$f'/f\in\mathbb{Z}[[x]]$ for polynomials vs. formal power series $f$

I am curious about the following problem from MIT's Problem Solving Seminar (#26 here, though the link may stop working after a few weeks):

Let $f(x) = a_0+a_1x+\cdots\in\mathbb{Z}[[x]]$ be a formal power series with integer coefficients, such that $a_0\ne0$. If $f'/f\in\mathbb{Z}[[x]]$, must $f/a_0\in\mathbb{Z}[[x]]$? (If I haven't made a mistake below, the answer is yes.)

The special case of integer polynomials, $f(x) = C\prod_{i=1}^{n}(1-r_i x)^{\alpha_i}\in\mathbb{Z}[x]$, translates (more or less) to the following classical problem:

If $\sum_{i=1}^{n} \alpha_i r_i^k \in\mathbb{Z}$ for all $k\ge0$, then $\prod_{i=1}^{n}(1-r_ix)^{\alpha_i}\in\mathbb{Z}[x]$, or equivalently (by considering minimal polynomials of the $r_i$ over $\mathbb{Q}$, with coefficients suitably reversed), $\prod_{i=1}^{n}(1-r_ix)\in\mathbb{Z}[x]$.

There are a few ways to do the polynomial version, which I'll outline below. However, the only solution I can find right now to the original goes as follows:

Yes. WLOG scale so that $\gcd(a_0,a_1,\ldots) = 1$.

For $k\ge1$, define $g_k = f^{(k)}f^{-1}$. Note that if we have $g_k\in\mathbb{Z}[[x]]$ for some $k\ge1$, then differentiating gives $f^{(k+1)}f^{-1} - f^{(k)}f'f^{-2}\in\mathbb{Z}[[x]]$. But $f^{(k)}f^{-1},f'f^{-1}\in\mathbb{Z}[[x]]$, so adding their product, $f^{(k)}f'f^{-2}\in\mathbb{Z}[[x]]$, yields $g_{k+1} = f^{(k+1)}f^{-1}\in\mathbb{Z}[[x]]$.

(A perhaps more natural, but equivalent, way to do the induction: If $f^{(k)} = f g_k$, then $$f^{(k+1)} = f'g_k + f g'_k = f[g_1g_k + g'_k],$$ and of course $g_1g_k + g'_k\in\mathbb{Z}[[x]]$.)

Thus by induction, $g_k\in\mathbb{Z}[[x]]$ for all $k\ge1$. But $$g_kf = f^{(k)} = k!\sum_{i\ge0}\binom{i+k}{k}a_{i+k}x^i,$$ so by Gauss's lemma (for formal power series) and the fact that $\gcd(a_0,a_1,\ldots) = 1$, we in fact have $g_k/k!\in\mathbb{Z}[[x]]$.

In particular, evaluating at $0$ (looking at constant terms) gives $a_0 \mid \binom{0+k}{k}a_{0+k} = a_k$ for all $k\ge1$, so $f/a_0$ indeed has integer coefficients, as desired.

Unfortunately, I don't really grok this proof, even in the simple case of integer polynomials $F = C\prod_{i=1}^{n}(1-r_i x)^{\alpha_i}\in\mathbb{Z}[x]$ described above. (I'll use capital $F$ for clarity.) For the polynomial case, I've seen two more intuitive methods that I feel might/should generalize, but unfortunately I can't see exactly how right now:

1. I think the standard proofs more or less use valuations in some way. Note that the conditions hold with $r_i$ replaced by $r_i^s$, for any fixed $s\ge1$. By Newton's identities, it's not hard to see that $k$th symmetric sum $\sigma_k(r_i^s)$, with multiplicity $\alpha_i$ for $r_i$, is divisible by $1/k!$ (i.e. $k!\sigma_k\in\mathbb{Z}$), so $(\sum\alpha_i)!\prod_{i=1}^{n}(1-r_i^s x)^{\alpha_i}\in\mathbb{Z}[x]$. Thus $(\sum\alpha_i)!r_i^s$ is an algebraic integer for every $s\ge1$ ($i$ fixed), and we can finish using valuations (as in the second post, or more elementarily but in the same spirit, as in the first post) here.

   Admittedly, it seems hard to directly extend Newton's identities to "infinitely many roots" for formal power series, but I imagine there could be something more indirect. I suppose I'm really wondering if we can define some fruitful kind of "valuation" (possibly for some sort of "roots") for arbitrary $f\in\mathbb{Z}[[x]]$.

2. Write $F'/F = \sum -\alpha_i r_i/(1-r_i x)$ as $P/Q$ with $P\in\mathbb{Z}[x]$ coprime to $Q = D\prod (1-r_ix)\in\mathbb{Z}[x]$ ($D\ne0$ chosen so $Q$, and therefore $P$, has integer coefficients). By Bezout's identity, there exist $A,B\in\mathbb{Z}[x]$ such that $AP+BQ = c$ for some nonzero integer $c$, so $$c/Q = (AP+BQ)/Q = A(P/Q)+B \in \mathbb{Z}[[x]].$$ By Gauss's lemma (for formal power series), the only way $c = Q(c/Q)$ can occur is if $Q(0)$ and $(c/Q)(0)$ (which multiply to $c$) are the gcd's of the coefficients of $Q$ and $c/Q$, respectively, so $Q/Q(0) = \prod(1-r_ix)$ has integer coefficients, as desired.

