Finding intermediate fields of the extension...












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I want to find the intermediate of the extension $mathbb{Q}=(sqrt{3+sqrt{7}},sqrt{3-sqrt{7}})/mathbb{Q}(sqrt{14})$. So far, I've been able to show that there are exactly two of them since the Galois group of $mathbb{Q}=(sqrt{3+sqrt{7}},sqrt{3-sqrt{7}})/mathbb{Q}$ is isomorphic to $D_4$ and I've been able to build the subfield lattice except for these two missing fields. I don't know how to proceed. Any help would be appreciated.






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  • $begingroup$
    How about $mathbb{Q}(sqrt{2},sqrt{7})$ first?
    $endgroup$
    – Mindlack
    Jan 9 at 11:52
















0












$begingroup$


I want to find the intermediate of the extension $mathbb{Q}=(sqrt{3+sqrt{7}},sqrt{3-sqrt{7}})/mathbb{Q}(sqrt{14})$. So far, I've been able to show that there are exactly two of them since the Galois group of $mathbb{Q}=(sqrt{3+sqrt{7}},sqrt{3-sqrt{7}})/mathbb{Q}$ is isomorphic to $D_4$ and I've been able to build the subfield lattice except for these two missing fields. I don't know how to proceed. Any help would be appreciated.






galois-theory extension-field







share|cite|improve this question









$endgroup$












  • $begingroup$
    How about $mathbb{Q}(sqrt{2},sqrt{7})$ first?
    $endgroup$
    – Mindlack
    Jan 9 at 11:52














0












0








0





$begingroup$


I want to find the intermediate of the extension $mathbb{Q}=(sqrt{3+sqrt{7}},sqrt{3-sqrt{7}})/mathbb{Q}(sqrt{14})$. So far, I've been able to show that there are exactly two of them since the Galois group of $mathbb{Q}=(sqrt{3+sqrt{7}},sqrt{3-sqrt{7}})/mathbb{Q}$ is isomorphic to $D_4$ and I've been able to build the subfield lattice except for these two missing fields. I don't know how to proceed. Any help would be appreciated.






galois-theory extension-field







share|cite|improve this question









$endgroup$




I want to find the intermediate of the extension $mathbb{Q}=(sqrt{3+sqrt{7}},sqrt{3-sqrt{7}})/mathbb{Q}(sqrt{14})$. So far, I've been able to show that there are exactly two of them since the Galois group of $mathbb{Q}=(sqrt{3+sqrt{7}},sqrt{3-sqrt{7}})/mathbb{Q}$ is isomorphic to $D_4$ and I've been able to build the subfield lattice except for these two missing fields. I don't know how to proceed. Any help would be appreciated.






galois-theory extension-field




galois-theory extension-field






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asked Jan 9 at 9:58









Ray BernRay Bern

20013




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  • $begingroup$
    How about $mathbb{Q}(sqrt{2},sqrt{7})$ first?
    $endgroup$
    – Mindlack
    Jan 9 at 11:52


















  • $begingroup$
    How about $mathbb{Q}(sqrt{2},sqrt{7})$ first?
    $endgroup$
    – Mindlack
    Jan 9 at 11:52
















$begingroup$
How about $mathbb{Q}(sqrt{2},sqrt{7})$ first?
$endgroup$
– Mindlack
Jan 9 at 11:52




$begingroup$
How about $mathbb{Q}(sqrt{2},sqrt{7})$ first?
$endgroup$
– Mindlack
Jan 9 at 11:52










1 Answer
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$begingroup$

Let $k=Q(sqrt 7)$. In the following, I'll refer to https://math.stackexchange.com/a/3066678/300700 (see also my comment on your other question), where it is shown that $k(sqrt {3pmsqrt 7})$ are two distinct quadratic extensions of $k$, hence $L=Q(sqrt {3+sqrt 7},sqrt {3-sqrt 7})$ is a biquadratic extension of $k$. The third quadratic subextension of $L/k$ is $k(sqrt 2)=Q(sqrt 7,sqrt 2)$ because $(3 +sqrt 7)(3-sqrt 7)=2$. It follows that $k(sqrt 2)/Q$ is a biquadratic extension, whose three quadratic subextensions are $k, Q(sqrt 2), Q(sqrt {14})$ (draw a Galois diagram).



