TITLE: When are roots of power series algebraic?
QUESTION [18 upvotes]: Let $K$ be a field and consider a power series $f(T) \in K[[T]]$. Under what conditions (on $K$ and/or on $f$) can we conclude that if $\alpha$ is a root of $f(T)$ then $\alpha$ is in fact algebraic over $K$?

This question is inspired by the following: In this paper of R. Pollack, in the proof of his Lemma 3.2, the author states that if $K$ is a finite extension of $\mathbb{Q}_p$ and $f(T)$ converges on the open unit disc of $\mathbb{C}_p$ and has finitely many roots, then each of these roots is algebraic over $K$.
I haven't yet come up with a proof of this statement (and any proofs offered would be appreciated), and perhaps having a proof in hand would help me to answer my question on my own, but I'm now curious about other situations in which this can occur.

REPLY [13 votes]: Here's another proof of the fact that any root of a (non-zero) $p$-adic power series (convergent on the open unit disc) is algebraic.  The proof is a little silly in that it uses Tate's theorem on Galois invariants of ${\mathbb C}_p$ (which is quite deep) instead of the Weierstrass preparation theorem (which is fairly elementary).  But here it is in any case.
The key fact I need is the following which I'll prove after explaining how it completes the proof.

Claim: Any transcendental element of ${\mathbb C}_p$ has uncountably
  many conjugates (under $\text{Gal}(\overline{\mathbb Q}_p/{\mathbb Q}_p))$.

Accepting this claim for the moment, let $\alpha$ be a transcendental zero of $f(x)$, our convergent power series.  Then all conjugates of $\alpha$ are also zeroes of $f(x)$ in the open unit disc.  Thus, by the above claim, $f(x)$ has uncountably many zeroes in the open unit disc of ${\mathbb C}_p$.  However, any uncountable set in a separable metric space (e.g. ${\mathbb C}_p$) has an accumulation point.  But then the zeroes of $f(x)$ have an accumulation point, necessarily in the open unit disc since that is closed in ${\mathbb C}_p$, which forces $f(x)$ to be identically zero.
Returning now to the claim, let $G=\text{Gal}(\overline{\mathbb Q}_p/{\mathbb Q}_p)$ and let $H$ denote the subgroup of $G$ which stabilizes $\alpha$.  Since the conjugates of $\alpha$ are in one-to-one correspondence with $G/H$, we will show that $H$ has uncountable index.
First note that $G$ acts continuously on ${\mathbb C}_p$ where we give ${\mathbb C}_p$ the $p$-adic topology (not the discrete topology).  Thus $H$ is a closed subgroup and hence of the form $\text{Gal}(\overline{\mathbb Q}_p/M)$ for some algebraic extension$M/{\mathbb Q}_p$.
Assume $G/H$ is finite, and thus that $M$ is a finite extension of ${\mathbb Q}_p$.  But then by Tate's theorem, $\alpha$ must be in $M$ as it is fixed by $\text{Gal}(\overline{\mathbb Q}_p/M)$.  This is impossible as $\alpha$ is transcendental.
Thus, $G/H$ is infinite, and we must now show that it is uncountable. We use the following (standard) fact:

Fact: Any infinite compact Hausdorff space with no isolated points is uncountable.

Since $G$ is compact, $G/H$ is compact.  The coset space $G/H$ is Hausdorff since $H$ is closed.  To see that $G/H$ has no isolated points, note that if it has one isolated point, then all of its points are isolated as $G$ acts transitively (by left multiplication) on $G/H$ by homeomorphisms.  But then $G/H$ is discrete which is impossible as it is infinite and compact.
Thus, $G/H$ is uncountable, and $\alpha$ has uncountably many conjugates.