integrality of certain rational numbers

Question

Let $P,Q\in\mathbb Q[X]$ be relatively prime polynomials ($X$ being an indeterminate). Assume that $Q(0)=1$ and that $P/Q$ is in $\mathbb Z[[X]]$.

Does this imply that $P$ and $Q$ are in $\mathbb Z[X]$?

Answer 1

This is a straightforward consequence of Fatou's Lemma on rational power series. Here is a statement and proof from R. P. Stanley's Enumerative Combinatorics, I (or see p. 629 of the free version here)

Answer 2

This is a minor complement to Bill's answer.

The answer to the question is yes. As indicated by Bill, this follows immediately from the statement below, known as Fatou's Lemma:

(1) If $f\in\mathbb Q[[x]]$ has integer coefficients, then there are $P,Q$ in $\mathbb Z[x]$ such that $f=P/Q$, $(P,Q)=1$ in $\mathbb Q[x]$, and $Q(0)=1$.

We start by quoting Stanley:

Define a formal power series $\sum_{n\ge0}a_nx^n$ with integer coefficients to be primitive if no integer $d > 1$ divides all the $a_i$. One easily shows that the product of primitive series is primitive.

We claim:

(2) The following conditions on a primitive polynomial $Q\in\mathbb Z[x]$ are equivalent:

(a) $Q(0)=\pm1$,

(b) $1/Q\in\mathbb Z[[x]]$,

(c) $Qg=m$ for some $g$ in $\mathbb Z[[x]]$ and some positive integer $m$.

It is clear that (a) and (b) are equivalent and imply (c). Assuming (c) and writing $g=dh$ where $d$ is a positive integer and $h$ a primitive element of $\mathbb Z[[x]]$, we get $Qh=m/d$. As $Q$ and $h$ are primitive, this yields $m=d$, and thus (b). This proves (2).

Let us prove (1).

We have $f=P/Q$ for some polynomials $P,Q$ in $\mathbb Z[x]$ such that $(P,Q)=1$ in $\mathbb Q[x]$ and no prime number divides every coefficient of $P$ and $Q$. The equality $Qf=P$, which holds in $\mathbb Z[[x]]$, shows that $Q$ is primitive. In view of (2), it suffices to prove (c). There are polynomials $A,B$ in $\mathbb Z[x]$ and a positive integer $m$ such that $AP+BQ=m$. Then (c) holds for $g:=Af+B$.