Ring of Integers Modulo Prime is a Field

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Theorem

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

Let $\left({\Z_m, +, \times}\right)$‎ be the ring of integers modulo $m$.


Then:

$m$ is prime

iff:

$\left({\Z_m, +, \times}\right)$ is a field.


Corollary

Let $\left({\Z_m, +, \times}\right)$‎ be the ring of integers modulo $m$.


Then:

$m$ is prime

iff:

$\left({\Z_m, +, \times}\right)$ is an integral domain.


Proof

Prime Modulus

$\left({\Z_m, +, \times}\right)$‎ is a commutative ring with unity by definition.

From Multiplicative Group of Integers Modulo m, $\left({\Z'_m, \times}\right)$ 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 Set of Coprime Integers:

$\Z'_m = \left\{{\left[\!\left[{1}\right]\!\right]_m, \left[\!\left[{2}\right]\!\right]_m, \ldots, \left[\!\left[{m-1}\right]\!\right]_m}\right\}$

That is:

$\Z'_m = \Z_m \setminus \left\{{\left[\!\left[{0}\right]\!\right]_m}\right\}$

where $\setminus$ denotes set difference.

Hence the result.

$\Box$


Composite Modulus

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

From Ring of Integers Modulo Composite is Not Integral Domain, $\left({\Z_m, +, \times}\right)$ is not an integral domain.

We have that a field is an integral domain.

Therefore $\left({\Z_m, +, \times}\right)$ is not a field.

$\blacksquare$


Proof of Corollary

We have that a field is an integral domain.

We also have that a finite integral domain is a field.

Hence the result.

$\blacksquare$


Sources

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