When is (D, K) an S-accr pair? Open Access

The rings considered in this article are commutative with identity. Modules considered are unitary modules over commutative rings. We use the abbreviation m.c. subset for multiplicatively closed subset. The m.c. subsets considered in this article are assumed that they do not contain the zero element of the ring. This article is motivated by the research work presented in Refs. [1–4]. Let R be a ring and let M be a module over R. Recall from [3, Definition 1] that M is said to satisfy (accr) (respectively, (accr*)) if for every submodule N of M and every finitely generated (respectively, principal) ideal B of R, the increasing sequence of residuals (N :_MB) ⊆ (N :_MB²) ⊆ (N :_MB³) ⊆⋯ terminates. We say that a ring satisfies (accr) (respectively, (accr*)) if it satisfies (accr) (respectively. (accr*)) as a module over itself. Various important properties of Noetherian modules and rings were generalized in Refs. [3, 4] to modules and rings satisfying (accr). It was proved in [3, Theorem 1] that for any R-module M, the properties (accr) and (accr*) are equivalent.

Let M be a module over a ring R. Let S be a m.c. subset of R. We use f.g. for finitely generated. Recall from [2, pp. 409 and 410] that M is said to be S-finite if sM ⊆ F for some s ∈ S and some f.g. submodule F of M. Also, recall from Ref. [2] that M is called S-Noetherian if every submodule of M is an S-finite module. We say that R is S-Noetherian if it is S-Noetherian as a module over itself. That is, R is S-Noetherian if each ideal of R is S-finite. In Ref. [2], D.D. Anderson and T. Dumitrescu stated and proved S-variant of several well-known results on Noetherian rings to S-Noetherian rings (see [2, Corollaries 5, 7 and Propositions 9, 10]).

Let S be a m.c. subset of a ring R and let M be an R-module. In Ref. [1], Hamed Ahmed and Hizem Sana introduced the following definition in order to generalize some known results about Noetherian modules. An increasing sequence of submodules of M, N₁ ⊆ N₂ ⊆ N₃ ⊆⋯ is said to be S-stationary if there exist s ∈ S and $k∈N$ such that sN_n ⊆ N_k for all n ≥ k [1, Definition 2.1]. A submodule N of M is said to be an extended submodule if there exists an ideal I of R such that N = IM. In [1, Theorem 2.1], it was shown that an S-finite R-module M is S-Noetherian if every extended submodule of M is S-finite. Also, in Ref. [1], the concept of S-accr modules and S-accr* modules were introduced and investigated. Recall from [1, Definition 3.1] that M is said to satisfy S-accr (respectively, S-accr*) if for every submodule N of M and every f.g. (respectively, principal) ideal B of R, the increasing sequence of submodules of M, (N :_MB) ⊆ (N :_MB²) ⊆ (N :_MB³) ⊆⋯ is S-stationary. In Ref. [1], several results on (accr) modules were generalized to S-accr modules (see [1, Theorems 3.1, 3.2, and 3.3]). It was shown in [1, Proposition 3.1] that the properties S-accr and S-accr* are equivalent.

Let M be a module over a ring R. Recall from Ref. [5] that M is said to satisfy strong accr* if for every submodule N of M and every sequence $<rn>$ of elements of R, the increasing sequence of submodules of M, (N :_Mr₁) ⊆ (N :_Mr₁r₂) ⊆ (N :_Mr₁r₂r₃) ⊆⋯ terminates. Let S be a m.c. subset of R. We say that M satisfies S-strong accr* if for every submodule N of M and every sequence $<rn>$ of elements of R, the increasing sequence of submodules of M, (N :_Mr₁) ⊆ (N :_Mr₁r₂) ⊆ (N :_Mr₁r₂r₃) ⊆⋯ is S-stationary [5]. The ring R is said to satisfy strong accr* (respectively, S-strong accr*) if R satisfies strong accr* (respectively, S-strong accr*) as a module over itself. In Ref. [5], some basic properties of rings and modules satisfying S-strong accr* were proved.

Let M be a module over a ring R. Recall from [6, Exercise 23, p. 295] that M is said to be a Laskerian R-module if M is a f.g. R-module and any proper submodule of M is a finite intersection of primary submodules of M. R is said to be a Laskerian ring if R is Laskerian as an R-module.

Let N be a $p$-primary submodule of an R-module M. N is said to be strongly primary if $pkM⊆N$ for some k ≥ 1. Recall from [6, Exercise 28, p. 298] that an R-module M is said to be a strongly Laskerian R-module if M is a f.g. R-module and any proper submodule of M is a finite intersection of strongly primary submodules of M. R is said to be a strongly Laskerian ring if R is strongly Laskerian as an R-module.

Let S be a m.c. subset of a ring R. Inspired by the research work presented on S-prime ideals of R in Ref. [7], the concept of S-primary ideal of R was introduced and its properties were investigated in Ref. [8]. Recall from Ref. [8] that an ideal $q$ of R with $q∩S=∅$ is said to be an S-primary ideal of R if there exists s ∈ S such that for all a, b ∈ R with $ab∈q$⁠, we have either $sa∈q$ or $sb∈q$⁠. An S-primary ideal $q$ is said to be S-strongly primary if there exist s′ ∈ S and $n∈N$ such that $s′(q)n⊆q$⁠. (In Ref. [8], an S-strongly primary ideal of a ring was referred to as a strongly S-primary ideal.) Let I be an ideal of R such that I ∩ S = ∅. We say that I is S-decomposable (respectively, S-strongly decomposable) if I can be expressed as a finite intersection of S-primary (respectively, S-strongly primary) ideals of R. Recall from Ref. [8] that R is said to be S-Laskerian (respectively, S-strongly Laskerian) if for any ideal I of R, either I ∩ S ≠ ∅ or (I :_Rs) is S-decomposable (respectively, S-strongly decomposable) for some s ∈ S (In Ref. [8], an S-strongly Laskerian ring was referred to as a strongly S-Laskerian ring.). Let f : R → S⁻¹R denote the usual homomorphism of rings given by $f(r)=r1$⁠. For an ideal I of R, f⁻¹(S⁻¹I) is called the saturation of I with respect to S and is denoted by Sat_S(I) or by S(I).

