Cancellation ideals of a ring extension

We study properties of cancellation ideals of ring extensions. Let R ⊆ S be a ring extension. A nonzero S-regular ideal I of R is called a (quasi)-cancellation ideal of the ring extension R ⊆ S if whenever IB = IC for two S-regular (finitely generated) R-submodules B and C of S, then B = C. We show...

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Date:2021
Main Author: Tchamna, S.
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Language:English
Published: Інститут прикладної математики і механіки НАН України 2021
Series:Algebra and Discrete Mathematics
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Cite this:Cancellation ideals of a ring extension / S. Tchamna // Algebra and Discrete Mathematics. — 2021. — Vol. 32, № 1. — С. 138–146. — Бібліогр.: 7 назв. — англ.

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spelling nasplib_isofts_kiev_ua-123456789-1887222025-02-09T15:55:12Z Cancellation ideals of a ring extension Tchamna, S. We study properties of cancellation ideals of ring extensions. Let R ⊆ S be a ring extension. A nonzero S-regular ideal I of R is called a (quasi)-cancellation ideal of the ring extension R ⊆ S if whenever IB = IC for two S-regular (finitely generated) R-submodules B and C of S, then B = C. We show that a finitely generated ideal I is a cancellation ideal of the ring extension R ⊆ S if and only if I is S-invertible. 2021 Article Cancellation ideals of a ring extension / S. Tchamna // Algebra and Discrete Mathematics. — 2021. — Vol. 32, № 1. — С. 138–146. — Бібліогр.: 7 назв. — англ. 1726-3255 DOI:10.12958/adm1424 2020 MSC: 13A15, 13A18, 13B02 https://nasplib.isofts.kiev.ua/handle/123456789/188722 en Algebra and Discrete Mathematics application/pdf Інститут прикладної математики і механіки НАН України
institution Digital Library of Periodicals of National Academy of Sciences of Ukraine
collection DSpace DC
language English
description We study properties of cancellation ideals of ring extensions. Let R ⊆ S be a ring extension. A nonzero S-regular ideal I of R is called a (quasi)-cancellation ideal of the ring extension R ⊆ S if whenever IB = IC for two S-regular (finitely generated) R-submodules B and C of S, then B = C. We show that a finitely generated ideal I is a cancellation ideal of the ring extension R ⊆ S if and only if I is S-invertible.
format Article
author Tchamna, S.
spellingShingle Tchamna, S.
Cancellation ideals of a ring extension
Algebra and Discrete Mathematics
author_facet Tchamna, S.
author_sort Tchamna, S.
title Cancellation ideals of a ring extension
title_short Cancellation ideals of a ring extension
title_full Cancellation ideals of a ring extension
title_fullStr Cancellation ideals of a ring extension
title_full_unstemmed Cancellation ideals of a ring extension
title_sort cancellation ideals of a ring extension
publisher Інститут прикладної математики і механіки НАН України
publishDate 2021
url https://nasplib.isofts.kiev.ua/handle/123456789/188722
citation_txt Cancellation ideals of a ring extension / S. Tchamna // Algebra and Discrete Mathematics. — 2021. — Vol. 32, № 1. — С. 138–146. — Бібліогр.: 7 назв. — англ.
