Galois group of a certain splitting field












3












$begingroup$



Let $f$ be the minimal polynomial for $sqrt{3+sqrt{2}}$. Find the Galois group of the splitting field $K$ over $mathbb{Q}$.




Here are the steps that I have taken.




  1. The minimum polynomial is $x^4-6x^2+7$.

  2. The roots of this are $pm sqrt{3 pm sqrt{2}}$.

  3. I am guessing that the Galois group is...maybe $mathbb{Z}/4mathbb{Z}$, analogous to how $mathbb{Z}/4mathbb{Z}$ is the Galois group for $mathbb{Q}(sqrt{2+sqrt{2}})$, but I am not sure how to show this.


Any hints appreciated, Thanks!






abstract-algebra galois-theory minimal-polynomials







share|cite|improve this question











$endgroup$

















    3












    $begingroup$



    Let $f$ be the minimal polynomial for $sqrt{3+sqrt{2}}$. Find the Galois group of the splitting field $K$ over $mathbb{Q}$.




    Here are the steps that I have taken.




    1. The minimum polynomial is $x^4-6x^2+7$.

    2. The roots of this are $pm sqrt{3 pm sqrt{2}}$.

    3. I am guessing that the Galois group is...maybe $mathbb{Z}/4mathbb{Z}$, analogous to how $mathbb{Z}/4mathbb{Z}$ is the Galois group for $mathbb{Q}(sqrt{2+sqrt{2}})$, but I am not sure how to show this.


    Any hints appreciated, Thanks!






    abstract-algebra galois-theory minimal-polynomials







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$



      Let $f$ be the minimal polynomial for $sqrt{3+sqrt{2}}$. Find the Galois group of the splitting field $K$ over $mathbb{Q}$.




      Here are the steps that I have taken.




      1. The minimum polynomial is $x^4-6x^2+7$.

      2. The roots of this are $pm sqrt{3 pm sqrt{2}}$.

      3. I am guessing that the Galois group is...maybe $mathbb{Z}/4mathbb{Z}$, analogous to how $mathbb{Z}/4mathbb{Z}$ is the Galois group for $mathbb{Q}(sqrt{2+sqrt{2}})$, but I am not sure how to show this.


      Any hints appreciated, Thanks!






      abstract-algebra galois-theory minimal-polynomials







      share|cite|improve this question











      $endgroup$





      Let $f$ be the minimal polynomial for $sqrt{3+sqrt{2}}$. Find the Galois group of the splitting field $K$ over $mathbb{Q}$.




      Here are the steps that I have taken.




      1. The minimum polynomial is $x^4-6x^2+7$.

      2. The roots of this are $pm sqrt{3 pm sqrt{2}}$.

      3. I am guessing that the Galois group is...maybe $mathbb{Z}/4mathbb{Z}$, analogous to how $mathbb{Z}/4mathbb{Z}$ is the Galois group for $mathbb{Q}(sqrt{2+sqrt{2}})$, but I am not sure how to show this.


      Any hints appreciated, Thanks!






      abstract-algebra galois-theory minimal-polynomials




      abstract-algebra galois-theory minimal-polynomials






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 26 '18 at 21:46









      user26857

      39.3k124183




      39.3k124183










      asked Dec 24 '18 at 22:12









      sarafisarafi

      884




      884






















          2 Answers
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          active

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          3












          $begingroup$

          Let $G=Gal(K/mathbb{Q})$.



          Hint : $mathbb{Q}(sqrt{2})$ is contained in $K$, and is a Galois subextension of $mathbb{Q}$. It is the fixed field of $H=Gal(K/mathbb{Q}(sqrt{2}))$; thus $G/H simeq Gal(mathbb{Q}(sqrt{2})/mathbb{Q})simeq mathbb{Z/2Z}$.



          The minimal polynomial of $sqrt{3+sqrt{2}}$ over $mathbb{Q}(sqrt{2})$ is $x^2-(3+sqrt{2})$, so it's pretty clear that $Hsimeq mathbb{Z/2Z}$. So $Gsimeq mathbb{Z/4Z}$ or $(mathbb{Z/2Z})^2$.



          Which one it is will depend on whether $sqrt{2}mapsto -sqrt{2}$ is a square in $G$ or not. Can you see if it can be a square ?