Let P be a property of rings. Whenever, a ring R is a subring of a ring T, we assume that R contains the identity element of T. We denote the collection of all intermediate rings between R and T by [R, T]. We say that (R, T) is a P-pair if A satisfies P for each A ∈ [R, T]. For example, we say that (R, T) is an accr pair (respectively, accr* pair) if A satisfies (accr) (respectively, (accr*)) for each A ∈ [R, T]. It follows from [3, Theorem 1] that (R, T) is an accr pair if and only if (R, T)is an accr* pair. Let S be a m.c. subset of a ring R. We say that (R, T) is an S-accr pair (respectively, S-accr* pair) if A satisfies S-accr (respectively, S-accr*) for each A ∈ [R, T]. It follows from [1, Proposition 3.1] that (R, T) is an S-accr pair if and only if (R, T) is an S-accr* pair. We use the abbreviation ACCRP (respectively, ACCR*P) for accr pair (respectively, accr* pair). We use the abbreviation S-ACCRP (respectively, S-ACCR*P) for S-accr pair (respectively, S-accr* pair). Similarly, one can define the concept of strong accr* pair (respectively, S-strong accr* pair). We use the abbreviation SACCR*P (respectively, S-SACCR*P) for strong accr* pair (respectively, S-strong accr* pair). We use the abbreviation LP (respectively, SLP) for Laskerian pair (respectively, strongly Laskerian pair). We use the abbreviation S-LP (respectively, S-SLP) for S-Laskerian pair (respectively, S-strongly Laskerian pair). We use the abbreviation NP (respectively, S-NP) for Noetherian pair (respectively, S-Noetherian pair). We know from [8, Corollary 3.9(1)] that any S-Laskerian ring satisfies S-accr. Therefore, it follows that any S-LP is an S-ACCRP. We know from [8, Corollary 3.9(2)] that any S-strongly Laskerian ring satisfies S-strong accr*. Hence, we obtain that any S-SLP is an S-SACCR*P. Let R be a subring of a ring T. In Ref. [9] (respectively [10]), for certain pairs of rings R ⊆ T, necessary and sufficient conditions were determined in order that (R, T) to be an LP (respectively, ACCRP). A ring R is said to satisfy ACCP if every increasing sequence of principal ideals of R is stationary. Let S be a m.c. subset of an integral domain D. We say that D satisfies S-ACCP if every increasing sequence of principal ideals of D is S-stationary [11]. Let T be an integral domain which contains D as a subring. We say that (D, T) is an S-ACCP pair if A satisfies S-ACCP for each A ∈ [D, T]. In Ref. [11], for certain pairs of domains D ⊆ T, necessary and sufficient conditions were determined in order that (D, T) to be an S-ACCP pair, where S is a m.c. subset of D.

Let D be an integral domain and let S be a m.c. subset of D. Let K be a field which contains D as a subring. The aim of this article is to investigate the conditions under which (D, K) is an S-ACCRP (respectively, S-SACCR*P). In Section 2 of this article, we focus on determining necessary and sufficient conditions in order that (D, K) to be an S-ACCRP. Recall from [7, Definition 2] that a m.c. subset S of a ring R is said to be a strongly multiplicatively closed if for any given elements $(sα)α∈Λ$ of S, $(⋂α∈ΛRsα)∩S≠∅$ (equivalently, $(⋂s∈SRs)∩S≠∅$⁠). If S is a strongly m.c. subset of D such that S⁻¹D is not a field, then it is proved in Theorem 2.12 that the statements (1) (D, K) is an S-ACCRP and (2) K is algebraic over D and the integral closure of S⁻¹D in K is a one-dimensional Prüfer domain are equivalent. Let D, S, K be as in the statement of Theorem 2.12. It is shown in Corollary 2.14 that the statements (1) (D, K) is an S-LP and (2) K is algebraic over D and the integral closure of S⁻¹D in K is a Laskerian Prüfer domain are equivalent. Let S be a m.c. subset of an integral domain D such that S⁻¹D is a field. Let us denote S⁻¹D by F. Let K be an extension field of F. If (D, K) is an S-ACCRP, then it is verified in Lemma 2.17 that tr. deg  K/F ≤ 1, where tr. deg  K/F denotes the transcendence degree of K over F. If K is algebraic over F, then it is shown in Proposition 2.19 that (D, K) is an S-SLP and hence, (D, K) is an S-SACCR*P. Several examples are given to illustrate the results proved in this section (see Examples 2.15, 2.16 and 2.20).

Let S be a m.c. subset of an integral domain D. Let K be a field which contains D as a subring. The aim of Section 3 of this article is to discuss some results regarding when (D, K) is an S-SACCR*P. Suppose that D is not a field. Let F denote the quotient field of D. If (D, K) is an SACCR*P, then it is proved in Theorem 3.3 that the following statements hold: (1) K is algebraic over D and the integral closure of D in K is a Dedekind domain and (2) The separable degree of K over F is finite and K has finite exponent over F. Suppose that S⁻¹D is not a field. If (D, K) is an S-SACCR*P, then it is deduced in Corollary 3.4 that the following statements hold: (1) K is algebraic over D and the integral closure of S⁻¹D in K is a Dedekind domain and (2) The separable degree of K over F is finite and K has finite exponent over F. It is verified in Example 3.5 that the field L, an infinite algebraic extension field of $Q$⁠, constructed by R. Gilmer [12, Example, p. 520] is such that $(Z,L)$ is an ACCRP but it is not an SACCR*P. Let S be a countable m.c. subset of an integral domain D such that S⁻¹D is integrally closed but not a field. Let F be the quotient field of D with char(F) = 0 (where char(F) denotes the characteristic of F). With the above hypotheses, it is proved in Corollary 3.6 that the following statements (1) (D, K) is an S- SLP; (2) (D, K) is an S-SACCR*P; (3) For any T ∈ [D, K] and any ideal I of T, there exists s ∈ S (depending on I) such that S(I) = (I :_Ts), K is a finite algebraic extension of F and moreover, S⁻¹D and the integral closure of S⁻¹D in K are Dedekind domains; and (4) (D, K) is an S-NP are equivalent. Let D be an integrally closed domain which is not a field. Let F be the quotient field of D with char(F) = 0. Let K be an extension field of F. With the above hypotheses, it is deduced in Corollary 3.7 that the statements (1) (D, K) is an SLP; (2) (D, K) is an SACCR*P; (3) [K : F] < ∞, and moreover D and the integral closure of D in K are Dedekind domains; and (4) (D, K) is an NP are equivalent. An integral domain T is provided in Example 3.8 such that the integral closure of T in its quotient field is a Dedekind domain but T does not satisfy strong accr*. Let S be a m.c. subset of an integral domain D. Suppose that S⁻¹D = F is the quotient filed of D. Let K be an extension field of F. If tr. deg  K/F = 1 and if (D, K) is an S-SACCR*P, then it is deduced in Corollary 3.9 that the following statements hold. (1) For each α ∈ K such that α is transcendental over F, the integral closure of F[α] in K is a Dedekind domain and (2) The separable degree of K over F(α) is finite and K has finite exponent over F(α). If F is a perfect field and K is an extension field of F such that tr. deg  K/F = 1, then it is shown in Corollary 3.10 that the statements (1) (F, K) is an SLP; (2) (F, K) is an SACCR*P; and (3) (F, K) is an NP are equivalent.

For a ring R, we denote the set of all prime ideals of R by Spec(R) and we denote the set of all maximal ideals of R by Max(R). Whenever a set A is a subset of a set B and A ≠ B, we denote it by A ⊂ B. For a ring R, we denote the group of units of R by U(R) and we denote the set of all zero-divisors of R by Z(R). The Krull dimension of a ring R is simply referred to as the dimension of R and is denoted by dim  R. For concepts and notations from commutative ring theory that are not specified in this article, the reader can refer standard text-books in commutative ring theory (for example [13, 14]).

