Proof
(related to Lemma: Any Positive Characteristic Is a Prime Number)
 By assumption, \((R, +,\cdot)\) is a ring with a multiplicative identity \(1\) and an additive identity \(0\), which is free of zero divisors.
 If the characteristic is \(\operatorname{char}( R )=0\), there is nothing to show.
 Suppose $\operatorname{char}( R )=n,$ $n > 0$ is composite.
 Therefore, there exist positive integers \(a,b\in\mathbb N\) with \(1 < a,b < n\) and \(a\cdot b=n\).
 It follows $n\cdot 1=(a\cdot b)\cdot 1=(a\cdot 1)\cdot(b\cdot 1)=0.$
 Since \(R\) is free of zero divisors, we have $(a\cdot 1)\cdot(b\cdot 1)=0\Longleftrightarrow (a\cdot 1)=0 \text{ or } (b\cdot 1)=0.$
 But this is a contradiction to the minimality of the characteristic, by its definition.
 Thus, $n > 0$ is a prime number.
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References
Bibliography
 Kramer Jürg, von Pippich, AnnaMaria: "Von den natürlichen Zahlen zu den Quaternionen", SpringerSpektrum, 2013