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












1















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
    Nov 19 '18 at 5:11










  • @JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
    – Idonknow
    Nov 19 '18 at 5:20










  • ${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
    – Gerry Myerson
    Nov 19 '18 at 5:31






  • 1




    @GerryMyerson Yes. Edited.
    – Idonknow
    Nov 19 '18 at 5:33










  • "The group is not Galois." I think you mean, "the extension is not Galois."
    – Gerry Myerson
    Nov 19 '18 at 5:34
















1















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
    Nov 19 '18 at 5:11










  • @JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
    – Idonknow
    Nov 19 '18 at 5:20










  • ${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
    – Gerry Myerson
    Nov 19 '18 at 5:31






  • 1




    @GerryMyerson Yes. Edited.
    – Idonknow
    Nov 19 '18 at 5:33










  • "The group is not Galois." I think you mean, "the extension is not Galois."
    – Gerry Myerson
    Nov 19 '18 at 5:34














1












1








1


1






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 Nov 19 '18 at 5:36

























asked Nov 19 '18 at 4:24









Idonknow

2,284749112




2,284749112












  • 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
    Nov 19 '18 at 5:11










  • @JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
    – Idonknow
    Nov 19 '18 at 5:20










  • ${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
    – Gerry Myerson
    Nov 19 '18 at 5:31






  • 1




    @GerryMyerson Yes. Edited.
    – Idonknow
    Nov 19 '18 at 5:33










  • "The group is not Galois." I think you mean, "the extension is not Galois."
    – Gerry Myerson
    Nov 19 '18 at 5:34


















  • 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
    Nov 19 '18 at 5:11










  • @JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
    – Idonknow
    Nov 19 '18 at 5:20










  • ${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
    – Gerry Myerson
    Nov 19 '18 at 5:31






  • 1




    @GerryMyerson Yes. Edited.
    – Idonknow
    Nov 19 '18 at 5:33










  • "The group is not Galois." I think you mean, "the extension is not Galois."
    – Gerry Myerson
    Nov 19 '18 at 5:34
















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
Nov 19 '18 at 5:11




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
Nov 19 '18 at 5:11












@JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
– Idonknow
Nov 19 '18 at 5:20




@JoelPereira only $3$ ways to permute $sqrt{13},$ $-sqrt{13}$ and $sqrt[3]{7}$ as other roots are not in the extension?
– Idonknow
Nov 19 '18 at 5:20












${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
– Gerry Myerson
Nov 19 '18 at 5:31




${bf Q}(sqrt{13}),root3of7)$ doesn't make sense. Do you mean ${bf Q}(sqrt{13},root3of7)$?
– Gerry Myerson
Nov 19 '18 at 5:31




1




1




@GerryMyerson Yes. Edited.
– Idonknow
Nov 19 '18 at 5:33




@GerryMyerson Yes. Edited.
– Idonknow
Nov 19 '18 at 5:33












"The group is not Galois." I think you mean, "the extension is not Galois."
– Gerry Myerson
Nov 19 '18 at 5:34




"The group is not Galois." I think you mean, "the extension is not Galois."
– Gerry Myerson
Nov 19 '18 at 5:34










1 Answer
1






active

oldest

votes


















3














Let $sigma in Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$, then $sigma$ is uniquely determine by $sigma(sqrt{13})$ and $sigma(sqrt[3]{7})$.



Since $sigma(alpha) in mathbb{Q}(sqrt{13},sqrt[3]{7}) ;forall alpha in mathbb{Q}(sqrt{13},sqrt[3]{7})$, we have that $sigma(sqrt[3]{7})$ is both a root of $x^3-7$ and an element of $mathbb{Q}(sqrt{13},sqrt[3]{7})$.



This implies that $sigma(sqrt[3]{7}) = sqrt[3]{7}$, since the other roots of $x^3-7$ do not lies in $mathbb{Q}(sqrt{13},sqrt[3]{7})$.



Clearly $sigma(sqrt{13}) in lbrace pm sqrt{13} rbrace$.



In conclusion, we see that $sigma$ is completely determined by its behaviour on $sqrt{13}$, and we can easily determine the automorphism group $Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$






share|cite|improve this answer























  • Just a quick question. How many element does the automorphism group have?
    – Idonknow
    Nov 23 '18 at 13:48












  • Exactlu two: one that maps $sqrt{13}$ to itself, and one which maps $sqrt{13}$ to $-sqrt{13}$. Hope it's clear.
    – Bilo
    Nov 23 '18 at 15:06













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1 Answer
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1 Answer
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active

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3














Let $sigma in Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$, then $sigma$ is uniquely determine by $sigma(sqrt{13})$ and $sigma(sqrt[3]{7})$.



Since $sigma(alpha) in mathbb{Q}(sqrt{13},sqrt[3]{7}) ;forall alpha in mathbb{Q}(sqrt{13},sqrt[3]{7})$, we have that $sigma(sqrt[3]{7})$ is both a root of $x^3-7$ and an element of $mathbb{Q}(sqrt{13},sqrt[3]{7})$.



This implies that $sigma(sqrt[3]{7}) = sqrt[3]{7}$, since the other roots of $x^3-7$ do not lies in $mathbb{Q}(sqrt{13},sqrt[3]{7})$.



Clearly $sigma(sqrt{13}) in lbrace pm sqrt{13} rbrace$.



