Prove that $sqrt{3pmsqrt{7}} notin mathbb{Q}(sqrt{3mpsqrt{7}})$.












5












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I'm currently solving a fairly long exercise related to Galois theory in which I've come across having to prove that $sqrt{3+sqrt{7}} notin mathbb{Q}(sqrt{3-sqrt{7}})$ and $sqrt{3-sqrt{7}} notin mathbb{Q}(sqrt{3+sqrt{7}})$. So far I haven't been able to find an "easy" or simple and understandable way to do so given that this isn't the main part of the problem.



Any help is appreciated!






abstract-algebra galois-theory algebraic-number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    $sqrt{3+sqrt{7}} = a+bsqrt{3-sqrt{7}} implies 3+sqrt{7} = a^2+2absqrt{3-sqrt{7}} + b(3-sqrt{7})$ $ implies c+dsqrt{7} = esqrt{3-sqrt{7}} implies c^2+2cdsqrt{7}+7d^2 = e^2(3-sqrt{7}) implies 2cd = -e^2$. But $-e^2 = -4a^2b^2$ and $2cd = 2(3-a^2-3b)(b+1)$, so we must have $2a^2b^2 = (a^2+3b-3)(b+1)$. Now solve for $a^2$ and hopefully this will give a contradiction.
    $endgroup$
    – mathworker21
    Jan 8 at 15:39






  • 3




    $begingroup$
    @mathworker21 What is $a$ and $b$? ${1,sqrt{3-sqrt{7}}}$ is not a basis for $mathbb{Q}(sqrt{3-sqrt{7}})$ over $mathbb{Q}$
    $endgroup$
    – mouthetics
    Jan 8 at 15:40










  • $begingroup$
    @mouthetics good point. imma bit rusty on my algebra. (they were supposed to be rationals)
    $endgroup$
    – mathworker21
    Jan 8 at 15:41
















5












$begingroup$


I'm currently solving a fairly long exercise related to Galois theory in which I've come across having to prove that $sqrt{3+sqrt{7}} notin mathbb{Q}(sqrt{3-sqrt{7}})$ and $sqrt{3-sqrt{7}} notin mathbb{Q}(sqrt{3+sqrt{7}})$. So far I haven't been able to find an "easy" or simple and understandable way to do so given that this isn't the main part of the problem.



Any help is appreciated!






abstract-algebra galois-theory algebraic-number-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    $sqrt{3+sqrt{7}} = a+bsqrt{3-sqrt{7}} implies 3+sqrt{7} = a^2+2absqrt{3-sqrt{7}} + b(3-sqrt{7})$ $ implies c+dsqrt{7} = esqrt{3-sqrt{7}} implies c^2+2cdsqrt{7}+7d^2 = e^2(3-sqrt{7}) implies 2cd = -e^2$. But $-e^2 = -4a^2b^2$ and $2cd = 2(3-a^2-3b)(b+1)$, so we must have $2a^2b^2 = (a^2+3b-3)(b+1)$. Now solve for $a^2$ and hopefully this will give a contradiction.
    $endgroup$
    – mathworker21
    Jan 8 at 15:39






  • 3




    $begingroup$
    @mathworker21 What is $a$ and $b$? ${1,sqrt{3-sqrt{7}}}$ is not a basis for $mathbb{Q}(sqrt{3-sqrt{7}})$ over $mathbb{Q}$
    $endgroup$
    – mouthetics
    Jan 8 at 15:40










  • $begingroup$
    @mouthetics good point. imma bit rusty on my algebra. (they were supposed to be rationals)
    $endgroup$
    – mathworker21
    Jan 8 at 15:41














5












5








5





$begingroup$


I'm currently solving a fairly long exercise related to Galois theory in which I've come across having to prove that $sqrt{3+sqrt{7}} notin mathbb{Q}(sqrt{3-sqrt{7}})$ and $sqrt{3-sqrt{7}} notin mathbb{Q}(sqrt{3+sqrt{7}})$. So far I haven't been able to find an "easy" or simple and understandable way to do so given that this isn't the main part of the problem.