We must now look at the nature of the extension $L/Q$. Since the conjugation $sigma: sqrt 7 to -sqrt 7$ of $k/Q$ permutes the two elements $3pmsqrt 7, L/Q$ is normal of degree 8, say with Galois group $G$. All groups $G$ of order 8 are known up to isomorphism: either $G$ is abelian, or $Gcong D_8$ (dihedral) or $H_8$ (quaternionic). Here $G$ is not abelian, as can be seen (a bit painfully) by lifting the conjugation $sigma$ of $Q(sqrt 7)/Q$ and the conjugation $tau$ of $Q(sqrt 2)/Q$, and computing their commutator (note that $Gal(L/k(sqrt 2))$ lies in the center of $G$. So $G$ is $cong D_8$ or $H_8$. A convenient way to distinguish between these two cases is: $G$ is quaternionic (resp. dihedral) iff its $3$ subgroups of order $4$ are (resp. one and only one is) cyclic. Here $G$ has $2$ subgroups of type $(2,2)$, which are $Gal(L/k)$ and $Gal(L/Q(sqrt 2))$, hence $Gcong D_8$ and its $3$-rd subgroup of order $4$, namely $Gal(L/Q(sqrt 14))$, must be cyclic.



Conclusion: the only strict subextension of $L/Q(sqrt 14)$ is $k(sqrt 2)=Q(sqrt 2, sqrt 7)$ ./.






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    $begingroup$

    Let $k=Q(sqrt 7)$. In the following, I'll refer to https://math.stackexchange.com/a/3066678/300700 (see also my comment on your other question), where it is shown that $k(sqrt {3pmsqrt 7})$ are two distinct quadratic extensions of $k$, hence $L=Q(sqrt {3+sqrt 7},sqrt {3-sqrt 7})$ is a biquadratic extension of $k$. The third quadratic subextension of $L/k$ is $k(sqrt 2)=Q(sqrt 7,sqrt 2)$ because $(3 +sqrt 7)(3-sqrt 7)=2$. It follows that $k(sqrt 2)/Q$ is a biquadratic extension, whose three quadratic subextensions are $k, Q(sqrt 2), Q(sqrt {14})$ (draw a Galois diagram).



    We must now look at the nature of the extension $L/Q$. Since the conjugation $sigma: sqrt 7 to -sqrt 7$ of $k/Q$ permutes the two elements $3pmsqrt 7, L/Q$ is normal of degree 8, say with Galois group $G$. All groups $G$ of order 8 are known up to isomorphism: either $G$ is abelian, or $Gcong D_8$ (dihedral) or $H_8$ (quaternionic). Here $G$ is not abelian, as can be seen (a bit painfully) by lifting the conjugation $sigma$ of $Q(sqrt 7)/Q$ and the conjugation $tau$ of $Q(sqrt 2)/Q$, and computing their commutator (note that $Gal(L/k(sqrt 2))$ lies in the center of $G$. So $G$ is $cong D_8$ or $H_8$. A convenient way to distinguish between these two cases is: $G$ is quaternionic (resp. dihedral) iff its $3$ subgroups of order $4$ are (resp. one and only one is) cyclic. Here $G$ has $2$ subgroups of type $(2,2)$, which are $Gal(L/k)$ and $Gal(L/Q(sqrt 2))$, hence $Gcong D_8$ and its $3$-rd subgroup of order $4$, namely $Gal(L/Q(sqrt 14))$, must be cyclic.



    Conclusion: the only strict subextension of $L/Q(sqrt 14)$ is $k(sqrt 2)=Q(sqrt 2, sqrt 7)$ ./.






    share|cite|improve this answer









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      0












      $begingroup$

      Let $k=Q(sqrt 7)$. In the following, I'll refer to https://math.stackexchange.com/a/3066678/300700 (see also my comment on your other question), where it is shown that $k(sqrt {3pmsqrt 7})$ are two distinct quadratic extensions of $k$, hence $L=Q(sqrt {3+sqrt 7},sqrt {3-sqrt 7})$ is a biquadratic extension of $k$. The third quadratic subextension of $L/k$ is $k(sqrt 2)=Q(sqrt 7,sqrt 2)$ because $(3 +sqrt 7)(3-sqrt 7)=2$. It follows that $k(sqrt 2)/Q$ is a biquadratic extension, whose three quadratic subextensions are $k, Q(sqrt 2), Q(sqrt {14})$ (draw a Galois diagram).



      We must now look at the nature of the extension $L/Q$. Since the conjugation $sigma: sqrt 7 to -sqrt 7$ of $k/Q$ permutes the two elements $3pmsqrt 7, L/Q$ is normal of degree 8, say with Galois group $G$. All groups $G$ of order 8 are known up to isomorphism: either $G$ is abelian, or $Gcong D_8$ (dihedral) or $H_8$ (quaternionic). Here $G$ is not abelian, as can be seen (a bit painfully) by lifting the conjugation $sigma$ of $Q(sqrt 7)/Q$ and the conjugation $tau$ of $Q(sqrt 2)/Q$, and computing their commutator (note that $Gal(L/k(sqrt 2))$ lies in the center of $G$. So $G$ is $cong D_8$ or $H_8$. A convenient way to distinguish between these two cases is: $G$ is quaternionic (resp. dihedral) iff its $3$ subgroups of order $4$ are (resp. one and only one is) cyclic. Here $G$ has $2$ subgroups of type $(2,2)$, which are $Gal(L/k)$ and $Gal(L/Q(sqrt 2))$, hence $Gcong D_8$ and its $3$-rd subgroup of order $4$, namely $Gal(L/Q(sqrt 14))$, must be cyclic.



