Let $M$ be an abelian group, $\mathfrak{B}$ a set of subgroups of $M$. Given $X,Y,Z$ in $\mathfrak{B}$ such that $X \oplus Y=Z$, we say that $X,Y$ are $\mathfrak{B}$ of $Z$. The set $\mathfrak{B}$ is to say to satisfy the $\textit{maximum condition with respect to summands}$ if every non empty subset $\mathfrak{C}$ of $\mathfrak{B}$ contains an element which is a an $\mathfrak{B}$-summand of no other element of $\mathfrak{C}$.
Let $M$ be right $R$-module. For $r\in R$ we write $l.ann_M(r)=\{m\in M|mr=0\}$. Let $\mathfrak{B}=\{l.ann_M(r) | r\in R\}$.If $\mathfrak{B}$ satisfies the maximum condition with respect to the summands. We say that in M $\textit{the left annihilators of elements of}$ $R$ $\textit{satisfy the maximum condition with respect to the summands} $.
There are many common situations where this holds.
For example it holds if M satisfies the maximum condition for left annihilators of elements of $R$.
It holds also if there exists a ring $S$ such that $M$ is an $S-R$-bimodule and $l.u.dim_S M$ is finite where $l.u.dim_S M$ denotes the uniform dimension of $M$ as left $S$-module, that is the supremum of independent sets of nonzero S-submodules of $M$.
If $\operatorname{ann}_M u $ is left annihilator of $u$; my question is to understand the implication $(a) \Rightarrow (b)$
(a) There exists a right $R$-module $M$ such that in $M$ the left annihilators of elements of $R$ satisfy the maximum condition with respect to summands and l.$ann_M(r)\neq 0$ for all $r \in R\setminus U(R)$.
(b) There exists a partial order $\geq$ on $R$ satisfying the minimum condition such that for all $a,b\in R$ if $1-ab \in R \setminus U(R)$ then $a>a-aba$.
The solution is to define the relation < by for $a,b\in R$ set $a>b$ in $R$ if $l.ann_M(a)<l.ann_M(b)$ in $\mathfrak{B}$. On $R$ the relation $>$ is irreflexive and transitive, so we have a partial order $\geq$ on $R$. It satisfies the minimum condition because $\mathfrak{B}$ satisfies the maximum condition with respect to $\mathfrak{B}$-summands. Moreover if $a,b \in R$ such that $1-ab \in R \setminus U(R)$ then $l.ann_M(1-ab)\neq 0$ and $l.ann_M(1-ab)\oplus l.ann_M(a)=l.\textit{ann}_M(a-aba)$, so $a>a-aba$.
My question is therefore why if $a,b \in R$ such that $1-ab\in R \setminus U(R)$ then $\operatorname{ann}_M(1-ab)\neq 0$ and $\operatorname{ann}_M(1-ab) \oplus \operatorname{ann}_M (a)=\operatorname{ann}_M(a-aba)$. One of the inclusion is easy, the other is not. Are there any supplementary hypothesis to satisfy this implication.