Polynomial residues

If p is a prime then m^p \equiv m (mod p) for all integers m. So the polynomial f = X^p-X \in \mathbb{Z}[X] has the property that f(m) \equiv 0 (mod p) for all integers m. Since \mathbb{Z}_p is an integral domain, any polynomial in \mathbb{Z}[X] of degree d that is not divisible by p has at most d roots in \mathbb{Z}_p. The degree of a polynomial that vanishes identically on \mathbb{Z}_p must therefore be at least p, so X^p-X is also minimal in that sense. What can we say if we replace the prime p by just any positive integer n? In other words, what can we say about f \in \mathbb{Z}[X] if f(m) \equiv 0 (mod n) for all integers m? This question was studied by Aubrey Kempner in the 1920’s. Here are some of his results:

“Suppose f \in \mathbb{Z}[X] has degree d and vanishes identically on \mathbb{Z}_n. Let \dfrac{n}{d!} = \dfrac{m}{q} for coprime integers m and q. Then f is divisible by m.”

In particular if the coefficients of f have no common factor then n must divide d!, the factorial of its degree. For prime n this implies that f is divisible by n unless d is at least n and this is in accordance with the first paragraph above. To see that this bound is sharp for any n define the binomial polynomials \binom{x}{d} by

\displaystyle \binom{x}{d} = \dfrac{x (x-1) \dotsc (x-d+1)}{d!}.

The binomial polynomials take integer values at integer arguments (the number of ways to choose d elements from a set of x). Therefore, if n divides d! then d! \binom{x}{d} is a monic polynomial of degree d in \mathbb{Z}[X] that vanishes identically on \mathbb{Z}_n. The other way around, suppose f \in \mathbb{Z}[X] has degree d and f(m) \equiv 0 (mod n) for all integers m. Let \dfrac{m}{q} = \dfrac{n}{d!} as before. Then there exist coefficients c_0, \dotsc, c_d \in \mathbb{Z} such that

f(x) = n \displaystyle \sum_{k=0}^d c_k \binom{x}{k} = \frac{m}{q} \sum_{k=0}^d c_k d! \binom{x}{k}.

For each k the polynomial c_k d! \binom{x}{k} has integer coefficients and since m and q are coprime the equation above implies that m divides f.

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