Ring of Integers Modulo Prime is Field

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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.


Corollary

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 an integral domain.


Proof 1

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.

$\Box$


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.

$\blacksquare$


Proof 2

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.

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

Hence 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.

$\blacksquare$


Proof 3

Let $m$ be prime.

From Ring of Integers Modulo Prime is Integral Domain, $\struct {\Z_m, +, \times}$ is an integral domain.

Let $\eqclass a m \ne \eqclass 0 m$ be a residue class modulo $m$.

We need to find a residue class modulo $m$ $\eqclass x m$ such that $\eqclass a m \eqclass x m = \eqclass 1 m$.

Because $m$ is prime and $m \nmid a$, we have:

$\gcd \set {a, m} = 1$

Hence from Bézout's Lemma:

$1 = x a + y m$

for some $x, y \in \Z$.

Thus we have:

\(\displaystyle \eqclass a m \eqclass x m\) \(=\) \(\displaystyle \eqclass {a x} m\)
\(\displaystyle \) \(=\) \(\displaystyle \eqclass {1 - y m} m\)
\(\displaystyle \) \(=\) \(\displaystyle \eqclass 1 m\)

So every non-zero residue class modulo $m$ has an inverse.

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

$\Box$


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.

$\blacksquare$


Proof 4

Let $m$ be prime.

From Ring of Integers Modulo Prime is Integral Domain, $\struct {\Z_m, +, \times}$ is an integral domain.

From Finite Integral Domain is Galois Field, $\struct {\Z_m, +, \times}$ is a field.

$\Box$


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.

$\blacksquare$


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