Duo, Bézout, and Distributive Rings of Skew Power Series

We give necessary and sufficient conditions on a ring R and an endomorphism σ of R for the skew power series ring R[[x; σ]] to be right duo right Bezout. In particular, we prove that R[[x; σ]] is right duo right Bezout if and only if R[[x; σ]] is reduced right distributive if and only if R[[x; σ]] is right duo of weak dimension less than or equal to 1 if and only if R is N0-injective strongly regular and σ is bijective and idempotent-stabilizing, extending to skew power series rings the Brewer-Rutter-Watkins characterization of commutative B´ezout power series rings.


Introduction and the main result
The aim of this paper is to characterize skew power series rings that are right duo right Bézout, which by the main result of the paper, Theorem 1.6, is the same as to characterize skew power series rings that are right duo right distributive as well as to characterize right duo skew power series rings of weak dimension less than or equal to 1. In this introductory section we recall those results on (skew) power series rings that motivated our studies and which we generalize in this paper. Here we also formulate our main result, Theorem 1.6, in which we give some equivalent conditions for a skew power series ring to be right duo right Bézout. This characterization theorem is proved in Section 3. In Section 2 we collect results on skew power series rings that are right Bézout, or right duo, or reduced, or whose coefficient ring is ℵ 0 -injective strongly regular, which we apply in the proof of the main result.
All rings in this paper are associative with unity. Recall that a ring R is right (left) Bézout if all its finitely generated right (left) ideals are principal, and R is a Bézout ring if it is right and left Bézout. A ring R is right (left) ℵ 0 -injective provided any homomorphism from a countably generated right (left) ideal of R into R extends to a right (left) R-module endomorphism of R. By an ℵ 0 -injective ring we mean a right and left ℵ 0 -injective ring.
In [1] J. W. Brewer, E. A. Rutter and J. J. Watkins proved that for any commutative power series ring R [[x]] the following equivalences hold: ] has weak dimension less than or equal to 1, where a ring R is said to be von Neumann regular if a ∈ aRa for any a ∈ R.
Bézout power series rings in the noncommutative setting were studied by D. Herbera in [3], where among other results she proved a generalization of the Brewer, Rutter and Watkins result, which we quote below (see also [10] for more general results).
Recall that a ring R is right (left) duo if all right (left) ideals of R are two-sided ideals, and R is duo if it is right and left duo. Recall also that a ring R is abelian if all idempotents of R are central, and a ring R is strongly regular if a ∈ a 2 R for any a ∈ R. Strongly regular rings are precisely abelian von Neumann regular rings. Since every element of a strongly regular ring R is a product of a unit and a central idempotent, it follows that strongly regular rings are duo, and for strongly regular rings the ℵ 0 -injectivity is a right-left symmetric condition (see Subsection 2.3). In [9] G. Marks studied the duo property for Ore extensions of rings, and among other results he proved the following theorem. Theorem 1.3 (Marks; see [9,Proposition 5]). Let R be a left or right self-injective von Neumann regular ring. The following conditions are equivalent: ( In [13] and [14] A. A. Tuganbaev extended the previously mentioned result of Brewer, Rutter and Watkins to skew power series rings, adding additionally the right distributivity property to the list of equivalences, and showing that the condition concerning the weak dimension can be replaced with a weaker property of 2-generated right ideals (i.e., right ideals generated by two elements). To quote those results of Tuganbaev, we need some definitions.
Let R be a ring. Then R is said to be right (left) distributive if the lattice of right (left) ideals of R is distributive, i.e., (I + J) ∩ K = (I ∩ K) + (J ∩ K) holds for any right (left) ideals I, J, K of R. We say that R is distributive if R is right and left distributive. The ring R is called semicommutative if the right annihilator of every element of R is a two-sided ideal (clearly the notion is left-right symmetric). Recall that R is said to be right (left) quasi-duo if every maximal right (left) ideal of R is a two-sided ideal. For an endomorphism σ of the ring R the skew power series ring over R in one variable x is denoted by R[[x; σ]]. Thus elements of R[[x; σ]] are power series of the form ∞ n=0 a n x n , where all the coefficients a n belong to R, with multiplication defined by the rule xa = σ(a)x for any a ∈ R. (1) R[[x; σ]] is right Bézout and R is semicommutative.
