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







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












$begingroup$


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
1






active

oldest

votes


















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











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054334%2fis-a-galois-extension-over-mathbbq-always-finite%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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












      $begingroup$

      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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054334%2fis-a-galois-extension-over-mathbbq-always-finite%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Wiesbaden

            Image
            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
            Read more

            Marschland

            Image
            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
            Read more

            Dieringhausen

            Image
            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
            Read more