DOI 10.1007/s10958-015-2335-3 Journal of Mathematical Sciences, Vol. 206, No. 5, May, 2015

EXTENSIONS OF AUTOMORPHISMS OF SUBMODULES A. A. Tuganbaev

UDC 512.55

Abstract. We study modules M such that all automorphisms of submodules in M can be extended to endomorphisms (automorphisms) of M .

All rings are assumed to be associative and with nonzero identity element; all modules are assumed to be unitary. A module M is said to be Noetherian if every properly ascending chain of submodules in M is finite. Expressions of the form “A is a Noetherian ring” mean that AA and A A are Noetherian modules. A module M is said to be automorphism-extendable [17] if for every submodule X in M , every automorphism of the module X can be extended to an endomorphism of the module M . A module M is said to be skew-injective if for every submodule X in M , every endomorphism of the module X can be extended to an endomorphism of the module M . Remark 1. It is clear that all direct summands of automorphism-extendable (skew-injective) modules are automorphism-extendable (respectively, skew-injective) modules, and every skew-injective module is automorphism-extendable. It follows from Examples 2.1, 2.6, and 2.8 in the present paper that an automorphism-extendable module MA is not necessarily skew-injective, even if M = A is a simple principal right (left) ideal domain, or M = A is a commutative regular ring, or A is a finite algebra over a field and M is a finite cyclic projective A-module. For two modules X and M , the module M is said to be injective with respect to X or X-injective if for every submodule X1 in X, each homomorphism X1 → M can be extended to a homomorphism X → M . A module M over the ring A is said to be injective if M is injective with respect to each A-module. A module M is said to be quasi-injective or self-injective if M is injective with respect to M . In [3, 9, 13], a module M is said to be automorphism-invariant if M is a characteristic submodule in the injective hull of M . Remark 2. It is clear that every injective module is quasi-injective, and every quasi-injective module is automorphism-extendable (skew-injective). It follows from Proposition 1.3 of the present paper that for an automorphism-invariant module M , every automorphism of any submodule of M can be extended to an automorphism of M ; in particular, M is automorphism-extendable. Every simple Abelian group M (for example, we can take M = Z/2Z) is a finite automorphism-invariant (quasi-injective) noninjective module over the ring Z of integers. In addition, Z is an example of an automorphism-extendable (skew-injective) Z-module that is not automorphism-invariant (quasi-injective). For a module E, a submodule M in E is said to be essential if M ∩ E1 = 0 for every nonzero submodule E1 in E. If E is an injective module and M is an essential submodule in E, then E is called the injective hull of the module M . The injective hull is unique up to isomorphism. Let A be a ring, and let M be a right A-module. We denote by Sing(M ) the submodule in M consisting of all elements of the module M whose annihilators in A are essential right ideals. A module M is said to be nonsingular (singular ) if Sing(M ) = 0 (respectively, Sing(M ) = M ). A ring A is said to be prime if the product of any two nonzero ideals of A does not equal zero. The main results of the present paper are Theorems 1–4. Translated from Fundamentalnaya i Prikladnaya Matematika, Vol. 18, No. 3, pp. 179–198, 2013. c 2015 Springer Science+Business Media New York 1072–3374/15/2065–0583 

