Theorem 1

Suppose $n\geqslant 2$.

a) The subgroup $\mathbf{S}\mathbf{L}_n(D)$ of $\mathbf{G}\mathbf{L}_n(D)$ is generated by the matrices $B_{ij}(\lambda )$, where $i$ and $j$ are distinct integers in the interval $[1, n]$ and $\lambda$ runs through D.

b) Suppose $n\geqslant 3$ or Card(D) $\geqslant 3$. The derived group of $\mathbf{G}\mathbf{L}_n(D)$ is equal to $\mathbf{S}\mathbf{L}_n(D)$.

c) Suppose $n\geqslant 3$ or Card(D) $\geqslant 4$. The derived group of $\mathbf{S}\mathbf{L}_n(D)$ is equal to $\mathbf{S}\mathbf{L}_n(D)$.

A) Denote by T the subgroup of $\mathbf{G}\mathbf{L}_n(D)$ generated by the matrices $B_{ij}(\lambda )$. By Example 1 of VIII, p. 454, we have det($B_{ij}(\lambda )$) $= 1$, and therefore $T\subset \mathbf{S}\mathbf{L}_n(D)$. To prove that these two groups are equal, it then suffices, by Example 3 of loc. cit., to prove that every matrix of the form diag(1$, . . . ,1, d$) with $\pi (d) = 1$ belongs to T. The matrix diag(1$, . . . ,1, d$) belongs to the image of the homomorphism $U\mapsto \begin{pmatrix} I_{n-2} & 0 \\ 0 & U \end{pmatrix}$ from $\mathbf{G}\mathbf{L}_2(D)$ to $\mathbf{G}\mathbf{L}_n(D)$; it therefore suffices to consider the case $n= 2$. Since the kernel of $\pi$ is the derived group of $D^*$, we may assume $d=uvu^{-1}v^{-1}$ with $u, v$ in $D^*$. Our assertion then follows from the equalities

(28) $\begin{pmatrix} 1 & 0 \\ 0 & d \end{pmatrix} = \begin{pmatrix} u^{-1} & 0 \\ 0 & u \end{pmatrix}\begin{pmatrix} v^{-1} & 0 \\ 0 & v \end{pmatrix}\begin{pmatrix} vu & 0 \\ 0 & u^{-1}v^{-1} \end{pmatrix}$

and

(29) $\begin{pmatrix} s & 0 \\ 0 & s^{-1} \end{pmatrix} = \begin{pmatrix} 1 & s \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 1-s^{-1} & 1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 1-s & 1 \end{pmatrix}$

for $s\in D^*$.

B) By construction, $\mathbf{S}\mathbf{L}_n(D)$ is the kernel of a homomorphism from $\mathbf{G}\mathbf{L}_n(D)$ to a commutative group, so it contains the derived group $(\mathbf{G}\mathbf{L}_n(D),\mathbf{G}\mathbf{L}_n(D))$ of $\mathbf{G}\mathbf{L}_n(D)$. By the above, we have

To prove c), it suffices to prove that the matrices $B_{ij}(\lambda )$ are commutators in $\mathbf{S}\mathbf{L}_n(D)$.

Suppose $n\geqslant 3$. If $i,j,k$ are distinct integers in the interval $[1, n]$ and $\mu, \nu$ are elements of D, then we have

By taking $\mu= 1$ and $\nu =\lambda$, we see that the matrix $B_{ij}(\lambda )$ is a commutator of elements of $\mathbf{S}\mathbf{L}_n(D)$.

Now suppose $n= 2$. Let $u$ and $v$ be elements of D with $u\not= 0$. We have the relations

(31) $\begin{pmatrix} 1 & v-uvu \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} u & 0 \\ 0 & u^{-1} \end{pmatrix}\begin{pmatrix} 1 & -v \\ 0 & 1 \end{pmatrix}\begin{pmatrix} u^{-1} & 0 \\ 0 & u \end{pmatrix}\begin{pmatrix} 1 & v \\ 0 & 1 \end{pmatrix}$ and

(32) $\begin{pmatrix} 1 & 0 \\ v-uvu & 1 \end{pmatrix} = \begin{pmatrix} u^{-1} & 0 \\ 0 & u \end{pmatrix}\begin{pmatrix} 1 & 0 \\ -v & 1 \end{pmatrix}\begin{pmatrix} u & 0 \\ 0 & u^{-1} \end{pmatrix}\begin{pmatrix} 1 & 0 \\ v & 1 \end{pmatrix}$.

We have det $\begin{pmatrix} u & 0 \\ 0 & u^{-1} \end{pmatrix}$ = 1, so the matrices $B_{12}(v-uvu)$ and $B_{21}(v-uvu)$ are commutators of elements of $\mathbf{S}\mathbf{L}_n(D)$.

Suppose that the field D has at least four elements. Let $\lambda$ be an element of D. If $\lambda$ is equal to 0, 1, or $-1$, choose an arbitrary element $u$ in $D-\{0,1,-1\}$; otherwise, set $u=\lambda$. In both cases, $u$ is a nonzero element of D, it commutes with $\lambda$, and we have $u^2\not= 1$. Set $v=\lambda (1-u^2)^{-1}$. We have $uv=vu$ and therefore $v-uvu=v(1-u^2) =\lambda$. It then follows from relations (31) and (32) that the matrices $B_{12}(\lambda )$ and $B_{21}(\lambda )$ are commutators in $\mathbf{S}\mathbf{L}_n(D)$. This completes the proof of c).

C) It remains to prove that $\mathbf{S}\mathbf{L}_2(D)$ is the derived group of $\mathbf{G}\mathbf{L}_2(D)$ when D has three elements. By A), the group $\mathbf{S}\mathbf{L}_2(D)$ is generated by the matrices $B_{12}(1) = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $B_{21}(1) = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$, and these matrices are commutators of elements of $\mathbf{G}\mathbf{L}_2(D)$ because we have $B_{21}(1) =^tB_{12}(1)$ and

(33) $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}^{-1}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{-1}$.