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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?

Eric Wofsey
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    @egreg There are more invertible elements in $\mathbb{C}[[X]]$ than just constant power series. – xyzzyz Mar 01 '18 at 21:34

1 Answers1

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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.

Eric Wofsey
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  • Nice. And it seems like there's no problem with iterating this method transfinitely through a transcendence basis for $\mathbb{C}$ over $\mathbb{Q}$, to obtain a subfield isomorphic to $\mathbb{C}$, right? Now I wonder if there's any chance that such an isomorphism extends to an automorphism of $\mathbb{C}[[X]]$... – Alex Kruckman Mar 03 '18 at 17:36
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    Yeah, I think that works. Or, rather than iterating, you could just start with an entire transcendence basis instead of just one transcendental element. – Eric Wofsey Mar 03 '18 at 18:09