Isomorphism between Ring of Integers Modulo 2 and Parity Ring

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Theorem

The ring of integers modulo $2$ and the parity ring are isomorphic.


Proof

To simplify the notation, let the elements of $\Z_2$ be identified as $0$ for $\eqclass 0 2$ and $1$ for $\eqclass 1 2$.


Let $f$ be the mapping from the parity ring $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ and the ring of integers modulo $2$ $\struct {\Z_2, +_2, \times_2}$:

$f: \struct {\set {\text{even}, \text{odd} }, +, \times} \to \struct {\Z_2, +_2, \times_2}$:
$\forall x \in R: \map f x = \begin{cases}

0 & : x = \text{even} \\ 1 & : x = \text{odd} \end{cases}$


The bijective nature of $f$ is apparent:

$f^{-1}: \struct {\Z_2, +_2, \times_2} \to \struct {\set {\text{even}, \text{odd} }, +, \times}$:
$\forall x \in \Z_2: \map {f^{-1} } x = \begin{cases}

\text{even} & : x = 0 \\ \text{odd} & : x = 1 \end{cases}$


Thus the following equations can be checked:

\(\ds 0 +_2 0 = \ \ \) \(\ds \map f {\text{even} } +_2 \map f {\text{even} }\) \(=\) \(\ds \map f {\text{even} + \text{even} }\) \(\ds = 0\)
\(\ds 0 +_2 1 = \ \ \) \(\ds \map f {\text{even} } +_2 \map f {\text{odd} }\) \(=\) \(\ds \map f {\text{even} + \text{odd} }\) \(\ds = 1\)
\(\ds 1 +_2 0 = \ \ \) \(\ds \map f {\text{odd} } +_2 \map f {\text{even} }\) \(=\) \(\ds \map f {\text{odd} + \text{even} }\) \(\ds = 1\)
\(\ds 1 +_2 1 = \ \ \) \(\ds \map f {\text{odd} } +_2 \map f {\text{odd} }\) \(=\) \(\ds \map f {\text{odd} + \text{odd} }\) \(\ds = 0\)


and:

\(\ds 0 \times_2 0 = \ \ \) \(\ds \map f {\text{even} } \times_2 \map f {\text{even} }\) \(=\) \(\ds \map f {\text{even} \times \text{even} }\) \(\ds = 0\)
\(\ds 0 \times_2 1 = \ \ \) \(\ds \map f {\text{even} } \times_2 \map f {\text{odd} }\) \(=\) \(\ds \map f {\text{even} \times \text{odd} }\) \(\ds = 0\)
\(\ds 1 \times_2 0 = \ \ \) \(\ds \map f {\text{odd} } \times_2 \map f {\text{even} }\) \(=\) \(\ds \map f {\text{odd} \times \text{even} }\) \(\ds = 0\)
\(\ds 1 \times_2 1 = \ \ \) \(\ds \map f {\text{odd} } \times_2 \map f {\text{odd} }\) \(=\) \(\ds \map f {\text{odd} \times \text{odd} }\) \(\ds = 1\)

$\blacksquare$


These results can be determined from their Cayley tables:

Cayley Tables for Parity Ring

$\begin{array}{r|rr}

+ & \text{even} & \text{odd} \\ \hline \text{even} & \text{even} & \text{odd} \\ \text{odd} & \text{odd} & \text{even} \\ \end{array} \qquad \begin{array}{r|rr} \times & \text{even} & \text{odd} \\ \hline \text{even} & \text{even} & \text{even} \\ \text{odd} & \text{even} & \text{odd} \\ \end{array}$


Cayley Tables for $\Z_2$

$\begin{array} {r|rr}

\struct {\Z_2, +_2} & \eqclass 0 2 & \eqclass 1 2 \\ \hline \eqclass 0 2 & \eqclass 0 2 & \eqclass 1 2 \\ \eqclass 1 2 & \eqclass 1 2 & \eqclass 0 2 \\ \end{array} \qquad \begin{array}{r|rr} \struct {\Z_2, \times_2} & \eqclass 0 2 & \eqclass 1 2 \\ \hline \eqclass 0 2 & \eqclass 0 2 & \eqclass 0 2 \\ \eqclass 1 2 & \eqclass 0 2 & \eqclass 1 2 \\ \end{array}$


They can be presented more simply as:

$\begin{array}{r|rr}

+ & 0 & 1 \\ \hline 0 & 0 & 1 \\ 1 & 1 & 0 \\ \end{array} \qquad \begin{array}{r|rr} \times & 0 & 1 \\ \hline 0 & 0 & 0 \\ 1 & 0 & 1 \\ \end{array}$


Sources

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