On the saturations of submodules
Let R ⊆ S be a ring extension, and let A be an R-submodule of S. The saturation of A (in S) by τ is set A[τ] = {x ∈ S : tx ∈ A for some t ∈ τ}, where τ is a multiplicative subset of R. We study properties of saturations of R-submodules of S. We use this notion of saturation to characterize star oper...
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Інститут прикладної математики і механіки НАН України
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irk-123456789-1883782023-02-27T01:27:51Z On the saturations of submodules Paudel, L. Tchamna, S. Let R ⊆ S be a ring extension, and let A be an R-submodule of S. The saturation of A (in S) by τ is set A[τ] = {x ∈ S : tx ∈ A for some t ∈ τ}, where τ is a multiplicative subset of R. We study properties of saturations of R-submodules of S. We use this notion of saturation to characterize star operations ⋆ on ring extensions R ⊆ S satisfying the relation (A ∩ B)⋆ = A⋆ ∩ B⋆ whenever A and B are two R-submodules of S such that AS = BS = S. 2018 Article On the saturations of submodules / L. Paudel, S. Tchamna // Algebra and Discrete Mathematics. — 2018. — Vol. 26, № 1. — С. 110–123. — Бібліогр.: 8 назв. — англ. 1726-3255 2010 MSC: 13A15, 13A18, 13B02. http://dspace.nbuv.gov.ua/handle/123456789/188378 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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Let R ⊆ S be a ring extension, and let A be an R-submodule of S. The saturation of A (in S) by τ is set A[τ] = {x ∈ S : tx ∈ A for some t ∈ τ}, where τ is a multiplicative subset of R. We study properties of saturations of R-submodules of S. We use this notion of saturation to characterize star operations ⋆ on ring extensions R ⊆ S satisfying the relation (A ∩ B)⋆ = A⋆ ∩ B⋆ whenever A and B are two R-submodules of S such that AS = BS = S. |
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Paudel, L. Tchamna, S. |
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Paudel, L. Tchamna, S. On the saturations of submodules Algebra and Discrete Mathematics |
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Paudel, L. Tchamna, S. |
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Paudel, L. |
| title |
On the saturations of submodules |
| title_short |
On the saturations of submodules |
| title_full |
On the saturations of submodules |
| title_fullStr |
On the saturations of submodules |
| title_full_unstemmed |
On the saturations of submodules |
| title_sort |
on the saturations of submodules |
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Інститут прикладної математики і механіки НАН України |
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2018 |
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http://dspace.nbuv.gov.ua/handle/123456789/188378 |
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On the saturations of submodules / L. Paudel, S. Tchamna // Algebra and Discrete Mathematics. — 2018. — Vol. 26, № 1. — С. 110–123. — Бібліогр.: 8 назв. — англ. |
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Algebra and Discrete Mathematics |
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AT paudell onthesaturationsofsubmodules AT tchamnas onthesaturationsofsubmodules |
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2025-07-16T10:24:30Z |
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2025-07-16T10:24:30Z |
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1837798755753000960 |
| fulltext |
“adm-n3” — 2018/10/20 — 9:02 — page 110 — #116
Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 26 (2018). Number 1, pp. 110–123
c© Journal “Algebra and Discrete Mathematics”
On the saturations of submodules
Lokendra Paudel and Simplice Tchamna∗
Communicated by D. Simson
Abstract. Let R ⊆ S be a ring extension, and let A be an
R-submodule of S. The saturation of A (in S) by τ is set A[τ ] =
{x ∈ S : tx ∈ A for some t ∈ τ}, where τ is a multiplicative subset
of R. We study properties of saturations of R-submodules of S. We
use this notion of saturation to characterize star operations ⋆ on
ring extensions R ⊆ S satisfying the relation (A ∩B)⋆ = A⋆ ∩B⋆
whenever A and B are two R-submodules of S such that AS =
BS = S.
1. Introduction
Throughout this paper, all rings considered are commutative with
identity. Let R ⊆ S be a ring extension, and let A be an R-submodule
of S. The saturation of A (in S) by τ is set
A[τ ] = {x ∈ S : tx ∈ A for some t ∈ τ} ,
where τ is a multiplicative subset of R [7, Definition 10, p. 18]. If p is
a prime ideal of R, then the saturation of A with respect to τ = R \ p
is denoted A[p]. If A is an R-submodule A of S and M is the set of
all maximal ideals of R, then A =
⋂
p∈M
(R[p]A[p]) and R =
⋂
p∈M
R[p]
[7, Remark 5.5, p. 50]. In their book [7], Knebusch and Zhang use this
∗Supported by the Georgia College faculty development travel grant-FY 16.
2010 MSC: 13A15, 13A18, 13B02.
Key words and phrases: saturation, star operation, ring extension, prime spec-
trum, localization, flat module.
“adm-n3” — 2018/10/20 — 9:02 — page 111 — #117
L. Paudel, S. Tchamna 111
notion of saturation to define and study Prüfer extension; see [7, pages 46-
73]. In her paper [8], McNair studies properties of saturations of the form
Iτ = {x ∈ S : tx ∈ I for some t ∈ τ}, when I is an ideal of R, and τ is a
multiplicative subset of R. In particular, McNair proved that if R ⊆ S is a
ring extension, and I is a radical ideal of R, then I[p]∩R = p and I[p] = p[p]
for any minimal prime ideal p of I [8, Proposition 2.1]. Furthermore, if
R ⊆ S ⊆ T and p is a prime ideal of S such that S[p] = R[p∩R] (saturation
in T ), then Sp = Rp∩R [8, Lemma 2.7]. In the present paper, we do not
limit our study to only ideals of R but to any R-submodule of S.
