More questions about irreducible...?

Question

Let p(x) be an element of F[x] with p(x) non-constant.

Show: p(x) is irreducible in F[x], if and only if, for every g(x) in F[x], either p(x) divides g(x) or p(x) is relatively prime to g(x).

Answer 1

The only if part is easy - suppose the contrary: then p(x) shares a nonconstant factor other than itself with g(x). Call this factor q(x) - then q(x) is a nonconstant divisor of p(x), and p(x) is not irreducible.

For the if part: suppose that p(x) is not irreducible. Then it has a nonconstant factor q(x) such that q(x)≠p(x). Then q(x) is not relatively prime to p(x) (since q(x) is a common factor of both), neither does p(x) divide q(x) (since q(x) will have strictly smaller degree than p(x)).

Therefore, p(x) is irreducible iff ∀g(x) ∈ F[x], either p(x) | g(x) or gcd (p(x), g(x)) ∈ F. Q.E.D.

Answer 2

Pretty much by definition. If F is a commutative ring with an identity and p(x) ε F, then if p(x) is prime it is said to be 'irreducible over F' (in the case where F, itself, is a subring of another ring, call it R[x], then p(x) may, or may *not*, be irreducible over R)


Doug