Characteristic and primitive roots of unity












2












$begingroup$


Let $F$ be a finite field.



If the characteristic of $F$ doesn’t divide $n$, then $F$ contains a primitive $n^{th}$ root of unity.



I believe the converse is true, too, but I can’t prove either direction?



I know that $F^{*}$ is cyclic and of order $p^n -1$, where $p$ is the characteristic of $F$ and $n >0$. So it contains a generator, $alpha$ with $alpha^{p^n -1} = 1$.



But how does this relate to the characteristic not dividing $n$?






abstract-algebra galois-theory finite-fields







share|cite|improve this question









$endgroup$












  • $begingroup$
    Not quite that way. If $n$ is not divisible by $p$ then $F$ always has an extension field $K$ such that $K$ contains a primitive $n$th root of unity. As the answerers explained, no extension field of $F$ will contain a primitive $p$th root of unity.
    $endgroup$
    – Jyrki Lahtonen
    Jan 18 at 10:23






  • 1




    $begingroup$
    More precisely, if $|F|=q$, $q$ some power of $p$, then the extension field $K$ containing an element of order $n$ has degree $k$, where $k$ is the smallest positive integer such that $nmid q^k-1$. This is implicit in Servaes' answer.
    $endgroup$
    – Jyrki Lahtonen
    Jan 18 at 10:25
















2












$begingroup$


Let $F$ be a finite field.



If the characteristic of $F$ doesn’t divide $n$, then $F$ contains a primitive $n^{th}$ root of unity.



I believe the converse is true, too, but I can’t prove either direction?



I know that $F^{*}$ is cyclic and of order $p^n -1$, where $p$ is the characteristic of $F$ and $n >0$. So it contains a generator, $alpha$ with $alpha^{p^n -1} = 1$.



But how does this relate to the characteristic not dividing $n$?






abstract-algebra galois-theory finite-fields







share|cite|improve this question









$endgroup$












  • $begingroup$
    Not quite that way. If $n$ is not divisible by $p$ then $F$ always has an extension field $K$ such that $K$ contains a primitive $n$th root of unity. As the answerers explained, no extension field of $F$ will contain a primitive $p$th root of unity.
    $endgroup$
    – Jyrki Lahtonen
    Jan 18 at 10:23






  • 1




    $begingroup$
    More precisely, if $|F|=q$, $q$ some power of $p$, then the extension field $K$ containing an element of order $n$ has degree $k$, where $k$ is the smallest positive integer such that $nmid q^k-1$. This is implicit in Servaes' answer.
    $endgroup$
    – Jyrki Lahtonen
    Jan 18 at 10:25














2












2








2





$begingroup$


Let $F$ be a finite field.



If the characteristic of $F$ doesn’t divide $n$, then $F$ contains a primitive $n^{th}$ root of unity.



I believe the converse is true, too, but I can’t prove either direction?



I know that $F^{*}$ is cyclic and of order $p^n -1$, where $p$ is the characteristic of $F$ and $n >0$. So it contains a generator, $alpha$ with $alpha^{p^n -1} = 1$.



But how does this relate to the characteristic not dividing $n$?






abstract-algebra galois-theory finite-fields







share|cite|improve this question









$endgroup$




Let $F$ be a finite field.



If the characteristic of $F$ doesn’t divide $n$, then $F$ contains a primitive $n^{th}$ root of unity.



I believe the converse is true, too, but I can’t prove either direction?



I know that $F^{*}$ is cyclic and of order $p^n -1$, where $p$ is the characteristic of $F$ and $n >0$. So it contains a generator, $alpha$ with $alpha^{p^n -1} = 1$.