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Here is the technique that I finally found works. There is a theorem in Hungerford's Section V, Chapter 4, Exercise 9: That allows us to classify biquadratic quartic extensions: Should the minimal polynomial be $x^4+ax^2+b$, we may classify the extension as such:




            1. If $b$ is square, then the Galois group is $mathbb{Z}/2mathbb{Z} times mathbb{Z} / 2mathbb{Z}$.

            2. If $b(a^2-4b)$ is square we have $mathbb{Z}/4mathbb{Z}$.

            3. If neither then we have $D_8$. (Dummit & Foote convention, symmetries of a square.)


            The proof is not relevant to answering my question so is discarded.



            An example of $1$ is the classic $mathbb{Q}(sqrt{2},sqrt{3})$, which is equal to $mathbb{Q}(sqrt{2}+sqrt{3})$, which has a minimal polynomial $x^4-10x^2+1$. 1, is trivially a square.



            An example of $2$ is $mathbb{Q}(sqrt{2+sqrt{2}})$, whose minimal polynomial is $x^4-4x^2+2$. Notice that $b(a^2-4b)= 16$.



            My question is the third kind, neither $b$ nor $b(a^2-4b)$ is a square. We can proceed similar to Dummit & Foote's Exercise 16 in 14.2.



            We will proceed, as the Exercise suggests by solving the polynomial and enumerating the roots, let:
            $alpha_1 = sqrt{3+sqrt{2}}$, $alpha_2 = -sqrt{3+sqrt{2}}$, $alpha_3 = sqrt{3-sqrt{2}}$, $alpha_4 = -sqrt{3-sqrt{2}}$. Two of these roots are real while two arent.



            It is easy to check that over $mathbb{Q}(sqrt{2})$, the following automorphisms:
            $sigma:alpha_1 mapsto alpha_2, alpha_3 mapsto alpha_3$ and $tau: alpha_1 mapsto alpha_1, alpha_3 mapsto alpha_4$ define the Klein-4 group ($V_4$, or $mathbb{Z}/2mathbb{Z} times mathbb{Z}/2mathbb{Z}$ if you prefer).



            But these in turn are over $mathbb{Q}(sqrt{2})$, which is degree two, so we have a Galois group of order $8$ (we have to show also show that $mathbb{Q}(alpha_1),mathbb{Q}(alpha_3)$, and their composite is Galois, because Galois over Galois is not Galois) in our hands, which is not Abelian. The only choice are $Q_8$ and $D_8$, but only $D_8$ has $V_4$ inside of it.






            share|cite|improve this answer









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              $begingroup$

              Let $G=Gal(K/mathbb{Q})$.



              Hint : $mathbb{Q}(sqrt{2})$ is contained in $K$, and is a Galois subextension of $mathbb{Q}$. It is the fixed field of $H=Gal(K/mathbb{Q}(sqrt{2}))$; thus $G/H simeq Gal(mathbb{Q}(sqrt{2})/mathbb{Q})simeq mathbb{Z/2Z}$.



              The minimal polynomial of $sqrt{3+sqrt{2}}$ over $mathbb{Q}(sqrt{2})$ is $x^2-(3+sqrt{2})$, so it's pretty clear that $Hsimeq mathbb{Z/2Z}$. So $Gsimeq mathbb{Z/4Z}$ or $(mathbb{Z/2Z})^2$.



              Which one it is will depend on whether $sqrt{2}mapsto -sqrt{2}$ is a square in $G$ or not. Can you see if it can be a square ?






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Let $G=Gal(K/mathbb{Q})$.



                Hint : $mathbb{Q}(sqrt{2})$ is contained in $K$, and is a Galois subextension of $mathbb{Q}$. It is the fixed field of $H=Gal(K/mathbb{Q}(sqrt{2}))$; thus $G/H simeq Gal(mathbb{Q}(sqrt{2})/mathbb{Q})simeq mathbb{Z/2Z}$.



                The minimal polynomial of $sqrt{3+sqrt{2}}$ over $mathbb{Q}(sqrt{2})$ is $x^2-(3+sqrt{2})$, so it's pretty clear that $Hsimeq mathbb{Z/2Z}$. So $Gsimeq mathbb{Z/4Z}$ or $(mathbb{Z/2Z})^2$.



