“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 zerodivisors, 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.
 We say that a divides b, and write ab, if ∃c ∈ R with b = ac.
 We say that a and be are associates if a = ub where u is a unit of D. In the ring of integers ℤ, the units are precisely the integers 1 and 1. If a, b are associates, then a = 1·b = b or a = (1)·b = b. Thus, two integers are associates in ℤ if and only if the’re the same up to sign.
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.
 A non zero, non unit element “a” of an integral domain D is irreducible if when a = bc, then b or c is a unit. In other words, it cannot be written as a non trivial (except by written it as a product of units) product of two elements of the ring.
 A non zero, non unit element a of an integral domain D is prime if when abc implies ab or ac, that is, if it divides a product, then it divides (at least) one of the terms of the product.
The concepts of irreducibles and primes are equivalent in the case of integers, but in general they are not.
Examples

R = ⟨x^{2}, y^{2}, xy⟩ is a subring of ℚ[x, y]. x^{2}, y^{2}, and xy are irreducible within R, but xy is not prime because xy  x^{2}y^{2} and yet xy ɫ x^{2} and xy ɫ y^{2}

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$\sqrt{5}$) = a^{2} + 5b^{2} with the following properties:
 N(x) = 0 ↭ x = 0.
 N(xy) = N(x)N(y) ∀x, y.
 u is a unit ↭ N(u) = 1.
 If N(x) is prime, then x is irreducible in $ℤ[\sqrt{5}]$
Proof.
 (1⇒) N(x) = 0 ⇒ a^{2} + 5b^{2} = 0 ⇒ [a^{2} ≥ 0, b^{2} ≥ 0] a^{2} = b^{2} = 0 ⇒ a = b = 0 ⇒ x = 0.
 (2) x = a + b$\sqrt{5}$, y = c + d$\sqrt{5}$ ⇒ xy = (ac 5bd) + (ad + bc)$\sqrt{5}$ ⇒ N(xy)= (ac 5bd)^{2} + 5(ad + bc)^{2} = a^{2}c^{2} + 5^{2}b^{2}d^{2} 10acbd + 5a^{2}d^{2} + 5b^{2}c^{2} + 10adbc= a^{2}(c^{2} +5d^{2}) + b^{2}(5c^{2} + 25d^{2}) = a^{2}(c^{2} +5d^{2}) + 5b^{2}(c^{2} + 5d^{2}) = (a^{2} + 5b^{2})(c^{2} + 5d^{2}) = N(x)N(y)
 (3⇒) Suppose u is a unit ⇒ ∃u^{1} ∈ $ℤ[\sqrt{5}]$ : u·u^{1} = u^{1}u = 1 ⇒ N(u·u^{1}) = N(1) = 1 ⇒ [N(xy) = N(x)N(y) ∀x, y] N(u)N(u^{1}) = 1 but all of these terms are naturals ⇒ N(u) = 1.
 (3⇐) Suppose u = a + b$\sqrt{5}$ and N(u) = a^{2} + 5b^{2} = 1 ⇒ b = 0, a = ± 1 ⇒ u = ± 1 and those are both units.
 (4) N(x) is prime, suppose x = yz. N(x) = [N(xy) = N(x)N(y) ∀x, y] N(y)N(z) and N(x) is prime in ℕ ⇒ Without losing generosity (you may need to swap or rename y and z), N(y) = 1 and N(z) = N(x) ⇒ [By 3] y is a unit, and therefore x is irreducible.
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 $ℤ[\sqrt{5}]$
 2 + $\sqrt{5}$ is irreducible. Suppose 2 + $\sqrt{5} = xy$ ⇒ N(x)N(y) = N(xy) = N(2 + $\sqrt{5})$ =[N(x) = a^{2} + 5b^{2}]) = 2^{2} + 5·1^{2} = 9. If x or y is a unit, then we are done∎. Let’s suppose that x and y are not units ⇒ their norms are not equal to one ⇒[N(x)N(y) = 9, N(x), N(y) ∈ ℕ] N(x) = N(y) = 3, but this is impossible because N(x) = a^{2} + 5b^{2} = 3, and a fast inspection yields that there are no integers a, b satisfying this equality ⊥
 We claim that 2 + $\sqrt{5}$ is not prime. (2 + $\sqrt{5})(2  \sqrt{5})$ =[(ab)(a+b) = a^{2}b^{2}] 4 + 5 = 9 ⇒ $2 + \sqrt{5}
3·3$, but $2 + \sqrt{5} ɫ 3$
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 noninvertible element has a unique factorization.
