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Motivation: A real number is rational iff its decimal expansion is periodic (by "periodic" I mean periodic after some steps). Similar, a p-adic number is rational iff its p-adic expansion is periodic. However, this isn't true in general for rational functions as the example $$\frac{1}{(1-X)^2}= \sum_{n=0}^\infty (n+1)X^n \in \mathbb{Q}(X)\subseteq \mathbb{Q}((X))$$ shows. But an inspection of the proof in the p-adic case (cf. Hasse: Number Theory, Chap. 9) shows:

If $F$ is a finite field, then $f \in F((X))$ is rational, i.e. $f \in F(X)$, iff its Laurent series is periodic.

Now let $f=g/h$ with polynomials $g,h \in \mathbb{Z}[X]$ such that the Laurent series of $f$ has integer coefficients. Reduction modulo p yields $\bar{f}=\bar{g}/\bar{h} \in\mathbb{F}_p(X)$. Hence the coefficients of $f$ are periodic modulo p. I wonder whether the converse is true:

Question 1: Let $f$ be a Laurent series with integer coefficients. Is $f \in Quot(\mathbb{Z}[X])$ iff the coefficients of $f$ are periodic for all primes (with period depending on the prime) ?

Suppose that the coefficients of $f$ are periodic modulo p. Hence the reduction modulo p is rational, i.e. $\bar{f} \in \mathbb{F}_p(X)$. Therefore an equivalent formulation of the question is:

Question 2: Let $f$ be a Laurent series with integer coefficients. Is $f$ rational, i.e. $f \in Quot(\mathbb{Z}[X])$ iff the reduction modulo p is rational for all primes ?

Edit: As shown by Felipe, the answer is in general negative. Since I'm mostly interested in convergent power series (with integer coefficients), I would like to ask in addition, if there are also counterexamples in this case ?

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No $\sum n! x^n$ is a counterexample. I think it's true for algebraic functions, though.

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  • $\begingroup$ Thanks for the counterexample. Do you have more information for the algebraic function case ? $\endgroup$
    – Ralph
    Nov 14, 2012 at 21:05
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    $\begingroup$ I didn't think too hard about the algebraic case, but an algebraic curve is parametrizable if and only if it has genus zero and the genus of a curve is the same as the genus of its reduction mod p for all but finitely many primes p. $\endgroup$
    – Felipe Voloch
    Nov 14, 2012 at 21:18
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    $\begingroup$ For algebraic functions, you'll get other counterexamples, for example $\sqrt{1+4x}$. In fact, the conjecture of Grothendieck-Katz claims that all the solutions of a linear o.d.e. with coefficients in ${\mathbf Q}[x]$ are algebraic if and only if for almost all prime numbers $p$, all the solutions of the reduced equation modulo $p$ are rational. My example comes from the equation $(1+4x)y'=2y$. $\endgroup$
    – ACL
    Nov 14, 2012 at 22:36
  • $\begingroup$ @Antoine: Your differential equation has solution $(1+4x)^{(p-1)/2}$ but that is not the same as the power series $\sum 4^n{1/2 \choose n}x^n$, which is not periodic modulo $p$, for odd $p$. $\endgroup$
    – Felipe Voloch
    Nov 14, 2012 at 22:52
  • $\begingroup$ The exponent should have been $(p+1)/2$. $\endgroup$
    – Felipe Voloch
    Nov 14, 2012 at 22:53

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