ORE EXTENSIONS OF 2-PRIMAL RINGS

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1 ORE EXTENSIONS OF 2-PRIMAL RINGS A. R. NASR-ISFAHANI A,B A Department of Mathematics, University of Isfahan, P.O. Box: , Isfahan, Iran B School of Mathematics, Institute for Research in Fundamental Sciences (IPM), P.O. Box: , Tehran, Iran a nasr isfahani@yahoo.com Abstract. Let R be a ring with an endomorphism α and an α-derivation δ. In this note we show that if R is (α, δ)-compatible then R is 2-primal if and only if the Ore extension R[x; α, δ] is 2-primal if and only if Nil(R) = Nil (R; α, δ) if and only if Nil(R)[x; α, δ] = Nil (R[x; α, δ]) if and only if every minimal (α, δ)-prime ideal of R is completely prime. Keywords : Ore extensions, 2-primal rings, (α, δ)-compatible rings. AMS Subject Classification: 16S36; 16N Introduction Throughout this paper R denotes an associative ring with unity, α is an endomorphism of R and δ an α-derivation of R, that is, δ is an additive map such that δ(ab) = δ(a)b + α(a)δ(b), for all a, b R. We denote by R[x; α, δ] the Ore extension whose elements are the polynomials over R, the addition is defined as usual and the multiplication subject to the relation xa = α(a)x + δ(a) for any a R. The lower nil radical (i.e., the intersection of all the prime ideals in R), the upper nil radical (i.e., sum of all nil ideals) and the set of all nilpotent elements of R are denoted by Nil (R), Nil (R) and Nil(R), respectively. A ring R is called 2-primal if Nil (R) = Nil(R). 2-primal rings are common generalization of commutative rings and rings without nilpotent elements. Shin in [12, Proposition 1.11] showed that a ring R is 2-primal if and only if every minimal prime ideal P of R is completely prime (i.e. R/P is a domain). Also he proved that the minimal-prime spectrum of a 2-primal ring is a Hausdorff space with a basis of clopen sets [12, Proposition 4.7]. G.F. Birkenmeier, H.E. Heatherly, and E.K. Lee in [2, Proposition 2.6] proved that the 2-primal condition is inherited by ordinary polynomial extensions. Example 2.1 of [3] shows that when R is 2-primal, differential polynomial ring R[x; δ] need not be 2-primal also Example 2.1 of [10] shows that for a 2-primal ring R the skew polynomial ring R[x; α] need not be 2-primal. In [10] and [11] G. Marks investigated conditions on ideals of a 2-primal ring R that will ensure that a skew polynomial ring R[x; α, δ] or a differential polynomial ring R[x; δ] be 2-primal. He proved that when R is a local ring with a nilpotent maximal ideal, the Ore extension R[x; α, δ] will or will not be 2-primal depending 0 This research was in part supported by a grant from IPM (No ). 1

2 2 A. R. NASR-ISFAHANI A,B on the δ-stability of the maximal ideal of R. An endomorphism α of R is called a rigid endomorphism if rα(r) = 0 implies r = 0 for each r R. A ring R is called to be α-rigid if there exists a rigid endomorphism α of R (for more details see [7]). According to Hong et al. [6], an α-ideal I is called an α-rigid ideal if aα(a) I implies a I for each a R. Hong et al. in [6] studied some connections between the α-rigid ideals of R and the related ideals of Ore extensions. They also studied the relationship of Nil (R) (resp. Nil (R)) and Nil (R[x; α, δ]) (resp. Nil (R[x; α, δ])) when Nil (R) (resp. Nil (R)) is an α-rigid ideal of R. They prove that if Nil (R) (resp. Nil (R)) is an α-rigid δ-ideal of R, then Nil (R[x; α, δ]) Nil (R)[x; α, δ] (resp. Nil (R[x; α, δ]) Nil (R)[x; α, δ]). Following [1], we say that R is α-compatible if for each a, b R, ab = 0 if and only if aα(b) = 0. Note that if