On the set of subrings wich are directed unions of Artinian subrings

ON THE SET OF SUBRINGS WHICH ARE DIRECTED UNIONS OF ARTINIAN SUBRINGS D. KARIM AND S. ZARZUELA Abstract. The paper contributes to the investigation of zero–dimensional rings which can be written as a directed union of Artinian subrings. We give conditions on DU (R) in order to be nonempty. Introduction Let Rbe a commutative ring with identity. We recall that Ris reduced if TP∈Min(R)P= (0), where Min(R) is the set of minimal prime ideals of R, and zero-dimensional if all prime ideals are maximal. The ring Ris said to be von Neumann regular if for each element r∈R, there exists r0∈Rsuch that r=r2r0. It is easily seen that Ris a von Neumann regular ring if and only if Ris reduced and zero-dimensional. Initially we note that if Ris the directed union of a family {Rα}α∈Aof subrings, each of dimension at most n, then dim(R)≤n; this follows from the fact that any chain P0⊂P1⊂ · · · ⊂ Pkof prime ideals of Rcontracts to a chain of distinct primes on some Rα. Thus, a ring that is a directed union of Artinian subrings is zero-dimensional. Recently many authors have been interested in zero-dimensionality (see [6, 7, 9]). The purpose of this paper is to pursue the study of the problem of whether a von Neumann regular ring Ris expressible as a directed union of Artinian subrings, raised by Gilmer and Heinzer in 1992 [3, Problem 42]. This fact leads us to consider the set of subrings of Rwhich are directed unions of Artinian subrings of Rdenoted DU(R) and the family of Artinian subrings A(R). Also we use Idem(R), Spec(R), and C(R), respectively, to denote the set of idempotent elements of R, the set of prime ideals of R, and the set {char(R/M): Mranges over the maximal ideals of R}. The present paper considers the following questions: (q1) If A(R) is a nonempty set is it true that DU(R)6=∅? 2000 Mathematics Subject Classification. Primary 13A99; Secondary 13A15, 13B02, 13E05. Key words and phrases. Zero–dimensional ring, von Neumann regular ring, infinite product, directed union of Artinian subrings, semi–quasilocal ring. 1 2 D. KARIM AND S. ZARZUELA (q2) If A(R)6=∅under what conditions is the set DU(R) nonempty? (q3) Let Rbe a von Neumann regular ring, under what conditions is Rexpressible as a directed union of Artinian subrings? It is worthwhile recalling that any zero-dimensional ring Rwith finite spectrum is expressible as a directed union of Artinian subrings [7, Theorem 5.4]. This leads us to examine the case where spectrum is infinite, i.e., we are interested with rings which are expressible as a directed unions of Artinian subrings but are not Artinians. We give a short overview of the paper. The first section is concerned with questions (q1) and (q2). We give a characterization of the set DU(R). Precisely, our main results of this section are Proposition 1.3, which gives the relationship between A(R) and DU(R), and Theorem 1.12 which shows that DU(R) can be a nonempty set and Ris not a directed union of Artinian subrings. In the second section we give a necessary and sufficient condition for a von Neumann regular ring to be a directed union of Artinian subrings that is an answer to question (q3). We probe this problem in connection with their families of residue fields say F(R) = {R M:Ma maximal ideal of R}. 