What are some examples of non isomorphic countable algebraically closed fields of characteristic zero? Or they don´t exist? I was reading some model theory, and saw stated that $ACF_p$ (the first order theory of algebraically closed fields of characteristic $p$) is $\kappa$-categorical for all $\kappa > \aleph_0$. Is this result is false for $\kappa=\omega$? 
In case that $ACF_p$ isn't $\omega$-categorical, this would mean that there exist at least two countable algebraically closed fields of characteristic $0$ that are not isomorphic. What would be some example of such fields? do we know all of them? (up to isomorphism, of course).
In case that $ACF_p$ is indeed $\omega$-categorical, how would a proof of that look like? The one provided for the case $\kappa>\aleph_0$ relies heavily on uncountability.
Thanks in advance to anyone who answers!
 A: Algebraically closed fields are determined up to isomorphism by their characteristic $c$ and their transcendence degree (over their prime subfield, i.e. over $\Bbb Q$ if $c=0$ and over $\Bbb Z_c$ if $c\ne0$). Hence the countable algebraically closed fields of characteristic $0$ are, up to isomorphism:
$$\overline{\Bbb Q(X_k,k\in F)}\quad F\text{ at most countable,}$$
where
$\Bbb Q(X_k,k\in F)$ denotes the field of rational functions with rational coefficients and set of indeterminates $\{X_k\mid k\in F\}.$
A: No, they are not all isomorphic. In particular, we may consider the field $k$ of algebraic numbers, which has the property that all elements of $k$ are the root of a polynomial with integer coefficients.
However, we may also consider the first order theory $T$ of an algebraically closed field with a transcendental number $c$. Since this theory is consistent (in particular, $\mathbb{C}$ is a model with $c = e$, for instance), it has a countable model $M$. $M$ is a countable, algebraically closed field with a transcendental number $c_M$. We could pick a particular $M$ as the smallest algebraically closed field containing $e$, which will be countable.
Since $M$ has a transcendental element while all elements of $k$ are algebraic, they are not isomorphic.
A: Suppose that $F$ is a countable algebraically closed field of characteristic zero. The field $F(x)$ of rational functions on $F$ is then also countable, and its algebraic closure $G=\overline{F(x)}$ is again countable.
If the transcendence degree of $F$ over $Q$ is finite and equal to $n$, then that of the new field $G$ is also finite and equal to $n+1$.
This allows you to construct countably many different countable algebraically closed fields of characteristic zero.
