Irreducible Polynomials and Relatively Prime Polynomials

Lemma 1:

Let g(x) be an irreducible polynomial with coefficients in a field K

Let h(x) be a polynomial with coefficients in a field K.

If g(x) does not divide h(x), then g(x) and h(x) are relatively prime

Proof:

(1) Let d(x) be the greatest common denominator for g(x) and h(x). [see Theorem 1, here for proof of the existence of d(x)]

(2) Since g(x) is irreducible, this means that d(x) must be of degree 0 or of the same degree as g(x). [see Definition 1, here]

(3) Assume that degree d(x) is nonzero.

(4) Then it follows that g(x)=C*d(x) where C is a constant. [since d(x) is a divisor of g(x) and since deg d(x) = deg g(x).]

(5) But then [1/C]*g(x) is a divisor of h(x) since d(x) is a divisor of h(x).

(6) But this is impossible since g(x) does not divide h(x).

(7) So we have a contradiction and we reject our assumption in step #3 and conclude that deg d(x) is 0.

(8) But then this means that g(x) and h(x) are relatively prime. [see Definition 3, here]

QED