On Steinitz-like conditions

Peer review under responsibility of Taibah University.

Abstract

In this paper, we investigate the transfer of Steinitz, semi-Steinitz, and weakly semi-Steinitz properties from a commutative ring to its subring retracts and fixed rings.

MSC:

• 13D05
• 13D02

Keywords:

• Steinitz ring
• Semi-Steinitz ring
• Weakly semi-Steinitz ring
• Subring retracts rings
• Fixed rings

1 Introduction

All rings considered in this paper are commutative with identity element and all modules are unital. We say that a ring R is Steinitz if any linearly independent subset of a free R-module F can be extended to a basis of F by adjoining elements of a given basis. We say that an ideal I of R is T-nilpotent if for any sequence {a_i} of elements of I, there is an integer n such that ${\prod }_{i=1}^n{a}_i=0$. In [1, Theorem 2], Chwe and Neggers proved that a ring is Steinitz if and only if it is a local ring with T-nilpotent maximal ideal. On the other hand, in [2, Proposition 5.4], Cox and Pendleton showed that Steinitz rings are precisely the perfect local rings. Recall that a ring R is perfect if every flat R-module is projective [3].

A ring R is said to be semi-Steinitz if every linearly independent finite subset of a finitely generated free R-module F can be extended to a basis of F, by adjoining elements of a given basis of F. In [4, Theorem 2.1], Nashier and Nichols showed that a ring is semi-Steinitz if and only if it is local and satisfies the (CH)-property, i.e., every finitely generated proper ideal of R has a non-zero annihilator.

A ring R is weakly semi-Steinitz if every linearly independent finite subset of a finitely generated free R-module F can be extended to a basis of F. An R-module P is said to be stably free if there exists a finitely generated R-module F such that P ⊕ F is free. We say that R is a Hermite ring if every finitely generated stably free R-module is free. In particular, local and semilocal rings are Hermite (see [5, I.4.7, p. 26]). In [5, Corollary I.4.5, p. 26], Lam showed that R is a Hermite ring if and only if for every r₁, …, r_n in R such that ${\sum }_{i=1}^n{r}_{i}R=R$, the row [r₁, …, r_n] can be completed to an invertible square matrix. It is not difficult to see that a ring R is Hermite if and only if it is so modulo its Jacobson radical. In [4, Theorem 2.2], Nashier and Nichols proved that R is a weakly semi-Steinitz ring if and only if R is a Hermite ring and verifies the (CH)-property.

For two rings A ⊆ B, we say that A is a module retract (or a subring retract) of B if there exists an A-module homomorphism ϕ : B ⟶ A such that ϕ|_A = id|_A; ϕ is called a module retraction map. If such a map ϕ exists, B contains A as an A-module direct summand.

Let G be a group of automorphisms of a ring R. We say that G is a locally finite group if the cardinality n_x of the orbit G_x of any element x of R is finite, where G_x = {g(x)/g ∈ G}. We denote by R^G = {x ∈ R/g(x) = x, ∀ g ∈ G} the fixed subring of R. In [6], Bergman remarks that R^G is a module retract of R in the following two cases:

(1)	G is a finite group and |G|, the order of G, is a unit in R where the module retraction map ϕ is defined by ϕ(x) = |G|−1∑g∈Gg(x) for every x of R.

(2)	G is a locally finite group and nx, the order of the orbit of any element x of R, is a unit of R where the module retraction map ϕ is defined by $\varphi (x)=1/n_x{\sum }_{y\in G_x}y$ for every x of R.

Considerable works have been concerned with the descent and ascent of a variety of finiteness and related homological properties between R and R^G (see [6–10]). In [7], Glaz proves that a necessary condition that makes the investigations presented in most of that work effectively solvable is the existence of a module retraction map from R and R^G. We can easily remark that some results of that work can be extended to any subring retract of R. In this paper we investigate the descent and ascent of Steinitz, semi-Steinitz and weakly semi-Steinitz properties between a ring R and its module retracts, also between R and its fixed rings.

