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Topic: Does polynomial P need to be an affine mapping
Replies: 7   Last Post: May 10, 2013 6:45 AM

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quasi

Posts: 10,188
Registered: 7/15/05
Re: Does polynomial P need to be an affine mapping
Posted: May 9, 2013 3:15 PM
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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.
>
>Choose a prime p which does not divide any of the
>coefficients.


It suffices for p to not divide the leading coefficient.

>Consider x = n/p^e where n is not divisible by p.

Where in this context, n is allowed to 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



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