Equivalence of Definitions of Integral Domain

Theorem
There are two definitions given for an integral domain:


 * $(1): \quad$ An integral domain $\left({D, +, \circ}\right)$ is a non-null, commutative ring with unity, but without zero divisors, i.e.:


 * $\forall x, y \in D: x \circ y = 0_D \implies x = 0_D \text{ or } y = 0_D$


 * $(2): \quad$ 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, $D^*$ is the set of non-zero elements of $D$.

These definitions are equivalent.

$(1)$ implies $(2)$
Assume $\left({D, +, \circ}\right)$ is an integral domain in sense $1$.

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

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

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

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

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

As $\left({D, +, \circ}\right)$ has no zero divisors, this follows from Zero Divisor Not Cancellable.

$(2)$ implies $(1)$
Assume $\left({D, +, \circ}\right)$ is an integral domain in sense $2$.

$\left({D, +, \circ}\right)$ is already a commutative ring.

Furthermore, as $\left({D^*, \circ}\right)$ is a monoid, it is nonempty.

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

It remains to show that $\left({D, +, \circ}\right)$ has no zero divisors.

We know all elements of $\left({D^*, \circ}\right)$ are cancellable.

From Zero Divisor Not Cancellable, we conclude that $\left({D, +, \circ}\right)$ cannot have zero divisors.