# Rings, Integral domains and Fields

Here in this post we learn briefly about rings, integral domains and fields in abstract algebra.

## Ring

A non-empty set *R* is said to be a ring if in *R* there are two binary operations (+) and (.) which we call addition and multiplication respectively, such that for a, b, c in *R*:

1. The algebraic system < *R*, + > is an abelian group

2. The associative law with respect to multiplication holds; that is, a . (b . c) = (a . b) . c

3. The two distributive laws a . (b + c) = a . b + a . c and (b + c) . a = b . a + c . a hold

In our definition of a ring, if there is an element 1 in *R* such that a . 1 = 1 . a = a for every a ∈ *R*, then we shall call *R* as a ring with **unity element**.

If the multiplication of *R* is such that a . b = b . a for every a, b in *R*, then R will be called a **commutative ring.**

Example |

The set of integers with usual addition and multiplication forms a commutative ring with unity element.

While the set of even integers under usual operations of addition and multiplication forms a commutative ring but the ring has no unity element.

**Note:**

In case of a ring, with unity element, the algebraic system < *R*, + > is an abelian group and the algebraic system < *R*, . > is a monoid.

## Some Elementary Properties

1. If 0 is the additive identity of the ring < *R*, +, . >, then for any a ∈ *R, *a . 0 = 0 . a = 0

2. For all a, b ∈ *R,* (-a) . b = -(a . b) = a . (-b)

3. For all elements a_{1}, a_{2}, …, a_{m} and b_{1}, b_{2}, …, b_{n} in *R*,

\[({{a}_{1}}+{{a}_{2}}+…+{{a}_{m}}).({{b}_{1}}+{{b}_{2}}+…+{{b}_{n}})=\sum\limits_{i=1}^{m}{\sum\limits_{j=1}^{n}{{{a}_{i}}.{{b}_{j}}}}\]

4. For all integers n and all a, b ∈ *R* n(a . b) = (na) . b = a . (nb)

5. In any commutative ring *R* the binomial formula

\[{{(a+b)}^{n}}=\sum\limits_{i+j=n}{\left( _{i}^{n} \right)}{{a}^{i}}{{b}^{j}},~a,b\in R\]

Holds for natural numbers n and non-negative integers i and j.

Example |

**If < R, +, . > is an algebraic system satisfying all the conditions for a ring with unity with possible exception of a + b = b + a, a, b ∈ R, prove that the axiom a + b = b + a must hold in R and that R is a ring.**

**Solution:**

Since 1 ∈ *R*, we have

\[(a+b).(1+1)\]

\[=(a+b).1+(a+b).1\]

\[=(a+b)+(a+b)\]

Also,

\[(a+b).(1+1)\]

\[=a.(1+1)+b.(1+1)\]

\[=(a+a)+(b+b)\]

\[Thus,~~(a+b)+(a+b)=(a+a)+(b+b)\]

Since associativity with respect to addition holds in R, we have, by left and right cancellation rules b + a = a + b

Hence < R, + > is an abelian group and < *R*, +, . > is a ring with unity.

## Some Special Classes of Rings:

Rings of integers where for two elements a and b in the ring a . b = 0 will imply either a = 0 or b = 0 and also a . b = b . a holds. But in general rings there is a possibility of a . b = 0 with neither a nor b being zero. Also there are rings where a . b = b . a does not hold. For example, in the ring of integers with addition and multiplication modulo b, we have 2 X 6^{3} = 0, but 2 and 3 are non-zero elements. Also it may be easily checked that in the ring M2(R) or 2 X 2 matrices with elements as real numbers the commutative property does not hold.

Accordingly, we have the following definitions

## Zero Divisor

If *R* is a commutative ring then a ≠ 0 ∈ *R* is said to be a **zero divisor** if there exists b ∈ *R*, b ≠ 0 such that a . b = 0.

In the example just considered, the integer 2 is a zero divisor in the ring of integers < *Z*, +_{6}, x_{6} >.

## Integral Domain

An **integral domain** is a commutative ring with unity element and without a zero divisor.

The ring of integers is an example of an integral domain.

## Division Ring

A ring is said to be a **division ring** or skew if its non-zero elements form a group under multiplication.

## Field

A **field** is a commutative ring with unity element such that the non-zero elements form a group under multiplication.

Thus, a field is a commutative division ring. Since the cancellation law holds in any group, any field is an integral domain.

We have just remarked that any field is an integral domain, but for the converse we have,

Theorem |

**A finite integral domain is a field.**

## Subring

A subset *S* of a ring *R* = < *R*, +, . > is said to be a **subring** of *R* if and only if *S* forms a ring under the operations (+) and (.) of *R*.;