“Hooray!” Apathicus was playing the Earth’s game and have just impressively pulverized the galaxy’s record by killing a pregnant woman, many babies and kids, differently-abled people, LGBTQ+ community members, a few senior citizens, and a nun in a wheelchair, all at once. Not to mention the grandma doing a double somersault and crashing with sickening thud onto the nun’s head. Obviously, he was displaying the empathy of an ATM machine, but who fucking really cares, Apocalypse, Anawim, #justtothepoint.
💍 A ring R is a non-empty set with two binary operations, addition (a + b) and multiplication (ab), such that ∀ a, b, c ∈ R:
Examples: ℤ, ℚ, ℝ, and ℂ under the usual operations of addition and multiplication are rings. ℤ/ℤn = {[0], [1], ··· [n-1]} under addition and multiplication modulo n are commutative rings with unity 1. GL(n, ℝ) is a ring with unity, but it is not commutative. The set of all polynomials in the variable x with real coefficients, ℝ[x] = {anxn + an-1xn-1 + ··· + a1x + a0 | ai ∈ ℝ}, +, .) is a commutative ring with the usual operations of addition and multiplication of polynomials. ℤ/pℤ is a field, where p is a prime number.
Let R be a ring, ∀a, b, c ∈ R. Then, the following statements hold true.
0 + a0 =[(R, +) group, 0 identity] a0 =[(R, +) group, 0 identity] a(0 + 0) =[R ring, distributive property] a0 + a0 ⇒ 0 + a0 = a0 + a0 ⇒ [By cancellation laws, (R, +) group] a0 = 0. Mutatis mutandis, 0a = 0.
[(R, +) group, there are additive inverses] a(b -b) = a0 =[Property 1] 0, a(b -b) =[Distributivity property] ab + a(-b) = 0 ⇒ -(ab) + (ab + a(-b)) = -(ab) ⇒[Associative] (-(ab) + ab) + a(-b) = -(ab) ⇒ a(-b) = -(ab). Mutatis mutandis, (-a)b = -(ab) = -ab.
The inverse of -a is an element x ∈ R s.t. -a +x = 0R. Consider -a + a = 0R, a is a solution, by inverses’ uniqueness and reinterpreting the previous equation [(R, +) is a group] -(-a) = a.
-(a +b) is the unique solution of (a+b) + x = 0, but -a + (-b) is a solution, too because (a + b) + ((-a) + (-b)) =[(R, +) Associativity and commutativity] (a + (-a)) + (b + (-b)) = 0R + 0R = 0R.
-(a -b) =[Property 4] -a + -(-b) =[Property 3] -a +b ∎
(-a)(-b) =[Property 2, a(-b) = (-a)b.] (-(-a))b = [Property 3] ab ∎
a(b -c) = a(b + (-c)) =[Distributivity] ab + a(-c) =[Property 2, a(-b) = (-a)b = -ab.] ab -ac. Mutatis mutandis, (b-c)a = ba - ca.
Suppose R has unity 1 ⇒ (-1)a + a =[1 ∈ R, 1a = a ∀a∈R] (-1)a + 1·a =[Distributivity] (-1 + 1)a = 0·a =[Property 1] 0 ⇒ (-1)a + a = 0 ⇒ (-1)a = -a ∎
Suppose R has a unit element 1 ⇒[Let’s apply the property 8 to a = -1] (-1)(-1) = -(-1) =[Property 3] 1 ∎
The proofs are identical to the ones given for groups.
⇒) Suppose (R, ·) is commutative. (a + b)2 = (a + b)(a +b) =[Distributive] (a +b)a + (a +b)b =[Distributive] a2 +ba +ab +b2 =[(R, ·) is commutative] a2 + 2ab + b2
⇐) Suppose a, b ∈ R, (a + b)2 = a2 + 2ab + b2, ab = ba?