R is α-compatible, then α is injective. Moreover, R is said to be δ-compatible if for each a, b R, ab = 0 implies that aδ(b) = 0. If R is both α-compatible and δ-compatible, we say that R is (α, δ)-compatible. Note that (α, δ)-compatible rings are a generalization of α-rigid rings. According to E. Hashemi [5], an ideal I of R is called an α-compatible ideal of R if for each a, b R, ab I if and only if aα(b) I. Moreover, I is called δ-compatible ideal if for each a, b R, ab I implies aδ(b) I. If I is both α-compatible and δ-compatible, we say that I is an (α, δ)-compatible ideal. Hashemi in [5] studied the connections of Nil (R) (resp. Nil (R)) and Nil (R[x; α, δ]) (resp. Nil (R[x; α, δ])) in case Nil (R) (resp. Nil (R)) is an (α, δ)-compatible ideal of R and provide a generalization of Hong et al. s results. In this note we continue the study of the radicals of Ore extensions in case R is (α, δ)-compatible. We show that if R is (α, δ)-compatible, then R[x; α, δ] is 2-primal if and only if R is 2-primal if and only if Nil (R, α, δ) = Nil(R) if and only if Nil(R)[x; α, δ] = Nil (R[x; α, δ]). We also show that for an (α, δ)- compatible ring R, R[x; α, δ] is 2-primal if and only if for each f R[x; α, δ], fα(f) Nil (R[x; α, δ]) f Nil (R[x; α, δ]) (for definition of α(f) see page 5). According to Shin s Theorem (R is 2-primal if and only if every minimal prime ideal P of R is completely prime), one of the attractions of 2-primal rings lies in the structure of their prime ideals. In this note we also provide a generalization of Shin s Theorem and prove that for (α, δ)-compatible ring R, R[x; α, δ] is 2-primal if and only if each minimal (α, δ)-prime ideal of R is completely prime. 2. Main results Lemma 2.1.([4, Lemma 2.1]) Let R be an (α, δ)-compatible ring. Then we have the following: (i) If ab = 0, then aα n (b) = α n (a)b = 0 for any positive integer n. (ii) If α k (a)b = 0 for some positive integer k, then ab = 0. (iii) If ab = 0, then α n (a)δ m (b) = 0 = δ m (a)α n (b) for any positive integers m, n. For 0 l n let us denote by fl n End(R; +) the sum of all words composed with l letters α and n l letters δ (e.g. fn n = α n ; f0 n = δ n ). Let I be an ideal of R. I is called δ-ideal if for each a I, δ(a) I, I is called α-ideal if for each a I, α(a) I. If I is both an α-ideal and δ-ideal, then I is

3 ORE EXTENSIONS OF 2-PRIMAL RINGS 3 called (α, δ)-ideal. Proposition 2.2. Assume that R is an α-compatible ring and Nil(R) is a δ-ideal of R. Then Nil(R[x; α, δ]) Nil(R)[x; α, δ]. Proof. Let f = a 0 + a 1 x + a n x n Nil(R[x; α, δ]). There exists a positive integer t such that f t = 0. Then a n α n (a n )α 2n (a n ) α (t 1)n (a n ) = 0 and so a n Nil(R). Let us write f = p + a n x n with p R[x; α, δ] and deg(p) < n. Then 0 = f t = p t + q, for some q R[x; α, δ]. Note that the coefficients of q can be written as sums of monomials in a i and f u v (a j ) where a i, a j {a 0, a 1,, a n } and u v 0 are positive integers and each monomial has a n or f u v (a n ). Since a n Nil(R) and Nil(R) is an (α, δ)-ideal, then f u v (a n ) Nil(R) for each positive integers u v 0. Thus q Nil(R)[x; α, δ], since Nil(R) is an ideal of R. Then p t Nil(R)[x; α, δ] and so a n 1 α n 1 (a n 1 ) α (n 1)(t 1) (a n 1 ) Nil(R) and by using Lemma 2.1, a n 1 Nil(R). Continuing in this way we can see that for each i, a i Nil(R) and the result follows. Corollary 2.3. If Nil(R) is an ideal of R then Nil(R[x]) Nil(R)[x]. Note that A. Smoktunowicz s Example [13] shows that there exists a ring R such that Nil(R) is an ideal of R but Nil(R)[x] Nil(R[x]). Proposition 2.4. Assume that R is an α-compatible ring. (1) If Nil(R) = Nil (R; α, δ), then R[x; α, δ] is 2-primal. (2) If Nil(R)[x; α, δ] = Nil (R[x; α, δ]), then R[x; α, δ] is 2-primal. Proof. 