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Chain conditions on composite Hurwitz series rings

  • Jung Wook Lim and Dong Yeol Oh EMAIL logo
Published/Copyright: September 22, 2017

Abstract

In this paper, we study chain conditions on composite Hurwitz series rings and composite Hurwitz polynomial rings. More precisely, we characterize when composite Hurwitz series rings and composite Hurwitz polynomial rings are Noetherian, S-Noetherian or satisfy the ascending chain condition on principal ideals.

MSC 2010: 13A15; 13E05; 13E99; 13F15

1 Introduction

1.1 Hurwitz series rings

The formal power series rings and polynomial rings have been of interest and have had important applications in many areas, one of which has been differential algebra. In [1], Keigher introduced a variant of the ring of formal power series and studied some of its properties. In [2], Keigher called such a ring the ring of Hurwitz series and examined its ring theoretic properties. Since then, many works on the ring of Hurwitz series have been done ([35]).

Let R be a commutative ring with identity, RX〛 (resp., R[X]) the formal power series ring (resp., polynomial ring) over R, and H(R) the set of formal expressions of the form n=0anXn, where anR. Define addition and *-product on H(R) as follows: for f=n=0anXn,g=n=0bnXnH(R),

f+g=n=0(an+bn)Xn and fg=n=0cnXn,

where cn=k=0n(nk)akbnk and (nk)=n!(nk)!k! for nonnegative integers nk. Then H(R) becomes a commutative ring with identity containing R under these two operations, i.e., H(R) = (RX〛, +, *). The ring H(R) is called the Hurwitz series ring over R. The Hurwitz polynomial ring h(R) over R is the subring of H(R) consisting of formal expressions of the form k=0nakXk, i. e., h(R) = (R[X], +, *).

Let RD be an extension of commutative rings with identity, and let H(R, D) = {fH(D)| the constant term of f belongs to R} (resp., h(R, D) = {fh(D)| the constant term of f belongs to R}). Then H(R, D) (resp., h(R, D)) is a commutative ring with identity. We call H(R, D) (resp., h(R, D)) a composite Hurwitz series ring (resp., composite Hurwitz polynomial ring). More precisely, H(R, D) (resp., h(R, D)) is a subring of H(D) (resp., h(D)) containing H(R) (resp., h(R)), i. e., H(R, D) = (R + XDX〛, +, *) (resp., h(R, D) = (R + XD[X], +, *)) where R + XDX〛 = {fDX〛| the constant term of f belongs to R} (resp., R + XD[X] = {fD[X]| the constant term of f belongs to R}). Hence if RD, then H(R, D) (resp., h(R, D)) gives algebraic properties of Hurwitz series (resp., Hurwitz polynomial) type rings strictly between two Hurwitz series rings (resp., Hurwitz polynomial rings). Also, it is easy to see that H(R, D) (resp., h(R, D)) is a pullback of R and H(D) (resp., h(D)).

1.2 Noetherian rings and related rings

Chain conditions have for many years been important tools in commutative algebra and algebraic geometry because of their use in producing many theorems and applications. For example, a relation between the ascending chain conditions on ideals and finitely generatedness of ideals in rings permits an interesting measure of the size and behavior of such rings, and the Noetherian condition plays a significant role to prove many results on varieties, homology and cohomology. Recently, Anderson and Dumitrescu [6] introduced the notion of S-Noetherian rings and gave a number of S-variants of well-known results for Noetherian rings. After them, S-Noetherian rings have been studied by some mathematicians (see [711]).

In [10, 1214], the authors characterized when composite rings R + XDX〛 and R + XD[X] are Noetherian rings, S-Noetherian rings, or satisfy the ascending chain condition on principal ideals. It was shown that R + XDX〛 (resp., R + XD[X]) is a Noetherian ring if and only if R is a Noetherian ring and D is a finitely generated R-module [13, Theorem 4] (or [12, Proposition 2.1]) (resp., [12, Proposition 2.1] (or [14, Corollary 2.2])); and that R + XD[X] (resp., R + XDX〛) is an S-Noetherian ring if and only if R is an S-Noetherian ring and D is an S-finite R-module [10, Theorems 3.6 and 4.4]. Also, it was shown that if D is a présimplifiable ring, then R + XDX〛 satisfies the ascending chain condition on principal ideals if and only if U(D) ∩ R = U(R) and for each sequence (dn)n≥ 1 of D with the property that for each n ≥ 1, dn = dn+1rn for some rnR, d1Dd2 D ⊆ ⋯ is stationary [12, Proposition 4.21].

