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

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 $$a$$.

• 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? – Alessandro Codenotti Dec 29 '18 at 12:54
• Use $\langle X\rangle$ for $\langle X\rangle$. – Shaun Dec 29 '18 at 12:54

An alternative to counting the roots is to find two rational numbers $$q so close to your algebraic number $$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$$.
Let $$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.
• How can I formulate the formula that the polynomial $p$ has exactly $k$ roots? – Gyt Dec 29 '18 at 13:05
• There exist $p_1, \cdots, p_k$ such that they are all unequal to each other and they all satisfy the condition. – Kenny Lau Dec 29 '18 at 13:06
• @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 Dec 29 '18 at 15:51