One of the things artificial intelligence algorithms had always found almost impossible to compute about people is their habit to continually stating and repeating the very obvious as if it were deeply profound and simultaneously, denying or ignoring obvious facts and truths -aka the elephant in the room-, lying to themselves, and even believing their own lies, Apocalypse, Anawim, #justtothepoint.
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:
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:
Assume that d is not a unit ⇒[R is a UFD] b = b_{1}b_{2}···b_{m}, c = c_{1}c_{2}···c_{n}, and d = d_{1}d_{2}···d_{p} where b_{i}, c_{j}, and d_{k} are all irreducibles ⇒ [bc = ad] b_{1}b_{2}···b_{m}c_{1}c_{2}···c_{n} = ad_{1}d_{2}···d_{p} ⇒[By the uniqueness of this factorization in R, R is a UFD] either a is associate with b_{i} for some i or a is associate with c_{j} for some j ⇒ a | b or a | c respectively ∎ If d is a unit, the argument is pretty similar.
a is associate with b_{i} for some i ⇒ ∃u unit, a = ub_{i} ⇒ b_{i} = u^{-1}a ⇒ a | b_{i} ⇒ a | b
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.
ℤ[i] = {a + bi| a, b ∈ ℤ} is a UFD.
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/a^{2}∈ ℤ ⇒ 1/a ∈ ℤ ⇒ a = ± 1 ⇒ u = ± 1 ⇒ 2 = ±$(1 + i\sqrt{3})$ ⊥
Notice that in ℂ, z^{-1} = $\frac{\overline z}{|z|}$
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:
Proof.
Let x ∈ ⟨b⟩, x ∈ ⟨a⟩?
x ∈ ⟨b⟩ ⇒ ∃y ∈ D: x = by ⇒ [b = ar] x = (ar)y =[Associativity] a(ry) ∈ ⟨a⟩∎
⟨b⟩ ⊆ ⟨a⟩ ⇒ b ∈ ⟨a⟩ ⇒ ∃r ∈ D: b = ar ⇒ 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⟩∎
⟨a⟩ ⊆ D is trivial. Let’s consider x ∈ D, x ∈ ⟨a⟩?
x ∈ D ⇒ x = x·1 = x·(a^{-1}a) = [Associativity, x ∈D, a^{-1}∈D ⇒ xa^{-1}∈D] (x·a^{-1})·a ∈ ⟨a⟩ = {ra | r ∈ D}. Therefore, D ⊆ ⟨a⟩ ⇒ D = ⟨a⟩
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
(⇐) 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
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∎