In [11], the author studied the transfer of Steinitz, semi-Steinitz and weakly semi-Steinitz in trivial extensions (see the definition in Section 2). Since any ring is a module retract of its trivial extensions, in this study we try to generalize some of the results presented in [11] to the module retracts. Then, we explore the conditions under which the Steinitz, semi-Steinitz and weakly semi-Steinitz properties are inherited from the module retracts and fixed rings. In a third section, we study the scopes and limits of the results achieved. Accordingly, we give some examples to show that the restrictions imposed to assure the descent of Steinitz, semi-Steinitz and weakly semi-Steinitz properties cannot be relaxed. We also prove that for two rings A ⊆ R, it is not possible – in general – to ascend Steinitz, semi-Steinitz or weakly semi-Steinitz from A to R.

2 Main results

We first study the transfer of Steinitz properties in module retracts and fixed rings.

Theorem 2.1

Let A ⊆ R be two rings. If R is a Steinitz ring then so is A in the following cases:

(1)	A is a module retract of R.

(2)	A = RG, where G is a locally finite group of automorphisms of R.

The proof of this theorem involves the following lemma:

Lemma 2.2

Let A ⊆ R be two rings such that A is a module retract of R.

(1)	If R is a local ring then so is A.

(2)	If R is a perfect ring then so is A.

Proof

(1)

Let M be the maximal ideal of R and x ∈ A ∖ (M ∩ A). Then x is an invertible element of R, i.e., there exists an element y of R such that xy = 1 and so xϕ(y) = 1, where ϕ is the module retraction map. Hence, y = ϕ(y) ∈ A which indicates that x is an invertible element of A. This means that A is a local ring with maximal ideal M ∩ A.

(2)

Let E be a flat A-module then E ⊗ AR is a flat R-module. E ⊗ AR is a projective R-module since R is a perfect ring. In addition, for any A-module N, we have:${\mathrm{Ext}}_A^1(E,N{\otimes }_{A}R)\cong {\mathrm{Ext}}_R^1(E{\otimes }_{A}R,N{\otimes }_{A}R)$(see [12, p. 118]) and then ${\mathrm{Ext}}_A^1(E,N{\otimes }_{A}R)=0$. As we note that N is a direct summand of N ⊗ AR since A is a direct summand of R, ${\mathrm{Ext}}_A^1(E,N)=0$ for all A-module N. This indicates that E is a projective A-module which in its turn proves that A is a perfect ring. □

Proof of Theorem 2.1

(1)	Follows immediately from Lemma 2.2.

(2)	R is an integral extension of RG since G is a locally finite group [13, Section 1, Proposition 22]. So, if M is the maximal ideal of R then RG is local with maximal ideal M ∩ RG [14, Corollary 3.2, p. 50] and also M ∩ RG is T-nilpotent since M is. This means that RG is a Steinitz ring. □

Let A be a ring, E be an A-module and R = A ∝ E be the set of pairs (a, e) with pairwise addition and multiplication given by: (a, e)(b, f) = (ab, af + be). R is called the trivial extension ring of A by E. Recall that R has always its Krull dimension equal to the Krull dimension of A [15, Theorem 25.1(3)]. The author of [15] confirmed in a private communication that [15, Theorem 25.1] is not true, i.e., an ideal J of R is not always of the form: J = I ∝ E′, where I = {a ∈ A/ (a, e) ∈ J for some e ∈ E} and E′ = {e ∈ E/ (a, e) ∈ J for some a ∈ A}. We only have that J ⊆ I ∝ E′ (see [16]). Nevertheless, it is easily seen that J = I ∝ E′ if and only if 0 ∝ E′ ⊆ J if and only if I ∝ 0 ⊆ J.

