Show that $x^2+\mathbf{1}=\mathbf{0}$ has two solutions in this ring homoporphism

Question

I have the following question here:

1. Given $a,b\in\mathbb{R}$, let $$[a,b]=\begin{pmatrix}a & -b \\ b & a \end{pmatrix}\in M_2(\mathbb{R})$$ Define $F=\{[a,b]\in M_2{(\mathbb{R})}|a,b\in \mathbb{R}\}$. Assume that $F$ is a ring.

a) Show that $[a,b][c,d]=[ac-bd,ad+bc]$ for all $a,b,c,d \in \mathbb{R}$.

b) Show that $F$ is commutative and has the identity element $[1,0]$.

c) Show that if $a$ and $b$ are both not zero then:
$\Big[\frac{a}{a^2+b^2},\frac{-b}{a^2+b^2} \Big][a,b]=[1,0]$.

I already did this entire question but the 2nd question relates to the first part.

2. Refer to question 1.

a) Show that $[a,b]+[c,d]=[a+c,b+d]$ for all $a,b,c,d \in \mathbb{R}$.

b) Show that the map $$\varphi: \mathbb{R} \rightarrow F$$
$$a \mapsto [a,0]$$
is an injective ring homoporphism.

c) If $\mathbf{0}=[0,0]$, the zero element of $F$, and $\mathbf{1}=[1,0]$, the identity element, show that the equation $x^2+\mathbf{1}=\mathbf{0}$ has exactly two solutions $x\in F$.

I already did parts a and b. It's part c I am not sure about.

$x^2+\mathbf{1}=\mathbf{0}$ implies we have $x^2+\begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix}$ or in other words, $x^2=\begin{pmatrix}-1 & 0 \\ 0 & -1 \end{pmatrix}=[-1,0]$.

I'm not sure how to solve for $x$. My mind is thinking complex numbers but I am not sure how to justify this or reason this out.

My idea was to go $[a,b][a,b]=[a^2-b^2,a^2-b^2]=[1,0]$. So this means $a^2-b^2=1$ and $a^2-b^2=0$ but the second equation implies $a=-b$ or $a=b$ but then both solutions imples the first equations have so solutions so this means we have an unique factorization and we only have 2 solutions.

Is that kind of the right idea?

Answer 1

You made a mistake calculating $[a,b][a,b]$. It should be $$ [a,b][a,b] = [a^2-b^2, 2ab] \stackrel{!}= [-1,0]. $$ Since $ab=0$ but $a^2-b^2=-1\neq 0$, exactly one of the real numbers $a$ and $b$ is zero.

If $a=0$, then $b^2=1$ implies $b=1$ or $b=-1$ and indeed $[0,1]$ and $[0,-1]$ both solve the equation.

If $b=0$ then $a^2=-1$ has no real solution so this case is excluded.

Hence the equation $[a,b]^2 + [1,0] = [0,0]$ has exactly the two solutions $[0,1]$ and $[0,-1]$.