Integer Multiples form Commutative Ring

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

Then $\left({n \Z, +, \times}\right)$ is a commutative ring.

Unless $n = 1$, $\left({n \Z, +, \times}\right)$ is not a ring with unity.

Proof
From Integer Multiples under Addition form Infinite Cyclic Group, $\left({n \Z, +}\right)$ is a cyclic group

From [Cyclic Group is Abelian]], $\left({n \Z, +}\right)$ is abelian.

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

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

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

So $\left({n \Z, +, \times}\right)$ is a commutative ring.

The unity of $\left({\Z, +, \times}\right)$ is $1$.

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

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

We have that Integer Multiples form Subring of Integers.

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