Ring of Integers Modulo Prime is Field/Proof 3
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.
Proof
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.
$\blacksquare$
Sources
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $4$: Fields: $\S 15$. Examples of Fields: Example $20$