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Theorem 1. Let A be a prime ring with the maximum condition on left annihilators, and let M be a right A-module that is not singular. The following conditions are equivalent: (1) M is an automorphism-extendable module; (2) every automorphism of an arbitrary submodule in M can be extended to an automorphism of the module M ; (3) either M is an injective module, or M = X ⊕ U , where X is an injective singular module and U is an automorphism-extendable nonsingular uniform nonzero module. A module M is said to be hereditary if all submodules of M are projective. A ring A is said to be bounded if every essential right or left ideal of A contains a nonzero ideal. It is proved in [10] that every hereditary Noetherian prime ring is either bounded or right and left primitive. Theorem 2. Let A be a bounded hereditary Noetherian prime ring, and let M be a right A-module. The following conditions are equivalent: (1) M is an automorphism-extendable module; (2) every automorphism of an arbitrary submodule of the module M can be extended to an automorphism of M ; (3) either M is a quasi-injective singular module, or M is an injective module that is not singular, or M = X ⊕ U , where X is an injective singular module and U is an automorphism-extendable nonsingular uniform nonzero module. A ring A is said to be right (left) invariant if every right (respectively, left) ideal of A is an ideal. A module M with injective hull E is said to be completely integrally closed if the following two equivalent conditions hold (the equivalence is proved in [16]): (1) for every submodule X in M , each homomorphism X → M that maps into itself some essential submodule in X can be extended to a homomorphism M → M . (2) f (M ) ⊆ M for every endomorphism f of the module E that maps into itself some essential submodule of the module M . Theorem 3. Let A be an invariant hereditary Noetherian domain, and let M be a right A-module. The following conditions are equivalent: (1) M is an automorphism-extendable module; (2) every automorphism of an arbitrary submodule of the module M can be extended to an automorphism of the module M ; (3) M is a skew-injective module; (4) M is a completely integrally closed module; (5) either M is a quasi-injective singular module, or M is an injective module that is not singular, or M = X ⊕ U , where X is an injective singular module and U is a nonsingular uniform nonzero module. Remark 3. In connection with Theorems 2 and 3, we note that there exists a bounded hereditary Noetherian domain A such that nonsingular uniform modules are not necessarily automorphism-extendable (see Example 2.4). In addition, Z is an automorphism-extendable (skew-injective) Z-module with injective hull QZ , and Z is not a characteristic submodule (for a module E, a submodule M in E is said to be characteristic if f (M ) ⊆ M for every automorphism f of the module E) of its injective hull Q, since f (Z) ⊆ Z, where f : q → q/2 is an automorphism of the Z-module Q. A ring A is called a right Goldie ring if A does not contain an infinite direct sum of nonzero right ideals and A is a ring with the maximum condition on right annihilators. Let A be a ring, and let M be a right A-module. We denote by t(M ) the set of all elements in M that are annihilated by some non-zero-divisors of the ring A. A module M is called a nontorsion (torsion, torsion-free) module if t(M ) = M (respectively, t(M ) = M , t(M ) = 0). 584

Remark 4. In [3, Theorem 16], it is proved that a module M is an automorphism-invariant module if and only if M is a pseudo-injective module, i.e., for any submodule X in M , every monomorphism X → M can be extended to an endomorphism of the module M . In [6, Theorem 6], it is proved that every pseudo-injective torsion-free module over a prime right Goldie ring is injective. In connection with Remark 4, we prove Theorem 4. Theorem 4. If A is a prime ring with the maximum condition on left annihilators and M is an automorphism-invariant right A-module that is not singular, then the module M is injective. The proofs of Theorems 1–4 are decomposed into a series of assertions, some of which are of independent interest. If X is an essential submodule of the module M , then M is called an essential extension of the module X. A submodule X of the module M is said to be closed in M if X = X  for every submodule X  in M that is an essential extension of the module X. A module M is said to be uniform if any two nonzero submodules of M have the nonzero intersection. A ring A is said to be regular if a ∈ aAa for every element a ∈ A. A module M is said to be distributive if X ∩ (Y + Z) = X ∩ Y + X ∩ Z for any submodules X, Y , and Z of M . A module M is said to be semi-Artinian if every nonzero factor module of M contains a simple submodule. A module M is said to be uniserial if any two submodules of M are comparable with respect to inclusion. Direct sums of uniserial modules are called serial modules. Other definitions and basic results used in this paper are contained, e.g., in [15, 18]. 1. Proof of Theorems 1–4 Remark 1.1. Let M be a module, and let X be a submodule in M . It is well known that by Zorn’s lemma, there exists at least one submodule Y in M such that X ∩ Y = 0 and X ⊕ Y is an essential submodule in M ; any such submodule Y is called a ∩-complement to X in M . Lemma 1.2. Let M be a module, E be the injective hull of M , X be a submodule in M , and let Y be an arbitrary ∩-complement to the module X in the module M . We denote by X  the essential submodule X ⊕ Y in M . (1) Every automorphism (endomorphism) f of the module X can be extended to an automorphism (respectively, the endomorphism) f  of the essential submodule X  in M by using the relation f  (x + y) = f (x) + y, where x ∈ X and y ∈ Y . In turn, the automorphism (endomorphism) f  of the module X  can be extended to an automorphism (respectively, the endomorphism) α of the injective module E. (2) Every homomorphism g : X → M with essential in X kernel K can be extended to a homomorphism g  : X  → M with essential in M kernel K  = K ⊕ Y , by using the relation g  (x + y) = g(x), where x ∈ X and y ∈ Y . In turn, the homomorphism g  can be extended to an endomorphism h of the injective module E with essential in E kernel, and 1E − h is an automorphism of the module E such that 1E − h coincides with the identity automorphism of the module K  on the essential submodule K  of the module M . Proof. Since X  is an essential submodule in M and M is an essential submodule in E, we have that X  is an essential submodule in E. (1) It is directly verified that f  is an automorphism (endomorphism) of the essential submodule X  of the module E. Since the module E is injective, the automorphism (respectively, endomorphism) f  of the module X  can be extended to an endomorphism α of E. We assume that f  is an automorphism of the module X  . Since X  is an essential submodule in E and X ∩ Ker(α) = 0, we have that α is a monomorphism, and the module α(E) is injective. Therefore, α(E) is a direct summand in E. In addition, X  = f  (X  ) = α(X  ), whence the module α(E) contains the essential submodule X  of the module E. Therefore, α(E) = E and α is an automorphism. 585

