Multiplicative Inverse in Ring of Integers Modulo m
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Theorem
Let $\struct {\Z_m, +_m, \times_m}$ be the ring of integers modulo $m$.
Then $\eqclass k m \in \Z_m$ has an inverse in $\struct {\Z_m, \times_m}$ if and only if $k \perp m$.
Proof 1
First, suppose $k \perp m$.
That is:
- $\gcd \set {k, m} = 1$
Then, by Bézout's Identity:
- $\exists u, v \in \Z: u k + v m = 1$
Thus:
- $\eqclass {u k + v m} m = \eqclass {u k} m = \eqclass u m \eqclass k m = \eqclass 1 m$
Thus $\eqclass u m$ is an inverse of $\eqclass k m$.
Suppose that:
- $\exists u \in \Z: \eqclass u m \eqclass k m = \eqclass {u k} m = 1$.
Then:
- $u k \equiv 1 \pmod m$
and:
- $\exists v \in \Z: u k + v m = 1$
Thus from Bézout's Identity:
- $k \perp m$
$\blacksquare$
Proof 2
From Ring of Integers Modulo m is Ring, $\left({\Z_m, +_m, \times_m}\right)$ is a commutative ring with unity $\left[\!\left[{1}\right]\!\right]_m$.
Thus by definition $\left({\Z_m, \times_m}\right)$ is a commutative monoid.
The result follows from Multiplicative Inverse in Monoid of Integers Modulo m.
$\blacksquare$