the key fact is that p_{i}(x) isirreducible. in particular, p_{i}(x) is NOT a unit, and so deg(p_{i}(x)) > 0.

(irreducible elements are non-units by definition: we want to eliminate "trivial factorizations" like u = (u^{2})(u^{-1}), or even worse: u = (-1)(-1)(u)).

p_{i}(x) clearly divides the "whole product" (that is, f(x)), and what we really want to show is that it divides "the q-part".

so consider (1/v)(f(x)). we have:

(1/v)f(x) = (u/v)p_{1}(x)....p_{s}(x) = (p_{i}(x))((u/v)[p_{1}(x)...p_{i-1}(x)p_{i+1}(x)...p_{s}(x)])

which shows (1/v)f(x) is a product of p_{i}(x) and some other polynomial in F[x] (namely: (u/v)[p_{1}(x)...p_{i-1}(x)p_{i+1}(x)...p_{s}(x)]).

but: (1/v)f(x) = (1/v)(vq_{1}(x).....q_{r}(x)) = (v/v)(q_{1}(x)...q_{r}(x)) = q_{1}(x)...q_{r}(x),

so p_{i}(x) must divide this polynomial as well.

then we apply theorem 8.4.5 to show that p_{i}(x) divides some q_{j}(x) (theorem 8.4.5 is KEY; it proves "irreducibles are prime" in F[x]).

this shows that F[x] is a unique factorization domain....actually we didn't even need "the polynomial-ness" (if that's a word) of F[x], the ONLY thing we need is the "euclidean function" d (deg(f) in this case), so that we have a division algorithm to find the gcd of 2 elements of F[x]: ALL euclidean domains are unique factorization domains, the proof is pretty much the same, and in fact, is the SAME proof used to show integers have a unique (up to sign, and order of prime factors) factorization into prime numbers (which are precisely the "irreducibles" of the ring Z).

note we used the fact that F is a field in one crucial step...asserting that 1/v exists. it is important that neither u nor v be 0 (the units of F[x], though, ARE the units of F, that is: the non-zero elements of F).