    # Unique Factorization Domains

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Recall. Let R be a commutative ring with unity and D an integral domain. An integral domain is a commutative ring with a multiplicative identity (1 ≠ 0) with no zero-divisors, that is, ab = 0 ⇒ a = 0 or b = 0.

A unique factorization domain is a ring in which a statement similar to the fundamental theorem of arithmetic holds. Specifically, a UFD is an integral domain in which every non-zero non-unit element can be written as a product of prime or irreducible elements uniquely up to order and multiplication by units.

Definition. We say an integral domain D is a unique factorization domain, UFD for short, if:

1. Every non-zero non-unit element of D can be factored, expressed or written as a product of irreducibles of D.
2. This factorization into irreducibles is unique in the following sense. If a = up1p2···pm = wq1q2···qn, u and w units, then m = n and there exists a bijective map Φ:{1, 2, …, n} → {1, 2, …, n} such that pi is associated to qΦ(i) ∀i ∈ {1, 2, …, n}. In other words, ∃Φ ∈ Sn: pi = qΦ(i)ui and ui is a unit ∀i ∈ {1, 2, …, n}. The uniqueness part of the definition is hard to verify, which is why the following equivalent definition is pretty handy. A unique factorization domain is an integral domain R in which every non-zero element can be written as a product of a unit and prime elements of R.

Theorem. Let R be a UFD and a ∈ R, a is an irreducible element iff a is a prime element.

Proof.

⇐) Suppose a ∈ R, a is a prime element. A unique factorization domain is an integral domain R where every non-zero element that is not a unit can be factorized uniquely as a product of irreducible elements. In particular, UFD ⇒ Integral domain ⇒ [In an integral domain, every prime element is irreducible.] a is irreducible.

⇒) Suppose a ∈ R, a is irreducible, and a | bc. Claim: either a | b or a | c.

There are three options:

1. Suppose b = 0 ⇒ b = a·0 ⇒ a | b.
2. Suppose b is a unit, c = (b-1b)c = b-1·(bc) ⇒[a | bc] a | c
3. We can assume that b, c are non-zero, non-units, a|bc ⇒∃d ∈ R: bc = ad.

Assume that d is not a unit ⇒[R is a UFD] b = b1b2···bm, c = c1c2···cn, and d = d1d2···dp where bi, cj, and dk are all irreducibles ⇒ [bc = ad] b1b2···bmc1c2···cn = ad1d2···dp ⇒[By the uniqueness of this factorization in R, R is a UFD] either a is associate with bi for some i or a is associate with cj for some j ⇒ a | b or a | c respectively ∎ If d is a unit, the argument is pretty similar.

a is associate with bi for some i ⇒ ∃u unit, a = ubi ⇒ bi = u-1a ⇒ a | bi ⇒ a | b

# Examples

1. ℤ is a UFD by the Fundamental Theorem of Arithmetic, e.g., 30 = 2·3·5 = 2·(-3)·(-5), but -3 = (-1)·3, -5 = (-1)·5 and -1 is a unit.

2. F is field, F[x] is a UFD.
3. ℤ[i] = {a + bi| a, b ∈ ℤ} is a UFD.

4. ℤ[$i\sqrt{3}$] = {$a + bi\sqrt{3}$| a, b ∈ ℤ} is not a UFD. 4 = 2·2 = $(1 + i\sqrt{3})(1 - i\sqrt{3})$

To be a UFD, 2 = $(1 + i\sqrt{3})u$, where u is a unit, u = $a + bi\sqrt{3}$ ∈ ℤ[$i\sqrt{3}$] ⊆ ℂ ⇒ u-1 = $\frac{a - bi\sqrt{3}}{a^2+3b^2}$ ∈ ℤ[$i\sqrt{3}$] ⇒ $\frac{a}{a^2+3b^2}$ ∈ ℤ ⇒ b = 0 ⇒ a/a2∈ ℤ ⇒ 1/a ∈ ℤ ⇒ a = ± 1 ⇒ u = ± 1 ⇒ 2 = ±$(1 + i\sqrt{3})$ ⊥

