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

Question

$\mathcal{M}_2(\Bbb R)$ is the ring of $2\times2$ matrices with real entries. Define a map $\phi:\Bbb C \to \mathcal{M}_2(\Bbb R)$ by

$$\phi(a+bi) = \begin{pmatrix} a & b \\ -b & a \end{pmatrix}$$

Prove that the mapping is a homomorphism and that it is injective.

Now, in order to show that it is a homomorphism is it enough if I simply show that $\phi(a)+\phi(b) = \phi(a+b)$ and $\phi(a)\phi(b) = \phi(ab)$?

Also, how do I show that it is injective? Does this have something to do with the kernel? If so, how do I find the kernel?

Answer 1

Hint:

This is a ''canonical '' isomorphism between $\mathbb{C}$ and a subring of $M_2(\mathbb{R})$ (here).

to show this fact the first step is to prove that the set $\mathcal{C}$ of matrices of the form $$ \begin{bmatrix} a&b\\ -b&a \end{bmatrix} \qquad a,b \in \mathbb{R} $$ is a field.

The associative and distributive properties of sum and product are inherited by the ring structure of $M_2(\mathbb{R}$, and so the existence of the neutral elements for the two operations. But we have to prove that any element of $\mathcal{C}$ has an inverse and that the product is commutative.

Commutativity is easy verified :

$$ \begin{bmatrix} a&b\\ -b&a \end{bmatrix} \begin{bmatrix} x&y\\ -y&x \end{bmatrix} = \begin{bmatrix} ax-by&ay+bx\\ -bx-ay&-by+ax \end{bmatrix}= \begin{bmatrix} x&y\\ -y&x \end{bmatrix} \begin{bmatrix} a&b\\ -b&a \end{bmatrix} $$

For the invertibility note that, for $A \in \mathcal{C} $, $\det(A)=a^2+b^2 \ne0 \forall a,b \in \mathbb{R} $ and $$ \begin{bmatrix} a&b\\ -b&a \end{bmatrix}^{-1}= \dfrac{1}{a^2+b^2} \begin{bmatrix} a&-b\\ b&a \end{bmatrix} $$

Now (second step) it is easy to see that for the given $\phi$ we have $$ \phi(1)=\begin{bmatrix} 1&0\\ 0&1 \end{bmatrix} $$ and $$ \phi(z)=A \Rightarrow \phi(\bar z)=A^T\Rightarrow \phi(z^{-1})=A^{-1}$$

and , obviously, $ \phi(z)=A \Rightarrow \phi(-z)=-A$.

The last step is to show that $ker (\phi)$ is trivial, but this is immediate since $$ \phi(a+ib)=\begin{bmatrix} 0&0\\ 0&0 \end{bmatrix} \iff a=b=0 $$

Finally: note that there are many (infinite) representations of $\mathbb{C}$ of this type, but is seems that there only two such that $\psi(\bar z)=A^T$ ( see the question: Matrix representation of $\mathbb{C}$ as $^*$Algebra.).

Answer 2

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 $d$.