Prove that all algebraic numbers are included in any elementary substructure of $mathbb R$












4












$begingroup$


Let $A$ be an elementary substructure of $mathbb R$ where $mathbb R$ is $langle mathbb R,+,cdot,0,1rangle$ . Show that $A$ contains any algebraic number.



What I tried to do was use the fact that if $a$ is an algebraic number than there exist some polynomial $p(x)$ such that $p(a)=0$. There exist a formula
$phi=exists{x}(x_ncdot x^n+ldots+x_0=0)$ That is both true in $mathbb R$ and in $A$ and thus there exists a number in $bin{A}$ that solves the polynomial. The problem is that I don't know if this number is $a$ or how to change the formula so that the number will be $a$.






logic algebraic-number-theory model-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    To answer your doubt $b$ is not necessarily $a$, for example if $a=sqrt{2}$ and $p=x^2-2$ then $b$ could be $-sqrt{2}$ or, in general, any Galois conjugate of $a$. However the formula $exists x_1exists x_2(x_1neq x_2land p(x_1)=p(x_2)=0)$ is also true in this case. Can you generalize this?
    $endgroup$
    – Alessandro Codenotti
    Dec 29 '18 at 12:54








  • 1




    $begingroup$
    Use $langle Xrangle$ for $langle Xrangle$.
    $endgroup$
    – Shaun
    Dec 29 '18 at 12:54
















4












$begingroup$


Let $A$ be an elementary substructure of $mathbb R$ where $mathbb R$ is $langle mathbb R,+,cdot,0,1rangle$ . Show that $A$ contains any algebraic number.



What I tried to do was use the fact that if $a$ is an algebraic number than there exist some polynomial $p(x)$ such that $p(a)=0$. There exist a formula
$phi=exists{x}(x_ncdot x^n+ldots+x_0=0)$ That is both true in $mathbb R$ and in $A$ and thus there exists a number in $bin{A}$ that solves the polynomial. The problem is that I don't know if this number is $a$ or how to change the formula so that the number will be $a$.






logic algebraic-number-theory model-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    To answer your doubt $b$ is not necessarily $a$, for example if $a=sqrt{2}$ and $p=x^2-2$ then $b$ could be $-sqrt{2}$ or, in general, any Galois conjugate of $a$. However the formula $exists x_1exists x_2(x_1neq x_2land p(x_1)=p(x_2)=0)$ is also true in this case. Can you generalize this?
    $endgroup$
    – Alessandro Codenotti
    Dec 29 '18 at 12:54








  • 1




    $begingroup$
    Use $langle Xrangle$ for $langle Xrangle$.
    $endgroup$
    – Shaun
    Dec 29 '18 at 12:54














4












4








4





$begingroup$


Let $A$ be an elementary substructure of $mathbb R$ where $mathbb R$ is $langle mathbb R,+,cdot,0,1rangle$ . Show that $A$ contains any algebraic number.



What I tried to do was use the fact that if $a$ is an algebraic number than there exist some polynomial $p(x)$ such that $p(a)=0$. There exist a formula
$phi=exists{x}(x_ncdot x^n+ldots+x_0=0)$ That is both true in $mathbb R$ and in $A$ and thus there exists a number in $bin{A}$ that solves the polynomial. The problem is that I don't know if this number is $a$ or how to change the formula so that the number will be $a$.






logic algebraic-number-theory model-theory







share|cite|improve this question











$endgroup$




Let $A$ be an elementary substructure of $mathbb R$ where $mathbb R$ is $langle mathbb R,+,cdot,0,1rangle$ . Show that $A$ contains any algebraic number.