As mentioned in the introduction, the m.c. subsets considered in this article are assumed that they do not contain 0. Let S be a m.c. subset of an integral domain D. Let K be a field which contains D as a subring and K is not necessarily the quotient field of D. The aim of this section is to determine necessary and sufficient conditions in order that (D, K) to be an S-ACCRP. In Proposition 2.4, we determine a necessary condition for (D, K) to be an S-ACCRP, where D is an integral domain such that S⁻¹D is not a field. We use Lemma 2.1 in the proof of Proposition 2.4. For a ring R, we denote the polynomial ring in one variable X over R by R[X].

Proof. We use some arguments similar to those that were used in the proof of [10, Proposition 1.3]. Let us denote the ideal (1 + rX)T of T by I. By hypothesis, T satisfies S-accr. Hence, the increasing sequence of ideals of T, (I :_Tr) ⊆ (I :_Tr²) ⊆ (I :_Tr³) ⊆⋯ is S-stationary. Therefore, there exist s ∈ S and $k∈N$ such that s(I :_Trⁿ) ⊆ (I :_Tr^k) for all n ≥ k. In particular, s(I :_Tr^k+1) ⊆ (I :_Tr^k). Notice that (1 + rX)X^k+1 ∈ T is such that (1 + rX)X^k+1 ∈ (I :_Tr^k+1). Hence, s(1 + rX)X^k+1 ∈ (I :_Tr^k). This implies that s(1 + rX)r^kX^k+1 = (1 + rX)t for some t ∈ T. Since there is no non-zero y ∈ R such that y(1 + rX) = 0, we obtain from [15, Theorem 2] that 1 + rX is a non-zero-divisor of R[X] and hence a non-zero divisor of T. Hence, it follows from s(1 + rX)r^kX^k+1 = (1 + rX)t, we get that sr^kX^k+1 ∈ T. Notice that sr^k−1X^k = s(1 + rX)r^k−1X^k − sr^kX^k+1. From (1 + rX)R[X] ⊂ T and sr^kX^k+1 ∈ T, we obtain that sr^k−1X^k ∈ T. This implies that sX ∈ T if k = 1. If k ≥ 2, then from sr^k−2X^k−1 = s(1 + rX)r^k−2X^k−1 − sr^k−1X^k, it follows that sr^k−2X^k−1 ∈ T. Proceeding like this, we obtain that sX ∈ T. Hence, sX = y + (1 + rX)f(X) for some y ∈ R and f(X) ∈ R[X]. It is clear that f(X) ≠ 0. Since r∉Z(R) by hypothesis, we get that deg((1 + rX)f(X)) = 1 +   deg(f(X)). From 1 = deg(sX) = deg(y + (1 + rX)f(X)), it follows that f(X) ∈ R. By comparing the coefficient of X on both sides of sX = y + (1 + rX)f(X), it follows that s = rf(X) ∈ Rr. This proves that s ∈ Rr ∩ S and so, Rr ∩ S ≠ ∅. □

Proof. By hypothesis, S⁻¹D is not a field. Let d ∈ D\{0} and s ∈ S be such that $ds∉U(S−1D)$⁠. As S ⊆ U(S⁻¹D), it follows that d∉U(S⁻¹D). We are assuming that K is not algebraic over D. Let α ∈ K be such that α is transcendental over D. Notice that $T=D+(1+dα)D[α]∈[D,K]$⁠. We claim that T does not satisfy S-accr. Suppose that T satisfies S-accr. Since d is a non-zero-divisor of D, we obtain from Lemma 2.1 that Dd ∩ S ≠ ∅. Therefore, there exist s₁ ∈ S and d₁ ∈ D such that s₁ = dd₁. This implies that d ∈ U(S⁻¹D) and this is in contradiction to the choice of d. This proves that T does not satisfy S-accr and so, (D, K) is not an S-ACCRP. □

Proof. If S⁻¹V is a field, then it is clear that dim  (S⁻¹V) = 0 < 1. Hence, we can assume that S⁻¹V is not a field. Hence, dim  (S⁻¹V) ≥ 1. We are assuming that V satisfies S-accr. Suppose that dim  (S⁻¹V) > 1. Then it follows that there exists $P∈Spec(S−1V)$ such that $heightP>1$ in S⁻¹V. It follows from [13, Proposition 3.11(iv)] that there exist non-zero prime ideals $p1⊂p2$ of V with $p2∩S=∅$ and $P=S−1p2$⁠. Let $x∈p2\p1$⁠. Let $y∈p1$ with y ≠ 0. Let us denote the ideal Vy by I. Notice that for any $n∈N$⁠, $xn∉p1$ and so, xⁿ∉Vy. Since the set of ideals of V is linearly ordered by inclusion, we get that y = xⁿv_n for some non-unit v_n of V. As we are assuming that V satisfies S-accr, the increasing sequence of ideals of V, (I :_Vx) ⊆ (I :_Vx²) ⊆ (I :_Vx³) ⊆⋯ is S-stationary. Hence, there exist s ∈ S and $k∈N$ such that s(I :_Vxⁿ) ⊆ (I :_Vx^k) for all n ≥ k. In particular, s(I :_Vx^k+1) ⊆ (I :_Vx^k). It is clear that for any $n∈N$⁠, (I :_Vxⁿ) = (Vxⁿv_n :_Vxⁿ) = Vv_n. From $y=xnvn∈p1$ and $xn∉p1$⁠, we get that $vn∈p1$⁠. From s(I :_Vx^k+1) ⊆ (I :_Vx^k), we obtain that sVv_k+1 ⊆ Vv_k. Hence, sv_k+1 = v_kw for some w ∈ V. It follows from y = x^k+1v_k+1 = x^kv_k that v_k = xv_k+1. From sv_k+1 = v_kw, we get that sv_k+1 = xv_k+1w and this implies that $s=xw∈p2$⁠. This is a contradiction, since $p2∩S=∅$⁠. Therefore, if V satisfies S-accr, then dim  (S⁻¹V) ≤ 1. □

Proof. We are assuming that (D.K) is an S-ACCRP. By hypothesis, S⁻¹D is not a field. Hence, we obtain from Corollary 2.2 that K is algebraic over D. Let F be the quotient field of D. Then F is also the quotient field of S⁻¹D. Let T ∈ [S⁻¹D, F] with T ≠ F. It is clear that dim  T ≥ 1. If dim  T > 1, then we obtain from [12, Corollary 19.7(1)] that there exists a valuation domain V ∈ [S⁻¹D, F] such that dim  V > 1. As S⁻¹D ⊆ V, it follows that V = S⁻¹V. By assumption, V satisfies S-accr. Hence, we obtain from Lemma 2.3 that dim  V = dim  (S⁻¹V) ≤ 1. This is a contradiction and so, dim  T ≤ 1. Therefore, dim  T = 1 for each T ∈ [S⁻¹D, F] with T ≠ F. Hence, we obtain from [16, Theorem 6] that the integral closure of S⁻¹D in F is a one-dimensional Prüfer domain. Since the field K is an algebraic extension of the field F, using [12, Theorem 22.3], it can be shown as in the proof of [9, Proposition 2.1] that the integral closure of S⁻¹D in K is a one-dimensional Prüfer domain. □