In conclusion, we see that $sigma$ is completely determined by its behaviour on $sqrt{13}$, and we can easily determine the automorphism group $Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$






share|cite|improve this answer























  • Just a quick question. How many element does the automorphism group have?
    – Idonknow
    Nov 23 '18 at 13:48












  • Exactlu two: one that maps $sqrt{13}$ to itself, and one which maps $sqrt{13}$ to $-sqrt{13}$. Hope it's clear.
    – Bilo
    Nov 23 '18 at 15:06


















3














Let $sigma in Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$, then $sigma$ is uniquely determine by $sigma(sqrt{13})$ and $sigma(sqrt[3]{7})$.



Since $sigma(alpha) in mathbb{Q}(sqrt{13},sqrt[3]{7}) ;forall alpha in mathbb{Q}(sqrt{13},sqrt[3]{7})$, we have that $sigma(sqrt[3]{7})$ is both a root of $x^3-7$ and an element of $mathbb{Q}(sqrt{13},sqrt[3]{7})$.



This implies that $sigma(sqrt[3]{7}) = sqrt[3]{7}$, since the other roots of $x^3-7$ do not lies in $mathbb{Q}(sqrt{13},sqrt[3]{7})$.



Clearly $sigma(sqrt{13}) in lbrace pm sqrt{13} rbrace$.



In conclusion, we see that $sigma$ is completely determined by its behaviour on $sqrt{13}$, and we can easily determine the automorphism group $Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$






share|cite|improve this answer























  • Just a quick question. How many element does the automorphism group have?
    – Idonknow
    Nov 23 '18 at 13:48












  • Exactlu two: one that maps $sqrt{13}$ to itself, and one which maps $sqrt{13}$ to $-sqrt{13}$. Hope it's clear.
    – Bilo
    Nov 23 '18 at 15:06
















3












3








3






Let $sigma in Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$, then $sigma$ is uniquely determine by $sigma(sqrt{13})$ and $sigma(sqrt[3]{7})$.



Since $sigma(alpha) in mathbb{Q}(sqrt{13},sqrt[3]{7}) ;forall alpha in mathbb{Q}(sqrt{13},sqrt[3]{7})$, we have that $sigma(sqrt[3]{7})$ is both a root of $x^3-7$ and an element of $mathbb{Q}(sqrt{13},sqrt[3]{7})$.



This implies that $sigma(sqrt[3]{7}) = sqrt[3]{7}$, since the other roots of $x^3-7$ do not lies in $mathbb{Q}(sqrt{13},sqrt[3]{7})$.



Clearly $sigma(sqrt{13}) in lbrace pm sqrt{13} rbrace$.



In conclusion, we see that $sigma$ is completely determined by its behaviour on $sqrt{13}$, and we can easily determine the automorphism group $Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$






share|cite|improve this answer














Let $sigma in Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$, then $sigma$ is uniquely determine by $sigma(sqrt{13})$ and $sigma(sqrt[3]{7})$.



Since $sigma(alpha) in mathbb{Q}(sqrt{13},sqrt[3]{7}) ;forall alpha in mathbb{Q}(sqrt{13},sqrt[3]{7})$, we have that $sigma(sqrt[3]{7})$ is both a root of $x^3-7$ and an element of $mathbb{Q}(sqrt{13},sqrt[3]{7})$.



This implies that $sigma(sqrt[3]{7}) = sqrt[3]{7}$, since the other roots of $x^3-7$ do not lies in $mathbb{Q}(sqrt{13},sqrt[3]{7})$.



Clearly $sigma(sqrt{13}) in lbrace pm sqrt{13} rbrace$.



In conclusion, we see that $sigma$ is completely determined by its behaviour on $sqrt{13}$, and we can easily determine the automorphism group $Aut(mathbb{Q}(sqrt{13},sqrt[3]{7})/mathbb{Q})$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 23 '18 at 13:45









Joel Pereira

65119




65119










answered Nov 23 '18 at 13:26









Bilo

1089




1089












  • Just a quick question. How many element does the automorphism group have?
    – Idonknow
    Nov 23 '18 at 13:48












  • Exactlu two: one that maps $sqrt{13}$ to itself, and one which maps $sqrt{13}$ to $-sqrt{13}$. Hope it's clear.
    – Bilo
    Nov 23 '18 at 15:06




















  • Just a quick question. How many element does the automorphism group have?
    – Idonknow
    Nov 23 '18 at 13:48












  • Exactlu two: one that maps $sqrt{13}$ to itself, and one which maps $sqrt{13}$ to $-sqrt{13}$. Hope it's clear.
    – Bilo
    Nov 23 '18 at 15:06


















Just a quick question. How many element does the automorphism group have?
– Idonknow
Nov 23 '18 at 13:48






Just a quick question. How many element does the automorphism group have?
– Idonknow
Nov 23 '18 at 13:48














Exactlu two: one that maps $sqrt{13}$ to itself, and one which maps $sqrt{13}$ to $-sqrt{13}$. Hope it's clear.
– Bilo
Nov 23 '18 at 15:06






Exactlu two: one that maps $sqrt{13}$ to itself, and one which maps $sqrt{13}$ to $-sqrt{13}$. Hope it's clear.
– Bilo
Nov 23 '18 at 15:06




















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