Rings

Different sets are naturally endowed with two binary operations: addition and multiplication. Examples that quickly come to mind are the integers, the integers modulo n, the real numbers, matrices, and polynomials. The notion of Rings describing this kind of sets was originated in the mid-19th century by Richard Dedekind.

6.1.1 Definition A ring (R, +, ⋅) is a set equipped with two binary operations indicated by + and ·,

satisfying the following three sets of axioms, called the ring axioms

1. 1. + is associative: (a + b) + c = a + (b + c) ∀a, b, c ∈ R

   2. There is an element 0 in R such that a + 0 = a ∀a ∈ R (that is, 0 is the additive identity).

   3. ∀a ∈ R there exists −a ∈ R such that a + (−a) = 0 (that is, −a is the additive inverse of a).

   4. + is commutative: a + b = b + a ∀a, b ∈ R

2. · is associative: (a · b) · c = a · (b · c) ∀a, b, c ∈ R

3. Multiplication is distributive with respect to addition, meaning that:

   a · (b + c) = a · b + a · c,   (a + b) · c = a · c + b · c   ∀a, b, c ∈ R

From now on we'll write ab instead of a · b. A set satisfying i) a), b), c) is called group: hence (R, +) is a group. If d) is also satisfied we have a commutative group or abelian group. (R, +) is oftentimes referred to as additive group of the ring.

A commutative ring is a ring in which the multiplication operation is commutative.

6.1.2. Examples of non-rings abd rings

• The set of natural numbers ℕ with the usual operations is not a ring, since (ℕ, +) is not even a group (the elements are not all invertible with respect to addition, e.g. there is no natural number which can be added to 2 to get 0 as a result).

• The set of odd integers with the usual addition and multiplication is not a ring. Among other things, Axiom 1 fails: The sum of two odd integers is not odd.

Examples of rings

• Let E be the set of even integers with the usual addition and multiplication. Since the sum or product of two even integers is also even, the closure axioms hold. Since 0 is an even integer, E has an additive identity element. If a is even, then − a is also even, and so Axiom i)c holds. The remaining axioms hold for all integers and, therefore, are true whenever a, b, c, are even. Consequently, E is a commutative ring. E does not have an identity, however, because no even integer e has the property that ae = a = ea for every even integer a.

• ℤ, ℚ, ℝ, ℂ, with respect to addition and multiplication.

• 𝕂[x], 𝕂[x₁, ..., x_n] with respect to addition and multiplication defined for this rings.

• The set ℤ_n of congruence class modulo n with the addition and the multiplication defined by [a] + [b] = [a + b] and [a] [b] = [ab]. Note that [0] is the 0 required by our axioms for a ring, and [1] is the unit element of R.

• The set M_n(𝕂) of square matrices n x n, over a field 𝕂, along with the standard matrix addition and multiplication.

• The power set of a set is a commutative ring under the natural operations of union and intersection

  A + B := (A ∪ B) ∩ 𝐶(A ∩ B),   A · B := A ∩ B

  it is easy to notice that it is an Abelian group with respect to the first operation (the neuter element is ∅ and the inverse of an A ∈ 𝓟(X) is A itself. The second operation as well is associative and the distributive properties are satisfied, hence the ring is commutative. In this ring every element is idempotent i.e. A² = A · A = A, ∀ A ∈ 𝓟(X).

• The set of continuous functions, in one variable x ∈ X

  R^X := {f: X ⟶ R}

  with the addition and multiplication defined by

  (f+g)(x) := f(x) + g(x),   (fg)(x) := f(x)g(x)   ∀f,g ∈ R^X, ∀x ∈ X

  (R^X, +, ·) is a ring: it is a commutative ring only if R is commutative; it is unitary only if R is unitary as well. The unit function is that mapping every element of X to the unit of R.

Let R₁, R₂, ..., R_n be rings. We can use these to construct a new ring by Cartesian product

R₁ x R₂ x R_n

and perform component addition and multiplication, defined as

(a₁, a₂, ..., a_n) + (b₁, b₂, ..., b_n) = (a₁ + b₁, a₂ + b₂, ..., a_n + b_n)

and

(a₁, a₂, ..., a_n) (b₁, b₂, ..., b_n) = (a₁ b₁, a₂b₂, ..., a_nb_n)

This ring is called the direct sum of R₁, R₂, ..., R_n and usually indicated as

R₁ ⊕ R₂ ⊕ ... ⊕ R_n

The following properties follow from ring axioms, we do not prove them since we've already proved it in the case of integers (without making it specific for integers alone).