Let f  be an endomorphism (automorphism) of the module X  , and let Y be a submodule in M that is maximal among submodules of the module M that have the zero intersection with X  . Then X = X  ⊕ Y is an essential submodule in M . Now we can define an endomorphism (respectively, an automorphism) f of the module X such that f (x + y) = f (x ) + y for any x ∈ X  and y ∈ Y . (2) It is directly verified that g  : X  → M is a homomorphism with essential in M kernel K  = K ⊕ Y . Since the module E is injective, the homomorphism g  can be extended to an endomorphism h of the module E. Since the module Ker(h) contains the essential submodule K  of M , we have that Ker(h) is an essential submodule in E. Then the restriction of the endomorphism 1E − h of the injective module E to Ker(h) is the identity automorphism of the essential submodule Ker(h) of the module E. Then 1E − h is an essential injective submodule of E. Therefore, 1E − h is an automorphism of the module E such that 1E − h coincides with the identity automorphism of the module K  on the essential submodule K  of the module M . Proposition 1.3. Let M be a module, and let E be the injective hull of M . The following conditions are equivalent: (1) M is an automorphism-extendable module, and for every essential submodule Y in M , each homomorphism Y → M with essential in Y kernel can be extended to an endomorphism of the module M with essential in M kernel ; (2) every automorphism of an arbitrary submodule X of M can be extended to an automorphism of the module M and for every submodule Y in M , each homomorphism Y → M with essential in Y kernel can be extended to an endomorphism of M with essential in M kernel ; (3) α(M ) ⊆ M for every automorphism α of the module E such that α(X) = X for some essential submodule X of the module M ; (4) α(M ) = M for every automorphism α of the module E such that α(X) = X for some essential submodule X of the module M . Proof. The implications (4) =⇒ (3) and (2) =⇒ (1) are directly verified. (1) =⇒ (3). Let α be an automorphism of the module E, and let α(X) = X for some essential submodule X of the module M . We denote by Y the submodule   α−1 M ∩ α(M ) = {y ∈ M | α(y) ∈ M } in M . Then α(Y ) ⊆ M , X ⊆ Y , and Y is an essential submodule in M . In addition, α induces the automorphism ϕ1 of the module X. Since M is an automorphism-extendable module, ϕ1 can be extended to an endomorphism ϕ2 of the module M . Since the module E is injective, ϕ2 can be extended to an endomorphism ϕ of E. We denote by g the restriction of the homomorphism α − ϕ to Y . We have ϕ(Y ) ⊆ M , α(Y ) ⊆ M , and g(X) = 0. Therefore, g is a homomorphism from Y into M with essential in Y kernel. By assumption, g can be extended to an endomorphism g1 of M . Since the module E is injective, g1 can be extended to an endomorphism β of E. Then (α − ϕ − β)(Y ) = (g − β)(Y ) = 0. We denote by Z the submodule {z ∈ M | (α − ϕ − β)(z) ∈ M } of M . Then Z is the complete pre-image in M of the module M ∩ (α − ϕ − β)(M ) under the action of the homomorphism α − ϕ − β, Y ⊆ Z, and α(Z) ⊆ (α − ϕ − β)(Z) + (ϕ + β)(Z) ⊆ M. Therefore, Y ⊆ Z ⊆ Y and Z = Y . Then (α − ϕ − β)(Z) = (α − ϕ − β)(Y ) = 0. If (α − ϕ − β)(M ) = 0, then α(M ) = (ϕ − β)(M ) ⊆ M , which is required. We assume that (α − ϕ − β)(M ) = 0. Since M is an essential submodule in E, we have that M ∩ (α − ϕ − β)(M ) is an essential submodule of the nonzero module (α − ϕ − β)(M ). Since Z is the complete pre-image in M of the nonzero module M ∩(α−ϕ−β)(M ) under the action of the homomorphism α − ϕ − β, we have that (α − ϕ − β)(Z) = 0. This is a contradiction. (3) =⇒ (4). Let X be an essential submodule of the module M , and let α be an automorphism of the module E with α(X) = X. It follows from (3) that α(M ) ⊆ M and α−1 (M ) ⊆ M . Then α(M ) = M . 586

