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Divisibility in Integral 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 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).

Divisibility in Integral Domains

Definitions. Suppose that we have two arbitrary elements from the ring, say a, b ∈ R.

Examples

  1. N(x) = 0 ↭ x = 0.
  2. N(xy) = N(x)N(y) ∀x, y.
  3. u is a unit ↭ N(u) = 1.
  4. If N(x) is prime, then x is irreducible in $ℤ[\sqrt{-5}]$

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 $ℤ[\sqrt{-5}]$

  1. 2 + $\sqrt{-5}$ is irreducible. Suppose 2 + $\sqrt{-5} = xy$ ⇒ N(x)N(y) = N(xy) = N(2 + $\sqrt{-5})$ =[N(x) = a2 + 5b2]) = 22 + 5·12 = 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) = a2 + 5b2 = 3, and a fast inspection yields that there are no integers a, b satisfying this equality ⊥
  2. We claim that 2 + $\sqrt{-5}$ is not prime. (2 + $\sqrt{-5})(2 - \sqrt{-5})$ =[(a-b)(a+b) = a2-b2] 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 non-invertible element has a unique factorization.

  1. N(x) = 0 ↭ x = 0.
  2. N(xy) = N(x)N(y) ∀x, y.
  3. u is a unit ↭ N(u) = 1.
  4. If N(x) is prime, then x is irreducible in $ℤ[i\sqrt{3}]$

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 $ℤ[i\sqrt{3}]$

  1. 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) = a2 + 3b2]) = 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) = a2 + 3b2 = 2 ⇒ b = 0, a2=2 where a ∈ ℤ ⊥
  2. We claim that 1 + i$\sqrt{3}$ is not prime. (1 + i$\sqrt{3})(1 - i\sqrt{3})$ =[(a-b)(a+b) = a2-b2] 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 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,

  1. 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∎
  2. 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 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, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Field and Galois Theory, by Patrick Morandi. Springer.
  5. Michael Penn, Andrew Misseldine, and MathMajor, YouTube’s channels.
  6. Contemporary Abstract Algebra, Joseph, A. Gallian.
  7. MIT OpenCourseWare, 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007, YouTube.
  8. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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