Any help is appreciated!






abstract-algebra galois-theory algebraic-number-theory







share|cite|improve this question











$endgroup$




I'm currently solving a fairly long exercise related to Galois theory in which I've come across having to prove that $sqrt{3+sqrt{7}} notin mathbb{Q}(sqrt{3-sqrt{7}})$ and $sqrt{3-sqrt{7}} notin mathbb{Q}(sqrt{3+sqrt{7}})$. So far I haven't been able to find an "easy" or simple and understandable way to do so given that this isn't the main part of the problem.



Any help is appreciated!






abstract-algebra galois-theory algebraic-number-theory




abstract-algebra galois-theory algebraic-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 15:59









Servaes

22.6k33793




22.6k33793










asked Jan 8 at 15:32









BBC3BBC3

368213




368213












  • $begingroup$
    $sqrt{3+sqrt{7}} = a+bsqrt{3-sqrt{7}} implies 3+sqrt{7} = a^2+2absqrt{3-sqrt{7}} + b(3-sqrt{7})$ $ implies c+dsqrt{7} = esqrt{3-sqrt{7}} implies c^2+2cdsqrt{7}+7d^2 = e^2(3-sqrt{7}) implies 2cd = -e^2$. But $-e^2 = -4a^2b^2$ and $2cd = 2(3-a^2-3b)(b+1)$, so we must have $2a^2b^2 = (a^2+3b-3)(b+1)$. Now solve for $a^2$ and hopefully this will give a contradiction.
    $endgroup$
    – mathworker21
    Jan 8 at 15:39






  • 3




    $begingroup$
    @mathworker21 What is $a$ and $b$? ${1,sqrt{3-sqrt{7}}}$ is not a basis for $mathbb{Q}(sqrt{3-sqrt{7}})$ over $mathbb{Q}$
    $endgroup$
    – mouthetics
    Jan 8 at 15:40










  • $begingroup$
    @mouthetics good point. imma bit rusty on my algebra. (they were supposed to be rationals)
    $endgroup$
    – mathworker21
    Jan 8 at 15:41


















  • $begingroup$
    $sqrt{3+sqrt{7}} = a+bsqrt{3-sqrt{7}} implies 3+sqrt{7} = a^2+2absqrt{3-sqrt{7}} + b(3-sqrt{7})$ $ implies c+dsqrt{7} = esqrt{3-sqrt{7}} implies c^2+2cdsqrt{7}+7d^2 = e^2(3-sqrt{7}) implies 2cd = -e^2$. But $-e^2 = -4a^2b^2$ and $2cd = 2(3-a^2-3b)(b+1)$, so we must have $2a^2b^2 = (a^2+3b-3)(b+1)$. Now solve for $a^2$ and hopefully this will give a contradiction.
    $endgroup$
    – mathworker21
    Jan 8 at 15:39






  • 3




    $begingroup$
    @mathworker21 What is $a$ and $b$? ${1,sqrt{3-sqrt{7}}}$ is not a basis for $mathbb{Q}(sqrt{3-sqrt{7}})$ over $mathbb{Q}$
    $endgroup$
    – mouthetics
    Jan 8 at 15:40










  • $begingroup$
    @mouthetics good point. imma bit rusty on my algebra. (they were supposed to be rationals)
    $endgroup$
    – mathworker21
    Jan 8 at 15:41
















$begingroup$
$sqrt{3+sqrt{7}} = a+bsqrt{3-sqrt{7}} implies 3+sqrt{7} = a^2+2absqrt{3-sqrt{7}} + b(3-sqrt{7})$ $ implies c+dsqrt{7} = esqrt{3-sqrt{7}} implies c^2+2cdsqrt{7}+7d^2 = e^2(3-sqrt{7}) implies 2cd = -e^2$. But $-e^2 = -4a^2b^2$ and $2cd = 2(3-a^2-3b)(b+1)$, so we must have $2a^2b^2 = (a^2+3b-3)(b+1)$. Now solve for $a^2$ and hopefully this will give a contradiction.
$endgroup$
– mathworker21
Jan 8 at 15:39




