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












4














Let $A$ be an elementary substructure of $R$ where $R$ is $<R,+,*,0,1>$ . 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
$phi=exists{x}(x_n*x^n+ldots+x_0=0)$ That is both true in $R$ 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
























  • 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?
    – Alessandro Codenotti
    6 hours ago








  • 1




    Use $langle Xrangle$ for $langle Xrangle$.
    – Shaun
    6 hours ago
















4














Let $A$ be an elementary substructure of $R$ where $R$ is $<R,+,*,0,1>$ . 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
$phi=exists{x}(x_n*x^n+ldots+x_0=0)$ That is both true in $R$ 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
























  • 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?
    – Alessandro Codenotti
    6 hours ago








  • 1




    Use $langle Xrangle$ for $langle Xrangle$.
    – Shaun
    6 hours ago














4












4








4







Let $A$ be an elementary substructure of $R$ where $R$ is $<R,+,*,0,1>$ . 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
$phi=exists{x}(x_n*x^n+ldots+x_0=0)$ That is both true in $R$ 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















Let $A$ be an elementary substructure of $R$ where $R$ is $<R,+,*,0,1>$ . 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
$phi=exists{x}(x_n*x^n+ldots+x_0=0)$ That is both true in $R$ 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 6 hours ago









Alessandro Codenotti

3,60311438




3,60311438










asked 7 hours ago









Gyt

578319




578319












  • 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?
    – Alessandro Codenotti
    6 hours ago








  • 1




    Use $langle Xrangle$ for $langle Xrangle$.
    – Shaun
    6 hours ago


















  • 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?
    – Alessandro Codenotti
    6 hours ago








  • 1




    Use $langle Xrangle$ for $langle Xrangle$.
    – Shaun
    6 hours ago
















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?
– Alessandro Codenotti
6 hours ago






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?
– Alessandro Codenotti
6 hours ago






1




1




Use $langle Xrangle$ for $langle Xrangle$.
– Shaun
6 hours ago




Use $langle Xrangle$ for $langle Xrangle$.
– Shaun
6 hours ago










2 Answers
2






active

oldest

votes


















5














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





























    6














    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





















    • How can I formulate the formula that the polynomial $p$ has exactly $k$ roots?
      – Gyt
      6 hours ago










    • There exist $p_1, cdots, p_k$ such that they are all unequal to each other and they all satisfy the condition.
      – Kenny Lau
      6 hours ago






    • 1




      @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 $k+1$".
      – Andreas Blass
      4 hours ago











    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









    5














    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


























      5














      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
























        5












        5








        5






        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












        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 3 hours ago









        Andreas Blass

        49.2k351106




        49.2k351106























            6














            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





















            • How can I formulate the formula that the polynomial $p$ has exactly $k$ roots?
              – Gyt
              6 hours ago










            • There exist $p_1, cdots, p_k$ such that they are all unequal to each other and they all satisfy the condition.
              – Kenny Lau
              6 hours ago






            • 1




              @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 $k+1$".
              – Andreas Blass
              4 hours ago
















            6














            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





















            • How can I formulate the formula that the polynomial $p$ has exactly $k$ roots?
              – Gyt
              6 hours ago










            • There exist $p_1, cdots, p_k$ such that they are all unequal to each other and they all satisfy the condition.
              – Kenny Lau
              6 hours ago






            • 1




              @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 $k+1$".
              – Andreas Blass
              4 hours ago














            6












            6








            6






            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












            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 6 hours ago









            Kenny Lau

            19.4k2158




            19.4k2158












            • How can I formulate the formula that the polynomial $p$ has exactly $k$ roots?
              – Gyt
              6 hours ago










            • There exist $p_1, cdots, p_k$ such that they are all unequal to each other and they all satisfy the condition.
              – Kenny Lau
              6 hours ago






            • 1




              @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 $k+1$".
              – Andreas Blass
              4 hours ago


















            • How can I formulate the formula that the polynomial $p$ has exactly $k$ roots?
              – Gyt
              6 hours ago










            • There exist $p_1, cdots, p_k$ such that they are all unequal to each other and they all satisfy the condition.
              – Kenny Lau
              6 hours ago






            • 1




              @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 $k+1$".
              – Andreas Blass
              4 hours ago
















