    # Rings II. Properties. Subrings.

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# Definition.

💍 A ring R is a non-empty set with two binary operations, 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.

Examples: ℤ, ℚ, ℝ, and ℂ under the usual operations of addition and multiplication are rings. ℤ/ℤn = {, , ··· [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.

# Properties of Rings

Let R be a ring, ∀a, b, c ∈ R. Then, the following statements hold true.

• 1: a0 = 0a = 0.

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.

• 2: a(-b) = (-a)b = -(ab) = -ab.

[(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.

• 3: -(-a) = a.

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.

• 4: −(a + b) = −a + (−b).

-(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.

• 5: −(a − b) = −a + b.

-(a -b) =[Property 4] -a + -(-b) =[Property 3] -a +b ∎

• 6: (-a)(-b) = ab.

(-a)(-b) =[Property 2, a(-b) = (-a)b.] (-(-a))b = [Property 3] ab ∎

• 7: a(b - c) = ab - ac and (b-c)a = ba - ca.

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.

• 8: If R has unity, then (-1)a = -a

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 ∎

• 9: If R has unity, then (-1)(-1) = 1.

Suppose R has a unit element 1 ⇒[Let’s apply the property 8 to a = -1] (-1)(-1) = -(-1) =[Property 3] 1 ∎

• 10: If R has unity, then it is unique. If an element has a multiplicative inverse, then it is unique.

The proofs are identical to the ones given for groups.

• 11: A ring R is commutative if and only if (a+b)2 = a2 +2ab + b2 ∀a, b ∈R. Futhermore, if R is commutative, the binomial theorem (a +b)n = $\sum_{k=0}^n {n \choose k}a^{n-k}b^k$ holds (it is left as an exercise).

⇒) 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 ∎

• 12: If R has unity, then R is commutative.

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 ∎

• 13: Let R be a ring. We say that R is boolean if for every a ∈ R, a2 = a. Every boolean ring is commutative.

∀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 ∎

• 14: Let R be a ring. We say that a ∈ R, a ≠ 0, is a zero-divisor, if there is an element b ∈ R, b ≠ 0, such that either ab = 0 or ba = 0. If a has a multiplicative inverse in R, then a is not a zero divisor.

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 ⊥

• 15: If a, b are any two arbitrary elements of a ring R and m and n are any two positive integers, then: (m + n)a = ma + na, m(a + b) = ma + mb, m(na) = (mn)a, (na)(mb) = (nm)(ab). aman = am+n, and (am)n = amn.

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.

• 16: Let R be a ring with unity. We say x ∈ R, x ≠ 0 is a left-zero divisor if ∃y ∈ R, y ≠ 0 such that x·y = 0. We say that u ∈ R has a right inverse if ∃v ∈ R such that u·v = 1. If an element u ∈ R has multiple right inverses, then it is a left zero divisor.

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 ∎

• 17: Suppose R is a commutative ring with unity. We say x ∈ R is nilpotent if ∃n ∈ ℕ such that xn = 0. If u is both unit and nilpotent, then u-x is a unit, too.

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 ∎

# Subrings

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.

1. {0} (the trivial subring) and R are subrings of any ring R.
2. {0, 2, 4} is a subring of the ring ℤ6 where 4 is the unity in {0, 2, 4}, e.g., 2·4 = 4·2 =6 2.
3. ∀n ∈ℤ, n>0, nℤ is a subring of ℤ.
4. Chain of subrings: ℤ ⊆ ℚ ⊆ ℝ ⊆ ℂ ⊆ $\mathbb{H}$
5. The set of Gaussian integer are the set of complex numbers whose real and imaginary parts are both integers, ℤ[i] = {a + bi | a, b ∈ ℤ} is a subring of ℂ.
6. The set of 2 x 2 diagonal matrices (those in which the entries outside the main diagonal are all zeroes), {$(\begin{smallmatrix}a & 0\\ 0 & b\end{smallmatrix})$| a, b ∈ ℤ } is a subring of the ring of all 2 x 2 matrices over ℤ.

# 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, 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.
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