$begingroup$
$sqrt{3+sqrt{7}} = a+bsqrt{3-sqrt{7}} implies 3+sqrt{7} = a^2+2absqrt{3-sqrt{7}} + b(3-sqrt{7})$ $ implies c+dsqrt{7} = esqrt{3-sqrt{7}} implies c^2+2cdsqrt{7}+7d^2 = e^2(3-sqrt{7}) implies 2cd = -e^2$. But $-e^2 = -4a^2b^2$ and $2cd = 2(3-a^2-3b)(b+1)$, so we must have $2a^2b^2 = (a^2+3b-3)(b+1)$. Now solve for $a^2$ and hopefully this will give a contradiction.
$endgroup$
– mathworker21
Jan 8 at 15:39




3




3




$begingroup$
@mathworker21 What is $a$ and $b$? ${1,sqrt{3-sqrt{7}}}$ is not a basis for $mathbb{Q}(sqrt{3-sqrt{7}})$ over $mathbb{Q}$
$endgroup$
– mouthetics
Jan 8 at 15:40




$begingroup$
@mathworker21 What is $a$ and $b$? ${1,sqrt{3-sqrt{7}}}$ is not a basis for $mathbb{Q}(sqrt{3-sqrt{7}})$ over $mathbb{Q}$
$endgroup$
– mouthetics
Jan 8 at 15:40












$begingroup$
@mouthetics good point. imma bit rusty on my algebra. (they were supposed to be rationals)
$endgroup$
– mathworker21
Jan 8 at 15:41




$begingroup$
@mouthetics good point. imma bit rusty on my algebra. (they were supposed to be rationals)
$endgroup$
– mathworker21
Jan 8 at 15:41










3 Answers
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Let $x_{pm}=sqrt{3 pm sqrt{7}}$.



It is easy to see that we have the following quadratic extensions:



$mathbb{Q} subset mathbb{Q}(sqrt{7}) subset mathbb{Q}(x_+)$,



$mathbb{Q} subset mathbb{Q}(sqrt{7}) subset mathbb{Q}(x_-)$.



Assume that $x_+ in K=mathbb{Q}(x_-)$. Then $sqrt{2} = x_+x_- in K$, thus $L=mathbb{Q}(sqrt{2},sqrt{7}) subset K$.



Since these fields have the same degree over $mathbb{Q}$, $K subset L$, ie $x_+=a+bsqrt{2}+csqrt{7}+dsqrt{14}$ for rationals $a,b,c,d$.



Taking squares, we get $3+sqrt{7}=(a^2+2b^2+7c^2+14d^2) + (2ab+14cd)sqrt{2} + (2da+2bc)sqrt{14} + (2ca+4bd)sqrt{7}$.



Thus $ab=-7cd$, $ad=-bc$, $2ca+4bd=1$, $a^2+2b^2+7c^2+14d^2=3$.



Assume $a=0$: then $d neq 0$ thus $c=0$ and $bd=1/4$, $2b^2+14d^2=3$. Usual quadratic theory yields then a contradiction.



Thus $b=-7cd/a$, and $ad=7c^2d/a$ thus $a^2=7c^2$ hence $a=0$. A contradiction, hence the result.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    In general, in this kind of problem, it is better not to mix the two operations + and $times$. Let me give an illustration here, using only $times$. Introduce the quadratic field $k=mathbf Q(sqrt 7)$. Adopting the notation $x_{pm}=sqrt {3 pm sqrt 7}$ suggested by @Mindlack, let us write $K_{pm}=k(x_{pm})$. These are two extensions of $k$ of degree at most $2$ :




    • if $K_{+}$ or $K_{-} =k$, i.e. $(3 pmsqrt 7)in {k^*}^2$, norming down to $mathbf Q$ shows that $N(3pmsqrt 7)=2$ is a square in $mathbf Q^*$: impossible


    • if both degrees are 2, $K_{pm}subset K_{mp}$ iff $K_{pm}= K_{mp}$, iff $k(x_{+})= k(x_{-})$, iff $2=(3+sqrt 7)(3-sqrt 7)in {k^*}^2$ (no specific calculation, this is rudimentary Kummer theory over $k$), iff $mathbf Q(sqrt 2)=mathbf Q(sqrt 7)$, iff $2.7$ is a square in $mathbf Q^*$(again by Kummer): impossible because $mathbf Z$ is a UFD ./.







