If p is a prime then (mod p) for all integers m. So the polynomial has the property that (mod p) for all integers m. Since is an integral domain, any polynomial in of degree d that is not divisible by p has at most d roots in . The degree of a polynomial that vanishes identically on must therefore be at least p, so 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 if (mod n) for all integers m? This question was studied by Aubrey Kempner in the 1920’s. Here are some of his results:

*“Suppose has degree d and vanishes identically on . Let 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 by

.

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 is a monic polynomial of degree d in that vanishes identically on . The other way around, suppose has degree d and (mod n) for all integers m. Let as before. Then there exist coefficients such that

.

For each k the polynomial has integer coefficients and since m and q are coprime the equation above implies that m divides f.