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







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$endgroup$












  • $begingroup$
    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











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    2 Answers
    2






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    2 Answers
    2






    active

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    active

    oldest

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    active

    oldest

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    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


















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