Degree of splitting field of $x^5-1$ over $mathbb{Q}$ and the order of its Galois group?












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$begingroup$


If I understand the result here correctly, the order of the Galois group is equal to the degree of the extension. That said, I am fairly certain that the splitting field $E$ of $x^5-1$ over $mathbb{Q}$ is
$$E={a_0+a_1omega+a_2omega^2+a_3omega^3+a_4omega^4}$$ where $omega$ is a complex root of unity. $E$, as a $mathbb{Q}$-vector space, has dimension 5.



However, if I understand this correctly, the Galois group for this polynomial is $mathbb{Z}/5mathbb{Z}^*$, which has order 4.



What is it that I'm missing? Am I misinterpreting or miscomputing the dimension of $E$ as a $mathbb{Q}$-vector space, or am I misinterpreting the post and the Galois group is in fact not the aforementioned group of units?






group-theory field-theory galois-theory







share|cite|improve this question











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    1












    $begingroup$


    If I understand the result here correctly, the order of the Galois group is equal to the degree of the extension. That said, I am fairly certain that the splitting field $E$ of $x^5-1$ over $mathbb{Q}$ is
    $$E={a_0+a_1omega+a_2omega^2+a_3omega^3+a_4omega^4}$$ where $omega$ is a complex root of unity. $E$, as a $mathbb{Q}$-vector space, has dimension 5.



    However, if I understand this correctly, the Galois group for this polynomial is $mathbb{Z}/5mathbb{Z}^*$, which has order 4.



    What is it that I'm missing? Am I misinterpreting or miscomputing the dimension of $E$ as a $mathbb{Q}$-vector space, or am I misinterpreting the post and the Galois group is in fact not the aforementioned group of units?






    group-theory field-theory galois-theory







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      If I understand the result here correctly, the order of the Galois group is equal to the degree of the extension. That said, I am fairly certain that the splitting field $E$ of $x^5-1$ over $mathbb{Q}$ is
      $$E={a_0+a_1omega+a_2omega^2+a_3omega^3+a_4omega^4}$$ where $omega$ is a complex root of unity. $E$, as a $mathbb{Q}$-vector space, has dimension 5.



      However, if I understand this correctly, the Galois group for this polynomial is $mathbb{Z}/5mathbb{Z}^*$, which has order 4.



      What is it that I'm missing? Am I misinterpreting or miscomputing the dimension of $E$ as a $mathbb{Q}$-vector space, or am I misinterpreting the post and the Galois group is in fact not the aforementioned group of units?






      group-theory field-theory galois-theory







      share|cite|improve this question











      $endgroup$




      If I understand the result here correctly, the order of the Galois group is equal to the degree of the extension. That said, I am fairly certain that the splitting field $E$ of $x^5-1$ over $mathbb{Q}$ is
      $$E={a_0+a_1omega+a_2omega^2+a_3omega^3+a_4omega^4}$$ where $omega$ is a complex root of unity. $E$, as a $mathbb{Q}$-vector space, has dimension 5.



      However, if I understand this correctly, the Galois group for this polynomial is $mathbb{Z}/5mathbb{Z}^*$, which has order 4.



      What is it that I'm missing? Am I misinterpreting or miscomputing the dimension of $E$ as a $mathbb{Q}$-vector space, or am I misinterpreting the post and the Galois group is in fact not the aforementioned group of units?






      group-theory field-theory galois-theory




      group-theory field-theory galois-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 11 '18 at 1:07







      Daniel

















      asked Dec 10 '18 at 21:24









      DanielDaniel

      599




      599






















          3 Answers
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          $begingroup$

          The polynomial is reducible: $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$.
          You do not need to add the element $1$ to the field (it is already there), only a root of the quaternary polynomial (which generates the rest of the roots). That is why the degree is $4$.






          share|cite|improve this answer









          $endgroup$





















            3












            $begingroup$

            While your $E$ is the right space, the set ${1,omega,omega^2,omega^3,omega^4}$ is not a basis for it, because that set isn't linearly independent over $mathbb{Q}$. We have the dependence relation $1+omega+omega^2+omega^3+omega^4$ (and nothing else), so any four of those suffice as a basis. As a vector space over $mathbb{Q}$, $E$ has dimension $4$ - so it's not a surprise when the Galois group has order $4$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Ah! I should've realized that the relation allowed us to generate the rationals using only $mathbb{Q}$-linear sums of the powers of $omega$! Thank you.
              $endgroup$
              – Daniel
              Dec 11 '18 at 1:53










            • $begingroup$
              One additional question if you don't mind: is E going to be equal to $mathbb{Q}(omega)$ or will it be something like $mathbb{Q}(omega)(omega^2)$?
              $endgroup$
              – Daniel
              Dec 11 '18 at 2:06












