Unique Factorization Domains

A unique factorization domain (UFD) is an integral domain R such that every a ≠ 0 in R can be written

a = up_{1} … p_{k}

where u is a unit and p_{1}, p_{2},…,p_{k} are primes in R.

Note that the factorization is essentially unique (by the same argument used to prove uniqueness of factorization in PIDs).

Note also that if R is a UFD, any finite collection a_{1}, …, a_{n} ∈ R has a highest common factor. For we can take out prime factors until we write a_{i} = rb_{i} where the b_{1}, …, b_{n} have no proper factors in common. Then r is the (unique up to units) highest common factor. We write r = hcf(a_{1}, …, a_{n}), but note that unless R is a PID we will not in general have r = λ_{1}a_{1} + … + λ_{n}a_{n} for λ_{i} ∈ R.

Observe that if R is an integral domain then R is a UFD iff it satisfies the following condition:

Every a ≠ 0 in R can be written as a product a = up_{1} … p_{k} where u is a unit and p_{i} is irreducible for each i. Moreover, this factorization is essentially unique in the sense that if we also have a = vq_{1} … q_{l} then k = l and, after renumbering the q_{i}, we have p_{i} ∼ q_{i} for all i.

Furthermore, every irreducible in a UFD is prime.

Our aim is to prove that the ring of polynomials over a unique factorization domain is itself a UFD. Along the way, we shall prove Gauss’ Lemma that the product of primitive polynomials in a UFD is itself primitive. (Recall that a polynomial over a UFD is said to be primitive if the greatest common divisor of its coefficients is 1.) The proof of the main theorem takes R a UFD and considers F its field of fractions. Now, R[X] is a subring of F [X] and F [X] is a PID and so is a UFD. We show that factorization in R[X] is determined by

(i) factorization in F [X]; and

(ii) factorization in R.

Lemma 1 (Gauss). A product of primitive polynomials is primitive.

Proof. Suppose

f(X) = a_{n}X^{n} + a_{n−1}X^{n−1} + … + a_{0}

and

g(X) = b_{m}X^{m} + b_{m−1}X^{m−1} + … + b_{0}

are primitive. Take p prime and i, j largest such that p ∤ a_{i} and p ∤ b_{j}. Now c_{i+j}, the (i + j)th coefficient of the product fg is

c_{i+j} = a_{i}b_{j} + (a_{i+1}b_{j−1} + …) + (a_{i−1}b_{j+1} + …),

a sum of a_{i}b_{j} and terms divisible by p. As p is prime, p ∤ a_{i}b_{j} and so p ∤ c_{i+j}. As p was an arbitrary prime, this shows that the product is primitive.

Lemma 2. (i)If u is a unit in R then u is a unit in R[X].

(ii) If p is a prime in R then p is a prime in R[X].

(iii) Suppose f(X) is a primitive polynomial in R[X] which is irreducible and so prime in F [X]. Then f is prime in R[X].

Proof. (i) Suppose u is a unit in R. Then there exists v ∈ R such that uv = 1. But then uv = 1 in R[X] so u is a unit in R[X].

(ii) The argument is the same as that used to prove Gauss’ Lemma. Suppose p is prime in R. To show that p is prime in R[X], it is enough to show that if

p ∤ a_{n}X^{n} + a_{n−1}X^{n−1} + … + a_{0}

and

p ∤ b_{m}X^{m} + b_{m−1}X^{m−1} + … + b_{0}

then

p ∤ (a_{n}X^{n} + a_{n−1}X^{n−1} + … + a_{0})(b_{m}X^{m} + b_{m−1}X^{m−1} + … + b_{0}).

So suppose

p ∤ a_{n}X^{n} + a_{n−1}X^{n−1} + … + a_{0}

and

p ∤ b_{m}X^{m} + b_{m−1}X^{m−1} + … + b_{0}.

Pick i greatest such that p ∤ a_{i} and j greatest such that p ∤ b_{j}. Then, writing c_{k} for the coefficient of X^{k} in the product,

p ∤ c_{i+j} = ∑_{r+s=i+j} a_{r}b_{s}

and so

p ∤ c_{n+m}X^{n+m} + c_{n+m−1}X^{n+m−1} + … + c_{0}.

(iii) Suppose f(X)|g(X)h(X) in R[X]; then f|gh in F [X] and so f|g or f|h in F [X]. Assume wlog f|g in F [X], so we have g = fk with k ∈ F [X]. Write g = ag̃ and k = \frac{b}{c} k̃ where a, b, c ∈ R and g̃, k̃ ∈ R[X] are primitive. Then we have ag̃ = \frac{b}{c}fk̃ or acg̃ = bfk̃. Now, as f, g̃ and k̃ (and hence fk̃ by Gauss) are primitive, we have ac ∼ b, i.e. b = uac for some unit u. Then g̃ = ufk̃ and so g = ag̃ = fuak̃ and so f|g in R[X]. This show that f is prime in R[X].

Theorem 3. If R is a UFD then so is R[X], the ring of polynomials over R.

Proof. Take a (non-zero) polynomial f ∈ R[X]. We can factorize it into irreducibles=primes in F [X] (as F [X] is a PID and so a UFD), and we may as well take the irreducibles to be primitive polynomials in R[X]. So we can write

f(X) = \frac{r}{s}g_{1}(X)g_{2}(X) … g_{k}(X)

where g_{i} ∈ R[X] is primitive and irreducible in F [X] (and so prime in R[X]). Now f(X) = af̃(X) for some a ∈ R and primitive f̃ ∈ R[X], and so

af̃ = \frac{r}{s}g_{1}g_{2} … g_{k}

or

asf̃ = rg_{1}g_{2} … g_{k}

so, as f̃, g_{1}, g_{2}, …, g_{k} are primitive, as ∼ r and we can write r = uas where u ∈ R is a unit. Then

f = uag_{1}g_{2} … g_{k}.

But now we can factorize a = va_{1}a_{2} … a_{l} where v ∈ R is a unit and a_{i} ∈ R is prime, and so we have a complete factorization:

f = \underbrace{(uv)}_{unit by 2(i)} \underbrace{a_{1}a_{2} … a_{l}}_{primes by 2(ii)} \underbrace{g_{1}g_{2} … g_{k}}_{primes by 2(iii)} .