Is every algebraically closed subfield of $\mathbb C[[X]]$ contained in $\mathbb C$? Let $F$ be a subring (with same unity) of $\mathbb C[[X]]$ such that $F$ is an algebraically closed field; then is it true that $F \subseteq \mathbb C$ ? 
Since $F, \mathbb C$ are algebraically closed field of the same characteristic , so I know that either $F$ embeds in $\mathbb C$ or $\mathbb C$ embeds in $F$ , but I am unable to say anything else. 
Arose in a comment here Given finite subset of $\mathbb C[[X]]$, is there a ring automorphism of $\mathbb C[[X]]$, fixing $A$, but not fixing $\mathbb C$ set wise?
 A: Let $a\in\mathbb{C}$ be any element that is transcendental over $\mathbb{Q}$, let $b=a+X$, and consider the subfield $K=\mathbb{Q}(b)\subset\mathbb{C}[[X]]$ (any nontrivial polynomial in $b$ with coefficients in $\mathbb{Q}$ has nonzero constant term and thus is invertible, so $b$ does generate a subfield of $\mathbb{C}[[X]]$).  Note that reduction mod $X$ map $\mathbb{C}[[X]]\to\mathbb{C}$ restricts to an isomorphism $K\to\mathbb{Q}(a)$ sending $b$ to $a$.  Now consider any irreducible polynomial $p(t)\in K[t]$.  When we reduce the coefficients of $p$ mod $X$, we get an irreducible polynomial $\tilde{p}(t)\in \mathbb{Q}(a)[t]$.  This irreducible polynomial has distinct roots in $\mathbb{C}$, and by Hensel's lemma each of these roots can be lifted to a root of $p(t)$ in $\mathbb{C}[[X]]$.
Now let $F$ be the algebraic closure of $K$ in $\mathbb{C}[[X]]$.  By the result of the previous paragraph, every irreducible polynomial over $K$ splits in $F$.  Since $F$ is algebraic over $K$, this means $F$ is algebraically closed.
