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












2














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







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  • 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
    4 hours ago








  • 1




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
















2














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
    4 hours ago








  • 1




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














2












2








2







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 4 hours ago









Alessandro Codenotti

3,60311438




3,60311438










asked 4 hours ago









Gyt

568319




568319












  • 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
    4 hours ago








  • 1




    Use $langle Xrangle$ for $langle Xrangle$.
    – Shaun
    4 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
    4 hours ago








  • 1




    Use $langle Xrangle$ for $langle Xrangle$.
    – Shaun
    4 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
4 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
4 hours ago






1




1




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




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










2 Answers
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2














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





























    4














    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
      4 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
      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
      1 hour ago











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    2 Answers
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    2 Answers
    2






    active

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    active

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    active

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    2














    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


























      2














      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
























        2












        2








        2






        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 1 hour ago









        Andreas Blass

        49.2k351106




        49.2k351106























            4














            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
              4 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
              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
              1 hour ago
















            4














            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
              4 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
              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
              1 hour ago














            4












            4








            4






            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 4 hours ago









            Kenny Lau

            19.4k2158




            19.4k2158












            • How can I formulate the formula that the polynomial $p$ has exactly $k$ roots?
              – Gyt
              4 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
              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
              1 hour ago


















            • How can I formulate the formula that the polynomial $p$ has exactly $k$ roots?
              – Gyt
              4 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
              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
              1 hour ago
















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




            How can I formulate the formula that the polynomial $p$ has exactly $k$ roots?
            – Gyt
            4 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
            4 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
            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
            1 hour 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
            1 hour ago


















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