NAGATA RINGS AND DIRECTED UNIONS OF ARTINIAN SUBRINGS

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NAGATA RINGS AND DIRECTED UNIONS OF ARTINIAN SUBRINGS

D. KARIM Received 24 October 2002

We investigate when a Nagata ringR(X)can be written as a directed union of Artinian subrings. For a family of zero-dimensional rings{R_α}α∈A, we show that α∈AR_α(X_α)is not a directed sum of Artinian subrings.

2000 Mathematics Subject Classification: 13A99, 13A15, 13B02, 13E05.

1. Introduction. All rings considered in this paper are assumed to be com- mutative with a unit element. IfR is a subring of a ringS, we assume that the unity element ofS belongs toR, and hence is the unit element ofR. Let Spec(R), Z(R), Inv(R), and Ann_R(I)denote, respectively, the spectrum ofR (the set of prime ideals ofR), the set of zero-divisors ofR, the set of invertible elements ofR, and the annihilator of a subsetIofR. By the dimension ofR, denoted as dimR, we mean the Krull dimension: dimRis the maximal length of a chain of proper prime idealP0⊂P1⊂ ··· ⊂PnofR. If there is no upper bound on the length of such chains, then we write dimR= ∞.

In this paper, we study zero-dimensional rings, in which each proper prime ideal is maximal, and Nagata rings. Our attention will be focused on proving that an infinite direct product ofRα(Xα), where{Rα}α∈A is a family of zero- dimensional rings and{X_α}α∈Ais a family of indeterminates, is not a directed union of Artinian subrings. Rings of Krull dimension zero have been studied intensively in the literature since the sixties. Directed unions of Artinian sub- rings have been investigated more recently, see [5,7].

In [1, Problem 42], Gilmer and Heinzer raised the following question.

(Q) Under which conditions is a von Neumann regular ring a directed union of Artinian subrings?

It would be interesting to consider this question for Nagata rings.

In 1992, Gilmer and Heinzer showed that a product of zero-dimensional rings has dimension zero or infinity, see [6, Theorem 11].

LetR be a commutative ring andf ∈R[X]. The content off is the ideal σ (f )ofRgenerated by the coefficients off. Then

S=

f∈R[X]|σ (f )=R

=R[X]\

MR[X]|Mis a maximal ideal ofR (1.1)

is a multiplicatively closed subset of R[X], and the localization R(X) = S⁻¹R[X]is called the Nagata ring in one variable overR. The Nagata ring inn variables with coefficients inRis the ring

R(n)=R

X1,...,X_n

=S_n⁻¹R

X1,...,X_n

, (1.2)

where

Sn= f∈R

X1,...,Xn

|σ (f )=R

. (1.3)

IfRis zero dimensional, then by [2, Proposition 1.21], dimR(X)=dimR[X]− 1=0 since dim(R[X])=1 (cf. [13, Theorem 2]). HenceR(X)is also zero di- mensional.

2. Nagata rings. We first fix notation. Data will consist of a directed system (Rj,fjk)of rings indexed by a directed set(I,≤)and its directed unionR=

j∈IRj, together with the canonical mapsfj:Rj→R. The ringRis a directed union ofR_j’s corresponding to thef_jk’s being inclusion maps. Thus directed unions can be treated by assuming allfjkto be monomorphisms. Notice that ifRjis a ring for eachj∈I, thenRis also a ring. However,Ris not necessarily Noetherian even if eachR_jis Noetherian. IfR= _j∈IR_jis a directed union of Artinian subrings, then we regard eachRias a subring ofR; in particular,Ri

andRhave the same unit element.

The proof ofLemma 2.1is straightforward and is left to the reader.

Lemma2.1. Let{R_i}^k_i=1be a finite family of rings andXa variable. Then (_k

i=1Ri)[X]_k

i=1Ri[X].

Let K and L be two fields. From Lemma 2.1, we know that(K×L)[X] is isomorphic toK[X]×L[X], and we can view a polynomial in(K×L)[X]as a pair of polynomials. Takeg/h∈(K×L)(X). Theng,h∈(K×L)[X]such that σ (h)=K×L. We have g=(g1,g2), whereg1∈K[X] and g2∈L[X];

alsoh=(h1,h2)such thath1∈K[X]\(0)andh2∈L[X]\(0). Henceg/h= (g1/h1,g2/h2)∈K(X)×L(X). Therefore, we haveK(X)×L(X)(K×L)(X), and hence we have the following result.

Proposition 2.2. Let _n

i=1Ki be a finite product of fields Ki. Then (n

i=1Ki)(X)n

i=1Ki(X)is a finite product of fields.

Proposition2.3. If{Rα}α∈Ais an infinite family of rings andXa variable over

α∈AR_α, thenϕ:(

α∈AR_α)(X)→

α∈AR_α(X)is an injective homomor- phism.