In [4], Benhissi and Koja studied when the Hurwitz rings H(R) and h(R) are Noetherian rings, S-Noetherian rings or satisfy the ascending chain condition on principal ideals. They showed that if char (R) = 0, then H(R) is a Noetherian ring if and only if h(R) is a Noetherian ring, if and only if R a Noetherian ring containing ℚ [4, Corollary 7.7]. Also, they proved that for an anti-Archimedean subset S of R with zero characteristic containing an element s0S divisible in R by all the nonzero positive integers, if R is an S-Noetherian ring, then h(R) is an S-Noetherian ring [4, Theorem 9.4]; and if R is an S-Noetherian ring and S consists of nonzerodivisors, then H(R) is an S-Noetherian ring [4, Theorem 9.6].

In this paper, we study chain conditions on composite Hurwitz series rings H(R, D) and composite Hurwitz polynomial rings h(R, D), where RD is an extension of commutative rings with identity. In Section 2, we give necessary and sufficient conditions for the rings H(R, D) and h(R, D) to be Noetherian rings. We show that if char (R) = 0, then H(R, D) is a Noetherian ring if and only if h(R, D) is a Noetherian ring, if and only R is a Noetherian ring and D is a finitely generated R-module containing ℚ. In Section 3, we give equivalent conditions for the rings H(R, D) and h(R, D) to be S-Noetherian rings, where S is an anti-Archimedean subset of R. We show that if char (R) = 0 and S is an anti-Archimedean subset of R consisting of nonzerodivisors of D which contains an element divisible in D by all the positive integers, then H(R, D) is an S-Noetherian ring if and only if R is an S-Noetherian ring and D is an S-finite R-module; and if char (R) = 0 and S is an anti-Archimedean subset of R which contains an element divisible in D by all the positive integers, then h(R, D) is an S-Noetherian ring if and only if R is an S-Noetherian ring and D is an S-finite R-module. In Section 4, we study when the rings H(R, D) and h(R, D) are présimplifiable. We prove that H(R, D) is présimplifiable if and only if Z(D) ∩ R ⊆ 1 + U(R), where Z(D) is the set of zero-divisors of D and U(R) is the set of units in R. We also prove that if D is a torsion-free ℤ-module, then h(R, D) is présimplifiable if and only if D is a domainlike ring. Finally, in Section 5, we characterize when the rings H(R, D) and h(R, D) satisfy the ascending chain condition on principal ideals. We show that if D is a présimplifiable ring, then H(R, D) satisfies the ascending chain condition on principal ideals if and only if U(D) ∩ R = U(R) and for each sequence (dn)n≥1 of D with the property that for each n ≥ 1, there exists an element rnR such that dn = dn+1rn, d1 Dd2 D ⊆ ⋯ is stationary; and if D is a présimplifiable ring with char (D) > 0, then h(R, D) satisfies the ascending chain condition on principal ideals if and only if U(D) ∩ R = U(R) and for each sequence (dn)n≥1 of D with the property that for each n ≥ 1, there exists an element rnR such that dn = dn+1rn, d1 Dd2 D ⊆ ⋯ is stationary.

2 Noetherian rings

Let R be a commutative ring with identity. Then the mapping ψ : RX〛 → H(R) (resp., ϕ : R[X] → h(R)) defined by

ψ(n=0anXn)=n=0n!anXn(resp.,ϕ(k=0nakXk)=k=0nk!akXk)

is a ring homomorphism [2, Proposition 2.3]; and ψ is an isomorphism if and only if ϕ is an isomorphism, if and only if the ℤ-module R is divisible and torsion-free, if and only if R contains ℚ, where ℚ is the field of rational numbers ([2, Proposition 2.4] and [4, Theorem 1.4 and Corollary 1.5]).

We start this section with the following simple observation without proof.

Lemma 2.1

Let RD be an extension of commutative rings with identity. Then the following conditions are equivalent.

  1. D contains ℚ.

  2. The ℤ-module D is divisible and torsion-free.

  3. The mapping ψ : R + XDX〛 → H(R, D) defined by ψ (n=0anXn)=n=0n!anXn is a ring isomorphism.

  4. The mapping ϕ : R + XD[X] → h(R, D) defined by (k=0nakXk)=k=0nk!akXk is a ring isomorphism.

Let RD be an extension of commutative rings with identity, and set XDX〛 = {fH(R, D)| the constant term of f is zero} (resp., XD[X] = {fh(R, D)| the constant term of f is zero Then it is easy to see that XDX〛 (resp., XD[X]) is an H(R, D)-module (resp., h(R, D)-module).

