Finding the units in $M_n(R)$

Let $R$ be a ring. Then I would like to know the units in $M_n(R)$.

Here is my idea. The determinant is a homomorphism from $M_n(R)$ to $R$. If $A$ is a unit, then there is a $B$ such that $AB = I$. Then taking determinants one finds that $\det(A)$ is a unit in $R$. So therefore the units in $M_n(R)$ are those matrices $A$ where $\det(A)$ is a unit in $R$.

Is this correct? Can this be done more explicitly?

• You want $R$ to be commutative. Nov 10 '15 at 14:08
• I don't know if det is still a homomorphism when $R$ is not commutative Nov 10 '15 at 14:08
• You seem to be confusing the statement with its converse. You've argued that if $A$ is a unit, then $\det(A)$ is also a unit. However, you need to actually argue the converse! Start with the assumption that $\det(A)$ is a unit and show that $A$ is as well. Nov 30 '15 at 15:07

When $R$ is commutative, you have $$\mathbf{A}\, \mathrm{adj}(\mathbf{A}) = \mathrm{adj}(\mathbf{A})\, \mathbf{A} = \det(\mathbf{A})\, \mathbf I_n \qquad$$ where $\mathrm{adj}(\mathbf{A})$ is the adjugate or classical adjoint of $\mathbf{A}$, that is, the transpose of its cofactor matrix.

This equation proves that

If $\det(\mathbf{A})$ is invertible in $R$, then $\mathbf{A}$ is invertible in $M_n(R)$.

because then $\mathbf{A}^{-1}=\det(\mathbf{A})^{-1}\mathrm{adj}(\mathbf{A})$.

Your argument based on $\det(\mathbf{AB})=\det(\mathbf{A})\det(\mathbf{B})$ proves the converse:

If $\mathbf{A}$ is invertible in $M_n(R)$, then $\det(\mathbf{A})$ is invertible in $R$.

Hint: Use the cofactor matrix if R is commutative

• So if my conclusion correct if $R$ is commutative? Nov 10 '15 at 14:24