Let p(x), q(x) be elements of F[x] with p(x) and q(x) nonassociate irreducibles.?

Question

Show: p(x) and q(x) are relatively prime (meaning that greatest common divisor of p(x) and q(x) are 1)

Answer 1

Suppose that they were not relatively prime. Then this means that a (non-constant) polynomial r(x) must divide both p(x) and q(x). However, since p(x) is irreducible, this means that either r(x)=kp(x) or r(x)=k, where k∈F (i.e. k is an arbitrary constant), and we already stipulated that r(x) is nonconstant. Thus, r(x)=kp(x). However, by the same logic, r(x)=kq(x), and this implies p(x)=kq(x). Since F is a field, k has a multiplicative inverse, which means k is a unit of F[x], and so p(x)=kq(x) implies that p(x) and q(x) are associates. But the problem says p(x) and q(x) are nonassociates. Therefore p(x) and q(x) are relatively prime. Q.E.D.

N.B. relatively prime in a ring of polynomials means that GCD of two polynomials is a constant, not that it is equal to 1. All polynomials are divisible by a constant.

Answer 2

Again, pretty much by definition. If p(x) is irreducible, the it must be prime (assuming we're still talking about a commutative ring with identity), This means that p(x) is neither 0 nor a unit, and that if p(x) = r(x)s(x) (where r(x), s(x) ε F[x]) then either r(x) or s(x) must be a unit. Now recall that for p(x), q(x), r(x) ε F[x], if r(x)|p(x) and r(x)|q(x), then r(x)|(a(x)p(x) + b(x)q(x)) for all a(x), b(x) ε F[x] and the *only* way this can happen is if r(x) = 1. (Note that '1' is just the usual convention. A polynomial can be divided by *any* constant)


Doug

Answer 3

f = p + q