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

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## 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 $q_i$s 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.