**Q: If, for elements $A$ and $B$ of a ring $R$,
we have<br>
(i) $AB=0$, and<br>
(ii) $T:=A+B$ is invertible,<br>
then how can we verify the existence of $C$ and $D$ in $R$
such that $CA+BD=1$?**<p><p>

Let me start by 
describing an explicit example of a ring $R$
having elements $A$ and $B$ satisfying (i) and (ii), yet
the desired elements $C$ and $D$ do not exist.
Then I will describe some extra conditions which
guarantee the existence of $C$ and $D$.<p>

Let $\Sigma=\mathbb Z^{\mathbb Z^+}$ be the
abelian group whose elements are the sequences
$\overline{m}=(m_1,m_2,m_3,\ldots)$
of integers indexed by the positive integers.
Define endomorphisms $A$ and $B$ of this abelian group by:<p>

$$A\overline{m} = (m_1,m_3,m_5,0,m_7,m_9,m_{11},0,m_{13},m_{15},m_{17},0,\ldots)$$

$$B\overline{m} = (0,0,0,m_2,0,0,0,m_4,0,0,0,m_6,0,0,0,m_8,0,\ldots).$$

That is, $A$ ignores the even coordinates of $\overline{m}$,
and orders the odd coordinates in the expected order, except
inserts a $0$ at the coordinates indexed by a multiple of $4$.
$B$ ignores the odd coordinates, orders the even coordinates
in the expected order, but inserts $0$'s everywhere EXCEPT
the coordinates indexed by a multiple of $4$.
The main points to note are that

($\alpha$) $\textrm{im}(B)$ is (properly) contained in
$\textrm{ker}(A)$, so in particular $AB$
is the zero endomorphism, and<br>
($\beta$) the endomorphism $T:=A+B$ acts on the group $\Sigma$
by permutation of coordinates.<p>

Thus, if $R=\textrm{End}(\Sigma)$, then $R$ meets all of
conditions of the question. But $R$ contains no elements
$C$ and $D$ such that $CA+BD=1$.
Such a relation
would force $\textrm{ker}(A)\subseteq \textrm{im}(B)$, which is false.
Specifically,
$$e_2=(0,1,0,0,\ldots)\in \textrm{ker}(A)\setminus \textrm{im}(B),$$
and this is in conflict with $CA+BD=1$, since
$(CA+BD)e_2 = BDe_2\in\textrm{im}(B)$,
while $1e_2=e_2\notin \textrm{im}(B)$.<p><p>

Now let me mention some finiteness conditions which can be
added to the problem to guarantee the existence of $C$ and $D$
so that $CA+BD=1$.<p>

(1) $T$ is not just invertible, but it is a unit of finite order. <br> OR, <br> 
(2) $T^{-1}$ belongs to the subring of $R$ generated by $A$ and $B$. <br> OR, <br>
(3) $T^{-1}B\in BR$. <br> OR, <br>
(4) $BR=B^2R$.<p>

Note that if (1) holds, and $T$ is a unit of order $n$, then
$T^{-1}=T^{n-1}=(A+B)^{n-1}\in \langle A, B\rangle$ and (2) holds.
Note that if (2) holds and if $R$ is generated by $A$ and $B$,
then the right ideal $BR$ is actually a $2$-sided ideal, since
it is closed under left multiplication by both $A$ and $B$.
In fact, $BR$ is the $2$-sided ideal $(B)$ generated by $B$.
Thus $T^{-1}B\in (B)=BR$, and (3) holds.
Now, if (3) holds, we have $T^{-1}B=BD$ for some $D\in R$,
so $B=TBD=(A+B)BD=B^2D\in B^2R$, which is enough to establish
that $BR=B^2R$, so (4) holds.<p>

Thus condition (4) is essentially the most general of these. Let me
add that condition to the conditions of original problem:<p>

**Claim.** If, for elements $A$ and $B$ of a ring $R$,
we have<br>
(i) $AB=0$, <br>
(ii) $T:=A+B$ is invertible, and<br>
(ii) $BR=B^2R$,<br>
then $R$ has elements $C$ and $D$ 
such that $CA+BD=1$<p><p>

**Reasoning.**
$A+B=T$, so $T^{-1}A+T^{-1}B=1$. 
We have added the assumption that $BR=B^2R$,
so $B=B^2D$ for some $D\in R$. Thus
$B=B^2D = (A+B)BD = TBD$, and we get $T^{-1}B=BD$.
Substituting this into the equation from the
first line of this argument we get $T^{-1}A+BD=1$.
Now let $C=T^{-1}$ to convert this to $CA+BD=1$. \\\\\\\\\\