Ring of Integers Modulo Prime is Field
Theorem
Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$ be the ring of integers modulo $m$.
Then:
- $m$ is prime
- $\struct {\Z_m, +, \times}$ 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
- $\struct {\Z_m, +, \times}$ 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 Identity:
- $1 = x a + y m$
for some $x, y \in \Z$.
Thus we have:
\(\ds \eqclass a m \eqclass x m\) | \(=\) | \(\ds \eqclass {a x} m\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \eqclass {1 - y m} m\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \eqclass 1 m\) |
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.
$\blacksquare$
Also see
Sources
- 1967: John D. Dixon: Problems in Group Theory ... (previous) ... (next): Introduction: Notation
- 1978: John S. Rose: A Course on Group Theory ... (previous) ... (next): $0$: Some Conventions and some Basic Facts
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 55$. Special types of ring and ring elements: $(4)$