In [11, Theorem 2.1(1)], the author proves that, for any ring A and any A-module E, the Steinitz property descends from the trivial ring extension A ∝ E to A. A direct application of Theorem 2.1(1) gives the same result, as shown in the following corollary:

Corollary 2.3

Let A be a ring, E an A-module and R = A ∝ E the trivial extension ring of A by E. If R is Steinitz then so is A.

Proof

A is a module retract of R, where the module retraction map ϕ is defined by ϕ(x, e) = x. So, the result follows from Theorem 2.1(1). □

Corollary 2.4

Let R = A + I, where A is a subring of R and I is an ideal of R such that I ∩ A = 0. If R is Steinitz then so is A.

Remark 2.5

In [17, Theorem 2.1(1)], under the same hypothesis of Corollary 2.4, the authors prove that if the ring R = A + I is Hermite then so is A. In Examples 3.2 and 3.3, we show that the condition “I is an ideal of R” cannot be removed.

In the sequel we study the transfer of semi-Steinitz and weakly semi-Steinitz properties. Particularly, Theorem 2.6(3) and Theorem 2.9 generalize the results of [11, Theorem 2.5(2,a) and Proposition 2.10] to module retracts, as we shown in Corollaries 2.7 and 2.11.

Theorem 2.6

Let A ⊆ R be two rings. If R is a semi-Steinitz ring then so is A in the following cases:

(1)	A is a module retract of R and R is a flat A-module.

(2)	A = RG, where G is a locally finite group of automorphisms of R and R is a flat RG-module.

(3)	A is a module retract of R where ϕ is the module retraction map and either (a) Kerϕ is a submodule of a free A-module or (b) Kerϕ = A/P, where P is a prime ideal of A with non zero annihilator.

(4)	R is reduced and A = RG, where G is a locally finite group of automorphisms of R.

Proof

Assume that R is a semi-Steinitz ring. Let M be the maximal ideal of R. In either cases A = R^G or A is a module retract of R, A is local with maximal ideal M ∩ A by Theorem 2.1. Hence, what remains to prove is that A verifies the (CH)-property. Let I be a proper finitely generated ideal of A, if A = R^G or A is a module retract of R, we have IR ⊆ (M ∩ A)R ⊆ M ⊂ R i.e., IR is a proper finitely generated ideal of R and so Ann_R(IR) ≠ 0 since R verifies the (CH)-property.

In cases (1) and (2), we have Ann_R(IR) = Ann_A(I)R since R is a flat A-module [18, p. 40]. Thus, Ann_A(I) ≠ 0. (3)(a) Assume that Kerϕ is a submodule of a free A-module F. Let x be a non-zero element of R such that xIR = 0, particularly xI = 0. Two cases are then possible:

• Case 1: ϕ(x) ≠ 0. So, ϕ(x)I = 0, which implies that Ann_A(I) ≠ 0.

• Case 2: ϕ(x) = 0, i.e., x ∈ Kerϕ. By hypothesis, there exist a subset ${\{f_i\}}_{i=1,\dots ,n}$ of a basis of F and ${\{a_i\}}_{i=1,\dots ,n}\subseteq A$, such that $x={\sum }_{i=1}^n{a}_i{f}_i$ and then a_iI = 0 for each i ∈ {1, …, n}. Since x ≠ 0, there exists i₀ ∈ {1, …, n} such that $a_{i_0}\ne 0$ which also satisfies $a_{i_0}I=0$. Accordingly, Ann_A(I) ≠ 0.

(b) Assume that Kerϕ = A/P, where P is a prime ideal of A with non-zero annihilator. Let x be a non-zero element of R such that xIR = 0 and so xI = 0. Two cases are then possible:

• Case 1: ϕ(x) ≠ 0. So, ϕ(x)I = 0, which proves that Ann_A(I) ≠ 0.

• Case 2: ϕ(x) = 0, i.e., $x=\overline{a}$ where a ∉ P and (a + P)I = 0, then aI ⊆ P. Since a ∉ P, I ⊆ P and thus I has a non-zero annihilator since P also does.

