb{F}_{p^n}$, how do I represent the subfield and work directly on integers?



Additionally if the subfield is itself an extension field (i.e. $mathbb{F}_{p^m}$ for some $m$ that divides $n$), how can I use a smaller data structure to represent it while at the same time keeping it also part of $mathbb{F}_{p^n}$?






number-theory field-theory finite-fields extension-field computational-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Not an answer, just two comments. (i) $mathbb{F}_p$ cannot be a (proper) extension field; (ii) equally important, though you don't ask about it, is that $mathbb{F}_{p^n}$ contains exactly one $mathbb{F}_p$, and there is only one way of setting up an isomorphism $mathbb{F}_ptomathbb{F}_{p^n}$.
    $endgroup$
    – ancientmathematician
    Jan 6 at 7:51












  • $begingroup$
    @ancientmathematician for (i), I meant some other extension field $mathbb{F}_{p^m}$, where $m$ divides $n$. Also, good point (ii).
    $endgroup$
    – Jus12
    Jan 6 at 8:09
















4












$begingroup$


I have read in few books (example Computational Number Theory, page 77) that any extension field $mathbb{F}_{p^n}$ "contains" as a subfield the prime field $mathbb{Z}_p$?



What exactly does "contains" mean? Since the representations of the two fields are not same (we have to use polynomials or some other structure to represent $mathbb{F}_{p^n}$, while we use modular arithmetic to represent $mathbb{Z}_p$), how can we say that one contains the other? Does it actually mean that it contains a subfield that is isomorphic to $mathbb{Z}_p$?



For instance, we can represent $mathbb{F}_{3^2}$ using arbitrary symbols ${0, 1, a, b, c, d, e, f, g}$ or by pairs ${(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)}$. How does this "contain" $mathbb{Z}_3$?



This is more related to the computational perspective, so for example if I am working in $mathbb{F}_{p^n}$, how do I represent the subfield and work directly on integers?



Additionally if the subfield is itself an extension field (i.e. $mathbb{F}_{p^m}$ for some $m$ that divides $n$), how can I use a smaller data structure to represent it while at the same time keeping it also part of $mathbb{F}_{p^n}$?






number-theory field-theory finite-fields extension-field computational-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Not an answer, just two comments. (i) $mathbb{F}_p$ cannot be a (proper) extension field; (ii) equally important, though you don't ask about it, is that $mathbb{F}_{p^n}$ contains exactly one $mathbb{F}_p$, and there is only one way of setting up an isomorphism $mathbb{F}_ptomathbb{F}_{p^n}$.
    $endgroup$
    – ancientmathematician
    Jan 6 at 7:51












  • $begingroup$
    @ancientmathematician for (i), I meant some other extension field $mathbb{F}_{p^m}$, where $m$ divides $n$. Also, good point (ii).
    $endgroup$
    – Jus12
    Jan 6 at 8:09














4












4








4





$begingroup$


I have read in few books (example Computational Number Theory, page 77) that any extension field $mathbb{F}_{p^n}$ "contains" as a subfield the prime field $mathbb{Z}_p$?



What exactly does "contains" mean? Since the representations of the two fields are not same (we have to use polynomials or some other structure to represent $mathbb{F}_{p^n}$, while we use modular arithmetic to represent $mathbb{Z}_p$), how can we say that one contains the other? Does it actually mean that it contains a subfield that is isomorphic to $mathbb{Z}_p$?



For instance, we can represent $mathbb{F}_{3^2}$ using arbitrary symbols ${0, 1, a, b, c, d, e, f, g}$ or by pairs ${(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)}$. How does this "contain" $mathbb{Z}_3$?



This is more related to the computational perspective, so for example if I am working in $mathbb{F}_{p^n}$, how do I represent the subfield and work directly on integers?



Additionally if the subfield is itself an extension field (i.e. $mathbb{F}_{p^m}$ for some $m$ that divides $n$), how can I use a smaller data structure to represent it while at the same time keeping it also part of $mathbb{F}_{p^n}$?






number-theory field-theory finite-fields extension-field computational-algebra







share|cite|improve this question











$endgroup$




I have read in few books (example Computational Number Theory, page 77) that any extension field $mathbb{F}_{p^n}$ "contains" as a subfield the prime field $mathbb{Z}_p$?



What exactly does "contains" mean? Since the representations of the two fields are not same (we have to use polynomials or some other structure to represent $mathbb{F}_{p^n}$, while we use modular arithmetic to represent $mathbb{Z}_p$), how can we say that one contains the other? Does it actually mean that it contains a subfield that is isomorphic to $mathbb{Z}_p$?



For instance, we can represent $mathbb{F}_{3^2}$ using arbitrary symbols ${0, 1, a, b, c, d, e, f, g}$ or by pairs ${(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)}$. How does this "contain" $mathbb{Z}_3$?



