Maps preserving algebraic numbers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T19:44:32Z http://mathoverflow.net/feeds/question/78526 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78526/maps-preserving-algebraic-numbers Maps preserving algebraic numbers Moosbrugger 2011-10-19T03:12:09Z 2011-10-19T04:07:48Z <p>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 $z$?)</p> <p>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.</p> http://mathoverflow.net/questions/78526/maps-preserving-algebraic-numbers/78527#78527 Answer by Faisal for Maps preserving algebraic numbers Faisal 2011-10-19T03:19:34Z 2011-10-19T03:26:36Z <p>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</p> <blockquote> <p>A.J. van der Poorten, <a href="http://journals.cambridge.org/action/displayAbstract?fromPage=online&amp;aid=4860012" rel="nofollow">Transcendental entire functions mapping every algebraic number field into itself</a>, J. Austral.Math. Soc. <strong>8</strong> (1968), 192–193.</p> </blockquote> http://mathoverflow.net/questions/78526/maps-preserving-algebraic-numbers/78528#78528 Answer by Ricky Demer for Maps preserving algebraic numbers Ricky Demer 2011-10-19T03:24:05Z 2011-10-19T03:24:05Z <p>$f : \mathbb{C} \to \mathbb{C} \:$ given by $\: f(z) = 0 \:$ shows that $f$ can be a polynomial. <br><br> <a href="http://mathoverflow.net/questions/42460/is-a-real-power-series-that-maps-rationals-to-rationals-defined-by-a-rational-fun/42465#42465" rel="nofollow">This answer</a> show that $f$ can be holomorphic and not a polynomial. <br><br><br> Unless I'm missing something, that answer works just as well <br> to show that $f$ can be an <a href="http://en.wikipedia.org/wiki/Entire_function" rel="nofollow">entire function</a> and not a polynomial.</p>