(4) Assume that R is reduced and that A = R^G. Let x be a non-zero element of Ann_R(IR) with orbit G_x of cardinality n_x< ∞, we have xI = 0, so σ_i(x)I = 0 for all 1 ≤ i ≤ n_x, where ${\sigma }_i(x)={\sum }_{\stackrel{1\le j_1<\cdots <j_i\le n_x}{y_{j_k}\in G_x}}{y}_{j_1}\dots y_{j_i}$ is the elementary symmetric function of x. By construction, σ_i(x) ∈ R^G for any i, x also satisfies the equation$x^{n_x}-{\sigma }_1(x)x^{n_x-1}+\cdots +{(-1)}^{n_x}{\sigma }_{n_x}(x)=\prod _{y\in G_x}(x-y)=0(★)$If σ_j(x) = 0 for all 1 ≤ j ≤ n_x, using (★), xⁿ = 0 and then x = 0 since R is a reduced ring, a contradiction. Therefore, there is an i such that 1 ≤ i ≤ n_x and σ_i(x) ≠ 0 which satisfies σ_i(x)I = 0. This means that ${\mathrm{Ann}}_{R^G}(I)\ne 0$ and this completes the proof of Theorem 2.6. □

Corollary 2.7

Let A be a ring and R the trivial extension ring of A by an A-module E such that either (a) E is a flat A-module or (b) E is a submodule of a free A-module or (c) E = A/P, where P is a prime ideal of A with non-zero annihilator. If R is semi-Steinitz then so is A.

Proof

We can easily show that A is a module retract of R, where the module retraction map ϕ is defined by ϕ(x, e) = x. Hence, the results follow immediately from Theorem 2.6 (1 and 3). □

Remark 2.8

In [11, Theorem 2.1 and 2.5(1)], under the same hypothesis of Corollary 2.3, the author proves that Steinitz, semi-Steinitz and weakly semi-Steinitz properties ascend from A to the trivial ring extension A ∝ E. In Examples 3.4–3.6, we show that for two rings A ⊆ R such that A is a module retract of R, we can ascend neither the Steinitz nor the semi-Steinitz or the weakly semi-Steinitz properties from A to R even under the same conditions as in Theorem 2.6 (1 and 3).

In the sequel, we need the following Lemma.

Lemma 2.9

[4, Corollary 2.5] and [19, Corollary 83]

Let R be a reduced ring. If R is Noetherian then the following are equivalent.

(1)	R is semisimple.

(2)	R is weakly semi-Steinitz.

(3)	Any non-zero element of R is invertible.

(4)	R verifies the (CH)-property.

Theorem 2.10

Let A ⊆ R be two rings such that A is Noetherian retract of R, where ϕ is the module retraction map. If R is a weakly semi-Steinitz ring which satisfies ak ≠ 0, for any non-zero divisor a of A and any non-zero element k of Kerϕ, then A is a weakly semi-Steinitz ring.

Proof

Since A is Noetherian, we only have to show that any non-zero divisor of A is invertible in A by Lemma 2.9. Let x be a non-zero divisor element of A and r = a + k an element of R where a ∈ A and k ∈ Ker(ϕ), such that rx = 0 (recall that R = A ⊕ Ker(ϕ) since A is a module retract of R). Then ax = kx = 0 which implies that a = 0 since x is a non-zero divisor element of A; and k = 0 by hypothesis, the result is that r = 0. Therefore, since R is weakly semi-Steinitz, x is invertible in R as well as in A since A is a module retract of R. □

Corollary 2.11

Let A be a Noetherian ring, E an A-module and R the trivial extension ring of A by E. If R is a weakly semi-Steinitz ring which satisfies ae ≠ 0, for any non-zero divisor element a of A and any non-zero element e of E, then A is a weakly semi-Steinitz ring.

Theorem 2.12

Let R be a Noetherian ring and G a locally finite group. If R^G is weakly semi-Steinitz then so is R. The converse is true if R is reduced.

