Continuous extensions reals and to p-adic numbers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T08:55:36Z http://mathoverflow.net/feeds/question/73792 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73792/continuous-extensions-reals-and-to-p-adic-numbers Continuous extensions reals and to p-adic numbers ε-δ 2011-08-26T18:57:41Z 2011-08-29T17:16:28Z <p>Assume $f\colon \mathbb Q\to \mathbb Q$ is a function which admits continuous extensions</p> <ul> <li>$f_0\colon\mathbb R\to \mathbb R$ and </li> <li>$f_p\colon \mathbb Q_p\to \mathbb Q_p$ for each prime $p$.</li> </ul> <blockquote> <p>Is it true that $f$ is a polynomial?</p> </blockquote> <p>I guess the answer is <strong>no</strong>, but I do not see a counterexample. </p> http://mathoverflow.net/questions/73792/continuous-extensions-reals-and-to-p-adic-numbers/73803#73803 Answer by Lavender Honey for Continuous extensions reals and to p-adic numbers Lavender Honey 2011-08-26T21:41:02Z 2011-08-26T21:41:02Z <p>The answer is no, and one can essentially use the same construction as in the answer: <a href="http://mathoverflow.net/questions/42460" rel="nofollow">http://mathoverflow.net/questions/42460</a></p> <p>Specifically, enumerate the non-zero rationals $\{r_1,r_2, \ldots\}$ in some way. Now consider the function: $$f(x) = \sum_{n=1}^{\infty} c_n x^{n^2} \prod_{i=1}^{n} (x - r_i).$$ If $c_n \in \mathbf{Q}$, then this is a well defined function from rationals to rationals. On the other hand, $f(x)$ converges to an analytic function in $\mathbf{Q}_v$ if and only if the coefficients of this power series converge to zero fast enough. Since the coefficients of the power series in the range $k = n^2$ to $k &lt; (n+1)^2$ are simply the cofficients of $c_n x^{n^2} \prod_{i=1}^{n} (x - r_i)$, this can be ensured by forcing these coefficients to be very highly divisible by the first $n$ primes, and small in the archimedean sense (by including a very very large prime in the denominator). </p>