Factorial domains

We now study integral domains in which the notion of irreducible element and prime element is the same.

4.9.1 Definition. An integral domain is said to be a factorial domain or a unique factorisation domain (UFD) or a factorial ring if every non-zero, non-invertible element a admits a factorisation into irreducible elements, that is to say, can be written as a product of irreducible elements

Moreover, this factorization is unique; that is if there is another factorisation into irreducibles

then

with m = n and, after reordering and relabeling is necessary with π_i and π_i' associates for i = 1,2,..,n.

Hence, this definition identifies the class of domains in which an analogue of the Fundamental Theorem of Arithmetic holds. The following proposition characterizes ring with unique factorization and it is often time handy in verifying that an integral domain is a unique factorization domain.

4.9.2 Proposition. An integral domain R with identity is a unique factorization domain if and only if the following properties are satisfied:

1. Every irreducible element is prime;

2. Given a sequence a₁, a₂, ..., a_i... of elements in R, such that for each i: a_i+1 | a_i, then there exists an index j such that, for each h,k ≥ j, a_h and a_k are associates.

Proof. Suppose that R is a unique factorization domain. We prove that (a) holds tue. Suppose a is irreducible, and let a | bc, i.e. bc = aq. By the unique factorization of every element of R we have

where b_i,c_i,q_i are irreducible. Owing to the uniqueness of the factorization, a must be associate to either b_i or c_i, and thus a divides either b or c, which is the condition required to be prime.

We now prove condition (b). Let a₁, a₂, ..., a_i... a sequence such that a_i+1|a_i for every i. Let n_i the number of irreducible factors in the unique factorization of each a_i. Then we have that

Since that the n_i make a non increasing sequence of positive integers, there must be an index j such that n_j = n_j+1 = n_j+2 = ⋅⋅⋅ , that is from a certain index j on, all a_i have the same number of irreducible factors. Given that a_i+1| a_i, i.e. a_i = a_i+1q, for i ≥ j, it must be that a_i+1 and a_i are associates.

For example: a₁ = 20 = 5*2; a₂ = 10 = 5*2; a₃ = 5 *1; a₄ = 5; a₅ = 5.

We suppose now valid (a) and (b). We show the existence of a factorization. Suppose by contradiction that there exists an a ∈ R which is not invertible and not a product of irreducible factors. In particular, there will be a non trivial factorization

where a₁ and b_i are not associates to a and at least one of the two factors is not the product of irreducible, say it is a_i. Then we have

where a₁ and b_i are not associates to a₁ and at least one of them is not the product of irreducible; Say it is a₂. Then we have

where a_i+1 and b_i+1 are not associates and at least one them is not the product of irreducible; Say it is a_i+1. This procedure is never ending. There exists thus a sequence (a_i) with a_i+1| a_i, such that the elements will never be associates. But this result contradicts (b).

We prove now the uniqueness of the factorization. Suppose two factorization of the same element are possible

with a₁, a₁ irreducible. We proceed by induction on the number n of the shortest factorization. For n = 1, (4.9.1) becomes

Being a₁ prime by condition (a), it divides one of the a_i say a₁'; this latter is an irreducible, so it must be that a₁ and a₁'are associates. If we cancel a₁ out, we have

hence all elements a_i for i = 1,2 ,.., m are invertible. Thus also the right side of 4.9.2 is composed by one irreducible factor a₁'. Suppose the theorem true for n − 1 and we prove it for n. Let b an element in R, having a factorization in n irreducible factors. Let

two factorization of b in irreducible factors. The left side with n irreducible factors; a₁ is a prime element dividing the left side, it must also divide one of the element on the right side, say a₁', since this latter is an irreducible factor, we have that a₁ and a₁' are associates. We cancel out this element and obtain

Now on the left side the number of irreducible factor is n − 1. By the induction hypothesis n − 1 = t − 1 (so that n = t) and, after possible reindexing, a_i' and a_i are associates for all i. This proves the theorem.□ (To understand why the use of 'associates' instead of equal element take 3·5 = (-3)(-5) both being prime factorization of 15 in Z)

4.9.3 Example. (Integral domain with 1 which is not a UFD). ℤ[√-3] is an example of integral domain, which is not a unique factorization domain. Consider the identity

these are two factorization of the same element. We already proved in example 4.8.8. that (1 + √−3) is irreducible we are left to show that 2 is irreducible too. Suppose that 2 = (a + b√−3)(c + d√−3), taking the norm, we have

The only possibility is that a² + 3b² = 1 or a² + 3b² = 4, since it's not possible the alternative a² + 3b² = 2. The first identity implies, b = 0 and a = ±1, that is a + b√−3 is invertible, in the second case c + d√−3 is invertible. Since 2 is not null and not invertible we proved that 2 is irreducible;

4.9.4 Proposition. If R is a UFD, every pair of elements a,b ∈ R, both non 0 have a GCD, d that is unique up to associates.

