# 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 $n 1_R$ the $n$th power of $1_R$.

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:

- $(1): \quad \left({0_R}\right)$, in which case $g$ is an isomorphism from $\Z$ onto $S$

or:

- $(2): \quad \left({p}\right)$ 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 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 $\left({p}\right)$ 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 $h \left({1}\right) = h \left({1^2}\right) = \left({h \left({1}\right)}\right)^2$, either $h \left({1}\right) = 1_R$ or $h \left({1}\right) = 0_R$ by Idempotent Elements of Ring with No Proper 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 $\forall n \in \Z: h \left({n}\right) = n 1 = g \left({n}\right)$.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 24$: Theorem $24.7$