Question

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?

Answer 1

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).

Answer 2

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.