Subring Generated by Unity of Ring with Unity

Theorem
Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Let the mapping $g: \Z \to R$ be defined as $\forall n \in \Z: \map g n = n 1_R$, where $n 1_R$ the $n$th power of $1_R$.

Let $\ideal x$ be the principal ideal of $\struct {R, +, \circ}$ generated by $x$.

Then $g$ is an epimorphism from $\Z$ onto the subring $S$ of $R$ generated by $1_R$.

If $R$ has no proper zero divisors, then $g$ is the only nonzero homomorphism from $\Z$ into $R$.

The kernel of $g$ is either:
 * $(1): \quad \ideal {0_R}$, in which case $g$ is an isomorphism from $\Z$ onto $S$

or:
 * $(2): \quad \ideal p$ for some prime $p$, in which case $S$ is isomorphic to the field $\Z_p$.

Proof
By the Index Law for Sum of Indices and Integral Multiple of Ring Element, we have:
 * $\paren {n 1_R} \paren {m 1_R} = n \paren {m 1_R} = \paren {n m} 1_R$

Thus $g$ is an epimorphism from $\Z$ onto $S$.

$R$ has no proper zero divisors.

By Kernel of Ring Epimorphism is Ideal, the kernel of $g$ is an ideal of $\Z$.

By Ring of Integers is Principal Ideal Domain, the kernel of $g$ is $\ideal p$ for some $p \in \Z_{>0}$.

By Kernel of Ring Epimorphism is Ideal (don't think this is the correct reference - check it), $S$ is isomorphic to $\Z_p$ and also has no proper zero divisors.

So from Integral Domain of Prime Order is Field either $p = 0$ or $p$ is prime.

Now we need to show that $g$ is unique.

Let $h$ be a non-zero (ring) homomorphism from $\Z$ into $R$.

As $\map h 1 = \map h {1^2} = \paren {\map h 1}^2$, either $\map h 1 = 1_R$ or $\map h 1 = 0_R$ by Idempotent Elements of Ring with No Proper Zero Divisors.

But, by Homomorphism of Powers: Integers, $\forall n \in \Z: \map h n = \map h {n 1} = n \map h 1$

So if $\map h 1 = 0_R$, then $\forall n \in \Z: \map h n = n 0_R = 0_R$.

Hence $h$ would be a zero homomorphism, which contradicts our stipulation that it is not.

So $\map h 1 = 1_R$, and thus $\forall n \in \Z: \map h n = n 1 = \map g n$.