Determine Gal$(mathbb{Q}(sqrt[8]{7},i)/mathbb{Q}(sqrt{7}))$ and...












1












$begingroup$


Let $K=mathbb{Q}(sqrt[8]{7},i)$, let $F_1 = mathbb{Q}(sqrt{7})$ and let $F_2=mathbb{Q}(sqrt{-7})$.



(a) Prove $K$ is Galois over $F_1$ and over $F_2$, and determine $[K:F_1]$ and $[K:F_2]$.



(b) Determine Gal$(K/F_1)$ and Gal$(K/F_2)$.



I first thought that $K/mathbb{Q}$ was a Galois extension so that I could apply the fundamental theorem of Galois Theory, but turned out that's only true when $K=mathbb{Q}(sqrt[8]{2},i)$. Now I'm having trouble to show $K$ is Galois over both $F_1$ and $F_2$, I got stuck on finding the separable minimal polynomial over $mathbb{Q}(sqrt{7})$ or over $mathbb{Q}(sqrt{-7})$. I feel like that $[K:F_1]=8=[K:F_2]$, since
$$mathbb{Q}(sqrt{7})subseteqmathbb{Q}(sqrt[8]{7})subseteqmathbb{Q}(sqrt[8]{7},i), $$ but I need the minimal polynomials to justify I think. I would appreciate for any help!






abstract-algebra galois-theory







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$endgroup$

















    1












    $begingroup$


    Let $K=mathbb{Q}(sqrt[8]{7},i)$, let $F_1 = mathbb{Q}(sqrt{7})$ and let $F_2=mathbb{Q}(sqrt{-7})$.



    (a) Prove $K$ is Galois over $F_1$ and over $F_2$, and determine $[K:F_1]$ and $[K:F_2]$.



    (b) Determine Gal$(K/F_1)$ and Gal$(K/F_2)$.



    I first thought that $K/mathbb{Q}$ was a Galois extension so that I could apply the fundamental theorem of Galois Theory, but turned out that's only true when $K=mathbb{Q}(sqrt[8]{2},i)$. Now I'm having trouble to show $K$ is Galois over both $F_1$ and $F_2$, I got stuck on finding the separable minimal polynomial over $mathbb{Q}(sqrt{7})$ or over $mathbb{Q}(sqrt{-7})$. I feel like that $[K:F_1]=8=[K:F_2]$, since
    $$mathbb{Q}(sqrt{7})subseteqmathbb{Q}(sqrt[8]{7})subseteqmathbb{Q}(sqrt[8]{7},i), $$ but I need the minimal polynomials to justify I think. I would appreciate for any help!






    abstract-algebra galois-theory







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $K=mathbb{Q}(sqrt[8]{7},i)$, let $F_1 = mathbb{Q}(sqrt{7})$ and let $F_2=mathbb{Q}(sqrt{-7})$.



      (a) Prove $K$ is Galois over $F_1$ and over $F_2$, and determine $[K:F_1]$ and $[K:F_2]$.



      (b) Determine Gal$(K/F_1)$ and Gal$(K/F_2)$.



      I first thought that $K/mathbb{Q}$ was a Galois extension so that I could apply the fundamental theorem of Galois Theory, but turned out that's only true when $K=mathbb{Q}(sqrt[8]{2},i)$. Now I'm having trouble to show $K$ is Galois over both $F_1$ and $F_2$, I got stuck on finding the separable minimal polynomial over $mathbb{Q}(sqrt{7})$ or over $mathbb{Q}(sqrt{-7})$. I feel like that $[K:F_1]=8=[K:F_2]$, since
      $$mathbb{Q}(sqrt{7})subseteqmathbb{Q}(sqrt[8]{7})subseteqmathbb{Q}(sqrt[8]{7},i), $$ but I need the minimal polynomials to justify I think. I would appreciate for any help!






      abstract-algebra galois-theory







      share|cite|improve this question











      $endgroup$




      Let $K=mathbb{Q}(sqrt[8]{7},i)$, let $F_1 = mathbb{Q}(sqrt{7})$ and let $F_2=mathbb{Q}(sqrt{-7})$.



      (a) Prove $K$ is Galois over $F_1$ and over $F_2$, and determine $[K:F_1]$ and $[K:F_2]$.



      (b) Determine Gal$(K/F_1)$ and Gal$(K/F_2)$.



      I first thought that $K/mathbb{Q}$ was a Galois extension so that I could apply the fundamental theorem of Galois Theory, but turned out that's only true when $K=mathbb{Q}(sqrt[8]{2},i)$. Now I'm having trouble to show $K$ is Galois over both $F_1$ and $F_2$, I got stuck on finding the separable minimal polynomial over $mathbb{Q}(sqrt{7})$ or over $mathbb{Q}(sqrt{-7})$. I feel like that $[K:F_1]=8=[K:F_2]$, since
      $$mathbb{Q}(sqrt{7})subseteqmathbb{Q}(sqrt[8]{7})subseteqmathbb{Q}(sqrt[8]{7},i), $$ but I need the minimal polynomials to justify I think. I would appreciate for any help!






      abstract-algebra galois-theory




      abstract-algebra galois-theory






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 4 at 23:12







      Alex

















      asked Jan 4 at 20:49









      AlexAlex

      757




      757






















          1 Answer
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          1












          $begingroup$

          If $F$ has characteristic zero, and $K=F(sqrt[4]a,i)$ where $ain F$, then $K$
          is Galois over $F$, being the splitting field of $x^4-a$. Your first
          example has $a=sqrt7$.



