Prove that the characteristic of a finite ring \( \mathcal{R} \) is finite and it is a divisor of \( |\mathcal{R}|\) where \( |\mathcal{R}|\) denotes the number of elements of \( \mathcal{R}\).
Solution:
Let \((\mathcal{R}, +, \cdot)\) is finite ring.
\(\colorbox{yellow}{First Part}\)
To prove, Char \(\mathcal{R}\) is finite.\(\\\)
Since, by definition, the characteristic of a ring is the least positive integer \(n\) (if exists) such that $$n\cdot\alpha=0 ~\forall~ \alpha\in \mathcal{R}$$
where \(0\) is the zero element of \(\mathcal{R}.\\\)
If no such positive integer \(n\) exists then Char \(\mathcal{R}=0.\\\)
Therefore, in both case, the characteristic of a ring is finite whether the ring is finite or not.
\(\colorbox{yellow}{Second Part}\)
To prove, Char \(\mathcal{R}\) is a divisor of \(|\mathcal{R}|.\\\)
Let \(|\mathcal{R}|=m\) and Char \(\mathcal{R}=n\\\)
Then we need to prove that \(n|m\\\)
Since \((\mathcal{R}, +, \cdot)\) is ring then \((\mathcal{R}, +)\) is a abelian group.\(\\\)
Therefore, the Lagrange’s theorem \(m\cdot\alpha=0 ~\forall~ \alpha\in \mathcal{R}.\\\)
By division algorithm, \(\exists\) two integers \(q\) and \(r\) such that
\begin{align*}
& m=nq+r~where~ 0 \le r \text{\textless} n \\
\implies & m\cdot\alpha=(nq+r)\cdot\alpha~\forall~ \alpha\in \mathcal{R} \\
\implies & 0=n\cdot(q\cdot\alpha)+r\cdot\alpha~\forall~ \alpha\in \mathcal{R}. \\
\implies & n\cdot 0+r\cdot\alpha=0~\forall~ \alpha\in \mathcal{R}. \\
\implies & r\cdot\alpha=0 ~\forall~ \alpha\in \mathcal{R}.
\end{align*}
But \(r \text{\textless} n\) and \(n\) is the least positive integer such that \( n\cdot\alpha=0 ~\forall~ \alpha\in \mathcal{R} \). Therefore the only possibility is \(r =0\).
