Subring Generated by Unity of Ring with Unity

Theorem
Let $$\left({R, +, \circ}\right)$$ 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 $$\forall n \in \Z: g \left({n}\right) = n 1_R$$, where $$n 1_R$$ is as defined in Power of an Element.

Let $$\left({x}\right)$$ be the principal ideal of $\left({R, +, \circ}\right)$ 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) $$\left({0_R}\right)$$, in which case $$g$$ is an isomorphism from $$\Z$$ onto $$S$$, or:
 * 2) $$\left({p}\right)$$ 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 Powers of Ring Elements, we have $$\left({n 1_R}\right) \left({m 1_R}\right) = n \left({m 1_R}\right) = \left({n m}\right) 1_R$$.

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


 * Assume that $$R$$ has no zero divisors.

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

By Ring of Integers is a Principal Ideal Domain, the kernel of $$g$$ is $$\left({p}\right)$$ for some $$p \in \Z^*_+$$.

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 zero divisors.

So from Integral Domain of Prime Order is a 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 $$h \left({1}\right) = h \left({1^2}\right) = \left({h \left({1}\right)}\right)^2$$, either $$h \left({1}\right) = 1_R$$ or $$h \left({1}\right) = 0_R$$ by Idempotent Elements of Ring with No Zero Divisors.

But, by Homomorphism of Powers: Integers, $$\forall n \in \Z: h \left({n}\right) = h \left({n 1}\right) = n h \left({1}\right)$$

So if $$h \left({1}\right) = 0_R$$, then $$\forall n \in \Z: h \left({n}\right) = n 0_R = 0_R$$.

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

So $$h \left({1}\right) = 1_R$$, and thus $$\forall n \in \Z: h \left({n}\right) = n 1 = g \left({n}\right)$$.