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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: 10,307
Registered: 7/15/05
Re: Does polynomial P need to be an affine mapping
Posted: May 9, 2013 7:58 AM
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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.
>
>Consider the set S of polynomials f in R[x] such that, for some
>nonempty intervals I,J of R, the set Q /\ J is a subset of
>f(Q /\ I).
>
>As Ken Pledger showed, if f is in S, f must have rational
>coefficients, that is, f must be an element of Q[x].
>
>It's clear that every element of S have must have degree at
>least 1.
>
>It's also clear that every element of Q[x] of degree 1 is in S.
>
>Moreover, if A,B in Q[x] are of degree 1, and f is in S, then
>(A o f o B) is in S.
>
>Claim all elements of S have degree exactly 1.
>
>Suppose otherwise, that is, suppose there exists f in S such
>that deg(f) = n, where n > 1.
>
>We proceed to transform f by a series of steps, where at each
>step, a new polynomial is obtained as a composition (sometimes
>on the left, sometimes on the right) of the previous polynomial
>with a polynomial in Q[x] of degree 1. Each new polynomial is
>still in S and the degree of each polynomial is still equal to
>n. The coefficients get progressively "better" in a sense that
>will become clear later.
>
>Let c_0 be the constant term of f,
>
>and let d be the lcm of the nonzero coefficients of f.


I meant:

and let d be the lcm of the denominators of the nonzero
coefficients of f.

>Let g(x) = A o B o f where B(x) = d*x and A(x) = x - d*(c_0)
>
>Then g is in S, still with degree n, but g has integer
>coefficients and zero constant term.
>
>Next, let p be a prime which does not divide the leading
>coefficient of g.
>
>Let h(x) = C o g o D where C(x) = (p^n)*x and D(x) = x/p.
>
>Then h is in S, still with degree n, still with integer
>coefficients, still with zero constant term, but in addition,
>all coefficients of h are multiples of p, except for the
>leading coefficient which is not a multiple of p.
>
>Since h is in S, there exist nonempty intervals I,J of R such
>that the set Q /\ J is a subset of h(Q /\ I).
>
>Now consider the set Y of rational numbers such that the
>denominator is not a multiple of p, while the numerator is a
>multiple of p but not a multiple of p^2.
>
>It's easily seen that Y is dense in R, hence Y /\ (Q /\ J)
>is nonempty.
>
>Let y1 be an element of Y /\ (Q /\ J).
>
>Write y1 = u/v where u,v are integers such that v is not a
>multiple of p, and u is a multiple of p but not a multiple
>of p^2.
>
>Since Q /\ J is a subset of h(Q /\ I), there exists x1 in
>Q /\ I such that h(x1) = y1.
>
>Then the polynomial
>
> k(x) = v*h(x) - u
>
>is a polynomial of degree n with integer coefficients and
>rational root x1.
>
>Since k has a rational root and deg(k) > 1, k is reducible in
>Q[x].
>
>Since k has integer coefficients, k is also reducible in Z[x].
>
>But k satisfies Eisenstein's criterion hence k is irreducible
>in Z[x], contradiction.
>
>This completes the proof.


quasi



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