Equivalence of Definitions of Characteristic of Ring

Theorem

Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.

The following definitions of the concept of Characteristic of Ring are equivalent:

Definition 1

For a natural number $n \in \N$, let $n \cdot x$ be defined as the power of $x$ in the context of the additive group $\struct {R, +}$:

$n \cdot x = \begin {cases}

0_R & : n = 0 \\ \paren {\paren {n - 1} \cdot x} + x & : n > 0 \end {cases}$

The characteristic $\Char R$ of $R$ is the smallest $n \in \N_{>0}$ such that $n \cdot 1_R = 0_R$.

If there is no such $n$, then $\Char R = 0$.

Definition 2

Let $g: \Z \to R$ be the initial homomorphism, with $\map g n = n \cdot 1_R$.

Let $\ideal p$ be the principal ideal of $\struct {\Z, +, \times}$ generated by $p$.

The characteristic $\Char R$ of $R$ is the positive integer $p \in \Z_{\ge 0}$ such that $\ideal p$ is the kernel of $g$.

Definition 3

The characteristic of $R$, denoted $\Char R$, is defined as follows.

Let $p$ be the order of $1_R$ in the additive group $\struct {R, +}$ of $\struct {R, +, \circ}$.

If $p \in \Z_{>0}$, then $\Char R := p$.

If $1_R$ is of infinite order, then $\Char R := 0$.

Proof

Definition 1 is equivalent to Definition 3

By definition of order, the order of $1_R$ is the smallest $p \in \Z_{> 0}$ such that $p \cdot 1_R = 0_R$, the identity of the additive group $\struct {R, +}$ of $\struct {R, +, \circ}$.

Hence if the order of $1_R$ is finite, the definitions coincide.

If the order of $1_R$ is infinite, there is no such $p$.

Then both definitions give $\Char R = 0$, so they coincide as well.

$\Box$

Definition 1 is equivalent to Definition 2

Let $g: \Z \to R$ be the initial homomorphism, with $\map g n = n \cdot 1_R$.

By Kernel of Ring Homomorphism is Ideal, $\ker g$ is an ideal of $R$.

By Ring of Integers is Principal Ideal Domain, there exists a unique $p \in \Z_{\ge 0}$ such that $\ker g$ is the principal ideal $\ideal p$.

Suppose there is a smallest $m$ such that $m \cdot 1_R = 0_R$.

Then for any $k \in \Z$:

$\map g {k m} = k m \cdot 1_R = k \cdot 0_R = 0_R$.

This gives $\ideal m \subseteq \ker g = \ideal p$.

Let $a \in \ker g$.

Then by the Division Theorem:

$\exists q, r \in \Z, 0 \le r < m: a = m q + r$

As $a, m \in \ker g = \ideal p$ so does $r = a - m q$.

Since $m$ is the smallest (strictly) positive integer such that $m \cdot 1_R = 0_R$:

$\forall n \in \Z: 0 < n < m \implies \map g n = n \cdot 1_R \ne 0_R$

This forces $r = 0$.

Thus $a = m q \in \ideal m$.

Hence we have $\ideal m = \ideal p$, giving $p = m$.

Now consider the case where $m$ does not exist.

Then for any $k \ne 0$:

$\map g k = k \cdot 1_R \ne 0_R$

And we have:

$\map g 0 = 0 \cdot 1_R = 0_R$

giving $\ker g = \set 0 = \ideal 0$.

Both definitions give $\Char R = 0$.

Therefore the definitions are equivalent in each case.

$\blacksquare$