Ultraproduct of the algebraic closure of finite fields Ok, I have done a little research on the next problem, and there are simple proofs of it using model theory, logic and the Łoś's theorem. But here, the idea is to prove it using only Field Theory, Galois Theory and the basic theory of transcendental extensions. This is the problem:
Let $W$ be the set of prime numbers and let $\mathfrak{U}$ be a non principal ultrafilter over $W$. For every prime $p$, let $\overline{\mathbb{F}_p}$ denote an algebraic closure of $\mathbb{F}_p$. In the following product
$$
\prod_{p\in W} \overline{\mathbb{F}_p},
$$
define the equivalence relation $(a_p)\sim (b_p)$ if and only if $\{p\in W: a_p=b_p\}$ belongs to $\mathfrak{U}$.
Let
$$
A=\prod_{p \in W} \overline{\mathbb{F}_p} ~/~ \sim
$$
denote the set of equivalence classes, such that $[a_p]$ is the class of $(a_p)$.


*

*Show that $A$ is not countable. 

*Show that $A$ is isomorphic to $\mathbb{C}$.


Proving that the addition and product ($[a_p]+[b_p]=[a_p+b_p]$, $[a_p][b_p]=[a_pb_p]$) are well defined on $A$ and that $\text{char}(A)=0$ is tedious but quiet straightforward. On the other hand proving (1) and (2) has been a rather difficult task. For (1), I honestly have no idea how to proceed; and for (2) I've tried using the fact that $\mathbb{C}$ has an infinite number of automorphism, but I haven't had any luck. Also I think that in order to prove (2) I actually need to show that $|A|=2^{\aleph_0}$.
Any suggestions on how to proceed will be really appreciated. Thanks in advance for any help you provide.
 A: In this link, you can find a proof that an ultraproduct of countable structures with respect to non-principal ultrafilters have size $2^{\aleph_0}$. 
Cardinality of ultraproduct
To show that $A$ is algebraically closed, note that by Los' theorem $A$ satisfies the sentences $$\theta_n:=\forall y_0,\ldots,y_n\left(y_n\neq 0\to\exists x(y_0+y_1\cdot x+\cdots+y_nx^n=0)\right)$$
for $n\geq 1$. Note that the set of sentences $\{\theta_n:n\geq 1\}$ states precisely that the field is algebraically closed. 
Finally, it remains to show that $A$ is isomorphic to $\mathbb{C}$, but it is probably easier to show a more general result:

If $K,L$ are two uncountable algebraically closed fields with the same characteristic and the same cardinality, then $K\cong L$.

The proof of this is as follows. Let $B_K,B_L$ be transcendence basis (maximal algebraically independent subsets of $K$ and $L$ respectively). Then $B_K,B_L$ have the same cardinality, and any bijection between them can be extended (e.g. using Zorn's lemma) to an isomorphism between $K$ and $L$. 
