A note on completeness in the theory of strongly clean rings
aa r X i v : . [ m a t h . R A ] J u l A note on completeness in the theory of strongly clean rings
Alexander J. Diesl Thomas J. DorseyOctober 27, 2018
Abstract
Many authors have investigated the behavior of strong cleanness under certain ring exten-sions. In this note, we prove that if R is a ring which is complete with respect to an ideal I andif x is an element of R whose image in R/I is strongly π -regular, then x is strongly clean in R . In this note, all rings are associative, with unity 1, which is preserved by homomorphisms.Recall that, following [Nic99], an element x of a ring R is said to be strongly clean if there isan idempotent e ∈ R , which commutes with x , such that x − e is a unit in R . If e is such anidempotent, we say that x is e -strongly clean. The element x ∈ R is called strongly π -regular ifthere is a positive integer n such that x n +1 ∈ x n R ∩ Rx n . Equivalently (see [Bor05, Section 10]),the element x is strongly π -regular if and only if x satisfies Fitting’s Lemma as an endomorphismof R R . This implies that x is strongly π -regular if and only if there is an idempotent e ∈ R thatcommutes with x such that x − e is a unit and exe is nilpotent. We therefore see that everystrongly π -regular element of a ring is strongly clean (a well-known fact that was originally shown,using different techniques, in [BM88]). A ring is said to be strongly clean (respectively, strongly π -regular) if each of its elements is strongly clean (respectively, strongly π -regular). As an aside,following a remarkable result of Dischinger (see [Dis76] or [Lam03, Exercise 23.5]), a ring R isstrongly π -regular if and only if for every x ∈ R there is a positive integer n such that x n +1 ∈ x n R .There are many examples of, and results about, strongly clean rings which happen to be com-plete with respect to an ideal I (e.g. [CYZ06a, Theorem 2.4], [CYZ06b, Theorem 9], [YZ07,Theorem 2.7], [FY06, Theorem 2.10], and [BDD08, Theorem 25, Corollary 26]). The aforemen-tioned results demonstrate that it is frequently true (though not always, as as seen in [BDD07,Example 45]) that, if R/I is strongly clean and R is I -adically complete, then R will be stronglyclean. Moreover, the use of completeness in this context often simplifies proofs greatly (e.g. see[BDD08, Theorem 25, Corollary 26] and the surrounding discussion). The present note continuesthis theme.Our investigation is motivated by [CZ07, Theorem 2.1] (see also [WC07, Theorem 3] for a resultin the nonunital case) which states that for a ring R and a ring endomorphism σ of R , an elementof the skew power series ring R [[ x ; σ ]] is strongly clean provided that its constant term is strongly π -regular in R . The proof is rather long and technical, and makes use of the added structure presentin the power series ring. The theorem in this note generalizes this result with a rather simple proofwhose main trick is to consider the appropriate Peirce decomposition when refining the relevantidempotent.We will need one definition which is a variant of the notion of “strong π -rad cleanness” definedin [Die05]. Definition.
Let R be a ring, let I be an ideal of R , and let e ∈ R be an idempotent. We saythat x ∈ R is e -strongly π -clean (of degree n ), with respect to I , if x is e -strongly clean and if, in1ddition, the image of exe is nilpotent (of degree n ) in R/I . We say that x ∈ R is strongly π -clean,with respect to I , if there exists an idempotent e ∈ R for which x is e -strongly π -clean, with respectto I .Clearly, if x is e -strongly π -clean, with respect to any ideal, then x is clearly e -strongly clean.The definition above, in the case when I is the Jacobson radical of R , agrees with the notion ofstrong π -rad cleanness defined in [Die05]; when I = 0, it agrees with the usual notion of strong π -regularity. In what follows, we will therefore call an element strongly π -regular of degree n if itis strongly π -clean of degree n with respect to the zero ideal. Also, if x is e -strongly π -clean withrespect to an ideal I , then it is e -strongly π -clean with respect to any ideal which contains I . Inparticular, if x ∈ R is strongly π -regular, it is clearly strongly π -clean with respect to any ideal I .Note, however, that requiring that x is strongly π -clean with respect to I is stronger than requiringthat the image of x ∈ R is strongly π -regular in R/I . The difference is that the idempotent in
R/I which witnesses the strong π -regularity of x in R/I need not lift to an idempotent of R , and evenif it does, it need not lift to one which commutes with x in R . Indeed, that is precisely what is atissue in the proof of the following result. Theorem.