Proof. We may write a and b as

where p₁, ..., p_n are pairwise non-associates irreducibles and each h₁ and h₁ is a nonnegative integer.

is a gcd, where m_i = min{h_j,k_j}.□

We write an element as (-2)(-3)(-5) or we prefer (-1)(2)(3)(5) with the unit. you could also use (1)(-2)(-3)(-5) for the first.

We prove now that every principal domain is a UFD.

4.9.4 Proposition. Let R a principal ideal domain. Then it is a UFD.

Proof. We show that hypothesis (a) and (b) of Proposition 4.9.2 hold true.

1. Every irreducible element is prime. Let a ∈ R irreducible and let a|bc. If I is the ideal generated by a and b then there exists a d such that I = (a,b) = (d), hence we have

   a = dh,   b = dk

   Since a is irreducible, we have either d ~ a or d ~ 1. In the first case a|b, in the second case letting d = 1, we have 1 = ra + sb for some s,r ∈ R, then c = cra + sbc which implies a | c, hence, a prime.

2. Let {a_i} a sequence of elements in R such that a_i+1 | a_i for every i. Let I = {a₁, a₂, ..., a_i...} the ideal generated by all the elements a_i. By the hypothesis

   I = {a₁, a₂, ..., a_i...} = (d)

   for every d ∈ R. Since that every element of I is a finite combination of a_i, we have

   d = r₁a₁ + r₂a₂ + r₃a₃ + ⋅⋅⋅ + r_ja_j

   for some j ∈ ℕ. Since every a_i is multiple of the next element in the sequence the previous relation can be written, as

   d = r₁a_jā₁ + r₂a_jā₂ + r₃a_jā₃ + ⋅⋅⋅ + r_ja_j

   we then know that for every i ≥ j, a_i | d. On the other hand d | a_i for every i, which both imply that a_i ~ d, ∀ i ≥ j, that is for i ≥ j all the a_i are associates.□

   4.9.6 Proposition. Every Euclidean domain is a UFD.

   4.9.7 Definition. Let R be a factorial ring and let f(x) ∈ R[x] be a non-zero polynomial. The divisor, or content, of f(x) is a greatest common divisor d(f(x)) of its coefficients.

   d(f(x)) := GCD(a₀, a₁, ..., a_n)

   This definition makes sense, as we have seen just above a UFD admits GCD.

   4.9.8 Definition. A polynomial f(x) ∈ R[x], such that d(f(x)) ~ 1, is said to be primitive, that is if its content is invertible (the coefficients of the polynomial are coprime).

   4.9.9 Proposition. Let R a unique factorization domain. Then the product of two primitive polynomials in R[x] is still a primitive polynomial.

   Proof. The proof is identical to Gauss's Lemma, which is the special case R = ℤ. The only difference is the prime element should be replace with irreducible element and that in a unique factorisation domain the notions of prime element and irreducible element coincide.□

   4.9.10 Corollary. Let R a unique factorization domain. If f(x) and g(x) are in R[x], then

   d(f(x) g(x)) = d(f(x)) d(g(x))

   that is the content of the product of two polynomials is equal to the product of their contents.

   Proof. We can write

   f(x) = αf*(x),  g(x) = βg*(x)

   where α and β indicate the content of respectively f(x) and g(x), and f*(x) and g*(x) are primitive polynomials. Then

   f(x)g(x) = αβ f*(x)g*(x)   ⇒   d(f(x) g(x)) = αβ = d(f(x)) d(g(x)).□

   We already seen that there's a strict connection between a polynomial of ℤ[x] and its irreducibility (or reducibility) of the same polynomial on ℚ. This is because ℚ is the quotient field of ℤ. We have that

   f(x) ∈ ℤ[x] primitive and irreducible on ℤ   ⇐⇒   f(x) is irreducible on ℚ.

   Moreover if f(x) is a polynomial of ℤ[x] (also non primitive), its reducibility on ℚ, implies the irreducibility on ℤ. We can state an analogous result of Gauss's Theorem on R[x].

