Is a Galois extension over $mathbb{Q}$ always finite?












0












$begingroup$


Let $K$ be a Galois extensions over $mathbb{Q}$.



Is $K$ always a splitting field of some $Pin mathbb{Q}[X]$? in which case K would be a finite extension.



I don't know where to start. I tried to use the primitive element theorem but I can't prove there are finitely many intermediate fields.



Thanks for your help, hints.






abstract-algebra







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$endgroup$








  • 5




    $begingroup$
    There are infinite Galois extensions, for instance the field of all algebraic numbers.
    $endgroup$
    – Lord Shark the Unknown
    Dec 27 '18 at 20:38






  • 1




    $begingroup$
    No, the whole algebraic closure is an example of an infinite Galois extension
    $endgroup$
    – Wojowu
    Dec 27 '18 at 20:38










  • $begingroup$
    Thank you, does it also imply that there are Galois extensions that are not splitting fields of some polynimials?
    $endgroup$
    – PerelMan
    Dec 27 '18 at 20:40








  • 4




    $begingroup$
    @PerelMan An extension is said to be normal (which is part of the definition of Galois) if it is the splitting field of a family of polynomials. If that family is not finite, the extension may not be expressible as splitting field of a single polynomial
    $endgroup$
    – Hagen von Eitzen
    Dec 27 '18 at 21:55










  • $begingroup$
    OP it would help if you told us your definition of Galois extension
    $endgroup$
    – D_S
    Dec 29 '18 at 16:23
















0












$begingroup$


Let $K$ be a Galois extensions over $mathbb{Q}$.



Is $K$ always a splitting field of some $Pin mathbb{Q}[X]$? in which case K would be a finite extension.



I don't know where to start. I tried to use the primitive element theorem but I can't prove there are finitely many intermediate fields.



Thanks for your help, hints.






abstract-algebra







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    There are infinite Galois extensions, for instance the field of all algebraic numbers.
    $endgroup$
    – Lord Shark the Unknown
    Dec 27 '18 at 20:38






  • 1




    $begingroup$
    No, the whole algebraic closure is an example of an infinite Galois extension
    $endgroup$
    – Wojowu
    Dec 27 '18 at 20:38










  • $begingroup$
    Thank you, does it also imply that there are Galois extensions that are not splitting fields of some polynimials?
    $endgroup$
    – PerelMan
    Dec 27 '18 at 20:40








  • 4




    $begingroup$
    @PerelMan An extension is said to be normal (which is part of the definition of Galois) if it is the splitting field of a family of polynomials. If that family is not finite, the extension may not be expressible as splitting field of a single polynomial
    $endgroup$
    – Hagen von Eitzen
    Dec 27 '18 at 21:55










  • $begingroup$
    OP it would help if you told us your definition of Galois extension
    $endgroup$
    – D_S
    Dec 29 '18 at 16:23














0












0








0





$begingroup$


Let $K$ be a Galois extensions over $mathbb{Q}$.



Is $K$ always a splitting field of some $Pin mathbb{Q}[X]$? in which case K would be a finite extension.



I don't know where to start. I tried to use the primitive element theorem but I can't prove there are finitely many intermediate fields.



Thanks for your help, hints.






abstract-algebra







share|cite|improve this question











$endgroup$




Let $K$ be a Galois extensions over $mathbb{Q}$.



Is $K$ always a splitting field of some $Pin mathbb{Q}[X]$? in which case K would be a finite extension.



I don't know where to start. I tried to use the primitive element theorem but I can't prove there are finitely many intermediate fields.



Thanks for your help, hints.






abstract-algebra




abstract-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 20:32









Pierre-Guy Plamondon

8,88511739




8,88511739










asked Dec 27 '18 at 20:30









PerelManPerelMan

669313




669313








  • 5




    $begingroup$
    There are infinite Galois extensions, for instance the field of all algebraic numbers.
    $endgroup$
    – Lord Shark the Unknown
    Dec 27 '18 at 20:38






  • 1




    $begingroup$
    No, the whole algebraic closure is an example of an infinite Galois extension
    $endgroup$
    – Wojowu
    Dec 27 '18 at 20:38










  • $begingroup$
    Thank you, does it also imply that there are Galois extensions that are not splitting fields of some polynimials?
    $endgroup$
    – PerelMan
    Dec 27 '18 at 20:40








  • 4




    $begingroup$
    @PerelMan An extension is said to be normal (which is part of the definition of Galois) if it is the splitting field of a family of polynomials. If that family is not finite, the extension may not be expressible as splitting field of a single polynomial
    $endgroup$
    – Hagen von Eitzen
    Dec 27 '18 at 21:55










  • $begingroup$
    OP it would help if you told us your definition of Galois extension
    $endgroup$
    – D_S
    Dec 29 '18 at 16:23














  • 5




    $begingroup$
    There are infinite Galois extensions, for instance the field of all algebraic numbers.
    $endgroup$
    – Lord Shark the Unknown
    Dec 27 '18 at 20:38






  • 1




    $begingroup$
    No, the whole algebraic closure is an example of an infinite Galois extension
    $endgroup$
    – Wojowu
    Dec 27 '18 at 20:38










  • $begingroup$
    Thank you, does it also imply that there are Galois extensions that are not splitting fields of some polynimials?
    $endgroup$
    – PerelMan
    Dec 27 '18 at 20:40








  • 4




    $begingroup$
    @PerelMan An extension is said to be normal (which is part of the definition of Galois) if it is the splitting field of a family of polynomials. If that family is not finite, the extension may not be expressible as splitting field of a single polynomial
    $endgroup$
    – Hagen von Eitzen
    Dec 27 '18 at 21:55










