Ring of Integers Modulo Prime is Field/Proof 1

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

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


$m$ is prime

if and only if:

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


Prime Modulus

$\struct {\Z_m, +, \times}$‎ is a commutative ring with unity by definition.

From Reduced Residue System under Multiplication forms Abelian Group, $\struct {\Z'_m, \times}$ is an abelian group.

$\Z'_m$ consists of all the elements of $\Z_m$ coprime to $m$.

Now when $m$ is prime, we have, from Reduced Residue System Modulo Prime:

$\Z'_m = \set {\eqclass 1 m, \eqclass 2 m, \ldots, \eqclass {m - 1} m}$

That is:

$\Z'_m = \Z_m \setminus \set {\eqclass 0 m}$

where $\setminus$ denotes set difference.

Hence the result.


Composite Modulus

Now suppose $m \in \Z: m \ge 2$ is composite.

From Ring of Integers Modulo Composite is not Integral Domain, $\struct {\Z_m, +, \times}$ is not an integral domain.

From Field is Integral Domain $\struct {\Z_m, +, \times}$ is not a field.