This is more related to the computational perspective, so for example if I am working in $mathbb{F}_{p^n}$, how do I represent the subfield and work directly on integers?



Additionally if the subfield is itself an extension field (i.e. $mathbb{F}_{p^m}$ for some $m$ that divides $n$), how can I use a smaller data structure to represent it while at the same time keeping it also part of $mathbb{F}_{p^n}$?






number-theory field-theory finite-fields extension-field computational-algebra




number-theory field-theory finite-fields extension-field computational-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 8:08







Jus12

















asked Jan 6 at 7:30









Jus12Jus12

1808




1808








  • 1




    $begingroup$
    Not an answer, just two comments. (i) $mathbb{F}_p$ cannot be a (proper) extension field; (ii) equally important, though you don't ask about it, is that $mathbb{F}_{p^n}$ contains exactly one $mathbb{F}_p$, and there is only one way of setting up an isomorphism $mathbb{F}_ptomathbb{F}_{p^n}$.
    $endgroup$
    – ancientmathematician
    Jan 6 at 7:51












  • $begingroup$
    @ancientmathematician for (i), I meant some other extension field $mathbb{F}_{p^m}$, where $m$ divides $n$. Also, good point (ii).
    $endgroup$
    – Jus12
    Jan 6 at 8:09














  • 1




    $begingroup$
    Not an answer, just two comments. (i) $mathbb{F}_p$ cannot be a (proper) extension field; (ii) equally important, though you don't ask about it, is that $mathbb{F}_{p^n}$ contains exactly one $mathbb{F}_p$, and there is only one way of setting up an isomorphism $mathbb{F}_ptomathbb{F}_{p^n}$.
    $endgroup$
    – ancientmathematician
    Jan 6 at 7:51












  • $begingroup$
    @ancientmathematician for (i), I meant some other extension field $mathbb{F}_{p^m}$, where $m$ divides $n$. Also, good point (ii).
    $endgroup$
    – Jus12
    Jan 6 at 8:09








1




1




$begingroup$
Not an answer, just two comments. (i) $mathbb{F}_p$ cannot be a (proper) extension field; (ii) equally important, though you don't ask about it, is that $mathbb{F}_{p^n}$ contains exactly one $mathbb{F}_p$, and there is only one way of setting up an isomorphism $mathbb{F}_ptomathbb{F}_{p^n}$.
$endgroup$
– ancientmathematician
Jan 6 at 7:51






$begingroup$
Not an answer, just two comments. (i) $mathbb{F}_p$ cannot be a (proper) extension field; (ii) equally important, though you don't ask about it, is that $mathbb{F}_{p^n}$ contains exactly one $mathbb{F}_p$, and there is only one way of setting up an isomorphism $mathbb{F}_ptomathbb{F}_{p^n}$.
$endgroup$
– ancientmathematician
Jan 6 at 7:51














$begingroup$
@ancientmathematician for (i), I meant some other extension field $mathbb{F}_{p^m}$, where $m$ divides $n$. Also, good point (ii).
$endgroup$
– Jus12
Jan 6 at 8:09




$begingroup$
@ancientmathematician for (i), I meant some other extension field $mathbb{F}_{p^m}$, where $m$ divides $n$. Also, good point (ii).
$endgroup$
– Jus12
Jan 6 at 8:09










2 Answers
2






active

oldest

votes


















6












$begingroup$

Technically it means that an isomorphic copy of $Bbb{Z}_p$ is in $Bbb{F}_{p^n}$. For example, in $Bbb{F}_9$, the constant polynomials will be an isomorphic copy of $Bbb{Z}_3$.



begin{align*}
Bbb{F}_9 & =Bbb{Z}_3[x]/langle x^2+1rangle\
& ={ax+b , | , a,b in Bbb{Z}_3, x^2+1 equiv 0}\
&={0,1,2,x,x+1,x+2,2x,2x+1,2x+2}
end{align*}



Then the set of constant polynomials ${0,1,2}$ is isomorphic to $Bbb{Z}_3$.




With your edited version:



$Bbb{F}_{p^n}$ can be thought of as a vector space over the field of scalars $Bbb{Z}_p$. Thus each object in $Bbb{F}_{p^n}$ is a $n-$tuple vector (this is very similar to the polynomial representation I have used above).