1. Preliminaries and general results Throughout this paper, all rings are assumed to be commutative with unit. Further, all ring-homomorphisms are unital. If Ris a subring of a ring S, we assume that the unity of Sbelongs to R. Turning to the Artinian case, two key properties of an Artinian ring Rthat come into play are that Spec(R) is finite and that Rhas only finitely many idempotents. The existence of Artinian subrings was among the primary focus of many mathematicians [6, 9, 10]. They were particularly interested in the question of whether a zero-dimensional ring Ris expressible as a directed union of Artinian subrings. Naturally hereditarily zero-dimensional rings 1are example. More generally each zero-dimensional ring with finite spectrum is a directed union of Artinian subrings. However not every zero-dimensional ring is expressible as a directed union of Artinian subrings. For instance, if R=Q∞ i=1 Qdenotes the infinite product of copies Q, the ring Ris a zero-dimensional ring that is not a directed union of Artinian subrings. This section was inspired by a desire to understand the relationship between A(R) and DU(R). Let (Rj, fjk) be a directed system of rings, indexed by a directed set (I, ≤). Let R=Sj∈IRj, together with the canonical maps fj:Rj−→ R. The ring Ris said to be a directed union of the Rj’s if the fjk’s are inclusion maps. 1A ring Ris hereditarily zero-dimensional if all subrings of Rare zero-dimensional. DIRECTED UNION OF ARTINIAN SUBRINGS 3 Thus, directed unions can be treated by assuming all fjk to be monomorphisms. Notice that Rneed not be Artinian even if each Rjis Artinian. Next, we use the fact that the Krull dimension is preserved under integral extensions (cf.[4, (11.8)]). In particular, an integral extension ring of a zero-dimensional ring is zero-dimensional. Lemma 1.1. If Sis integral over an Artinian ring A, then Sis written as a directed union of Artinian subrings, and hence DU(S)6=∅. Proof. we can write S=[{A[F] : Fis a finite subset of S}. The ring A[F] is intermediate between Aand S, so dim(A[F]) = 0 and A[F] is a finitely generated A-module. It follows that A[F] is a Noetherian ring, and hence A[F] is Artinian (see [2, Theorem 8.5]). Now, let F1and F2be two finite subsets of S, then F=F1∪F2⊂Sand A[F1]⊆A[F] and A[F2]⊆ A[F]. It follows that {A[F] : Fis a finite subset of S}is a directed family of Artinian subrings and hence S=S{A[F] : Fis a finite subset of S}is a directed union of Artinian subrings. Thus, DU(S)6=∅.¤ In general, if Ris von Neumann regular which is integral over a Noetherian subring, then Ris a directed union of Artinian subrings. Lemma 1.2. Let Rbe a ring. Then (i) If DU(R)6=∅then C(R)is finite. (ii) If C(R)is infinite then A(R) = ∅. Proof. (i) Let S∈ DU(R), then S=Si∈ISiis a directed union of Artinian subrings Si. Then Si∈ A(R) and hence A(R)6=∅. Therefore C(R) is finite [5, Proposition 1]. (ii) Assume that A(R)6=∅, let S∈ A(R), then C(R)⊆ C(S) and hence C(S) is infinite, a contradiction with Sis Artinian (C(R) is finite). ¤ For the next results, we need the following definition. Let {Rα}α∈Abe a nonempty family of rings and Qα∈ARαtheir product. We frequently consider Qα∈ARαas the set of all functions f:A−→ Sα∈ARα, such that f(α)∈Rαfor each α∈A, with addition and multiplication defined pointwise: (f+g)(α) = f(α)+g(α) and (fg)(α) = f(α)g(α). In this perspective, the direct sum ideal of Qα∈ARα, denoted Lα∈ARα, is the set of functions f∈Qα∈ARαthat are finitely nonzero (i.e., {α∈A: f(α)6= 0 in Rα}is finite). We use Zand N, respectively, to denote the set of integers and the set of natural numbers. 