**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.13**

Solution: Define $$\varphi : \mathbb{C} \rightarrow M_2(\mathbb{R}) by a+bi \mapsto \begin{bmatrix} a & b \\ -b & a \end{bmatrix}.$$ This mapping is clearly well defined. Moreover, for all $a_1 + b_1i, a_2 + b_2i \in \mathbb{C}$, note the following.\begin{align*}\varphi((a_1 + b_1i) + (a_2 + b_2i)) =&\ \varphi((a_1+a_2) + (b_1+b_2)i)\\

=&\ \begin{bmatrix} a_1+a_2 & b_1+b_2 \\ -b_1-b_2 & a_1+a_2 \end{bmatrix}\\

=&\ \begin{bmatrix} a_1 & b_1 \\ -b_1 & a_1 \end{bmatrix} + \begin{bmatrix} a_2 & b_2 \\ -b_2 & a_2 \end{bmatrix}\\

=&\ \varphi(a_1 + b_1i) + \varphi(a_2 + b_2i)\end{align*} \begin{align*}\varphi((a_1 + b_1i)(a_2 + b_2i)) =&\ \varphi((a_1a_2 - b_1b_2) + (a_1b_2 + a_2b_1)i)\\

=&\ \begin{bmatrix} a_1a_2 - b_1b_2 & a_1b_2 + a_2b_1 \\ -a_1b_2 - a_2b_1 & a_1a_2 - b_1b_2 \end{bmatrix}\\

=&\ \begin{bmatrix} a_1 & b_1 \\ -b_1 & a_1 \end{bmatrix} \cdot \begin{bmatrix} a_2 & b_2 \\ -b_2 & a_2 \end{bmatrix}\\

=&\ \varphi(a_1 + b_1i) \varphi(a_2 + b_2i)\end{align*}Thus $\varphi$ is a ring homomorphism. Now suppose $a+bi$ is in the kernel of $\varphi$. Then $$\begin{bmatrix} a & b \\ -b & a \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix},$$ so that $a+bi = 0$. Since the kernel of $\varphi$ is trivial, it is injective.

By Proposition 5 in the text, $\mathsf{im}\ \varphi$ is a subring of $M_2(\mathbb{R})$ to which $\mathbb{C}$ is isomorphic.