Theorem 1.5 (Tuganbaev; see [13,Theorem 1]). Let R be an abelian ring and σ an automorphism of R such that σ(e) = e for any idempotent e = e 2 ∈ R. Then the following conditions are equivalent: (1) R[[x; σ]] has weak dimension less than or equal to 1.
In this paper we show that in a sense all the conditions appearing in the results quoted above (starting from this of Brewer, Rutter and Watkins) are equivalent. More precisely, these results are consequences of the following theorem, which being the main result of the paper, specifies eleven equivalent conditions for a skew power series ring R[[x; σ]] to be right duo right Bézout. We would like to emphasize that besides conditions suggested by the quoted results, the theorem also contains some new ones. We would also like to stress that according to the second part of this theorem, each of the equivalent conditions implies its left analogue (but need not be equivalent to it; see Example 3.3), and if σ is an automorphism, then any condition in this theorem is equivalent to its left analogue. To prove this theorem we need some results on special classes of skew power series rings which we assemble in Section 2, and we apply them in Section 3 in the proof of the theorem.
Recall that a ring R is reduced if it contains no nonzero nilpotent element, i.e., a 2 = 0 implies a = 0 for any a ∈ R. Theorem 1.6. Let σ be an endomorphism of a ring R. Then the following conditions are equivalent: is right Bézout and right duo. (11) R[[x; σ]] has weak dimension less than or equal to 1, R is abelian, σ is bijective and σ(e) = e for any e = e 2 ∈ R. (12) R is ℵ 0 -injective strongly regular, σ is bijective and σ(e) = e for any e = e 2 ∈ R.
Furthermore, these conditions imply the left analogues of (1)- (10), and in particular, if any of these conditions holds, then R[[x; σ]] is duo. Moreover, if σ is bijective, then these conditions are equivalent to the left analogues of (1)-(10).

Special classes of skew power series rings
In this section we prove some results on special classes of skew power series rings that will be used in the proof of Theorem 1.6 in the next section.
2.1. Right Bézout rings of skew power series. In Section 3 we will need the following generalization of Herbera's result [3, Lemma 2.2]. Recall that a ring R is Dedekind-finite if for any a, b ∈ R, ab = 1 implies ba = 1.
We claim that (cf. [3, Proposition 2.1]) To see this, note that since A is right Bézout, the right ideal gA + xA is principal and thus there exist Multiplying the first part of (2.1) on the left by f , we obtain x = (xh + f xk)α, and equating x-coefficients, we see that Since R is Dedekind-finite, it follows that σ(α 0 ) is a unit of R, and thuŝ α is a unit of A (see [7, p. 10]). Since (2.1) implies that (2.2)ĝ = (gh + xk)α and x = (gh + xk)β, we deduce that By assumption f g = x, and thusfĝ = x. Hence (2. 3) implies that f x = xγ =γx, andf =γ follows. Since σ is injective, we have f = γ, which proves our claim. Now we are in a position to prove that R is von Neumann regular. Let a ∈ R. Since A is right Bézout, the right ideal aA + xA is principal, and thus for some a = (ap + xq)δ and x = (ap + xq)ǫ.
2.2. Right duo rings of skew power series. Observe that for any ring we have the following implication relations: right duo ⇒ semicommutative ⇒ abelian.
We will use this observation in the proof of the following proposition.
Proposition 2.2. Let R be a ring and σ an endomorphism of R. If R[[x; σ]] is right duo, then R is right duo, σ is bijective and σ(e) = e for any e = e 2 ∈ R.
Since by assumption A is right duo, for any a, b ∈ R there exists f = ∞ n=0 f n x n ∈ A such that ba = af . Hence ba = af 0 ∈ aR, proving that R is right duo.
