Characteristic of Finite Ring with No Zero Divisors/Proof 2

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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 $R$.


$(1): \quad n$ 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 $n$.


Suppose $\Char R = n$ where $n$ is composite.

Let $n = r s$, where $r, s \in \Z, r > 1, s > 1$.

First note that:

\(\displaystyle \paren {r \cdot 1_R} \circ \paren {s \cdot 1_R}\) \(=\) \(\displaystyle \paren {r s} \paren {1_R \circ 1_R}\) Powers of Ring Elements
\(\displaystyle \) \(=\) \(\displaystyle \paren {r s} 1_R\)


\(\displaystyle \paren {r \cdot 1_R} \circ \paren {s \cdot 1_R}\) \(=\) \(\displaystyle n \cdot 1_R\)
\(\displaystyle \) \(=\) \(\displaystyle 0_R\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle r \cdot 1_R = 0_R\) \(\lor\) \(\displaystyle s \cdot 1_R = 0_R\)

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.