“How much water should I drink” I asked the doctor (undergoing iodine therapy). “Have you ever seen a camel?” “I have seen them in tobacco packages”, I replied. “No, I mean, a real camel…” “No, I have not. Have you ever gone to Egypt? I thought camels have already become extinct because of lung cancer,” I reasoned.
“No, I have never gone to Egypt. I mean, a video about camels,” he was getting upset. “No, of course not. I only watch videos about people humiliating themselves, cute cats, and porn like everyone else, don’t you?” “OK, you win! Drink as much water as fucking possible. I am starting to consider very seriously to change careers from physician to rent boy,” he reply and went away, fuming. 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 ring R is called a principal ideal domain, PID for short, if it is an integral domain such that every ideal is principal, that is, has the form ⟨a⟩ = {ra | r ∈ R}, e.g., ℤ, F (the only ideals are the trivial ideal ⟨0⟩ and the whole field ⟨1⟩ = F) and F[x] (where F is a field).
Definitions. Suppose that we have two arbitrary elements from the ring, say a, b ∈ R.
A unit of a ring, u ∈ R, is an invertible element for the multiplication operation of the ring, that is, ∃v ∈ R, vu = uv = 1.
The concepts of irreducibles and primes are equivalent in the case of integers, but in general they are not.
R = ⟨x2, y2, xy⟩ is a subring of ℚ[x, y]. x2, y2, and xy are irreducible within R, but xy is not prime because xy | x2y2 and yet xy ɫ x2 and xy ɫ y2
Consider the ring $ℤ[\sqrt{-5}]$ = {a + b$\sqrt{-5}$ | a, b ∈ ℤ} and a function N, called the norm, N: $ℤ[\sqrt{-5}]$ → ℕ, N(a + b$\sqrt{-5}$) = a2 + 5b2 with the following properties:
Proof.
The concepts of irreducibles and primes are equivalent in the case of integers, but in general they are not. 2 + $\sqrt{-5}$ is irreducible but not a prime in $ℤ[\sqrt{-5}]$
It follows that the factorization of the element 9 into irreducible elements are not unique, 9 = 3·3 = (2 + $\sqrt{-5})(2 - \sqrt{-5})$ ⇒ The ring $ℤ[\sqrt{-5}]$ is not a unique factorization domain, that is, an integral domain in which every nonzero non-invertible element has a unique factorization.
Proof.
The concepts of irreducibles and primes are equivalent in the case of integers, but in general they are not. 1 + i$\sqrt{3}$ is irreducible but not a prime in $ℤ[i\sqrt{3}]$
Theorem. In an integral domain, every prime element is irreducible.
Proof.
Suppose that a is a prime in an integral domain D and a = bc. Is a irreducible? ↭ Is b or c a unit?
a is prime, a = bc ⇒ a | b or a | c. Let’s assume without any loss of generality a | b ⇒ ∃t ∈ D such that at = b.
An integral domain is a commutative ring with a multiplicative identity with no zero-divisors, therefore 1 ∈ D ⇒ b·1 = b = at = (bc)t =[Associativity] b(ct) ⇒ [In an integral domain, the cancellation property holds] 1 = ct ⇒ c is a unit ∎
Theorem. Let R be a ring which is also a principal ideal domain. Let a ≠ 0 be an element of R. Then, a is irreducible (it cannot be written as a non trivial product -except units- of two elements of the ring) if and only if a is prime. Hence, 💡primes and irreducible elements in a PID are the same.
Proof.
⇐) A ring R is a Principal ideal domain, PID for short, if it is an integral domain such that every ideal is principal ⇒ R is an integral domain ⇒[Theorem. In an integral domain, every prime element is irreducible.] In an integral domain, every prime is irreducible and we are done.
⇒) Let a be an irreducible element of a principal ideal domain. We claim that a is prime.
Let’s suppose that a | bc, a | b or a | c?
We are going to consider the ideal I defined as I = {ax + by | x, y ∈ D} ⇒ [D is a PID, therefore every ideal has the form ⟨a⟩ for some a ∈ D] I = {ax + by | x, y ∈ D} = ⟨d⟩.
a ∈ I = ⟨d⟩ ⇒ a = dr ⇒ [By assumption, a is irreducible] d or r is a unit, so we are faced with just two options,
Proposition. ℤ is a principal ideal domain.
Proof.
Let I be an ideal. If I is the trivial ideal, I = {0}, then I = ⟨0⟩. Let us assume that this is not the case, I ≠ {0}, so there is a smallest positive number on I, say n. We claim that I = ⟨n⟩, obviously ⟨n⟩ ⊆ I.
∀m ∈ I, by the Euclid’s division theorem, m = qn + r, 0 ≤ r < n ⇒ r = m - qn ∈ I ⇒ [By assumption n is the smallest positive number on I, 0 ≤ r < n] r = 0 ⇒ m = qn ⇒ m ∈ ⟨n⟩∎
Proposition. ℤ and F[x] where F is a field are principal ideal domains, but ℤ[x] is not a principal ideal domain.
Proof.
The ideal I = {f(x) ∈ ℤ[x] | f(0) is even} is not principal, that is, of the form ⟨h(x)⟩
By reduction to the absurd, I = {f(x) ∈ ℤ[x] | f(0) is even} = ⟨h(x)⟩
x, 2 ∈ I ⇒ ∃f(x), g(x)∈ ℤ[x] such that 2 = h(x)f(x) and x = h(x)g(x) ⇒ 0 = deg(2) = deg(h(x)) + deg(f(x)) ⇒ deg(h(x)) = deg(f(x)) = 0 ⇒ h(x) is a constant polynomial.
2 = h(x)f(x) ⇒ 2 = h(1)f(1) ⇒ h(1) = ±1 or ±2. Since the constant polynomial 1 is not in I, h(1) = ±2 ⇒ x = h(x)g(x) =[h(x) is a constant polynomial, h(1) = ±2] ±2g(x) for some g(x) ∈ ℤ[x] ⊥