Characteristic of Finite Ring with No Zero Divisors/Proof 2
Theorem
Let $\struct {R, +, \circ}$ be a finite ring with unity with no proper zero divisors whose zero is $0_R$ and whose unity is $1_R$.
Let $n \ne 0$ be the characteristic of $R$.
Then:
- $(1): \quad n$ must be a prime number
- $(2): \quad n$ is the order of all non-zero elements in $\struct {R, +}$.
It follows that $\struct {R, +} \cong C_n$, where $C_n$ is the cyclic group of order $n$.
Proof
Suppose $\Char R = n$ where $n$ is composite.
Let $n = r s$, where $r, s \in \Z, r > 1, s > 1$.
First note that:
\(\ds \paren {r \cdot 1_R} \circ \paren {s \cdot 1_R}\) | \(=\) | \(\ds \paren {r s} \paren {1_R \circ 1_R}\) | Integral Multiple of Ring Element | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {r s} 1_R\) |
Then:
\(\ds \paren {r \cdot 1_R} \circ \paren {s \cdot 1_R}\) | \(=\) | \(\ds n \cdot 1_R\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0_R\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds r \cdot 1_R = 0_R\) | \(\lor\) | \(\ds s \cdot 1_R = 0_R\) |
as $R$ has no proper zero divisors.
But both $r$ and $s$ are less than $n$ which contradicting the minimality of $n$.
So if $\Char R = n$ it follows that $n$ must be prime.
Now let $x \in R^*$.
Then by Characteristic times Ring Element is Ring Zero, $n \cdot x = 0_R$.
It follows from Element to Power of Multiple of Order is Identity that:
- $\order x \divides n$
Since $n$ is prime, either $\order x = 1$ or $\order x = n$.
It cannot be $1$, from Null Ring iff Characteristic is One, so the result follows.
$\blacksquare$
Sources
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $4$: Fields: $\S 17$. The Characteristic of a Field: Theorem $30$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 61.1$ Characteristic of an integral domain or field