4 D. KARIM AND S. ZARZUELA Proposition 1.3. If R=Ln j=1 Rjis the direct product of zero-dimensional rings Rj, then the following conditions are equivalent. (i) A(R)6=∅; (ii) A(Rj)6=∅, for each j= 1, . . . , n; (iii) DU(Rj)6=∅, for each j= 1, . . . , n; (iv) DU(R)6=∅. We use the following lemmas to prove Proposition 1.3. Lemma 1.4. Let Rbe a zero-dimensional ring with only finitely many idempotent elements, then Ris expressible as a directed union of Artinian subrings. In order to prove this result, we need the following lemmas. Lemma 1.5. Any zero–dimensional ring Rwith only finitely many idempotents is semi-quasilocal. Proof. Let M1, . . . , Mr+1 be distinct maximal ideals of R. Let x∈Mr+1(∪r i=1Mi), since dimR= 0 by [8, Theorem 3.1], there exists t∈Z+and ean idempotent element of Rsuch that xtR=eR. Hence e∈Mr+1(∪r i=1Mi). It follows that if Rhas nmaximal ideals, it has at least n−1 idempotents. Therefore, Ris necessarily semi-quasilocal. ¤ Lemma 1.6. [7, Corollary 5.5] If Ris a zero–dimensional semi-quasilocal ring then there exists an Artinian subring of R. Proof of Lemma 1.4. An immediate consequence of Lemma 1.5 and Lemma 1.6. ¤ Proof of Proposition 1.3. (i) ⇔(ii). Let S∈ A(R) and ejbe the idempotent element of Rassociated with {j}. Then S[e1, . . . , en] is an integral extension of S. Moreover, S[e1, . . . , en] is finitely generated over Sand hence S[e1, . . . , en] = Se1⊕ · · ·⊕ Senis Artinian. Therefore, Sejis an Artinian subring of Rj, for each j. Conversely, if Sj∈ A(Rj) for each j, then ⊕n j=1Sjis an Artinian subring of R. (ii) ⇔(iii). Let jo∈ {1, . . . , n}and suppose that A(Rjo)6=∅. Let A∈ A(Rjo) and Idem(Rjo) = {el}l∈Lbe the set of all idempotent elements of Rjo. Two cases are then possible: Case 1: If {el}l∈Lis a finite set, by Lemma 1.4, Rjois a directed union of Artinian subrings and hence DU(R)6=∅. Case 2: If {el}l∈Lis infinite, then A[{el}l∈L] is integral over A. According to Lemma 1.1, A[{el}l∈L] is a directed union of Artinian subrings. Consequently, DU(R)6=∅. Conversely, if S=Si∈ISi∈ DU(Rjo) for some DIRECTED UNION OF ARTINIAN SUBRINGS 5 jo∈ {1, . . . , n}, then Si∈ A(Rjo) and hence A(Rjo)6=∅. (iii) ⇔(iv). Let Sj∈ DU(Rj) for each j= 1, . . . , n and S=Ln j=1 Sj. To show that S∈ DU(R) it suffices to show it for n= 2. Now suppose that S=S1⊕S2, where S1=Si∈JViand S2=Sk∈KWkare directed unions of Artinian subrings. It is easy to see that S=S1⊕S2=S(i,k)∈J×K(Vi⊕Wk) and {Vi⊕Wk}(i,k)∈J×Kis a directed family. We notice also that Vi⊕Wkis Artinian for each (i, k)∈J×K. It follows that S=S(i,k)∈J×K(Vi⊕Wk) is a directed union of Artinian subrings. Conversely, let W=Sj∈IWj∈ DU(R) and eibe the idempotent element of Rassociated with {i}. Then Wei=Sj∈IWjei. From (i)⇔(ii)Wjei∈ A(Ri) and if Wl⊆Ws then Wlei⊆Wseiand hence the family {Wjei}j∈Jis directed. Thus, Wei∈ DU(Ri). ¤ Proposition 1.3 does not hold for infinite direct products as the following example shows. Example 1.7. Let R=Q∞ i=1 Z piZ, where {pi}∞ i=1 is an infinite family of distinct prime integers. Clearly, A(Z piZ)6=∅(respectively, DU(R)6=∅) for each i. However, since {pi:i= 1,2, . . . } ⊆ C(R), we have that C(R) is infinite. Thus, by [10, Theorem 2.1], A(R) = ∅(respectively, DU(R) = ∅). This means that Rhas no Artinian subring. Proposition 1.8. Let Rbe a ring and {Lλ}λ∈Λits residue fields. Then A(Qλ∈ΛLλ)6=∅if and only if DU(Qλ∈ΛLλ)6=∅. Proof. Suppose that DU(Qλ∈ΛLλ)6=∅, let S∈ DU(Qλ∈ΛLλ) then S=Si∈ISiis a directed union of Artinian subrings, and hence Si∈ A(S)⊆ A(Qλ∈ΛLλ). Thus, A(Qλ∈ΛLλ)6=∅. Conversely, assume that A(Qλ∈ΛLλ)6=∅, then Sλ∈ΛC(Lλ)⊆ C(Qλ∈ΛLλ) is a finite set of prime numbers. In other words, Sλ∈ΛC(Lλ) = {p1, . . . , pk}, where piis a prime number for each i= 1, . . . , k. We can write Λ = Λ1∪ · · · ∪ Λkas a finite partition, where Λi={λ∈Λ/char(Lλ) = pi}, for each i= 1, . . . , k, and hence Qλ∈ΛLλ=Qλ∈Λ1LλLQλ∈Λ2LλL· · · LQλ∈ΛkLλ. Without loss of generality, we assume that Λ = Λi, for some i∈ {1, . . . , k}. In other words, we suppose that char(Lλ) = pifor each λ∈Λ. In this case, Zw pi⊆Qλ∈ΛLλ, where wis the cardinality of Λ, Zw piis the infinite direct product of copies of Zpi, and Zpiis the prime field of characteristic pi. Since lim →Z(i) pi Zω pi, where Z(i) pi={{xj}∞ j=1 ∈Zw0 pi:xi−1=xi=. . . }is a subring of R, and Z(i) pi≃Zi pi, the product of icopies of Zpi, an Artinian von Neumann regular ring. Therefore, the direct limit lim →Z(i) pi∈ DU(Zω pi) and DU(Zω pi)⊆ DU(Qλ∈ΛLλ), we must have DU(Qλ∈ΛLλ)6=∅.¤ 6 D. KARIM AND S. ZARZUELA Remark 1.9. If C(R) is finite, then DU(Qλ∈ΛLλ)6=∅. This follows from the fact that the finiteness of C(R) implies that A(Qλ∈ΛLλ)6=∅. Proposition 1.10. Let Rbe a zero-dimensional ring and N(R)its nilradical. Then the following conditions are equivalent: (i) DU(R)6=∅; (ii) DU(R N(R))6=∅. Proof. (i) ⇒(ii). Let ϕ:R³R N(R)be the canonical projection, and T∈ DU(R). Then ϕ(T) is a directed union of Artinian subrings of R N(R). Indeed, if T=Si∈ITiis a directed union of Artinian subrings, then it is not difficult to see that ϕ(T) = Si∈Iϕ(Ti), where ϕ(Ti) = Ti Ti∩N(R)=Ti N(Ti). Notice that if Ti⊆Tjthen Ti N(Ti)⊆Tj N(Tj), in other words, ϕ(Ti)⊆ϕ(Tj) and it is easy to see that Ti N(Ti)∈ A(R N(R)) for each i∈I. Therefore, ϕ(T) = Si∈Iϕ(Ti) is a directed union of Artinian subrings. Thus, DU(R N(R))6=∅. (ii) ⇒(i). Suppose that DU(R N(R))6=∅and let S=Si∈ISi∈ DU(R N(R)) then Si∈ A(R N(R)). Let σ:R³R N(R)the canonical projection. Let i0∈I and Ri0=σ−1(Si0), the inverse image of Si0. It is not difficult to show that Ri0 N(Ri0)≃Si0and hence Ri0is a semi-quasilocal zero-dimensional subring of R. By [7, Corollary 5.5], Ri0is a directed union of Artinian subrings. In other words, Ri0∈ DU(R). ¤ Now we observe that if Ris a directed union of Artinian subrings, then DU(R) is a nonempty set. This raises the question of what the relationship is between DU(R)6=∅and the property that Ris a directed union of Artinian subrings. Example 1.11. Let Ω = Q(X) be a simple transcendental extension of Q, where Qdenotes the field of rational numbers. Let R= ΩNbe a countable direct product of copies of Ω. We consider Ω(i)={{xj}∞ j=1 ∈ΩN:xi−1= xi=...