            How can I formulate the formula that the polynomial $p$ has exactly $k$ roots?
            – Gyt
            6 hours ago




            How can I formulate the formula that the polynomial $p$ has exactly $k$ roots?
            – Gyt
            6 hours ago












            There exist $p_1, cdots, p_k$ such that they are all unequal to each other and they all satisfy the condition.
            – Kenny Lau
            6 hours ago




            There exist $p_1, cdots, p_k$ such that they are all unequal to each other and they all satisfy the condition.
            – Kenny Lau
            6 hours ago




            1




            1




            @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 $k+1$".
            – Andreas Blass
            4 hours ago




            @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 $k+1$".
            – Andreas Blass
            4 hours ago


















            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.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • 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.


            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

            Morgemoulin

            Image
            Morgemoulin — miejscowość i gmina — Państwo   Francja Region Grand Est Departament Moza Okręg Verdun Kod INSEE 55357 Powierzchnia 6,82 km² Populacja  (1990) • liczba ludności 75 • gęstość 11 os./km² Kod pocztowy 55400 Położenie na mapie Francji Morgemoulin 49°14′N   5°35′E / 49,233333   5,583333 Portal Francja Morgemoulin – miejscowość i gmina we Francji, w regionie Grand Est, w departamencie Moza. Według danych na rok 1990 gminę zamieszkiwało 75 osób, a gęstość zaludnienia wynosiła 11 osób/km² (wśród 2335 gmin Lotaryngii Morgemoulin plasuje się na 970. miejscu pod względem liczby ludności, natomiast pod względem powierzchni na miejscu 848.). Bibliografia | Francuski urząd statystyczny ( fr. ) . p   •   d   •   e Gminy okręgu Verdun Abaucourt-Hautecourt • Aincreville • Ambly-sur-Meuse • Amel-sur-l'Étang • Ancemont • Arrancy-sur-Crusne • Aubréville • Autré...
            Read more

            Scott Moir

            Image
            Scott Moir Tessa Virtue i Scott Moir w 2016 r. Reprezentacja   Kanada Data i miejsce urodzenia 2 września 1987 London Wzrost 180 cm Konkurencja Pary taneczne Partner sportowy Tessa Virtue Trener Marie-France Dubreuil, Patrice Lauzon, Romain Haguenauer Poprzednio: Marina Zujewa, Oleg Epstein, Igor Szpilband, Johnny Johns, Carol Moir, Paul MacIntosh, Suzanne Killing Klub Montreal International School of Skating Poprzednio: Arctic Edge FSC Ilderton Skating Club Lokalizacja treningowa Montreal Poprzednio: Canton Kitchener Rekordy życiowe ISU przed sezonem 2018/2019 Nota łączna 206,07 Igrzyska Olimpijskie 2018 Taniec krótki 83,67 Igrzyska Olimpijskie 2018 Taniec dowolny 122,40 Igrzyska Olimpijskie 2018 Dorobek medalowy Reprezentacja   Kanada Igrzyska olimpijskie Złoto Vancouver 2010 pary taneczne Złoto Pjongczang 2018 pary taneczne Złoto Pjongczang 2018 drużynowo Srebro S...
            Read more

            Souastre

            Image
            Souastre — miejscowość i gmina — Herb Państwo   Francja Region Hauts-de-France Departament Pas-de-Calais Okręg Arras Kod INSEE 62800 Powierzchnia 7,18 km² Populacja  (1990) • liczba ludności 352 • gęstość 49 os./km² Kod pocztowy 62111 Położenie na mapie Francji Souastre 50°09′N   2°34′E / 50,150000   2,566667 Portal Francja Souastre – miejscowość i gmina we Francji, w regionie Hauts-de-France, w departamencie Pas-de-Calais. Według danych na rok 1990 gminę zamieszkiwały 352 osoby, a gęstość zaludnienia wynosiła 49 osób/km² (wśród 1549 gmin regionu Nord-Pas-de-Calais Souastre plasuje się na 887. miejscu pod względem liczby ludności, natomiast pod względem powierzchni na miejscu 507.). Bibliografia | Francuski urząd statystyczny ( fr. ) . p   •   d   •   e Gminy okręgu Arras Ablain-Saint-Nazaire Ablainzevelle Acheville Achicourt Achiet-le-Grand Achiet-le-Pe...
            Read more