We are now ready to study when composite Hurwitz rings H(R, D) and h(R, D) are Noetherian rings.

Theorem 2.2

Let RD be an extension of commutative rings with identity. If char (R) = 0, then the following statements are equivalent.

  1. H(R, D) (resp., h(R, D)) is a Noetherian ring.

  2. R is a Noetherian ring and D is a finitely generated R-module containing ℚ.

  3. R is a Noetherian ring and XDX〛 (resp., XD[X]) is a Noetherian H(R, D)-module (resp., h(R, D)-module).

Proof

(1) ⇒(2) Suppose that H(R, D) (resp., h(R, D)) is a Noetherian ring, and let p be any prime number. Since (X, X2, …) is finitely generated, there exists a positive integer n such that Xpn ∈ (X, X2, …, Xpn−1); so we can find suitable elements g1, …, gpn−1H(R, D) (resp., g1, …, gpn−1h(R, D)) such that Xpn = X * g1 + ⋯ + Xpn−1 * gpn−1. Comparing the coefficients of Xpn in both sides, we get

1=pn1b1++pnpn1bpn1

for some b1, …, bpn−1D. Note that p divides (pnk) for all k = 1, …, pn − 1[4, Lemma 7.3]; so p is a unit in D. Since all the prime numbers are units in D, all the nonzero integers are also units in D. Therefore D contains ℚ, and hence by Lemma 2.1, R + XDX〛 (resp., R + XD[X]) is a Noetherian ring. Thus R is a Noetherian ring and D is a finitely generated R-module [13, Theorem 4] (or [12, Proposition 2.1]) (resp., [12, Proposition 2.1] (or [14, Corollary 2.2])).

(2) ⇒ (1) Assume that R is a Noetherian ring and D is a finitely generated R-module. Then R + XDX〛 (resp., R + XD[X]) is Noetherian [13, Theorem 4] (or [12, Proposition 2.1]) (resp., [12, Proposition 2.1] (or [14, Corollary 2.2])). Since D contains ℚ, Lemma 2.1 forces H(R, D) (resp., h(R, D)) to be a Noetherian ring.

(1) ⇔ (3) We first show the composite Hurwitz series ring case. Let u : RD be the natural injection and v : H(D) → D the canonical projection. Consider the following commutative diagram

H(R,D)=R×DH(D)RuH(D)vDH(D)/XD[[X]].

Then H(R, D) is the pullback of u and v. Thus the equivalence follows from [15, Proposition 4.10].

The proof for the composite Hurwitz polynomial ring case is the same as that for the composite Hurwitz series ring case. □ 

When R = D in Theorem 2.2, we obtain

Corollary 2.3

([4, Corollary 7.7]). Let R be a commutative ring with identity. If char (R) = 0, then the following assertions are equivalent.

  1. R is a Noetherian ring containing ℚ.

  2. H(R) is a Noetherian ring.

  3. h(R) is a Noetherian ring.

We next show that in Theorem 2.2, the condition that char (R) = 0 is essential.

Theorem 2.4

Let RD be an extension of commutative rings with identity and let E be either H(R, D) or h(R, D). If char (R) > 0, then E is never a Noetherian ring.

Proof

Suppose on the contrary that E is a Noetherian ring. Then (X, X2, …) is a finitely generated ideal of E; so there exists a positive integer q such that (X, X2, …) = (X, X2, …, Xq). Let char (R)=p1k1pmkm, where p1, …, pm are distinct prime numbers. Then we can take a positive integer n such that pin > q for all i = 1, …, m; so Xpin(X,X2,,Xpin1) for all i = 1, …, m. Therefore for each i = 1, …, m, there exist suitable elements gi1, …, gi(pin1) E such that Xpin=Xgi1++Xpin1gi(pin1). By comparing the coefficients of Xpin in both sides, we get

1=pin1bi1++pinpin1bi(pin1)

for some bi1, …, bi(pin1) D. Hence we obtain

1=i=1m((pin1)bi1++(pinpin1)bi(pin1))ki.