To prove that σ is bijective, we use an argument which is essentially due to Marks (see [9, Proof of Theorem 1]). Since A is right duo, for any a ∈ R there exists g ∈ A such that ax = xg. Equating x-coefficients in this equality, we obtain a ∈ σ(R), and thus σ is surjective. To show that σ is injective, consider any a ∈ R with σ(a) = 0. Since σ is surjective, Equating constant terms, x-, and x 3 -coefficients in (2.5), we obtain the following equations: From the first part of (2.6) we obtain 0 = σ(bc 0 ) = aσ(c 0 ), and thus the second part of (2.6) implies that a = bc 1 , which leads to 0 = aσ(c 1 ).

Skew power series rings over ℵ 0 -injective rings.
In the proof of Theorem 1.6 we will need the following well known characterization of ℵ 0 -injective strongly regular rings (see [13,Lemmas 3,4]). Recall that a ring R is said to be right ℵ 0 -algebraically compact if for any system of a countable number of linear equations with a countable number of indeterminates and with coefficients from R written on the left, if every finite subsystem of this system has a solution in R, then the whole system has a solution in R. This is to say that if A is an ℵ 0 × ℵ 0 row-finite matrix over R, indexed by N × N, X is a column of ℵ 0 indeterminates and B is a column of ℵ 0 elements of R, both indexed by N, then the system AX = B is finitely solvable if and only if it is solvable. Left ℵ 0 -algebraically compact rings are defined analogously. It is well known that for a von Neumann regular ring R, the ring R is right (left) ℵ 0 -algebraically compact if and only if it is left (right) ℵ 0 -injective (see [3, Proposition 1.1]). Proposition 2.3. For a strongly regular ring R, the following conditions are equivalent.
n=0 of orthogonal idempotents of R and for each sequence (a n ) ∞ n=0 of elements of R there exists b ∈ R such that be n = a n e n for all n ∈ N ∪ {0}.
In the proof of Theorem 1.6 we will need also the following proposition.
Proposition 2.4. If R is an ℵ 0 -injective strongly regular ring and σ is an automorphism of R such that σ(e) = e for every e = e 2 ∈ R, then any principal right ideal of the ring R[[x; σ]] is generated by a power series whose coefficients are orthogonal idempotents of R.
Since R is a strongly regular ring, for any n ∈ N ∪ {0} there exist an idempotent d n ∈ R and a unit u n ∈ R such that f n = d n u n . Set e n x n ∈ A.
Since e 0 , e 1 , e 2 , . . . are orthogonal idempotents, to complete the proof it suffices to show that f A = gA.
Since R is ℵ 0 -injective strongly regular, by Proposition 2.3 there exists a sequence (b n ) ∞ n=0 of elements of R such that e m b n = e m σ −m (f m+n ) for any m, n ∈ N ∪ {0}, and since σ(e) = e for any idempotent e ∈ R, we obtain We claim that f = gh, where h = ∞ n=0 b n x n ∈ A. For this, write gh = ∞ n=0 c n x n . Then applying (2.7) and the equality f n = d n f n , we obtain It is easy to see that n−1 i=0 e i + n−1 i=0 (1 − d i ) = 1 for any n ∈ N, and thus n i=0 Hence c n = d n f n = f n , and thus gh = f , which proves that f A ⊆ gA.
To prove the opposite inclusion, we have to show that there exists p = ∞ n=0 p n x n ∈ A such that g = f p. Equating x n -coefficients of g and f p for all n ∈ N ∪ {0}, we obtain a system of a countable number of linear equations Since σ is an automorphism and σ(e) = e for any idempotent e ∈ R, the system (2.8) can be written in the equivalent form Hence to prove the existence of p ∈ A with g = f p, it suffices to show that the system has a solution in R. Since R is ℵ 0 -injective strongly regular, Proposition 2.3 implies that the system (2.9) has a solution if and only if any its finite subsystem has a solution. Thus it suffices to show that for any m ∈ N ∪ {0} there exist y 0 , y 1 , . . . , y m ∈ R such that (2.10) e n = i+j=n σ −n (f i )y j for n = 0, 1, . . . , m.
