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.