I'm trying to solve this question:
Prove that the set of $2 \times 2$-matrices of the form
$$
\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right), \quad a, b, c, d \in \mathbb{R}
$$
with the usual matrix addition and multiplication is a non-commutative ring (that is, the multiplication is not commutative). Can you find conditions on $a, b, c, d$ to make it into a field?
I already did the first part and showed that this is a non-commutative ring. Now, I'm trying to find the conditions to make it be a field. For this, we just have to find the multiplicative inverse, which from linear algebra we know that a $2\times 2$ matrix is invertible if $\operatorname{det} (A) \neq 0$, which means $ad-bc \neq 0$ or equivalently, one row isn't a multiple of the other. Now, I just want to know that if this is enough!?
Best Answer
This is how I finally did:
if $a=d$ and $b=-c$, the set $SM_2(\mathbb{R})=\bigg\{\begin{pmatrix} a & -b \\ b & a\\ \end{pmatrix} \bigg| a, b \in \mathbb{R}\bigg\}$ turns into a field because:
$SM_2(\mathbb{R})$ is commutative, because: \begin{equation*} \begin{aligned} \begin{pmatrix} a & -b \\ b & a\\ \end{pmatrix} \begin{pmatrix} c & -d \\ d & c\\ \end{pmatrix} &= \begin{pmatrix} ac-bd & -ad-bc \\ bc+ad & -bd+ac\\ \end{pmatrix} \\ &= \begin{pmatrix} c & -d \\ d & c\\ \end{pmatrix} \begin{pmatrix} a & -b \\ b & a\\ \end{pmatrix} \end{aligned} \end{equation*}
The multiplicative inverse exists and is the inverse of matrix $A$, because $\det(A)=a^2-(-b)b=a^2+b^2\neq 0$, so $A$ is invertible and $A^{-1}=\frac{1}{a^2+b^2}\begin{pmatrix} a & b \\ -b & a\\ \end{pmatrix}$ and \begin{equation*} \begin{aligned} \begin{pmatrix} a & -b \\ b & a\\ \end{pmatrix} \begin{pmatrix} \frac{a}{a^2+b^2} & \frac{b}{a^2+b^2} \\ \frac{-b}{a^2+b^2} & \frac{a}{a^2+b^2}\\ \end{pmatrix} &= \begin{pmatrix} \frac{a^2+b^2}{a^2+b^2} & \frac{ab-ba}{a^2+b^2} \\ \frac{ba-ab}{a^2+b^2} & \frac{b^2+a^2}{a^2+b^2}\\ \end{pmatrix} \\ &= \begin{pmatrix} 1 & 0 \\ 0 & 1\\ \end{pmatrix} \end{aligned} \end{equation*}
Thus, $SM_2(\mathbb{R})$ is a field. Plus, it cannot be an ordered field. For this, consider $\varphi: SM_2(\mathbb{R}) \longrightarrow \mathbb{C}$ such that $\varphi\left( \begin{pmatrix} a & -b \\ b & a\\ \end{pmatrix}\right)=a+bi$. We have:
For any $A, B \in SM_2(\mathbb{R})$, we have: \begin{equation*} \begin{aligned} \varphi(AB)&= \varphi\left( \begin{pmatrix} a & -b \\ b & a\\ \end{pmatrix}\begin{pmatrix} c & -d \\ d & c\\ \end{pmatrix}\right)= \varphi\left( \begin{pmatrix} ac-bd & -ad-bc \\ bc+ad & -bd+ac\\ \end{pmatrix} \right) \\ &= (ac-bd)+(ad+bc)i= ac+adi+bci+bdi^2\\ &= a(c+di)+bi(c+di)= (a+bi)(c+di)\\ &= \varphi\left( \begin{pmatrix} a & -b \\ b & a\\ \end{pmatrix}\right) \varphi\left(\begin{pmatrix} c & -d \\ d & c\\ \end{pmatrix}\right)=\varphi(A)\varphi(B) \end{aligned} \end{equation*} So, $\varphi$ is a homomorphism.
For any $A, B \in SM_2(\mathbb{R})$, assume that $\varphi(A)=\varphi(B)$, then $a+bi=c+di$, so $(a-c)+(b-d)i=0$, and hence, $a=c$ and $b=d$ which means that $A=\begin{pmatrix} a & -b \\ b & a\\ \end{pmatrix}= \begin{pmatrix} c & -d \\ d & c\\ \end{pmatrix} =B$. So, $\varphi$ is injective.
Now, let $a+bi\in \mathbb{C}$. Then, since we defined matrices in $SM_2(\mathbb{R})$ to be all the $2\times 2$ matrices with $a=d$, and $b=-c$, there exists $A=\begin{pmatrix} a & -b \\ b & a\\ \end{pmatrix} \in SM_2(\mathbb{R})$ such that $\varphi\left( \begin{pmatrix} a & -b \\ b & a\\ \end{pmatrix}\right)=a+bi$.
So $\varphi$ is an isomorphism. Hence, $SM_2(\mathbb{R})$ is isomorphic to $\mathbb{C}$ and since no order can be defined on $\mathbb{C}$, the aforementioned field cannot take any orders as well.