    share|cite|improve this answer









    $endgroup$





















      2












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      HINT: Show that both have minimal polynomial $f:=X^4-6X^2+2$ over $Bbb{Q}$, and hence that $[Bbb{Q}(sqrt{3pmsqrt{7}}):Bbb{Q}]=4$, but that the splitting field of $f$ over $Bbb{Q}$ has degree greater than $4$.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        My goal is to justify that the splitting field is $mathbb{Q}(sqrt{3+sqrt{7}},sqrt{3-sqrt{7}})$, can I see that its degree is grater than 4 without having to prove what I was asking?
        $endgroup$
        – BBC3
        Jan 8 at 16:27










      • $begingroup$
        How to prove that the splitting field of $f$ over $mathbb{Q}$ has degree greater than $4$? Any other method?
        $endgroup$
        – mouthetics
        Jan 8 at 16:28










      • $begingroup$
        Do you mean $X^4 - 6X^2 + 2$?
        $endgroup$
        – Connor Harris
        Jan 8 at 17:08










      • $begingroup$
        @ConnorHarris Indeed I do, edited.
        $endgroup$
        – Servaes
        Jan 8 at 22:46






      • 1




        $begingroup$
        @mouthetics As the question is tagged Galois theory, determining the order of the Galois group was the approach I had in mind.
        $endgroup$
        – Servaes
        Jan 8 at 22:48











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      3 Answers
      3






      active

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      3 Answers
      3






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

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      7












      $begingroup$

      Let $x_{pm}=sqrt{3 pm sqrt{7}}$.



      It is easy to see that we have the following quadratic extensions:



      $mathbb{Q} subset mathbb{Q}(sqrt{7}) subset mathbb{Q}(x_+)$,



      $mathbb{Q} subset mathbb{Q}(sqrt{7}) subset mathbb{Q}(x_-)$.



      Assume that $x_+ in K=mathbb{Q}(x_-)$. Then $sqrt{2} = x_+x_- in K$, thus $L=mathbb{Q}(sqrt{2},sqrt{7}) subset K$.



      Since these fields have the same degree over $mathbb{Q}$, $K subset L$, ie $x_+=a+bsqrt{2}+csqrt{7}+dsqrt{14}$ for rationals $a,b,c,d$.



      Taking squares, we get $3+sqrt{7}=(a^2+2b^2+7c^2+14d^2) + (2ab+14cd)sqrt{2} + (2da+2bc)sqrt{14} + (2ca+4bd)sqrt{7}$.



      Thus $ab=-7cd$, $ad=-bc$, $2ca+4bd=1$, $a^2+2b^2+7c^2+14d^2=3$.



      Assume $a=0$: then $d neq 0$ thus $c=0$ and $bd=1/4$, $2b^2+14d^2=3$. Usual quadratic theory yields then a contradiction.



      Thus $b=-7cd/a$, and $ad=7c^2d/a$ thus $a^2=7c^2$ hence $a=0$. A contradiction, hence the result.






      share|cite|improve this answer









      $endgroup$


















        7












        $begingroup$

        Let $x_{pm}=sqrt{3 pm sqrt{7}}$.



        It is easy to see that we have the following quadratic extensions:



        $mathbb{Q} subset mathbb{Q}(sqrt{7}) subset mathbb{Q}(x_+)$,



        $mathbb{Q} subset mathbb{Q}(sqrt{7}) subset mathbb{Q}(x_-)$.



        Assume that $x_+ in K=mathbb{Q}(x_-)$. Then $sqrt{2} = x_+x_- in K$, thus $L=mathbb{Q}(sqrt{2},sqrt{7}) subset K$.



        Since these fields have the same degree over $mathbb{Q}$, $K subset L$, ie $x_+=a+bsqrt{2}+csqrt{7}+dsqrt{14}$ for rationals $a,b,c,d$.



        Taking squares, we get $3+sqrt{7}=(a^2+2b^2+7c^2+14d^2) + (2ab+14cd)sqrt{2} + (2da+2bc)sqrt{14} + (2ca+4bd)sqrt{7}$.