          The Galois group of $K/F_1$ has order $8$, and computing it is very similar
          to standard examples such as $Bbb Q(sqrt[4]2,i)/Bbb Q$. The group is
          dihedral of order $8$.



          I don't think $K/F_2$ is Galois.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            You only need $operatorname{char}(F) ne 2$
            $endgroup$
            – Kenny Lau
            Jan 4 at 22:08










          • $begingroup$
            I also feel like that $K$ is not Galois over $F_2$ but I don't think there's anything wrong with the question, though. Does anyone else see why $K/F_2$ should be Galois?
            $endgroup$
            – Alex
            Jan 5 at 7:31












          • $begingroup$
            @Alex You asked what the Galois group of a non-Galois extension is....
            $endgroup$
            – Lord Shark the Unknown
            Jan 5 at 7:34










          • $begingroup$
            @LordSharktheUnknown Haha, you could be right. But this question was literally on my past exam, and I still don't know how to solve it yet. I wish there's a misprint in the problem.
            $endgroup$
            – Alex
            Jan 5 at 7:38










          • $begingroup$
            Exams have misprints, just as all other types of writing. @Alex
            $endgroup$
            – Lord Shark the Unknown
            Jan 5 at 7:40













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          1 Answer
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          active

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          $begingroup$

          If $F$ has characteristic zero, and $K=F(sqrt[4]a,i)$ where $ain F$, then $K$
          is Galois over $F$, being the splitting field of $x^4-a$. Your first
          example has $a=sqrt7$.



          The Galois group of $K/F_1$ has order $8$, and computing it is very similar
          to standard examples such as $Bbb Q(sqrt[4]2,i)/Bbb Q$. The group is
          dihedral of order $8$.



          I don't think $K/F_2$ is Galois.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            You only need $operatorname{char}(F) ne 2$
            $endgroup$
            – Kenny Lau
            Jan 4 at 22:08










          • $begingroup$
            I also feel like that $K$ is not Galois over $F_2$ but I don't think there's anything wrong with the question, though. Does anyone else see why $K/F_2$ should be Galois?
            $endgroup$
            – Alex
            Jan 5 at 7:31












          • $begingroup$
            @Alex You asked what the Galois group of a non-Galois extension is....
            $endgroup$
            – Lord Shark the Unknown
            Jan 5 at 7:34










          • $begingroup$
            @LordSharktheUnknown Haha, you could be right. But this question was literally on my past exam, and I still don't know how to solve it yet. I wish there's a misprint in the problem.
            $endgroup$
            – Alex
            Jan 5 at 7:38










          • $begingroup$
            Exams have misprints, just as all other types of writing. @Alex
            $endgroup$
            – Lord Shark the Unknown
            Jan 5 at 7:40


















          1












          $begingroup$

          If $F$ has characteristic zero, and $K=F(sqrt[4]a,i)$ where $ain F$, then $K$
          is Galois over $F$, being the splitting field of $x^4-a$. Your first
          example has $a=sqrt7$.



          The Galois group of $K/F_1$ has order $8$, and computing it is very similar
          to standard examples such as $Bbb Q(sqrt[4]2,i)/Bbb Q$. The group is
          dihedral of order $8$.



          I don't think $K/F_2$ is Galois.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            You only need $operatorname{char}(F) ne 2$
            $endgroup$
            – Kenny Lau
            Jan 4 at 22:08










          • $begingroup$
            I also feel like that $K$ is not Galois over $F_2$ but I don't think there's anything wrong with the question, though. Does anyone else see why $K/F_2$ should be Galois?
            $endgroup$
            – Alex
            Jan 5 at 7:31












          • $begingroup$
            @Alex You asked what the Galois group of a non-Galois extension is....
            $endgroup$
            – Lord Shark the Unknown
            Jan 5 at 7:34










          • $begingroup$
            @LordSharktheUnknown Haha, you could be right. But this question was literally on my past exam, and I still don't know how to solve it yet. I wish there's a misprint in the problem.
            $endgroup$
            – Alex
            Jan 5 at 7:38










          • $begingroup$
            Exams have misprints, just as all other types of writing. @Alex
            $endgroup$
            – Lord Shark the Unknown
            Jan 5 at 7:40
















          1












          1








          1





          $begingroup$

          If $F$ has characteristic zero, and $K=F(sqrt[4]a,i)$ where $ain F$, then $K$
          is Galois over $F$, being the splitting field of $x^4-a$. Your first
          example has $a=sqrt7$.



