Stable general linear and elementary groups for right modules

Throughout, $R$ is an associative unital ring, not assumed commutative (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides). For $n\ge 0$ let $R^n$ be the set of column vectors $v=(v_1,\dots ,v_n)$ with $v_i\in R$, made into a right $R$-module by entrywise addition and the right action $(v\cdot r{)}_i:=v_{i}r(r\in R)$ (Unital left and right modules over a ring; unqualified module means left module); for $n=0$ this is the zero module, whose unique element is the empty column. A map $f:R^n\to R^m$ is right $R$-linear when $f(v+v^{\mathrm{\prime }})=f(v)+f(v^{\mathrm{\prime }})$ and $f(v\cdot r)=f(v)\cdot r$.

Matrices. Every right-linear $f:R^n\to R^m$ has a unique matrix $A\in M_{m\times n}(R)$, written $A=(A_{ij})$, such that $(Av{)}_i={\sum }_{j=1}^n{A}_{ij}{v}_j,$ the sum being the finite sum in the additive group of $R$; uniqueness uses that the standard basis vectors $e_1,\dots ,e_n$ of $R^n$ generate it and that columns of $A$ are the coordinate vectors of $f(e_j)$. If $g:R^m\to R^p$ has matrix $B$, then $g\circ f:R^n\to R^p$ has matrix $BA$, with $(BA{)}_{ik}={\sum }_j{B}_{ij}{A}_{jk}$; this is the displayed order and it uses only associativity and distributivity.

General linear group. Let ${\mathrm{G}\mathrm{L}}_n(R)$ be the group of $n\times n$ matrices over $R$ possessing a two-sided inverse, with multiplication of matrices as the group operation and $I_n$ as the identity; by the previous paragraph ${\mathrm{G}\mathrm{L}}_n(R)$ is exactly the group of right-linear automorphisms of $R^n$. The stabilization $A\mapsto \mathrm{diag}(A,1)$ identifies ${\mathrm{G}\mathrm{L}}_n(R)$ with a subgroup of ${\mathrm{G}\mathrm{L}}_{n+1}(R)$, and $\mathrm{G}\mathrm{L}(R):={\bigcup }_{n\ge 0}{\mathrm{G}\mathrm{L}}_n(R)$ is the stable general linear group, in which every element is represented by some $n\times n$ matrix. A matrix $A$ belongs to ${\mathrm{G}\mathrm{L}}_n(R)$ when there is a matrix $B$ satisfying both $AB=I_n$ and $BA=I_n$; either equation alone need not imply the other over an arbitrary unital ring. No determinant, commutativity, or rank function is used anywhere in this definition.

For a concrete one-sided inverse, take $R={\mathrm{End}}_k(V)$ where $V$ has basis $e_0,e_1,\dots$. Let $S(e_i)=e_{i+1}$ and let $L(e_0)=0$, $L(e_{i+1})=e_i$. Then $LS=1_V$, while $SL(e_0)=0\mathrm{\ne }{e}_0$, so the $1\times 1$ matrices $A=L$ and $B=S$ satisfy $AB=I_1$ but not $BA=I_1$.

Elementary matrices. For $n\ge 1$, $1\le i,j\le n$ with $i\mathrm{\ne }j$ and $r\in R$, let $E_{ij}$ be the matrix with entry $1$ in position $(i,j)$ and $0$ elsewhere, and let $e_{ij}(r):=I_n+r{E}_{ij}$ be the corresponding elementary matrix. It is invertible with two-sided inverse $e_{ij}(-r)$, because $E_{ij}^2=0$ and hence $e_{ij}(r)e_{ij}(-r)=I_n$. Let ${\mathrm{E}}_n(R)$ be the subgroup of ${\mathrm{G}\mathrm{L}}_n(R)$ generated by all elementary matrices in ${\mathrm{G}\mathrm{L}}_n(R)$ for $n\ge 1$, and set ${\mathrm{E}}_0(R)=\{I_0\}$. Let $\mathrm{E}(R):={\bigcup }_{n\ge 0}{\mathrm{E}}_n(R)$ be the stable elementary subgroup, the subgroup of $\mathrm{G}\mathrm{L}(R)$ generated by the images of all elementary matrices. Elementary matrices are stable in the same way: $e_{ij}(r)$ in ${\mathrm{G}\mathrm{L}}_n(R)$ is the image of the elementary matrix with the same indices in ${\mathrm{G}\mathrm{L}}_{n+1}(R)$.