Subrings of Integers are Sets of Integer Multiples

Theorem
Let $$\left({\Z, +, \times}\right)$$ be the integral domain of integers.

The subrings of $$\left({\Z, +, \times}\right)$$ are the rings of integer multiples:
 * $$\left({n \Z, +, \times}\right)$$

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

There are no other subrings of $$\left({\Z, +, \times}\right)$$ but these.

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

We also note that when $$n = 0$$, we have:
 * $$\left({n \Z, +, \times}\right) = \left({0, +, \times}\right)$$

which is the null ring.

When $$n = 1$$, we have:
 * $$\left({n \Z, +, \times}\right) = \left({\Z, +, \times}\right)$$

From Null Ring and Ring Itself Subrings, these extreme cases are subrings of $$\left({\Z, +, \times}\right)$$.

From Subgroups of the Integers, the only additive subgroups of $$\left({\Z, +, \times}\right)$$ are $$\left({n \Z, +}\right)$$.

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

Hence the result.