What I tried to do was use the fact that if $a$ is an algebraic number than there exist some polynomial $p(x)$ such that $p(a)=0$. There exist a formula
$phi=exists{x}(x_ncdot x^n+ldots+x_0=0)$ That is both true in $mathbb R$ and in $A$ and thus there exists a number in $bin{A}$ that solves the polynomial. The problem is that I don't know if this number is $a$ or how to change the formula so that the number will be $a$.






logic algebraic-number-theory model-theory




logic algebraic-number-theory model-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 0:06









Asaf Karagila♦

305k33435765




305k33435765










asked Dec 29 '18 at 12:40









GytGyt

547419




547419












  • $begingroup$
    To answer your doubt $b$ is not necessarily $a$, for example if $a=sqrt{2}$ and $p=x^2-2$ then $b$ could be $-sqrt{2}$ or, in general, any Galois conjugate of $a$. However the formula $exists x_1exists x_2(x_1neq x_2land p(x_1)=p(x_2)=0)$ is also true in this case. Can you generalize this?
    $endgroup$
    – Alessandro Codenotti
    Dec 29 '18 at 12:54








  • 1




    $begingroup$
    Use $langle Xrangle$ for $langle Xrangle$.
    $endgroup$
    – Shaun
    Dec 29 '18 at 12:54


















  • $begingroup$
    To answer your doubt $b$ is not necessarily $a$, for example if $a=sqrt{2}$ and $p=x^2-2$ then $b$ could be $-sqrt{2}$ or, in general, any Galois conjugate of $a$. However the formula $exists x_1exists x_2(x_1neq x_2land p(x_1)=p(x_2)=0)$ is also true in this case. Can you generalize this?
    $endgroup$
    – Alessandro Codenotti
    Dec 29 '18 at 12:54








  • 1




    $begingroup$
    Use $langle Xrangle$ for $langle Xrangle$.
    $endgroup$
    – Shaun
    Dec 29 '18 at 12:54
















$begingroup$
To answer your doubt $b$ is not necessarily $a$, for example if $a=sqrt{2}$ and $p=x^2-2$ then $b$ could be $-sqrt{2}$ or, in general, any Galois conjugate of $a$. However the formula $exists x_1exists x_2(x_1neq x_2land p(x_1)=p(x_2)=0)$ is also true in this case. Can you generalize this?
$endgroup$
– Alessandro Codenotti
Dec 29 '18 at 12:54






$begingroup$
To answer your doubt $b$ is not necessarily $a$, for example if $a=sqrt{2}$ and $p=x^2-2$ then $b$ could be $-sqrt{2}$ or, in general, any Galois conjugate of $a$. However the formula $exists x_1exists x_2(x_1neq x_2land p(x_1)=p(x_2)=0)$ is also true in this case. Can you generalize this?
$endgroup$
– Alessandro Codenotti
Dec 29 '18 at 12:54






1




1




$begingroup$
Use $langle Xrangle$ for $langle Xrangle$.
$endgroup$
– Shaun
Dec 29 '18 at 12:54




$begingroup$
Use $langle Xrangle$ for $langle Xrangle$.
$endgroup$
– Shaun
Dec 29 '18 at 12:54










2 Answers
2






active

oldest

votes


















9












$begingroup$

An alternative to counting the roots is to find two rational numbers $q<r$ so close to your algebraic number $a$ that $a$ is the only root of $p$ between $q$ and $r$. Then use the fact that an elementary submodel of the real field must also have a solution of $p$ between $q$ and $r$ and that this solution has to be $a$.






share|cite|improve this answer









$endgroup$





















    6












    $begingroup$

    Let $a in Bbb R$ be algebraic with $k$ real conjugates, and use the sentence saying that the minimal polynomial of $a$ (after clearing denominators to make sure coefficients in $Bbb Z$) has $k$ roots.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      How can I formulate the formula that the polynomial $p$ has exactly $k$ roots?
      $endgroup$
      – Gyt
      Dec 29 '18 at 13:05






    • 1




      $begingroup$
      There exist $p_1, cdots, p_k$ such that they are all unequal to each other and they all satisfy the condition.
      $endgroup$
      – Kenny Lau
      Dec 29 '18 at 13:06