            • $begingroup$
              I suppose I am wondering how I'm supposed to symbolically represent $E$. My suspicion is that it's $E(omega)(omega^2)$ because elements will end up looking like $${f+f'omega^2 ; | ; f,f'in mathbb{Q}(omega)}={a+bomega + c omega^2 +domega^3 ; | ; a,b,c,d in mathbb{Q}}$$
              $endgroup$
              – Daniel
              Dec 11 '18 at 2:12










            • $begingroup$
              Sorry, $mathbb{Q}(omega)(omega^2)$
              $endgroup$
              – Daniel
              Dec 11 '18 at 2:28










            • $begingroup$
              How you represent it depends on whether you're looking for a vector space basis or an algebraic generator. As a vector space, pick four of the roots to be a basis. If you're allowing multiplication, it's just $mathbb{Q}(omega)$.
              $endgroup$
              – jmerry
              Dec 11 '18 at 2:29



















            0












            $begingroup$

            Order of Galois extension need not equal to degree of extension it is less than equal to degree of extension example is |Aut (Q(2^{1/3})/Q)| =1 but [ Q(2^{1/3}) : Q ] is 3 and equality hold when extension is splitting field of separable polynomial.






            share|cite|improve this answer









            $endgroup$













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              3 Answers
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              $begingroup$

              The polynomial is reducible: $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$.
              You do not need to add the element $1$ to the field (it is already there), only a root of the quaternary polynomial (which generates the rest of the roots). That is why the degree is $4$.






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                The polynomial is reducible: $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$.
                You do not need to add the element $1$ to the field (it is already there), only a root of the quaternary polynomial (which generates the rest of the roots). That is why the degree is $4$.






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  The polynomial is reducible: $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$.
                  You do not need to add the element $1$ to the field (it is already there), only a root of the quaternary polynomial (which generates the rest of the roots). That is why the degree is $4$.






                  share|cite|improve this answer









                  $endgroup$



                  The polynomial is reducible: $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$.
                  You do not need to add the element $1$ to the field (it is already there), only a root of the quaternary polynomial (which generates the rest of the roots). That is why the degree is $4$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 21:30









                  A. PongráczA. Pongrácz

                  5,9631929




                  5,9631929























                      3












                      $begingroup$

                      While your $E$ is the right space, the set ${1,omega,omega^2,omega^3,omega^4}$ is not a basis for it, because that set isn't linearly independent over $mathbb{Q}$. We have the dependence relation $1+omega+omega^2+omega^3+omega^4$ (and nothing else), so any four of those suffice as a basis. As a vector space over $mathbb{Q}$, $E$ has dimension $4$ - so it's not a surprise when the Galois group has order $4$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Ah! I should've realized that the relation allowed us to generate the rationals using only $mathbb{Q}$-linear sums of the powers of $omega$! Thank you.
                        $endgroup$
                        – Daniel
                        Dec 11 '18 at 1:53










                      • $begingroup$
                        One additional question if you don't mind: is E going to be equal to $mathbb{Q}(omega)$ or will it be something like $mathbb{Q}(omega)(omega^2)$?
                        $endgroup$
                        – Daniel
                        Dec 11 '18 at 2:06












                      • $begingroup$
                        I suppose I am wondering how I'm supposed to symbolically represent $E$. My suspicion is that it's $E(omega)(omega^2)$ because elements will end up looking like $${f+f'omega^2 ; | ; f,f'in mathbb{Q}(omega)}={a+bomega + c omega^2 +domega^3 ; | ; a,b,c,d in mathbb{Q}}$$
                        $endgroup$
                        – Daniel
                        Dec 11 '18 at 2:12










                      • $begingroup$
                        Sorry, $mathbb{Q}(omega)(omega^2)$
                        $endgroup$
                        – Daniel
                        Dec 11 '18 at 2:28










                      • $begingroup$
                        How you represent it depends on whether you're looking for a vector space basis or an algebraic generator. As a vector space, pick four of the roots to be a basis. If you're allowing multiplication, it's just $mathbb{Q}(omega)$.
                        $endgroup$
                        – jmerry
                        Dec 11 '18 at 2:29
















                      3












                      $begingroup$

                      While your $E$ is the right space, the set ${1,omega,omega^2,omega^3,omega^4}$ is not a basis for it, because that set isn't linearly independent over $mathbb{Q}$. We have the dependence relation $1+omega+omega^2+omega^3+omega^4$ (and nothing else), so any four of those suffice as a basis. As a vector space over $mathbb{Q}$, $E$ has dimension $4$ - so it's not a surprise when the Galois group has order $4$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Ah! I should've realized that the relation allowed us to generate the rationals using only $mathbb{Q}$-linear sums of the powers of $omega$! Thank you.
                        $endgroup$
                        – Daniel
                        Dec 11 '18 at 1:53










