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

Question

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 $a$.

Answer 1

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 $a$.

Answer 2

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.