    • 3




      $begingroup$
      @Gyt The suggested formulation says that $p$ has at least $k$ roots, which is adequate for the problem at hand. If you really want to say "exactly $k$ roots" then combine this "at least $k$" with the negation of "at least $k+1$".
      $endgroup$
      – Andreas Blass
      Dec 29 '18 at 15:51











    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%2f3055807%2fprove-that-all-algebraic-numbers-are-included-in-any-elementary-substructure-of%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    9












    $begingroup$

    An alternative to counting the roots is to find two rational numbers $q<r$ so close to your algebraic number $a$ that $a$ is the only root of $p$ between $q$ and $r$. Then use the fact that an elementary submodel of the real field must also have a solution of $p$ between $q$ and $r$ and that this solution has to be $a$.






    share|cite|improve this answer









    $endgroup$


















      9












      $begingroup$

      An alternative to counting the roots is to find two rational numbers $q<r$ so close to your algebraic number $a$ that $a$ is the only root of $p$ between $q$ and $r$. Then use the fact that an elementary submodel of the real field must also have a solution of $p$ between $q$ and $r$ and that this solution has to be $a$.






      share|cite|improve this answer









      $endgroup$
















        9












        9








        9





        $begingroup$

        An alternative to counting the roots is to find two rational numbers $q<r$ so close to your algebraic number $a$ that $a$ is the only root of $p$ between $q$ and $r$. Then use the fact that an elementary submodel of the real field must also have a solution of $p$ between $q$ and $r$ and that this solution has to be $a$.






        share|cite|improve this answer









        $endgroup$



        An alternative to counting the roots is to find two rational numbers $q<r$ so close to your algebraic number $a$ that $a$ is the only root of $p$ between $q$ and $r$. Then use the fact that an elementary submodel of the real field must also have a solution of $p$ between $q$ and $r$ and that this solution has to be $a$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 29 '18 at 15:54









        Andreas BlassAndreas Blass

        50k451108




        50k451108























            6












            $begingroup$

            Let $a in Bbb R$ be algebraic with $k$ real conjugates, and use the sentence saying that the minimal polynomial of $a$ (after clearing denominators to make sure coefficients in $Bbb Z$) has $k$ roots.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              How can I formulate the formula that the polynomial $p$ has exactly $k$ roots?
              $endgroup$
              – Gyt
              Dec 29 '18 at 13:05






            • 1




              $begingroup$
              There exist $p_1, cdots, p_k$ such that they are all unequal to each other and they all satisfy the condition.
              $endgroup$
              – Kenny Lau
              Dec 29 '18 at 13:06






            • 3




              $begingroup$
              @Gyt The suggested formulation says that $p$ has at least $k$ roots, which is adequate for the problem at hand. If you really want to say "exactly $k$ roots" then combine this "at least $k$" with the negation of "at least $k+1$".
              $endgroup$
              – Andreas Blass
              Dec 29 '18 at 15:51
















            6












            $begingroup$

            Let $a in Bbb R$ be algebraic with $k$ real conjugates, and use the sentence saying that the minimal polynomial of $a$ (after clearing denominators to make sure coefficients in $Bbb Z$) has $k$ roots.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              How can I formulate the formula that the polynomial $p$ has exactly $k$ roots?
              $endgroup$
              – Gyt
              Dec 29 '18 at 13:05






            • 1




              $begingroup$
              There exist $p_1, cdots, p_k$ such that they are all unequal to each other and they all satisfy the condition.
              $endgroup$
              – Kenny Lau
              Dec 29 '18 at 13:06






            • 3




              $begingroup$
              @Gyt The suggested formulation says that $p$ has at least $k$ roots, which is adequate for the problem at hand. If you really want to say "exactly $k$ roots" then combine this "at least $k$" with the negation of "at least $k+1$".
              $endgroup$
              – Andreas Blass
              Dec 29 '18 at 15:51














