Sum of subfields of $\mathbb{C}$

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.

• 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 $j\neq i$". – YCor Dec 22 '18 at 22:49
• I'm guessing you want to take only finitely many algebraically closed subfields? – user44191 Dec 26 '18 at 20:37
• @user44191 why "I'm guessing"? you're not guessing, you're reading. It's explicit. – 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 $$F_1+F_2=\mathbb{C}$$.

• 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. – YCor Dec 28 '18 at 9:11
• @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? – Jeremy Rickard Dec 31 '18 at 9:08
• 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 $K$. – YCor Dec 31 '18 at 9:15
• @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? – Jeremy Rickard Dec 31 '18 at 10:57
• 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. – YCor Dec 31 '18 at 11:15