Let R be a ring, complete with respect to an ideal I , which is necessarily contained inthe Jacobson radical J ( R ). Let x ∈ R . If the image of x is strongly π -regular of degree n in R/I ,then x is strongly π -clean of degree n , with respect to I . In particular, x is strongly clean in R . Proof.
For i ≥
1, let π i : R → R/I i denote the natural quotient map. These maps will simplify theexposition, since we will be dealing simultaneously with several different quotients of R . We willproduce a sequence of idempotents ( e i ) i ≥ of R with the following properties:(1) For each i ≥ π i ( x ) is π i ( e i )-strongly π -clean of degree n , with respect to π i ( I ),(2) e i − e i − ∈ I i − for each i ≥ e = lim i →∞ e i specifies a well-defined element of R , which is idempotent since e − e = lim i →∞ ( e i − e i ) = 0 . Finally, we show that x is e -strongly π -clean, with respect to I , in R . Note first that xe − ex = lim i →∞ ( e i x − xe i ) = 0 . Since π ( x − e ) = π ( x ) − π ( e ) is a unit in R/I , and I ⊆ J ( R ), x − e is a unit in R . Finally,( exe ) n ∈ I , since π ( exe ) n = π ( e xe ) n = 0 in R/I by (1).To begin, since π ( x ) is strongly π -regular of degree n in R/I , there is an idempotent z in R/I such that π ( x ) − z ∈ U ( R/I ), π ( x ) z = zπ ( x ) and ( zπ ( x ) z ) n = 0. Since R is I -adically complete,idempotents lift modulo I (e.g., by [Lam01, Theorem 21.31]), and we may lift z to an idempotentof R , which we call e . Note that z = π ( e ), and that condition (1) is satisfied.Inductively, suppose that e , . . . , e m have been constructed satisfying conditions (1) and (2) for1 ≤ i ≤ m . Write f m = 1 − e m . Since π m ( x ) is π m ( e m )-strongly π -clean with respect to I/I m in R/I m , the Peirce decomposition of π m ( x m ) with respect to π m ( e m ) is π m ( x m ) = (cid:18) π m ( e m x m e m ) 00 π m ( f m x m f m ) (cid:19) , π m ( f m x m f m ) is a unit in π m ( f m Rf m ) and π m ( e m xe m ) n ∈ π m ( I ). Note that, since f m I m f m ⊆ f m J ( R ) f m = J ( f m Rf m ) and units lift modulo the Jacobson radical, we see that f m xf m is a unitin f m Rf m . Note further that, since ( e m xe m ) n ∈ I , it follows that π m +1 ( e m xe m ) is nilpotent in thering R/I m +1 .We now consider the Peirce decomposition of π m +1 ( x ) with respect to the idempotent π m +1 ( e m )in R/I m +1 . The elements x and e m , which commute modulo I m , need not commute modulo I m +1 ,so the Peirce decomposition of π m +1 ( x ) in R/I m +1 need not be diagonal. Nevertheless, if we writeit as π m +1 ( x ) = (cid:18) a bc d (cid:19) , then, by the discussion in the previous paragraph, a is nilpotent in π m +1 ( e m Re m ), d is a unit in π m +1 ( f m Rf m ), and b, c ∈ π m +1 ( I m ).We will now perturb the idempotent e m to an idempotent e m +1 which commutes with x modulo I m +1 . Working in R/I m +1 , the Peirce decomposition of π m +1 ( e m ), with respect to itself, is simply (cid:18) (cid:19) . Now, for any r ∈ π m +1 ( e m Rf m ∩ I m ) and any s ∈ π m +1 ( f m Re m ∩ I m ), the element π m +1 ( e m )+ r + s ,whose Peirce decomposition is (cid:18) rs (cid:19) , is idempotent in R/I m +1 since ( I m ) ⊆ I m +1 . We seek r and s as above such that (cid:18) rs (cid:19) (cid:18) a bc d (cid:19) = (cid:18) a bc d (cid:19) (cid:18) rs (cid:19) . Since π m +1 ( I m ) = 0 in R/I m +1 and b, c, r, s ∈ π m +1 ( I m ), the above equation holds if and only if ar − rd = b and ds − sa = − c . Let k be the index of nilpotence of a . Set r = − P ki =1 a i − bd − i , whichbelongs to π m +1 ( e m Rf m ∩ I m ), and set s = P ki =1 d − i ca i − , which belongs to π m +1 ( f m Re m ∩ I m ).It is easy to check (as in [BDD07, Example 13]), that ds − sa = − c and ar − rd = b , and it followsthat the idempotent (cid:18) rs (cid:19) commutes with π m +1 ( x ) in R/I m +1 . Lift (cid:18) rs (cid:19) to an idempotent e m +1 of R , which clearly agrees, modulo I m , with e m . Since π m ( x ) − π m ( e m ) is a unit in R/I m and x − e m +1 agrees with x − e m , modulo I m ⊆ J ( R ), π m +1 ( x ) − π m +1 ( e m +1 ) is a unit in R/I m +1 .Since e m +1 xe m +1 agrees with e xe , modulo I , and π ( e xe ) is nilpotent of index n in R/I , π m +1 ( e m +1 xe m +1 ) is nilpotent of index n modulo π m +1 ( I ). We therefore conclude that π m +1 ( x )is π m +1 ( e m +1 )-strongly π -clean of degree n , with respect to π m +1 ( I ) in R/I m +1 . By induction, wehave produced a sequence of idempotents satisfying conditions (1) and (2), which completes theproof.A few remarks are in order. The hypothesis that the image of x in R/I is strongly π -regularcannot be weakened to the hypothesis that the image of x is strongly clean. Let R be a ring witha ring endomorphism σ such that the triangular matrix ring T ( R ) is strongly clean but such that T ( R [[ x ; σ ]]) is not strongly clean (such an example is constructed in [BDD07, Example 45]). View T ( R [[ x ; σ ]]) ∼ = T ( R )[[ x, σ ]] as a skew power series ring over T ( R ) (using σ to also denote theobvious extension of σ to a ring endomorphism of T ( R )), and let I be the ideal of T ( R [[ x ; σ ]])generated by x . Then T ( R [[ x ; σ ]]) /I ∼ = T ( R ) is strongly clean, and T ( R [[ x ; σ ]]) is complete with3espect to I , but T ( R [[ x ; σ ]]) is not strongly clean and is therefore certainly not strongly π -cleanwith respect to I .There is, however, some room for improvement on the hypotheses. The hypothesis that theimage of x in R/I is strongly π -regular can be weakened to the (substantially less pleasant hypoth-esis) that there be an idempotent e ∈ R such that x is e -strongly clean in R/I and such that forany idempotent e ′ which agrees with e modulo I , the maps l e ′ xe ′ − r (1 − e ′ ) x (1 − e ′ ) : e ′ ( R/I m )(1 − e ′ ) → e ′ ( R/I m )(1 − e ′ )and l (1 − e ′ ) x (1 − e ′ ) − r e ′ xe ′ : (1 − e ′ )( R/I m ) e ′ → (1 − e ′ )( R/I m ) e ′ are surjective and have the additional property that the preimage of I m − is contained in I m − ,for each m . This is precisely what is used in the proof, and is guaranteed when (1 − e ) x (1 − e ) is aunit and exe is nilpotent in the respective corner rings (e.g. see [BDD07, Example 13]). We havenot, however, found a hypothesis which is simultaneously more pleasant and more general, and weexpect that the main use of this result will be in the case we explicitly stated above. References [BDD07] Gautam Borooah, Alexander J. 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Linear AlgebraAppl. , 425(1):119–129, 2007.Alexander J. DieslDepartment of MathematicsWellesley CollegeWellesley, MA 02481 USAEmail: [email protected]
Thomas J. DorseyCenter for Communications Research4320 Westerra CourtSan Diego, CA 92126-1967USAEmail: [email protected]@ccrwest.org