            6












            6








            6





            $begingroup$

            Let $a in Bbb R$ be algebraic with $k$ real conjugates, and use the sentence saying that the minimal polynomial of $a$ (after clearing denominators to make sure coefficients in $Bbb Z$) has $k$ roots.






            share|cite|improve this answer









            $endgroup$



            Let $a in Bbb R$ be algebraic with $k$ real conjugates, and use the sentence saying that the minimal polynomial of $a$ (after clearing denominators to make sure coefficients in $Bbb Z$) has $k$ roots.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 29 '18 at 12:55









            Kenny LauKenny Lau

            19.9k2159




            19.9k2159












            • $begingroup$
              How can I formulate the formula that the polynomial $p$ has exactly $k$ roots?
              $endgroup$
              – Gyt
              Dec 29 '18 at 13:05






            • 1




              $begingroup$
              There exist $p_1, cdots, p_k$ such that they are all unequal to each other and they all satisfy the condition.
              $endgroup$
              – Kenny Lau
              Dec 29 '18 at 13:06






            • 3




              $begingroup$
              @Gyt The suggested formulation says that $p$ has at least $k$ roots, which is adequate for the problem at hand. If you really want to say "exactly $k$ roots" then combine this "at least $k$" with the negation of "at least $k+1$".
              $endgroup$
              – Andreas Blass
              Dec 29 '18 at 15:51


















            • $begingroup$
              How can I formulate the formula that the polynomial $p$ has exactly $k$ roots?
              $endgroup$
              – Gyt
              Dec 29 '18 at 13:05






            • 1




              $begingroup$
              There exist $p_1, cdots, p_k$ such that they are all unequal to each other and they all satisfy the condition.
              $endgroup$
              – Kenny Lau
              Dec 29 '18 at 13:06






            • 3




              $begingroup$
              @Gyt The suggested formulation says that $p$ has at least $k$ roots, which is adequate for the problem at hand. If you really want to say "exactly $k$ roots" then combine this "at least $k$" with the negation of "at least $k+1$".
              $endgroup$
              – Andreas Blass
              Dec 29 '18 at 15:51
















            $begingroup$
            How can I formulate the formula that the polynomial $p$ has exactly $k$ roots?
            $endgroup$
            – Gyt
            Dec 29 '18 at 13:05




            $begingroup$
            How can I formulate the formula that the polynomial $p$ has exactly $k$ roots?
            $endgroup$
            – Gyt
            Dec 29 '18 at 13:05




            1




            1




            $begingroup$
            There exist $p_1, cdots, p_k$ such that they are all unequal to each other and they all satisfy the condition.
            $endgroup$
            – Kenny Lau
            Dec 29 '18 at 13:06




            $begingroup$
            There exist $p_1, cdots, p_k$ such that they are all unequal to each other and they all satisfy the condition.
            $endgroup$
            – Kenny Lau
            Dec 29 '18 at 13:06




            3




            3




            $begingroup$
            @Gyt The suggested formulation says that $p$ has at least $k$ roots, which is adequate for the problem at hand. If you really want to say "exactly $k$ roots" then combine this "at least $k$" with the negation of "at least $k+1$".
            $endgroup$
            – Andreas Blass
            Dec 29 '18 at 15:51




            $begingroup$
            @Gyt The suggested formulation says that $p$ has at least $k$ roots, which is adequate for the problem at hand. If you really want to say "exactly $k$ roots" then combine this "at least $k$" with the negation of "at least $k+1$".
            $endgroup$
            – Andreas Blass
            Dec 29 '18 at 15:51


















            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%2f3055807%2fprove-that-all-algebraic-numbers-are-included-in-any-elementary-substructure-of%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

            Bressuire

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
            Read more

            Cabo Verde

            Image
              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
            Read more

            Gyllenstierna

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
            Read more