In Section 2, we study properties of saturations, and we show that the
saturation is distributive with respect to the finite intersection (and finite
sum ). Note that when R is a domain with quotient field S, the notion of
saturation coincides with the notion of localization, and AR[p] = A[p] for
any R-submodule A of S. In Example 2.5, we construct a ring extension
R ⊆ S and an R-submodule A of S such that AR[p] ( A[p]. In Proposition
2.7, Proposition 2.9 and Proposition 2.11, we investigate conditions under
which the equality AR[p] = A[p] is satisfied for a ring extension R ⊆ S and
an R-submodule A of S.
In Section 3, we use the notion of saturation to study properties of star
operations on ring extensions. In Theorem 3.10, which is a generalization
of [2, Theorem 4] due to D.D Anderson, we characterize star operations
⋆ on a ring extension R ⊆ S satisfying the relation (A ∩B)⋆ = A⋆ ∩B⋆,
where A and B are S-regular R-submodules of S.
2. Some properties of saturations of submodules
In this section, we consider a ring extension R ⊆ S and we study proper-
ties of the saturations of R-submodules of S. Let A be an R-submodule of S.
The saturation of A (in S) by τ is set A[τ ] = {x ∈ S : tx∈A for some t∈τ},
where τ is a multiplicative subset of R [7, Definition 10, p. 18]. If p is a
prime ideal of R, then the saturation of A with respect to τ = R \ p is
denoted A[p]. For each R-submodule A of S, the set [R :S A] denotes the
set of all x ∈ S such that xA ⊆ R, and (R :A A) denotes the set of all
x ∈ R such that xA ⊆ R [7, Definition 2, p. 85]. We also use the notations
[R : A] and (R : A) when the context is clear. The R-submodule A of S
is called S-invertible, if there exists an R-submodule B of S such that
AB = R [7, Definition 3, p. 90]. In this case, we write B = A−1, and
A−1 = [R :S A] = {x ∈ S : xA ⊆ R} [7, Remarks 1.10, p. 90].
“adm-n3” — 2018/10/20 — 9:02 — page 112 — #118
112 On the saturations of submodules
Proposition 2.1. Let R ⊆ S be a ring extension, and let τ be a multi-
plicative subset of R. Let A and B be two R-submodules of S. Then
(1) AR[τ ] ⊆ A[τ ].
(2) (A ∩B)[τ ] = A[τ ] ∩B[τ ].
(3) (A+B)[τ ] = (A[τ ] +B[τ ])[τ ].
(4) If B is finitely generated, then (A[τ ] : B[τ ]) = (A : B)[τ ].
Proof. (1) Let x ∈ AR[τ ]. Then x =
∑ℓ
i=1 aiui with ai ∈ A and ui ∈ R[τ ]
for 1 6 i 6 ℓ. Let ti ∈ τ such that tiui ∈ R for each 1 6 i 6 ℓ, and let
t =
∏ℓ
i=1 ti. Then tx =
∑ℓ
i=1 ai(tui) ∈ A. It follows that x ∈ A[τ ]. This
shows that AR[τ ] ⊆ A[τ ].
(2) Let x ∈ (A ∩B)[τ ]. Then there exists t ∈ τ such that tx ∈ A ∩B.
It follows that tx ∈ A and tx ∈ B. Therefore, x ∈ A[τ ] and x ∈ B[τ ]. Thus
x ∈ A[τ ]∩B[τ ]. This shows that (A∩B)[τ ] ⊆ A[τ ]∩B[τ ]. On the other hand,
if x is an element of A[τ ]∩B[τ ], then there exist t1, t2 ∈ τ such that t1x ∈ A
and t2x ∈ B. It follows that t1t2x ∈ A∩B. So x ∈ (A∩B)[τ ], since t1t2 ∈ τ .
This shows that A[τ ]∩B[τ ] ⊆ (A∩B)[τ ]. Therefore, (A∩B)[τ ] = A[τ ]∩B[τ ].
(3) By the definition, the containment A ⊆ A[τ ] is always true. Thus
(A+B)[τ ] ⊆ (A[τ ]+B[τ ])[τ ], since A+B ⊆ A[τ ]+B[τ ]. Now let x ∈ (A[τ ]+
B[τ ])[τ ]. Then there exists t ∈ τ such that tx = x1 + x2 for some x1 ∈ A[τ ]
and x2 ∈ B[τ ]. Let t1, t2 ∈ τ such that t1x1 ∈ A and t2x2 ∈ B. Then
tt1t2x ∈ A+B. It follows that x ∈ (A+B)[τ ], since tt1t2 ∈ τ . This shows
that (A[τ ] +B[τ ])[τ ] ⊆ (A+B)[τ ]. Therefore, (A+B)[τ ] = (A[τ ] +B[τ ])[τ ].
(4) Let u ∈ (A : B)[τ ]. Then tu ∈ (A : B) for some t ∈ τ . It
follows that tuB ⊆ A. Let x ∈ B[τ ]. Then there exists t′ ∈ τ such that
t′x ∈ B. Thus tt′xu = (tu)(t′x) ∈ A. Hence xu ∈ A[τ ]. It follows that
u ∈ (A[τ ] : B[τ ]), since x was chosen arbitrarily in B[τ ]. This shows that
(A : B)[τ ] ⊆ (A[τ ] : B[τ ]). It remains to show that (A[τ ] : B[τ ]) ⊆ (A : B)[τ ].
Let y1, . . . , yn ∈ S such that B = (y1, . . . , yn)R.
Step 1. If n = 1, then B is generated by the element y1 of S. Let x ∈ (A[τ ] :
B[τ ]). Then xB[τ ] ⊆ A[τ ]. It follows that xy1 ∈ A[τ ], since y1 ∈ B ⊆ B[τ ].
So there exists t ∈ τ such that txy1 = a for some a ∈ A. Let b ∈ B.
Then b = ry1 for some r ∈ R. Therefore, t(xb) = r(txy1) = ra ∈ A. It
follows that tx ∈ (A : B), since b was chosen arbitrarily in B. Hence
x ∈ (A : B)[τ ]. This shows that (A[τ ] : B[τ ]) ⊆ (A : B)[τ ].