Proof of Theorem 2.11

Since R is Noetherian, it suffices to show that any non-zero divisor element of R is invertible in R by Lemma 2.9. Let x be a non-zero divisor element of R. Then g(x) is a non-zero divisor element of R for every g ∈ G. Let x₁ = x, x₂, …, x_n be the distinct elements of G_x and let y = x₁, …, x_n. Then y ∈ R^G and is a non-zero divisor element of R^G. Therefore, y is invertible in R^G, hence x = x₁ is invertible in R.

Conversely, assume that R is reduced. Then by Lemma 2.9 it is semisimple. In particular it is a Von Neumann regular ring and so is R^G by [8, Corollary 4]. Then R is a faithfully flat R^G-module by [7, Theorem 7] which implies that R^G is Noetherian. Accordingly, R^G is a semisimple ring. □

Remark 2.13

In Example 3.1 we show that the hypothesis “R reduced” cannot be removed in Theorem 2.12.

3 Examples

In this section, Example 3.1 shows that the hypothesis “R reduced” cannot be removed in Theorem 2.12 and that the hypotheses “G is a locally finite group and R^G is a module retract of R” are not sufficient to transfer the semi-Steinitz (resp., the weakly semi-Steinitz) property from R to R^G. In [17, Theorem 2.1(1)], the authors prove that for two rings A ⊆ R such that R = A + I and I is an ideal of R which satisfies I ∩ A = {0}, if R is Hermite then so is A. In Examples 3.2 and 3.3, we show that the hypothesis “I is an ideal of R” is necessary. We prove also that generally we cannot descend the weakly semi-Steinitz property from a ring R to a module retract A of R even if R is reduced and a faithfully flat A-module. In Examples 3.4–3.6 we prove that, in general, we can ascend neither the weakly semi-Steinitz nor the semi-Steinitz or the Steinitz properties from a ring R to its subring retracts and fixed rings.

Example 3.1

Let (A, M) be a valuation Noetherian domain such that its maximal ideal M does not contain 2 (e.g. A = (Z Z)_p where p ≠ 2). Let R = A ∝ (A/M) be the trivial extension ring of A by A/M and G = 〈σ〉 where σ is defined by σ(x, e) = (x, − e) for all (x, e) ∈ R. Then

(1)	R is Noetherian.

(2)	G is a finite group where |G| = 2 and RG = A is a module retract of R.

(3)	R is a semi-Steinitz ring.

(4)	RG is not weakly semi-Steinitz. In particular, RG is neither semi-Steinitz nor Steinitz.

