What is the basis of a field $mathbbF_p^k$ for a prime $p$ and a positive integer $k$? [closed]

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What is the basis of a field $mathbbF_p^k$ for a prime $p$ and a positive integer $k$?




vector-spaces field-theory




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closed as off-topic by José Carlos Santos, Brahadeesh, amWhy, Shailesh, Lord Shark the Unknown Aug 23 at 11:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, amWhy, Shailesh
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  • Basis as what??
    – Bernard
    Aug 23 at 9:33










  • Any element in $mathbbF_p^k$can be represented as $sum_i=0^k-1 a_i b_i$ where $a_i in F_p$ and $b_i$ is a basis. What is the particular value of $b_i$??
    – mallea
    Aug 23 at 10:01











  • So as an $mathbf F_p$-vector space.
    – Bernard
    Aug 23 at 10:06






  • 1




    @mallea yes, because $mathbbF_p^k$ is a vector space over $mathbbF_p^l$ whenever $l$ divides $k$.
    – Brahadeesh
    Aug 23 at 10:14






  • 1




    try to edit your question rather just clarifying in the comment. also include thoughts and attempts if possible.
    – Siong Thye Goh
    Aug 23 at 10:23














up vote
0
down vote

favorite












What is the basis of a field $mathbbF_p^k$ for a prime $p$ and a positive integer $k$?




vector-spaces field-theory




share|cite|improve this question












closed as off-topic by José Carlos Santos, Brahadeesh, amWhy, Shailesh, Lord Shark the Unknown Aug 23 at 11:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, amWhy, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.












  • Basis as what??
    – Bernard
    Aug 23 at 9:33










  • Any element in $mathbbF_p^k$can be represented as $sum_i=0^k-1 a_i b_i$ where $a_i in F_p$ and $b_i$ is a basis. What is the particular value of $b_i$??
    – mallea
    Aug 23 at 10:01











  • So as an $mathbf F_p$-vector space.
    – Bernard
    Aug 23 at 10:06






  • 1




    @mallea yes, because $mathbbF_p^k$ is a vector space over $mathbbF_p^l$ whenever $l$ divides $k$.
    – Brahadeesh
    Aug 23 at 10:14






  • 1




    try to edit your question rather just clarifying in the comment. also include thoughts and attempts if possible.
    – Siong Thye Goh
    Aug 23 at 10:23












up vote
0
down vote

favorite









up vote
0
down vote

favorite











What is the basis of a field $mathbbF_p^k$ for a prime $p$ and a positive integer $k$?




vector-spaces field-theory




share|cite|improve this question












What is the basis of a field $mathbbF_p^k$ for a prime $p$ and a positive integer $k$?





vector-spaces field-theory





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 23 at 9:21









mallea

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closed as off-topic by José Carlos Santos, Brahadeesh, amWhy, Shailesh, Lord Shark the Unknown Aug 23 at 11:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, amWhy, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by José Carlos Santos, Brahadeesh, amWhy, Shailesh, Lord Shark the Unknown Aug 23 at 11:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, amWhy, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.











  • Basis as what??
    – Bernard
    Aug 23 at 9:33










  • Any element in $mathbbF_p^k$can be represented as $sum_i=0^k-1 a_i b_i$ where $a_i in F_p$ and $b_i$ is a basis. What is the particular value of $b_i$??
    – mallea
    Aug 23 at 10:01











  • So as an $mathbf F_p$-vector space.
    – Bernard
    Aug 23 at 10:06






  • 1




    @mallea yes, because $mathbbF_p^k$ is a vector space over $mathbbF_p^l$ whenever $l$ divides $k$.
    – Brahadeesh
    Aug 23 at 10:14






  • 1




    try to edit your question rather just clarifying in the comment. also include thoughts and attempts if possible.
    – Siong Thye Goh
    Aug 23 at 10:23
















  • Basis as what??
    – Bernard
    Aug 23 at 9:33










  • Any element in $mathbbF_p^k$can be represented as $sum_i=0^k-1 a_i b_i$ where $a_i in F_p$ and $b_i$ is a basis. What is the particular value of $b_i$??
    – mallea
    Aug 23 at 10:01











