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Every binomial polynomial is irreducible but not prime in the ring of integer-valued polynomials over rational integers

Contents

Statement

In the ring of integer-valued polynomials over rational integers, every binomial polynomial, i.e., every polynomial of the form:

\binom{x}{r}

is an irreducible element but not a prime element.

Proof

The binomial polynomial is irreducible

For this, we need a lemma: the largest number that we can guarantee divides the value of any polynomial of the form \prod_{i=1}^r (x - a_i) for all x \in \mathbb{Z} is r!.

With this lemma, we observe that if \binom{x}{r} is expressed as a product of polynomials of smaller degree, each of them is of the form (p_i/q_i) times a product of linear polynomials. In each of the cases, we get that q_i is bounded from above by the factorial of the number of linear polynomials in that factor. Thus, the product of all the q_is is bounded by the product of the factorials, which is strictly less than r!, a contradiction.

The binomial polynomial is not prime

For r > 1, we have:

r!\binom{x}{r} = x(x-1)(x-2) \dots (x-r + 1).

This yields:

\binom{x}{r} | x(x-1)(x-2) \dots (x-r+1).

On the other hand, we have that since r > 1, \binom{x}{r} does not divide any of the linear polynomials x - i.

For r = 1, use that:

x | x(x-1) = 2\binom{x}{2}

but x does not divide \binom{x}{2}, since the ratio, (x-1)/2, is not an integer-valued polynomial.