                      • $begingroup$
                        One additional question if you don't mind: is E going to be equal to $mathbb{Q}(omega)$ or will it be something like $mathbb{Q}(omega)(omega^2)$?
                        $endgroup$
                        – Daniel
                        Dec 11 '18 at 2:06












                      • $begingroup$
                        I suppose I am wondering how I'm supposed to symbolically represent $E$. My suspicion is that it's $E(omega)(omega^2)$ because elements will end up looking like $${f+f'omega^2 ; | ; f,f'in mathbb{Q}(omega)}={a+bomega + c omega^2 +domega^3 ; | ; a,b,c,d in mathbb{Q}}$$
                        $endgroup$
                        – Daniel
                        Dec 11 '18 at 2:12










                      • $begingroup$
                        Sorry, $mathbb{Q}(omega)(omega^2)$
                        $endgroup$
                        – Daniel
                        Dec 11 '18 at 2:28










                      • $begingroup$
                        How you represent it depends on whether you're looking for a vector space basis or an algebraic generator. As a vector space, pick four of the roots to be a basis. If you're allowing multiplication, it's just $mathbb{Q}(omega)$.
                        $endgroup$
                        – jmerry
                        Dec 11 '18 at 2:29














                      3












                      3








                      3





                      $begingroup$

                      While your $E$ is the right space, the set ${1,omega,omega^2,omega^3,omega^4}$ is not a basis for it, because that set isn't linearly independent over $mathbb{Q}$. We have the dependence relation $1+omega+omega^2+omega^3+omega^4$ (and nothing else), so any four of those suffice as a basis. As a vector space over $mathbb{Q}$, $E$ has dimension $4$ - so it's not a surprise when the Galois group has order $4$.






                      share|cite|improve this answer









                      $endgroup$



                      While your $E$ is the right space, the set ${1,omega,omega^2,omega^3,omega^4}$ is not a basis for it, because that set isn't linearly independent over $mathbb{Q}$. We have the dependence relation $1+omega+omega^2+omega^3+omega^4$ (and nothing else), so any four of those suffice as a basis. As a vector space over $mathbb{Q}$, $E$ has dimension $4$ - so it's not a surprise when the Galois group has order $4$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 10 '18 at 21:53









                      jmerryjmerry

                      6,602718




                      6,602718












                      • $begingroup$
                        Ah! I should've realized that the relation allowed us to generate the rationals using only $mathbb{Q}$-linear sums of the powers of $omega$! Thank you.
                        $endgroup$
                        – Daniel
                        Dec 11 '18 at 1:53










                      • $begingroup$
                        One additional question if you don't mind: is E going to be equal to $mathbb{Q}(omega)$ or will it be something like $mathbb{Q}(omega)(omega^2)$?
                        $endgroup$
                        – Daniel
                        Dec 11 '18 at 2:06












                      • $begingroup$
                        I suppose I am wondering how I'm supposed to symbolically represent $E$. My suspicion is that it's $E(omega)(omega^2)$ because elements will end up looking like $${f+f'omega^2 ; | ; f,f'in mathbb{Q}(omega)}={a+bomega + c omega^2 +domega^3 ; | ; a,b,c,d in mathbb{Q}}$$
                        $endgroup$
                        – Daniel
                        Dec 11 '18 at 2:12










                      • $begingroup$
                        Sorry, $mathbb{Q}(omega)(omega^2)$
                        $endgroup$
                        – Daniel
                        Dec 11 '18 at 2:28










                      • $begingroup$
                        How you represent it depends on whether you're looking for a vector space basis or an algebraic generator. As a vector space, pick four of the roots to be a basis. If you're allowing multiplication, it's just $mathbb{Q}(omega)$.
                        $endgroup$
                        – jmerry
                        Dec 11 '18 at 2:29


















                      • $begingroup$
                        Ah! I should've realized that the relation allowed us to generate the rationals using only $mathbb{Q}$-linear sums of the powers of $omega$! Thank you.
                        $endgroup$
                        – Daniel
                        Dec 11 '18 at 1:53










                      • $begingroup$
                        One additional question if you don't mind: is E going to be equal to $mathbb{Q}(omega)$ or will it be something like $mathbb{Q}(omega)(omega^2)$?
                        $endgroup$
                        – Daniel
                        Dec 11 '18 at 2:06












                      • $begingroup$
                        I suppose I am wondering how I'm supposed to symbolically represent $E$. My suspicion is that it's $E(omega)(omega^2)$ because elements will end up looking like $${f+f'omega^2 ; | ; f,f'in mathbb{Q}(omega)}={a+bomega + c omega^2 +domega^3 ; | ; a,b,c,d in mathbb{Q}}$$
                        $endgroup$
                        – Daniel
                        Dec 11 '18 at 2:12










