describe the ring $R=M_2(\Bbb F)$, where $\Bbb F$ is a field Let the ring $R=M_2(\Bbb F)$, where $\Bbb F$ is a field.


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*What is the description $R^*$.

*Find order $(M_2(\Bbb Z_3))^*$.
 A: *

*Given a ring $A$, $A^\star$ is the group of invertible elements of $A$ (that is, the group of units). Here, they are invertible matrices, that is matrices $M=\left(\begin{matrix} a & b\\c & d\end{matrix}\right)$ with $ad-bc\neq0$, because in a field, every nonzero element has an inverse.
The inverse of $M$ is then $\dfrac{1}{ad-bc}\left(\begin{matrix} d & -c\\-b & a\end{matrix}\right)$.

*It's easier to count singular matrices. Since $\Bbb Z_3$ is a field (with three elements), equation $ad-bc=0$ has always a solution in $a$ if $d\neq0$, thus for each $d \in \{1,2\}$, you have $3^2$ solutions, so $18$. And if $d=0$, any $a$ is valid, and you must have $bc=0$, thus $b=0$ or $c=0$ (that is, $3+3-1=5$ solutions for each $a$, so $5\times3$ total solutions when $d=0$). Therefore, there are $18+15=33$ matrices which are not invertible, among $3^4=81$ matrices, thus $|R^\star|=81-33=48$. That is $GL_2(3)$ is a group of order $48$.

You can generalize the above result, to a field with $n$ elements (where necessarily $n=p^k$ for some prime $p$ and some integer $k\geq0$).


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*The equation $ad-bc$ has a solution in $a$ for each $d\neq0$, that is $n-1$ values of $d$ and $n^2$ values for $b$ and $c$, thus $n^2(n-1)$ such matrices.

*When $d=0$, all $n$ values of $a$ are valid, and you must have $b=0$ or $c=0$. This amounts to $n(2n-1)$ solutions. (pay attention to the fact that $2n$ instead of $2n-1$ would count $b=c=0$ twice)
All in all, there are $n^2(n-1)+n(2n-1)$ singular matrices, and the number of regular matrices is thus
$$n^4-n^2(n-1)-n(2n-1)=n^4-n^3-n^2+n=n[n^2(n-1)-(n-1)]=n(n-1)(n^2-1)$$
Therefore, the group $GL(2,n)$ has order $n(n-1)(n^2-1)$.
But notice the formula is valid for a group of matrices with coefficients in a finite field. More generally, you may also consider matrices over a ring, such as $\Bbb Z/4$, which is not a field. Then the matrix ring has also a group of units, but it's more complicated to compute its order. The determinant has then to be a unit of the ring of coefficients.
