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.

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    $\begingroup$ @Qiaochu: I think he is implicitly applying the evaluation homomorphism, and also implicitly saying that this power series must (as least locally) converge, to give a well-defined function. $\endgroup$ Commented Oct 21, 2011 at 22:02
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    $\begingroup$ For K equal to any complete extension field of Q_p, this is a consequence of the Weierstrass Preparation Theorem for formal power series whose coefficients tend to 0 (which is what convergence at 1 requires). Or it's a consequence of Strassman's theorem for power series. The point is that you can factor the power series as a polynomial multiplied by a power series which is invertible on the closed unit disc. $\endgroup$
    – KConrad
    Commented Oct 21, 2011 at 22:17
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    $\begingroup$ @Jeff H: The Weierstrass preparation theorem isn't just an example of this phenomenon, it is the phenomenon. $\endgroup$ Commented Oct 22, 2011 at 7:41
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    $\begingroup$ @KConrad: This is an answer, not a comment. $\endgroup$ Commented Oct 22, 2011 at 9:50
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    $\begingroup$ In the case where $K$ is a finite extension of ${\bf Q}_p$, the argument I had in mind was the following: for every $\sigma$ in $Aut({\bf C}_p/K)$, $\sigma(\alpha)$ is again a root of $f$. Thus, $\alpha$ only has finitely many conjugates and thus must be algebraic. $\endgroup$
    – sibilant
    Commented Oct 23, 2011 at 4:11

3 Answers 3


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 $r$.

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    $\begingroup$ @KConrad: you don't even need to assume anything about your original power series other than that it is convergent on the open unit disc. Indeed, your argument shows that any root of a convergent power series which is contained in some closed disc is necessarily algebraic. But of course any point in the open disc is contained in some closed disc! Thus all roots of any convergent power series on the open unit disc are algebraic. $\endgroup$
    – sibilant
    Commented Oct 23, 2011 at 13:56
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    $\begingroup$ That's almost true. The zero series converges on the open unit disc and has quite a few non-algebraic roots. And you want the radius of a closed disc to be in $|K^\times|^{1/∞}=\{\sqrt[n]{|a|}:a\in K,,n≥1\}$. But I agree the proof shows for any nonzero series $f(x)$ with coefficients in $K$ and any complete extension field $F$ of $K$, the roots of $f(x)$ are algebraic over $K$ in any open disc of $F$ where $f(x)$ converges. That's also true for closed discs in $F$ with radius in $|K^\times|^{1/∞}$. Is there be a counterexample for a closed disc in $F$ with radius not in $|K^\times|^{1/∞}$? $\endgroup$
    – KConrad
    Commented Oct 23, 2011 at 14:56
  • $\begingroup$ Is it true that any transcendental element of Cp has uncountably many conjugates? If so, then the argument I gave in the comments above would also show that all zeroes of convergent power series are algebraic. The argument would be, if there were a transcendental root then there would be uncountably many roots. But then in some closed disc there would be infinitely many zeroes. $\endgroup$
    – sibilant
    Commented Oct 23, 2011 at 15:24
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    $\begingroup$ Actually it's clear for a very simple reason why there's not going to be a root at an "inaccessible" radius: in the notation of my previous comment, say $f(x)$ converges on $\{z∈F:|z|≤R\}$ and $R$ is in $|F^\times|$ and not in $|K^\times|^{1/∞}$ (a better notation for that would be $|K^\times|^{\mathbf Q}$). If $|z|=R$ in $F$ then no two nonzero terms in the series for $f(z)$ have equal absolute value, and the terms in the series are tending to 0, so by the ultrametric inequality $|f(z)|$ is the absolute value of its largest term. So in particular $f(z) \not= 0$. $\endgroup$
    – KConrad
    Commented Oct 23, 2011 at 16:04
  • $\begingroup$ @KConrad: Thank you, this is great! $\endgroup$
    – Jeff H
    Commented Oct 23, 2011 at 18:05

I feel considerable trepidation in muddying the waters that Keith has so nicely clarified, but it seems to me, in view of the discreteness of the valuation on $K$, that first, the hypothesis that there are only finitely many roots is unnecessary, and second, it's unnecessary to go by way of the closed unit disk.

As an example of a series with infinitely many roots, hold in mind the logarithmic series, $\sum_n (-1)^{n+1}x^n/n$, convergent on the open unit disk. Its roots are the numbers $\zeta-1$, $\zeta$ running through the $p$-power roots of unity. For a series to be convergent on the open disk of ${\bf C}_p$, it's necessary and sufficient that the limit slope of the Newton polygon be nonnegative. You can define this as $\lim_n v(a_n)/n$ if you like. In case the series has positive limit slope, then there's a segment with positive slope, and you can use Weierstrass Prep directly to show that there are only finitely many roots of the series in the open disk, and these are roots of an integral monic $K$-polynomial. If, as in the case of the log, your series has limit slope zero, then you need a suitably jazzed-up version of W-Prep saying every vertex $V$ of the polygon gives you a monic polynomial factor, still with $K$-integers for coefficients. Or, since in this case we're only worried about the roots being algebraic, you can just replace $f(x)$ by $f(p^\lambda x)$, where $\lambda$ is a positive rational between the negatives of the slopes on either side of $V$. And then apply the first case, 'cause now the limit slope is $\lambda$. In any case, and by whatever method, the roots of $f$ are all algebraic.

  • $\begingroup$ @Lubin: This is a great (counter?)example. Thanks for this alternate perspective! $\endgroup$
    – Jeff H
    Commented Oct 24, 2011 at 1:14
  • $\begingroup$ Some sloppiness on my part: that should have been the lim inf of those numbers in the def. of the limit slope. $\endgroup$
    – Lubin
    Commented Oct 24, 2011 at 17:53

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.

  • $\begingroup$ Rob, this is wonderful...thank you!! $\endgroup$
    – Jeff H
    Commented Oct 28, 2011 at 15:09

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