   So in that vein, I would be interested in some version of Bezout's identity for integer formal power series. (Especially one that helps for the formal power series version of the problem.) Of course, to actually use such a fact, we would have to first factor out a gcd of $f$ and $f'$ (in $\mathbb{Z}[[x]]$), as when we have multiple roots in the polynomial case.

Assuming you are familiar with $p$-adic numbers, this one is vulnerable to a big tool: the "$p$-adic Weierstrass preparation lemma". Intuitively, this lemma lets you treat questions about $p$-adic power series as if they were questions about symmetric polynomials in infinitely many roots. We fix a prime $p$ for which we will prove that $v_p(a_0) \leq v_p(a_n)$ for all $n$.

Notation Let $f(z) = \sum a_n z^n$ be a power series with coefficients in $\mathbb{Q}_p$. We define the Newton polygon of $f$ to be the lower convex hull of the points $(i, v_p(a_i))$. Matt Baker has a blog post with a nice images of Newton polygon's for polynomials.

With power series, the polygon goes indefinitely to the right, but there is a subtlety: The last segment of the polygon might be a line segment of infinite length which doesn't actually pass through the $(i, v_p(a_i))$. For example, let $f(x) = 1+px + p x^2 + p x^3 + \cdots$. Then we are considering the convex hull of the points $(0,0)$ and $(i,1)$ for all $i>0$. This is the positive orthant, and its lower edge is the $x$-axis. However, the points $(i,1)$ aren't actually on the $x$-axis. The Newton polygon of the polynomial $1+p x + p x^2 + \cdots + p x^N$ is a line segment from $(0,0)$ to $(N, 1)$, of slope $1/N$, and the Newton polygon of the limiting series reflects the fact that this line segment approaches the $x$-axis as $N \to \infty$. In the end, this subtlety won't get in our way; it will just require us to phrase things more carefully.

We will write $\rho$ for the slope of the unbounded ray at the right end of the Newton polygon, or $\rho = \infty$ if there is no such ray. Let $\lambda < \rho$ and let $P_{\leq \lambda}$ be the finite portion of the Newton polygon made up of line segments of slope $\leq \lambda$; let $P_{> \lambda}$ be the infinite part of the Newton polygon made up of segments of slope $> \lambda$.

The lemma We can factor $f(z) = f_{\leq \lambda}(z) \cdot f_{>\lambda}(z)$ where

• $f_{\leq \lambda}(z)$ is a polynomial with coefficients in $\mathbb{Q}_p$ and Newton polygon $P_{\leq \lambda}$ and

• $f_{>\lambda}(z)$ is a power series with coefficients in $\mathbb{Q}_p$ and Newton polygon $P_{>\lambda}$ with initial vertex translated back to $(0,0)$.

See Chapters IV.3 and IV.4 of Koblitz's book p-adic numbers, p-adic analysis and zeta functions. The specific result is Theorem IV.4.14, but you'll want to read the surrounding material to put it into context.

Application to your question Let $f(z)$ be the power series which interests you. Since the coefficients of $f$ are integers, we have $v_p(a_i) \geq 0$, so the Newton polygon is in the first quadrant. Thus $\rho \geq 0$. Your goal is to show that $v_p(a_i) \geq v_p(a_0)$ for all $i$; i.e., that the Newton polygon has no segments of negative slope. Suppose otherwise and choose $\lambda$ which is $<0$ and greater than all negative slopes of the Newton polygon.

Then we can write $f(z) = f_{\leq \lambda}(z) f_{>\lambda}(z)$ so $\frac{f'}{f} = \frac{f'_{\leq \lambda}}{f_{\leq \lambda}} + \frac{f'_{> \lambda}}{f_{> \lambda}}$. Since $P_{> \lambda}$ has no segments of negative slope, the constant term of $f_{> \lambda}$ must $p$-adically divide all other terms of $f_{> \lambda}$. So the coefficients of $\frac{f'_{> \lambda}}{f_{> \lambda}}$ are $p$-adically integral. We see that, if the coefficients of $f'/f$ are $p$-adically integral, so are the coefficients of $f'_{\leq \lambda}/f_{\leq \lambda}$. But $f_{\leq \lambda}$ is a polynomial (over $\mathbb{Z}_p$, not $\mathbb{Z}$) and can be handled by the methods you already sketch. $\square$

In general, $p$-adic Weierstrass preparation allows you to write a $p$-adic power series as a product of a polynomial and a power series whose coefficients are $p$-adically small enough not to worry about.