                Which one it is will depend on whether $sqrt{2}mapsto -sqrt{2}$ is a square in $G$ or not. Can you see if it can be a square ?






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Let $G=Gal(K/mathbb{Q})$.



                  Hint : $mathbb{Q}(sqrt{2})$ is contained in $K$, and is a Galois subextension of $mathbb{Q}$. It is the fixed field of $H=Gal(K/mathbb{Q}(sqrt{2}))$; thus $G/H simeq Gal(mathbb{Q}(sqrt{2})/mathbb{Q})simeq mathbb{Z/2Z}$.



                  The minimal polynomial of $sqrt{3+sqrt{2}}$ over $mathbb{Q}(sqrt{2})$ is $x^2-(3+sqrt{2})$, so it's pretty clear that $Hsimeq mathbb{Z/2Z}$. So $Gsimeq mathbb{Z/4Z}$ or $(mathbb{Z/2Z})^2$.



                  Which one it is will depend on whether $sqrt{2}mapsto -sqrt{2}$ is a square in $G$ or not. Can you see if it can be a square ?






                  share|cite|improve this answer









                  $endgroup$



                  Let $G=Gal(K/mathbb{Q})$.



                  Hint : $mathbb{Q}(sqrt{2})$ is contained in $K$, and is a Galois subextension of $mathbb{Q}$. It is the fixed field of $H=Gal(K/mathbb{Q}(sqrt{2}))$; thus $G/H simeq Gal(mathbb{Q}(sqrt{2})/mathbb{Q})simeq mathbb{Z/2Z}$.



                  The minimal polynomial of $sqrt{3+sqrt{2}}$ over $mathbb{Q}(sqrt{2})$ is $x^2-(3+sqrt{2})$, so it's pretty clear that $Hsimeq mathbb{Z/2Z}$. So $Gsimeq mathbb{Z/4Z}$ or $(mathbb{Z/2Z})^2$.



                  Which one it is will depend on whether $sqrt{2}mapsto -sqrt{2}$ is a square in $G$ or not. Can you see if it can be a square ?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 24 '18 at 22:26









                  MaxMax

                  13.3k11040




                  13.3k11040























                      0












                      $begingroup$

                      Here is the technique that I finally found works. There is a theorem in Hungerford's Section V, Chapter 4, Exercise 9: That allows us to classify biquadratic quartic extensions: Should the minimal polynomial be $x^4+ax^2+b$, we may classify the extension as such:




                      1. If $b$ is square, then the Galois group is $mathbb{Z}/2mathbb{Z} times mathbb{Z} / 2mathbb{Z}$.

                      2. If $b(a^2-4b)$ is square we have $mathbb{Z}/4mathbb{Z}$.

                      3. If neither then we have $D_8$. (Dummit & Foote convention, symmetries of a square.)


                      The proof is not relevant to answering my question so is discarded.



                      An example of $1$ is the classic $mathbb{Q}(sqrt{2},sqrt{3})$, which is equal to $mathbb{Q}(sqrt{2}+sqrt{3})$, which has a minimal polynomial $x^4-10x^2+1$. 1, is trivially a square.



                      An example of $2$ is $mathbb{Q}(sqrt{2+sqrt{2}})$, whose minimal polynomial is $x^4-4x^2+2$. Notice that $b(a^2-4b)= 16$.



                      My question is the third kind, neither $b$ nor $b(a^2-4b)$ is a square. We can proceed similar to Dummit & Foote's Exercise 16 in 14.2.



                      We will proceed, as the Exercise suggests by solving the polynomial and enumerating the roots, let:
                      $alpha_1 = sqrt{3+sqrt{2}}$, $alpha_2 = -sqrt{3+sqrt{2}}$, $alpha_3 = sqrt{3-sqrt{2}}$, $alpha_4 = -sqrt{3-sqrt{2}}$. Two of these roots are real while two arent.



                      It is easy to check that over $mathbb{Q}(sqrt{2})$, the following automorphisms:
                      $sigma:alpha_1 mapsto alpha_2, alpha_3 mapsto alpha_3$ and $tau: alpha_1 mapsto alpha_1, alpha_3 mapsto alpha_4$ define the Klein-4 group ($V_4$, or $mathbb{Z}/2mathbb{Z} times mathbb{Z}/2mathbb{Z}$ if you prefer).



