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
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Theorem $24.8 \ 1^\circ$