Before provingProposition 2.3, we need the following lemma.

Lemma2.4. Iff=(fα)_α∈A∈

α∈ARα[X], thenσ (f )=

α∈Aσ (fα).

Proof. Let f =hnXⁿ+ ··· +h1X+h0, with hi ∈

α∈ARα for each i= 0,1,...,n. Thenσ (f )=(hn,...,h1,h0)

α∈ARα, which implies that σ (fα)= (h_n(α),...,h1(α),h0(α))R_α, whereh_i(α)is theαth component ofh_i. We can conclude thatσ (f )⊆

α∈Aσ (fα). The proof of the converse is similar.

Proof ofProposition2.3. The mapϕ:(

α∈ARα)[X]→

α∈ARα[X]de- fined byϕ(F)=(F(α))_α∈A, whereF(α)is the polynomial projection ofF over Rα[X], is an injective homomorphism. Now, we consider

ψ:s

α∈A

R_α

(X)→

α∈A

R_α(X), (2.1)

defined byψ(F/H)=(F(α)/H(α))_α∈A, whereF,H∈(

α∈ARα)[X]such that σ (H) =

α∈ARα. It is clear that ψ is a homomorphism. Now, let F/H ∈ (

α∈AR_α)(X)such thatψ(F/H)=0. We haveσ (H)=

α∈AR_α, and, byLemma 2.4, σ (H(α)) = Rα. Then ψ(F/H) = (F(α)/H(α))_α∈A = 0 implies that F(α)/H(α)=0 inRα(X)for eachα∈A. For eachα∈A, there existshα∈ R_α[X]such thatσ (h_α)=R_αandF(α)h_α=0. We setG=(h_α)_α∈A. ThenFG=0, G∈(

α∈AR_α)[X], and, byLemma 2.4,σ (G)=

α∈Aσ (h_α)=

α∈AR_α. It fol- lows thatF/H=0.

The converse ofProposition 2.3is not true in general, as shown in the next example.

Example2.5. For eachi∈Z⁺, letR_i=Fpⁱ be the Galois field withpⁱ el- ements, where p is a positive prime integer. From [10, Theorem 5.5, page 247], Fpⁱ =GF(p)(ξ_i), where ξ_i is a pⁱth primitive root of unity, for each i∈Z⁺, and GF(p)is the Galois field withp elements. Let ξ= {ξ_i}^∞_i=1 be an element of _∞

i=1Ri such that [GF(p)(ξi): GF(p)] =i, for each i∈Z⁺, and f_i=Ir r (ξ_i,GF(p)) the minimal polynomial of ξ_i over GF(p). Clearly, f_i is of degreei. It follows that there exists no monic polynomialF ∈_∞

i=1R_i[X]

such thatF= {fi}^∞_i=1, otherwiseξis an integral element over_∞

i=1GF(p)since F(ξ)=0.

Our next result will be useful later.

Lemma2.6. LetR be a zero-dimensional ring with finite spectrum. ThenR can be expressed as a finite product of zero-dimensional quasilocal subrings.

Proof. Let Spec(R)= {M_i}ⁿ_i=1be the spectrum ofR. LetS_Mi(0)=Kerϕ_i, for each i=1,...,n, whereϕi :R →RM_i and ϕi(r )=r /1 is the canonical homomorphism. Since Rad(SM_i(0))=Mi∈Max(R), then SM_i(0)is a primary ideal. Note that ∩ⁿ_i=1S_Mi(0)=(0)and S_Mi(0)+S_Mj(0)=R for eachi≠j in {1,...,n}. Therefore, RR/∩ⁿ_i=1SM_i(0). By the Chinese remainder theorem,

R _n

i=1R/SM_i(0), where R/SM_i(0)is quasilocal and zero-dimensional, for i=1,...,n.

We note that ifRis a von Neumann regular ring (i.e.,Ris reduced and zero dimensional), thenRis an Artinian if and only ifRis a finite product of fields if and only ifRis Noetherian. Indeed, ifRis von Neumann regular and Artinian, then, by [3, Corollary 8.2], Spec(R)is finite, and hence,R=R1⊕···⊕Rn, where eachR_iis a quasilocal and zero-dimensional ring, fori=1,...,n. SinceRis a von Neumann regular ring, eachR_iis a von Neumann regular ring, by [4, Result 3.2]. AsRiis a quasilocal ring, by [4, Theorem 3.1],Riis a field fori=1,...,n, and it follows thatRis a finite product of fields.

Proposition2.7. LetRbe a ring andXa variable overR. Then (1) ifRis a directed union of Artinian subrings, then so isR(X);

(2) ifRis a reduced ring andR(X)is a directed union of Artinian subrings, thenRhas the same property;

(3) Ris a directed union of zero-dimensional subrings with finite spectra if and only ifR(X)has the same property.

