Rings

“Hell is empty, all the devils and motherfuckers are here, and you are the greatest whore ever. Your cunt is a common shore, but your brain is a deserted island. It is full of palm trees, but no coconuts, Apocalypse, Anawim, #justtothepoint.

Motivation

In mathematics, addition (+), subtraction (-), multiplication (×), and division (:) are the four basic arithmetic operations. In Abstract Algebra, the key structures are groups, rings, fields, vector spaces, and modules.

1. If the natural numbers are taken to be {1,2,3,…}, then the natural numbers are closed and associative under addition, but there is no identity and no inverses. Therefore, the set ℕ of natural numbers is not a group.
2. The set of integers is a group under addition (ℤ, +). The set of integers is not a group under multiplication. It is closed, associative, and has the identity element 1. However, most of the integers do not have an inverse, e.g., 3, 4 or 7 has no inverses.

   In Abstract Algebra, subtraction is not an operation itself. It is just addition of the first term with the additive inverse of the second term, e.g., 4 -3 = 4 +(-3).

3. (ℚ, +), (ℝ, +), and (ℂ, +) are groups with respect to addition.
4. (ℚ*, ·), (ℝ*, ·), (ℂ*, ·) the sets of nonzero rational numbers, the set of nonzero real numbers, and the set of non-zero complex numbers are all groups under multiplication.

   Division is not considered a basic operation in abstract algebra. It is the multiplication of the dividend with the reciprocal or multiplicative inverse of the divisor, that is, a ÷ b = a x 1/b, 4 ÷ 5 = 4 x 1/5.

5. GL(n, ℝ), the set of invertible n × n matrices, is a group under matrix multiplication.
6. The permutation group S₃ is the group of all permutations of three elements, also known as the symmetric group on three.

In abstract algebra, subtraction is difficult to conceptualize, what does $(\begin{smallmatrix}1 & 2 & 3\\ 2 & 3 & 1\end{smallmatrix})-(\begin{smallmatrix}1 & 2 & 3\\ 3 & 2 & 1\end{smallmatrix})$ mean?

Definition.

💍 A ring R is a non-empty set with two binary operations, namely addition (a + b) and multiplication (ab), such that ∀ a, b, c ∈ R:

1. Both operations are closed: a + b ∈ R, a·b ∈ R.
2. Commutative under addition: a + b = b + a.
3. Associative under addition: (a + b) + c = a + (b + c).
4. There is an additive identity 0 ∈ R such that a + 0 = a, ∀ a ∈ R.
5. There are inverse elements for addition,, ∃-a ∈ R: a + (-a) = (-a) + a = 0.
6. Associative under product: a(bc) = (ab)c.
7. Multiplication is distributive over addition: a(bc) = ab + ac, (b + c)a = ba + ca.

Facts, types, and definitions

• A ring is an Abelian group under addition.

• A ring is not necessarily commutative under multiplication, e.g., GL(n, ℝ) is not commutative, $(\begin{smallmatrix}1 & 1\\ 0 & 0\end{smallmatrix})(\begin{smallmatrix}1 & 0\\ 1 & 0\end{smallmatrix})=(\begin{smallmatrix}2 & 0\\ 0 & 0\end{smallmatrix})≠(\begin{smallmatrix}1 & 0\\ 1 & 0\end{smallmatrix})(\begin{smallmatrix}1 & 1\\ 0 & 0\end{smallmatrix})=(\begin{smallmatrix}1 & 1\\ 1 & 1\end{smallmatrix})$. When the ring is commutative under multiplication, we say that the ring is a commutative ring, that is, ab = ba, ∀a, b ∈ R, e.g., 2ℤ, 3ℤ or xℤ[x] = {a₁x + a₂x² + ···| a_i ∈ ℤ} are commutative rings, but they do not have identity.

• A ring does not necessarily have an identity under multiplication. When it does have a multiplicative identity, it is called as ring with unity or 1, e.g., the ring of functions R = {ℝ → ℝ}, f + g: ℝ → ℝ, x → f(x) + g(x), and f · g: ℝ → ℝ, x → f(x) · g(x) is a ring with unity (1: ℝ → ℝ, x → 1, ∀f ∈ R, 1·f = f·1 = 1) and it has zero divisors, e.g, f(x) = x or x², and $g(x) = \begin{cases} 0, &x ≠ 0 \\ 1, &x = 0 \end{cases}$, f·g = 0, but f ≠ 0, g≠ 0.

• If a ring R is commutative with unity and ∀a, b ∈ R, ab = 0 ⇒ a= 0 or b = 0, then R is called an integral domain. In other words, an integral domain is a commutative ring with unity and no zero divisors.

Let’s solve the equation x² + x -6 = 0 in ℤ₁₂. The left side of this equation can be factor as (x-2)(x+3) = 0, so the solutions are 2 (x -2 = 0) and -3 (x -3 = 0). There is a big assumption that we are taken for granted: if the product of two terms is zero then common sense 🐒 says at least one of the two terms has to be zero to start with.

However, this is not always the case, e.g., x² +5x + 6 ≡ 0 (mod 12) ↭ (x +2)(x + 3) ≡ 0 (mod 12). Consider the following facts,

1. x + 2 ≡ 0 (mod 12) ⇒ x ≡ 10 (mod 12).
2. x + 3 ≡ 0 (mod 12) ⇒ x ≡ 9 (mod 12).
3. We can see that this equation has not two, but four solutions: 1, 6, 9 and 10, that’s because 3·4 ≡ 0 (mod 12) even though 3 ≇ 0 (mod 12) and 4 ≇ 0 (mod 12). In other words, ℤ₁₂ has zero divisors.