                      But these in turn are over $mathbb{Q}(sqrt{2})$, which is degree two, so we have a Galois group of order $8$ (we have to show also show that $mathbb{Q}(alpha_1),mathbb{Q}(alpha_3)$, and their composite is Galois, because Galois over Galois is not Galois) in our hands, which is not Abelian. The only choice are $Q_8$ and $D_8$, but only $D_8$ has $V_4$ inside of it.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Here is the technique that I finally found works. There is a theorem in Hungerford's Section V, Chapter 4, Exercise 9: That allows us to classify biquadratic quartic extensions: Should the minimal polynomial be $x^4+ax^2+b$, we may classify the extension as such:




                        1. If $b$ is square, then the Galois group is $mathbb{Z}/2mathbb{Z} times mathbb{Z} / 2mathbb{Z}$.

                        2. If $b(a^2-4b)$ is square we have $mathbb{Z}/4mathbb{Z}$.

                        3. If neither then we have $D_8$. (Dummit & Foote convention, symmetries of a square.)


                        The proof is not relevant to answering my question so is discarded.



                        An example of $1$ is the classic $mathbb{Q}(sqrt{2},sqrt{3})$, which is equal to $mathbb{Q}(sqrt{2}+sqrt{3})$, which has a minimal polynomial $x^4-10x^2+1$. 1, is trivially a square.



                        An example of $2$ is $mathbb{Q}(sqrt{2+sqrt{2}})$, whose minimal polynomial is $x^4-4x^2+2$. Notice that $b(a^2-4b)= 16$.



                        My question is the third kind, neither $b$ nor $b(a^2-4b)$ is a square. We can proceed similar to Dummit & Foote's Exercise 16 in 14.2.



                        We will proceed, as the Exercise suggests by solving the polynomial and enumerating the roots, let:
                        $alpha_1 = sqrt{3+sqrt{2}}$, $alpha_2 = -sqrt{3+sqrt{2}}$, $alpha_3 = sqrt{3-sqrt{2}}$, $alpha_4 = -sqrt{3-sqrt{2}}$. Two of these roots are real while two arent.



                        It is easy to check that over $mathbb{Q}(sqrt{2})$, the following automorphisms:
                        $sigma:alpha_1 mapsto alpha_2, alpha_3 mapsto alpha_3$ and $tau: alpha_1 mapsto alpha_1, alpha_3 mapsto alpha_4$ define the Klein-4 group ($V_4$, or $mathbb{Z}/2mathbb{Z} times mathbb{Z}/2mathbb{Z}$ if you prefer).



                        But these in turn are over $mathbb{Q}(sqrt{2})$, which is degree two, so we have a Galois group of order $8$ (we have to show also show that $mathbb{Q}(alpha_1),mathbb{Q}(alpha_3)$, and their composite is Galois, because Galois over Galois is not Galois) in our hands, which is not Abelian. The only choice are $Q_8$ and $D_8$, but only $D_8$ has $V_4$ inside of it.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Here is the technique that I finally found works. There is a theorem in Hungerford's Section V, Chapter 4, Exercise 9: That allows us to classify biquadratic quartic extensions: Should the minimal polynomial be $x^4+ax^2+b$, we may classify the extension as such:




                          1. If $b$ is square, then the Galois group is $mathbb{Z}/2mathbb{Z} times mathbb{Z} / 2mathbb{Z}$.

                          2. If $b(a^2-4b)$ is square we have $mathbb{Z}/4mathbb{Z}$.

                          3. If neither then we have $D_8$. (Dummit & Foote convention, symmetries of a square.)


                          The proof is not relevant to answering my question so is discarded.



                          An example of $1$ is the classic $mathbb{Q}(sqrt{2},sqrt{3})$, which is equal to $mathbb{Q}(sqrt{2}+sqrt{3})$, which has a minimal polynomial $x^4-10x^2+1$. 1, is trivially a square.