Step 2. More generally, for n > 1, we have B = B1 + · · ·+Bn, where each
Bi is the R-submodule of S generated by yi, 1 6 i 6 n. So
(A : B)[τ ] = (A : B1 + · · ·+Bn)[τ ] = (
n⋂
i=1
(A : Bi))[τ ].
“adm-n3” — 2018/10/20 — 9:02 — page 113 — #119
L. Paudel, S. Tchamna 113
But by (2), we have (
⋂n
i=1(A : Bi))[τ ] =
⋂n
i=1(A : Bi)[τ ]. Furthermore,
by Step 1, we have
⋂n
i=1(A : Bi)[τ ] =
⋂n
i=1(A[τ ] : Bi[τ ]). So
(A : B)[τ ] =
n⋂
i=1
(A[τ ] : Bi[τ ]) = (A[τ ] : B1[τ ] + · · ·+Bn[τ ]).
But by (3), we have
B1[τ ] + · · ·+Bn[τ ] ⊆ (B1[τ ] + · · ·+Bn[τ ])[τ ] = (B1 + · · ·+Bn)[τ ] = B[τ ].
It follows that (A[τ ] : B[τ ]) ⊆ (A[τ ] : B1[τ ] + · · ·+Bn[τ ]) = (A : B)[τ ].
Lemma 2.2. Let R ⊆ S be a ring extension, and suppose that p is a
prime ideal of R. Then p = pR[p] ∩R = p[p] ∩R.
Proof. The relation p ⊆ pR[p] ∩ R ⊆ p[p] ∩ R is a direct consequence of
Proposition 2.1(1). Now we show that p[p] ∩ R ⊆ p. Let x ∈ p[p] ∩ R.
Then there exists t ∈ R \ p such that tx ∈ p. It follows that x ∈ p, since
p is a prime ideal of R and t /∈ p. This shows that p[p] ∩ R ⊆ p. Thus
p = pR[p] ∩R = p[p] ∩R.
Remark 2.3. In [8, Proposition 2.1], McNair showed that if R ⊆ S is
a ring extension and I is a radical ideal of R, then I[p] ∩ R = p for any
minimal prime ideal p of I. So by taking I = p, we obtain Lemma 2.2.
For the rest of the article, if R is a ring, we denote by Spec(R) the set
of all prime ideals of R. The intersection of all prime ideal of R be denoted
by Nil(R). A ring extension is said to be tight if for every x ∈ S \R, there
exists an S-invertible ideal I of R such that xI ⊆ R [7, Definition 1, p. 94].
Proposition 2.4. Let R ⊆ S be a tight ring extension. If Nil(R) = 0
then, ⋃
p∈Spec (R)
R[p] = S.
Proof. Let x ∈ S, then there exists an S-invertible ideal I of R such
that xI ⊆ R. Furthermore, we have IS = S. Thus I 6= 0. It follows that
there exists p ∈ Spec(R) such that I * p. Let t ∈ I \ p. Then tx ∈ R.
Hence x ∈ R[p]. This shows that S ⊆ ∪
p∈Spec(R)R[p]. The containment
∪
p∈Spec(R)R[p] ⊆ S is clear. Thus S =
⋃
p∈Spec(R)R[p].
In the next example, we construct a ring extension R ⊆ S, a prime
ideal p of R and an R-submodule A of S such that A[p] 6= AR[p].
“adm-n3” — 2018/10/20 — 9:02 — page 114 — #120
114 On the saturations of submodules
Example 2.5. Let D be an integral domain, and let X, Y be two in-
determinates over D. Let R = D[X2, Y ] and S = D[X,Y ]. Let p be the
ideal of R generated by Y , and let A be the R-submodule of S generated
by X2Y . Then A[p] 6= AR[p].
Proof. Let t = X4 ∈ R. Then t /∈ p and tY = Y X4 = (X2Y )X2 ∈ A.
This shows that Y ∈ A[p]. Now we show that Y /∈ AR[p]. By contradiction,
suppose that Y ∈ AR[p]. Then Y =
∑ℓ
i=1X
2Y UiVi with Ui ∈ R and
Vi ∈ R[p] for 1 6 i 6 ℓ. It follows that 1 =
∑ℓ
i=1X
2UiVi. This last
equality is not possible, since X is not a unit in S. Thus Y /∈ AR[p].
Therefore, A[p] 6= AR[p].
For the rest of this section, we investigate conditions under which
A[p] = AR[p], when A is an R-submodule of S and p is a prime ideal
of R. First we recall some definitions. Let T be a ring, and let Γ be an
additive totally ordered abelian group. Let Γ ∪ ∞ = Γ ∪ {∞}, where
∞ + g = g + ∞ = ∞ for all g ∈ Γ ∪ ∞, and g < ∞ for all g ∈ Γ. A
valuation on T with values in Γ is a map v : T → Γ ∪∞ such that:
(1) v(xy) = v(x) + v(y) for all x, y ∈ T .
(2) v(x+ y) > min {v(x), v(y)} for all x, y ∈ T .
(3) v(1) = 0 and v(0) = ∞.
If v(T ) = Γ ∪∞, then v is called a Manis valuation on S [7, Definition 4,
p. 12]. In this case, V = {x ∈ T : v(x) > 0} is called a Manis subring of T ,
the extension V ⊆ T is called a Manis extension, and (V, p) is called a
Manis pair in T , where p = {x ∈ T : v(x) > 0} [7, Definition 1, p. 22].
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 if AS = S [7, Definition 1,
p. 84]. The ring S is called a Prüfer extension of R if (R[p], p[p]) is a Manis
pair in S for every maximal ideal p of R. In this case, we say that R is
Prüfer subring of S. The ring extension R ⊆ S is called Bézout extension if
R ⊆ S is a Prüfer extension and every S-invertible ideal of R is a principal
ideal. More on Manis valuations, Prüfer extensions and Bézout extensions
can be found in [7].