1) Assume that R is α-compatible and Nil(R) = Nil (R; α, δ). Since Nil (R; α, δ) is an (α, δ)-ideal then by Proposition 2.2, Nil(R[x; α, δ]) Nil(R)[x; α, δ] = Nil (R; α, δ)[x; α, δ]. By [8, Lemma 5.1], Nil (R; α, δ)[x; α, δ] Nil (R[x; α, δ]) and so R[x; α, δ] is 2-primal. 2) Assume that Nil(R)[x; α, δ] = Nil (R[x; α, δ]), then Nil(R) is an ideal of R. Obviously Nil(R) is an α-ideal. Now let a Nil(R), then α(a)x + δ(a) = xa Nil(R)[x; α, δ] and so δ(a) Nil(R). Thus by Proposition 2.2, Nil(R[x; α, δ]) Nil(R)[x; α, δ] = Nil (R[x; α, δ]) and so R[x; α, δ] is 2-primal. Let R be a ring, End(R; +) the ring of additive endomorphisms of R and Φ a subset of End(R; +). A sequence (a 0, a 1,, a n, ) of elements of R is called a Φ-m-sequence if for any i N there exist ϕ i, ϕ i Φ and r i R such that a i+1 = ϕ i (a i )r i ϕ i (a i). An element a R is called strongly Φ-nilpotent if every Φ-m-sequence starting with a eventually vanishes. If Φ = {id R } we recover the corresponding classical notions (for more details see [8]). Theorem 2.5. Let R be an (α, δ)-compatible ring. Then R[x; α, δ] is 2-primal if and only if Nil(R) = Nil (R; α, δ) if and only if Nil(R)[x; α, δ] = Nil (R[x; α, δ]). Proof. If Nil(R) = Nil (R; α, δ) then by Proposition 2.4, R[x; α, δ] is 2-primal. Assume that R[x; α, δ] is 2-primal. By [8, Lemma 5.1], Nil (R; α, δ)[x; α, δ] Nil (R[x; α, δ]) = Nil(R[x; α, δ]), and so Nil (R; α, δ) Nil(R). Since R[x; α, δ]

4 4 A. R. NASR-ISFAHANI A,B is 2-primal, R is 2-primal by [2]. Now let a Nil(R) = Nil (R), then a is strongly nilpotent. So each m-sequence starting with a eventually vanishes. By using Lemma 2.1, each {α, δ}-m-sequence starting with a eventually vanishes. Thus a is strongly {α, δ}-nilpotent and by [8, Proposition 1.11], a Nil (R; α, δ). Then Nil(R) Nil (R; α, δ). If Nil(R)[x; α, δ] = Nil (R[x; α, δ]) then by Proposition 2.4, R[x; α, δ] is 2-primal. Assume that R[x; α, δ] is 2-primal. Then Nil(R) = Nil (R; α, δ) and by [8, Lemma 5.1], Nil(R)[x; α, δ] = Nil (R; α, δ)[x; α, δ] Nil (R[x; α, δ]). Also by Proposition 2.2, Nil (R[x; α, δ]) = Nil(R[x; α, δ]) Nil(R)[x; α, δ] and the result follows. Corollary 2.6. Assume that R is an (α, δ)-compatible ring. Then R[x; α, δ] is 2-primal if and only if R is 2-primal and Nil(R)[x; α, δ] = Nil(R[x; α, δ]). Proof. If R[x; α, δ] is 2-primal, then R is 2-primal and by Theorem 2.5, Nil(R)[x; α, δ] = Nil(R[x; α, δ]). Now assume that R is 2-primal and Nil(R)[x; α, δ] = Nil(R[x; α, δ]). By the same argument as in the proof of Theorem 2.5, Nil(R) Nil (R; α, δ). Then by [8, Lemma 5.1], Nil(R[x; α, δ]) = Nil(R)[x; α, δ] Nil (R; α, δ)[x; α, δ] Nil (R[x; α, δ]) and the result follows. Lemma 2.7. Assume that R is a reduced (α, δ)-compatible ring. Let P be a minimal (α, δ)-prime ideal of R then P is completely prime. Proof. Let Φ be the subsemigroup of End(R, +) generated by α, δ, S = R\P and S be the multiplicative monoid generated by Φ(S). We claim that 0 S. For otherwise we have an equation s 1 s 2 s n = 0, with s i S and with n minimal. Since R is reduced and (s n Rs 1 s 2 s n 1 ) 2 = 0 we have s n Rs 1 s 2 s n 1 = 0. P is (α, δ)-prime, so there exists an element s = ϕ 1 (s n )rϕ 2 (s 1 ) S, for some r R and ϕ 1, ϕ 2 Φ. We have s n rs 1 s 2 s n 1 = 0 and since R is (α, δ)-compatible, ss 2 s n 1 = ϕ 1 (s n )rϕ 2 (s 1 )s 2 s n 1 = 0 which is contradiction to the minimality of n and so 0 S. By [8, Proposition 1.10], we can enlarge (0) to an (α, δ)-prime