Proof. (1) ⇒ (2) We are assuming that there exist s ∈ S and an ideal J of R such that J is S-decomposable (respectively, S-strongly decomposable) and sI ⊆ J ⊆ I. Thus, there exist $n∈N$ and S-primary (respectively, S-strongly primary) ideals $q1,…,qn$ of R such that $J=⋂i=1nqi$⁠. Let i ∈ {1, …, n}. It follows from (1) ⇒ (2) of [8, Proposition 2.6] that $S−1qi$ is a primary (respectively, strongly primary) ideal of S⁻¹R and there exists t_i ∈ S such that $S(qi)=(qi:Rti)$⁠. It follows from [13, Proposition 3.11(v)] that $S−1J=⋂i=1nS−1qi$⁠. This proves that S⁻¹J admits a primary (respectively. strong primary) decomposition in S⁻¹R. As $s1∈U(S−1R)$⁠, we obtain from sI ⊆ J ⊆ I that S⁻¹I = S⁻¹J. Therefore, we get that the ideal S⁻¹I of S⁻¹R admits a primary (respectively, strong primary) decomposition in S⁻¹R. Let $t=∏i=1nti$⁠. Then t ∈ S. Let i ∈ {1, …, n}. Notice that $(qi:Rti)$ is a primary (respectively strongly primary) ideal of R with $(qi:Rti)∩S=∅$⁠. It is clear that $t=ti(∏j∈{1,…,n}\{i}tj)$⁠. Hence, we obtain from [13, Lemma 4.4 (iii)] that $(qi:Rti)=(qi:Rt)$⁠. This shows that $S(qi)=(qi:Rt)$ is a primary (respectively. strongly primary) ideal of R. From $J=⋂i=1nqi$⁠, it follows that $S(J)=⋂i=1nS(qi)=⋂i=1n(qi:Rt)=((⋂i=1nqi):Rt)=(J:Rt)$⁠. Let i ∈ {1, …, n}. As $s∉(qi:Rt)$⁠, it follows from [13, Lemma 4.4 (iii)] that $(qi:Rt)=(qi:Rst)$⁠. Therefore, we obtain that S(J) = (J :_Rst). From S⁻¹I = S⁻¹J and J ⊆ I, we get that S(I) = S(J) = (J :_Rst) ⊆ (I :_Rst) ⊆ S(I). Thus with s′ = st and $Qi=(qi:Rs′)$ for each i ∈ {1, …, n}, we obtain that s′ ∈ S and $S(I)=(I:Rs′)=⋂i=1nQi$ is a primary (respectively. strong primary) decomposition of S(I) in R with $Qi∩S=∅$ for each i ∈ {1, …, n}.

(2) ⇒ (1) By assumption, there exist s′ ∈ S and primary (respectively strongly primary) ideals $Q1,…,Qn$ of R such that $S(I)=(I:Rs′)=⋂i=1nQi$ with $Qi∩S=∅$ for each i ∈ {1, …, n}. Let i ∈ {1, …, n}. Let us denote $s′Qi$ by $qi$⁠. It is clear that $qi∩S=∅$⁠. As $Qi$ is a primary ideal of R with $Qi∩S=∅$⁠, it follows that $(s′Qi:Rs′)=Qi$⁠. Hence, we obtain from (2) ⇒ (1) of [8, Proposition 2.4] that $qi$ is an S-primary (respectively, S-strongly primary) ideal of R. Let us denote s′(I :_Rs′) by J. Since $Qi$ is a primary ideal of R and $Qi∩S=∅$⁠, we obtain that $J=s′(I:Rs′)=s′(⋂i=1nQi)=⋂i=1ns′Qi=⋂i=1nqi$ is S-decomposable (respectively. S-strongly decomposable). Notice that s′I ⊆ s′(I :_Rs′) = J ⊆ I. □

Proof. (1) ⇔ (3) This is (1) ⇔ (2) of [8, Proposition 3.2].

(2) ⇒ (3) Let A be any proper ideal of S⁻¹R. Then it follows from [13, Proposition 3.11 (i) and (ii)] that there exists an ideal I of R such that I ∩ S = ∅ and A = S⁻¹I. By (2), there exist s ∈ S and an ideal J of R such that J is S-decomposable (respectively. S-strongly decomposable) with sI ⊆ J ⊆ I. Hence, we obtain from (1) ⇒ (2) of Proposition 2.5 that A = S⁻¹I admits a primary (respectively, strong primary) decomposition in S⁻¹R and there exists s′ ∈ S such that S(I) = (I :_Rs′) admits a primary (respectively, strong primary) decomposition in R. This proves that S⁻¹R is Laskerian (respectively, strongly Laskerian) and given an ideal I of R with I ∩ S = ∅, there exists s′ ∈ S such that S(I) = (I :_Rs′).

(3) ⇒ (2) Let I be an ideal of R with I ∩ S = ∅. It was shown in the proof of (2) ⇒ (1) of [8, Proposition 3.2] that there exist n ≥ 1 and primary (respectively, strongly primary) ideals $q1,…,qn$ of R such that $S(I)=(I:Rs′)=⋂i=1nqi$ with $qi∩S=∅$ for each i ∈ {1, …, n}. Let J = s′(I :_Rs′). Then it is already verified in the proof of (2) ⇒ (1) of Proposition 2.5 that J is S-decomposable (respectively, S-strongly decomposable) and s′I ⊆ J ⊆ I.

□

Recall from Ref. [7] that a m.c. subset S of a ring R is said to be strongly multiplicatively closed if $(⋂s∈SRs)∩S≠∅$⁠. In Ref. [17], strongly multiplicatively closed subsets are referred to as m.c. subsets satisfying maximal multiple condition. Let S be a strongly m.c. subset of an integral domain D such that S⁻¹D is not a field. In Theorem 2.12, we provide a necessary and sufficient condition in order that (D, K) to be an S-ACCRP, where K is a field which contains D as a subring.

Proof. By hypothesis, S is a strongly m.c. subset of R. Hence, there exists s ∈ S such that s ∈ Rt for all t ∈ S. Let I be any ideal of R. It is clear that (I :_Rs) ⊆ S(I). Let r ∈ S(I). Then there exists t ∈ S such that tr ∈ I. As s ∈ Rt, it follows that sr ∈ I. This shows that S(I) ⊆ (I :_Rs) and so, S(I) = (I :_Rs). □

Proof. It can be proved using arguments similar to those that were used in the proof of [5, Lemma 2.6] that R satisfies S-accr*. We know from [1, Proposition 3.1] that the properties S-accr and S-accr* are equivalent. Therefore, we obtain that R satisfies S-accr. □

Proof. This follows from (3) ⇒ (1) of Corollary 2.6. □

Proof. (1) ⇒ (2) We are assuming that R satisfies S-accr. If a ring T satisfies M-accr, where M is any m.c. subset of T, then we know from [1, Example 3.1(3)] that M⁻¹T satisfies (accr*). Hence, we get that S⁻¹R satisfies (accr*). Since the properties (accr) and (accr*) are equivalent by [3, Theorem 1], we obtain that S⁻¹R satisfies (accr).

