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 $a,b\ne \pm 1,\pm n$ and then $ab\equiv 0(\mathrm{mod}n)$ gives a pair of non-zero zero divisors.

If $n$ is prime then $ab=0(\mathrm{mod}n)$ 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).