Proof. (1) IfR = i∈IRi is a directed union of Artinian subrings, then R(X)= i∈IRi(X). Since eachRi is Noetherian, by [12, (6.17)],Ri(X)is also Noetherian and eachR_i(X)is zero dimensional since eachR_i is Noetherian, (cf. [2, Proposition 1.21]). By [3, Theorem 8.5], Ri(X)is an Artinian ring for eachi∈I. The family{Ri(X)}i∈I is directed because the family{Ri}i∈I is di- rected. ThenR(X)is a directed union of Artinian subrings.

(2) IfR(X)is a directed union of Artinian subrings, then, by [8, Theorem 2.4(a)], eachRj=Sj∩Ris zero dimensional. SinceRj⊆Sjand Spec(Sj)is finite (cf. [3, Theorem 8.3]), this yields that each Spec(R_j)is finite. AsRis reduced, and by [4, Theorem 3.1], eachR_j is a von Neumann regular ring with finite spectrum. It follows thatRj is Artinian, and hence,R= j∈IRjis a directed union of Artinian subrings.

(3) The proof of this result follows from the fact that Spec(R)= {m(X)|m∈ Spec(R)}, and hence,|Spec(R)| = |Spec(R(X))|.

Remark2.8. (1) LetRbe a hereditary zero-dimensional ring, that is, a ring for which all subrings are zero dimensional. Then R is a directed union of Artinian subrings, and therefore,R(X)is a directed union of Artinian subrings that is not hereditarily zero dimensional, sinceR[X]⊂R(X)and dim(R[X])

=1 (cf. [13, Theorem 2]).

(2) LetRbe a von Neumann regular ring andX1,...,Xnvariables overR. We denoteR(X1,...,X_n)=R(n)for eachn∈Z⁺. Then, byProposition 2.7and [9, Lemma 15.3],Ris a directed union of Artinian subrings if and only ifR(n)is a directed union of Artinian subrings, for eachn∈Z⁺.

(3) If, inProposition 2.7(3), we takeX= {Y_i}i∈Ian infinite family of indeter- minates overR, thenRis a directed union of zero-dimensional subrings with finite spectra if and only if so isR(X).

Letη(x)be the index of nilpotency ofx∈R. We define

η(R)=supη(x)|x∈N(R), (2.2) whereN(R)is the set of nilpotent elements ofR. From [7, Theorem 3.4], we know that dim(

α∈ATα)=0 if and only if{α∈A|η(Tα) > k}is finite for somek∈Z⁺, where{T_α}α∈Ais a family of zero-dimensional rings.

LetR be a ring such thatη(R) < kfor somek∈Z⁺ and letXbe a variable overR. Thenη(R[X])need not be bounded. Also, we note that if dim(R)=0, then dim(R(X))=0.

Let {R_α}α∈A be a family of zero-dimensional rings andX a variable, and suppose that dim(

α∈ARα)=0. We added that, if eachRαis a directed union of Artinian subrings, thenRα(X)is also a directed union of Artinian subrings for each α∈A, see Proposition 2.7. Assume that there existsk∈Z⁺ such that{α∈A|η(Rα) > kor there existsM∈Spec(Rα):|Rα/M|> k}is finite.

Then, by [7, Theorem 6.7],

α∈ARαis a directed union of Artinian subrings.

However, for eachk∈Z⁺,{α∈A|η(R_α(X)) > k}is an infinite set. This means that

α∈ARα(X)is not zero dimensional. Now, we suppose that eachRαis a von Neumann regular ring, and we show thatRα(X)is also a von Neumann regular ring, and hence,

α∈AR_α(X_α)is a von Neumann regular ring, where eachX_αis a variable overR_α.

Theorem2.9. Let{R_α}α∈Abe a family of von Neumann regular rings and Xαan indeterminate overRα, for eachα∈A. Then

α∈ARα(Xα)is not a di- rected union of Artinian subrings.

The proof ofTheorem 2.9requires the following two lemmas.

Lemma2.10. LetRbe a ring andU a multiplicatively closed subset ofR. If Ris reduced, thenU⁻¹Ris also reduced.

Proof. Letr /s∈N(U⁻¹R), whereN(U⁻¹R)is the nilradical ofU⁻¹R. Then there existsn0∈N^∗such that(r /s)ⁿ⁰=0; this means that there existsu∈U such that(r u)ⁿ⁰=0. SinceRis reduced, we haver u=0, and hence,r /s=0.

In other words,N(U⁻¹R)=(0). Thus,U⁻¹Ris reduced.

Lemma2.11. LetRbe a ring andXa variable overR. ThenRis reduced if and only ifR(X)is reduced.