(2) ⇒ (1) We are assuming that S⁻¹R satisfies (accr). By hypothesis, S is a strongly m.c. subset of R. Hence, we obtain from Lemma 2.7 that there exists s ∈ S such that S(I) = (I :_Rs) for all ideals I of R. It now follows from Lemma 2.8 that R satisfies S-accr. □

Proof. (1) ⇒ (2) This follows from (1) ⇒ (3) of Corollary 2.6.

(2) ⇒ (1) With the help of Lemmas 2.7 and 2.9, this can be proved as in the proof of (2) ⇒ (1) of Corollary 2.10. □

Proof. (1) ⇒ (2) This follows from Proposition 2.4. The proof of (1) ⇒ (2) does not need the assumption that S is a strongly m.c. subset of D.

(2) ⇒ (1) Let T ∈ [D, K]. We claim that S⁻¹T satisfies (accr). This is clear if S⁻¹T is a field. Hence, we can assume that S⁻¹T is not a field. Therefore, dim  (S⁻¹T) ≥ 1. Let D₁ denote the integral closure of S⁻¹D in K and let D₂ denote the integral closure of S⁻¹T in K. As S⁻¹D ⊆ S⁻¹T, it is clear that D₁ is a subring of D₂. From K is algebraic over D, we obtain that K is the quotient field of D_i for each i ∈ {1, 2}. Thus D₂ is an overring of D₁. By assumption, D₁ is a one-dimensional Prüfer domain. Hence, we obtain from [12, Theorem 26.1(1)] that D₂ is a Prüfer domain and   dim  D₂ ≤ dim  D₁ = 1. As D₂ is integral over S⁻¹T, we obtain from [12, 11.8] that dim  (S⁻¹T) = dim  D₂. From   dim  (S⁻¹T) ≥ 1, we get that dim  D₂ ≥ 1 and so, dim  D₂ = 1. Therefore, it follows that dim  (S⁻¹T) = 1. As S⁻¹T is a one-dimensional integral domain, we obtain from [3, Theorem 6] that S⁻¹T satisfies (accr). Since S is a strongly m.c. subset of D, it follows that S is a strongly m.c.subset of T. As S⁻¹T satisfies (accr), we obtain from (2) ⇒ (1) of Corollary 2.10 that T satisfies S-accr. This proves that (D, K) is an S-ACCRP. □

Let D be an integral domain which is not a field. It is clear that S = U(D) is a strongly m.c. subset of D. Let K be a field which contains D as a subring. Notice that S⁻¹D = D and if T ∈ [D, K], then T satisfies S-accr if and only if T satisfies (accr). Hence, the following Corollary 2.13 is an immediate consequence of Theorem 2.12.

Proof. (1) ⇒ (2) Let T ∈ [D, K]. By hypothesis, T is S-Laskerian. Hence, we obtain from [8, Corollary 3.9(1)] that T satisfies S-accr* and so, we obtain from [1, Proposition 3.1] that T satisfies S-accr. This shows that (D, K) is an S-ACCRP. Therefore, it follows from (1) ⇒ (2) of Theorem 2.12 that K is algebraic over D and the integral closure of S⁻¹D in K is a one-dimensional Prüfer domain. Moreover, for any T ∈ [D, K], T is S-Laskerian and so, we obtain from (1) ⇒ (3) of Corollary 2.6 that S⁻¹T is Laskerian. If A ∈ [S⁻¹D, K], then A = S⁻¹T for some T ∈ [D, K]. Hence, we get that A is Laskerian and so, (S⁻¹D, K) is an LP. Therefore, we obtain that the integral closure of S⁻¹D in K is a Laskerian Prüfer domain. Notice that the proof of (1) ⇒ (2) of this proposition does not need the assumption that the m.c. subset S is strongly multiplicatively closed.

(2) ⇒ (1) Let $S−1D‾$ denote the integral closure of S⁻¹D in K. By (2), K is algebraic over D and $S−1D‾$ is a Laskerian Prüfer domain. By hypothesis, S⁻¹D is not a field. Hence, we obtain from the if part of [9, Proposition 2.1] that (S⁻¹D, K) is an LP. Let T ∈ [D, K]. Then S⁻¹T is Laskerian. As S is a strongly m.c. subset of D, we get that S is a strongly m.c. subset of T. Hence, we obtain from (2) ⇒ (1) of Corollary 2.11 that T is S-Laskerian. This proves that (D, K) is an S-LP. □

□

In Example 2.16, we provide an example of a domain T and a m.c. subset S of T such that S⁻¹T is a one-dimensional valuation domain but (T, L) is not an S-ACCRP (where L is the quotient field of T) thereby illustrating that (2) ⇒ (1) of Theorem 2.12 can fail to hold if the hypothesis in Theorem 2.12 that S is a strongly m.c. subset is omitted.

Proof. It is well-known that V = K(X)[[Y]] is a discrete valuation ring (for example, refer [18, p. 322]). Notice that $m=VY$ is the only non-zero prime ideal of V. It is clear that $V=K(X)+m$⁠. We know from [13, Example (2), p. 94] that D = K[X]_XK[X] is a discrete valuation ring. Since D is a valuation domain with quotient field K(X), it follows from [19, Theorem 2.1(h)] that $T=D+m$ is a valuation domain and as dim  D = dim  V = 1, we obtain from [19, Theorem 2.1(f)] that dim  T = 2. Since $Tm≅D$ as rings and D is an integral domain, we get that $m∈Spec(T)$⁠. Let $S=T\m$⁠. From $m∈Spec(T)$⁠, it follows that S is a m.c. subset of T. Notice that $S−1T=Tm$⁠. We claim that $Tm=V$⁠. It is clear that $Tm⊆V$⁠. Let v ∈ V. As $V=K(X)+m$ and K(X) is the quotient field of D, it follows that $v=ab+m$ for some a, b ∈ D with b ≠ 0 and $m∈m$⁠. This implies that $v=a+bmb$⁠. From a + bm ∈ T and $b∈T\m$⁠, we get that $v∈Tm$⁠. This proves that $V⊆Tm$ and so, $V=Tm$⁠. Hence, S⁻¹T is a one-dimensional valuation domain. We know from [13, Proposition 5.18 (iii)] that S⁻¹T is integrally closed. Thus if L is the quotient field of T, then S⁻¹T is the integral closure of S⁻¹T in L and it is a one-dimensional valuation domain. We claim that T does not satisfy S-accr. Suppose that T satisfies S-accr. Let us denote the ideal TY by I. As T satisfies S-accr by assumption, the increasing sequence of ideals of T, (I :_TX) ⊆ (I :_TX²) ⊆ (I :_TX³) ⊆⋯ is S-stationary. Hence, there exist s ∈ S and $k∈N$ such that s(I :_TXⁿ) ⊆ (I :_TX^k) for all n ≥ k. Let $r∈N$⁠. Notice that $Y=YXrXr$ and hence, it follows that $(I:TXr)=(T(YXr)Xr:TXr)=TYXr$⁠. From s(I :_TXⁿ) ⊆ (I :_TX^k) for all n ≥ k, we obtain that $sTYXn⊆TYXk$⁠. This implies that $s∈⋂r=1∞TXr$⁠. It is clear that $⋂r=1∞TXr=(⋂r=1∞DXr)+m$⁠. Since D is a discrete valuation ring, we get that $⋂r=1∞DXr=(0)$⁠. Hence, we obtain that $s∈⋂r=1∞TXr=m$⁠. This is impossible, since $S=T\m$⁠. Therefore, we obtain that T does not satisfy S-accr and so, (T, L) is not an S-ACCRP. □

Let F be a field and let K be an extension field of F. Let S be a m.c. subset of F. As each element of S is a unit in F, it follows that (F, K) is an S-ACCRP (respectively, S-LP) if and only if (F, K) is an ACCRP (respectively, LP). The reader can refer [10, p. 320] (respectively, [9, pp. 94 and 95]) for the solution to the problem of when (F, K) is an ACCRP (respectively, LP).