                      • $begingroup$
                        Sorry, $mathbb{Q}(omega)(omega^2)$
                        $endgroup$
                        – Daniel
                        Dec 11 '18 at 2:28










                      • $begingroup$
                        How you represent it depends on whether you're looking for a vector space basis or an algebraic generator. As a vector space, pick four of the roots to be a basis. If you're allowing multiplication, it's just $mathbb{Q}(omega)$.
                        $endgroup$
                        – jmerry
                        Dec 11 '18 at 2:29
















                      $begingroup$
                      Ah! I should've realized that the relation allowed us to generate the rationals using only $mathbb{Q}$-linear sums of the powers of $omega$! Thank you.
                      $endgroup$
                      – Daniel
                      Dec 11 '18 at 1:53




                      $begingroup$
                      Ah! I should've realized that the relation allowed us to generate the rationals using only $mathbb{Q}$-linear sums of the powers of $omega$! Thank you.
                      $endgroup$
                      – Daniel
                      Dec 11 '18 at 1:53












                      $begingroup$
                      One additional question if you don't mind: is E going to be equal to $mathbb{Q}(omega)$ or will it be something like $mathbb{Q}(omega)(omega^2)$?
                      $endgroup$
                      – Daniel
                      Dec 11 '18 at 2:06






                      $begingroup$
                      One additional question if you don't mind: is E going to be equal to $mathbb{Q}(omega)$ or will it be something like $mathbb{Q}(omega)(omega^2)$?
                      $endgroup$
                      – Daniel
                      Dec 11 '18 at 2:06














                      $begingroup$
                      I suppose I am wondering how I'm supposed to symbolically represent $E$. My suspicion is that it's $E(omega)(omega^2)$ because elements will end up looking like $${f+f'omega^2 ; | ; f,f'in mathbb{Q}(omega)}={a+bomega + c omega^2 +domega^3 ; | ; a,b,c,d in mathbb{Q}}$$
                      $endgroup$
                      – Daniel
                      Dec 11 '18 at 2:12




                      $begingroup$
                      I suppose I am wondering how I'm supposed to symbolically represent $E$. My suspicion is that it's $E(omega)(omega^2)$ because elements will end up looking like $${f+f'omega^2 ; | ; f,f'in mathbb{Q}(omega)}={a+bomega + c omega^2 +domega^3 ; | ; a,b,c,d in mathbb{Q}}$$
                      $endgroup$
                      – Daniel
                      Dec 11 '18 at 2:12












                      $begingroup$
                      Sorry, $mathbb{Q}(omega)(omega^2)$
                      $endgroup$
                      – Daniel
                      Dec 11 '18 at 2:28




                      $begingroup$
                      Sorry, $mathbb{Q}(omega)(omega^2)$
                      $endgroup$
                      – Daniel
                      Dec 11 '18 at 2:28












                      $begingroup$
                      How you represent it depends on whether you're looking for a vector space basis or an algebraic generator. As a vector space, pick four of the roots to be a basis. If you're allowing multiplication, it's just $mathbb{Q}(omega)$.
                      $endgroup$
                      – jmerry
                      Dec 11 '18 at 2:29




                      $begingroup$
                      How you represent it depends on whether you're looking for a vector space basis or an algebraic generator. As a vector space, pick four of the roots to be a basis. If you're allowing multiplication, it's just $mathbb{Q}(omega)$.
                      $endgroup$
                      – jmerry
                      Dec 11 '18 at 2:29











                      0












                      $begingroup$

                      Order of Galois extension need not equal to degree of extension it is less than equal to degree of extension example is |Aut (Q(2^{1/3})/Q)| =1 but [ Q(2^{1/3}) : Q ] is 3 and equality hold when extension is splitting field of separable polynomial.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Order of Galois extension need not equal to degree of extension it is less than equal to degree of extension example is |Aut (Q(2^{1/3})/Q)| =1 but [ Q(2^{1/3}) : Q ] is 3 and equality hold when extension is splitting field of separable polynomial.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Order of Galois extension need not equal to degree of extension it is less than equal to degree of extension example is |Aut (Q(2^{1/3})/Q)| =1 but [ Q(2^{1/3}) : Q ] is 3 and equality hold when extension is splitting field of separable polynomial.






                          share|cite|improve this answer









                          $endgroup$



                          Order of Galois extension need not equal to degree of extension it is less than equal to degree of extension example is |Aut (Q(2^{1/3})/Q)| =1 but [ Q(2^{1/3}) : Q ] is 3 and equality hold when extension is splitting field of separable polynomial.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 14 '18 at 12:22









                          user499117user499117

                          409




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