“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.
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.
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).
(ℚ, +), (ℝ, +), and (ℂ, +) are groups with respect to addition.
(ℚ*, ·), (ℝ*, ·), (ℂ*, ·) the sets of nonzero rational numbers, the set of nonzero real numbers, and the set of nonzero 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.
GL(n, ℝ), the set of invertible n × n matrices, is a group under matrix multiplication.
The permutation group S_{3} is the group of all permutations of three elements, also known as the symmetric group on three.
+

x
÷
ℕ
✅
❌ 2 3
✅
❌
ℤ
✅
✅
✅
❌ 2/3
ℝ
✅
✅
✅
✅ except by zero
M_{mxn}(ℝ)
✅
✅
❌ (rxs)·(sxt)
❌
ℝ[x]
✅
✅
✅
❌
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 nonempty set with two binary operations, namely addition (a + b) and multiplication (ab), such that ∀ a, b, c ∈ R:
Both operations are closed: a + b ∈ R, a·b ∈ R.
Commutative under addition: a + b = b + a.
Associative under addition: (a + b) + c = a + (b + c).
There is an additive identity 0 ∈ R such that a + 0 = a, ∀ a ∈ R.
There are inverse elements for addition,, ∃a ∈ R: a + (a) = (a) + a = 0.
Associative under product: a(bc) = (ab)c.
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] = {a_{1}x + a_{2}x^{2} + ··· 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 x^{2}, 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 x^{2} + x 6 = 0 in ℤ_{12}. The left side of this equation can be factor as (x2)(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., x^{2} +5x + 6 ≡ 0 (mod 12) ↭ (x +2)(x + 3) ≡ 0 (mod 12). Consider the following facts,
x + 2 ≡ 0 (mod 12) ⇒ x ≡ 10 (mod 12).
x + 3 ≡ 0 (mod 12) ⇒ x ≡ 9 (mod 12).
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, ℤ_{12} 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
ℤ, ℚ, ℝ, and ℂ under the usual operations of addition and multiplication are rings.
Are there any finite rings? Yes, of course. ℤ/ℤ_{n} = {[0], [1], ··· [n1]} 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.
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.
The set of all polynomials in the variable x with real coefficients, ℝ[x] = {a_{n}x^{n} + a_{n1}x^{n1} + ··· + a_{1}x + a_{0}  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_{n}x^{n} + a_{n1}x^{n1} + ··· a_{1}x a_{0}, the additive and multiplicative identity are 0 (the constant polynomial with a_{n} = a_{n1} = ··· = a_{1} = 0, a_{0} = 0) and 1 (the constant polynomial with a_{n} = a_{n1} = ··· = a_{1} = 0, a_{0} = 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.
Let R_{1}, R_{2},…, R_{n} be a finite collection of rings, R_{1}⊕R_{2}⊕···⊕ R_{n} = {(a_{1}, a_{2},…, a_{n}): a_{i} ∈ R_{i}} is a ring called the direct sum of R_{1}, R_{2},…, R_{n} under componentwise addition and multiplication; that is, (a_{1}, a_{2},…, a_{n}) + (b_{1}, b_{2},…, b_{n}) = (a_{1} + b_{1}, a_{2} + b_{2},…, a_{n} + b_{n})
(a_{1}, a_{2},…, a_{n})(b_{1}, b_{2},…, b_{n}) = (a_{1}b_{1}, a_{2}b_{2},…, a_{n}b_{n})
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 ∈ ℤ_{6}, 2·3 = 0. 4, 3 ∈ ℤ_{12}, 4·3 = 0.
However, if p is a prime number, ℤ_{p} is a field, that is, every nonzero 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 noncommutative 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} = $\frac{abicjdk}{a^2+b^2+c^2+d^2}$. $(a +bi +cj +dk)\frac{abicjdk}{a^2+b^2+c^2+d^2} = \frac{(a^2+b^2+c^2+d^2)+(cd+cd)i + (bd+bd)j + (bc +cb)k}{a^2+b^2+c^2+d^2}$ = 1.
Bibliography
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.
This website uses cookies to improve your navigation experience. By continuing, you are consenting to our use of cookies, in accordance with our Cookies Policy and Website Terms and Conditions of use.