Note that pi divides (pinj) for all i = 1, …, m and all j = 1, …, pin − 1[4, Lemma 7.3]; so char (R) divides 1 in D. Hence 1 = 0 in D, which is a contradiction. Thus E is not a Noetherian ring.□ 

3 S-Noetherian rings

Let R be a commutative ring with identity, S a (not necessarily saturated) multiplicative subset of R, and M a unitary R-module. Recall from [6, Definition 1] that an ideal I of R is S-finite if there exist an element sS and a finitely generated ideal J of R such that sIJI; and R is an S-Noetherian ring if each ideal of R is S-finite. Also, we say that the R-module M is S-finite if sMFM for some sS and some finitely generated R-module F; and M is S-Noetherian if each R-submodule of M is S-finite.

Our first result in this section gives a necessary condition for composite Hurwitz rings H(R, D) and h(R, D) to be S-Noetherian rings, where RD is an extension of commutative rings with identity and S is a multiplicative subset of R.

Proposition 3.1

Let RD be an extension of commutative rings with identity, S be a (not necessarily saturated) multiplicative subset of R, and E be either H(R, D) or h(R, D) . If E is an S-Noetherian ring, then the following assertions hold.

  1. S contains an element s divisible in D by all the prime numbers.

  2. char (R) = 0.

Proof

(1) Suppose that E is an S-Noetherian ring. Then (X, X2, …) is S-finite; so there exist sS and f1, …, fm ∈ (X, X2, …) such that s * (X, X2, …) ⊆ (f1, …, fm). Let p be any prime number. Since f1, …, fm ∈ (X, X2, …), we can find a positive integer n such that s * (X, X2, …) ⊆ (X, X2, …, Xpn−1). Therefore sXpn ∈ (X, X2, …, Xpn−1), and hence we can write sXpn = X * g1 + ⋯ + Xpn−1 * gpn−1 for some g1, …, gpn−1E. Comparing the coefficients of Xpn in both sides, we obtain

s=(pn1)d1++(pnpn1)dpn1

for some d1, …, dpn−1D. Note that p divides (pnk) for all k = 1, …, pn − 1[4, Lemma 7.3]; so p divides s in D. Thus s is divisible in D by all the prime numbers.

(2) Suppose on the contrary that char (R) ≠ 0, and let char

(R)=p1α1pmαm,

where p1, …, pm are distinct prime numbers. Fix an i ∈ {1, …, m}. Since E is an S-Noetherian ring, by (1), there exist elements SjS and djD such that Sj = pidj. Since char (R) = char(D), we get

s1α1smαm= char(R)d1α1dmαm=0,

which indicates that 0 ∈ S. However this is absurd. Thus char (R) = 0.□ 

Let R be a commutative ring with identity and S a (not necessarily saturated) multiplicative subset of R. We say that S is anti-Archimedean if (⋂n≥1 snR) ∩ S ≠ ∅ for every sS. We also say that an integral domain R is an anti-Archimedean domain if ⋂n≥1 anR ≠ 0 for each 0 ≠ aR (see [16]). Thus R is an anti-Archimedean domain if and only if R\{0} is an anti-Archimedean subset of R. Clearly, every multiplicative subset consisting of units is anti-Archimedean. Also, if V is a valuation domain with no height-one prime ideal (or equivalently, every nonzero prime ideal of V has infinite height), then V\{0} is an anti-Archimedean subset of V [16, Proposition 2.1]. We next characterize when a composite Hurwitz series ring H(R, D) is an S-Noetherian ring under the assumption that S is an anti-Archimedean subset of R.

Theorem 3.2

Let RD be an extension of commutative rings with identity and char (R) = 0, and let S be an anti-Archimedean subset of R consisting of nonzerodivisors of D. If S contains an element divisible in D by all the positive integers, then the following statements are equivalent.

  1. H(R, D) is an S-Noetherian ring.

  2. R is an S-Noetherian ring and XDXis an S-Noetherian H(R, D)-module.

  3. R is an S-Noetherian ring and D is an S-finite R-module.

Proof

(1) ⇔ (2) Consider the following commutative diagram

H(R,D)=R×DH(D)RuH(D)vDH(D)/XD[[X]],

where u is the natural injection and v is the canonical projection. Then H(R, D) is the pullback of u and v. Thus the equivalence follows from [9, Proposition 2.3].

