Suppose $f:\mathbb{C}\to\mathbb{C}$ is a map with your favorite smoothness condition (say, $C^1, C^{\infty}$ or holomorphic) and suppose that $f(\overline{\mathbb{Q}})\subset\overline{\mathbb{Q}}$. Is $f$ a polynomial? (I.e., if you chose $C^1$ or $C^{\infty}$ in the beginning, then is $f$ a polynomial in $z$ and $\overline{z}$, and if you chose holomorphic, then is $f$ a polynomial in $z$?)


I imagine this is a well-known problem, but I don't remember having seen it before. A holomorphic counter-example would be quite impressive, though I doubt such a thing exists.



Maps preserving algebraic numbers
This is indeed very well-known. There are plenty of counterexamples. In fact, there's a remarkable construction due to van der Poorten of a "transcendental" entire function $f$ such that $f$ and all its derivatives map any algebraic number $\alpha$ into $\mathbb Q (\alpha)$. See



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> A.J. van der Poorten, [Transcendental entire functions mapping every algebraic number field into itself](http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=4860012), J. Austral.Math. Soc. **8** (1968), 192–193.
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