series Algebra and Discrete Mathematics
work_keys_str_mv AT tchamnas cancellationidealsofaringextension
first_indexed 2025-11-27T17:25:12Z
last_indexed 2025-11-27T17:25:12Z
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fulltext “adm-n3” — 2021/11/8 — 20:27 — page 138 — #140 © Algebra and Discrete Mathematics RESEARCH ARTICLE Volume 32 (2021). Number 1, pp. 138–146 DOI:10.12958/adm1424 Cancellation ideals of a ring extension S. Tchamna Communicated by E. I. Zelmanov Abstract. We study properties of cancellation ideals of ring extensions. Let R ⊆ S be a ring extension. A nonzero S-regular ideal I of R is called a (quasi)-cancellation ideal of the ring extension R ⊆ S if whenever IB = IC for two S-regular (finitely generated) R-submodules B and C of S, then B = C. We show that a finitely generated ideal I is a cancellation ideal of the ring extension R ⊆ S if and only if I is S-invertible. 1. Introduction and background Throughout this article, we assume that all rings are commutative with identity. The notion of cancellation ideal for a ring has been studied in [1] and [2]. An ideal I of a ring R is called cancellation ideal if whenever IB = IC for two ideals B and C of R, then B = C [2]. A finitely generated ideal is a cancellation ideal if and only if for each maximal ideal M of R, IM is a regular principal ideal of RM [1, Theorem 1]. D.D Anderson and D.F Anderson used the notion of cancellation ideal to characterize Prüfer domain. A ring R is a Prüfer domain if and only if every finitely generated nonzero ideal of R is a cancellation ideal[1, Theorem 6]. In this paper, we study the notion of cancellation ideal for ring extensions; which is a generalization of the notion of cancellation ideal for rings. Let R ⊆ S be a ring extension, and let A be an R-submodule of S. The R-submodule A is said to be S-regular ifAS = S[5, Definition 1, p. 84]. For two R-submodules E, F of S, denote by [E : F ] the set of all x ∈ S such that xF ⊆ E. 2020 MSC: 13A15, 13A18, 13B02. Key words and phrases: ring extension, cancellation ideal, pullback diagram. https://doi.org/10.12958/adm1424 “adm-n3” — 2021/11/8 — 20:27 — page 139 — #141 S. Tchamna 139 An R-submodule A of S is said to be S-invertible, if there exists an R- submodule B of S such that AB = R[5, Definition 3, p 90]. In this case, we write B = A−1, and A−1 = [R : A] = {x ∈ S : xA ⊆ R} [5, Remark 1.10, p. 90]. For the R-submodule A of S, and for a multiplicative subset τ of R, we denote by A[τ ] the set of all x ∈ S such that tx ∈ A for some t ∈ τ . If p is a prime ideal of R, and τ = R \ p, then A[p] denotes the set of all x ∈ S such that tx ∈ A for some t ∈ τ . The set A[τ ] is called the saturation of A by τ . Properties of the saturation of a submodule are studied in [5, p. 18] and [6]. An S-regular ideal I of R is called (quasi)-cancellation ideal of the ring extension R ⊆ S if whenever IB = IC for two S-regular (finitely generated) R-submodules B and C of S, then B = C. In section 2, we study properties of (quasi)-cancellation ideals of ring extensions. In Proposition 2.4, we prove that a finitely generated S-regular ideal I of R is a cancellation ideal if and only it is a quasi-cancellation ideal. In Theorem 2.12, we show that for an S-regular finitely generated ideal I of R, the followings are equivalent: (1) I is a cancellation ideal of the ring extension R ⊆ S. (2) I is an S-invertible ideal of R. (3) IR[X] is a cancellation ideal of the ring extension R[X] ⊆ S[X]. Remark 1.1. Let R ⊆ S be a