
Re: Does polynomial P need to be an affine mapping
Posted:
May 9, 2013 9:40 AM


quasi wrote:
> quasi wrote:
>>>Let P be a polynomial with real coefficients. Suppose there >>>are nonempty intervals I and J such that P maps surjectively >>>the rationals of I into the rationals of J. Does this imply >>>P is an affine mapping? >> >>Yes.
My argument is similar to yours, but possibly simpler.
As has been established, P(x) has rational coefficients. Choose a prime p which does not divide any of the coefficients. Consider x = n/p^e where n is not divisible by p. Then it is easy to see that P(x) = m/p^{re}, where p does not divide m, and r is the degree of P. It follows that if x is rational, and r > 1, P(x) cannot be of the form c/p (with c not divisible by p).
 Timothy Murphy email: gayleard /at/ eircom.net tel: +353862336090, +35312842366 smail: School of Mathematics, Trinity College Dublin