But how does this relate to the characteristic not dividing $n$?






abstract-algebra galois-theory finite-fields




abstract-algebra galois-theory finite-fields






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 18 at 8:49









the manthe man

772715




772715












  • $begingroup$
    Not quite that way. If $n$ is not divisible by $p$ then $F$ always has an extension field $K$ such that $K$ contains a primitive $n$th root of unity. As the answerers explained, no extension field of $F$ will contain a primitive $p$th root of unity.
    $endgroup$
    – Jyrki Lahtonen
    Jan 18 at 10:23






  • 1




    $begingroup$
    More precisely, if $|F|=q$, $q$ some power of $p$, then the extension field $K$ containing an element of order $n$ has degree $k$, where $k$ is the smallest positive integer such that $nmid q^k-1$. This is implicit in Servaes' answer.
    $endgroup$
    – Jyrki Lahtonen
    Jan 18 at 10:25


















  • $begingroup$
    Not quite that way. If $n$ is not divisible by $p$ then $F$ always has an extension field $K$ such that $K$ contains a primitive $n$th root of unity. As the answerers explained, no extension field of $F$ will contain a primitive $p$th root of unity.
    $endgroup$
    – Jyrki Lahtonen
    Jan 18 at 10:23






  • 1




    $begingroup$
    More precisely, if $|F|=q$, $q$ some power of $p$, then the extension field $K$ containing an element of order $n$ has degree $k$, where $k$ is the smallest positive integer such that $nmid q^k-1$. This is implicit in Servaes' answer.
    $endgroup$
    – Jyrki Lahtonen
    Jan 18 at 10:25
















$begingroup$
Not quite that way. If $n$ is not divisible by $p$ then $F$ always has an extension field $K$ such that $K$ contains a primitive $n$th root of unity. As the answerers explained, no extension field of $F$ will contain a primitive $p$th root of unity.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 10:23




$begingroup$
Not quite that way. If $n$ is not divisible by $p$ then $F$ always has an extension field $K$ such that $K$ contains a primitive $n$th root of unity. As the answerers explained, no extension field of $F$ will contain a primitive $p$th root of unity.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 10:23




1




1




$begingroup$
More precisely, if $|F|=q$, $q$ some power of $p$, then the extension field $K$ containing an element of order $n$ has degree $k$, where $k$ is the smallest positive integer such that $nmid q^k-1$. This is implicit in Servaes' answer.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 10:25




$begingroup$
More precisely, if $|F|=q$, $q$ some power of $p$, then the extension field $K$ containing an element of order $n$ has degree $k$, where $k$ is the smallest positive integer such that $nmid q^k-1$. This is implicit in Servaes' answer.
$endgroup$
– Jyrki Lahtonen
Jan 18 at 10:25










3 Answers
3






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3












$begingroup$

fresh mans dream: for an integral domain of characteristic $p$ we have $(x-y)^p=x^p-y^p$, in particular $x^{p}-1$ has only $x-1$ as linear factors, which means that there is no $p$-root of unity except for $1$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Consider $n=pm$. Then $x^n-1 = x^{mp}-1 = (x^m-1)^p$ and so $n$-th roots of unity are actually $m$-th roots of unity.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      If the order of $F^*$ is $p^k-1$ then the units of $F$ have orders dividing $p^k-1$. This means that for any primitive $n$-th root of unity in $F$, the order $n$ divides $p^k-1$. Of course $p^k-1$ and $p$ are coprime, so $n$ is coprime to $p$, that is to say $p$ does not divide $n$.



      The converse is clearly false; there are infinitely many $n$ for which the characteristic of $F$ doesn't divide $n$, but only finitely many roots of unity in $F$. So $F$ certainly does not contain a primitive $n$-th root of unity for all those $n$.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        fresh mans dream: for an integral domain of characteristic $p$ we have $(x-y)^p=x^p-y^p$, in particular $x^{p}-1$ has only $x-1$ as linear factors, which means that there is no $p$-root of unity except for $1$.






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          fresh mans dream: for an integral domain of characteristic $p$ we have $(x-y)^p=x^p-y^p$, in particular $x^{p}-1$ has only $x-1$ as linear factors, which means that there is no $p$-root of unity except for $1$.