Lemma 2.6 ([7, Theorem 1.13, p. 91]). If R ⊆ S is a Prüfer extension,
then every finitely generated S-regular R-submodule of S is S-invertible.
Proposition 2.7. Let R ⊆ S be a ring extension, and let A be an R-
submodule of S. If A is S-invertible, then for any prime ideal p of R,
AR[p] = A[p].
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L. Paudel, S. Tchamna 115
Proof. Let A be an S-invertible R-submodule of S. By Proposition 2.1(1),
we have AR[p] ⊆ A[p]. Thus it suffices to show that A[p]⊆AR[p]. Let x∈A[p].
Then there exists t ∈ R \ p such that tx ∈ A. Since AA−1 = R, there
exist a1, . . . , aℓ ∈ A and b1, . . . , bℓ ∈ A−1 such that a1b1 + · · ·+ aℓbℓ = 1.
Moreover, for 1 6 i 6 ℓ, we have txbi ∈ R, since tx ∈ A and bi ∈ A−1. Thus
xbi ∈ R[p] for 1 6 i 6 ℓ. It follows that x =
∑ℓ
i=0 ai(xbi) ∈ AR[p].
Remark 2.8. Let R ⊆ S be a Prüfer extension, and let A be an S-regular
finitely generated R-submodule of S. Then by Lemma 2.6,A is S-invertible.
It follows from Proposition 2.7 that AR[p] = A[p] for any prime ideal p
of R.
We recall from [3, Definition 2, p.13] that and R-module M is flat, if for
every injective homomorphism v : M1 → M2 of R-modules, the R-module
homomorphism 1M ⊗ v : M ⊗R M1 → M ⊗R M2 is also injective.
Theorem 2.9. Let R ⊆ S be a ring extension, and suppose that R =⋂n
i=1 Vi. If the extension Vi ⊆ S is Prüfer for each 1 6 i 6 n, then
AR[p] = A[p] for any S-regular flat R-submodule A of S and any prime
ideal p of R.
Proof. Let A be an S-regular flat R-submodule of S. Since AS = S, there
exist u1, . . . , uℓ ∈ A and s1, . . . , sℓ ∈ S such that u1s1 + · · · + uℓsℓ = 1.
Let Ai be the Vi-submodule of S generated by u1, . . . , uℓ. Then it is clear
that AiS = S. Hence Ai is an S-regular finitely generated Vi-submodule
of S.
Let x ∈ A[p]. Then there exists t ∈ R \ p such that tx ∈ A. Let Bi =
txVi +Ai. Then Bi is also an S-regular finitely generated Vi-submodule
of S. It follows from Lemma 2.6 that Bi is an S-invertible Vi-submodule
of S. Therefore, there exists a Vi-submodule Ci of S such that BiCi = Vi.
Let ci0, ci1, . . . , ciℓ ∈ Ci such that ci0tx + ci1u1 + · · · + ciℓuℓ = 1. Then
x = (tx)(xci0) + u1(xci1) + · · · + uℓ(xciℓ). Now notice that t(cikx) =
(tx)cik ∈ BiCi = Vi for 0 6 k 6 ℓ and 1 6 i 6 ℓ. It follows that cikx ∈ Vi[p]
for 0 6 k 6 ℓ and 1 6 i 6 ℓ. Therefore, for 1 6 i 6 ℓ, we have x ∈ AVi[p],
since tx, u1, . . . , uℓ ∈ A. This shows that x ∈
⋂n
i=1AVi[p]. Since A is a flat
R-module, we have
⋂n
i=1AVi[p] = A(
⋂n
i=1 Vi[p]) [3, Proposition 6, p. 18].
Furthermore, by Proposition 2.1(2), we have
⋂n
i=1 Vi[p] = (
⋂n
i=1 Vi)[p] =
R[p]. It follows that x ∈ AR[p]. This shows that A[p] ⊆ AR[p], and hence
A[p] = AR[p].
Remark 2.10. Let R ⊆ S be a Prüfer extension, and let A be an S-regular
R-submodule of S.
“adm-n3” — 2018/10/20 — 9:02 — page 116 — #122
116 On the saturations of submodules
(1) If A is finitely generated, then A[p] = AR[p]; see Remark 2.8.
(2) If A is a flat R-submodule of S, then A[p] = AR[p]. This is obtained
by taking n = 1 in the hypothesis of Proposition 2.9.
In the next result, we show that in a Prüfer extension R ⊆ S, the
equality AR[p] = A[p] is always true for any S-regular R-submodule A
of S and any prime ideal p of R (no other condition is needed). The proof
of Proposition 2.11 is a modification of Proposition 2.9.
Proposition 2.11. If R ⊆ S is a Prüfer extension, then for any S-regular
R-submodule of S, we have AR[p] = A[p] for each prime ideal p of R.
Proof. Let A be an S-regular R-submodule of S, and let x ∈ A[p]. Then
there exists t ∈ R \ p such that tx ∈ A. Since AS = S, there exist
a1, . . . , aℓ ∈ A and s1, . . . , sℓ ∈ S such that a1s1 + · · ·+ aℓsℓ = 1. Let A′
be the R-submodule of S generated by a1, · · · , aℓ. Then it is clear that A′ is
an S-regular finitely generated R-submodule of S. Let B = txR+A′. Then
B is also an S-regular finitely generated R-submodule of S. It follows from
Lemma 2.6 that B is S-invertible. Hence, there exist b0, b1, . . . , bℓ ∈ B−1
such that b0tx+ b1a1 + · · ·+ bℓaℓ = 1. But for 0 6 i 6 ℓ, we have t(bix) =
bi(tx) ∈ R, since tx ∈ B and bi ∈ B−1. It follows that bix ∈ R[p] for each
0 6 i 6 ℓ. Therefore, x = (tx)(xb0)+a1(xb1)+ · · ·+aℓ(xbℓ) ∈ AR[p]. This
shows that A[p] ⊆ AR[p]. Hence A[p] = AR[p].