 Consider the ring $ℤ[i\sqrt{3}]$ = {a + bi$\sqrt{3}$  a, b ∈ ℤ} and a function N, called the norm, N: $ℤ[i\sqrt{3}]$ → ℕ, N(a + bi$\sqrt{3}$) = a^{2} + 3b^{2} with the following properties:
 N(x) = 0 ↭ x = 0.
 N(xy) = N(x)N(y) ∀x, y.
 u is a unit ↭ N(u) = 1.
 If N(x) is prime, then x is irreducible in $ℤ[i\sqrt{3}]$
Proof.
 (1⇒) N(x) = 0 ⇒ a^{2} + 3b^{2} = 0 ⇒ [a^{2} ≥ 0, b^{2} ≥ 0] a^{2} = b^{2} = 0 ⇒ a = b = 0 ⇒ x = 0.
 (2) x = a + bi$\sqrt{3}$, y = c + di$\sqrt{3}$ ⇒ xy = (ac 3bd) + (ad + bc)i$\sqrt{3}$ ⇒ N(xy)= (ac 3bd)^{2} + 3(ad + bc)^{2} = […] = (a^{2} + 3b^{2})(c^{2} + 3d^{2}) = N(x)N(y)
 (3⇒) Suppose u is a unit ⇒ ∃u^{1} ∈ $ℤ[i\sqrt{3}]$ : u·u^{1} = u^{1}u = 1 ⇒ N(u·u^{1}) = N(1) = 1 ⇒ [N(xy) = N(x)N(y) ∀x, y] N(u)N(u^{1}) = 1 but all of these terms are naturals ⇒ N(u) = 1.
 (3⇐) Suppose u = a + bi$\sqrt{3}$ and N(u) = a^{2} + 3b^{2} = 1 ⇒ b = 0, a = ± 1 ⇒ u = ± 1 and those are both units.
 (4) N(x) is prime, suppose x = yz. N(x) = [N(xy) = N(x)N(y) ∀x, y] N(y)N(z) and N(x) is prime in ℕ ⇒ Without losing generosity (you may need to swap or rename y and z), N(y) = 1 and N(z) = N(x) ⇒ [By 3] y is a unit, and therefore x is irreducible.
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 $ℤ[i\sqrt{3}]$
 1 + i$\sqrt{3}$ is irreducible. Suppose 1 + i$\sqrt{3} = xy$ ⇒ N(x)N(y) = N(xy) = N(1 + i$\sqrt{3})$ =[N(x) = a^{2} + 3b^{2}]) = 4. If x or y is a unit, then we are done∎. Let’s suppose that x and y are not units ⇒ their norms are not equal to one ⇒[N(x)N(y) = 4, N(x), N(y) ∈ ℕ] N(x) = N(y) = 2, but this is impossible because N(x) = a^{2} + 3b^{2} = 2 ⇒ b = 0, a^{2}=2 where a ∈ ℤ ⊥
 We claim that 1 + i$\sqrt{3}$ is not prime. (1 + i$\sqrt{3})(1  i\sqrt{3})$ =[(ab)(a+b) = a^{2}b^{2}] 4 = 2·2 ⇒ $1 + i\sqrt{3}
2·2$, but $1 + i\sqrt{3} ɫ 2$
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 zerodivisors, 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,
 Let’s assume that d is a unit, then I = {ax + by  x, y ∈ D} = ⟨d⟩ =[d is a unit, ∃d^{1}∈ D, d·d^{1} = 1 ∈ ⟨d⟩ and consequently I = ⟨d⟩ = D] D ⇒ ∃x, y ∈ D: ax + by = 1 ⇒ c = cax + bcy ⇒ [a  cax, a  bcy (By assumption, a  bc)] a  c∎
 Let’s assume that r is a unit, then I = {ax + by  x, y ∈ D} = ⟨d⟩ =[a = dr ⇒ ⟨a⟩⊆⟨d⟩. Besides, r is a unit, ∃r^{1}∈ D, d = ar^{1} ∈ ⟨a⟩ ⇒ ⟨d⟩⊆⟨a⟩] ⟨a⟩, b ∈ I = ⟨a⟩ ⇒ a  b∎
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] ⊥
Bibliography
This content is licensed under a Creative Commons AttributionNonCommercialShareAlike 4.0 International License. This post relies heavily on the following resources, specially on
NPTELNOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.
 NPTELNOC IITM, Introduction to Galois Theory.
 Algebra, Second Edition, by Michael Artin.
 LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
 Field and Galois Theory, by Patrick Morandi. Springer.
 Michael Penn, Andrew Misseldine, and MathMajor, YouTube’s channels.
 Contemporary Abstract Algebra, Joseph, A. Gallian.
 MIT OpenCourseWare, 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007, YouTube.
 Calculus Early Transcendentals: Differential & MultiVariable Calculus for Social Sciences.