        Thus $ab=-7cd$, $ad=-bc$, $2ca+4bd=1$, $a^2+2b^2+7c^2+14d^2=3$.



        Assume $a=0$: then $d neq 0$ thus $c=0$ and $bd=1/4$, $2b^2+14d^2=3$. Usual quadratic theory yields then a contradiction.



        Thus $b=-7cd/a$, and $ad=7c^2d/a$ thus $a^2=7c^2$ hence $a=0$. A contradiction, hence the result.






        share|cite|improve this answer









        $endgroup$
















          7












          7








          7





          $begingroup$

          Let $x_{pm}=sqrt{3 pm sqrt{7}}$.



          It is easy to see that we have the following quadratic extensions:



          $mathbb{Q} subset mathbb{Q}(sqrt{7}) subset mathbb{Q}(x_+)$,



          $mathbb{Q} subset mathbb{Q}(sqrt{7}) subset mathbb{Q}(x_-)$.



          Assume that $x_+ in K=mathbb{Q}(x_-)$. Then $sqrt{2} = x_+x_- in K$, thus $L=mathbb{Q}(sqrt{2},sqrt{7}) subset K$.



          Since these fields have the same degree over $mathbb{Q}$, $K subset L$, ie $x_+=a+bsqrt{2}+csqrt{7}+dsqrt{14}$ for rationals $a,b,c,d$.



          Taking squares, we get $3+sqrt{7}=(a^2+2b^2+7c^2+14d^2) + (2ab+14cd)sqrt{2} + (2da+2bc)sqrt{14} + (2ca+4bd)sqrt{7}$.



          Thus $ab=-7cd$, $ad=-bc$, $2ca+4bd=1$, $a^2+2b^2+7c^2+14d^2=3$.



          Assume $a=0$: then $d neq 0$ thus $c=0$ and $bd=1/4$, $2b^2+14d^2=3$. Usual quadratic theory yields then a contradiction.



          Thus $b=-7cd/a$, and $ad=7c^2d/a$ thus $a^2=7c^2$ hence $a=0$. A contradiction, hence the result.






          share|cite|improve this answer









          $endgroup$



          Let $x_{pm}=sqrt{3 pm sqrt{7}}$.



          It is easy to see that we have the following quadratic extensions:



          $mathbb{Q} subset mathbb{Q}(sqrt{7}) subset mathbb{Q}(x_+)$,



          $mathbb{Q} subset mathbb{Q}(sqrt{7}) subset mathbb{Q}(x_-)$.



          Assume that $x_+ in K=mathbb{Q}(x_-)$. Then $sqrt{2} = x_+x_- in K$, thus $L=mathbb{Q}(sqrt{2},sqrt{7}) subset K$.



          Since these fields have the same degree over $mathbb{Q}$, $K subset L$, ie $x_+=a+bsqrt{2}+csqrt{7}+dsqrt{14}$ for rationals $a,b,c,d$.



          Taking squares, we get $3+sqrt{7}=(a^2+2b^2+7c^2+14d^2) + (2ab+14cd)sqrt{2} + (2da+2bc)sqrt{14} + (2ca+4bd)sqrt{7}$.



          Thus $ab=-7cd$, $ad=-bc$, $2ca+4bd=1$, $a^2+2b^2+7c^2+14d^2=3$.



          Assume $a=0$: then $d neq 0$ thus $c=0$ and $bd=1/4$, $2b^2+14d^2=3$. Usual quadratic theory yields then a contradiction.



          Thus $b=-7cd/a$, and $ad=7c^2d/a$ thus $a^2=7c^2$ hence $a=0$. A contradiction, hence the result.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 8 at 16:06









          MindlackMindlack

          2,78717




          2,78717























              2












              $begingroup$

              In general, in this kind of problem, it is better not to mix the two operations + and $times$. Let me give an illustration here, using only $times$. Introduce the quadratic field $k=mathbf Q(sqrt 7)$. Adopting the notation $x_{pm}=sqrt {3 pm sqrt 7}$ suggested by @Mindlack, let us write $K_{pm}=k(x_{pm})$. These are two extensions of $k$ of degree at most $2$ :




