Let K be the splitting field of the polynomial $x^4-2x^2-2$. Find an automorphism $sigma in Gal(K/mathbb{Q})$...












3












$begingroup$


My attempt:
The roots of the polynomials are $x=sqrt{1+sqrt{3}}, -sqrt{1+sqrt{3}}, -sqrt{1-sqrt{3}}, -sqrt{1-sqrt{3}}$. and the automorphisms $sigma in Gal(K/mathbb{Q})$ permutes the root of the polynomial. But none of such automorphisms are of order 4! I don't know how to proceed further.



Thanks for any help!






group-theory field-theory galois-theory







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    3












    $begingroup$


    My attempt:
    The roots of the polynomials are $x=sqrt{1+sqrt{3}}, -sqrt{1+sqrt{3}}, -sqrt{1-sqrt{3}}, -sqrt{1-sqrt{3}}$. and the automorphisms $sigma in Gal(K/mathbb{Q})$ permutes the root of the polynomial. But none of such automorphisms are of order 4! I don't know how to proceed further.



    Thanks for any help!






    group-theory field-theory galois-theory







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      My attempt:
      The roots of the polynomials are $x=sqrt{1+sqrt{3}}, -sqrt{1+sqrt{3}}, -sqrt{1-sqrt{3}}, -sqrt{1-sqrt{3}}$. and the automorphisms $sigma in Gal(K/mathbb{Q})$ permutes the root of the polynomial. But none of such automorphisms are of order 4! I don't know how to proceed further.



      Thanks for any help!






      group-theory field-theory galois-theory







      share|cite|improve this question









      $endgroup$




      My attempt:
      The roots of the polynomials are $x=sqrt{1+sqrt{3}}, -sqrt{1+sqrt{3}}, -sqrt{1-sqrt{3}}, -sqrt{1-sqrt{3}}$. and the automorphisms $sigma in Gal(K/mathbb{Q})$ permutes the root of the polynomial. But none of such automorphisms are of order 4! I don't know how to proceed further.



      Thanks for any help!






      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




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      asked Jan 8 at 1:29









      InfinityInfinity

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          1 Answer
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          $begingroup$

          First observe that $K$ must be a degree 8 extension over $Bbb Q$. Here $Bbb Q(sqrt{1+sqrt{3}})$ is a degree 4 extension, but note that it cannot contain $sqrt{1-sqrt{3}}$ since $Bbb Q(sqrt{1+sqrt{3}}) subset Bbb R$ whereas $sqrt{1-sqrt{3}} not in Bbb R$. Thus, we can only have all of the roots of $x^4-2x^2-2$ by a further quadratic extension. (Here $sqrt{1-sqrt{3}}$ is only degree 2 over $Bbb Q(sqrt{1+sqrt{3}})$ since it satisfies $x^2 -1+sqrt{3} = 0$ and the latter field contains $sqrt{3}$.)



          Now we know that automorphisms in the Galois group $text{Gal}(F/Bbb Q)$ are determined by how they permute the elements of $x^4-2x^2-2$, so the only possibility for a group of order 8 must be given by the Sylow 2-subgroups of $S_4$, which contain two 4-cycles apiece. The only suitable candidates for these 4-cycles are
          $$
          sqrt{1+sqrt{3}} to
          sqrt{1-sqrt{3}} to
          -sqrt{1+sqrt{3}} to
          -sqrt{1-sqrt{3}} to
          sqrt{1+sqrt{3}}
          $$
          or its inverse. This follows since if $sqrt{1+sqrt{3}}$ is sent to $-sqrt{1+sqrt{3}}$ that will force a 2-cycle, and similarly for the pair $pmsqrt{1-sqrt{3}}$.