(4) =⇒ (2). Let X be a submodule of the module M , and let ϕ be an automorphism of the module X. By Lemma 1.2(1) and the property that M is an essential submodule in E, we can assume that X is an essential submodule in E. By Lemma 1.2(1), the automorphism ϕ of the module X can be extended to an automorphism α of the injective module E. By assumption, α(M ) = M . Therefore, ϕ can be extended to an automorphism of the module M . Let h1 : Y → M be a homomorphism with essential in Y kernel K1 . By Lemma 1.2(2), we can assume that Y and K1 are essential submodules in E. The homomorphism h1 can be extended to an endomorphism h of the injective module E, and h has the essential kernel K. We denote by α the endomorphism 1E − h of the module E. Then the restriction of the endomorphism 1E − h to K is the identity automorphism of the essential submodule K of the injective module E. Therefore, Ker(1E − h) = 0, and (1E − h)(E) is an injective essential submodule in E. Therefore, 1E − h is an automorphism of the module E and (1E − h)(K) = K. By (4), (1E − h)(M ) = M . Therefore, (1E − h)|M is an automorphism of the module M . Then 1M − (1E − h)|M is an endomorphism of the module M with essential in M kernel, and 1M − (1E − h)|M coincides with h1 on Y . Proposition 1.4. Let M be a module, E be the injective hull of the module M , and let M = T ⊕ U , where T is an injective module and U is a nonsingular module. The following conditions are equivalent: (1) M is an automorphism-extendable module; (2) every automorphism of an arbitrary submodule X of the module M can be extended to an automorphism of the module M ; (3) α(M ) ⊆ M for every automorphism α of the module E such that α(X) = X for some essential submodule X of the module M . Proof. The implication (3) =⇒ (2) follows from Proposition 1.3. The implication (2) =⇒ (1) is obvious. (1) =⇒ (3). Let Y be an essential submodule in M , h : Y → M be a homomorphism with essential in Y kernel, and let π : M = T ⊕ U → U be the projection with kernel T . Then the module πh(Y ) is singular, and it is contained in the nonsingular module U . Therefore, πh(Y ) = 0. Then h(Y ) ⊆ T and h is a homomorphism from the module Y into the module T . Since the module T is injective, h can be extended to a homomorphism M → T ⊆ M . This homomorphism is the required endomorphism of the module M that extends h. Corollary 1.5. Let M be a nonsingular module, and let E be the injective hull of M . The following conditions are equivalent: (1) M is an automorphism-extendable module; (2) every automorphism of an arbitrary submodule X of the module M can be extended to an automorphism of the module M ; (3) α(M ) ⊆ M for every automorphism α of the module E such that α(X) = X for some essential submodule X of the module M . Corollary 1.5 follows from Proposition 1.4 and the property that the zero module is injective. Proposition 1.6. Let M = T ⊕U , where T is an injective module, and let U be a nonsingular module such that Hom(T  , U ) = 0 for every submodule T  of the module T . The following conditions are equivalent: (1) M is an automorphism-extendable module; (2) every automorphism of an arbitrary submodule X of the module M can be extended to an automorphism of the module M ; (3) U is an automorphism-extendable module; (4) every automorphism of an arbitrary submodule X of the module U can be extended to an automorphism of the module U . Proof. The equivalence (1) ⇐⇒ (2) follows from Proposition 1.4. The implication (1) =⇒ (3) is directly verified. 587