              • if $K_{+}$ or $K_{-} =k$, i.e. $(3 pmsqrt 7)in {k^*}^2$, norming down to $mathbf Q$ shows that $N(3pmsqrt 7)=2$ is a square in $mathbf Q^*$: impossible


              • if both degrees are 2, $K_{pm}subset K_{mp}$ iff $K_{pm}= K_{mp}$, iff $k(x_{+})= k(x_{-})$, iff $2=(3+sqrt 7)(3-sqrt 7)in {k^*}^2$ (no specific calculation, this is rudimentary Kummer theory over $k$), iff $mathbf Q(sqrt 2)=mathbf Q(sqrt 7)$, iff $2.7$ is a square in $mathbf Q^*$(again by Kummer): impossible because $mathbf Z$ is a UFD ./.







              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                In general, in this kind of problem, it is better not to mix the two operations + and $times$. Let me give an illustration here, using only $times$. Introduce the quadratic field $k=mathbf Q(sqrt 7)$. Adopting the notation $x_{pm}=sqrt {3 pm sqrt 7}$ suggested by @Mindlack, let us write $K_{pm}=k(x_{pm})$. These are two extensions of $k$ of degree at most $2$ :




                • if $K_{+}$ or $K_{-} =k$, i.e. $(3 pmsqrt 7)in {k^*}^2$, norming down to $mathbf Q$ shows that $N(3pmsqrt 7)=2$ is a square in $mathbf Q^*$: impossible


                • if both degrees are 2, $K_{pm}subset K_{mp}$ iff $K_{pm}= K_{mp}$, iff $k(x_{+})= k(x_{-})$, iff $2=(3+sqrt 7)(3-sqrt 7)in {k^*}^2$ (no specific calculation, this is rudimentary Kummer theory over $k$), iff $mathbf Q(sqrt 2)=mathbf Q(sqrt 7)$, iff $2.7$ is a square in $mathbf Q^*$(again by Kummer): impossible because $mathbf Z$ is a UFD ./.







                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  In general, in this kind of problem, it is better not to mix the two operations + and $times$. Let me give an illustration here, using only $times$. Introduce the quadratic field $k=mathbf Q(sqrt 7)$. Adopting the notation $x_{pm}=sqrt {3 pm sqrt 7}$ suggested by @Mindlack, let us write $K_{pm}=k(x_{pm})$. These are two extensions of $k$ of degree at most $2$ :




                  • if $K_{+}$ or $K_{-} =k$, i.e. $(3 pmsqrt 7)in {k^*}^2$, norming down to $mathbf Q$ shows that $N(3pmsqrt 7)=2$ is a square in $mathbf Q^*$: impossible


                  • if both degrees are 2, $K_{pm}subset K_{mp}$ iff $K_{pm}= K_{mp}$, iff $k(x_{+})= k(x_{-})$, iff $2=(3+sqrt 7)(3-sqrt 7)in {k^*}^2$ (no specific calculation, this is rudimentary Kummer theory over $k$), iff $mathbf Q(sqrt 2)=mathbf Q(sqrt 7)$, iff $2.7$ is a square in $mathbf Q^*$(again by Kummer): impossible because $mathbf Z$ is a UFD ./.







                  share|cite|improve this answer









                  $endgroup$



                  In general, in this kind of problem, it is better not to mix the two operations + and $times$. Let me give an illustration here, using only $times$. Introduce the quadratic field $k=mathbf Q(sqrt 7)$. Adopting the notation $x_{pm}=sqrt {3 pm sqrt 7}$ suggested by @Mindlack, let us write $K_{pm}=k(x_{pm})$. These are two extensions of $k$ of degree at most $2$ :




                  • if $K_{+}$ or $K_{-} =k$, i.e. $(3 pmsqrt 7)in {k^*}^2$, norming down to $mathbf Q$ shows that $N(3pmsqrt 7)=2$ is a square in $mathbf Q^*$: impossible


                  • if both degrees are 2, $K_{pm}subset K_{mp}$ iff $K_{pm}= K_{mp}$, iff $k(x_{+})= k(x_{-})$, iff $2=(3+sqrt 7)(3-sqrt 7)in {k^*}^2$ (no specific calculation, this is rudimentary Kummer theory over $k$), iff $mathbf Q(sqrt 2)=mathbf Q(sqrt 7)$, iff $2.7$ is a square in $mathbf Q^*$(again by Kummer): impossible because $mathbf Z$ is a UFD ./.