Notice that in ℂ, z-1 = $\frac{\overline z}{|z|}$

5. ℤ[$\sqrt{5}$] ≤ ℝ is not a UFD. 4 = 2·2 = $(3 + \sqrt{5})(3- \sqrt{5})$. To be a UFD, $2 = (3 + \sqrt{5})u$ and u ∈ ℤ[$\sqrt{5}$] is a unit. say u = $(a + b\sqrt{5})$. $2 = (3 + \sqrt{5})u = (3 + \sqrt{5})(a + b\sqrt{5}) = (3a +5b)+(a +3b)\sqrt{5}$

It helps to consider that all these terms live in ℚ[$\sqrt{5}$], a 2-dimensional vector space with standard basis {1, $\sqrt{5}$}, where 2 = $(3a +5b)+(a +3b)\sqrt{5}$ ⇒ 3a + 5b = 2, a + 3b = 0 ⇒ a = -3b, 3(-3b) +5b = 2, -9b +5b = 2, -4b = 2, b = -2/4 = -1/2 ∈ ℤ ⊥ u = $(a + b\sqrt{5})$ ∉ ℤ[$\sqrt{5}$]⊥

Lemma. Let D be a principal integral domain or PID for short (an integral domain for which where every ideal is principal, i.e., can be generated by a single element), a,b ∈ D, then:

1. a|b ↭ ⟨b⟩ ⊆ ⟨a⟩. Recall ⟨a⟩ = {ra | r ∈ D}
2. a and b are associates ↭ ⟨b⟩ = ⟨a⟩
3. a is a unit ↭ ⟨a⟩ = D

Proof.

• (1⇒) Suppose a|b ⇒ ∃r ∈ D: b = ar. We claim that ⟨b⟩ ⊆ ⟨a⟩.

Let x ∈ ⟨b⟩, x ∈ ⟨a⟩?

x ∈ ⟨b⟩ ⇒ ∃y ∈ D: x = by ⇒ [b = ar] x = (ar)y =[Associativity] a(ry) ∈ ⟨a⟩∎

• (1⇐) Suppose ⟨b⟩ ⊆ ⟨a⟩, a|b?

⟨b⟩ ⊆ ⟨a⟩ ⇒ b ∈ ⟨a⟩ ⇒ ∃r ∈ D: b = ar ⇒ a | b

• (2⇒) Suppose a and b are associates ⇒ [Associates means that they differ by a unit] ∃u ∈D, u is unit, that is, an element with a multiplicative inverse such that a = bu ⇒ b | a ⇒ [1. a|b ↭ ⟨b⟩ ⊆ ⟨a⟩] ⟨a⟩ ⊆ ⟨b⟩

Besides a = bu ⇒ [u is a unit] au-1 = b ⇒ a | b ⇒ [1. a|b ↭ ⟨b⟩ ⊆ ⟨a⟩] ⟨b⟩ ⊆ ⟨a⟩ ⇒ [But previously, we have proved that ⟨a⟩ ⊆ ⟨b⟩] ⟨a⟩ = ⟨b⟩∎

• (2⇐) Suppose ⟨b⟩ = ⟨a⟩ ⇒ ⟨a⟩ ⊆ ⟨b⟩ and ⟨b⟩ ⊆ ⟨a⟩ ⇒ [1. a|b ↭ ⟨b⟩ ⊆ ⟨a⟩] b|a and a|b ⇒ ∃x, y ∈ D: b = ax and a = by ⇒ b = ax = (by)x = b(yx) ⇒ b = b·1 = b(yx) ⇒ [D is a PID, that is, an integral domain in which every ideal is principal, and the cancellation laws hold in integral domains] 1 = yx, hence x is a unit, so b = ax means that a and b are associates
• (3⇒) Suppose that a is a unit, that is, ∃a-1 ∈ D (a has a multiplicative inverse)