Let K be an extension field of a field F. We denote the transcendence degree of K over F by the notation tr. deg  K/F.

Let S be a m.c. subset of an integral domain D such that S⁻¹D is a field. Then S⁻¹D is necessarily the quotient field of D. Let K be a field which contains D as a subring. Let us denote S⁻¹D by F. It is clear that K is an extension field of F. If (D, K) is an S-ACCRP, then we verify in Lemma 2.17 that tr. deg  K/F ≤ 1. If K is algebraic over F, then we verify in Proposition 2.19 that (D, K) is an S-SACCR*P. We use Lemma 2.18 in the proof of Proposition 2.19.

Proof. Let us denote S⁻¹D by F. Then F is necessarily the quotient field of D. Suppose that tr. deg  K/F > 1. Then it is possible to find X, Y ∈ K such that {X, Y} is algebraically independent over F. Observe that S is a m.c. subset of F[X] and S ⊆ U(F) = U(F[X]). As (D, K) is an S-ACCRP by hypothesis, it follows that (F[X], F(X, Y)) is an S-ACCRP. From S ⊆ U(F[X]), we get that (F[X], F(X, Y)) is an ACCRP. Since F[X] is not a field, we obtain from (1) ⇒ (2) of Corollary 2.13 that F(X, Y) is algebraic over F[X]. This is a contradiction and so, we obtain that tr. deg  K/S⁻¹D ≤ 1.

Assume that (D, K) is an S-LP. We know from [8, Corollary 3.9(1)] and [1, Proposition 3.1] that any S-Laskerian ring satisfies S-accr. Hence, (D, K) is an S-ACCRP and therefore, we obtain that tr. deg  K/S⁻¹D ≤ 1.

□

Proof. As S⁻¹D is a field, we get that S⁻¹D is strongly Laskerian. Let I be any ideal of D with I ∩ S = ∅. Then S⁻¹I = (0) and so, I = (0). It is clear that S(I) = (0) = ((0) :_Ds) for any s ∈ S. Hence, we obtain from (3) ⇒ (1) of Corollary 2.6 that D is S-strongly Laskerian. We know from [8, Corollary 3.9(2)] that any S-strongly Laskerian ring satisfies S-strong accr*. Hence, we obtain that D satisfies S-strong accr*. □

Proof. Let T ∈ [D, K]. Then S is a m.c. subset of T. If T is a field, then it is clear that T is S-strongly Laskerian. Suppose that T is not a field. By hypothesis, K is algebraic over D and as S⁻¹D is a field, it follows that K is integral over S⁻¹D. Notice that S⁻¹T ∈ [S⁻¹D, K]. As S⁻¹T is integral over S⁻¹D, we obtain from [13, Proposition 5.7] that S⁻¹T is a field. It now follows from Lemma 2.18 that T is S-strongly Laskerian. This proves that (D, K) is an S-SLP. As any S-strongly Laskerian ring satisfies S-strong accr* by [8, Corollary 3.9(2)], we obtain that (D, K) is an S-SACCR*P. □

In Example 2.20, we provide an example of a domain T and a m.c. subset S of T such that (T, L) is an S-SACCR*P but (T, L) is not an ACCRP, where L is the quotient field of T.

Proof. In the notation of Example 2.16, $m=VY$ and $m$ is the only non-zero prime ideal of V. The integral domain $T=D+m$ is such that $m$ is an ideal of both T and V. Now, $S={Yn∣n∈N∪{0}}$ is a m.c. subset of T. Notice that S⁻¹T = S⁻¹V = L, where L is the quotient field of T. It now follows from Proposition 2.19 that (T, L) is an S-SLP and so, (T, L) is an S-SACCR*P. It is clear that if a ring satisfies S-strong accr*, then it satisfies S-accr*. We know from [1, Proposition 3.1] that the properties S-accr and S-accr* are equivalent. Therefore, we obtain that (T, L) is an S-ACCRP. As T is a two-dimensional valuation domain, it follows from Lemma 2.3 that T does not satisfy (accr). Indeed, it is already observed in the proof of Example 2.16 that T does not satisfy S₁-accr, where $S1=T\m$⁠. Thus (T, L) is an S-SACCR*P but (T, L) is not an ACCRP. □

Proof. (1) ⇒ (2) We are assuming that (D.K) is an S-ACCRP. Let α ∈ K be such that α is transcendental over F. Notice that F[α] is not a field. Now, (F[α], K) is an S-ACCRP. As S ⊆ U(F[α]), it follows that (F[α], K) is indeed an ACCRP. Therefore, we obtain from (1) ⇒ (2) of Corollary 2.13 that the integral closure of F[α] in K is a one-dimensional Prüfer domain. It is clear that (1) ⇒ (2) of this proposition does not need the assumption that the m.c. subset S of D is a strongly m.c. subset of D.

(2) ⇒ (1) Let T ∈ [D, K]. Let L denote the quotient field of T. By hypothesis, tr. deg  K/F = 1. Hence, either L is algebraic over F or tr. deg  L/F = 1. If L is algebraic over F, then it follows from Proposition 2.19 that (D, L) is an S-SLP and as T ∈ [D, L], we get that T satisfies S-strong accr* and so, T satisfies S-accr. Suppose that tr. deg  L/F = 1. Let t ∈ T be such that t is transcendental over F. Notice that T ∈ [D[t], K]. It is clear that S is a strongly m.c. subset of D[t] and S⁻¹(D[t]) = F[t] is not a field. From tr. deg  K/F = 1, it follows that K is algebraic over D[t]. By assumption, the integral closure of S⁻¹(D[t]) in K is a one-dimensional Prüfer domain. Hence, we obtain from (2) ⇒ (1) of Theorem 2.12 that (D[t], K) is an S-ACCRP. Since T ∈ [D[t], K], we get that T satisfies S-accr. This proves that (D.K) is an S-ACCRP. □

Proceeding as in the proof of Proposition 2.21, the following Proposition 2.22 can be proved with the help of Corollary 2.14 and Proposition 2.19.