ideal P disjoint from S. But P is minimal (α, δ)-prime and so we must have P = P. Then S = S and so S is closed under multiplication. Thus R/P is a domain and the result follows. Now we give a generalization of Shin s Theorem [12] for Ore extensions. Theorem 2.8. Assume that R is an (α, δ)-compatible ring. Then R[x; α, δ] is 2- primal if and only if every minimal (α, δ)-prime ideal of R is completely prime. Proof. Assume that R[x; α, δ] is 2-primal, then by Theorem 2.5, R = R Nil (R;α,δ) is reduced. Let P be a minimal (α, δ)-prime ideal of R, then P is a minimal (α, δ)- prime ideal of R. Since R/P = R/P, then by Lemma 2.7, P is completely prime. Now assume that every minimal (α, δ)-prime ideal P of R is completely prime. Let {P i } i I be a family of all minimal (α, δ)-prime ideals of R. Then Nil (R; α, δ) = i I P R i and so Nil (R;α,δ) embeds in i I R/P R i. Then Nil (R;α,δ) is reduced and so Nil(R) Nil (R; α, δ). Since Nil (R; α, δ)[x; α, δ] Nil (R[x; α, δ]) Nil(R[x; α, δ]), then Nil (R; α, δ) Nil(R). Thus Nil(R) = Nil (R; α, δ) and

5 ORE EXTENSIONS OF 2-PRIMAL RINGS 5 the result follows by Theorem 2.5. Let R be a ring with an endomorphism α and α-derivation δ such that αδ = δα. Then there exists an endomorphism ᾱ on R[x; α, δ] which extends α, given by ᾱ( n i=0 a ix i ) = n i=0 α(a i)x i for all a i R. We denote ᾱ by α. Theorem 2.9. Assume that R is an (α, δ)-compatible ring and αδ = δα. Then R[x; α, δ] is 2-primal if and only if for each f R[x; α, δ], fα(f) Nil (R[x; α, δ]) implies that f Nil (R[x; α, δ]). Proof. Assume that R[x; α, δ] is 2-primal and let f = a 0 + a 1 x + + a n x n R[x; α, δ] such that fα(f) Nil (R[x; α, δ]). Then by Theorem 2.5, fα(f) Nil(R)[x; α, δ] and so a n α n+1 (a n ) Nil(R). Since R is α-compatible, a n Nil(R). fα(f) = (a 0 +a 1 x+ +a n 1 x n 1 )(α(a 0 )+α(a 1 )x+ +α(a n 1 )x n 1 )+ (a 0 + a 1 x + + a n 1 x n 1 )α(a n )x n + a n x n (α(a 0 ) + α(a 1 )x + + α(a n 1 )x n 1 ) + a n x n α(a n )x n and so (a 0 +a 1 x+ +a n 1 x n 1 )(α(a 0 )+α(a 1 )x+ +α(a n 1 )x n 1 ) Nil(R)[x; α, δ] and by the same argument a n 1 Nil(R). Continuing in this way we have a i Nil(R) for each i. Then f Nil(R)[x; α, δ] and the result follows by Theorem 2.5. Conversely assume that fα(f) Nil (R[x; α, δ]) implies that f Nil (R[x; α, δ]) for each f R[x; α, δ]. Let f R[x; α, δ] such that f 2 = 0. Then fα(f)α(fα(f)) = 0 Nil (R[x; α, δ]) and so fα(f) Nil (R[x; α, δ]), hence f Nil (R[x; α, δ]). Then Nil(R[x; α, δ]) Nil (R[x; α, δ]) and the result follows. Theorem Assume that R is an (α, δ)-compatible ring. Then R[x; α, δ] is 2-primal if and only if R is 2-primal. Proof. Assume that R is 2-primal and let a Nil(R) = Nil (R). Then a is strongly nilpotent and since R is (α, δ)-compatible then a is strongly {α, δ}- nilpotent and by [8, Proposition 1.11], a Nil (R; α, δ). Also we have Nil (R; α, δ) Nil(R), then the result follows by Theorem 2.5. Corollary Assume that R is an (α, δ)-compatible ring. Then the following are equivalent: (1) R is 2-primal. (2) R[x; α, δ] is 2-primal. (3) Nil(R) = Nil (R; α, δ). (4) Nil(R)[x; α, δ] = Nil (R[x; α, δ]). (5) R is 2-primal and Nil(R)[x; α, δ] = Nil(R[x; α, δ]). (6) Every minimal (α, δ)-prime ideal of R is completely prime. ACKNOWLEDGEMENT. The author would like to thank the Banach Algebra Center of Excellence for Mathematics, University of Isfahan. Special thanks are due to the referee who read this paper very carefully and made many useful suggestions. REFERENCES [1] S. Annin, Associated primes over skew polynomial rings, Comm. Algebra 30(5) (2002)

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