          The Galois group of $K/F_1$ has order $8$, and computing it is very similar
          to standard examples such as $Bbb Q(sqrt[4]2,i)/Bbb Q$. The group is
          dihedral of order $8$.



          I don't think $K/F_2$ is Galois.






          share|cite|improve this answer









          $endgroup$



          If $F$ has characteristic zero, and $K=F(sqrt[4]a,i)$ where $ain F$, then $K$
          is Galois over $F$, being the splitting field of $x^4-a$. Your first
          example has $a=sqrt7$.



          The Galois group of $K/F_1$ has order $8$, and computing it is very similar
          to standard examples such as $Bbb Q(sqrt[4]2,i)/Bbb Q$. The group is
          dihedral of order $8$.



          I don't think $K/F_2$ is Galois.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 21:20









          Lord Shark the UnknownLord Shark the Unknown

          106k1161133




          106k1161133












          • $begingroup$
            You only need $operatorname{char}(F) ne 2$
            $endgroup$
            – Kenny Lau
            Jan 4 at 22:08










          • $begingroup$
            I also feel like that $K$ is not Galois over $F_2$ but I don't think there's anything wrong with the question, though. Does anyone else see why $K/F_2$ should be Galois?
            $endgroup$
            – Alex
            Jan 5 at 7:31












          • $begingroup$
            @Alex You asked what the Galois group of a non-Galois extension is....
            $endgroup$
            – Lord Shark the Unknown
            Jan 5 at 7:34










          • $begingroup$
            @LordSharktheUnknown Haha, you could be right. But this question was literally on my past exam, and I still don't know how to solve it yet. I wish there's a misprint in the problem.
            $endgroup$
            – Alex
            Jan 5 at 7:38










          • $begingroup$
            Exams have misprints, just as all other types of writing. @Alex
            $endgroup$
            – Lord Shark the Unknown
            Jan 5 at 7:40




















          • $begingroup$
            You only need $operatorname{char}(F) ne 2$
            $endgroup$
            – Kenny Lau
            Jan 4 at 22:08










          • $begingroup$
            I also feel like that $K$ is not Galois over $F_2$ but I don't think there's anything wrong with the question, though. Does anyone else see why $K/F_2$ should be Galois?
            $endgroup$
            – Alex
            Jan 5 at 7:31












          • $begingroup$
            @Alex You asked what the Galois group of a non-Galois extension is....
            $endgroup$
            – Lord Shark the Unknown
            Jan 5 at 7:34










          • $begingroup$
            @LordSharktheUnknown Haha, you could be right. But this question was literally on my past exam, and I still don't know how to solve it yet. I wish there's a misprint in the problem.
            $endgroup$
            – Alex
            Jan 5 at 7:38










          • $begingroup$
            Exams have misprints, just as all other types of writing. @Alex
            $endgroup$
            – Lord Shark the Unknown
            Jan 5 at 7:40


















          $begingroup$
          You only need $operatorname{char}(F) ne 2$
          $endgroup$
          – Kenny Lau
          Jan 4 at 22:08




          $begingroup$
          You only need $operatorname{char}(F) ne 2$
          $endgroup$
          – Kenny Lau
          Jan 4 at 22:08












          $begingroup$
          I also feel like that $K$ is not Galois over $F_2$ but I don't think there's anything wrong with the question, though. Does anyone else see why $K/F_2$ should be Galois?
          $endgroup$
          – Alex
          Jan 5 at 7:31






          $begingroup$
          I also feel like that $K$ is not Galois over $F_2$ but I don't think there's anything wrong with the question, though. Does anyone else see why $K/F_2$ should be Galois?
          $endgroup$
          – Alex
          Jan 5 at 7:31














          $begingroup$
          @Alex You asked what the Galois group of a non-Galois extension is....
          $endgroup$
          – Lord Shark the Unknown
          Jan 5 at 7:34




          $begingroup$
          @Alex You asked what the Galois group of a non-Galois extension is....
          $endgroup$
          – Lord Shark the Unknown
          Jan 5 at 7:34












          $begingroup$
          @LordSharktheUnknown Haha, you could be right. But this question was literally on my past exam, and I still don't know how to solve it yet. I wish there's a misprint in the problem.
          $endgroup$
          – Alex
          Jan 5 at 7:38




          $begingroup$
          @LordSharktheUnknown Haha, you could be right. But this question was literally on my past exam, and I still don't know how to solve it yet. I wish there's a misprint in the problem.
          $endgroup$
          – Alex
          Jan 5 at 7:38












          $begingroup$
          Exams have misprints, just as all other types of writing. @Alex
          $endgroup$
          – Lord Shark the Unknown
          Jan 5 at 7:40






          $begingroup$
          Exams have misprints, just as all other types of writing. @Alex
          $endgroup$
          – Lord Shark the Unknown
          Jan 5 at 7:40




















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