Proof. Assume thatR is reduced and takef =anXⁿ+ ··· +a1X+a0∈ N(R[X]). Then there existsl∈Z⁺ such thatf^l=0. Therefore,a^l_n=a^l_n−1=

··· =a^l₀=0. SinceRis reduced, we havea_n=a_n−1= ··· =a0=0, and hence, f=0. It follows thatN(R[X])=(0). ByLemma 2.10,R(X)is a also reduced becauseR(X)=S⁻¹R[X]is a localization ofR[X].

The converse implication follows from the fact that every subring of a re- duced ring is reduced.

Note that the two equivalent conditions ofLemma 2.11are also equivalent to R(n)being reduced sinceR(n)=R(n−1)(X), for eachn∈Z⁺, see [9, Lemma 15.3].

Proof oftheorem2.9. By [7, Theorem 6.7],

α∈ARα(Xα)is a directed union of Artinian subrings if and only if there existsk∈Z⁺such that{α∈A| there existsM∈Spec(R_α(X_α))with|R_α/M|> k}is finite. It was shown in [12, (6.17)], that Spec(Rα(Xα))= {M(Xα)|M∈Spec(Rα)}for eachα∈A. More- over,Rα(Xα)/Mα(Xα)(Rα/Mα)(X) and (Rα/Mα)(Xα)is an at least count- ably denumerable field. Therefore, for eachα∈A, for eachM_α∈Spec(R_α), for each k∈Z⁺ we have |(Rα/Mα)(Xα)|> k. Then, for each k∈Z⁺, the set {α∈A|there existsMα∈Spec(Rα)with|(Rα/Mα)(Xα)|> k}is infinite. Thus,

α∈AR_α(X_α)is not a directed union of Artinian subrings.

Corollary2.12. Let{R_α}α∈A be a family of von Neumann regular rings andXa variable. Then

α∈AR_α(X)is not a directed union of Artinian subrings.

LetRbe a ring and{R_α}α∈Aan infinite family of nonzero rings such thatR is, up to isomorphism, a subring of eachRα. We useR^∗to denote the image of Runder the diagonal imbedding, that is,R^∗=ϕ(R), whereϕ:R

α∈AR_α is the monomorphism defined byϕ(x)= {x_α}α∈Asuch thatx_α=xfor each α∈A.

Proposition2.13. LetF be an absolutely algebraic field andR=_ω0 F a countable direct product of copies ofF. Define

᏿= x_i∞

i=1∈R|x_i∞

i=1has only finitely many distinct coordinates .

(2.3) Then᏿is the maximal subring ofRwhich can be expressed as a directed union of Artinian subrings.

Proof. First, we claim that᏿is a directed union of Artinian subrings. For each j ∈ Z⁺, we define Sj as the subring of ᏿ consisting of all sequences {x_i}^∞_i=0∈᏿such thatx_j=x_j+1= ···. If we denote byπ the prime subring ofR, then each S_j contains π, S0is the diagonal imbedding ofF in R, and SjF^j+1is an Artinian von Neumann regular ring. Clearly,Sj⊆S_j+1for each j∈Z⁺. Therefore,᏿= ^∞_j=1S_jand᏿is a directed union of Artinian subrings.

Now, letT be a subring ofω0F withT= j∈JT_ja directed union of Artinian subrings andt= {ti}^∞_i=1∈T. There existsj0∈Jsuch thatt∈Tj₀andTj₀is a finite product of fields; hencet∈᏿.

Example2.14. Letpbe a prime integer andXa variable over GF(p), where GF(p)is the Galois field withpelements. LetR=(GF(p)(X))^ω0be a countable direct product of copies of GF(p)(X). We note thatRis a von Neumann regular ring as a direct product of fields. ByTheorem 2.9, the ringRis not a directed union of Artinian subrings. Let ᏿= {{x_i}^∞_i=1∈R :{x_i}^∞_i=1 has only finitely many distinct components}andV =(GF(p)(X))^∗+I, whereI=_∞

i=1GF(p)

is the direct sum ideal of R and (GF(p)(X))^∗ is the diagonal imbedding of GF(p)(X)inR. The ring᏿is the biggest subring ofRwhich is a directed union of Artinian subrings. We remark that, ifV⊂᏿, then, by [11, Corollary 4],V is also a directed union of Artinian subrings. Since GF(p)is a finite field, we haveω₀GF(p)⊂᏿and, by [11, Corollary 4], is a directed union of Artinian subrings.

Acknowledgment. The author would like to thank the referee for a num- ber of helpful comments and suggestions.

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D. Karim: Department of Mathematics, Faculty of Sciences Semlalia, University of Cadi Ayyad, P.O. Box 2390 Marrakech, Morocco

E-mail address:[email protected]