So consider the set of the vectors
$$S={(0,0,ldots ,c) , | , c in Bbb{Z}_p}.$$
This set in $Bbb{F}_{p^n}$ is isomorphic to $Bbb{Z}_p$.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    The mapping ${Bbb Z}_prightarrow {Bbb F}_{p^n}$ given by $amapsto acdot 1$, where $acdot 1 = 1+ldots+1$ ($a$-times) is the $a$-multiple of $1$ and $1$ is the unit element of ${Bbb F}_{p^n}$, is a (ring) monomorphism. In this way, ${Bbb F}_{p^n}$ contains an isomorphic copy of ${Bbb Z}_p$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I wanted a representation-independent way to define this subset. Hence, this useful. We can use "double and repeat" if $a$ is large.
      $endgroup$
      – Jus12
      Jan 6 at 8:28












    • $begingroup$
      Indeed, ''double and repeat'' can be used.
      $endgroup$
      – Wuestenfux
      Jan 6 at 8:28











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    Technically it means that an isomorphic copy of $Bbb{Z}_p$ is in $Bbb{F}_{p^n}$. For example, in $Bbb{F}_9$, the constant polynomials will be an isomorphic copy of $Bbb{Z}_3$.



    begin{align*}
    Bbb{F}_9 & =Bbb{Z}_3[x]/langle x^2+1rangle\
    & ={ax+b , | , a,b in Bbb{Z}_3, x^2+1 equiv 0}\
    &={0,1,2,x,x+1,x+2,2x,2x+1,2x+2}
    end{align*}



    Then the set of constant polynomials ${0,1,2}$ is isomorphic to $Bbb{Z}_3$.




    With your edited version:



    $Bbb{F}_{p^n}$ can be thought of as a vector space over the field of scalars $Bbb{Z}_p$. Thus each object in $Bbb{F}_{p^n}$ is a $n-$tuple vector (this is very similar to the polynomial representation I have used above).



    So consider the set of the vectors
    $$S={(0,0,ldots ,c) , | , c in Bbb{Z}_p}.$$
    This set in $Bbb{F}_{p^n}$ is isomorphic to $Bbb{Z}_p$.






    share|cite|improve this answer











    $endgroup$


















      6












      $begingroup$

      Technically it means that an isomorphic copy of $Bbb{Z}_p$ is in $Bbb{F}_{p^n}$. For example, in $Bbb{F}_9$, the constant polynomials will be an isomorphic copy of $Bbb{Z}_3$.



      begin{align*}
      Bbb{F}_9 & =Bbb{Z}_3[x]/langle x^2+1rangle\
      & ={ax+b , | , a,b in Bbb{Z}_3, x^2+1 equiv 0}\
      &={0,1,2,x,x+1,x+2,2x,2x+1,2x+2}
      end{align*}



      Then the set of constant polynomials ${0,1,2}$ is isomorphic to $Bbb{Z}_3$.




      With your edited version:



      $Bbb{F}_{p^n}$ can be thought of as a vector space over the field of scalars $Bbb{Z}_p$. Thus each object in $Bbb{F}_{p^n}$ is a $n-$tuple vector (this is very similar to the polynomial representation I have used above).



      So consider the set of the vectors
      $$S={(0,0,ldots ,c) , | , c in Bbb{Z}_p}.$$
      This set in $Bbb{F}_{p^n}$ is isomorphic to $Bbb{Z}_p$.






      share|cite|improve this answer











      $endgroup$
















        6












        6








        6





        $begingroup$

        Technically it means that an isomorphic copy of $Bbb{Z}_p$ is in $Bbb{F}_{p^n}$. For example, in $Bbb{F}_9$, the constant polynomials will be an isomorphic copy of $Bbb{Z}_3$.



        begin{align*}
        Bbb{F}_9 & =Bbb{Z}_3[x]/langle x^2+1rangle\
        & ={ax+b , | , a,b in Bbb{Z}_3, x^2+1 equiv 0}\
        &={0,1,2,x,x+1,x+2,2x,2x+1,2x+2}
        end{align*}



        Then the set of constant polynomials ${0,1,2}$ is isomorphic to $Bbb{Z}_3$.




        With your edited version:



        $Bbb{F}_{p^n}$ can be thought of as a vector space over the field of scalars $Bbb{Z}_p$. Thus each object in $Bbb{F}_{p^n}$ is a $n-$tuple vector (this is very similar to the polynomial representation I have used above).



        So consider the set of the vectors
        $$S={(0,0,ldots ,c) , | , c in Bbb{Z}_p}.$$
        This set in $Bbb{F}_{p^n}$ is isomorphic to $Bbb{Z}_p$.






        share|cite|improve this answer











        $endgroup$



        Technically it means that an isomorphic copy of $Bbb{Z}_p$ is in $Bbb{F}_{p^n}$. For example, in $Bbb{F}_9$, the constant polynomials will be an isomorphic copy of $Bbb{Z}_3$.



        begin{align*}
        Bbb{F}_9 & =Bbb{Z}_3[x]/langle x^2+1rangle\
        & ={ax+b , | , a,b in Bbb{Z}_3, x^2+1 equiv 0}\
        &={0,1,2,x,x+1,x+2,2x,2x+1,2x+2}
        end{align*}



        Then the set of constant polynomials ${0,1,2}$ is isomorphic to $Bbb{Z}_3$.