                          An example of $2$ is $mathbb{Q}(sqrt{2+sqrt{2}})$, whose minimal polynomial is $x^4-4x^2+2$. Notice that $b(a^2-4b)= 16$.



                          My question is the third kind, neither $b$ nor $b(a^2-4b)$ is a square. We can proceed similar to Dummit & Foote's Exercise 16 in 14.2.



                          We will proceed, as the Exercise suggests by solving the polynomial and enumerating the roots, let:
                          $alpha_1 = sqrt{3+sqrt{2}}$, $alpha_2 = -sqrt{3+sqrt{2}}$, $alpha_3 = sqrt{3-sqrt{2}}$, $alpha_4 = -sqrt{3-sqrt{2}}$. Two of these roots are real while two arent.



                          It is easy to check that over $mathbb{Q}(sqrt{2})$, the following automorphisms:
                          $sigma:alpha_1 mapsto alpha_2, alpha_3 mapsto alpha_3$ and $tau: alpha_1 mapsto alpha_1, alpha_3 mapsto alpha_4$ define the Klein-4 group ($V_4$, or $mathbb{Z}/2mathbb{Z} times mathbb{Z}/2mathbb{Z}$ if you prefer).



                          But these in turn are over $mathbb{Q}(sqrt{2})$, which is degree two, so we have a Galois group of order $8$ (we have to show also show that $mathbb{Q}(alpha_1),mathbb{Q}(alpha_3)$, and their composite is Galois, because Galois over Galois is not Galois) in our hands, which is not Abelian. The only choice are $Q_8$ and $D_8$, but only $D_8$ has $V_4$ inside of it.






                          share|cite|improve this answer









                          $endgroup$



                          Here is the technique that I finally found works. There is a theorem in Hungerford's Section V, Chapter 4, Exercise 9: That allows us to classify biquadratic quartic extensions: Should the minimal polynomial be $x^4+ax^2+b$, we may classify the extension as such:




                          1. If $b$ is square, then the Galois group is $mathbb{Z}/2mathbb{Z} times mathbb{Z} / 2mathbb{Z}$.

                          2. If $b(a^2-4b)$ is square we have $mathbb{Z}/4mathbb{Z}$.

                          3. If neither then we have $D_8$. (Dummit & Foote convention, symmetries of a square.)


                          The proof is not relevant to answering my question so is discarded.



                          An example of $1$ is the classic $mathbb{Q}(sqrt{2},sqrt{3})$, which is equal to $mathbb{Q}(sqrt{2}+sqrt{3})$, which has a minimal polynomial $x^4-10x^2+1$. 1, is trivially a square.



                          An example of $2$ is $mathbb{Q}(sqrt{2+sqrt{2}})$, whose minimal polynomial is $x^4-4x^2+2$. Notice that $b(a^2-4b)= 16$.



                          My question is the third kind, neither $b$ nor $b(a^2-4b)$ is a square. We can proceed similar to Dummit & Foote's Exercise 16 in 14.2.



                          We will proceed, as the Exercise suggests by solving the polynomial and enumerating the roots, let:
                          $alpha_1 = sqrt{3+sqrt{2}}$, $alpha_2 = -sqrt{3+sqrt{2}}$, $alpha_3 = sqrt{3-sqrt{2}}$, $alpha_4 = -sqrt{3-sqrt{2}}$. Two of these roots are real while two arent.



                          It is easy to check that over $mathbb{Q}(sqrt{2})$, the following automorphisms:
                          $sigma:alpha_1 mapsto alpha_2, alpha_3 mapsto alpha_3$ and $tau: alpha_1 mapsto alpha_1, alpha_3 mapsto alpha_4$ define the Klein-4 group ($V_4$, or $mathbb{Z}/2mathbb{Z} times mathbb{Z}/2mathbb{Z}$ if you prefer).



                          But these in turn are over $mathbb{Q}(sqrt{2})$, which is degree two, so we have a Galois group of order $8$ (we have to show also show that $mathbb{Q}(alpha_1),mathbb{Q}(alpha_3)$, and their composite is Galois, because Galois over Galois is not Galois) in our hands, which is not Abelian. The only choice are $Q_8$ and $D_8$, but only $D_8$ has $V_4$ inside of it.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 3 at 22:15









                          sarafisarafi

                          884




                          884






























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