# Integer Multiples form Commutative Ring

## Theorem

Let $n \Z$ be the set of integer multiples of $n$.

Then $\struct {n \Z, +, \times}$ is a commutative ring.

Unless $n = 1$, $\struct {n \Z, +, \times}$ is not a ring with unity.

## Proof

From Integer Multiples under Addition form Infinite Cyclic Group, $\struct {n \Z, +}$ is a cyclic group

From Cyclic Group is Abelian, $\struct {n \Z, +}$ is abelian.

From Integer Multiples Closed under Multiplication and Integer Multiplication is Associative, we have that $\struct {n \Z, \times}$ is a semigroup.

From Integer Multiplication Distributes over Addition it follows that $\struct {n \Z, +, \times}$ is a ring.

From Integer Multiplication is Commutative we have that $\struct {n \Z, +, \times}$ is a commutative ring.

So $\struct {n \Z, +, \times}$ is a commutative ring.

$\Box$

The unity of $\struct {\Z, +, \times}$ is $1$.

But unless $n = 1$, we have that $1 \notin n \Z$.

As $\struct {\Z, +, \times}$ is an integral domain, all its elements are cancellable, by the Cancellation Law of Ring Product of Integral Domain.

We have that Subrings of Integers are Sets of Integer Multiples.

From Identity of Cancellable Monoid is Identity of Submonoid it follows that $\struct {n \Z, +, \times}$ has no unity, because if it did, that unity would be $1$.

$\blacksquare$

## Examples

### Even Integers form Commutative Ring

Let $2 \Z$ be the set of even integers.

Then $\struct {2 \Z, +, \times}$ is a commutative ring.

However, $\struct {2 \Z, +, \times}$ is not an integral domain.