          If you don't want to appeal to abstract Galois theory and want a concrete justification that this permutation of elements corresponds to a well-defined field automorphism, this can be done directly by modeling $K$ as $Bbb Q[x,y]/(x^4-2x^2-2, y^2+x^2-2)$, and checking that the self-map given by $xmapsto y, ymapsto -x$ is well-defined on the quotient. Here the justification for the $y^2+x^2-2$ is given by $(sqrt{1+sqrt{3}})^2 + (sqrt{1-sqrt{3}})^2 = 1+sqrt{3}+1-sqrt{3} = 2$.






          share|cite|improve this answer









          $endgroup$













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            $begingroup$

            First observe that $K$ must be a degree 8 extension over $Bbb Q$. Here $Bbb Q(sqrt{1+sqrt{3}})$ is a degree 4 extension, but note that it cannot contain $sqrt{1-sqrt{3}}$ since $Bbb Q(sqrt{1+sqrt{3}}) subset Bbb R$ whereas $sqrt{1-sqrt{3}} not in Bbb R$. Thus, we can only have all of the roots of $x^4-2x^2-2$ by a further quadratic extension. (Here $sqrt{1-sqrt{3}}$ is only degree 2 over $Bbb Q(sqrt{1+sqrt{3}})$ since it satisfies $x^2 -1+sqrt{3} = 0$ and the latter field contains $sqrt{3}$.)



            Now we know that automorphisms in the Galois group $text{Gal}(F/Bbb Q)$ are determined by how they permute the elements of $x^4-2x^2-2$, so the only possibility for a group of order 8 must be given by the Sylow 2-subgroups of $S_4$, which contain two 4-cycles apiece. The only suitable candidates for these 4-cycles are
            $$
            sqrt{1+sqrt{3}} to
            sqrt{1-sqrt{3}} to
            -sqrt{1+sqrt{3}} to
            -sqrt{1-sqrt{3}} to
            sqrt{1+sqrt{3}}
            $$
            or its inverse. This follows since if $sqrt{1+sqrt{3}}$ is sent to $-sqrt{1+sqrt{3}}$ that will force a 2-cycle, and similarly for the pair $pmsqrt{1-sqrt{3}}$.



            If you don't want to appeal to abstract Galois theory and want a concrete justification that this permutation of elements corresponds to a well-defined field automorphism, this can be done directly by modeling $K$ as $Bbb Q[x,y]/(x^4-2x^2-2, y^2+x^2-2)$, and checking that the self-map given by $xmapsto y, ymapsto -x$ is well-defined on the quotient. Here the justification for the $y^2+x^2-2$ is given by $(sqrt{1+sqrt{3}})^2 + (sqrt{1-sqrt{3}})^2 = 1+sqrt{3}+1-sqrt{3} = 2$.






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              First observe that $K$ must be a degree 8 extension over $Bbb Q$. Here $Bbb Q(sqrt{1+sqrt{3}})$ is a degree 4 extension, but note that it cannot contain $sqrt{1-sqrt{3}}$ since $Bbb Q(sqrt{1+sqrt{3}}) subset Bbb R$ whereas $sqrt{1-sqrt{3}} not in Bbb R$. Thus, we can only have all of the roots of $x^4-2x^2-2$ by a further quadratic extension. (Here $sqrt{1-sqrt{3}}$ is only degree 2 over $Bbb Q(sqrt{1+sqrt{3}})$ since it satisfies $x^2 -1+sqrt{3} = 0$ and the latter field contains $sqrt{3}$.)



              Now we know that automorphisms in the Galois group $text{Gal}(F/Bbb Q)$ are determined by how they permute the elements of $x^4-2x^2-2$, so the only possibility for a group of order 8 must be given by the Sylow 2-subgroups of $S_4$, which contain two 4-cycles apiece. The only suitable candidates for these 4-cycles are
              $$
              sqrt{1+sqrt{3}} to
              sqrt{1-sqrt{3}} to
              -sqrt{1+sqrt{3}} to
              -sqrt{1-sqrt{3}} to
              sqrt{1+sqrt{3}}
              $$
              or its inverse. This follows since if $sqrt{1+sqrt{3}}$ is sent to $-sqrt{1+sqrt{3}}$ that will force a 2-cycle, and similarly for the pair $pmsqrt{1-sqrt{3}}$.



              If you don't want to appeal to abstract Galois theory and want a concrete justification that this permutation of elements corresponds to a well-defined field automorphism, this can be done directly by modeling $K$ as $Bbb Q[x,y]/(x^4-2x^2-2, y^2+x^2-2)$, and checking that the self-map given by $xmapsto y, ymapsto -x$ is well-defined on the quotient. Here the justification for the $y^2+x^2-2$ is given by $(sqrt{1+sqrt{3}})^2 + (sqrt{1-sqrt{3}})^2 = 1+sqrt{3}+1-sqrt{3} = 2$.