        With your edited version:



        $Bbb{F}_{p^n}$ can be thought of as a vector space over the field of scalars $Bbb{Z}_p$. Thus each object in $Bbb{F}_{p^n}$ is a $n-$tuple vector (this is very similar to the polynomial representation I have used above).



        So consider the set of the vectors
        $$S={(0,0,ldots ,c) , | , c in Bbb{Z}_p}.$$
        This set in $Bbb{F}_{p^n}$ is isomorphic to $Bbb{Z}_p$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 6 at 7:50

























        answered Jan 6 at 7:42









        Anurag AAnurag A

        26.3k12251




        26.3k12251























            2












            $begingroup$

            The mapping ${Bbb Z}_prightarrow {Bbb F}_{p^n}$ given by $amapsto acdot 1$, where $acdot 1 = 1+ldots+1$ ($a$-times) is the $a$-multiple of $1$ and $1$ is the unit element of ${Bbb F}_{p^n}$, is a (ring) monomorphism. In this way, ${Bbb F}_{p^n}$ contains an isomorphic copy of ${Bbb Z}_p$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I wanted a representation-independent way to define this subset. Hence, this useful. We can use "double and repeat" if $a$ is large.
              $endgroup$
              – Jus12
              Jan 6 at 8:28












            • $begingroup$
              Indeed, ''double and repeat'' can be used.
              $endgroup$
              – Wuestenfux
              Jan 6 at 8:28
















            2












            $begingroup$

            The mapping ${Bbb Z}_prightarrow {Bbb F}_{p^n}$ given by $amapsto acdot 1$, where $acdot 1 = 1+ldots+1$ ($a$-times) is the $a$-multiple of $1$ and $1$ is the unit element of ${Bbb F}_{p^n}$, is a (ring) monomorphism. In this way, ${Bbb F}_{p^n}$ contains an isomorphic copy of ${Bbb Z}_p$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I wanted a representation-independent way to define this subset. Hence, this useful. We can use "double and repeat" if $a$ is large.
              $endgroup$
              – Jus12
              Jan 6 at 8:28












            • $begingroup$
              Indeed, ''double and repeat'' can be used.
              $endgroup$
              – Wuestenfux
              Jan 6 at 8:28














            2












            2








            2





            $begingroup$

            The mapping ${Bbb Z}_prightarrow {Bbb F}_{p^n}$ given by $amapsto acdot 1$, where $acdot 1 = 1+ldots+1$ ($a$-times) is the $a$-multiple of $1$ and $1$ is the unit element of ${Bbb F}_{p^n}$, is a (ring) monomorphism. In this way, ${Bbb F}_{p^n}$ contains an isomorphic copy of ${Bbb Z}_p$.






            share|cite|improve this answer









            $endgroup$



            The mapping ${Bbb Z}_prightarrow {Bbb F}_{p^n}$ given by $amapsto acdot 1$, where $acdot 1 = 1+ldots+1$ ($a$-times) is the $a$-multiple of $1$ and $1$ is the unit element of ${Bbb F}_{p^n}$, is a (ring) monomorphism. In this way, ${Bbb F}_{p^n}$ contains an isomorphic copy of ${Bbb Z}_p$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 6 at 8:10









            WuestenfuxWuestenfux

            4,8321513




            4,8321513












            • $begingroup$
              I wanted a representation-independent way to define this subset. Hence, this useful. We can use "double and repeat" if $a$ is large.
              $endgroup$
              – Jus12
              Jan 6 at 8:28












            • $begingroup$
              Indeed, ''double and repeat'' can be used.
              $endgroup$
              – Wuestenfux
              Jan 6 at 8:28


















            • $begingroup$
              I wanted a representation-independent way to define this subset. Hence, this useful. We can use "double and repeat" if $a$ is large.
              $endgroup$
              – Jus12
              Jan 6 at 8:28












            • $begingroup$
              Indeed, ''double and repeat'' can be used.
              $endgroup$
              – Wuestenfux
              Jan 6 at 8:28
















            $begingroup$
            I wanted a representation-independent way to define this subset. Hence, this useful. We can use "double and repeat" if $a$ is large.
            $endgroup$
            – Jus12
            Jan 6 at 8:28






            $begingroup$
            I wanted a representation-independent way to define this subset. Hence, this useful. We can use "double and repeat" if $a$ is large.
            $endgroup$
            – Jus12
            Jan 6 at 8:28














            $begingroup$
            Indeed, ''double and repeat'' can be used.
            $endgroup$
            – Wuestenfux
            Jan 6 at 8:28




            $begingroup$
            Indeed, ''double and repeat'' can be used.
            $endgroup$
            – Wuestenfux
            Jan 6 at 8:28


















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