Principal Ideal of Characteristic of Ring is Subset of Kernel of Multiple Function

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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:

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

where:

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


Proof

We have from Multiple Function on Ring is Homomorphism that $g_a$ is a group homomorphism.

By definition of kernel:

$x \in \map \ker {g_a} \iff \map {g_a} x = 0_R$

Hence to show that $\ideal p \subseteq \map \ker {g_a}$, we need to show that:

$\forall x \in \ideal p: \map {g_a} x = 0_R$


By definition of characteristic, $p \in \Z_{\ge 0}$ is such that $\ideal p$ is the kernel of $g_1$:

$\map {g_1} n = n \cdot 1_R$

So:

\(\ds x\) \(\in\) \(\ds \ideal p\)
\(\ds \leadsto \ \ \) \(\ds \map {g_1} x\) \(=\) \(\ds 0_R\) Definition 2 of Characteristic of Ring: $\ideal p$ is the kernel of $g_1$
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds x \cdot 1_R\) \(=\) \(\ds 0_R\) Definition of $g_a$ where here $a = 1$


Then:

\(\ds \map {g_a} x\) \(=\) \(\ds x \cdot a\) Definition of $g_a$
\(\ds \) \(=\) \(\ds x \cdot \paren {a \circ 1_R}\) Definition of Multiplicative Identity
\(\ds \) \(=\) \(\ds a \circ \paren {x \cdot 1_R}\) Multiple of Ring Product
\(\ds \) \(=\) \(\ds a \circ 0_R\) from $(1)$
\(\ds \) \(=\) \(\ds 0_R\) Definition of Ring Zero


So:

$x \in \map \ker {g_a}$

and so:

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

$\blacksquare$


Sources

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