}a subring of R. It is easy to see that Ω(i)≃Ωi, the finite product of icopies of Ω, an Artinian von Neumann regular ring. It follows that A(R)6=∅, and lim →Ω(i)=T⊆ΩN. The ring Tis a directed union of Artinian subrings, in other words, T∈ DU(R). However, Ris not a directed union of Artinian subrings: indeed, consider y= (yi)i∈N∗∈Rsuch that yi6=yjfor i6=j. Then, for each i∈N∗,y /∈Ω(i). It follows that lim →Ω(i) ΩN. Hence A(R)6=∅does not imply that Ris a directed union of Artinian subrings. Let Rbe a ring and {Rα}α∈Aan infinite family of nonzero rings such that Ris, up to isomorphism, a subring of each Rα. We use R∗to denote DIRECTED UNION OF ARTINIAN SUBRINGS 7 the diagonal imbedding of Rin Qα∈ARα, that is R∗=ϕ(R), where ϕ: R ,→Qα∈ARαis a monomorphism defined by ϕ(x) = {xα}α∈Asuch that xα=xfor each α∈A. Theorem 1.12. Let Rbe a zero–dimensional ring and F(R) = {Li}i∈I its set of residue fields. Assume that for each pair (i, j)∈I2we have Lj∩Lk=L /∈ F(R). Then (1) DU(R)6=∅. (2) Ris not a directed union of Artinian subrings. To prove this result, we need the following Lemma. Lemma 1.13. [11, Proposition 2.1] Let Rbe a Von Neumann regular ring and R⊂T⊂Qλ∈Λ R Mλsuch that T=Si∈ITiis a directed union of Artinian subrings and {Mλ}λ∈Λ=Spec(R). Then F(R) = F(T). Proof of Theorem 1.12. (1) According to Lemma 1.2, we assume that C(R) is finite, otherwise, DU(R) = ∅. Since C(R) = {p1, . . . , pl}is a finite set of prime integers, we can write I=I1∪ · · · ∪ Ilas a partition of Iwhere Ij={i∈I/char(Li) = pj}for j= 1, . . . , l. Let Tj=Qi∈IjLifor each j= 1, . . . , l. Thus Ql i=1 Tiis isomorphic to Qi∈ILi. For j= 1, . . . , l, let Zpjbe the prime subfield of characteristic pjand ZIj pjthe direct product of copies Zpj. Then Ql j=1 ZIj pjis a subring of Qi∈ILi. By [7, Theorem 6.7], Ql j=1 ZIj pj=Sk∈ASkis a directed union of Artinian subrings. Let Rk=R∩Skis a von Neumann regular subring of R, for each k∈I. Since each Rkis a subring of Skand |Idem(Rk)| ≤ |Idem(Sk)|, it follows that each Rkis Artinian. The family {Rk}k∈Ais directed because the family {Sk}k∈Ais so. It follows that Sk∈ARk⊆Ris a directed union of Artinian subrings. Thus DU(R)6=∅, and hence DU(R)6=∅. (2) Since C(R) = {p1, . . . , pl}and F(R) = F1(R)∪ F2(R)∪ · ·· ∪ Fl(R), where Fj(R) = {F∈ F(R)/char(F) = pj}, to prove that the condition of theorem 1.12 is satisfied for F(R) it suffices to show that it is satisfied for each Fj(R). In this case we suppose that F(R) = Fj(R) for some j∈ {1, . . . , l}. Now, assume that L /∈ Fj(R). Let Sbe the subring of Qj∈ILjconsisting of eventually constant sequences. Thus S ≃ L∗+J, the L-subalgebra of Qj∈ILjgenerated by the direct sum ideal J=Li∈ILi, where L∗is the diagonal imbedding of Lin Qj∈ILj. First, we claim that Sis the maximal subring of Qj∈ILjwith respect to being a directed union of Artinian subrings. Let T=Sj∈JTjbe a subring of Qα∈ALαthat is a directed union of Artinian subrings, and let t={tα}α∈A∈T. There exists jo∈Jsuch that t∈Tjoand Tjois a finite product of fields. Hence 8 D. KARIM AND S. ZARZUELA t∈ S. We claim that F(S) = {L}S{Lj}j∈I. Let pi:T=Qj∈ILj→Libe the canonical projection and pi|S its restriction on S, which is a surjective homomorphism. We have that Kerpi|S = (1 −ei)TTS= (1 −ei)S=Mi, with eithe primitive idempotent with support {i}, and S/Mi≃Lifor each i∈I. Also, Jis a maximal ideal of Sand S/J ≃(L)∗≃L. Thus {J} ∪ {Mi}i∈I⊆Max(S). Let P∈Spec(S), if J⊆P, then J=P. If J*Pthen ei/∈P, for some i∈I, and hence 1 −ei∈P. Therefore, Mi⊆Pand P=Mi. Consequently, Max(S) = {J}S{Mi}i∈I. Thus F(S) = {L}S{Lj}j∈I. If Ris a directed union of Artinian subrings, then R⊂ S. By Lemma 1.13, F(R) = F(S), a contradiction to the fact that L /∈ F(R). ¤ Corollary 1.14. Let Rbe a ring and Sis a multiplicative closed