Proof

(1)	Let J be an ideal of R, I = {a ∈ A/(a, e) ∈ J for some e ∈ E} and E′ = {e ∈ E/(a, e) ∈ J for some a ∈ A}. Then, two cases are possible: Case 1: Either I ∝ 0 ⊆ J or 0 ∝ E′ ⊆ J. As mentioned earlier, J = I ∝ E′ where I = aA and a ∈ A (since A is a valuation Noetherian domain), and either E′ = 0 or E′ = E. Clearly, J is a finitely generated ideal of R. Case 2: I ∝ 0 ⊈ J and 0 ∝ E′ ⊈ J. There exists an element a of A such that I = aA since A is a valuation Noetherian domain. There exists also an element e ∈ E′ such that (a, e) ∈ J. If e = 0 then I ∝ 0 = R(a, 0) ⊆ J, a contradiction. Thus, e ≠ 0, i.e., there exists an invertible element c in A such that $e=\overline{c}$, and $(c,0)({\mathrm{ac}}^{-1},\overline{1})=(a,\overline{c})\in J$, which implies that $({\mathrm{ac}}^{-1},\overline{1})\in J$ since (c, 0) is invertible in R. Hence, if we change a by ac−1 we find that $(a,\overline{1})\in J$. Our aim is to prove that $J=R(a,\overline{1})$. Let $(b,\overline{c})\in J$. Hence, b ∈ Aa, i.e., b = λa where λ ∈ A. Two cases are then possible: (1) $\overline{c}=\overline{0}$. If λ ∉ M, it is invertible in A, which implies that I ∝ 0 = (Aa) ∝ 0 = (Aλa) ∝ 0 = R(b, 0) ⊆ J, a contradiction. Therefore, λ ∈ M and so $(b,\overline{0})=(\lambda a,\overline{0})=(\lambda ,\overline{0})(a,\overline{1})\in R(a,\overline{1})$. (2) $\overline{c}\ne \overline{0}$, i.e., c is an invertible element in A. Assume that λ ∈ M, then $(b,\overline{c})=(\lambda a,\overline{c})=(c,\overline{0})(\lambda c^{-1}a,\overline{1})\in J$. So, $(\lambda c^{-1}a,\overline{1})\in J$ since $(c,\overline{0})$ is invertible in R, which implies that $((1-\lambda c^{-1})a,\overline{0})\in J$ since $(a,\overline{1})\in J$. Therefore, since (1 − λc−1) ∉ M, I ∝ 0 ⊆ J (the same proof as in (1), we only replace λ by 1 − λc−1), which in its turn is a contradiction. Hence, λ ∉ M and then $(b,\overline{c})=(\lambda a,\overline{c})=(\lambda ,0)(a,\overline{{\lambda }^{-1}c})$, which implies that $(a,\overline{{\lambda }^{-1}c})\in J$ since (λ, 0) is invertible in R. In addition, since $(a,\overline{1})\in J$, $(0,\overline{1-{\lambda }^{-1}c})\in J$. If we note that $\overline{1}\ne \overline{{\lambda }^{-1}c}$, then $0\propto E^{\prime }=0\propto E=R(0,\overline{1-{\lambda }^{-1}c})\subseteq J$, a contradiction. Hence, $\overline{{\lambda }^{-1}c}=\overline{1}$ and $(b,\overline{c})=(\lambda ,0)(a,\overline{{\lambda }^{-1}c})=(\lambda ,0)(a,\overline{1})\in R(a,\overline{1})$. This means that $J=R(a,\overline{1})$, and this completes the proof.
(2)	RG = A since 2 ∉ M. It is also a module retract of R since R is a trivial extension ring of A.
(3)	R is a semi-Steinitz ring since A is local (see [20, Proposition 5]).
(4)	RG = A is not a weakly semi-Steinitz ring since it is a domain which is not a field. So, it is neither semi-Steinitz nor Steinitz. □

Next, we show that we cannot descend the weakly semi-Steinitz and Hermite properties from a ring R to its subring retract even if R is a faithfully flat A-module (Example 3.2) or R is reduced (Example 3.3).

Let A be a ring and f(X) a polynomial of A[X]. Let c(f(X)) the ideal of A generated by the coefficients of f(X). Recall that the localization A(X) = S⁻¹A[X] is called the Nagata ring, where S is the multiplicative set in A[X] formed by the polynomials ring f(X) such that C(f(X)) = A.

Example 3.2

Let A be a non Hermite ring (e.g. A = IR[X, Y, Z]/(X² + Y² + Z² − 1)), E the A-module oplus _m∈Max(A)A/m and R the trivial extension ring of A by E.

(1)	R is not a weakly semi-Steinitz ring [20, Proposition 5].

(2)	R is a module retract of R(X) since R(X) = R ⊕ XR[X] ⊕ B, where B = {(f(x)/g(x))/f(x), g(x) ∈ R[X], dof(X) < dog(x) and C(g(x)) = R} [5, Proposition 1.4(1)].

(3)	R(X) is a faithfully flat R-module. Indeed, by construction, R(X) is a flat R[X]-module and R[X] is a free R-module. So, R(X) is a flat R-module and, since R is a direct summand of R(X), R(X) is a faithfully flat R-module.

(4)	R(X) satisfies the (CH)-property since so does R (see [19, Proposition 116]). Also, it is a Hermite ring by [21, Theorem 2.11 and Lemma 3.1]. Therefore, it is a weakly semi-Steinitz ring.

