How can I prove that every uncountable algebraically closed field of characteristic $0$ is saturated?

Question

The definition of saturated that I am working with is that for every subset of parameters of cardinality strictly inferior to that of the field, every type on that set of parameters is realized in that field. I guess I should use somehow quantifier elimination or the fact that $ACF_0$ is complete, but I still have no good idea. Moreover, I don't know how I should use the fact that the field is uncountable, and I know that the hypothesis on characteristic $0$ is actually superfluous.

Answer 1

Let $F$ be our field. Because of Quantifier Elimination, we can assume that the type only contains equalities of the kind $p(x)=0 $ and inequalities of the kind $p(x) \neq 0$, where $p(x)$ is a polynomial such that its coefficients are closed terms taken from the constants $0, 1$ together with the set of parameters $B$ and functions $\cdot$ and $+$. Since these polynomials are $\vert B \vert$ many, if the type only contains inequalities, because of $\vert B \vert <\vert F \vert $ there are elements that realize every inequality. If, on the other hand, we have an equality of the kind $p(x) = 0$ in our type, and let $a_1,...,a_n$ be the roots of the polynomial. If the type is not realized that means that the intersection of the sets defined by the formulas is $\emptyset$. But then for each $a_i$ there is a formula $\phi_i(x)$ which it does not realize. Then, if you take the conjunction $ p(x)=0 \land \phi_1(x)\land ...\land \phi_n(x)$ , it is not realized in the structure.