  • So as an $mathbf F_p$-vector space.
    – Bernard
    Aug 23 at 10:06






  • 1




    @mallea yes, because $mathbbF_p^k$ is a vector space over $mathbbF_p^l$ whenever $l$ divides $k$.
    – Brahadeesh
    Aug 23 at 10:14






  • 1




    try to edit your question rather just clarifying in the comment. also include thoughts and attempts if possible.
    – Siong Thye Goh
    Aug 23 at 10:23















Basis as what??
– Bernard
Aug 23 at 9:33




Basis as what??
– Bernard
Aug 23 at 9:33












Any element in $mathbbF_p^k$can be represented as $sum_i=0^k-1 a_i b_i$ where $a_i in F_p$ and $b_i$ is a basis. What is the particular value of $b_i$??
– mallea
Aug 23 at 10:01





Any element in $mathbbF_p^k$can be represented as $sum_i=0^k-1 a_i b_i$ where $a_i in F_p$ and $b_i$ is a basis. What is the particular value of $b_i$??
– mallea
Aug 23 at 10:01













So as an $mathbf F_p$-vector space.
– Bernard
Aug 23 at 10:06




So as an $mathbf F_p$-vector space.
– Bernard
Aug 23 at 10:06




1




1




@mallea yes, because $mathbbF_p^k$ is a vector space over $mathbbF_p^l$ whenever $l$ divides $k$.
– Brahadeesh
Aug 23 at 10:14




@mallea yes, because $mathbbF_p^k$ is a vector space over $mathbbF_p^l$ whenever $l$ divides $k$.
– Brahadeesh
Aug 23 at 10:14




1




1




try to edit your question rather just clarifying in the comment. also include thoughts and attempts if possible.
– Siong Thye Goh
Aug 23 at 10:23




try to edit your question rather just clarifying in the comment. also include thoughts and attempts if possible.
– Siong Thye Goh
Aug 23 at 10:23










1 Answer
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The finite field $mathbf F_p^k$ is generated, as an $mathbf F_p$-algebra, by a root $omega$ of an irreducible polynomial of degree $k$.



Such polynomials are irreducible factors of degree $k$, $ne X$, of the polynomial $X^p^k-X$.



A basis of $mathbf F_p^k$ over $mathbf F_p$ is then the set
$$mathcal B=1,omega, omega^2,dots omega^k-1.$$






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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote













    The finite field $mathbf F_p^k$ is generated, as an $mathbf F_p$-algebra, by a root $omega$ of an irreducible polynomial of degree $k$.



    Such polynomials are irreducible factors of degree $k$, $ne X$, of the polynomial $X^p^k-X$.



    A basis of $mathbf F_p^k$ over $mathbf F_p$ is then the set
    $$mathcal B=1,omega, omega^2,dots omega^k-1.$$






    share|cite|improve this answer
























      up vote
      3
      down vote













      The finite field $mathbf F_p^k$ is generated, as an $mathbf F_p$-algebra, by a root $omega$ of an irreducible polynomial of degree $k$.



      Such polynomials are irreducible factors of degree $k$, $ne X$, of the polynomial $X^p^k-X$.



      A basis of $mathbf F_p^k$ over $mathbf F_p$ is then the set
      $$mathcal B=1,omega, omega^2,dots omega^k-1.$$






      share|cite|improve this answer






















        up vote
        3
        down vote










        up vote
        3
        down vote









        The finite field $mathbf F_p^k$ is generated, as an $mathbf F_p$-algebra, by a root $omega$ of an irreducible polynomial of degree $k$.



        Such polynomials are irreducible factors of degree $k$, $ne X$, of the polynomial $X^p^k-X$.



        A basis of $mathbf F_p^k$ over $mathbf F_p$ is then the set
        $$mathcal B=1,omega, omega^2,dots omega^k-1.$$






        share|cite|improve this answer












        The finite field $mathbf F_p^k$ is generated, as an $mathbf F_p$-algebra, by a root $omega$ of an irreducible polynomial of degree $k$.



        Such polynomials are irreducible factors of degree $k$, $ne X$, of the polynomial $X^p^k-X$.



        A basis of $mathbf F_p^k$ over $mathbf F_p$ is then the set
        $$mathcal B=1,omega, omega^2,dots omega^k-1.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 23 at 10:25









        Bernard

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        111k635103












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