                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 8 at 20:32









                  nguyen quang donguyen quang do

                  8,4791723




                  8,4791723























                      2












                      $begingroup$

                      HINT: Show that both have minimal polynomial $f:=X^4-6X^2+2$ over $Bbb{Q}$, and hence that $[Bbb{Q}(sqrt{3pmsqrt{7}}):Bbb{Q}]=4$, but that the splitting field of $f$ over $Bbb{Q}$ has degree greater than $4$.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        My goal is to justify that the splitting field is $mathbb{Q}(sqrt{3+sqrt{7}},sqrt{3-sqrt{7}})$, can I see that its degree is grater than 4 without having to prove what I was asking?
                        $endgroup$
                        – BBC3
                        Jan 8 at 16:27










                      • $begingroup$
                        How to prove that the splitting field of $f$ over $mathbb{Q}$ has degree greater than $4$? Any other method?
                        $endgroup$
                        – mouthetics
                        Jan 8 at 16:28










                      • $begingroup$
                        Do you mean $X^4 - 6X^2 + 2$?
                        $endgroup$
                        – Connor Harris
                        Jan 8 at 17:08










                      • $begingroup$
                        @ConnorHarris Indeed I do, edited.
                        $endgroup$
                        – Servaes
                        Jan 8 at 22:46






                      • 1




                        $begingroup$
                        @mouthetics As the question is tagged Galois theory, determining the order of the Galois group was the approach I had in mind.
                        $endgroup$
                        – Servaes
                        Jan 8 at 22:48
















                      2












                      $begingroup$

                      HINT: Show that both have minimal polynomial $f:=X^4-6X^2+2$ over $Bbb{Q}$, and hence that $[Bbb{Q}(sqrt{3pmsqrt{7}}):Bbb{Q}]=4$, but that the splitting field of $f$ over $Bbb{Q}$ has degree greater than $4$.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        My goal is to justify that the splitting field is $mathbb{Q}(sqrt{3+sqrt{7}},sqrt{3-sqrt{7}})$, can I see that its degree is grater than 4 without having to prove what I was asking?
                        $endgroup$
                        – BBC3
                        Jan 8 at 16:27










                      • $begingroup$
                        How to prove that the splitting field of $f$ over $mathbb{Q}$ has degree greater than $4$? Any other method?
                        $endgroup$
                        – mouthetics
                        Jan 8 at 16:28










                      • $begingroup$
                        Do you mean $X^4 - 6X^2 + 2$?
                        $endgroup$
                        – Connor Harris
                        Jan 8 at 17:08










                      • $begingroup$
                        @ConnorHarris Indeed I do, edited.
                        $endgroup$
                        – Servaes
                        Jan 8 at 22:46






                      • 1




                        $begingroup$
                        @mouthetics As the question is tagged Galois theory, determining the order of the Galois group was the approach I had in mind.
                        $endgroup$
                        – Servaes
                        Jan 8 at 22:48














                      2












                      2








                      2





                      $begingroup$

                      HINT: Show that both have minimal polynomial $f:=X^4-6X^2+2$ over $Bbb{Q}$, and hence that $[Bbb{Q}(sqrt{3pmsqrt{7}}):Bbb{Q}]=4$, but that the splitting field of $f$ over $Bbb{Q}$ has degree greater than $4$.






                      share|cite|improve this answer











                      $endgroup$



                      HINT: Show that both have minimal polynomial $f:=X^4-6X^2+2$ over $Bbb{Q}$, and hence that $[Bbb{Q}(sqrt{3pmsqrt{7}}):Bbb{Q}]=4$, but that the splitting field of $f$ over $Bbb{Q}$ has degree greater than $4$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jan 8 at 22:46

























                      answered Jan 8 at 15:48









                      ServaesServaes

                      22.6k33793




                      22.6k33793












                      • $begingroup$
                        My goal is to justify that the splitting field is $mathbb{Q}(sqrt{3+sqrt{7}},sqrt{3-sqrt{7}})$, can I see that its degree is grater than 4 without having to prove what I was asking?
                        $endgroup$
                        – BBC3
                        Jan 8 at 16:27