Let S be a m.c. subset of an integral domain D such that S⁻¹D is not a field. Let K be a field which contains D as a subring. The aim of this section is to determine when (D, K) is an S-SACCR*P. Let S be a m.c. subset of a ring R such that R satisfies S-strong accr*. Then it is clear that R satisfies S-accr* and hence, it follows from [1, Proposition 3.1] that R satisfies S-accr. Thus if (D, K) is an S-SACCR*P, then (D, K) is an S-ACCRP. In Corollary 3.4, we determine some necessary conditions for (D, K) to be an S-SACCR*P. First, we state and prove some preliminary results that are useful for solving some of the problems considered in this section.

Let R be a ring. It is well-known that the set of all nilpotent elements of R forms an ideal of R and is called the nilradical of R [13, Proposition 1.7]. We denote the nilradical of R by nil(R). Recall from [20, p. 466] that a sequence $<xn>$ of elements of R is said to be T-nilpotent if there exists $k∈N$ such that $∏i=1kxi=0$⁠.

Proof. Let I be any ideal of R and let $<xn>$ be any sequence of elements of $I$⁠. We consider the following cases.

• Case(1): I ∩ S ≠ ∅.

Let s ∈ I ∩ S. Then $s∏j=1nxj∈I$ for all n ≥ 1.

• Case(2): I ∩ S = ∅.

As I ∩ S = ∅, we obtain from [13, Proposition 3.11 (ii)] that S⁻¹I ≠ S⁻¹R. As R satisfies S-strong accr* by hypothesis, it follows from [5, Lemma 2.4] that S⁻¹R satisfies strong accr*. That is, S⁻¹R satisfies (C) in the notation of [21]. From S⁻¹R satisfies (C), it follows that $S−1RS−1I$ satisfies (C). Hence, we obtain from [21, Lemma 1.2] that $nil(S−1RS−1I)$ is T-nilpotent. Notice that $nil(S−1RS−1I)=S−1IS−1I$⁠. We know from [13, Proposition 3.11 (v)] that $S−1I=S−1I$⁠. It is clear that for each $n∈N$⁠, $xn1+S−1I∈nil(S−1RS−1I)$⁠. From $nil(S−1RS−1I)$ is T-nilpotent, we get that there exists $k∈N$ such that $∏j=1k(xj1+S−1I)=0+S−1I$⁠. This implies that $∏j=1kxj1∈S−1I$⁠. Hence, there exists s ∈ S such that $s∏j=1kxj∈I$⁠. Let n ≥ k. Then it is clear that $s∏j=1nxj∈I$⁠. □

Let S be a m.c. subset of a ring R such that R is S-Laskerian. In Lemma 3.2, we determine necessary and sufficient conditions in order that R to satisfy S-strong accr*.

Proof. (1) ⇒ (2) This follows from Lemma 3.1.

(2) ⇒ (3) This is clear.

(3) ⇒ (1) Let I be any ideal of R and $<rn>$ be any sequence of elements of R. We verify that the increasing sequence of ideals of R, (I :_Rr₁) ⊆ (I :_Rr₁r₂) ⊆ (I :_Rr₁r₂r₃) ⊆⋯ is S-stationary. First, we verify the above assertion in the case $I=q$ is a primary ideal of R. Let $q$ be $p$-primary. Observe that $p=q$⁠. We consider the following cases:

• Case(i): There exists $k∈N$ such that $ri∉p$ for each $i∈N$ with i ≥ k.

In such a case, for any $n∈N$ with n ≥ k, $∏i=knri∉p$⁠. Hence, we obtain from [13, Lemma 4.4 (iii)] that $(q:R∏i=knri)=q$ for all n ≥ k. Therefore, for all n ≥ k, $(q:R∏i=1nri)=q$ in the case k = 1. If k ≥ 2, then for all n ≥ k, $(q:R∏i=1nri)=(q:R∏i=1k−1ri)$⁠.

• Case(ii): There exist positive integers k₁ < k₂ < k₃ < ⋯ such that $rkj∈p$ for each j ≥ 1.

By (3), there exist s ∈ S and $j0∈N$ such that $s∏i=1jrki∈q$ for all j ≥ j₀. Hence, for all $n≥kj0$⁠, $s(q:R∏i=1nri)⊆Rs⊆(q:R∏i=1kj0ri)$⁠.

This shows that for any primary ideal $q$ of R and for any sequence $<rn>$ of elements of R, the increasing sequence of ideals of R, $(q:Rr1)⊆(q:Rr1r2)⊆(q:Rr1r2r3)⊆⋯$ is S-stationary.

Let I be any ideal of R and let $<rn>$ be any sequence of elements of R. Suppose that I ∩ S ≠ ∅. Let s ∈ I ∩ S. Then for all n ≥ 1, $s(I:R∏i=1nri)⊆Rs⊆I⊆(I:Rr1)$⁠. Suppose that I ∩ S = ∅. Since R is S-Laskerian by hypothesis, we obtain from the proof of (2) ⇒ (3) of Corollary 2.6 that there exists s′ ∈ S such that (I :_Rs′) admits a primary decomposition in R. Hence, there exist $t∈N$ and primary ideals $q1,…,qt$ of R such that $(I:Rs′)=⋂i=1tqi$⁠. Let i ∈ {1, …, t}. Since $qi$ is a primary ideal of R, it follows as shown above that there exist s_i ∈ S and $ki∈N$ such that $si(qi:R∏j=1nrj)⊆(qi:R∏j=1kirj)$ for all n ≥ k_i. Let $s=∏i=1tsi$ and let k = max(k₁, …, k_t). Now, s ∈ S and for all n ≥ k, $s((I:Rs′):R∏j=1nrj)=s((⋂i=1tqi):R∏j=1nrj)⊆⋂i=1tsi(qi:R∏j=1nrj)⊆⋂i=1t(qi:R∏j=1krj)=((⋂i=1tqi):R∏j=1krj)=((I:Rs′):R∏j=1krj)$⁠. This implies that for all n ≥ k, $ss′(I:R∏i=1nri)⊆(I:R∏i=1kri)$⁠.

This shows that for any ideal I of R and for any sequence $<rn>$ of elements of R, the increasing sequence of ideals of R, (I :_Rr₁) ⊆ (I :_Rr₁r₂) ⊆ (I :_Rr₁r₂r₃) ⊆⋯ is S-stationary. Therefore, we obtain that R satisfies S-strong accr*. □

Proof. (1) If a ring T satisfies strong accr*, then T satisfies (accr*) and hence by [3, Theorem 1], we get that T satisfies (accr). We are assuming that (D, K) is an SACCR*P. Therefore, (D, K) is an ACCRP. Hence, we obtain from (1) ⇒ (2) of Corollary 2.13 that K is an algebraic extension of F and the integral closure of D in K is a one-dimensional Prüfer domain. Let $D¯$ denote the integral closure of D in K. Notice that $D¯$ is a Prüfer domain, $dimD¯=1$⁠, and it satisfies strong accr*. That is, the Prüfer domain $D¯$ satisfies (C) in the notation of [21]. Hence, we obtain from [21, Proposition 2.3] that $D¯$ is Noetherian. Therefore, each non-zero fractional ideal of $D¯$ is f.g. and as $D¯$ being a Prüfer domain, it follows that each non-zero fractional ideal of $D¯$ is invertible. Hence, we obtain from [13, Theorem 9.8] that $D¯$ is a Dedekind domain. Let S = {1}. It follows from (2) ⇒ (1) of Corollary 2.14 that (D, K) is an S-LP. As S = {1}, we get that (D, K) is an LP.