3. Star operations induced by saturations of submodules
In this section, we use the notion of saturation along with the notion
of star operation to investigate properties of ring extensions. First we
recall the definition of a star operation as given by Knebusch and Kaiser
in [6, Definition 1, p. 139]. Let R ⊆ S be a ring extension. A map ⋆:
J (R,S) → J (R,S), where J (R,S) is the set of all R-submodules of S,
is called star operation on R ⊆ S if the following conditions are satisfied
for all A, B ∈ J (R,S).
(c1) A ⊆ A⋆.
(c2) If A ⊆ B, then A⋆ ⊆ B⋆.
(c3) (A⋆)⋆ = A⋆.
(c4) AB⋆ ⊆ (AB)⋆.
A star operation ⋆ on a ring extension R ⊆ S is said to be strict if R⋆ = R.
For more on star operation of ring extension, see [6, pages 139 - 164].
In the next two results, we give examples of star operations induced
by saturations.
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L. Paudel, S. Tchamna 117
Proposition 3.1. Let R ⊆ S be a ring extension, and let ⋆ : J (R,S) →
J (R,S) be the map defined by A⋆ =
⋂
p∈P
A[p], where P is a set of prime
ideals of R. Then
(1) The map ⋆ is a star operation on the extension R ⊆ S.
(2) A[p] = (A⋆)[p] for each A ∈ J (R,S) and each p ∈ P.
(3) If R =
⋂
p∈P
R[p], then p⋆ = p for each p ∈ P.
Proof. (1) Let A,B and J be elements of J (R,S). From the definition
of ⋆, it is clear that A ⊆ A⋆, and A⋆ ⊆ B⋆ if A ⊆ B. Furthermore,
JA⋆ = J(
⋂
p∈P
A[p]) ⊆ JA[p] for each p ∈ P. It follows that JA⋆ ⊆⋂
p∈P
JA[p] = (JA)⋆.
Now we show that (A⋆)⋆ = A⋆. Let x ∈ (A⋆)⋆. Then for each p0 ∈ P , we
have x ∈ (A⋆)[p0]. So there exists t0 ∈ R\p0 such that t0x ∈ A⋆ =
⋂
pP
A[p].
In particular t0x ∈ A[p0]. Therefore, t0s0x ∈ A for some s0 ∈ R \ p0. It
follows that x ∈ A[p0], since t0s0 ∈ R \ p0. Since p0 was chosen arbitrarily
in P, we have x ∈
⋂
p
A[p] = A⋆. Thus, (A⋆)⋆ ⊆ A⋆. The containment
A⋆ ⊆ (A⋆)⋆ is clear. Hence (A⋆)⋆ = A⋆.
(2) Let A ∈ J (R,S), and let p ∈ P . By part (1), we have A ⊆ A⋆. So
A[p] ⊆ (A⋆)[p]. It remains to show that (A⋆)[p] ⊆ A[p]. Notice that for each
p ∈ P, we have A⋆ ⊆
⋂
q∈P
A[q] ⊆ A[p]. Thus (A⋆)[p] ⊆ (A[p])[p] = A[p].
(3) Let p0 ∈ P . Then by part (1), we have p0 ⊆ p⋆0. Now we show the
containment p⋆0 ⊆ p0. Let x ∈ p⋆0. Then x ∈
⋂
p∈P
p0[p] ⊆
⋂
p∈P
R[p] = R.
Furthermore, there exists t0 ∈ R \ p0 such that t0x ∈ p0. It follows that
x ∈ p0, since p0 is prime and t0 /∈ p. This shows that p⋆0 ⊆ p0. Hence
p = p⋆.
Proposition 3.2. Let R ⊆ S be a ring extension, and let ⋆ : J (R,S) →
J (R,S) be the map defined by A⋆ =
⋂
p∈P
AR[p], where P is a set of
prime ideals of R. Then
(1) The map ⋆ is a star operation on the extension R ⊆ S.
(2) A[p] = (A⋆)[p] for each A ∈ J (R,S) and each p ∈ P.
(3) If R =
⋂
p∈P
R[p], then p⋆ = p for each p ∈ P.
Proof. (1) Let A, B and J be elements of J (R,S). From the definition
of ⋆, it is clear that A ⊆ A⋆. Also, if A ⊆ B, then AR[p] ⊆ BR[p] for
each prime ideal p of P. Hence A⋆ =
⋂
p∈P
AR[p] ⊆
⋂
p∈P
BR[p] = B⋆.
Furthermore, JA⋆ = J(
⋂
p∈P
AR[p]) ⊆ JAR[p] for each p ∈ P. It follows
that JA⋆ ⊆
⋂
p∈P
JAR[p] = (JA)⋆. Notice that if q ∈ P, then (A⋆)⋆ =⋂
p∈P
A⋆R[p] ⊆ A⋆R[q] = (
⋂
p∈P
AR[p])R[q] ⊆ (AR[q])R[q] ⊆ AR[q]. It
follows that (A⋆)⋆ ⊆ AR[q] for each q ∈ P . Thus (A⋆)⋆ =
⋂
q∈P
ARq = A⋆.
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118 On the saturations of submodules
(2) Let A ∈ J (R,S), and let p ∈ P. By part (1), we have A ⊆ A⋆.
It follows that A[p] ⊆ (A⋆)[p]. It remains to show that (A⋆)[p] ⊆ A[p]. We
have (A⋆)[p] = (
⋂
q∈P
AR[q])[p] ⊆ (AR[p])[p]. But by Proposition 2.1(1),
we have AR[p] ⊆ A[p]. Thus (A⋆)[p] ⊆ (A[p])[p] = A[p]. This shows that
(A⋆)[p] = A[p].