ring extension, and let A,B be two R- submodules of S. Then A = B if and only if A[m] = B[m] for each maximal ideal m of R. In fact, if A = B, then it clear that A[m] = B[m] for each maximal ideal m of R. Conversely, if A[m] = B[m] for each m ∈ M, where M is the set of all maximal ideals of R, then by [5, Remark 5.5, p. 50], we have A = ∩m∈MA[m] = ∩m∈MBm = B. Let R ⊆ S and L ⊆ T be two ring extensions, and consider the following commutative diagram R L S T Ψ α where kerΨ is an ideal of R, Ψ : S −→ T is surjective, the restriction α : R −→ L of Ψ is also surjective and the vertical mappings are inclusions. When kerΨ is a maximal ideal of S, the previous commutative diagram is called a pullback diagram a type �. Pullback diagrams of type � are studied by S. Gabelli and E. Houston in [4]. “adm-n3” — 2021/11/8 — 20:27 — page 140 — #142 140 Cancellation ideals of a ring extension Lemma 1.2. Consider the above pullback diagram of type �. If A,B are two S-regular ideals of R such that Ψ(A) = Ψ(B), then A = B. Proof. Let A,B be two S-regular ideals of R such that Ψ(A) = Ψ(B). By [7, Remark 1.1], we have kerΨ ⊆ A and kerΨ ⊆ B. Let a ∈ A. Then there exists b ∈ B such that Ψ(a) = Ψ(b). Hence a− b ∈ kerΨ ⊆ B. Thus a ∈ B. This shows that A ⊆ B. With the same argument, B ⊆ A. Thus A = B. 2. Cancellation ideals of ring extensions In this section, we define and study properties of cancellation ideals of ring extensions. Definition 2.1. Let R ⊆ S be a ring extension. A nonzero S-regular ideal I of R is called a (quasi)-cancellation ideal of the ring extension R ⊆ S if whenever IB = IC for two S-regular (finitely generated) R-submodules B and C of S, then B = C. The following proposition studies cancellation ideals in pullback diagram of type �. In this article, the Jacobson radical of a ring is denoted Jac(R). Proposition 2.2. Suppose that the following diagram R L S T Ψ is a pullback diagram of type � such that kerΨ ⊆ Jac(R). Then an S- regular ideal I of R is a cancellation ideal of the extension R ⊆ S if and only if Ψ(I) is a cancellation ideal of the extension L ⊆ T . Proof. Suppose that I is a cancellation ideal of the extension R ⊆ S. Since IS = S, we have Ψ(I)Ψ(S) = Ψ(S). It follows that Ψ(I)T = T . Hence Ψ(I) is a T -regular ideal of L. Let E and F be two T -regular L-submodules of T such that Ψ(I)E = Ψ(I)F . Let B = Ψ−1(E) and C = Ψ−1(F ). Then by [7, Lemma 2.8(1)] B and C are two S-regular ideals of R. Furthermore, E = Ψ(B) and F = Ψ(C) since Ψ is surjective. It follows from the equality Ψ(I)E = Ψ(I)F that Ψ(I)Ψ(B) = Ψ(I)Ψ(C). Hence Ψ(IB) = Ψ(IC). Furthermore, (IB)S = IS = S and (IC)S = IS = S. Therefore, by “adm-n3” — 2021/11/8 — 20:27 — page 141 — #143 S. Tchamna 141 Lemma 1.2, we have IB = IC. Hence B = C since I is a cancellation ideal of the extension R ⊆ S. It follows that E = Ψ(B) = Ψ(C) = F . This shows that Ψ(I) is a cancellation ideal of the extension L ⊆ T . Conversely, suppose that Ψ(I) is a cancellation ideal of the extension L ⊆ T . Let B and C be two S-regular R-submodules of S such that IB = IC. Then Ψ(I)Ψ(B) = Ψ(I)Ψ(C). Since BS = S, we have Ψ(B)T = T . Hence Ψ(B) is a T -regular ideal of L. With the same argument, Ψ(C) is a T -regular ideal of L. It follows that Ψ(B) = Ψ(C) since Ψ(I) is a cancellation ideal of the extension L ⊆ T . Therefore, by Lemma 1.2, we have B = C. This shows that I is a cancellation ideal of the extension R ⊆ S. In the next proposition, we give a characterization of a cancellation ideal of a ring extension. This