(2) Let L be the maximal separable subfield of K over F. We claim that [L: F] < ∞. Suppose that [L: F] is not finite. Let α₁ ∈ L\F. Let [F(α₁): F] = n₁. Then n₁ > 1. Notice that L is an infinite algebraic and separable extension of F(α₁). Hence, we obtain from [22, Lemma 1, p. 194] that there exists α₂ ∈ L such that [F(α₁, α₂): F(α₁)] = n₂ > n₁. Since the separable algebraic extension L over F is assumed to be an infinite extension, by repeated use of [22, Lemma 1, p. 194], it is possible to find positive integers 1 < n₁ < n₂ < n₃ < ⋯ and an infinite sequence $<αk>$ of elements from L such that [F(α₁): F] = n₁ and for each k ≥ 2, [F(α₁, α₂, …, α_k): F(α₁, …, α_k−1)] = n_k.

The remaining part of the proof is suggested by the proof of [23, Lemma 3] and the proof of [9, Proposition 2.12]. Notice that as F is the quotient field of D, there exists y₁ ∈ D\{0} such that the irreducible polynomial of y₁α₁ over F has coefficients in D and its degree is n₁. It is clear that y₁α₁ ∈ F(α₁) is integral over D. Set z₁ = y₁α₁. It is clear that D[z₁] is a free D-module with basis ${1,z1,…,z1n1−1}$⁠. By hypothesis, D is not a field. Hence, it is possible to find a non-zero element d ∈ D such that d ∉ U(D). Let us denote the ring D[dz₁] by D₁. It is clear that F(α₁) is the quotient field of D₁, D₁ is a free D-module with basis ${1,dz1,…,(dz1)n1−1}$⁠, and D₁ is an integral extension of D. Observe that the irreducible polynomial of α₂ over F(α₁) is of degree n₂ and it is possible to find y₂ ∈ D₁\{0} such that the irreducible polynomial of y₂α₂ over F(α₁) has coefficients in D₁. Set z₂ = y₂α₂. It is clear that z₂ is integral over D₁ and D₁[z₂] is a free D₁-module with basis ${1,z2,…,z2n2−1}$⁠. Let us denote D₁[dz₂] by D₂. Notice that F(α₁, α₂) is the quotient field of D₂, D₂ is a free D₁-module with basis ${1,dz2,…,(dz2)n2−1}$⁠, and D₂ is integral over D₁. Proceeding like this, we obtain a strictly increasing sequence of subrings of L, D₁ ⊂ D₂ ⊂ D₃ ⊂ ⋯ such that for each k ≥ 1, D_k = D_k−1[dz_k] (with D₀ = D) is a free D_k−1- module with basis ${1,dzk,…,(dzk)nk−1}$⁠, and D_k is integral over D_k−1. Also, F(α₁, …, α_k) is the the quotient field of D_k for each $k∈N$ and by the choice of z_k, it is clear that z_k is integral over D_k−1 for each k ≥ 1. Let us denote the ring $⋃k=1∞Dk$ by T. Since D₁ is integral over D and D₂ is integral over D₁, it follows from [13, Corollary 5.4] that D₂ is integral over D. Let k ≥ 2. Assume it is verified that D_k is integral over D. As D_k+1 is integral over D_k, we obtain from [13, Corollary 5.4] that D_k+1 is integral over D. This proves that $T=⋃k=1∞Dk$ is integral over D. Also, observe that z_k is integral over D for each k ≥ 1. Notice that $⋃k=1∞F(α1,…,αk)$ is the quotient field of T. Now, T ∈ [D, L] and as L is a subfield of K and (D, K) is an SACCR*P by hypothesis, it follows that T satisfies strong accr*. Let $I=Td+∑k=1∞Tdzk$⁠. We assert that $I=Td$⁠. As Td ⊆ I, it is clear that $Td⊆I$⁠. Let $p∈Spec(T)$ be such that $d∈p$⁠. Let us denote the ring $⋃k=1∞D[zk]$ by T₁. Notice that T is a subring of T₁. From z_k is integral over D for each k ≥ 1, it follows that T₁ is integral over T. Now, it follows from [13, Theorem 5.10] that there exists $q∈Spec(T1)$ such that $q∩T=p$⁠. Notice that $d∈q$ and as z_k ∈ T₁ for each k ≥ 1, it follows that IT₁ = T₁d. Therefore, $IT1⊆q$⁠. Hence, $I⊆IT1∩T⊆q∩T=p$⁠. Let $V(Td)={p∈Spec(T)∣p⊇Td}$⁠. We know from [13, Proposition 1.14] that $Td=⋂p∈V(Td)p$ and so, we get that $I⊆Td$⁠. It follows from [13, Exercise 1.13 (ii), page 9] that $Td=Td$⁠. Hence, we obtain that $I⊆Td=Td$⁠. Therefore, $I=Td$⁠. Now, T satisfies strong accr* and $<dzn>$ is a sequence of elements of T such that $dzn∈I=Td$ for each $n∈N$⁠. Hence, by applying Lemma 3.1 with S = {1}, we obtain that there exists $k∈N$ such that $∏i=1ndzi∈Td$ for each n ≥ k. Thus $∏i=1kdzi∈Td∩Dk$⁠. Observe that D_j is a free D_j−1-module with basis ${1,dzj,…,(dzj)nj−1}$ for each j ≥ 1 (where, D₀ = D) and from d ∈ D, it follows that Djd ∩ D_j−1 = D_j−1d for each j ≥ 1. Hence, we obtain from $∏i=1kdzi∈Td∩Dk$ that $∏i=1kdzi∈Dkd$⁠. If k = 1, then we get that dz₁ ∈ D₁d. Observe that $D1d=Dd+Dd(dz1)+⋯+Dd(dz1)n1−1$⁠. Therefore, $dz1∈Dd+Dd(dz1)+⋯+Dd(dz1)n1−1$ —— (1). Hence, by comparing the coefficient of dz₁ on both sides of (1), it follows that 1 ∈ Dd. This is impossible, since d is not a unit of D. Suppose that k ≥ 2. Notice that $∏i=1k−1dzi∈Dk−1$⁠. It is clear that $(∏i=1k−1dzi)dzk∈Dkd=Dk−1d+Dk−1d(dzk)+⋯+Dk−1d(dzk)nk−1$——- (2). Hence, by comparing the coefficient of dz_k on both sides of (2), it follows that $∏i=1k−1dzi∈Dk−1d$⁠. Proceeding like this, we get that dz₁ ∈ D₁d and this is already verified to be impossible. Therefore, T does not satisfy strong accr*. This is in contradiction to the assumption that (D, K) is an SACCR*P. Therefore, the separable degree of K over F must be finite.