Solution: In order to prove $\mathcal{C} $ is a subring of $M_2(\mathbb{R} )$, we need to check that it is closed under addition and multiplication and each element has an additive inverse. Let $C_i = \begin{pmatrix} a_i & b_i \\ -b_i & a_i \\ \end{pmatrix}\in \mathcal{C} $ for $i=1,2$. Then we have \begin{align*} C_1 + C_2 & = \begin{pmatrix} a_1 & b_1 \\ -b_1 & a_1 \\ \end{pmatrix} + \begin{pmatrix} a_2 & b_2 \\ -b_2 & a_2 \\ \end{pmatrix} \\ & = \begin{pmatrix} a_1+a_2 & b_1+b_2 \\ -\left( b_1 + b_2 \right) & a_1+a_2 \end{pmatrix} \in \mathcal{C} . \end{align*}

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Also, \begin{align*} C_1 \cdot C_2 & = \begin{pmatrix} a_1 & b_1 \\ -b_1 & a_1 \\ \end{pmatrix} \cdot \begin{pmatrix} a_2 & b_2 \\ -b_2 & a_2 \\ \end{pmatrix} \\ & = \begin{pmatrix} a_1a_2 - b_1b_2 & a_1 b_2 + b_1a_2 \\ -b_1a_2-a_1b_2 & -b_1a_2 + a_1b_2 \\ \end{pmatrix} \in \mathcal{C} . \end{align*} Now note that \[ \begin{pmatrix} a & b \\ -b & a \\ \end{pmatrix} + \begin{pmatrix} -a & -b \\ b & -a \\ \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \\ \end{pmatrix}. \] Thus, every element has an additive inverse and hence, we proved that $\mathcal{C} $ is a subring of $M_2(\mathbb{R} )$.

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2. We need to prove that $\mathcal{C}$ is a field. We need to show that every nonzero element has a multiplicative inverse. At first note that the multiplicative identity is \[ \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} \in \mathcal{C} . \] Now consider, \begin{align*} \begin{pmatrix} a & b \\ -b & a \\ \end{pmatrix} \cdot \begin{pmatrix} a & -b \\ b & a \\ \end{pmatrix} & = \begin{pmatrix} a^2 + b^2 & 0 \\ 0 & b^2 + a^2 \\ \end{pmatrix} \\ & = \left( a^2 + b^2 \right) \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}. \end{align*} Therefore, \[ \begin{pmatrix} a & b \\ -b & a \\ \end{pmatrix}^{-1} = \frac{1}{a^2 + b^2} \begin{pmatrix} a & -b \\ b & a \\ \end{pmatrix}. \] Therefore, $\mathcal{C} $ is a field.

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3. It is easy to prove that $\mathcal{R} $ is a subring of $\mathcal{C} $. Let us denote $A_r= \begin{pmatrix} r & 0 \\ 0 & r \\ \end{pmatrix}$. Note that $\mathcal{R} \subseteq \mathcal{C} $. Note that \begin{align*} A_{r_1}+A_{r_2} & = \begin{pmatrix} r_1 & 0 \\ 0 & r_1 \\ \end{pmatrix} + \begin{pmatrix} r_2 & 0 \\ 0 & r_2 \\ \end{pmatrix} \\ & = \begin{pmatrix} r_1+r_2 & 0 \\ 0 & r_1+r_2 \\ \end{pmatrix} = A_{r_1+r_2},\text{ and } \\ A_{r_1}\cdot A_{r_2} & = \begin{pmatrix} r_1 & 0 \\ 0 & r_1 \\ \end{pmatrix} \cdot \begin{pmatrix} r_2 & 0 \\ 0 & r_2 \\ \end{pmatrix} \\ & = \begin{pmatrix} r_1r_2 & 0 \\ 0 & r_1r_2 \\ \end{pmatrix} = A_{r_1r_2}. \end{align*}

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Also, the additive inverse of $A_r$ is $A_{-r}$. \vf \hf Consider the following map \[ \varphi : \mathcal{R} \to \mathbb{R},~A_r \mapsto r. \] We claim that the above map is an isomorphism. For that, \begin{align*} & \varphi \left( A_{r_1} + A_{r_2} \right) = \varphi \left( A_{r_1+r_2} \right) = r_1 + r_2 = \varphi \left( A_{r_1} \right) + \varphi \left( A_{r_2} \right). \\ & \varphi \left( A_{r_1}\cdot \varphi \left( A_{r_2} \right) \right) = \varphi \left( A_{r_1r_2} \right) = r_1r_2 = \varphi \left( A_{r_1} \right) \cdot \varphi \left( A_{r_2} \right). \end{align*} Thus, $\varphi $ is a homomorphism. Observe that $\varphi^{-1}(r) = A_r$. Thus, $\varphi$ is an isomorphism and hence $\mathcal{R} \cong \mathbb{R} $.

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Now it remains to show that every element of $\mathcal{C} $ can be uniquely written as $\alpha +\mathbf{i} \beta $ where $\iota =\begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix}$. Let $X = \begin{pmatrix} a & b \\ -b & a \\ \end{pmatrix}\in \mathcal{C} $. We claim that $X=A_a + \mathbf{i} A_b$. Observe that \begin{align*} A_a + \mathbf{i} A_b & = \begin{pmatrix} a & 0 \\ 0 & a \\ \end{pmatrix} + \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix} \begin{pmatrix} b & 0 \\ 0 & b \\ \end{pmatrix} \\ & = \begin{pmatrix} a & 0 \\ 0 & a \\ \end{pmatrix} + \begin{pmatrix} 0 & b \\ -b & 0 \\ \end{pmatrix} \\ & = \begin{pmatrix} a & b \\ -b & a \\ \end{pmatrix}. \end{align*}

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Now we will prove the uniqueness of the choice $A_a$ and $A_b$. Suppose that $X= A_c+\mathbf{i} A_d$. Note that \[ A_c + \mathbf{i} A_d = \begin{pmatrix} c & d \\ -d & c \\ \end{pmatrix}. \] Since \[ X = A_a+\mathbf{i} A_b = A_c + \mathbf{i} A_d \implies a=c\text{ and } b=d. \] Thus, the representation is unique. We also observe that \[ \mathbf{i} ^2 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix}^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \\ \end{pmatrix}= -\textbf{1} _\mathcal{R} . \]