The proof I am talking about goes like this: Given $k$ algebraically closed and $(f_1,..,f_k)=I\neq (1)$ an ideal in $A=k[x_1,..,x_n]$, let $m$ be a maximal ideal with $I\subseteq m$ and observe that $k$ embeds into the algebraic closure $k'$ of $A/m$ , and that $f_1,..,f_k$ have the common zero $(\overline x_1,..,\overline x_n)$ in the extension field. Now use model-completeness of the theory $\Phi$ of $k$, which is the same as the theory of $k'$, and conclude that $\Phi \models \exists y_1,..,y_n f_1(y_1,..,y_n)\land ..\land f_k(y_1,..,y_n)$, because this holds in $k'$, so it must hold in $k$ also.
The proof is described for example here on page 89 and here on page 8 and 9.
My problem is: $\exists y_1,..,y_n f_1(y_1,..,y_n)\land ..\land f_k(y_1,..,y_n)$ is not immediately a formula over the laguage of fields, since it contains the symbols/functions $f_1,..,f_k$ which are not part of that language. I can see how to restate this formula using only symbols of the language of fields in the case that $f_1,..,f_k$ have integral coefficients, but not in the general case. (e.g. for $k=\mathbb{C}$ there are uncountably many possiblities for coefficients)