1. a ⋅ 0 = 0 ⋅ a = 0;

2. (−a) ⋅ b = −(a ⋅ b);

6.1.2 Definition. Let [R, +, ⋅] and [R', +', ⋅'] be rings. An isomorphism φ between R and R' is a biunivoque correspondence between R and R' preserving the operations, i.e. such that

and

If an isomorphism from R to R' exists, we say that R is isomorphic to R' and use the notation R ≅ R'. The relation of being isomorphic is an equivalence relation.

Any algebraic property that holds on R is also true on R' and viceversa owning to the bijective map, hence from an algebraic point of view they are identical. When we'll study a particular ring we'll refer to its isomorphism class containing all rings isomorphic to the first.

Examples of ring isomorphism

• The isomorphism between the ring 𝕂[x] of polynomials with coefficients over 𝕂 and the ring of sequences having from a point on all 0 terms. We the operations we defined when discussing polynomials

• The map from ℂ to ℂ of complex numbers to their complex conjugates.

Some properties satisfied by rings do not appear among ring axioms and are used to further characterize rings

• A commutative ring satisfies the commutative property for multiplication.

• A commutative ring with identity is a ring with a nonzero element, 1 ∈ R, that is an identity under multiplication i.e. if 1 ⋅ a = a ∀a ∈ R.

  Let R be a ring with identity. The element a ∈ R is a unit or invertible elements, if it has a multiplicative inverse in R, that is if there exists b ∈ R such that ab = ba = 1, that is a divides 1. The element b is called the inverse of a, and is written a⁻¹. It is unique; for if b and c are both inverses of a, then

  c = 1c = (ba)c = b(ac) = b ⋅ 1 = b.

  The units in ℤ are 1 and −1.

  Remark Let u ∈ R be a unit with inverse v, so that uv = 1. For any b ∈ R we have u(vb) = (uv)b = 1b = b. Therefore

  a unit divides every element of R

• An integral domain is a commutative ring with identity and no ­zero-­divisors i.e. ab = 0 ⇒ a = 0 or b = 0.

A ring F, is a field provided these conditions hold:

1. F is a commutative ring.

2. F has a unity e, and e ≠ 0.

3. Every nonzero element of F has a multiplicative inverse.

Another way to look at the definition of a field is t his: It is a commutative ring in which one can always solve equations of the form ax = b (when a ≠ 0). The solution is, of course, x = a⁻¹ b. In other words: In all rings we can add, subtract, and multiply, but in fields we can also divide. Or to say this yet another way, a field is a commutative ring with unity for which all non-zero elements are associates of one another. ℚ, ℝ and ℂ, are familiar examples of fields. We have seen that if p is a prime, then ℤ_p is a field.

• A division ring is a ring with identity (not necessarily commutative).

It is clear that only rings sharing the properties mentioned above can be considered isomorphic e.g. a ring with unit cannot be isomorphic to a ring without unit. For example for any integer n ∈ ℕ, let

be the set of integer multiplies of n. The set nℤ is a subset of ℤ and inherits addition and multiplication from ℤ. If we take two elements from nℤ, say nh and nk we can see that

by the distributive and associative laws. Addition and multiplication are both associative in nℤ and multiplication is distributive with respect to addition. In nℤ, 0 is the additive identity. However, there is no multiplicative identity in nℤ unless n = ± 1. There are also no multiplicative inverses for nk unless n = ± 1 and k = ± 1.

As a consequence ℤ is not isomorphic to 2ℤ. We can view nℤ as a "substructure" of ℤ.

6.1.3 Proposition. Let R be a ring with identity.

1. The identity is a unit; it is equal to its inverse.

2. If a is a unit, then so is a⁻¹; its inverse is a.

3. If a and b are units, then so is ab; its inverse is b⁻¹ a⁻¹.

Proof.

1. 1 ⋅ 1 = 1

2. This is shown by the equation aa⁻¹ = a⁻¹ a = 1

3. We have (ab)(b⁻¹a⁻¹) = a (bb⁻¹) a⁻¹ = aa⁻¹ = 1.

   and, similarly, (b⁻¹ a⁻¹)(ab) = 1.□

6.1.6 Proposition. Every finite Integral domain R is a field.

Proof. Since R is a commutative ring with identity, we need only show that for each a ≠ 0_R, the equation ax = 1_R has a solution. Let a₁, a₂, ..., a_n. be the distinct elements of R and suppose a_s ≠ 0_R. To show that a_sx ≠ 1_R has a solution, consider the products a_sa₁, a_sa₂, ..., a_sa_n. If a_i ≠ a_j, then we must have a_sa_i ≠ a_sa_j (because a_sa_j = a_sa_j would imply that a_s = a_j by cancellation). Therefore, a_sa₁, ..., a_sa_n, are n distinct elements of R. However, R has exactly n elements all together, and so these must be all the elements of R in some order. In particular, for some j, a_sa_j = 1_R. Therefore, the equation a_sx = 1_R has a solution and R is a field.  □