# If the sum of a right (principal) ideal with a left one contains an invertible element and the product is zero then do they contain idempotents?

I am trying to solve a problem on additive categories, that gives the following question on (non-commutative unital associative) rings: if for elements $$a$$ and $$b$$ of a ring $$R$$ we have $$ab=0$$ and $$a+b=t$$ is invertible then how can one verify the existence of certain $$c$$ and $$c'$$ in $$R$$ such that $$ca+bc'=1$$ (and so, $$ca$$ and $$bc'$$ are idempotents)?

It appears that this implication does not hold unconditionally; yet I would be deeply grateful for any hints that would allow to study it (note however that the case where $$t-1$$ is nilpotent is certainly simple). In this case I am interested in there is an extra conservativity assumption; in particular, in the quotient of $$R$$ by the two-sided ideal generated by $$t-1$$ non-invertible elements of $$R$$ do not become invertible. I have tried to relate my question with von Neumann regularity of elements of rings, but was not able to do this.

• What do you mean by the case where $t$ is nilpotent is certainly simple"? By assumption, $t$ is invertible, so it cannot be nilpotent. – Victor Protsak Nov 4 '18 at 19:13
• I am sorry; $t-1$ may be nilpotent instead.:) – Mikhail Bondarko Nov 4 '18 at 19:26

## 1 Answer

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

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 $$D$$.

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:

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

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

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
($$\beta$$) the endomorphism $$T:=A+B$$ acts on the group $$\Sigma$$ by permutation of coordinates.

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 $$1e_2=e_2\notin \textrm{im}(B)$$.

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

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

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.

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

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

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 $$CA+BD=1$$. \\\\\

• Thank you very much! I will try to relate your sufficient conditions to the conservativity assumption. – Mikhail Bondarko Nov 7 '18 at 8:06