The equivalence (3) ⇐⇒ (4) follows from Corollary 1.5. (3) =⇒ (1). Let E be the injective hull of the module M , X be an essential submodule in M , and let α be an automorphism of the module E with α(X) = X. By Proposition 1.4, it is sufficient to prove that α(M ) ⊆ M . For the injective hull E of the module M = T ⊕ U , there exists a direct decomposition E = T ⊕ U1 , where U1 is the injective hull of the nonsingular module U . Since α is an automorphism of the module T ⊕ U1 and Hom(T  , U ) = 0 for every submodule T  of the module T , it is directly verified that α(T ) = T . Let h : E → E/T be the natural epimorphism. (We can assume that h is the projection α1 from the module E = T ⊕ U1 onto the module U1 with kernel T .) Then α induces the automorphism  of the injective hull h(E) of the module h(U ). Since α(X) = X, we have α1 h(X) = h(X). By applying 1.5 to the automorphism-extendable nonsingular module h(M ) = h(U ), we obtain  Corollary  that α1 h(M ) ⊆ h(M ). Then α(M ) ⊆ M + T = M . Corollary 1.7. Let M = T ⊕U , where T is an injective module that is an essential extension of a singular module, and U is a nonsingular module. The following conditions are equivalent: (1) M is an automorphism-extendable module; (2) every automorphism of an arbitrary submodule X of the module M can be extended to an automorphism of the module M ; (3) U is an automorphism-extendable module; (4) every automorphism of an arbitrary submodule X of the module U can be extended to an automorphism of the module U . Proof. Since the module T is an essential extension of a singular module, we have that Hom(T  , U ) = 0 for every submodule T  of the module T . Therefore, Corollary 1.7 follows from Proposition 1.6. Lemma 1.8. Let M be a module, and let E be the injective hull of M . (1) M is a completely integrally closed module if and only if the module M is skew-injective and h(M ) ⊆ M for every endomorphism h of the module E with essential in E kernel. (2) If the module M is nonsingular, then M is skew-injective if and only if M is a completely integrally closed module. Proof. (1). The assertion is known; e.g., see [15, Assertion 10.16(3)]. (2). The assertion follows from (1) and the property that the essential extension E of the nonsingular module M does not have nonzero endomorphisms with essential kernels. Proposition 1.9. Let M = T ⊕ U , where T is an injective module, U is a nonsingular module, and Hom(T  , U ) = 0 for every submodule T  of the module T . The following conditions are equivalent: (1) M is a skew-injective module; (2) M is a completely integrally closed module; (3) U is a skew-injective module; (4) U is a completely integrally closed module. Proof. The implication (2) =⇒ (1) follows from Lemma 1.8(1). The implication (1) =⇒ (3) is directly verified. The equivalence (3) ⇐⇒ (4) follows from Lemma 1.8(2). (4) =⇒ (2). Let E be the injective hull of the module M , X be an essential submodule of M , and let α be an endomorphism of the module E with α(X) ⊆ X. By Lemma 1.8(1), it is sufficient to prove that α(M ) ⊆ M . For the injective hull E of the module M = T ⊕ U , there exists a direct decomposition E = T ⊕ U1 , where U1 is the injective hull of the nonsingular module U . Since α is an endomorphism of the module T ⊕U1 and Hom(T  , U ) = 0 for every submodule T  of the module T , it is directly verified that α(T ) ⊆ T . Let h : E → E/T be the natural epimorphism. (We can assume that h is the projection from the module E = T ⊕ U1 onto the module U1 with kernel T .) Then α induces  the endomorphism α1 of the injective hull h(E) of the module h(U ). Since α(X)⊆ X, we have α1 h(X) ⊆ h(X). Since h(M ) = h(U ) is a completely integrally closed module, α1 h(M ) ⊆ h(M ). Then α(M ) ⊆ M + T = M . 588

Lemma 1.10. Let A be a ring, M be an automorphism-extendable right A-module, and let X and Y be two submodules in M with X ∩ Y = 0. (1) If f : Y → X is a homomorphism, then there exists an endomorphism g of the module M that coincides with f : Y → X on Y . (2) If M = X ⊕ Y , then the module X is injective with respect to Y . (3) If the module M/X is nonsingular, then the module X is injective with respect to Y . Proof. (1). We define an endomorphism α of the module X ⊕ Y by the relation α(x + y) = x + f (y) + y for all x ∈ X and y ∈ Y . We assume that 0 = α(x + y) = x + f (y) + y,

x ∈ X, y ∈ Y.

