Homomorphism from Integers into 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 characteristic of $$R$$ be $$p$$.

For any $$a \in R$$, we define the mapping $$g_a : \Z \to R$$ from the integers into $$R$$ as:
 * $$\forall n \in \Z: g_a \left({n}\right) = n \cdot a$$

Then $$g_a$$ is a group homomorphism from $$\left({\Z, +}\right)$$ to $$\left({R, +}\right)$$.

Also:
 * $$\left({p}\right) \subseteq \ker \left({g_a}\right)$$

where:
 * $$\ker \left({g_a}\right)$$ is the kernel of $$g_a$$;
 * $$\left({p}\right)$$ is the principal ideal of $$\Z$$ generated by $$p$$.

Also:
 * $$p \backslash n \implies n \cdot a = 0$$

where $$p \backslash n$$ denotes that $$p$$ is a divisor of $$n$$.

Proof
The fact that $$g_a$$ is a group homomorphism follows directly from Index Laws for Monoids.

By General Distributivity Theorem, we have that:
 * $$\forall n \in \Z^*: \left({n \cdot x} \right) \circ y = n \cdot \left({x \circ y}\right) = x \circ \left({n \cdot y}\right)$$

So:
 * $$\forall n \in \Z^*: n \cdot a = \left({n \cdot a} \right) \circ 1_R = a \circ \left({n \cdot 1_R}\right)$$

So when $$n \cdot 1 = 0$$ we have $$n \cdot a = 0$$.