(a + b)2 = (a +b)(a +b) =[Distributive] (a +b)a + (a +b)b =[Distributive] a2 +ba +ab +b2 =[By assumption equals a2 + 2ab + b2 and we expand 2ab] = a2 + ab + ab + b2 ⇒[By applying cancellation laws, (R, +) group] ba = ab ∎
a + b =[(R, +) have inverses] ((-a) + a) + a + b + (b + (-b)) =[By assumption, R has a multiplicative unity] (-a) + a·1 + a·1 + b·1 + b·1 + (-b) =[Distributivity] (-a) + a(1 + 1) + b(1 + 1) + (-b) =[Distributivity] (-a) +(a + b)(1 + 1) + (-b) =[Distributivity] (-a) + a + b + a + b + (-b) =[Associativity] ((-a) + a) + b + a + (b + (-b)) = b + a ∎
∀a, b ∈ R, a + b ∈ R ⇒[R is a boolean ring] (a +b)2 = a +b. Besides, (a + b)2 = (a +b)(a +b) =[Distributive] (a +b)a + (a +b)b =[Distributive] a2 +ba +ab +b2 =[R is a boolean ring, a2= a, b2 = b] a + ba +ab + b.
Therefore, a + ba +ab + b = a + b ⇒[By applying cancellation laws] ba + ab = 0 ⇒ ab = -(ba) = -ba. Futhermore, a + a ∈ R ⇒ a + a = (a + a)2 = a2 + 2a2 + a2 =[Boolean ring, a2 = a] a +2a +a = a + a + a + a ⇒ a + a = a + a + a + a ⇒[By cancellation laws] a +a = 0 ⇒ a = -a, i.e., each element is its own additive inverse, and we have previously demonstrated that ab = -ba =[ba ∈ R, so ba is also its own additive inverse] ba ∎
Suppose for the sake of contradiction ∃b ∈ R, b ≠ 0, ba = 0, and c is the multiplicative inverse of a. Let’s compute bac. bac =[Associativity] (ba)c =[By assumption, ba = 0] 0·c = 0. On the other hand, bac =[By associativity] b(ac) =[By assumption, c is the multiplicative inverse of a] b·1 = b. Thus, b = bac = 0 ⊥
Let’s prove a few, e.g. (m+n)a = a +··m+n times··+a = a +··m times··+a + a +··n times··+a = ma +na. (na)(mb) = (a +··n times··+a)(b +··m times··+b) =[By repeated applications of the distribution law]ab +··nm times··+ab = (nm)(ab). aman = a ···m times···a · a +··n times··+a = a ···m+n times··a = am+n.
Suppose u ∈ R has multiple right inverses, say v1, v2 ∈ R, v1 ≠ v2, and uv1 = 1 = uv2.
0 =[R is a ring with unity] 1 -1 = uv1 -uv2 =[Distributivity] u(v1 -v2). Therefore, v1-v2 ≠ 0 (u has multiple right inverses, v1 ≠ v2), v1-v2 ∈ R, u·(v1 -v2) = 0 ⇒ u is a left zero divisor ∎
Suppose x is nilpotent, ∃n: xn = 0, un = un -xn =[R is a commutative ring with unity] (u -x)(un-1 + un-2x + un-3x2 + ··· + uxn-2 + xn-1). Let’s rename v = un-1 + un-2x + un-3x2 + ··· + uxn-2 + xn-1, so un = (u-x)v ⇒[R is a commutative ring, Associativity - unu-n = (u-x)vu-n = (u-x)u-nv -] 1 = (u-x)(u-nv). Hence, u-x is a unit, its inverse (u-x)-1 = u-nv ∎
A subset S of a ring R is a subring of R if S is itself a ring when binary operations of addition and multiplication on R are restricted to the subset.
Subring Test. A non-empty subset S of a ring R is a subring if S is closed under subtraction and multiplication, that is, ∀a, b ∈ S, a - b and ab ∈ S.
Proof: Trivial. S is an Abelian group under addition because S is closed under subtraction and R is commutative. Besides, multiplication in R is associative as well as distributive over addition, and this properties are obviously inherited in S.
Examples.