⟨a⟩ ⊆ D is trivial. Let’s consider x ∈ D, x ∈ ⟨a⟩?

x ∈ D ⇒ x = x·1 = x·(a-1a) = [Associativity, x ∈D, a-1∈D ⇒ xa-1∈D] (x·a-1)·a ∈ ⟨a⟩ = {ra | r ∈ D}. Therefore, D ⊆ ⟨a⟩ ⇒ D = ⟨a⟩

• (3⇐) Suppose D = ⟨a⟩, a is a unit?

D = ⟨a⟩. 1 ∈ D = ⟨a⟩ ⇒ ∃b ∈ D: 1 = ab ⇒ a is a unit∎

Theorem. Let D be a principal ideal domain or PID for short and consider the ideal ⟨p⟩ generated by p in D, ⟨p⟩ ≠ {0}, then ⟨p⟩ is a maximal ideal ↭ p is irreducible, that is, if p can be factored into two elements p = ab, one of them (a or b) needs to be a unit.

Proof:

(⇒) Suppose ⟨p⟩ is a maximal ideal and p = ab, we claim that a or b needs to be a unit. p = ab ⇒ a | p ⇒ [a|b ↭ ⟨b⟩ ⊆ ⟨a⟩] ⟨p⟩ ⊆ ⟨a⟩ ⊆ D ⇒ [⟨p⟩ is a maximal ideal] ⟨p⟩ = ⟨a⟩ or ⟨a⟩ = D

1. ⟨p⟩ = ⟨a⟩ ⇒ [2. a and b are associated ↭ ⟨b⟩ = ⟨a⟩] p and a are associates, p = ab ⇒ b is a unit ∎
2. ⟨a⟩ = D ⇒ [3. a is a unit ↭ ⟨a⟩ = D] a is a unit ∎

(⇐) Suppose p is irreducible. Claim: ⟨p⟩ is a maximal ideal.

Let consider an ideal I (it is generated by an element a ∈ D because D is a PID) such that ⟨p⟩ ⊆ ⟨a⟩ ⊆ D ⇒ [1. a|b ↭ ⟨b⟩ ⊆ ⟨a⟩.] a | p ⇒ ∃b ∈ D: p = ab ⇒ [By assumption, p is irreducible] a or b is a unit

1. a is a unit ⇒ [3. a is a unit ↭ ⟨a⟩ = D] ⟨a⟩ = D ∎
2. b is a unit and p = ab ⇒ p and a are associates ⇒ [2. a and b are associated ↭ ⟨b⟩ = ⟨a⟩] ⟨p⟩ = ⟨a⟩ ∎

Corollary. Let D be a principal ideal domain or PID for short. If p ∈ D is irreducible, then it is prime.

Proof.

Suppose p is irreducible ⇒[Theorem. Let D be a principal ideal domain or PID, ⟨p⟩ ≠ {0}, then ⟨p⟩ is a maximal ideal ↭ p is irreducible,] ⟨p⟩ is a maximal ideal and let’s suppose that p|ab and we claim that p|a or p|b.

p|ab ⇒ ∃r ∈ D: ab = pr ⇒ ab ∈ ⟨p⟩ ⇒ [Theorem. Let R be a UFD and a ∈ R, a is an irreducible element iff a is a prime element. ⟨p⟩ is maximal ⇒ p is irreducible ⇒ p is prime. Recall: A is a maximal ideal ⇒ R/A is a field ⇒ R/A is an integral domain ⇒ A is prime] a ∈ ⟨p⟩ or b ∈⟨p⟩ ⇒ p|a or p|b∎

# Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.
1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn (Abstract Algebra), and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. Andrew Misseldine: College Algebra and Abstract Algebra.
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