Let $\mathcal{L}$ be the language of field theory and $\mathfrak{A}$ and $\mathfrak{B}$ be two $\mathcal{L}$ structures such that $\mathfrak{B}$ is a field. If $\mathfrak{A}\prec\mathfrak{B}$ and $p(x)$ is a polynomial with coefficients in the universe of $\mathfrak{A}$ such that $p(x)=0$ has a solution in $\mathfrak{B}$, then $p(x)$ has a solution in $\mathfrak{A}$?
I know that $\mathfrak{A}\prec\mathfrak{B}$ implies that $\mathfrak{A}\equiv\mathfrak{B}$. And I would like to justify:
$$\mathfrak{B}\models(\exists x)(p(x)=0)\Rightarrow\mathfrak{A}\models(\exists x)(p(x)=0)$$
However, in order to do so, I would have to express $(\exists x)(p(x)=0)$ with the language $\mathcal{L}$ but I have not succeeded since I am not considering that $\mathfrak{B}$ is algebraically closed. In fact, now I suppose the statement is false but I don't have any counterexample.
Somebody could help me?
$(\exists x)(p(x)=0)$ is a formula in the language of field theory with parameters from $\mathfrak{A}$ (the coefficients of $p$). So if it is true in $\mathfrak{B}$, it is true in $\mathfrak{A}$.
Note that because the formula has parameters, we need to use the fact that $\mathfrak{A}$ is an elementary substructure of $\mathfrak{B}$, not merely that they are elementarily equivalent.