# 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 $p$.

For any $a \in R$, we define the mapping $g_a: \Z \to R$ from the integers into $R$ as:

$\forall n \in \Z: \map {g_a} n = n \cdot a$

Then $g_a$ is a group homomorphism from $\struct {\Z, +}$ to $\struct {R, +}$.

Also:

$\ideal p \subseteq \map \ker {g_a}$

where:

$\map \ker {g_a}$ is the kernel of $g_a$;
$\ideal p$ is the principal ideal of $\Z$ generated by $p$.

Also:

$p \divides n \implies n \cdot a = 0$

where $p \divides n$ 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:

 $\ds \map {g_a} m + \map {g_a} n$ $=$ $\ds m \cdot a + n \cdot a$ $\ds$ $=$ $\ds \paren {m + n} \cdot a$ $\ds$ $=$ $\ds \map {g_a} {m + n}$

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:

 $\ds n$ $\in$ $\ds \ideal p$ $\ds \leadsto \ \$ $\ds p$ $\divides$ $\ds n$ Definition of Ideal of Ring $\ds \leadsto \ \$ $\ds n \cdot a$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds n$ $\in$ $\ds \map \ker {g_a}$ Definition of Kernel of Group Homomorphism

By definition of subset:

$\ideal p \subseteq \map \ker {g_a}$

$\blacksquare$