quasi
Posts:
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Registered:
7/15/05


Re: Does polynomial P need to be an affine mapping
Posted:
May 9, 2013 7:58 AM


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