   4.9.9 Proposition. Let R be a unique factorization domain, and let f(x) ∈ R[x]. Then if f(x) can be decomposed in the product of two polynomials with coefficients in Q(R), then it can be decomposed in the product of two polynomials of the same degree with coefficients in R.

   Proof. It is identical to that provided for Gauss's Theorem; Just remember that any element of R can be written as quotient of elements of Q(R) as it was the case for ℚ and ℤ.

   We consider a field as a UFD, since any element is equal to a unit (itself) times an empty product of irreducible.

   4.9.11 Proposition. If R is a unique factorization domain then so is R[x].

   Proof. We must show that every nonzero non-unit f(x) ∈ R[x] has a unique factorization into irreducible. We write f(x) as follows

   f(x) = α ⋅ f*(x)   (4.9.3)

   We want to factorize each term on its own. We consider first the polynomials f*(x) thought as a polynomial in Q(R)[x], which is a UFD (since it's a field) then admits the unique factorization

   f*(x) = p₁(x)p₂(x)p₃(x) ⋅⋅⋅ p_n(x)

   with p_i ∈ Q(R)[x], irreducible on Q(R). Every p_i in turn can be written as in the proof of Gauss' theorem as

   p_i = d_i/m_i ⋅ p_i*(x),   d_i, m_i ∈ R

   with primitives p_i ∈ R[x]. Thus

   f*(x) = d₁/m₁ ⋅ d₂/m₂ ⋅⋅⋅ d_n/m₂ ⋅ p₁*(x) p₂*(x) ⋅⋅⋅ p_n*(x)

   from which

   m₁m₂⋅⋅⋅m_n f*(x) = d₁ ⋅ d₂ ⋅⋅⋅ d_n ⋅ p₁*(x) p₂*(x) ⋅⋅⋅ p_n*(x)

   The left-hand-side polynomial and the right-hand-side one have the same content. And as f*(x) and p₁*(x) ⋅⋅⋅ p_n*(x) are primitives, we have m₁m₂⋅⋅⋅ m_n = d₁ ⋅ d₂ ⋅⋅⋅ d_n, hence we can write

   f*(x) = p₁*(x)p₂*(x)p₃*(x) ⋅⋅⋅ p_n*(x)

   which represents a factorization of f*(x) in R in terms of irreducible, since p₁*(x) and p_i(x) are associates with the latter in turn irreducible; since they are primitives they are irreducible also in R.

   It is left to prove the uniqueness of such a factorization. Suppose

   f*(x) = p₁*(x)p₂*(x)p₃*(x) ⋅⋅⋅ p_n*(x) = q₁(x)q₂(x)q₃(x) ⋅⋅⋅ q_n(x)

   with q_i irreducible in R. Now, these are two factorisation also in Q(R) where we know the factorization to be unique; hence it must be n = r with q_i and p_i*(x) associates in Q(R)[x]. Since the q_i are primitives, as it can be easily verified, they must be also associates to the p_j*(x) in R, hence the factorization is unique.

   We consider now the factorization of α in (4.9.3). It easily verified that α can be factorized only as a product of elements in R, α = α₁α₂⋅⋅⋅α_s and in such case the factorization is unique by hypothesis. Thus

   f(x) = α = α₁α₂⋅⋅⋅α_s   ⋅   p₁*(x)p₂*(x)p₃*(x) ⋅⋅⋅ p_n*(x)

   thus the theorem is completely proved.□

   4.9.12 Corollary. Let R a unique factrorization domain, then R[x₁, x₂, ..., x_n] is a unique factorization domain.

   4.9.13 Example (UFD which is not principal). ℤ[x] is UFD but it is not principal; indeed consider the ideal I =(2,x) made by all polynomials of ℤ[x] with event constant term. There cannot be a polynomial f(x) ∈ ℤ[x] such that I = (f(x)) because we would have 2 = f(x)h(x) and x = f(x)k(x), with h(x),k(x) ∈ ℤ[x], which is impossible.

   Througout this chapter we encountered different types of rings: Euclidean domain, principal domains, UFDs. All these rings are integral domain with 1 and moreover in each of them there exists the GCD of two non 0 elements. To sum up we have the following

   𝓕 = {fields}

   𝓔 = {euclidean domains}

   𝓟 = {principal domains}

   𝓕 = {unique factorization domains}

   𝓓 = {integral domains with 1}

   We saw that a field is an Euclidean domain, an Euclidean domain is a principal domain, that a principal domain is a unique factorization domain. We have thus the following inclusions

   𝓕 ⊂ 𝓔 ⊂ 𝓟 ⊂ 𝓕 ⊂ 𝓓

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