Rational function over Q

Assume that f is a rational function with rational coefficients. Suppose moreover that f takes integer values at the integers. Then in fact f is a polynomial with rational coefficients.

Rational function over Q

2 thoughts on “Rational function over Q

  1. nachodarago says:

    The main ingredient to prove this is the following:
    Let F\in\mathbb{Z}[X] a polynomial with integer coefficients and consider the set S = \{p\text{ prime such that } p\mid F(n)\text{ for some }n\in\mathbb{Z}\}. Then S is an infinite set.

    The proof of this fact is easy. An sketch of a proof is the following: suppose S is finite and enumerate p_1,\ldots,p_r its elements. Consider F((p_1\cdots p_r)^\ell) for sufficiently large \ell such that F((p_1\cdots p_r)^\ell) is greater than the independent coefficient of F and also greater the valuation of each prime p_i of the independent coefficient. This will give the desired contradiction.

    Now, suppose f is a rational function which is not a polynomial, then there exist g,h\in\mathbb{Q}[X] such that f=g/h. Moreover, we can suppose that g,h are coprime and thus by Euclid algorithm, there are polynomials A,B\in\mathbb{Q}[X] such that Ag+Bh=1. By clearing denominators we obtain polynomials \tilde{A},\tilde{B},\tilde{g},\tilde{h}\in\mathbb{Z}[X] such that \tilde{A}\tilde{g}+\tilde{B}\tilde{h}=m for a certain integer m and \tilde{g}/\tilde{h} is an integral multiple of g/h. Since h(n)\mid g(n) for every n\in\mathbb{Z} we must have that every prime divisor of \tilde{h}(n) divides m. But we saw that this set must be infinite. This is a contradiction and we are done.

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  2. Let r= f/g be a rational function and suppose that it takes integer values and g has degree greater than f. Then r=0. Indeed, f/g\to 0, so f must have infinitely many roots, and f=0. Taking now f/g we can assume g has degree at most that of f. Writing f/g = A + S/g with A,S rational polynomials and S of degree less than g, we can assume by multiplying by a common denominator that A is a polynomial in integers. Then S/g takes integral values at the integers, and we conclude that it is zero by the above.

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