Units in $M_2(\mathbb Z_2)$

Detailed Solution

A matrix in $M_2(\mathbb Z_2)$ is a **unit** if and only if it is invertible. Since $\mathbb Z_2=\{0,1\}$, there are $$ 2^4=16 $$ matrices in $M_2(\mathbb Z_2)$ altogether. To count the invertible matrices, we can choose their columns successively. For an invertible $2\times2$ matrix over $\mathbb Z_2$: The first column can be any nonzero vector in $\mathbb Z_2^2$. There are $$ 2^2-1=3 $$ choices: $$ \begin{pmatrix}1\\0\end{pmatrix}, \quad \begin{pmatrix}0\\1\end{pmatrix}, \quad \begin{pmatrix}1\\1\end{pmatrix}. $$ The second column must be linearly independent of the first column. The vector space $\mathbb Z_2^2$ has $4$ vectors in total. Once a nonzero first column is chosen, its span contains $$ 2 $$ vectors: $$ \{0,v\}. $$ Therefore, there are $$ 4-2=2 $$ choices for the second column. Hence the number of invertible matrices is $$ (2^2-1)(2^2-2) =3\cdot2 =6. $$ Therefore, $$ \boxed{|R^\times|=6}. $$ Thus the correct answer is $$ \boxed{\mathrm{A}}. $$ For $$ A= \begin{pmatrix} a&b\\ c&d \end{pmatrix}, $$ the matrix is invertible over $\mathbb Z_2$ precisely when $$ \det(A)=ad-bc\neq0. $$ Since the only nonzero element of $\mathbb Z_2$ is $1$, this means $$ \det(A)=1. $$ Exactly $6$ of the $16$ matrices satisfy this condition. **Option B:** $8$ counts too many matrices; not every matrix with nonzero entries is invertible. **Option C:** $15$ is the number of nonzero matrices, but a nonzero matrix need not be invertible. For example, $$ \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix} $$ is nonzero but singular. **Option D:** $16$ is the total number of elements of $M_2(\mathbb Z_2)$, not the number of units. Therefore, $$ \boxed{\text{The ring }M_2(\mathbb Z_2)\text{ has exactly }6\text{ units}.} $$