result is an analogue of [3, Proposition 2.1, p. 10] in the case of cancellation ideal of a ring. Proposition 2.3. Let R ⊆ S be a ring extension, and let I be an S-regular ideal of R. The following statements are equivalent. (1) I is a (quasi)-cancellation ideal of the ring extension R ⊆ S. (2) [IJ : I] = J for any S-regular (finitely generated) R-submodule J of S. (3) If IJ ⊆ IK for two S-regular (finitely generated) R-submodules J and K of S, then J ⊆ K. Proof. (1) ⇒ (2) Suppose that I is a cancellation ideal of the extension R ⊆ S, and let J be an S-regular R-submodule of S. The containment J ⊆ [IJ : I] is always true. Let x ∈ [IJ : I]. Then xI ⊆ IJ . It follows that (x, J)I ⊆ IJ , where (x, J) is the R-submodule of S generated by x and J . Therefore, (x, J)I = IJ since the containment IJ ⊆ (x, J)I is always true. Furthermore, (x, J) is an S-regular R-submodule of S since J ⊆ (x, J). It follows from the definition of a cancellation ideal that (x, J) = J . This shows that x ∈ J , and thus [IJ : I] ⊆ J . Therefore [IJ : I] = J . (2) ⇒ (3) Suppose that the statement (2) is true. Let J and K be two S-regular R-submodules of S. Then by (2), we have [IK : I] = K. If IJ ⊆ IK, then J ⊆ [IK : I] = K. (3) ⇒ (1) This implication is obvious. Proposition 2.4. Let R ⊆ S be a ring extension, and let I be a finitely generated S-regular ideal of R. Then I is a cancellation ideal of R ⊆ S if and only if I is a quasi-cancellation ideal of R ⊆ S. “adm-n3” — 2021/11/8 — 20:27 — page 142 — #144 142 Cancellation ideals of a ring extension Proof. Let I be a finitely generated S-regular ideal of R. If I is a can- cellation ideal of the extension R ⊆ S, then obviously I is an quasi- cancellation ideal of the extension R ⊆ S. Conversely, suppose that I is a quasi-cancellation ideal of the extension R ⊆ S. Let a1, . . . , an ∈ R be a set of generators of I. Let B,C be two S-regular R-submodules of S such that IB ⊆ IC. Let b ∈ B. Then bI ⊆ IC. So, for 1 6 i 6 n, we have bai = ∑k j=1 ajcij with cij ∈ C for 1 6 j 6 k. Furthermore, since CS = S, there exist u1, . . . , uℓ ∈ C and s1, . . . , sℓ ∈ S such that u1s1+· · ·+uℓsℓ = 1. Let C ′ be the R-submodule of S generated by the elements of the set {u1, . . . , un, cij : 1 6 i 6 n, 1 6 j 6 k}. Let B0 be the R-submodule of S generated by b. Then (B0 + (u1, . . . , un)R) I ⊆ IC ′. It follows from the the equivalence (1) ⇔ (3) of Proposition 2.3 that B0+(u1, . . . , un)R ⊆ C ′ since B0 + (u1, . . . , un) and C ′ are finitely generated S-regular ideal of S. Therefore, b ∈ C ′ ⊆ C. Hence B ⊆ C since b was arbitrary chosen in B. This shows that I is a cancellation ideal of the extension R ⊆ S. Lemma 2.5. Let R ⊆ S be a ring extension, and let u1, . . . , uℓ ∈ S. Define the sets E = (u1, . . . , uℓ)R[p] and A = (u1, . . . , uℓ)R, where p is a prime ideal of R. For any ideal I of R, we have: (1) (AI)[p] = (EI)[p]. In particular, A[p] = E[p]. (2) (EI)[p] = (EI[p])[p]. Proof. (1) First, observe that AI ⊆ EI. So (AI)[p] ⊆ (EI)[p]. Let x ∈ (EI)[p]. Then there exists t ∈ R \ p such that tx ∈ EI. Therefore, tx = ∑n i=1 eixi for some ei ∈ E and xi ∈ I, 1 6 i 6 n. For each 1 6 i 6 n, write ei = ∑ℓ j=1 ujyij with yij ∈ R[p] for 1 6 j 6 ℓ. Let sij ∈ R \ p such that sijyij ∈ R, si = ∏ℓ j=1 sij and s = ∏n i=1. Then siei ∈ A. It follows that (st)x = ∑ℓ i=1(sei)xi ∈ AI. Thus x ∈ (AI)[p] since st ∈ R \ p. This shows that (EI)[p] ⊆ (AI)[p]. Hence (AI)[p] = (EI)[p]. In particular, if we take I = R, then we get A[p] = E[p]. (2) The containment (EI)[p] ⊆ ( EI[p] ) [p] is clear since EI ⊆ EI[p]. Let x ∈ ( EI[p] ) [p] . Then tx ∈ EI[p] for some t ∈ R \ p. Thus tx = ∑k i=1 viyi with vi ∈ E and yi ∈ I[p] for 1 6 i 6 k. Let si ∈ R \ p such that siyi ∈ I, and let s = ∏k i=1 si. Then (st)x = ∑k i=1 vi(syi) ∈ EI. It follows that x ∈ (EI)[p]. Therefore, (EI)[p] = (EI[p])[p] Theorem 2.6. Let R ⊆ S be a ring extension, and let I be a finitely generated S-regular ideal of R. The following statements are equivalent. (1) I is a quasi-cancellation ideal of the extension R ⊆ S. “adm-n3” — 2021/11/8 — 20:27 — page 143 — #145 S. Tchamna 143 (2) For each prime ideal p of R, and for each S-regular finitely generated R[p]-submodule E of S, we have [(EI)[p] : I[p]] = E[p]. Proof. (1) ⇒ (2) Suppose that I is a quasi-cancellation ideal of the extension R ⊆ S, and let p be a prime ideal of R. Let E be a finitely generated S-regular R[p]-submodule of S. Then E = (u1, . . . , uℓ)R[p] for some elements u1, . . . , uℓ of S. Let A be the R-submodule of S generated by u1, . . . , uℓ. Then by Proposition 2.3 and Proposition 2.4, we have [AI : I] = A. It follows from [6, Proposition 2.1(4)] that [(AI)p : I[p]] = A[p]. Hence by Lemma 2.5, we have [(EI[p])[p] : I[p]] = [(EI)p : I[p]] = [(AI)p : I[p]] = A[p] = E[p]. (2) ⇒ (1) Suppose that the statement (2) is true. Let A be an S-regular finitely generated R-submodule of S, and let p be a prime ideal of R. Let E = AR[p]. Then by Lemma 2.5, we have (AI)[p] = (EI)[p] and A[p] = E[p]. So, by hypothesis we have A[p] = E[p] = [(EI)[p] : I[p]] = [(AI)[p] : I[p]]. But by [6, Proposition 2.1(4)], we have [(AI)[p] : I[p]] = [(AI) : I][p]. Therefore, A[p] = [(AI) : I][p] for each prime ideal p of R. It follows from Remark 1.1 that [AI : I] = A. Therefore, by the equivalence (1) ⇔ (2) of Proposition 2.3, I is a quasi-cancellation ideal of the extension R ⊆ S. In their book [5], Knebusch and Zhang defined the notion of Prüfer exten- sion using valuation ring [5, Definition 1, p. 46]. Several characterizations of a Prüfer extension are given in [5, Theorem 5.2, p. 47]. For the purpose of this work, we will use the following: a ring extension R ⊆ S is called Prüfer extension if R is integrally closed in S and R[α] = R[αn] for any α ∈ S and any n ∈ N. Lemma 2.7. [5, Theorem 1.13, p. 91] If a ring extension R ⊆ S is a Prüfer extension, then every finitely generated S-regular R-submodule of S is S-invertible. Proposition 2.8. Let R ⊆ S be a ring extension, and let I be an S-regular ideal of R. (1) If I is a cancellation ideal of the extension R ⊆ S, then [I : I] = R. (2) If the extension R ⊆ S is Prüfer, then the converse of statement (1) is also true (i.e. in a Prüfer extension R ⊆ S, if I is an S-regular ideal satisfying [I : I] = R, then I is a quasi-cancellation ideal). Proof. (1) The proof follows directly from the equivalence (1) ⇔ (2) of Theorem 2.3. It suffices to take J = R. (2) Suppose that the extension R ⊆ S is Prüfer, and let I be an S-regular ideal of R such that [I : I] = R. Let A be an S-regular finitely “adm-n3” — 2021/11/8 — 20:27 — page 144 — #146 144 Cancellation ideals of a ring extension generated R-submodule of S. Then by Lemma 2.7, A is S-invertible. We show that A[I : I] = [AI : I]. Let x ∈ [AI : I]. Then xI ⊆ AI. Hence xIA−1 ⊆ I. Thus xA−1 ⊆ [I : I]. It follows that x ∈ A[I : I]. On the other hand, let y = ∑k i=1 aivi ∈ A[I : I] with ai ∈ A and vi ∈ [I : I] for 1 6 i 6 k. Then viI ⊆ I. Hence aiviI ⊆ AI. Therefore, aivi ∈ [AI : I]. So y = ∑k i=1 aivi ∈ [AI : I]. This shows that [AI : I] = A[I : I]. Hence [AI : I] = A[I : I] = AR = A. Hence, by the equivalence (1) ⇔ (2) of Proposition 2.3, I is a quasi-cancellation ideal of the extension R ⊆ S. Let R ⊆ S be a ring extension. A nonzero S-regular ideal I of R is called m-canonical ideal of the extension R ⊆ S if [I : [I : J ]] = J for all S-regular ideal J of R. Properties of m-canonical ideals of a ring extension are studied in [7]. Corollary 2.9. Any m-canonical ideal of a Prüfer extension is a quasi- cancellation ideal. Proof. If I is an m-canonical ideal of a Prüfer extension R ⊆ S, then by [7, Proposition 2.3], we have [I : I] = R. It follows from Proposition 2.8(2) that I is a quasi-cancellation ideal of the extension R ⊆ S. Lemma 2.10. Let R ⊆ S be a ring extension, and let I be an S-regular ideal of R which is a cancellation ideal of R ⊆ S. If I = (x, y) +A, where A is an ideal of R containing mI for some maximal ideal m of R, then I = (x) +A or I = (y) +A. Proof. Let J = (x2 + y2, xy, xA, yA,A2)R. Then IJ = I3. Observe that I2 is S-regular since I2S = I(IS) = IS = S. Also, from the equality IJ = I3 we have (IJ)S = I3S = I(IS) = IS = S. So JS = S. This shows that J is an S-regular ideal of R. It follows from the equation IJ = I3 and the fact that I is a cancellation ideal of the extension R ⊆ S that J = I2. Thus x2 = t(x2 + y2) + terms from (xy, xA, yA,A2), with t ∈ R. Suppose that t ∈ m. Then x2 ∈ (y2, xy, xA, yA,A2), since tx ∈ mI ∈ A. Let K = (y)+A. Then I2 = IK. Furthermore, from the equality IK = I2, we have K(IS) = I2S. Hence KS = S. Therefore, K is an S-regular ideal of S. It follows that I = K since I is a cancellation ideal of the extension R ⊆ S. The rest of the proof is similar to the proof of [2, Lemma]. Proposition 2.11. Let R ⊆ S be a ring extension, and let I be a nonzero S-regular ideal of R. If I is a cancellation ideal of the extension R ⊆ S, then for each maximal ideal m of R, there exists a ∈ R such that I[m] = (a)[m]. “adm-n3” — 2021/11/8 — 20:27 — page 145 — #147 S. Tchamna 145 Proof. Suppose that I is a cancellation ideal of the ring extension R ⊆ S, and let m be a maximal ideal of R. Suppose that I ⊆ m. Then by Lemma 2.10, and the proof of [2, Theorem], there exists a ∈ I such that for each b ∈ I, (1 − u)b = ra for some u ∈ m and r ∈ R. Therefore, (1− u)b ∈ (a). So b ∈ (a)[m]. This shows that I[m] ∈ (a)[m]. On the other hand, the equality (a)[m] ⊆ I[m] is always true. Thus I[m] = (a)[m]. If I * m, then I[m] = (1)[m] = R[m]. In fact, for x ∈ R[m], there exists s ∈ R \m such that sx ∈ R. Thus (st)x ∈ I for each t ∈ I \m. It follows that x ∈ I[m]. Theorem 2.12. Let R ⊆ S be a ring extension, and let I be a nonzero finitely generated S-regular ideal of R. The following statements are equiv- alent. (1) I is a cancellation ideal of the extension R ⊆ S. (2) I is a quasi-cancellation ideal of the extension R ⊆ S. (3) I is an S-invertible ideal of R. (4) IR[X] is a cancellation ideal of the extension R[X] ⊆ S[X]. Proof. The equivalence (1) ⇔ (2) is the result of Theorem 2.4. (1) ⇒ (3) Suppose that I is a cancellation ideal of the extension R ⊆ S, and let m be a maximal ideal of R. By the previous proposition, I[m] = (a)[m] for some a ∈ R. It follows that ( I[m] ) m[m] = ( (a)[m] ) m[m] . But by [5, Lemma 2.9(b), p. 28], we have Im = ( I[m] ) m[m] and (a)m = ( (a)[m] ) m[m] . Hence Im = (a)m. This shows that I is locally principal. It follows from [5, Proposition 2.3, p. 97] that I is S-invertible. (3) ⇒ (1) This implication is obvious. (3) ⇒ (4) Suppose