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            fresh mans dream: for an integral domain of characteristic $p$ we have $(x-y)^p=x^p-y^p$, in particular $x^{p}-1$ has only $x-1$ as linear factors, which means that there is no $p$-root of unity except for $1$.






            share|cite|improve this answer









            $endgroup$



            fresh mans dream: for an integral domain of characteristic $p$ we have $(x-y)^p=x^p-y^p$, in particular $x^{p}-1$ has only $x-1$ as linear factors, which means that there is no $p$-root of unity except for $1$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 18 at 8:54









            EnkiduEnkidu

            1,36719




            1,36719























                2












                $begingroup$

                Consider $n=pm$. Then $x^n-1 = x^{mp}-1 = (x^m-1)^p$ and so $n$-th roots of unity are actually $m$-th roots of unity.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  Consider $n=pm$. Then $x^n-1 = x^{mp}-1 = (x^m-1)^p$ and so $n$-th roots of unity are actually $m$-th roots of unity.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Consider $n=pm$. Then $x^n-1 = x^{mp}-1 = (x^m-1)^p$ and so $n$-th roots of unity are actually $m$-th roots of unity.






                    share|cite|improve this answer









                    $endgroup$



                    Consider $n=pm$. Then $x^n-1 = x^{mp}-1 = (x^m-1)^p$ and so $n$-th roots of unity are actually $m$-th roots of unity.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 18 at 8:56









                    WuestenfuxWuestenfux

                    4,7291513




                    4,7291513























                        2












                        $begingroup$

                        If the order of $F^*$ is $p^k-1$ then the units of $F$ have orders dividing $p^k-1$. This means that for any primitive $n$-th root of unity in $F$, the order $n$ divides $p^k-1$. Of course $p^k-1$ and $p$ are coprime, so $n$ is coprime to $p$, that is to say $p$ does not divide $n$.



                        The converse is clearly false; there are infinitely many $n$ for which the characteristic of $F$ doesn't divide $n$, but only finitely many roots of unity in $F$. So $F$ certainly does not contain a primitive $n$-th root of unity for all those $n$.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          If the order of $F^*$ is $p^k-1$ then the units of $F$ have orders dividing $p^k-1$. This means that for any primitive $n$-th root of unity in $F$, the order $n$ divides $p^k-1$. Of course $p^k-1$ and $p$ are coprime, so $n$ is coprime to $p$, that is to say $p$ does not divide $n$.



                          The converse is clearly false; there are infinitely many $n$ for which the characteristic of $F$ doesn't divide $n$, but only finitely many roots of unity in $F$. So $F$ certainly does not contain a primitive $n$-th root of unity for all those $n$.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            If the order of $F^*$ is $p^k-1$ then the units of $F$ have orders dividing $p^k-1$. This means that for any primitive $n$-th root of unity in $F$, the order $n$ divides $p^k-1$. Of course $p^k-1$ and $p$ are coprime, so $n$ is coprime to $p$, that is to say $p$ does not divide $n$.



                            The converse is clearly false; there are infinitely many $n$ for which the characteristic of $F$ doesn't divide $n$, but only finitely many roots of unity in $F$. So $F$ certainly does not contain a primitive $n$-th root of unity for all those $n$.






                            share|cite|improve this answer









                            $endgroup$



                            If the order of $F^*$ is $p^k-1$ then the units of $F$ have orders dividing $p^k-1$. This means that for any primitive $n$-th root of unity in $F$, the order $n$ divides $p^k-1$. Of course $p^k-1$ and $p$ are coprime, so $n$ is coprime to $p$, that is to say $p$ does not divide $n$.



                            The converse is clearly false; there are infinitely many $n$ for which the characteristic of $F$ doesn't divide $n$, but only finitely many roots of unity in $F$. So $F$ certainly does not contain a primitive $n$-th root of unity for all those $n$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 18 at 10:15









                            ServaesServaes

                            25.9k33996




                            25.9k33996






























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