Prove that the characteristic of a finite ring \( \mathcal{R} \) with no zero divisors is a prime number.
Solution:
Let \((\mathcal{R}, +, \cdot)\) is finite ring having no divisors of zero. Therefoer Char \(\mathcal{R}=n.\) is finite since the characteristic of a finite ring is finite. Let Char \(\mathcal{R}=n.\)
To prove, \(n\) is a prime.
Since, \( \mathcal{R} \) is a finite ring having no divisors of zero then \( \mathcal{R} \) is a ring with unity \( I \). Then, by definition of the the characteristic of a ring, \(n\) is the least positive integer such that \begin{equation} n\cdot I=0 \end{equation}
where \(0\) is the zero element of \(\mathcal{R}.\\ \)
If possible let, \(n\) is composite. Then \( \exists \) two integers \(p\text{\textless} n\) and \(q \text{\textless} n\) such that \(n=pq\). Then from (1), we have
\begin{align*}
& (pq) \cdot I=0 \\
\implies & (p\cdot I)\cdot (q\cdot I)=0\\
\end{align*}
This implies either \(p\cdot I=0\) or \(q\cdot I=0\), since \( \mathcal{R} \) has no divisors of zero. \( \\ \)
But \(p\cdot I \neq 0\) since \(n\) is the least positive integer such that \(n\cdot I=0\) and \(p\text{\textless} n\). Also \(q\cdot I \neq 0\) since \(n\) is the least positive integer such that \(n\cdot I=0\) and \(q \text{\textless} n\). \( \\ \)
Therefore our assumption is wrong.
