# Subrings of Integers are Sets of Integer Multiples

## Theorem

Let $\struct {\Z, +, \times}$ be the integral domain of integers.

The subrings of $\struct {\Z, +, \times}$ are the rings of integer multiples:

$\struct {n \Z, +, \times}$

where $n \in \Z: n \ge 0$.

There are no other subrings of $\struct {\Z, +, \times}$ but these.

## Proof

From Integer Multiples form Commutative Ring, it is clear that $\struct {n \Z, +, \times}$ is a subring of $\struct {\Z, +, \times}$ when $n \ge 1$.

We also note that when $n = 0$, we have:

$\struct {n \Z, +, \times} = \struct {0, +, \times}$

which is the null ring.

When $n = 1$, we have:

$\struct {n \Z, +, \times} = \struct {\Z, +, \times}$

From Null Ring and Ring Itself Subrings, these extreme cases are also subrings of $\struct {\Z, +, \times}$.

From Subgroups of Additive Group of Integers, the only additive subgroups of $\struct {\Z, +, \times}$ are $\struct {n \Z, +}$.

So there can be no subrings of $\struct {\Z, +, \times}$ which do not have $\struct {n \Z, +}$ as their additive group.

Hence the result.

$\blacksquare$

## Examples

### Even Integers

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

Then $\struct {2 \Z, +, \times}$ is a subring of $\struct {\Z, +, \times}$.