4 added 50 characters in body

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.

3 added 16 characters in body

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 it $\sigma$ 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 it follows that $P(z)$ and $Q(z)$ have rational coefficients.

2 added 65 characters in body

Well, I think there is a simpler argument. For a power series $g(x)\in\mathbb{C}[[x]]$ and $\sigma\in Aut(\mathbb{C})$ (not continuous outside except for the complex conjugation and or the identity!identity it 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 it follows that $P(z)$ and $Q(z)$ have rational coefficients.

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