Solution: Recall that \[ R\left[ \mathbb{Z} / 2\mathbb{Z} \right] \coloneqq \left\{ a \cdot\bar{0} + b\cdot \bar{1}: a,b \in R \right\} . \] Define a map \begin{gather*} \phi : R \left[ \mathbb{Z} / 2 \mathbb{Z} \right] \to R \times R, \\ a \cdot\bar{0} + b\cdot \bar{1} \mapsto (a,b). \end{gather*} Then we will show that $\phi $ is an isomorphism.

At first note that for $a \cdot\bar{0} + b\cdot \bar{1}, c \cdot\bar{0} + d\cdot \bar{1} \in R\left[ \mathbb{Z} / 2\mathbb{Z} \right] $, we have \begin{align*} \phi \left( \left( a \cdot\bar{0} + b\cdot \bar{1} \right) \left( c \cdot\bar{0} + d\cdot \bar{1} \right) \right) & = \phi \left( (ac + bd)\cdot \bar{0} + (ad + bc) \bar{1} \right) \\ & = (ac + bd, ad + bc) \\ & = (a,b) \cdot (c,d) \\ & = \phi \left( a \cdot\bar{0} + b\cdot \bar{1} \right) \cdot \phi \left( c \cdot\bar{0} + d\cdot \bar{1} \right) . \end{align*} Thus, $\phi $ is a homomorphism.

The kernel of $\phi $ is \begin{align*} \ker \phi & = \left\{ x = a \cdot\bar{0} + b\cdot \bar{1} \in R\left[ \mathbb{Z} /2\mathbb{Z} : \phi (x) = (0,0) \right] \right\} \\ & = \left\{ a \cdot\bar{0} + b\cdot \bar{1} \in R\left[ \mathbb{Z} /2\mathbb{Z} : (a,b) = (0,0) \right] \right\} \\ & = \{ 0 \} . \end{align*} Thus, $\phi $ is injective.

Also for any $(a,b) \in R \times R$, \[ \phi \left( a \cdot\bar{0} + b\cdot \bar{1} \right) = (a,b). \] Thus, $\phi $ is surjective and hence an isomorphism.