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.

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    $\begingroup$ Remark: for given $i$, the condition "$F_i$ is not contained in $\bigcup_{j\neq i}F_j$ is equivalent to "$F_i$ is not contained in $F_j$ for any $j\neq i$". $\endgroup$ – YCor Dec 22 '18 at 22:49
  • $\begingroup$ I'm guessing you want to take only finitely many algebraically closed subfields? $\endgroup$ – user44191 Dec 26 '18 at 20:37
  • $\begingroup$ @user44191 why "I'm guessing"? you're not guessing, you're reading. It's explicit. $\endgroup$ – YCor Dec 26 '18 at 23:54

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 $F_1+F_2=\mathbb{C}$.

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    $\begingroup$ Here's a way to deduce alg. closed fields of char 0 and arbitrary infinite card. First, the above construction works for arbitrary uncountable regular card. Second, given a large alg. closed field $C=A\oplus B$, sum of two subfields, choose a (set-wise) projection $p$ onto $A$ such that $p(x)-x\in B$ for all $x$. Then the operations of taking the field generated by a subset, taking its relative alg. closure, and taking union with the proj. to $A$, are finitary and preserve infinite card, and hence every infinite subset is contained in a $p$-stable alg. closed subfield of the same cardinal. $\endgroup$ – YCor Dec 28 '18 at 9:11
  • $\begingroup$ @YCor I don’t think I can work out the details of your construction. In particular, how do you make sure that you end up with proper subfields? $\endgroup$ – Jeremy Rickard Dec 31 '18 at 9:08
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    $\begingroup$ I should have written $C=A+B$ (not $A\oplus B$), with $A,B\neq C$, and pick $a\in A-B$, $b\in B-A$, and assume that the subfield contains $a,b$. I construct a subfield $K$ of $C$, stable under $p$, containing $a$ and $b$, and of any prescribed cardinal in $[\omega,|C|]$. So $K=(A\cap K)+(B\cap K)$ and both are proper subfields of $K$. $\endgroup$ – YCor Dec 31 '18 at 9:15
  • $\begingroup$ @YCor Ah,right! For some reason (unrelated to anything you actually wrote!) I thought you were trying to decompose a fixed subfield of $C$. So this deals with any algebraically closed field with infinite transcendence degree, I think. I don't think the characteristic matters? $\endgroup$ – Jeremy Rickard Dec 31 '18 at 10:57
  • $\begingroup$ It doesn't matter in my construction. So if you can produce such decomposition in any characteristic and unbounded cardinals (as it seems), it works. $\endgroup$ – YCor Dec 31 '18 at 11:15

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