Dieudonné determinant

Remarks

Such a homomorphism exists, by Dieudonné's theorem, and it is unique because $GL_n(D)$ is generated by $E_n(D)$ and the matrices $\mathrm{diag}(1, \dots, 1, a)$. For a diagonal matrix, $\det \mathrm{diag}(a_1, \dots, a_n)$ is the class of $a_1 a_2 \cdots a_n$. For $n = 2$ and $a \ne 0$, $\det \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is the class of $a d - a c a^{-1} b$; for $a = 0$ it is the class of $-cb$. Example: if $D$ is a field, then $D^{\times}_{\mathrm{ab}} = D^\times$ and the Dieudonné determinant is the usual determinant. Non-example: over the quaternions $\mathbb{H}$, the map $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto ad - bc$ is not multiplicative, so it is not the Dieudonné determinant; the order of the factors matters. For $n \ge 2$ the kernel of $\det$ is $E_n(D)$, and $\det$ induces an isomorphism $K_1(D) \cong D^{\times}_{\mathrm{ab}}$.