(1) ⇒ (3) Suppose that H(R, D) is an S-Noetherian ring. Since R is a homomorphic image of H(R, D), R is an S-Noetherian ring [11, Lemma 2.2]. Note that XDX〛 is an ideal of H(R, D); so there exist sS and f1, …, fnDX〛 such that s * XDX〛 ⊆ (Xf1, …, Xfn). Therefore for any dD, we have

sdX=Xf1g1++Xfngn

for some g1, …, gnH(R, D). Comparing the coefficients of X in both sides, we get

sd=f1(0)g1(0)++fn(0)gn(0),

where for each i = 1, …, n, fi(0) and gi(0) denote the constant terms of fi and gi, respectively. Note that fi(0) ∈ D and gi(0) ∈ R for all i = 1, …, n. Hence sDf1(0)R + ⋯ + fn(0)R. Thus D is an S-finite R-module.

(3) ⇒ (1) Suppose that R is an S-Noetherian ring and D is an S-finite R-module. Then D is an S-Noetherian R-module [10, Lemma 3.5(2)]; so D is an S-Noetherian ring. Since D is an S-finite R-module, there exist sS and d1, …, dmD such that sDd1R + ⋯ + dmR; so we have

sH(D)d1H(R)++dmH(R)d1H(R,D)++dmH(R,D).

Hence H(D) is an S-finite H(R, D)-module. Clearly, S is an anti-Archimedean subset of D. Note that char (D)= char(R) = 0; so by our assumption, H(D) is an S-Noetherian ring [4, Theorem 9.6]. Since H(D) is an S-finite H(R, D)-module, H(R, D) is an S-Noetherian ring [6, Corollary 7] (or [10, Lemma 3.5(3)]).□ 

Recall that if R is an S-Noetherian ring with char (R) = 0 and S is an anti-Archimedean subset of R which contains an element divisible in R by all the positive integers, then h(R) is also an S-Noetherian ring [4, Theorem 9.4]. By combining this result with a similar argument as in the proof of Theorem 3.2, we obtain equivalent conditions for a composite Hurwitz polynomial ring h(R, D) to be an S-Noetherian ring.

Theorem 3.3

Let RD be an extension of commutative rings with identity and char (R) = 0, and let S be an anti-Archimedean subset of R. If S contains an element divisible in D by all the positive integers, then the following statements are equivalent.

  1. h(R, D) is an S-Noetherian ring.

  2. R is an S-Noetherian ring and XD[X] is an S-Noetherian h(R, D)-module.

  3. R is an S-Noetherian ring and D is an S-finite R-module.

Remark 3.4

Let RD be an extension of commutative rings with identity and S a multiplicative subset of R. If R contains ℚ, then all the integers are units in D; so every element in S is divisible in D by all the positive integers. Hence if R contains ℚ, then it follows from Lemma 2.1 that Theorems 3.2 and 3.3 are nothing but parts of [10, Theorems 4.4 and 3,6].

By Proposition 3.1 and Theorems 3.2 and 3.3, we may ask the following question.

Question 3.5

Do Theorems 3.2 and 3.3 still hold if S contains an element divisible in D by all the prime numbers?

We end this section with an example satisfying the conditions in Theorems 3.2 and 3.3. More precisely, we construct an integral domain R, not containing ℚ, with char (R) = 0 such that there exists an anti-Archimedean subset S of R containing an element divisible in R by all the positive integers.

Example 3.6

Letbe the ring of integers and G the weak direct sum of {Zi}i=1 which has the reverse lexicographic order, where Zi = ℤ for all positive integers i. Let {Xi}i=1{Yi}i=1 be a set of indeterminates overand v be the valuation on Q({Xi}i=1,{Yi}i=1) induced by the mapping Xi ↦ 0 and Yiei, of {Xi}i=1{Yi}i=1 into G, where ei is an element of G whose i-th component is 1 and j-th component is 0 for ji.

  1. Let V be the valuation ring of v and set V* = V\{0}. Then V is a valuation domain containingwith no height-one prime ideals [17, page 254, Exercise 20] and V* is an anti-Archimedean subset of V[16, Proposition 2.1]. Clearly, V is a V*-Noetherian ring; so VX(resp., V[X]) is a V*-Noetherian ring [6, Proposition 10] (resp., [6, Proposition 9]). Since V contains ℚ, H(V) (resp., h(V)) is isomorphic to VX(resp., V[X]) [4, Theorem 1.4] (resp., [4, Corollary 1.5]). Hence H(V) (resp., h(V)) is a V*-Noetherian ring.

