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Let $f(z)=\sum_{n\geq 0}a_n z^n$ be a Taylor series with rational coefficients with infinitely non-zero $a_n$ which converges in a small neighboorhood around $0$. Furthermore, assume that \begin{align*} f(z)=\frac{P(z)}{Q(z)}, \end{align*} where $P(z)$ and $Q(z)$ are coprime monic complex polynomials. By developing $\frac{P(z)}{Q(z)}$ as a power sereis around $0$ and comparing it with $f(z)$ we obtain infinitely many polynomial equations in the roots of $P(z)$ and $Q(z)$ which are equal to rational numbers so this seems to force the roots of $P(z)$ and $Q(z)$ to be algebraic numbers.

Q: How does one prove this rigourously?

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A minor observations: It seems some condition is needed to avoid 'degenerate' cases; e.g., what if P = Q, or P = S Q with S a rational polynomial. – user9072 Jan 22 '11 at 18:33
Don't the coefficients of the Taylor series satisfy a recurrence relation (or difference equation)? – Mark Bennet Jan 22 '11 at 18:48
I think what is being asked is to prove that if a sequence of rational numbers satisfies a recurrence with complex coefficients, then it must satisfy a recurrence with rational coefficients. – Gjergji Zaimi Jan 22 '11 at 18:51
Yes Gjergji, I think that you may rephrase the problem in these terms – Hugo Chapdelaine Jan 22 '11 at 18:53
up vote 6 down vote accepted

Let there be two fields $k\subset K$, and let $f\in k[[x]]$ be a formal power series with coefficients in $k$. If $f\in K(x)$ (rational functions with coefficients in $K$) then $f\in k(x)$. A proof of this is given in J.S. Milne's notes on Etale Cohomology (lemma 27.9).

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Thanks a lot for the reference. (Actually the proof was taken from Bourbaki Algebre IV) – Hugo Chapdelaine Jan 22 '11 at 20:13
The result is classical, but I couldn't find a proper reference, even though I remember there is an article of Polya that discusses this. – Gjergji Zaimi Jan 22 '11 at 20:18

Well, I think there is a simpler argument. For a power series $g(x)\in\mathbb{C}[[x]]$ and $\sigma\in Aut(\mathbb{C})$ (note that except for the complex conjugation or the identity $\sigma$ is not continuous!) we may define define the power series with coefficients twisted by $\sigma$ which we denote by $g^{\sigma}(x)$. Now an element in $Aut(\mathbb{C})$ respect finite sum and products so it follows from that, that for all $\sigma\in Aut(\mathbb{C})$ one has $$ f^{\sigma}(z)=\frac{P^{\sigma}(z)}{Q^{\sigma}(z)}. $$ From this (and the unique factorization of $\mathbf{C}[x]$) it follows that $P(z)$ and $Q(z)$ have rational coefficients.

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Yes, and this works in Gjergji's more general setting, too. I would emphasize that $K[[x]]$ is a UFD, it is implicitly used in your proof. – GH from MO Jan 23 '11 at 12:35
Yes, I did not check it but the proof might carry over to power series ring in many variables. – Hugo Chapdelaine Jan 23 '11 at 14:54
There is something which bothers me. You use the fact that the subfield of $\C$ fixed by $Aut(\C)$ is $\Q$. Why is it true ? For exemple it is false if you replace $\C$ by $\R$. – Auguste Hoang Duc Jan 27 '11 at 22:06
Yes you are right, I use the fact that the fixed field of $Aut(C)$ is $Q$. The fact that $Aut(R)={Id}$ is not a problem since you may work in a suitable algebraic closure and as you know $C$ is an algebraic closure of $R$. I guess that in general if you have a field $k\subseteq K$ then you want to show the existence of a field $L$ which contains $K$ such that $Aut_k(L)=k$. Once you have that the proof works. – Hugo Chapdelaine Jan 28 '11 at 0:46
To Hugo: You probably want that $L^{Aut_k(L)}=k$. But such an $L$ only exists if the field extension $K/k$ is separable. – ACL Nov 15 '12 at 22:55

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