Kernel of Multiple Function on Ring with Characteristic Zero is Trivial

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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 $g_a: \Z \to R$ 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.


Let $a \in R$ such that $a$ is not a zero divisor of $R$.

Let the characteristic of $R$ be $0$.

Then:

$\map \ker {g_a} = \set {0_R}$

where $\ker$ denotes the kernel of $g_a$.


That is:

$n \cdot a = 0_R$

if and only if:

$n = 0$


Proof

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


Aiming for a contradiction, suppose $\exists n \ne 0: n \cdot a = 0_R$.

Then:

\(\ds n\) \(\in\) \(\ds \map \ker {g_a}\) Definition of Kernel of Group Homomorphism
\(\ds \leadsto \ \ \) \(\ds n\) \(\in\) \(\ds \ideal p\) Kernel of Non-Zero Divisor Multiple Function is Primary Ideal of Characteristic
where $p$ is the characteristic of $R$
\(\ds \leadsto \ \ \) \(\ds p\) \(\divides\) \(\ds n\) Definition of Integral Ideal
\(\ds \leadsto \ \ \) \(\ds p\) \(\ne\) \(\ds 0\)

This contradicts our assertion that the characteristic of $R$ is $0$.

Hence by Proof by Contradiction there can be no such $n \ne 0$ such that $n \cdot a = 0_R$.

$\blacksquare$


Sources

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