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Let $A$ be an elementary substructure of $\mathbb R$ where $\mathbb R$ is $\langle \mathbb R,+,\cdot,0,1\rangle$ . 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 $\phi=\exists{x}(x_n\cdot x^n+\ldots+x_0=0)$ That is both true in $\mathbb R$ and in $A$ and thus there exists a number in $b\in{A}$ 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 $a$.

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  • $\begingroup$ 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 $\exists x_1\exists x_2(x_1\neq x_2\land p(x_1)=p(x_2)=0)$ is also true in this case. Can you generalize this? $\endgroup$ – Alessandro Codenotti Dec 29 '18 at 12:54
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    $\begingroup$ Use $\langle X\rangle$ for $\langle X\rangle$. $\endgroup$ – Shaun Dec 29 '18 at 12:54
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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 $a$.

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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.

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