Prove that a mapping from C to M2(R) is injective and a homomorphismm

Solution 1:

A ring homomorphism from a field to any ring is necessarily injective (assuming that ring homomorphisms map $1$ to $1$), because its kernel is a proper ideal and a field has only $\{0\}$ as proper ideal.

In this case it's also easy to verify it directly, because if $\phi(a+bi)$ is the null matrix, then necessarily $a=b=0$.

The verification that $\phi$ is a homomorphism consists in showing that $$ \phi(x+y)=\phi(x)+\phi(y),\qquad \phi(xy)=\phi(x)\phi(y),\qquad \phi(1)=\begin{bmatrix}1 & 0\\0&1\end{bmatrix} $$ for all $x,y\in\mathbb{C}$. For multiplication, consider $x=a+bi$ and $y=c+di$; then $xy=(ac-bd)+(ad+bc)i$, so $$ \phi(xy)=\begin{bmatrix}ac-bd & ad+bc \\ -(ad+bc) & ac-bd \end{bmatrix} $$ whereas $$ \phi(x)\phi(y)= \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} c & d \\ -d & c \end{bmatrix} $$ Can you finish doing the matrix product? The check for the addition should be carried on similarly.

Note. In the definition of $\phi$ it is implicitly assumed that $a,b\in\mathbb{R}$ and the same assumption is made in the check, also for $c$ and $d$.