(3) Let p0 ∈ P. Then by part (1), we have p0 ⊆ p⋆0. Now we show
the containment p⋆0 ⊆ p0. We have p⋆0 =
⋂
p∈P
p0R[p]. But by Proposi-
tion 2.1(1), we have p0R[p] ⊆ p0[p] for each p ∈ P. Thus p⋆0 ⊆
⋂
p∈P
p0[p].
Furthermore, by Proposition 3.1(3), we have
⋂
p∈P
p0[p] = p0. Therefore,
p⋆[0] = p[0].
Let R ⊆ S be a ring extension, and let X be an indeterminate over S.
For each element f of S[X], and each subring L of S containing R, we
denote by cL(f) the L-submodule of S generated by the coefficients
of f . Let T = {g ∈ S[X] : cS(g) = S}. Then T is a multiplicative subset
of S[X] [5, ]. We denote by S(X) the quotient ring (S[X])T . Hence
S(X) is the set of all elements of the form f
g
with f, g ∈ S[X] such that
cS(g) = S (for properties of the ring S(X), see [5, p. 410]). A subring B
of S(X) is said to be Kronecker subring of S(X) if X ∈ B, and for every
f = a0 + a1X + · · · anX
n ∈ S[X] with cS(f) = S the elements
aj
f
∈ S(X)
with j = o, . . . n, are contained in B [6, Definition 4, p. 127].
For each star operation ⋆ on a ring extension R ⊆ S, we denote by
Kr(⋆) the set of all elements f
g
∈ S(X) with (cR(f)H)⋆ ⊆ (cR(g)H)⋆ for
some finitely generated S-regular R-submodule H of S. The set Kr(⋆) is
a Kronecker subring of S(X)[6, Theorem 3.10, p. 143], and the extension
Kr(⋆) ⊆ S(X) is Bézout [6, Theorem 1.5, p. 128].
Remark 3.3. Let R ⊆ S be a ring extension, and let P be a set of
prime ideals of R. Let ⋆1, ⋆2 : J (R,S) → J (R,S) be two maps defined
for each A ∈ J (R,S) by A⋆1 =
⋂
[p]∈P A[p] and A⋆2 =
⋂
[p]∈P AR[p]
respectively. Then by Proposition 3.1, ⋆1 is a star operation on R ⊆ S.
Also, by Proposition 3.2, the map ⋆2 is also a star operation on R ⊆ S. Let
f
g
∈ Kr(⋆2). Then there exists a finitely generated S-regular R-submodule
H of S such that cR(f)H ⊆ (cR(f)H)⋆2 ⊆ (cR(g)H)⋆2 . But by Proposition
2.1(1), (cR(g)H)⋆2 ⊆ (cR(g)H)⋆1 . It follows that (cR(f)H) ⊆ (cR(g)H)⋆1 ,
and hence (cR(f)H)⋆1 ⊆ (cR(g)H)⋆1 . Therefore, f
g
∈ Kr(⋆1). This shows
that Kr(⋆2) ⊆ Kr(⋆1).
Remark 3.4. Let R ⊆ S be a ring extension, and let M be the set
of all maximal ideals of R. Then by [7, Remark 5.5, p. 50], we have
Ab1 =
⋂
p∈M
A[p] =
⋂
p∈M
AR[p] = Ab2 . Hence Kr(b1) = Kr(b2).
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L. Paudel, S. Tchamna 119
Proposition 3.5. Let R ⊆ S be a ring extension, and let P be a set of
prime ideals of R. Let ⋆1 and ⋆2 be the two star operations on R ⊆ S defined
for each R-submodule A of S by A⋆1 =
⋂
p∈M
A[p] and A⋆2 =
⋂
p∈M
AR[p].
If the extension R ⊆ S is Prüfer, then Kr(⋆1) = Kr(⋆2).
Proof. Suppose that the extension R ⊆ S is Prüfer. Let f
g
∈ Kr(⋆1). Then
there exists a finitely generated S-regular R-submodule H of S such that
cR(f)H ⊆ (cR(f)H)⋆1 ⊆ (cR(g)H)⋆1 . Furthermore, (cR(g)H)S = S, since
cR(g)S = S and HS = S. Thus cR(g)H is an S-regular finitely generated
R-submodule of S. Therefore, by Proposition 2.11, we have (cR(g)H)⋆2 =
(cR(g)H)⋆1 . It follows from the relation cR(f)H ⊆ (cR(g)H)⋆1 that
cR(f)H ⊆ (cR(g)H)⋆2 . Therefore,
(cR(f)H)⋆2 ⊆ ((cR(g)H)⋆2)⋆2 = (cR(g)H)⋆2 .
Thus f
g
∈ Kr(⋆2). This shows that Kr(⋆1) ⊆ Kr(⋆2). But by Remark 3.3,
we have Kr(⋆2) ⊆ Kr(⋆1). Hence Kr(⋆1) = Kr(⋆2).
Remark 3.6. Let R ⊆ S be a ring extension, and let A, B be two
R-submodules of S. By definition of (A : B), we have (A : B)B ⊆ A.
Furthermore, the containment (A : B) ⊆ R implies that (A : B)B ⊆ B.
Thus (A : B)B ⊆ A ∩B.
Lemma 3.7. Let R ⊆ S be a ring extension, and let A be an R-submodule
of S. For each S-invertible R-submodule I of S, we have A∩ I = (A : I)I.
