Determine the automorphism group $Aut(mathbb{Q}(sqrt{13}, sqrt[3]{7})/mathbb{Q})$











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Question: Determine the automorphism group $$Aut(mathbb{Q}(sqrt{13}, sqrt[3]{7})/mathbb{Q}).$$




My attempt:
Since the polynomial $(x^2-13)(x^3-7)$ has roots
$$sqrt{13}, -sqrt{13}, sqrt[3]{7}, sqrt[3]{7}omega, sqrt[3]{7}omega^2$$
where $omega$ is the cube root of unity.
Since the extension does not contain all roots, so the extension is not Galois.



However, I do not know how to determine the automorphism group.



Any hint is appreciated.






abstract-algebra field-theory galois-theory







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  • An automorphism of the field must send a root to another root of the same multiplicity. How may ways can the roots be permuted?
    – Joel Pereira
    10 hours ago










  • @JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
    – Idonknow
    10 hours ago










  • ${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
    – Gerry Myerson
    10 hours ago






  • 1




    @GerryMyerson Yes. Edited.
    – Idonknow
    10 hours ago










  • "The group is not Galois." I think you mean, "the extension is not Galois."
    – Gerry Myerson
    10 hours ago















up vote
1
down vote

favorite













Question: Determine the automorphism group $$Aut(mathbb{Q}(sqrt{13}, sqrt[3]{7})/mathbb{Q}).$$




My attempt:
Since the polynomial $(x^2-13)(x^3-7)$ has roots
$$sqrt{13}, -sqrt{13}, sqrt[3]{7}, sqrt[3]{7}omega, sqrt[3]{7}omega^2$$
where $omega$ is the cube root of unity.
Since the extension does not contain all roots, so the extension is not Galois.



However, I do not know how to determine the automorphism group.



Any hint is appreciated.






abstract-algebra field-theory galois-theory







share|cite|improve this question
























  • An automorphism of the field must send a root to another root of the same multiplicity. How may ways can the roots be permuted?
    – Joel Pereira
    10 hours ago










  • @JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
    – Idonknow
    10 hours ago










  • ${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
    – Gerry Myerson
    10 hours ago






  • 1




    @GerryMyerson Yes. Edited.
    – Idonknow
    10 hours ago










  • "The group is not Galois." I think you mean, "the extension is not Galois."
    – Gerry Myerson
    10 hours ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite












Question: Determine the automorphism group $$Aut(mathbb{Q}(sqrt{13}, sqrt[3]{7})/mathbb{Q}).$$




My attempt:
Since the polynomial $(x^2-13)(x^3-7)$ has roots
$$sqrt{13}, -sqrt{13}, sqrt[3]{7}, sqrt[3]{7}omega, sqrt[3]{7}omega^2$$
where $omega$ is the cube root of unity.
Since the extension does not contain all roots, so the extension is not Galois.



However, I do not know how to determine the automorphism group.



Any hint is appreciated.






abstract-algebra field-theory galois-theory







share|cite|improve this question
















Question: Determine the automorphism group $$Aut(mathbb{Q}(sqrt{13}, sqrt[3]{7})/mathbb{Q}).$$




My attempt:
Since the polynomial $(x^2-13)(x^3-7)$ has roots
$$sqrt{13}, -sqrt{13}, sqrt[3]{7}, sqrt[3]{7}omega, sqrt[3]{7}omega^2$$
where $omega$ is the cube root of unity.
Since the extension does not contain all roots, so the extension is not Galois.



However, I do not know how to determine the automorphism group.



Any hint is appreciated.






abstract-algebra field-theory galois-theory




abstract-algebra field-theory galois-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 10 hours ago

























asked 11 hours ago









Idonknow

2,205746110




2,205746110












  • An automorphism of the field must send a root to another root of the same multiplicity. How may ways can the roots be permuted?
    – Joel Pereira
    10 hours ago










  • @JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
    – Idonknow
    10 hours ago










  • ${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
    – Gerry Myerson
    10 hours ago






  • 1




    @GerryMyerson Yes. Edited.
    – Idonknow
    10 hours ago










  • "The group is not Galois." I think you mean, "the extension is not Galois."
    – Gerry Myerson
    10 hours ago


















  • An automorphism of the field must send a root to another root of the same multiplicity. How may ways can the roots be permuted?
    – Joel Pereira
    10 hours ago










  • @JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
    – Idonknow
    10 hours ago










  • ${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
    – Gerry Myerson
    10 hours ago






  • 1




    @GerryMyerson Yes. Edited.
    – Idonknow
    10 hours ago










  • "The group is not Galois." I think you mean, "the extension is not Galois."
    – Gerry Myerson
    10 hours ago
















An automorphism of the field must send a root to another root of the same multiplicity. How may ways can the roots be permuted?
– Joel Pereira
10 hours ago




An automorphism of the field must send a root to another root of the same multiplicity. How may ways can the roots be permuted?
– Joel Pereira
10 hours ago












@JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
– Idonknow
10 hours ago




@JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
– Idonknow
10 hours ago












${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
– Gerry Myerson
10 hours ago




${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
– Gerry Myerson
10 hours ago




1




1




@GerryMyerson Yes. Edited.
– Idonknow
10 hours ago




@GerryMyerson Yes. Edited.
– Idonknow
10 hours ago












"The group is not Galois." I think you mean, "the extension is not Galois."
– Gerry Myerson
10 hours ago




"The group is not Galois." I think you mean, "the extension is not Galois."
– Gerry Myerson
10 hours ago















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