Proving $\text{GL}_{2}(R): Showing Homomorphism & Isomorphism

[NoName3]

Let $R$ be a commutative ring and let $\text{M}_2(R)$ denote the ring of $2 \times 2$ matrices with coefficients in $R$.

(a) Show that the group of units in $\text{M}_2(R)$ is $\text{GL}_2(R) = \left\{A \in \text{M}_2(R): \text{det}(A) \in R^{\times} \right\}$;
(b) show tha $\text{GL}_2(R) \to R^{\times}$ is a homomorphism;
(c) prove that $\text{GL}_2(R)$ is isomorphic to $S_3.$

(a) $\text{GL}_{2}(R) = \left\{\dfrac{1_R}{\text{det}(A)}A^{-1}: \det(A) \ne 0_R\right\}$. But $\dfrac{1_R}{\text{det}(A)} = \det(A^{-1})$ and $\det(A^{-1})\cdot A^{-1}$ $ = A \in \text{M}_2(R).$
Moreover, $\text{det}(A) \ne 0 \implies \text{det}(A) \in R^{\times}$ thus $\text{GL}_{2}(R) = \left\{A \in \text{M}_2(R): \det(A) \in R^{\times}\right\}, $ is that correct?

 

[Deveno]

This looks a bit backwards to me. You want to show that if $\det(A) \in R^{\times}$, that there exists $B$ such that $AB = I_2$.

You can do this by calculating, for:

$A = \begin{bmatrix}a&b\\c&d\end{bmatrix}$

$B = (\det(A))^{-1}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$

(note this only makes sense if $\det(A) \in R^{\times}$).

This shows that $\text{GL}_2(R) \subseteq (M_2(R))^{\times}$.

To show the reverse inclusion, it is easiest to use part (b) first: we have for a matrix $A$ in $(M_2(R))^{\times}$ that there is a matrix $B$ such that $AB = I_2$.

Hence $1 = \det(I_2) = \det(AB) = \det(A)\det(B)$, which shows that $\det(A)$ is a unit of $R$.

It's not clear what you mean by "integers in $R$", for a general commutative ring $R$.