quasi
Posts:
11,740
Registered:
7/15/05


Re: Does polynomial P need to be an affine mapping
Posted:
May 9, 2013 4:24 PM


quasi wrote: >Timothy Murphy wrote: >>quasi wrote: >>>quasi wrote: >>>>steinerartur 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.
Moreover, by applying affine transformations, we can assume P has integer coefficients.
>>Choose a prime p which does not divide any of the >>coefficients. > >It suffices for p to not divide the leading coefficient.
I think my comment above is valid provided P has integer coefficients (which is ok to assume).
>>Consider x = n/p^e where n is not divisible by p. > >Where in this context, n is allowed to be rational. > >>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. > >Yes, where m too is allowed to be rational. > >>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). > >Nice, and definitely simpler.
quasi