Proof. Let A be an R-submodule of S, and let I be an S-invertible R-
submodule of S. Let p be a prime ideal of R. By [7, Proposition 2.3, p. 97],
there exists wp ∈ I such that the Rp-module Ip is generated by
wp
1 . Let
x ∈ (A ∩ I)p = Ap ∩ Ip. Then x =
rwp
t
with r ∈ R and t ∈ R \ p. Since
rwp
t
∈ Ap, we have r
t
∈ (Ap :Rp
Ip). Furthermore, I is a finitely generated
R-submodule of S [7, Remark 1.10 (a)]. Hence (Ap :Rp
Ip) = (A :R I)p
[1, Corollary 3.15 p.43]. Thus r
t
∈ (A :R I)p. It follows that x =
rwp
t
∈
(A :R I)pIp = ((A :R I)I)p. This shows that (A ∩ I)p ⊆ ((A :R I)I)p. On
the other hand, the containment (A :R I)I ⊆ A∩I follows from Remark 3.6.
Hence ((A :R I)I)p ⊆ (A ∩ I)p. Therefore, ((A :R I)I)p = (A ∩ I)p. It
follows from [7, Lemma 1.1, p. 85] that A ∩ I = (A : I)I, since p was
chosen arbitrarily in R.
Remark 3.8. Let R ⊆ S be a tight ring extension, and let A be an
S-regular R-submodule of A. Then for each element x of A, there exists
an S-invertible R-submodule E of S such that such that x ∈ E ⊆ A.
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120 On the saturations of submodules
Proof. Let A be an S-regular R-submodule of S. Then AS = S. Therefore,
there exist a1, . . . , an ∈ A and s1, . . . , sn∈S such that a1s1+· · ·+ansn=1.
Let x ∈ A. Since the extension R ⊆ S is tight, there exists u ∈ S such
that xu ∈ R. Let E be the R-submodule of S generated by x, a1, . . . , an.
Then E is S-invertible, and E−1 is the R-submodule of S generated by
u, s1, . . . , sn. Furthermore, x ∈ E ⊆ A.
Remark 3.9. Let R ⊆ S be a tight ring extension, and let B be a subring
of S containing R. Let x ∈ B. If x ∈ R, then xR ⊆ R. Thus R ⊆ (R : x),
and hence (R : x)S = S. If x ∈ S \ R, then by the definition of a tight
extension, there exists a S-invertible ideal I of R such that xI ⊆ R. Hence
I ⊆ (R : x). Thus IS ⊆ (R : x)S. But by [7, Remarks 1.10(d), p. 90],
we have IS = S. Hence (R : x)S = S. It follows from [7, Theorem 3.13,
p. 37] and [7, Proposition 1.15, p. 91] that if A and B are two S-regular R-
submodules of S, then A∩B and (A : B) are also S-regular R-submodules
of S.
The following theorem is inspired by [2, Theorem 4], which is a result
due to D.D. Anderson. We consider a tight ring extension R ⊆ S and
we characterize star operations ⋆ on R ⊆ S satisfying the condition
(A ∩B)⋆ = A⋆ ∩B⋆ where A and B are S-regular R-submodule of S.
Theorem 3.10. Let R ⊆ S be a tight ring extension, and let P be a set
of prime ideals of R such that each S-regular ideal of R is contained in
some p ∈ P. Then for each star operation ⋆ on the ring extension R ⊆ S
satisfying R = R⋆, the following statements are equivalent.
(1) A⋆ =
⋂
p∈P
A[p] for each S-regular R-submodules A of R.
(2) (a) R =
⋂
p∈P
R[p].
(b) (A∩B)⋆ = A⋆ ∩B⋆ for all S-regular R-submodules A, B of R.
(3) (a) R =
⋂
p∈P
R[p].
(b) (A :R B)⋆ = (A⋆ :R B⋆) for all S-regular R-submodule A, B
of R with B finitely generated.
Proof. (1) ⇒ (2) Since R is an S-regular R-submodule of R, we have
R⋆ =
⋂
p∈P
R[p]. It follows that R =
⋂
p∈P
R[p], since by hypothesis, we
have R = R⋆. Let A and B be two S-regular R-submodules of S. Then
by Remark 3.9, A ∩ B is also an S-regular R-submodule of S. Thus by
hypothesis, we have (A∩B)⋆ =
⋂
p∈P
(A∩B)[p]. But by Proposition 2.1(2),
(A ∩B)[p] = A[p] ∩B[p]. Therefore,
(A∩B)⋆ =
⋂
p∈P
(A∩B)[p] =
⋂
p∈P
(A[p]∩B[p]) =
⋂
p∈P
A[p]∩
⋂
p∈P
B[p] = A⋆∩B⋆.
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L. Paudel, S. Tchamna 121
(2) ⇒ (1) Let A be an S-regular R-submodule of S. Then AS = S. It
follows that A⋆S = S, since A ⊆ A⋆. Let x ∈ A⋆. Then by Remark 3.8,
there exists an S-invertible R-submodule I of S such that x ⊆ I ⊆
A⋆. Since I ⊆ A⋆, we have I⋆ = A⋆ ∩ I⋆. But by hypothesis, we have
A⋆ ∩ I⋆ = (A ∩ I)⋆. Also, by Lemma 3.7, we have A ∩ I = (A : I)I. It
follows that I⋆ = ((A : I)I)⋆. But by [6, Proposition 4.1(b), p. 146], we
have ((A : I)I)⋆ = (A : I)⋆I. This shows that I⋆ = (A : I)⋆I, and so
I ⊆ (A : I)⋆I. By multiplying the last relation by I−1, we get R ⊆ (A : I)⋆.
Furthermore, by [7, Remarks 1.10(d)], I is an S-regular R-submodule of S.
It follows from Remark 3.9 that (A : I) is an S-regular ideal of R. If (A : I)
is a proper ideal of R, then by hypothesis, (A : I) ⊆ p0 for some p0 ∈ P of R.
In this case S = (A : I)S ⊆ p0S. Hence p0S = S. So p0 is an S-regular
ideal of R. It follows that p⋆0 =
⋂
p∈P
p0[p]. Therefore, as in the the proof
of Proposition 3.1(3), we have p0 = p⋆0. Hence R ⊆ (A : I)⋆ ⊆ p⋆0 = p0;
which is impossible. Thus R = (A : I). Let p ∈ P and let t ∈ R \ p. Then
t ∈ (A : I) = R. It follows that tI ⊆ A. In particular, tx ∈ A. Thus
x ∈ A[p]. This shows that A⋆ ⊆
⋂
[p∈P]A[p].
