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> 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 $K$?
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This question is inspired by the following: In [this paper](http://math.bu.edu/people/rpollack/Papers/On_the_padic_Lfunction_of_a_modular_form_at_a_ss_prime.pdf) 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.



When are roots of power series algebraic?
At first I thought the question concerned the closed unit disc. Since it involves the open unit disc in ${\mathbf C}\_p$ we need a little detail to see why the Weierstrass preparation theorem for series on the closed unit disc can be used. We'll pass to a suitable finite extension of K to pull this off. 


Since there are assumed to be only finitely many zeros of $f(x)$ in the open unit disc of ${\mathbf C}\_p$, let $\alpha$ be one of those zeros with largest absolute value. A priori 
$\alpha$ might be transcendental over $K$, but note that $|\alpha|\_p$ is a rational power of $p$ and there are definitely $t$ in $\overline{K}$ with $|t|\_p = |\alpha|\_p$ (just take $t$ to be a suitable rational power of $p$ in ${\mathbf C}\_p$). Let $L = K(t)$, which is a finite extension of $K$. Since $|t|\_p < 1$, the power series $g(x) = f(tx)$ has coefficients in $L$ and converges on the closed unit disc in $L$ (equivalently, in ${\mathbf C}\_p$). By the Weierstrass Preparation Theorem for (nonzero!) power series in $L[[x]]$ which converge on the closed unit disc, we can write $g(x) = W(x)U(x)$ where $W(x)$ is a polynomial in $L[x]$ and $U(x)$ is a power series in $L[[x]]$ with nonzero constant term which converges on the closed unit disc along with its inverse formal power series. Therefore on any complete extension field of $K$, $U(x)$ converges on its closed unit disc and is nonvanishing. For any root $r$ of $f(x)$ in ${\mathbf C}\_p$, we have $|r|\_p \leq |t|\_p$, so $0 = f(r) = g(r/t) = W(r/t)U(r/t)$. The number $U(r/t)$ is nonzero, so $W(r/t) = 0$, which implies $r/t$ is algebraic over $L$, and thus over $K$. Since $t$ is algebraic over $K$, $r$ is algebraic over $K$.


Instead of assuming $f(x)$ has finitely many zeros in the open unit disc of ${\mathbf C}\_p$ we only need the (superficially) weaker assumption that the roots of $f(x)$ in the open unit disc of ${\mathbf C}\_p$ are uniformly bounded in absolute value away from 1. That is, assume that for some $\varepsilon > 0$ the two conditions $|r|\_p < 1$ and $f(r) = 0$ in ${\mathbf C}\_p$ imply $|r|\_p \leq 1 - \varepsilon$. Then we can conclude the roots of $f(x)$ in the open unit disc of ${\mathbf C}\_p$ are both algebraic over $K$ and finite in number. Picking $t$ in $\overline{K}$ so that $1 - \varepsilon \leq |t|\_p < 1$, which is definitely possible, we can run through the previous argument and see that if $|r|\_p < 1$ and $f(r) = 0$ in ${\mathbf C}\_p$ then $W(r/t) = 0$, so not only does $r$ lie in a finite extension of $K$ but there are finitely many choices for $r$.


