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Let $z$ be a complex variable and $f(z)$ be a formal power series with rational coefficients (an element in $\mathbb Q[[z]]$), with a finite radius of convergence, and assume $f(z)$ has a meromorphic continuation to the whole complex plane (so it has at most countably many poles). What do we know about the number-theoretic property of the roots and poles? Are they algebraic numbers? If they are, are they stable under the action of the Galois group of the rationals?

More generally, if the coefficients of $f(z)$ are algebraic numbers, and let $\sigma$ be an automorphism of the algebraic closure of the rationals, then what is the relation between roots of $f(z)$ and roots of $f^{\sigma}(z),$ where $f^{\sigma}(z)$ is the power series obtained by applying sigma to the coefficients of $f(z)?$

Without assuming the finiteness of radius of convergence, $\sin(z)$ gives a counter-example.

Edit: Let me give a second try, by imposing more requirements on $f(z).$ I'm thinking about the case where $f(z)$ is the zeta function of an algebraic stack over a finite field, so let's assume $f(z)$ has an infinite product expansion over $\ell$-adic numbers, like $\prod P_{odd}(z)/\prod P_{even}(z),$ where each $P_i(z)$ is a polynomial over $\mathbb Q_{\ell}$ with constant term 1. Assume they have distinct weights, e.g. reciprocal roots of $P_i(z)$ have weights $i.$ Then can we conclude that all coefficients of $P_i(z)$ are rational numbers? Thanks.

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2 Answers 2

The power series for (1/(z+1))*sin(z) has rational coefficients and a finite radius of convergence, but its zeros are not algebraic. You can construct more interesting examples too : for instance, consider sin(z)/cos(z+2i)

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Thanks. Let me give a second try, by imposing more requirements on f(z). I'm thinking about the case where f(z) is the zeta function of an algebraic stack over a finite field, so let's assume f(z) has an infinite product expansion over \ell-adic numbers, like \prod P_{odd}(z)/\prod P_{even}(z), where each P_i(z) is a polynomial over Q_{\ell} with constant term 1. Assume they have distinct weights, e.g. reciprocal roots of P_i(z) have weights i. Then can we conclude that all coefficients of P_i(z) are rational numbers? Thanks. –  shenghao Oct 16 '09 at 17:30
    
OK, that was something that should have gone in an edit of the question, not a comment. –  Ben Webster Oct 16 '09 at 18:22

Allowing the coefficients to be rational (with unbounded denominators) allows for too many possibilities, since for any power series with real coefficients, we may approximate the terms by terms with rational coefficients and with errors that tend as fast to zero as we wish in any bounded region of the plane (use the Dirichlet theorem on Diophantine approximation). Thus you can get a power series with rational coefficients from any power series with real coefficients by adding a suitable entire function. The class of power series with real coefficients is so wide that your requirement of meromorphic continuation does not meaningfully constrain. Unbounded denominators is the problem here.

If the denominators are bounded, your question ties in with a very nice theorem of Fritz Carlson from the early 1920s: A power series with integer coefficients and radius of convergence 1 either sums to a rational function or else has the unit circle as a natural boundary (analytic continuation across the unit circle is not possible anywhere on the circle). Of course, a natural boundary is the generic case in Carlson´s theorem.

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