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                First observe that $K$ must be a degree 8 extension over $Bbb Q$. Here $Bbb Q(sqrt{1+sqrt{3}})$ is a degree 4 extension, but note that it cannot contain $sqrt{1-sqrt{3}}$ since $Bbb Q(sqrt{1+sqrt{3}}) subset Bbb R$ whereas $sqrt{1-sqrt{3}} not in Bbb R$. Thus, we can only have all of the roots of $x^4-2x^2-2$ by a further quadratic extension. (Here $sqrt{1-sqrt{3}}$ is only degree 2 over $Bbb Q(sqrt{1+sqrt{3}})$ since it satisfies $x^2 -1+sqrt{3} = 0$ and the latter field contains $sqrt{3}$.)



                Now we know that automorphisms in the Galois group $text{Gal}(F/Bbb Q)$ are determined by how they permute the elements of $x^4-2x^2-2$, so the only possibility for a group of order 8 must be given by the Sylow 2-subgroups of $S_4$, which contain two 4-cycles apiece. The only suitable candidates for these 4-cycles are
                $$
                sqrt{1+sqrt{3}} to
                sqrt{1-sqrt{3}} to
                -sqrt{1+sqrt{3}} to
                -sqrt{1-sqrt{3}} to
                sqrt{1+sqrt{3}}
                $$
                or its inverse. This follows since if $sqrt{1+sqrt{3}}$ is sent to $-sqrt{1+sqrt{3}}$ that will force a 2-cycle, and similarly for the pair $pmsqrt{1-sqrt{3}}$.



                If you don't want to appeal to abstract Galois theory and want a concrete justification that this permutation of elements corresponds to a well-defined field automorphism, this can be done directly by modeling $K$ as $Bbb Q[x,y]/(x^4-2x^2-2, y^2+x^2-2)$, and checking that the self-map given by $xmapsto y, ymapsto -x$ is well-defined on the quotient. Here the justification for the $y^2+x^2-2$ is given by $(sqrt{1+sqrt{3}})^2 + (sqrt{1-sqrt{3}})^2 = 1+sqrt{3}+1-sqrt{3} = 2$.






                share|cite|improve this answer









                $endgroup$



                First observe that $K$ must be a degree 8 extension over $Bbb Q$. Here $Bbb Q(sqrt{1+sqrt{3}})$ is a degree 4 extension, but note that it cannot contain $sqrt{1-sqrt{3}}$ since $Bbb Q(sqrt{1+sqrt{3}}) subset Bbb R$ whereas $sqrt{1-sqrt{3}} not in Bbb R$. Thus, we can only have all of the roots of $x^4-2x^2-2$ by a further quadratic extension. (Here $sqrt{1-sqrt{3}}$ is only degree 2 over $Bbb Q(sqrt{1+sqrt{3}})$ since it satisfies $x^2 -1+sqrt{3} = 0$ and the latter field contains $sqrt{3}$.)



                Now we know that automorphisms in the Galois group $text{Gal}(F/Bbb Q)$ are determined by how they permute the elements of $x^4-2x^2-2$, so the only possibility for a group of order 8 must be given by the Sylow 2-subgroups of $S_4$, which contain two 4-cycles apiece. The only suitable candidates for these 4-cycles are
                $$
                sqrt{1+sqrt{3}} to
                sqrt{1-sqrt{3}} to
                -sqrt{1+sqrt{3}} to
                -sqrt{1-sqrt{3}} to
                sqrt{1+sqrt{3}}
                $$
                or its inverse. This follows since if $sqrt{1+sqrt{3}}$ is sent to $-sqrt{1+sqrt{3}}$ that will force a 2-cycle, and similarly for the pair $pmsqrt{1-sqrt{3}}$.



                If you don't want to appeal to abstract Galois theory and want a concrete justification that this permutation of elements corresponds to a well-defined field automorphism, this can be done directly by modeling $K$ as $Bbb Q[x,y]/(x^4-2x^2-2, y^2+x^2-2)$, and checking that the self-map given by $xmapsto y, ymapsto -x$ is well-defined on the quotient. Here the justification for the $y^2+x^2-2$ is given by $(sqrt{1+sqrt{3}})^2 + (sqrt{1-sqrt{3}})^2 = 1+sqrt{3}+1-sqrt{3} = 2$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 8 at 2:20









                Rolf HoyerRolf Hoyer

                11.2k31629




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