To show that the finite system (2.10) has a solution, note that for any i ∈ N ∪ {0}, Since σ −i (u i ) is a unit of R, the power series in the brackets is a unit of A, and thus e i x i = f k i for some k i ∈ A. Therefore, setting q m = m i=0 k i ∈ A, we obtain e 0 + e 1 x + e 2 x 2 + · · · + e m x m = f q m . Now it is easy to see that if q m = ∞ n=0 a n x n and y j = σ −j (a j ) for any j ∈ N∪{0}, then y 0 , y 1 , . . . , y m is a solution of (2.10).

Reduced rings of skew power series.
Recall that an endomorphism σ of a ring R is said to be rigid if aσ(a) = 0 for every nonzero a ∈ R. This notion proved to be very useful in characterizing reduced skew polynomial rings and skew power series rings in the work of J. Krempa in [6] and C. Y. Hong, N. K. Kim and T. K. Kwak in [4], to whom the first part of the following proposition is essentially due. Proof: is reduced, then for any nonzero a ∈ R we have aσ(a)x 2 = (ax) 2 = 0, which proves that σ is rigid. Conversely, if σ is rigid, then it can be easily shown that R is reduced (see [4, p. 218 (ii) Obviously, σ is injective by (i). Let e = e 2 ∈ R. Since the ring R[[x; σ]] is reduced, it is abelian, and thus ex = xe = σ(e)x, proving that e = σ(e).

Proof of the main result
We are now in a position to prove the main result of the paper as stated in Section 1. In the proof we will use the following well known implication relations that hold for any ring: Proof of Theorem 1.6: From Theorem 1.4 we obtain (8) ⇔ (12), and by Theorem 1.5 we have (11) ⇔ (9) ⇔ (12). Thus the conditions (8), (9), (11) and (12)  ] and any f ∈ A we have Af ⊆ f A. By Proposition 2.4 we can assume that f = ∞ n=0 e n x n , where (e n ) ∞ n=0 is a sequence of orthogonal idempotents of R, and we have to show that gf ∈ f A for any g = ∞ n=0 g n x n ∈ A. Since R is ℵ 0 -injective strongly regular, by Proposition 2.3 there exists a sequence (t m ) ∞ m=0 of elements of R such that for any m, n ∈ N ∪ {0} we have e n t m = e n σ −n (g m ), or equivalently, e n σ n (t m ) = e n g m . Set h = ∞ m=0 t m x m ∈ A. Now it is easy to verify that gf = f h, which completes the proof of (3.1).
To complete the proof it suffices to show that if σ is bijective and for some i ∈ {1, 2, . . . , 10} the left analogue of (i) is satisfied, then (12) is satisfied. But if the left analogue of (i) is satisfied, then from the isomorphism R[[x; σ]] op ≃ R op [[x; σ −1 ]] it follows that (i) is satisfied for R op [[x; σ −1 ]], R op and σ −1 . Hence, by the first part of the theorem, (12) is satisfied for R op and σ −1 . Thus (12) is also satisfied for R and σ.
In the following corollary we continue the list of characterizations for the ring R[[x; σ]] to be right duo right Bézout, given in Theorem 1.6. (19) R is right duo right ℵ 0 -injective von Neumann regular, σ is bijective and σ(e) = e for any e = e 2 ∈ R.
Proof: As an immediate consequence of the implication chart ( * ) given at the beginning of this section, we obtain (13) ⇒ (15) and (14) ⇒ (16) ⇒ (17). Moreover, by ( * ), (15) as well as (17) imply that R is Dedekindfinite, and thus R is von Neumann regular by Proposition 2.1. Since a von Neumann regular ring that is right quasi-duo or abelian must be strongly regular, it follows that (15) as well as (17)  . This proves the equivalence of (13) through (18). Furthermore, from Theorem 1.4 it follows that (16) is equivalent to the condition (8) of Theorem 1.6. Since strongly regular rings are precisely right duo von Neumann regular rings, it follows that (19) is equivalent to the condition (12) of Theorem 1.6.