Integers Modulo Prime Finite Field

Finite fields of prime order can be constructed very simply using \(\mathbb{Z}_p\) with addition and multiplication modulo \(p\) where \(p\) is prime.

Proof

We will take it as given that \(\mathbb{Z}_p\) is a commutative ring with with identity, although this is routine to prove.

This is a corollary of every finite integral domain is a field, and the proof of that fact is rather elementary and can be substituted directly here without the use of any advanced techniques. That means, we must prove that \(\mathbb{Z}_n\) is an integral domain if and only if \(n\) is prime.

If \(n\) is composite then clearly \(n = ab\) for some \(a, b \neq \pm 1, \pm n\) and then \(ab \equiv 0 \pmod n\) gives a pair of non-zero zero divisors.

If \(n\) is prime then \(ab = 0 \pmod n\) would imply that \(n \mid ab\) which means that \(n \mid a\) or \(n \mid b\) from the definition of prime elements in rings, of which prime numbers are.

Invertibility of non-zero elements can also be taken as a corollary of the Bezout identity. Similarly we can prove this by proving that \(p \mathbb{Z}\) is a maximal ideal of \(\mathbb{Z}\), a fact that follows directly from the lack of divisors. That is, \(m\mathbb{Z} \subseteq n \mathbb{Z}\) if and only if \(n \mid m\), and the only divisors of prime numbers are \(1\) (which yields the whole ring) and the prime itself (which yields the same ideal).