  • $begingroup$
    OP it would help if you told us your definition of Galois extension
    $endgroup$
    – D_S
    Dec 29 '18 at 16:23








5




5




$begingroup$
There are infinite Galois extensions, for instance the field of all algebraic numbers.
$endgroup$
– Lord Shark the Unknown
Dec 27 '18 at 20:38




$begingroup$
There are infinite Galois extensions, for instance the field of all algebraic numbers.
$endgroup$
– Lord Shark the Unknown
Dec 27 '18 at 20:38




1




1




$begingroup$
No, the whole algebraic closure is an example of an infinite Galois extension
$endgroup$
– Wojowu
Dec 27 '18 at 20:38




$begingroup$
No, the whole algebraic closure is an example of an infinite Galois extension
$endgroup$
– Wojowu
Dec 27 '18 at 20:38












$begingroup$
Thank you, does it also imply that there are Galois extensions that are not splitting fields of some polynimials?
$endgroup$
– PerelMan
Dec 27 '18 at 20:40






$begingroup$
Thank you, does it also imply that there are Galois extensions that are not splitting fields of some polynimials?
$endgroup$
– PerelMan
Dec 27 '18 at 20:40






4




4




$begingroup$
@PerelMan An extension is said to be normal (which is part of the definition of Galois) if it is the splitting field of a family of polynomials. If that family is not finite, the extension may not be expressible as splitting field of a single polynomial
$endgroup$
– Hagen von Eitzen
Dec 27 '18 at 21:55




$begingroup$
@PerelMan An extension is said to be normal (which is part of the definition of Galois) if it is the splitting field of a family of polynomials. If that family is not finite, the extension may not be expressible as splitting field of a single polynomial
$endgroup$
– Hagen von Eitzen
Dec 27 '18 at 21:55












$begingroup$
OP it would help if you told us your definition of Galois extension
$endgroup$
– D_S
Dec 29 '18 at 16:23




$begingroup$
OP it would help if you told us your definition of Galois extension
$endgroup$
– D_S
Dec 29 '18 at 16:23










1 Answer
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$begingroup$

There are infinite algebraic Galois extensions of $mathbb{Q}$, simply take a splitting field $F$ of a infinite family of polinomials like $x^2-p$ where $pinmathbb{Z}$ is a prime. Remember that, since $mathbb{Q}$ has characteristic zero every extension is separable, and a splitting field of a family of polynomials is normal, so is Galois.



Now, if $K$ is a splitting field of a (only one) polynomial $p(x)inmathbb{Q}[x]$, then $K/mathbb{Q}$ is finite. In fact, using basic Galois Theory $[K:mathbb{Q}]leq n!$, where $n=deg p(x)$.



Edit: In the last question. Using Galois theory, there are finitely many intermediate fields as there are finite subgroups.






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    $begingroup$

    There are infinite algebraic Galois extensions of $mathbb{Q}$, simply take a splitting field $F$ of a infinite family of polinomials like $x^2-p$ where $pinmathbb{Z}$ is a prime. Remember that, since $mathbb{Q}$ has characteristic zero every extension is separable, and a splitting field of a family of polynomials is normal, so is Galois.



    Now, if $K$ is a splitting field of a (only one) polynomial $p(x)inmathbb{Q}[x]$, then $K/mathbb{Q}$ is finite. In fact, using basic Galois Theory $[K:mathbb{Q}]leq n!$, where $n=deg p(x)$.



    Edit: In the last question. Using Galois theory, there are finitely many intermediate fields as there are finite subgroups.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      There are infinite algebraic Galois extensions of $mathbb{Q}$, simply take a splitting field $F$ of a infinite family of polinomials like $x^2-p$ where $pinmathbb{Z}$ is a prime. Remember that, since $mathbb{Q}$ has characteristic zero every extension is separable, and a splitting field of a family of polynomials is normal, so is Galois.



      Now, if $K$ is a splitting field of a (only one) polynomial $p(x)inmathbb{Q}[x]$, then $K/mathbb{Q}$ is finite. In fact, using basic Galois Theory $[K:mathbb{Q}]leq n!$, where $n=deg p(x)$.



      Edit: In the last question. Using Galois theory, there are finitely many intermediate fields as there are finite subgroups.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        There are infinite algebraic Galois extensions of $mathbb{Q}$, simply take a splitting field $F$ of a infinite family of polinomials like $x^2-p$ where $pinmathbb{Z}$ is a prime. Remember that, since $mathbb{Q}$ has characteristic zero every extension is separable, and a splitting field of a family of polynomials is normal, so is Galois.



        Now, if $K$ is a splitting field of a (only one) polynomial $p(x)inmathbb{Q}[x]$, then $K/mathbb{Q}$ is finite. In fact, using basic Galois Theory $[K:mathbb{Q}]leq n!$, where $n=deg p(x)$.



        Edit: In the last question. Using Galois theory, there are finitely many intermediate fields as there are finite subgroups.






        share|cite|improve this answer











        $endgroup$



        There are infinite algebraic Galois extensions of $mathbb{Q}$, simply take a splitting field $F$ of a infinite family of polinomials like $x^2-p$ where $pinmathbb{Z}$ is a prime. Remember that, since $mathbb{Q}$ has characteristic zero every extension is separable, and a splitting field of a family of polynomials is normal, so is Galois.



        Now, if $K$ is a splitting field of a (only one) polynomial $p(x)inmathbb{Q}[x]$, then $K/mathbb{Q}$ is finite. In fact, using basic Galois Theory $[K:mathbb{Q}]leq n!$, where $n=deg p(x)$.



        Edit: In the last question. Using Galois theory, there are finitely many intermediate fields as there are finite subgroups.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 29 '18 at 16:20

























        answered Dec 29 '18 at 15:53









        José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda

        802110




        802110






























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