Problem: Define $\varphi :\mathbb{C} \to M_2(\mathbb{R} )$ as \[ a + \iota b \mapsto \begin{pmatrix} a & b \\ -b & a \\ \end{pmatrix}. \] Show that $\varphi $ is an injective ring homomorphism.
Solution: We recall that $\varphi : R_1 \to R_2 $ is said to be a ring homomorphism between two rings $R_1$ and $R_2$ if for $r_1,r_2 \in R$
Let $z_1 , z_2 \in \mathbb{C} $ and assume that \[ z_j = a_j + \iota b_j,~ j = 1,2. \] Then we have \begin{align*} \varphi \left( z_1 + z_2 \right) & = \varphi \left( \left( a_1 + a_2 \right) + \iota \left( b_1 + b_2 \right) \right) \\ & = \begin{pmatrix} a_1 + a_2 & b_1 + b_2 \\ -\left( b_1 + b_2 \right) & a_1 + a_2 \\ \end{pmatrix} \\ & = \begin{pmatrix} a_1 & b_1 \\ -b_1 & a_1 \\ \end{pmatrix} + \begin{pmatrix} a_2 & b_2 \\ -b_2 & a_2 \\ \end{pmatrix} \\ & = \varphi \left( a_1 + \iota b_1 \right) + \varphi \left( a_2 + \iota b_2 \right) \\ & = \varphi \left( z_1 \right) + \varphi \left( z_2 \right) . \end{align*}
Similarly, \begin{align*} z_1 \cdot z_2 & = \left( a_1 + \iota b_1 \right) \cdot \left( a_2 + \iota b_2 \right) \\ & = \left( a_1 a_2 - b_1 b_2 \right) + \iota \left( a_1 b_2 + a_2 b_1 \right) . \end{align*} Therefore, \begin{align*} \varphi \left( z_1 \cdot z_2 \right) & = \varphi \left( a_1 a_2 - b_1 b_2 \right) + \iota \left( a_1 b_2 + a_2 b_1 \right) \\ & = \begin{pmatrix} a_1 a_2 - b_1 b_2 & a_1 b_2 + a_2 b_1 \\ - \left( a_1 b_2 + a_2 b_1 \right) & a_1 a_2 - b_1 b_2 \\ \end{pmatrix} \\ & = \begin{pmatrix} a_1 & b_1 \\ -b_1 & a_1 \\ \end{pmatrix} \begin{pmatrix} a_2 & b_2 \\ -b_2 & a_2 \\ \end{pmatrix} \\ & = \varphi \left( z_1 \right) \cdot \varphi \left( z_2 \right) . \end{align*}
Finally we need to show that $\varphi (1) = \text{Id}_2$, where $\text{Id}_2$ is the identity matrix of size $2$. Note that \[ \varphi \left( 1 \right) = \varphi (1 + \iota 0) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} = \text{Id}_2. \] Therefore, $\varphi $ is a ring homomorphism.
Now we need to show that it is injective. Let $z_1, z_2 \in \mathbb{C} $ such that $\varphi (z_1) = \varphi (z_2)$. We need to show that $z_1 = z_2$. Note that \begin{align*} \varphi (z_1) = \varphi (z_2) & \implies \varphi \left( a_1 + \iota b_1 \right) = \varphi \left( a_2 + \iota b_2 \right) \\ & \implies \begin{pmatrix} a_1 & b_1 \\ -b_1 & a_1 \\ \end{pmatrix} = \begin{pmatrix} a_2 & b_2 \\ -b_2 & a_2 \\ \end{pmatrix} \\ & \implies a_1 = a_2 \text{ and } b_1 = b_2 . \end{align*} Thus, $\varphi $ is injective.