that I is an S-invertible ideal of the extension R ⊆ S. First, note that (IR[X])(S[X]) = S[X] since IS = S. Hence IR[X] is an S[X]-regular ideal of R[X]. Let J be the R-submodule of S such that IJ = R. Then (IR[X])(JR[X]) = R[X]. This shows that IR[X] is an S[X]-invertible ideal of R[X]. It follows from the equivalence (1) ⇔ (3) that IR[X] is a cancellation ideal of the extension R[X] ⊆ S[X]. (4) ⇒ (1) Suppose that IR[X] is a cancellation ideal of the extension R[X] ⊆ S[X]. Let J be an S-regular ideal of R. Then by the equivalence (1) ⇔ (2) of Proposition 2.3, we have [(IR[X])(JR[X]) : IR[X]] = JR[X]. We show that [IJ : I] = J . First, note that the containment J ⊆ [IJ : I] is always true. Let u ∈ [IJ : I]. Then uI ⊆ IJ . Therefore, uIR[X] ⊆ (IJ)R[X] ⊆ (IR[X])(JR[X]). Hence u ∈ [(IR[X])(JR[X]) : IR[X]] = JR[X]. It follows that u ∈ JR[X] ∩ S = J . This shows that [IJ : I] ⊆ J . Hence [IJ : I] = J . It follows from the equivalence (1) ⇔ (2) of Proposition 2.3 that I is a cancellation ideal of the extension R ⊆ S. “adm-n3” — 2021/11/8 — 20:27 — page 146 — #148 146 Cancellation ideals of a ring extension Corollary 2.13. Let R ⊆ S be a ring extension, and let I be a finitely generated S-regular ideal of R. If I is a cancellation ideal of the extension R ⊆ S, then I[m] is a cancellation ideal of the extension R[m] ⊆ S for each maximal ideal m of R. Proof. Let I be a finitely generated S-regular ideal of R, and let m be a maximal ideal of R. Suppose that I is a cancellation ideal of the extension R ⊆ S. Then by the previous theorem, I is S-invertible. Let J be an R-submodule of S such that IJ = R. Then I[m]J[m] ⊆ (IJ)[m] ⊆ R[m]. Furthermore, since IS = S, there exist xi ∈ I and yi ∈ J , 1 6 i 6 ℓ, such that 1 = ∑ℓ i=1 xiyi. Let u ∈ R[m]. There exists t ∈ R \m such that tu ∈ R and u = ∑ℓ i=1(uxi)yi. But for 1 6 i 6 ℓ, t(uxi) = (tu)xi ∈ I since tu ∈ R and xi ∈ I. It follows that uxi ∈ I[m]. Therefore, u = ∑ℓ i=1(uxi)yi ∈ I[m]J ⊆ I[m]J[m]. This shows that R[m] ⊆ I[m]J[m]. Thus I[m]J[m] = R[m]. Hence I[m] is an S-invertible R[m]-submodule of S. It follows that I[m] is a cancellation ideal of the extension R[m] ⊆ S, since an invertible ideal of ring extension is always a cancellation ideal. References [1] Anderson, D.D., Anderson, D. F. (1984). Some remarks on cancellation ideals. Math Japonica. 29 (6), pp 879-886. [2] Anderson, D.D., Roitman, M. (1997). A characterization of cancellation ideals. Proc. Amer. Math. Soc. No. 10, pp 2853 - 2854. [3] Fuchs, L.; Salce, L. (2001) Modules over non-Noetherian domains. Mathematical Surveys and Monographs, Vol. 84. American Mathematical Society, Providence, RI. xvi+613 pp. [4] Gabelli, S; Houston, E. (2000). Ideal theory in pullbacks. In Chapman S. T; Glaz S., eds. Non-Noetherian commutative ring theory. Math. Appl., Vol 520. Dordrecht: Kluwer Acad. Publ., pp. 199-227. [5] Knebusch, M., Zhang, D. (2002). Manis valuations and Prüfer extensions I. Lecture Notes in Mathematics, Vol. 1791. Berlin: Springer-Verlag. [6] Paudel, L., Tchamna, S. (2018). On the saturation of submodules. Algebra and Discrete Mathematics, Vol. 26, No. 1, pp. 110 - 123 [7] Tchamna, S. (2017). Multiplicative canonical ideals of ring extensions. Journal of Algebra and Its Appl. Vol. 16, No. 4:170069. Contact information Simplice Tchamna Department of Mathematics Georgia College, Milledgeville, GA, USA E-Mail(s): simplice.tchamna@gcsu.edu Received by the editors: 26.07.2019 and in final form 30.10.2020. mailto:simplice.tchamna@gcsu.edu S. Tchamna