## Nil clean rings.(English)Zbl 1296.16016

Given a ring $$R$$, one says that an element $$r\in R$$ is clean if $$r=e+u$$ where $$e\in R$$ is an idempotent and $$u$$ is a unit of $$R$$. Moreover, the element $$r$$ is strongly clean if the idempotent and unit can be chosen to commute. When all elements of a ring are clean (or strongly clean), then we say that the ring is clean (respectively, strongly clean). These classes of rings were first introduced by Keith Nicholson as interesting examples of exchange rings, but their study has blossomed in recent years as connections have been found to strongly $$\pi$$-regular rings (and elements), direct sum decomposition properties, and linear algebra applications.
The paper under review contains a very thorough and readable introduction to this subject. The author then generalizes the notion of clean elements as follows. Given any property $$\mathcal P$$ on elements of rings, one says that $$r\in R$$ is $$\mathcal P$$-clean if $$r=e+z$$ where $$e\in R$$ is an idempotent and $$z\in R$$ has property $$\mathcal P$$. The element is strongly $$\mathcal P$$-clean if additionally $$e$$ and $$z$$ can be chosen to commute. When $$\mathcal P$$ satisfies three simple conditions, then it turns out that an element, say, of an endomorphism ring of a module $$M$$ is strongly $$\mathcal P$$-clean if and only if there is a corresponding direct sum decomposition of the module $$M$$; this generalizes what happens in the case of the strongly clean elements. The author also points out an interesting disconnect, not mentioned in other sources, between the usual definition of clean elements and the definition given for non-unital rings.
The remainder of the paper is focused on the case where $$\mathcal P$$ is the property “is a nilpotent”. Thus, an element $$r\in R$$ is called nil clean if $$r=e+z$$ where $$e\in R$$ is an idempotent and $$z\in R$$ is a nilpotent, and strongly nil clean when $$e$$ and $$z$$ can be chosen to commute. The sum $$r=e+z$$ is called a nil clean decomposition of $$r$$. Nil clean rings and strongly nil clean rings are defined similarly. Many examples are given which clarify the relationship these properties have to other well-known ring and element properties. For example, every strongly nil clean element is strongly $$\pi$$-regular, and the Jacobson radical $$J(R)$$ is always a nil ideal in a nil clean ring. Additionally, there are multiple structure theorems proven relative to the behavior of these properties under matrix extensions, corner rings, direct products, and quotients; with four open questions raised in the paper on these topics.
The final theorem of the paper gives five equivalent characterizations of the uniquely nil clean rings (i.e. those rings whose elements each have exactly one nil clean decomposition). It appears from the proof that the author meant to include in characterization (3) the additional assumption that $$R/J(R)$$ is reduced. However, the statement of the theorem is true as it stands; indeed one can slightly generalize Proposition 3.18 to get that if $$R$$ is a strongly nil clean ring, then every nilpotent element of $$R$$ is contained in $$J(R)$$. As the short proof of this extension demonstrates many of the nice results of the paper, we include it here.
Let $$R$$ be a strongly nil clean ring. By Proposition 3.16 we know $$J(R)$$ is a nil ideal, and by Proposition 3.13 we know that strong nil cleanness passes to factor rings, so we can reduce to the case when $$J(R)=0$$. By Proposition 3.14, we then must have $$2=0$$ in $$R$$. Suppose, by way of contradiction, there is an element $$y\in R$$ with $$y^2=0\neq y$$. Since $$J(R)=0$$ there are no nil left ideals, and so there is some element $$x\in R$$ with $$xy$$ not nilpotent. As $$R$$ is strongly nil clean, write $$xy=e+z$$ for some (nonzero) idempotent $$e\in R$$ and nilpotent $$z\in R$$ which commute. Replacing $$x$$ by $$(1+z)^{-1}ex$$, we have $$xy=e$$. Since $$0=xy^2=ey$$, we have $$ex(1-e)ye=e$$. Thus replacing $$x$$ by $$ex(1-e)$$ and $$y$$ by $$(1-e)ye$$, we may assume that $$x\in eR(1-e)$$ and $$y\in (1-e)Re$$ with $$xy=e$$.
At this point set $$f:=(1-e)-yx$$, which is an idempotent in $$(1-e)R(1-e)$$. Also set $$r:=e+x+y$$, $$s:=x+y+yx$$, and $$u:=r+f$$. It is easy to compute that $$r^2=s$$, $$rs=e+yx=1-f$$ is an idempotent, and $$u^3=1$$ so that $$u$$ is a unit. By Corollary 3.11, we know that all units in strongly nil clean rings are unipotent. As $$u=1+s$$ (remember $$2=0$$) this means that $$s$$ must be nilpotent. But we find that $$s^3=e+yx$$ is a nonzero idempotent (since $$e\neq 0$$), giving the needed contradiction.

### MSC:

 16N40 Nil and nilpotent radicals, sets, ideals, associative rings 16D70 Structure and classification for modules, bimodules and ideals (except as in 16Gxx), direct sum decomposition and cancellation in associative algebras) 16U60 Units, groups of units (associative rings and algebras) 16E50 von Neumann regular rings and generalizations (associative algebraic aspects) 16U80 Generalizations of commutativity (associative rings and algebras)
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### References:

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