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

Let $g_a: \Z \to R$ be the mapping from the integers into $R$ defined as:
 * $\forall n \in \Z:\forall a \in R: \map {g_a} n = n \cdot a$

where $\cdot$ denotes the multiple operation.

Then the following hold:

Proof
By Multiple of Ring Product, we have that:
 * $\forall n \in \Z_{>0}: \paren {n \cdot x} \circ y = n \cdot \paren {x \circ y} = x \circ \paren {n \cdot y}$

So:
 * $\forall n \in \Z_{>0}: n \cdot a = \paren {n \cdot a} \circ 1_R = a \circ \paren {n \cdot 1_R}$

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

For $n \in \Z_{<0}$, we have $-n \in \Z_{>0}$.

So:
 * $n \cdot a = -\paren {-n} \cdot a = -\paren {\paren {-n \cdot a} \circ 1_R} = -\paren {a \circ \paren {-n \cdot 1_R}}$

so when $-n \cdot 1_R = 0$ we have $n \cdot a = 0$.

For $n = 0$, we trivially have $n \cdot a = 0$.

By definition of characteristic, we have:
 * $p \divides n \iff n \cdot 1_R = 0$

by the above, this implies $n \cdot a = 0$.

Therefore:
 * $p \divides n \implies n \cdot a = 0$

We also have:

By definition of subset:
 * $\ideal p \subseteq \map \ker {g_a}$