# Equivalence of Definitions of Integral Domain

## Theorem

The following definitions of the concept of Integral Domain are equivalent:

### Definition 1

An integral domain $\struct {D, +, \circ}$ is:

a commutative ring which is non-null
with a unity
in which there are no (proper) zero divisors, that is:
$\forall x, y \in D: x \circ y = 0_D \implies x = 0_D \text{ or } y = 0_D$

that is, in which all non-zero elements are cancellable.

### Definition 2

An integral domain $\left({D, +, \circ}\right)$ is a commutative ring such that $\left({D^*, \circ}\right)$ is a monoid, all of whose elements are cancellable.

In this context, $D^*$ denotes the ring $D$ without zero: $D \setminus \left\{{0_D}\right\}$.

## Proof

### $(1)$ implies $(2)$

Let $\struct {D, +, \circ}$ be an integral domain in sense $1$.

As $\struct {D, +, \circ}$ is already a commutative ring, it remains to show that $\struct {D^*, \circ}$ is a monoid.

Because $\circ$ is a ring product, and $\struct {D, +, \circ}$ has no zero divisors, we conclude Closure and Associativity.

Furthermore, $\struct {D, +, \circ}$ is non-null, hence $0_D \ne 1_D$, and we conclude $1_D \in D^*$.

Therefore, we also have an Identity for $\struct {D^*, \circ}$, and hence it is a monoid.

It remains to show that all elements of $\struct {D^*, \circ}$ are cancellable.

As $\struct {D, +, \circ}$ has no zero divisors, this follows from Ring Element is Zero Divisor iff not Cancellable.

$\Box$

### $(2)$ implies $(1)$

Let $\struct {D, +, \circ}$ be an integral domain in sense $2$.

$\struct {D, +, \circ}$ is already a commutative ring.

Furthermore, as $\struct {D^*, \circ}$ is a monoid, it is nonempty.

Also, we conclude that $\struct {D, +, \circ}$ is a non-null ring with unity.

It remains to show that $\struct {D, +, \circ}$ has no zero divisors.

We know all elements of $\struct {D^*, \circ}$ are cancellable.

From Ring Element is Zero Divisor iff not Cancellable, we conclude that $\struct {D, +, \circ}$ cannot have zero divisors.

$\blacksquare$