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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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Posts: 12,067
Registered: 7/15/05
Re: Does polynomial P need to be an affine mapping
Posted: May 9, 2013 4:24 PM
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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?


>>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

>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.


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