subset of R. (i) If Ris a directed union of Artinian subrings, then S−1Rso is. (ii) If DU(R)6=∅, then DU(S−1R)6=∅. (iii) We have DU(R(X)) 6=∅if DU(R)6=∅, where R(X)is the Nagata ring and Xis an indeterminate over R. Proof. (i) Suppose that R=Sα∈ARαis a directed union of Artinian subrings and Sα=S∩Rαbe a multiplicative closed subset of Rα. It is not difficult to show that S−1R=Sα∈ASαRα. Since Rαis Artinian, the localization S−1 αRαis also Artinian. The family {S−1 αRα}α∈Ais directed since {Rα}α∈Aso is. It follows that S−1R=Sα∈ASαRαis a directed union of Artinian subrings. (ii) Suppose that DU(R)6=∅, let T=Sj∈JTj∈ DU(R) and U=S∩Tbe a multiplicative closed subset of T. By (i), U−1T⊆S−1Ris a directed union of Artinian subrings. It follows that DU(S−1R)6=∅. As R(X) = S−1R[X], where S=R[X]S{MR[X] : Mis a maximal ideal of R}, we have DU(R(X)) 6=∅if DU(R)6=∅. (iii) If R=Si∈IRiis a directed union of Artinian subrings, then R(X) = Si∈IRi(X). Since each Riis Nœtherian, by [12, (6.17)], Ri(X) is also Nœtherian and each Ri(X) is zero-dimensio- nal as each Riis zero-dimensional (cf.[1, Proposition 1.21]). By [2, Theorem 8.5], Ri(X) is an Artinian ring for each i∈I. The family {Ri(X)}i∈I is directed because so is the family {Ri}i∈I. Then R(X) is a directed union of Artinian subrings. ¤ DIRECTED UNION OF ARTINIAN SUBRINGS 9 2. Von Neumann Regular rings as a directed union of Artinian subrings Let Rbe a von Neumann regular ring and {Mi}i∈Iits spectrum. Since R is reduced, we have Ti∈IMi= (0) and hence the homomorphism ϕ:R→ Qi∈I R Mi, defined by ϕ(x) = x+Mi, is injective. This allows us to view R as a subring of Qi∈I R Mi. We identify xwith its image {xi}i∈I∈Qi∈I R Mi. Finally, we denote Fi=R Mifor each i∈I. We assume that there is a field Ω containing each Fi. We can always make this assumption if the Fihave the same characteristic. Given x∈R, let xi∈Ω be the coset x+Mi∈Fi, but viewed as an element of Ω. We identify xwith its image {xi}i∈I∈ΩI, the infinite direct product of copies of Ω. Given y={yi}i∈I∈ΩI, let ||y|| ={yi:i∈I} ⊆ Ω. Finally, we put S={y∈ΩI:||y|| is finite} Proposition 2.1. With the notation and assumptions above, we have the following: (1) Sis von Neumann regular. (2) Sis a directed union of Artinian subrings. (3) If R⊂ S then Ris a directed union of Artinian subrings. To prove this result, we require the following two lemmas. Lemma 2.2. Let {Sα}α∈Abe a family of von Neumann regular subrings of a von Neumann regular ring T. Then R:= Tα∈ASαis von Neumann regular. Proof. Given x∈R, let ybe the unique element of Tsatisfying xyx =x and yxy =y. For the existence, choose any zsuch that xzx =xand put y:= zxz. For the uniqueness, just note that uniqueness is clear for fields, so it is true locally and therefore globally. The existence and uniqueness, applied to each Sα, shows that y∈Sαfor each α. Thus y∈Ras desired. ¤ Lemma 2.3. Let Rand Sbe von Neumann regular rings, with Ra subring of S. (i) If Sis a directed union of Artinian subrings, so is R. (ii) If DU(S)6=∅, then DU(R)6=∅. Proof. (i) Write S=Si∈ISias the directed union of Artinian rings Si. Since R=Si∈IRiis the directed limit of the rings Ri=R∩Si, it will suffice to show that each Riis Artinian. By Lemma 2.2, we know at least that Ri is von Neumann regular. Since Rihas only finitely many idempotents (as Ri⊆Si), Riis a direct product of finitely many fields. ¤