For the containment
⋂
[p∈P]A[p] ⊆ A⋆, let A′ =
⋂
p∈P
A[p]. Then
A ⊆ A′. It follows that A′S = S. Let y ∈ A′. Then by Remark 3.8,
there exists an S-invertible R-submodule J of S such that y ∈ J ⊆ A′.
Let p ∈ P. Then by Proposition 3.1(2), we have A′
[p] = A[p]. Thus
J ⊆ A′ ⊆ A′
[p] = A[p]. But J is a finitely generated R-submodule of S
[7, Remarks 1.10(a), p. 90 ]. So there exist u1, . . . , uℓ ∈ S such that
J = (u1, . . . , uℓ)R. For 1 6 i 6 ℓ, let ti ∈ R \ p such that tiui ∈ A, and let
t =
∏ℓ
i=1 ti. Then t ∈ R\p and tJ ⊆ A. It follows that (A : J)∩(R\p) 6= ∅.
Notice that J is an S-regular R-submodule of S, since each S-invertible
R-submodule of S is S-regular[7, Remarks 1.10(d), p. 90]. Therefore, by
Remark 3.9, we conclude that (A : J) is an S-regular ideal of R. It follows
from the hypothesis and the relation (A : J) ∩ (R \ p) 6= ∅ for each
p ∈ P, that (A : J) is not a proper ideal of R. Hence (A : J) = R, and
thus (A : J)⋆ = R⋆. Furthermore, J⋆ = (JR)⋆ = JR⋆ [6, Proposition 4.1,
p. 146]. Therefore, J⋆ = JR⋆ = J(A : J)⋆ = (J(A : J))⋆ = (J ∩ A)⋆. It
follows that y ∈ J ⊆ J⋆ ⊆ A⋆. This shows that A′ =
⋂
p∈P
A[p] ⊆ A⋆.
Therefore, A⋆ =
⋂
p∈P
A[p].
(1) ⇒ (3) We have R⋆ =
⋂
p∈P
R[p], since R is an S-regular R-
submodule of S. It follows that R =
⋂
p∈P
R[p], since by hypothesis,
we have R = R⋆. Let A and B be two S-regular R-submodules of S,
and suppose that B is finitely generated. Then by Remark 3.9, the
ideal (A : B) of R is also S-regular. Thus by the hypothesis, we have
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122 On the saturations of submodules
(A : B)⋆ =
⋂
p∈P
(A : B)[p]. It follows from Proposition 2.1(4) that
(A : B)⋆ =
⋂
p∈P
(A[p] : B[p]) ⊆ (
⋂
p∈P
A[p] :
⋂
p∈P
B[p]) = (A⋆ : B⋆).
Now we show that (A⋆ : B⋆) ⊆ (A : B)⋆. First notice that the contain-
ment (A⋆ : B⋆) ⊆ (A⋆ : B) is always true. Let d ∈ (A⋆ : B⋆). Then
for each p ∈ P, we have d ∈ (A⋆ : B) ⊆ (A⋆ : B)[p]. But by Proposi-
tion 2.1(4), we have (A⋆ : B)[p] = ((A⋆)[p] : B[p]). Thus d ∈ ((A⋆)[p] : B[p]).
Furthermore, A[p] ⊆ (A⋆)[p] = (
⋂
q∈P
A[q])[p] ⊆ (A[p])[p] = A[p]. Thus
(A⋆)[p] = A[p]. Hence d ∈ (A[p] : B[p]). It follows from Proposition 2.1(4)
that d ∈ (A : B)[p]. This shows that d ∈
⋂
p∈P
(A : B)[p] = (A : B)⋆. Hence
(A⋆ : B⋆) ⊆ (A : B)⋆, since d was chosen arbitrarily in (A⋆ : B⋆). Thus
(A⋆ : B⋆) = (A : B)⋆.
(3) ⇒ (1) Let A be an S-regular R-submodule of S. Let x ∈ A⋆. The
R-submodule A⋆ is S-regular, since A ⊆ A⋆. Therefore, by Remark 3.8,
there exits an S-invertible R-submodule I of S such that x ∈ I ⊆ A⋆. The
containment I ⊆ A⋆ implies I⋆ ⊆ A⋆. Thus R = (A⋆ : I⋆). Furthermore,
I is a finitely generated R-submodule of S [7, Remark 1.10 (a)]. Thus
R = (A⋆ : I⋆) = (A : I)⋆. Suppose that (A : I) is a proper ideal
of R. Then by hypothesis, (A : I) ⊆ p0 for some p0 ∈ P of R. Thus
S = (A : I)S ⊆ p0S. Hence p0S = S. So p0 is an S-regular ideal of R. It
follows that p⋆0 =
⋂
p∈P
p0[p]. Thus, as in the the proof of Proposition 3.1(3),
we have p0 = p⋆0. Hence R ⊆ (A : I)⋆ ⊆ p⋆0 = p0; which is impossible. Thus
R = (A : I). The rest of the proof is identical to the proof of (2) ⇒ (1).
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“adm-n3” — 2018/10/20 — 9:02 — page 123 — #129
L. Paudel, S. Tchamna 123
Contact information
Lokendra Paudel Department of Mathematics, The University of
Akron, Akron, OH 44325, USA
E-Mail(s): lpaudel1@uakron.edu
Simplice Tchamna Department of Mathematics, Georgia College &
State University, Campus Box 017,
Milledgeville, GA 31061, USA
E-Mail(s): simplice.tchamna@gcsu.edu
Received by the editors: 13.12.2016
and in final form 17.01.2017.
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