      Conclusion: the only strict subextension of $L/Q(sqrt 14)$ is $k(sqrt 2)=Q(sqrt 2, sqrt 7)$ ./.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Let $k=Q(sqrt 7)$. In the following, I'll refer to https://math.stackexchange.com/a/3066678/300700 (see also my comment on your other question), where it is shown that $k(sqrt {3pmsqrt 7})$ are two distinct quadratic extensions of $k$, hence $L=Q(sqrt {3+sqrt 7},sqrt {3-sqrt 7})$ is a biquadratic extension of $k$. The third quadratic subextension of $L/k$ is $k(sqrt 2)=Q(sqrt 7,sqrt 2)$ because $(3 +sqrt 7)(3-sqrt 7)=2$. It follows that $k(sqrt 2)/Q$ is a biquadratic extension, whose three quadratic subextensions are $k, Q(sqrt 2), Q(sqrt {14})$ (draw a Galois diagram).



        We must now look at the nature of the extension $L/Q$. Since the conjugation $sigma: sqrt 7 to -sqrt 7$ of $k/Q$ permutes the two elements $3pmsqrt 7, L/Q$ is normal of degree 8, say with Galois group $G$. All groups $G$ of order 8 are known up to isomorphism: either $G$ is abelian, or $Gcong D_8$ (dihedral) or $H_8$ (quaternionic). Here $G$ is not abelian, as can be seen (a bit painfully) by lifting the conjugation $sigma$ of $Q(sqrt 7)/Q$ and the conjugation $tau$ of $Q(sqrt 2)/Q$, and computing their commutator (note that $Gal(L/k(sqrt 2))$ lies in the center of $G$. So $G$ is $cong D_8$ or $H_8$. A convenient way to distinguish between these two cases is: $G$ is quaternionic (resp. dihedral) iff its $3$ subgroups of order $4$ are (resp. one and only one is) cyclic. Here $G$ has $2$ subgroups of type $(2,2)$, which are $Gal(L/k)$ and $Gal(L/Q(sqrt 2))$, hence $Gcong D_8$ and its $3$-rd subgroup of order $4$, namely $Gal(L/Q(sqrt 14))$, must be cyclic.



        Conclusion: the only strict subextension of $L/Q(sqrt 14)$ is $k(sqrt 2)=Q(sqrt 2, sqrt 7)$ ./.






        share|cite|improve this answer









        $endgroup$



        Let $k=Q(sqrt 7)$. In the following, I'll refer to https://math.stackexchange.com/a/3066678/300700 (see also my comment on your other question), where it is shown that $k(sqrt {3pmsqrt 7})$ are two distinct quadratic extensions of $k$, hence $L=Q(sqrt {3+sqrt 7},sqrt {3-sqrt 7})$ is a biquadratic extension of $k$. The third quadratic subextension of $L/k$ is $k(sqrt 2)=Q(sqrt 7,sqrt 2)$ because $(3 +sqrt 7)(3-sqrt 7)=2$. It follows that $k(sqrt 2)/Q$ is a biquadratic extension, whose three quadratic subextensions are $k, Q(sqrt 2), Q(sqrt {14})$ (draw a Galois diagram).



        We must now look at the nature of the extension $L/Q$. Since the conjugation $sigma: sqrt 7 to -sqrt 7$ of $k/Q$ permutes the two elements $3pmsqrt 7, L/Q$ is normal of degree 8, say with Galois group $G$. All groups $G$ of order 8 are known up to isomorphism: either $G$ is abelian, or $Gcong D_8$ (dihedral) or $H_8$ (quaternionic). Here $G$ is not abelian, as can be seen (a bit painfully) by lifting the conjugation $sigma$ of $Q(sqrt 7)/Q$ and the conjugation $tau$ of $Q(sqrt 2)/Q$, and computing their commutator (note that $Gal(L/k(sqrt 2))$ lies in the center of $G$. So $G$ is $cong D_8$ or $H_8$. A convenient way to distinguish between these two cases is: $G$ is quaternionic (resp. dihedral) iff its $3$ subgroups of order $4$ are (resp. one and only one is) cyclic. Here $G$ has $2$ subgroups of type $(2,2)$, which are $Gal(L/k)$ and $Gal(L/Q(sqrt 2))$, hence $Gcong D_8$ and its $3$-rd subgroup of order $4$, namely $Gal(L/Q(sqrt 14))$, must be cyclic.



        Conclusion: the only strict subextension of $L/Q(sqrt 14)$ is $k(sqrt 2)=Q(sqrt 2, sqrt 7)$ ./.







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        answered Jan 10 at 5:29









        nguyen quang donguyen quang do

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