  2. Let R = ℤ + M, where M is the maximal ideal of V. Then R is an anti-Archimedean ring [16, Proposition 2.6] and each ideal of R is comparable with M under the set theoretic inclusion [17, page 202, Exercise 12]. For any integer n, nR = n ℤ + nM = n ℤ + M contains M (sinceM = M, nM = M). Hence every element in M is divisible in R by all the positive integers. Note that H(R) (resp., h(R)) is not isomorphic to RX(resp., R[X]) because R does not contain ℚ. Since R* = R\{0} is anti-Archimedean subset of R and R is R*-Noetherian, it follows from [4, Theorem 9.6] (resp., [4, Theorem 9.4]) that H(R) (resp., h(R)) is an R*-Noetherian ring.

  3. Let RD be an extension of integral domains, where R is defined in (2). If D is a finitely generated R-module, then D is an R*-finite R-module. Hence H(R, D) (resp., h(R, D)) is an R*-Noetherian ring.

4 Présimplifiable rings

Let R be a commutative ring with identity, U(R) the set of units of R, and Z(R) the set of zero-divisors of R. Recall that R is preésimplifiable if whenever a, bR satisfy ab = a, either a = 0 or bU(R). It was shown in [18] that R is présimplifiable if and only if Z(R) ⊆ 1 + U(R). In [4], the authors studied when Hurwitz rings H(R) and h(R) are présimplifiable. In this section, we modify some properties of elements (units and nilpotent) of H(R) and h(R) in [4] to give equivalent conditions for composite Hurwitz series rings and composite Hurwitz polynomial rings to be présimplifiable.

Our first result in this section is a necessary and sufficient condition for a composite Hurwitz series ring H(R, D) to be présimplifiable, where RD is an extension of commutative rings with identity.

Proposition 4.1

Let RD be an extension of commutative rings with identity. Then H(R, D) is présimplifiable if and only if Z(D) ∩ R ⊆ 1 + U(R).

Proof

(⇒) Let rZ(D) ∩ R. Since H(R, D) is présimplifiable, we obtain

rZ(H(R,D))1+U(H(R,D)).

Thus r ∈ 1 + U(R) [19, Lemma 2.2(1)].

(⇐) Let f = i=0aiXi Z(H(R, D Then by the assumption, we obtain

a0Z(R)Z(D)R1+U(R).

Therefore f − 1 ∈ U(H(R, D)) [19, Lemma 2.2(1)], and hence f ∈ 1 + U(H(R, D)). Thus H(R, D) is présimplifiable.□ 

Remark 4.2

For an extension RD of commutative rings with identity, Z(R) ⊆ Z(D) ∩ R. Hence if H(R, D) is présimplifiable, then so is H(R). However, the converse does not hold in general. Consider R = ℤ ⊆ D = ℤ[Y]/(3Y). Then clearly H(R) is présimplifiable but H(R, D) is notprésimplifiable because 3 ∈ Z(H(R, D)) and 3 ∉ 1 + U(H(R, D)).

We next study when a composite Hurwitz polynomial ring h(R, D) is présimplifiable, where RD is an extension of commutative rings with identity. To do this, we need two lemmas.

Lemma 4.3

Let RD be an extension of commutative rings with identity and f = i=0naiXi h(R, D). Then the following assertions hold.

  1. f is nilpotent if and only if a0 is nilpotent and for each i = 1, …, n, ai is nilpotent or some power of ai is with torsion.

  2. f is a unit if and only if a0 is a unit in R and for each i = 1, …, n, ai is nilpotent or some power of ai is with torsion.

Proof

  1. The proof is identical to that of [4, Theorem 2.4].

  2. (⇒) Assume that f is a unit in h(R, D). Then we can find an element g = i=0maiXi h(R, D) such that f * g = 1; so a0b0 = 1. Hence a0 is a unit in R. Since h(R, D) ⊆ h(D), f is a unit in h(D); so for each i = 1, …, n, ai is nilpotent or some power of ai is with torsion [4, Theorem 3.1].

(⇐) Assume that for each i = 1, …, n, ai is nilpotent or some power of ai is with torsion. Then by (1), i=0naiXi is nilpotent in h(R, D). Since a0 is a unit in h(R, D), f is a unit in h(R, D).□ 

Lemma 4.4

Let RD be an extension of commutative rings with identity. Then the following assertions are equivalent.

  1. For each fZ(h(R, D)), there exists an element dD\{0} such that d * f = 0.

  2. D is a torsion-free ℤ-module.

Proof

(1) ⇒ (2) Let a be any nonzero element in D. If there exists a positive integer n such that na = 0, then 0 = naXn = aX * Xn−1; so Xn−1 is a zero-divisor of h(R, D). By the assumption, we can find an element dD\{0} such that dXn−1 = 0, which is absurd. Thus D is a torsion-free ℤ-module.

