be a monic polynomial with integer coefficients.

. Then in fact

.

The proof is quite simple using algebra. Suppose

; then there is a least positive integer

and a least positive integer

such that

and

are in

. It's easy to see that there is no integer

which divides all the coefficients of

(or else

could be made smaller), and similarily for

.

Take any prime

which divides the product

, and let

denote the canonical projection

. Then

. Since

is an integral domain, we must have

or

. But

and

or else all the coefficients of

or all the coefficients of

would be divisible by

, which contradicts the choice of

, as described above.