                      • $begingroup$
                        How to prove that the splitting field of $f$ over $mathbb{Q}$ has degree greater than $4$? Any other method?
                        $endgroup$
                        – mouthetics
                        Jan 8 at 16:28










                      • $begingroup$
                        Do you mean $X^4 - 6X^2 + 2$?
                        $endgroup$
                        – Connor Harris
                        Jan 8 at 17:08










                      • $begingroup$
                        @ConnorHarris Indeed I do, edited.
                        $endgroup$
                        – Servaes
                        Jan 8 at 22:46






                      • 1




                        $begingroup$
                        @mouthetics As the question is tagged Galois theory, determining the order of the Galois group was the approach I had in mind.
                        $endgroup$
                        – Servaes
                        Jan 8 at 22:48


















                      • $begingroup$
                        My goal is to justify that the splitting field is $mathbb{Q}(sqrt{3+sqrt{7}},sqrt{3-sqrt{7}})$, can I see that its degree is grater than 4 without having to prove what I was asking?
                        $endgroup$
                        – BBC3
                        Jan 8 at 16:27










                      • $begingroup$
                        How to prove that the splitting field of $f$ over $mathbb{Q}$ has degree greater than $4$? Any other method?
                        $endgroup$
                        – mouthetics
                        Jan 8 at 16:28










                      • $begingroup$
                        Do you mean $X^4 - 6X^2 + 2$?
                        $endgroup$
                        – Connor Harris
                        Jan 8 at 17:08










                      • $begingroup$
                        @ConnorHarris Indeed I do, edited.
                        $endgroup$
                        – Servaes
                        Jan 8 at 22:46






                      • 1




                        $begingroup$
                        @mouthetics As the question is tagged Galois theory, determining the order of the Galois group was the approach I had in mind.
                        $endgroup$
                        – Servaes
                        Jan 8 at 22:48
















                      $begingroup$
                      My goal is to justify that the splitting field is $mathbb{Q}(sqrt{3+sqrt{7}},sqrt{3-sqrt{7}})$, can I see that its degree is grater than 4 without having to prove what I was asking?
                      $endgroup$
                      – BBC3
                      Jan 8 at 16:27




                      $begingroup$
                      My goal is to justify that the splitting field is $mathbb{Q}(sqrt{3+sqrt{7}},sqrt{3-sqrt{7}})$, can I see that its degree is grater than 4 without having to prove what I was asking?
                      $endgroup$
                      – BBC3
                      Jan 8 at 16:27












                      $begingroup$
                      How to prove that the splitting field of $f$ over $mathbb{Q}$ has degree greater than $4$? Any other method?
                      $endgroup$
                      – mouthetics
                      Jan 8 at 16:28




                      $begingroup$
                      How to prove that the splitting field of $f$ over $mathbb{Q}$ has degree greater than $4$? Any other method?
                      $endgroup$
                      – mouthetics
                      Jan 8 at 16:28












                      $begingroup$
                      Do you mean $X^4 - 6X^2 + 2$?
                      $endgroup$
                      – Connor Harris
                      Jan 8 at 17:08




                      $begingroup$
                      Do you mean $X^4 - 6X^2 + 2$?
                      $endgroup$
                      – Connor Harris
                      Jan 8 at 17:08












                      $begingroup$
                      @ConnorHarris Indeed I do, edited.
                      $endgroup$
                      – Servaes
                      Jan 8 at 22:46




                      $begingroup$
                      @ConnorHarris Indeed I do, edited.
                      $endgroup$
                      – Servaes
                      Jan 8 at 22:46




                      1




                      1




                      $begingroup$
                      @mouthetics As the question is tagged Galois theory, determining the order of the Galois group was the approach I had in mind.
                      $endgroup$
                      – Servaes
                      Jan 8 at 22:48




                      $begingroup$
                      @mouthetics As the question is tagged Galois theory, determining the order of the Galois group was the approach I had in mind.
                      $endgroup$
                      – Servaes
                      Jan 8 at 22:48


















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