(2) ⇒ (1) Since h(R, D) ⊆ h(D), the implication follows directly from [4, Theorem 4.1].□ 

Let R be a commutative ring with identity. Recall that R is a domainlike ring if every zero-divisor of R is nilpotent. It is easy to see that R is domainlike if and only if (0) is primary.

Proposition 4.5

Let RD be an extension of commutative rings with identity. If D is a torsion-free ℤ-module, then h(R, D) is présimplifiable if and only if D is a domainlike ring.

Proof

(⇒) Let dZ(D). Since h(R, D) is présimplifiable, we have

dXZ(h(R,D))1+U(h(R,D));

so −1 + dXU(h(R, D)). Since D is a torsion-free ℤ-module, d is nilpotent by Lemma 4.3(2). Thus D is a domainlike ring.

(⇐) Let f = i=0naiXi Z(h(R, D)). Since D is a torsion-free ℤ-module, by Lemma 4.4, there exists an element dD\{0} such that d * f = 0. Hence for all i = 0, …, n, ai is a zero-divisor of D. By the assumption, ai is nilpotent for all i = 0, …, n; so by Lemma 4.3(1), f is nilpotent. Therefore f − 1 is a unit in h(R, D), and hence f ∈ 1 + U(h(R, D)). Thus h(R, D) is présimplifiable.□ 

We next show that in Proposition 4.5, the condition that D is a torsion-free ℤ-module is essential.

Proposition 4.6

Let RD be an extension of commutative rings with identity and char (D) = 0. If D is not a torsion-free ℤ-module, then h(R, D) is never présimplifiable.

Proof

Assume that D is not a torsion-free ℤ-module. Then we can find an element dD\{0} and a positive integer n such that nd = 0; so X * dXn−1 = ndXn = 0. Hence XZ(h(R, D)). If h(R, D) is présimplifiable, then Z(h(R, D)) ⊆ 1 + U(h(R, D)); so − 1 + X is a unit in h(R, D). Hence by Lemma 4.3(2), 1 is with torsion. However, this is impossible because char (D) = 0. Thus h(R, D) is not présimplifiable.□ 

Let RD be an extension of commutative rings with identity. Then it follows directly from Lemma 4.3(2) that if char (R) > 0, then i=0naiXi h(R, D) is a unit if and only if a0 is a unit in R. Hence a similar argument as in the proof of Proposition 4.1 shows the following result.

Proposition 4.7

Let RD be an extension of commutative rings with identity. If char (D) > 0, then h(R, D) is présimplifiable if and only if Z(D) ∩ R ⊆ 1 + U(R).

5 Rings satisfying ascending chain condition on principal ideals

Let R be a commutative ring with identity. We say that R satisfies the ascending chain condition on principal ideals (ACCP) if there does not exist a strict ascending chain of principal ideals of R. It was shown in [19, Theorem 2.4] that if RD is an extension of integral domains with char (D) = 0, then H(R, D) satisfies ACCP if and only if h(R, D) satisfies ACCP, if and only if ⋂n≥1r1rnD = (0) for each infinite sequence (rn)n≥1 consisting of nonzero nonunits of R. In this section, we study an equivalent condition for H(R, D) and h(R, D) to satisfy ACCP, where RD is an extension of présimplifiable rings with identity.

Theorem 5.1

Let RD be an extension of commutative rings with identity. If D is a présimplifiable ring, then the following statements are equivalent.

  1. H(R, D) satisfies ACCP.

  2. U(D) ∩ R = U(R) and for each sequence (dn)n≥1 of D with the property thatfor each n ≥ 1, there exists an element rnR such that dn = dn+1rn, d1Dd2D ⊆ ⋯ is stationary.

Proof

(1) ⇒ (2) Let uU(D) ∩ R. Then 1uXH(R,D)1u2XH(R,D) ⊆ ⋯ is an ascending chain of principal ideals of H(R, D). Since H(R, D) satisfies ACCP, there exists a positive integer m such that 1um+1 X* H(R, D) = 1um X * H(R, D). Hence uU(R). Clearly, U(R) ⊆ U(D) ∩ R, which shows that U(D) ∩ R = U(R).

