Ring of Integers Modulo Prime is Field/Proof 2

Theorem

Let $m \in \Z: m \ge 2$.

Let $\struct {\Z_m, +, \times}$‎ be the ring of integers modulo $m$.

Then:

$m$ is prime

if and only if:

$\struct {\Z_m, +, \times}$ is a field.

Proof

Let $p$ be prime.

From Irreducible Elements of Ring of Integers, we have that $p$ is irreducible in the ring of integers $\struct {\Z, +, \times}$.

From Ring of Integers is Principal Ideal Domain, $\struct {\Z, +, \times}$ is a principal ideal domain.

Let $\ideal p$ denote the principal ideal of $\struct {\Z, +, \times}$ generated by $p$.

By Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal, $\ideal p$ is a maximal ideal of $\struct {\Z, +, \times}$.

Let $\Z / \ideal p$ denote the quotient ring of $\Z$ by $\paren p$.

By Maximal Ideal iff Quotient Ring is Field, $\Z / \ideal p$, is a field.

But $\Z / \ideal p$ is exactly $\struct {\Z_p, +, \times}$.

$\Box$

Let $p$ be composite.

Then $p$ is not irreducible in $\struct {\Z, +, \times}$.

Thus by Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal, $\ideal p$ is not a maximal ideal of $\struct {\Z, +, \times}$.

Hence by Maximal Ideal iff Quotient Ring is Field, $\Z / \ideal p$ is not a field.

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 63$. Construction of fields as factor rings