Recall that in an irreducible element in an integral domain is a non-zero, non-unit element that admits no decomposition as the product of two non-unit elements. In particular, irreducible elements in polynomial rings are called irreducible polynomials; examples include in , and or in .
Ideal-theoretic characterization of irreducibility. Let be an integral domain. An element is irreducible if, and only if, the ideal is non-zero, and is maximal among the principal ideals of . More precisely, we have , and if is any element such that , then either or .
(Proof is a good exercise, easy & omitted.)
Recall that in any integral domain, two elements are relatively prime if, whenever divides both elements (i.e. is a common divisor), then is a unit. It’s not too hard to show the following “ideal-theoretic” characterization of this concept: in an integral domain , the elements and are relatively prime if, and only if, implies .
Proposition. Let be an integral domain, and let and be two irreducible elements in . Either , or and are relatively prime.
Proof. Let be a common divisor of and . Then and are both elements of , hence and . Because and are irreducible, either and , or . In the first case, we obtain . In the second case, we obtain that is a unit, whence and are relatively prime.
Corollary. Let be a unique factorization domain with fraction field . Suppose and are two relatively prime polynomials in , and that is non-constant and irreducible. Then the inclusions of and in the larger ring are relatively prime polynomials.
Proof. By the previous proposition, it suffices to show that as ideals in . Suppose on the contrary that the ideals are equal, so that for some . We may put all monomials in over the same denominator and write for some non-zero and some . Hence we have the equation in . In particular, this means . Because is a UFD, the ideal generated by an irreducible element is prime, so either or . Since the latter is impossible by relative primality, we must have that divides in . But is a non-zero constant, so is a constant polynomial as well. This contradicts our hypothesis that was non-constant.
Notice that by Gauss’ Lemma on polynomials, the polynomial of the previous corollary is irreducible in as well.
EDIT A way better proof is made in a more recent post.