The reason why an integral domain is required to be a commutative ring with unity and non zero divisors is to be able to use the cancellation property, a·x = a·y (a ≠ 0) ⇒[(R, +) Abelian group] a·x -a·y = 0 ⇒[Distributive] a(x -y) = 0 ⇒[No zero divisors, a ≠ 0] x -y = 0 ⇒ x = y.

• A nonzero element of a commutative ring with unity does not necessarily have a multiplicative inverse. When it does, it is called a unit of the ring.
• A division ring is a ring with unity in which 0 ≠ 1 and every nonzero element has a multiplicative inverse, ∀a ∈ R, a ≠ 0, ∃a^-1: a^-1·a = a·a^-1 = 1_R.
• A commutative division ring is a field.

Examples

1. ℤ, ℚ, ℝ, and ℂ under the usual operations of addition and multiplication are rings.
2. Are there any finite rings? Yes, of course. ℤ/ℤ_n = {[0], [1], ··· [n-1]} under addition and multiplication modulo n are commutative rings with unity 1 -we are abusing notation, the unity is indeed [1]-. U(n) is the set of units. Counterexample: 2ℤ = {even integers}, 3ℤ are rings without multiplicative unity, 1 ∉ 2ℤ and 1 ∉ 3ℤ.
3. Given a positive integer n, the set of all n×n matrices with real coefficients is a ring under addition and matrix multiplication. GL(n, ℝ) is a ring with unity, $(\begin{smallmatrix}1 & 0 & ···& 0\\ 0 & 1 & ··· & 0\\ · & · & ··1·· & 0\\ 0 & 0 & ··· & 1\end{smallmatrix})$ but it is not commutative.
4. The set of all polynomials in the variable x with real coefficients, ℝ[x] = {a_nxⁿ + a_n-1x^n-1 + ··· + a₁x + a₀ | a_i ∈ ℝ}, +, .) is a commutative ring with the usual operations of addition and multiplication of polynomials. f(x) + g(x), f(x) -g(x), and f(x)·g(x) are polynomials, -f(x) = -a_nxⁿ + -a_n-1x^n-1 + ··· -a₁x -a₀, the additive and multiplicative identity are 0 (the constant polynomial with a_n = a_n-1 = ··· = a₁ = 0, a₀ = 0) and 1 (the constant polynomial with a_n = a_n-1 = ··· = a₁ = 0, a₀ = 1) respectively.

   More generally, let R be a ring (ℤ, ℝ, ℂ, ℤ/nℤ, GL(n, ℝ)), we can use it to build a new ring R[x], the ring of polynomials with coefficients in R.

5. Let R₁, R₂,…, R_n be a finite collection of rings, R₁⊕R₂⊕···⊕ R_n = {(a₁, a₂,…, a_n): a_i ∈ R_i} is a ring called the direct sum of R₁, R₂,…, R_n under component-wise addition and multiplication; that is, (a₁, a₂,…, a_n) + (b₁, b₂,…, b_n) = (a₁ + b₁, a₂ + b₂,…, a_n + b_n)
   (a₁, a₂,…, a_n)(b₁, b₂,…, b_n) = (a₁b₁, a₂b₂,…, a_nb_n)
6. ℤ/pℤ is a field, where p is a prime number.

• Exercise. If n is composite, then ℤ_n is a commutative ring with unity, but not an integral domain. ℤ_p with p prime is a field.

Proof.

If n is composite ⇒ ∃a, b ∈ ℤ: n = a·b, 1 < a, b < n ⇒ a, b ∈ ℤ_n, a ≠ 0, b ≠ 0, a·b = n =_ℤn 0 ∎ Some examples: 2, 3 ∈ ℤ₆, 2·3 = 0. 4, 3 ∈ ℤ₁₂, 4·3 = 0.

However, if p is a prime number, ℤ_p is a field, that is, every non-zero element has an inverse. It is obviously a commutative ring with unity, ∀x ∈ ℤ_p - {0} ⇒[x < p, p prime] gcd(x, p) = 1 ⇒ ∃a, b: ax + bp = 1, 1 ≡ ax + bp (mod p) ≡ ax (mod p), and therefore a is a multiplicative inverse of x.

Let’s prove that it is an integral domain. Suppose that there are two elements a, b ∈ ℤ_p with a·b = 0 ∈ ℤ_p ⇒ ab ≡ 0 (mod p) ⇒ p | ab ⇒[p is prime] p | a or p | b ⇒ a = [0] in ℤ_p or b = [0] in ℤ_p.

• Exercise. The quaternions form a non-commutative division ring. Q₈ ≤ GL₂(ℂ). Q₈ = {1, -1, i, -i, j, -j, k, -k} = ⟨-1, i, j, k | (-1⟩² = 1, i² = j² = k² = ijk = -1⟩.

The identity is 1 + (0i +0j +0k).
∀x ∈ Q₈, x = a +bi +cj +dk, let’s prove that x^-1 = $\frac{a-bi-cj-dk}{a^2+b^2+c^2+d^2}$. $(a +bi +cj +dk)\frac{a-bi-cj-dk}{a^2+b^2+c^2+d^2} = \frac{(a^2+b^2+c^2+d^2)+(-cd+cd)i + (-bd+bd)j + (-bc +cb)k}{a^2+b^2+c^2+d^2}$ = 1.

Bibliography

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.