Prime Number

The notion of a prime number is a special case of the ideas of a prime element and irreducible element in the ring of integers \(\mathbb{Z}\).

This definition looks much like that of irreducibility in rings.

Intuitively we may invoke the fact that every irreducible element is prime in a unique factorisation domain to prove that such elements are prime elements. However typically the defining property of prime elements is used to prove that \(\mathbb{Z}\) is a unique factorisation domain. Therefore, we must prove this property of prime numbers prior to proving the equivalence of prime element and irreducible elements. The relevant result proving this is Euclid's lemma.

Once this is known, we can prove the following.

Proof

Suppose \(n > 1\) is a prime number and \(n = ab\) for some \(a, b \in \mathbb{Z}\). Since \(n > 1\) we know \(n \neq 0\) and hence \(a, b \neq 0\). As such, we assume without loss of generality that \(a\) and \(b\) are greater than zero, since if they are both negative, we can work with their corresponding absolute values. Since \(n = ab\), \(a \mid n\) and \(b \mid n\). As such, \(a, b \in \{1, n\}\) because \(n\) is prime. It is impossible for \(a = b = n\) otherwise \(ab = n^2 \neq n\) because \(n > 1\). Therefore either \(a = 1\) or \(b = 1\) and we have that \(n\) is irreducible because \(\pm 1\) are units.

For the reverse implication, assume \(n > 1\) is irreducible. Then if \(n = ab\), either \(a\) or \(b\) is a unit. If \(a\) is a unit then \(a \in \{\pm 1\}\) and therefore \(b = \{\pm n\}\); likewise for \(b\). Since we may choose \(a\) to be any divisor of \(n\) and adjust \(b\) accordingly, we can conclude that the only divisors of of \(n\) are \(\{\pm 1, \pm n\}\) and hence the only positive divisors are \(1\) and \(n\), thus \(n\) is a prime number.

Since \(\mathbb{Z}\) is a unique factorisation domain by the fundamental theorem of arithmetic we know that the set of irreducible elements is equivalent to the set of prime elements.