Let (dn)n≥1 be a sequence of D with the property that for each n ≥ 1, there exists an element rnR such that dn = dn+1rn. Then d1X * H(R, D) ⊆ d2X * H(R, D) ⊆ ⋯ is an ascending chain of principal ideals of H(R, D). Since H(R, D) satisfies ACCP, the chain d1X * H(R, D) ⊆ d2 X * H(R, D) ⊆ ⋯ is stationary; so we can find a positive integer m such that dnX * H(R, D) = dmX * H(R, D) for all nm. Hence dnD = dmD for all nm. Thus the chain d1Dd2D ⊆ ⋯ stops.

(2) ⇒ (1) Let f1 * H(R, D) ⊆ f2 * H(R, D) ⊆ ⋯ be an ascending chain of nonzero principal ideals of H(R, D). Then for each n ≥ 1, fn = fn+1 * gn for some gnH(R, D). If fn is a unit for some n ≥ 1, then there is nothing to prove; so we assume that fn is a nonunit for all n ≥ 1. For each n ≥ 1, write fn = m=knanmXm and gn = m=0bnmXm , where ankn ≠ 0. Since fn is a multiple of fn+1, k1k2 ≥ ⋯ ≥ 0; so there exists a positive integer q such that kn = kq for all nq. Hence ankn = an+1kn+1bn0 for all nq. By the assumption, the chain aqkqDaq+1kq+1 D ⊆ ⋯ is stationary; so we can find an integer pq such that amkmD = apkpD for all mp. Therefore for each np, there exists an element dnD such that an+1kn+1 = ankndn. Hence ankn = ankndnbn0 for all np. Since D is présimplifiable and ankn ≠ 0, dnbn0 is a unit in D, which indicates that bn0U(D) ∩ R = U(R). Hence gn is a unit in H(R, D) [19, Lemma 2.2(1)], which shows that fn * H(R, D) = fp * H(R, D) for all np. Thus H(R, D) satisfies ACCP.□ 

Let RD be an extension of commutative rings with identity. Note that by Lemma 4.3(2), if char (R) > 0, then i=0naiXi h(R, D) is a unit if and only if a0 is a unit in R. Hence a similar argument as in the proof of Theorem 5.1 shows the following result.

Theorem 5.2

Let RD be an extension of commutative rings with identity. If D is a présimplifiable ring with char (D) > 0, then the following statements are equivalent.

  1. h(R, D) satisfies ACCP.

  2. U(D) ∩ R = U(R) and for each sequence (dn)n≥1 of D with the property thatfor each n ≥ 1, there exists an element rnR such that dn = dn+1rn, d1Dd2 D ⊆ ⋯ is stationary.

When R = D in Theorems 5.1 and 5.2, we obtain

Corollary 5.3

Let R be a présimplifiable ring with identity. Then the following assertions hold.

  1. R satisfies ACCP if and only if H(R) satisfies ACCP.

  2. If char (R) > 0, then R satisfies ACCP if and only if h(R) satisfies ACCP.

We are closing this paper with an example which shows that if a ring has characteristic zero, then ACCP property does not ascend into the Hurwitz polynomial ring extension.

Example 5.4

Let K be a field with char (K) = 0, {An}n=1 a set of indeterminates over K, and set S = K[{An}n=1]/({An+1(AnAn+1)}n=1). Let an be the image of An in S and R be the localization of S at the ideal (a1, a2,…)S.

  1. R is a présimplifiable ring which satisfies ACCP ([12, Remark 4.17] and [20, Example]).

  2. Note that for all n ≥ 1, anX + 1 = (an+1X + 1) * ((anan+1)X + 1); so (a1X + 1) * h(R) ⊆ (a2X + 1) * h(R) ⊆ ⋯ is an ascending chain of principal ideals of h(R). Suppose on the contrary that the chain stops. Then there exists a positive integer m such that am+1X + 1 = (amX + 1) * f for some fh(R). Now, an easy calculation shows that f = n=0bnXn , where b0 = 1 and bn = (−1)n+1 n!amn1(am+1am) for all n ≥ 1. Since char (K) = 0, bn ≠ 0 for all nonnegative integers n [20, Example]. Hence f2h(R), which is absurd. Thus h(R) does not satisfy ACCP.

Acknowledgement

The authors would like to thank referees for several valuable comments and suggestions. The corresponding author (D.Y.Oh) was supported by Research Fund from Chosun University, 2014.

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Received: 2017-4-14
Accepted: 2017-7-25
Published Online: 2017-9-22

© 2017 Lim and Oh

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

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