Let $M_2(\mathbb{R})$ denote the set of all $2 \times 2$ matrices obviously with all Real entries.
let $R =$
$$
\begin{bmatrix}
a & b \\
-b & a \\
\end{bmatrix}
$$
Prove that $R$ is subring of $M_2(\mathbb{R})$
By definition a subrinng S of a ring R is a subgroup of R under the addition it inherits from 𝑅 which is closed under the inherited multiplication and has the same multiplicative identity as 𝑅. In order to show $R$ is a subgroup I need to show the following:
1) R is closed under addition and multiplication
2) R is associative
3) R has an identity element
4) Every element has an inverse
Number 1)
Well $R+R$ yields:
$$
\begin{bmatrix}
2a & 2b \\
-2b & 2a \\
\end{bmatrix}
$$
and
$R \times R$ yields:
$$
\begin{bmatrix}
a^2 – b^2 & 2ab \\
-2ab & b^2-a^2 \\
\end{bmatrix}
$$
Since $2a,2b,-2b \in \mathbb{R}$ and since $a^2-b^2,2ab,-2ab,b^2-a^2 \in \mathbb{R}$ It follows that R is closed under addition and multiplication
Number 2: I am not sure how to say that it is associative. I should be able to imply this from the fact that it is in the set of all $2 \times 2$ matrices
Number 3: $R$ has the identity element $I_2$ which is denoted as:
$$
\begin{bmatrix}
1 & 0 \\
0 & 1 \\
\end{bmatrix}
$$
Number 4: By definition the inverse of a $2 \times 2$ matrix is:
$\frac{1}{ad-bc} \times$
$$
\begin{bmatrix}
d & -b \\
-c & a \\
\end{bmatrix}
$$
So in the case of $R+R$ the inverse is $\frac{1}{4a^2+4b^2} \times$:
$$
\begin{bmatrix}
2a & -2b \\
2b & 2a \\
\end{bmatrix}
$$
and in the case of $R \times R$ the inverse is $\frac{1}{((a^2-b^2)(b^2-a^2))-(4a^2b^2)} \times$
$$
\begin{bmatrix}
b^2-a^2 & -2ab \\
2ab & a^2-b^2 \\
\end{bmatrix}
$$
Does this work properly show that $R$ is a subgroup of $M_2(\mathbb{R})$
Clearly $I_2$ which is the multiplicative dentity of $R$ is also the same multiplicative identity of $M_2(\mathbb{R})$ which is mentioned in the definition.
Does adding this last part in with the rest of the work show that $R$ is infact a subring?
Best Answer
You do way too much work. By the subring criterion, it is both sufficient and necessary to check $$1 \in R$$ $$x,y \in R \implies x-y \in R$$ $$x,y \in R \implies xy \in R$$
all of which follow after an easy computation.