Let D be a reduced ring and P = {p_∝, ∝ ∈ Ω} ⊆ Spec(D). Let I = Ω × I N. For any i = (∝ , n), set p_i = p_∝ and D_i = D/p_i. Let B = oplus _i∈ID_i and $\varphi :D⟶\prod _{i\in I}{D}_i$ defined by $\varphi (d)={\{d+p_i\}}_{i\in I}$. Let R = A + B, where A = ϕ(D). R is a reduced subring of $\prod _{i\in I}{D}_i$ and A ∩ B = {0} since D is reduced (see [15, p. 169]).

Example 3.3

Let D be a reduced non-Hermite ring, P = Max(D) and R = A + B.

(1)	R is not Hermite. Indeed, suppose that R is Hermite, then it is not difficult to show that A also is. In addition, Kerϕ = Rad(D), so D/(Rad(D)) ≅ A. D is then Hermite which contradicts the above hypothesis. Hence, R is not a weakly semi-Steinitz ring.

(2)	R satisfies the (CH)-property (see [15, Theorem 26.4 and 27.1] and [20, Section 2, Rem 1]). As in Example 3.2(4), we prove that R(X) is a weakly semi-Steinitz ring.

(3)	R(X) is a faithfully flat R-module and it is reduced since R is reduced. Example 3.2(2) proves also that R is a module retract of R(X).

In the sequel, we give some examples concerning the ascent of Steinitz, semi-Steinitz and weakly semi-Steinitz properties.

Example 3.4

Let A be a Steinitz ring and A[X] the polynomial ring. A is a module retract of A[X] since A[X] = A ⊕ XA[X]. In addition, XA[X] ≅ A^IN which means that it is a free A-module. Nevertheless, A[X] is not weakly semi-Steinitz since Ann_A[X](X) = 0. So, it is neither semi-Steinitz nor Steinitz.

In Theorem 2.12, we have proved that, for G a locally finite group of automorphisms of a ring R, the Noetherianity is sufficient to ascend the weakly semi-Steinitz property from the fixed ring R^G to R. In the following example we prove that under the same hypothesis it is not possible to ascend neither the semi-Steinitz nor the Steinitz properties from R^G to R.

Example 3.5

Let R be a Noetherian Steinitz ring, n ∈ IN and $G=\{{\sigma }_{{\pi }_i}/1\le i\le n\}$ such that for any 1 ≤ i ≤ n, π_i ∈ S_n defined as follows$\left\{\begin{array}{cc}{\pi }_i(k)=i+k\hfill & \mathsf{if}i+k\le n\hfill \\ {\pi }_i(k)=i+k-n\hfill & \mathrm{if}i+k>n\hfill \end{array}\right.$(π_n = id) and ${\sigma }_{{\pi }_i}(x_1,\dots ,x_n)=(x_{{\pi }_i(1)},\dots ,x_{{\pi }_i(n)})$, for any element (x₁, …, x_n) ∈ Rⁿ. G is a finite group of automorphisms of R of cardinality n.

(1)	Rn is Noetherian but not semi-Steinitz since it is not local.

(2)	${(R^n)}^G=R$ is Steinitz by hypothesis, it is also a module retract of Rn, which is a faithfully flat R-module.

Example 3.6

Let A be a Steinitz ring and R = A × (A/P) the direct product of A by A/P, where P is a prime ideal of A. We can see easily that A is a module retract of R. R is not weakly semi-Steinitz since ${\mathrm{Ann}}_A(1,\overline{a})=0$, for any non-zero element a of A (since A/P is a domain). Hence, R is not weakly semi-Steinitz. In particular, it is neither semi-Steinitz nor Steinitz.

Acknowledgements

The authors would like to express their sincere thanks for the referee for his/her helpful suggestions and comments, which have greatly improved this paper.

Notes

Peer review under responsibility of Taibah University.