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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: 10,232
Registered: 7/15/05
Re: Does polynomial P need to be an affine mapping
Posted: May 9, 2013 5:40 AM
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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.

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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