Do there exist algebraically closed subfields $F_1, F_2, \dots, F_n$ ($n \geq 2$) of the field of complex numbers such that no $F_i$ is contained in $\bigcup_{j \neq i} F_j$ and $F_1 + F_2 + \dots F_n = \mathbb{C}$?
The answer ought to be "No". For example, if $x_i \in F_i \setminus \bigcup_{j \neq i} F_j$, then it doesn't seem possible to write the product $x_1 x_2 \dots x_n$ as a sum $a_1 + a_2 + \dots + a_n$ where each $a_i \in F_i$. But I don't see how to derive a contradiction from this.
I think I can show that if the cardinality of the continuum is a regular cardinal (for example, if the continuum hypothesis is true, or more generally if $2^{\aleph_0}=\aleph_n$ for any natural number $n$), then $\mathbb{C}$ is the sum of two proper algebraically closed subfields.
I'll use the convention that a "cardinal" is the least ordinal of a given cardinality, and regard the continuum $\mathfrak{c}$ as an ordinal.
Let $\mathcal{X}$ be a transcendence basis of $\mathbb{C}$ over $\mathbb{Q}$.
Both $\mathcal{X}$ and $\mathbb{C}$ have continuum cardinality and may be well-ordered, $\mathcal{X}=\left\{X_\alpha\mid\alpha<\mathfrak{c}\right\}$ and $\mathbb{C}=\left\{z_\alpha\mid\alpha<\mathfrak{c}\right\}$, with the order type of $\mathfrak{c}$. In the case of $\mathbb{C}$ we shall do this so that $z_0=0$.
For each ordinal $\alpha\leq\mathfrak{c}$, let $K_\alpha$ be the algebraic closure in $\mathbb{C}$ of the field extension of $\mathbb{Q}$ generated by $\left\{X_\beta\mid\beta<\alpha\right\}$. Note that every $z\in\mathbb{C}$ is in the algebraic closure of the transcendental extension of $\mathbb{Q}$ generated by finitely many $X_\beta$, so $z\in K_\alpha$ for some $\alpha<\mathfrak{c}$.
By transfinite induction we can define a strictly increasing function f:$\mathfrak{c}\to\mathfrak{c}$ such that $f(0)=0$ and $z_\alpha\in K_{f(\alpha)}$ for every $\alpha$. [For a successor ordinal $\alpha+$ let $f(\alpha+)$ be the smallest ordinal such that $f(\alpha+)>f(\alpha)$ and $z_{\alpha+}\in K_{f(\alpha+)}$. For a limit ordinal $\alpha$ let $f(\alpha)$ be the smallest ordinal greater than or equal to $\bigcup_{\beta<\alpha}f(\beta)$ such that $z_\alpha\in K_\alpha$. Since we are assuming that $\mathfrak{c}$ is a regular cardinal, we never have to assign the value $\mathfrak{c}$ to $f(\alpha)$.]
For each $\alpha$, let $Y_\alpha=X_{f(\alpha)}+z_\alpha$.
Let $F_1$ be the algebraic closure in $\mathbb{C}$ of the extension of $\mathbb{Q}$ generated by $\left\{X_{f(\alpha)}\mid 0<\alpha<\mathfrak{c}\right\}$. Since $X_0\not\in F_1$, this is a proper subfield of $\mathbb{C}$.
Let $F_2$ be the algebraic closure in $\mathbb{C}$ of the extension of $\mathbb{Q}$ generated by $\left\{Y_\alpha\mid 0<\alpha<\mathfrak{c}\right\}$. I claim that the set $\left\{Y_\alpha\mid \alpha<\mathfrak{c}\right\}$ is algebraically independent over $\mathbb{Q}$, and so $Y_0\not\in F_2$, and so $F_2$ is also a proper subfield of $\mathbb{C}$.
Assuming the claim is false, there is some $\alpha$ such that $Y_\alpha$ is algebraic over the field generated by $\left\{Y_\beta\mid\beta<\alpha\right\}$. Since $Y_\beta\in K_{f(\alpha)}$ for $\beta<\alpha$ and $Y_\alpha=X_{f(\alpha)}+z_\alpha$, where $z_\alpha\in K_{f(\alpha)}$, this implies that $X_{f(\alpha)}$ is algebraic over $K_{f(\alpha)}$, which is false.
Finally, for any $\alpha>0$, $z_\alpha=-X_{f(\alpha)}+Y_\alpha\in F_1+F_2$, and clearly $z_0=0\in F_1+F_2$. Therefore $F_1+F_2=\mathbb{C}$.