Then α is a monomorphism, since y = −x − f (y) ∈ X ∩ Y = 0,

f (y) = 0,

x = x + f (y) + y = α(x + y) = 0. In addition, we have that for any x ∈ X and y ∈ Y ,     x + y = x − f (y) + (f (y) + y) = α x − f (y) + α(y) ∈ α(X ⊕ Y ). Therefore, α is an automorphism of the module X ⊕ Y . In addition, the endomorphism α − 1X⊕Y of the module X ⊕ Y coincides with the homomorphism f : Y → X on Y . Since M is an automorphism-extendable module, the automorphism α of the module X ⊕ Y can be extended to an endomorphism β of the module M . We denote by g the endomorphism β − 1M of the module M . Then g coincides with f on Y . (2). Let Y1 be a submodule in Y , and let f1 be a homomorphism from Y1 into X. By (1), there exists an endomorphism g of the module M that coincides with f1 : Y1 → X on Y1 . Let π be the projection from the module M = X ⊕ Y onto X with kernel Y , and let u : Y → M be the natural embedding. We denote by f the homomorphism πgu from Y into X. Then f coincides with f1 on Y1 . Therefore, the module X is injective with respect to Y . (3). Let Y1 be a submodule in Y , and let f1 be a homomorphism from Y1 into X. By Zorn’s lemma, there exists a submodule Z in Y such that Y1 ∩ Z = 0 and Y1 ⊕ Z is an essential submodule in Y . We set Y2 = Y1 ⊕ Z. The homomorphism f1 : Y1 → X can be extended to a homomorphism f2 : Y2 → X with the use of the relation f2 (y1 + z) = f1 (y1 ). By (1), there exists an endomorphism g of the module M that coincides with f2 : Y2 → X on Y2 . It remains to prove that g(y) ∈ X for every element y of the module Y . Let h : M → M/X be the natural epimorphism. Since Y2 is an essential submodule in Y , we have that yB ⊆ Y2 for some essential right ideal B of the ring A (see, e.g., [15, 4.4(1)]). Then g(y)B = g(yB) ⊆ g(Y2 ) = f2 (Y2 ) = f1 (Y1 ) ⊆ X,   h g(y) B = h(g(y)B) ⊆ h(X) = 0.

  Therefore, h g(y) ∈ Sing(M/X) = 0 and g(y) ∈ Ker(h) = X.

Remark 1.11. If A is a right nonsingular ring, then every essential extension of an arbitrary singular right A-module is a singular module (see, e.g., [4, Proposition 1.23]). Remark 1.12. Let A be a ring with the maximum condition on left annihilators. It is well known (see, e.g., [15, Assertion 4.27]) that A is a ring with the minimum condition on right annihilators, and every subset B in A contains a finite set {b1 , . . . , bn } such that r(B) = r({b1 , . . . , bn }) = r(b1 ) ∩ · · · ∩ r(bn )}. In addition, if A is a semiprime ring with the maximum condition on left annihilators, then the ring A is right and left nonsingular (see, e.g., [15, Proposition 4.28]). Lemma 1.13. Let A be a prime ring with the maximum condition on left annihilators, and let M be a right A-module that is not singular. (1) M is a finitely faithful module. 589

(2) Every right A-module that is injective with respect to M is an injective module. In particular, if the module M is quasi-injective, then the module M is injective. (3) If M is an automorphism-extendable module, then M = X ⊕ U , where X is an injective singular module, U is a nonsingular nonzero module, and either U is a nonuniform injective module (in this case, the module M is injective), or U is an automorphism-extendable nonsingular uniform nonzero module. (4) If M is an automorphism-invariant module, then the module M is injective. Proof. By Remark 1.12, the ring A is right nonsingular. By Remark 1.11, all essential extensions of any singular right A-module are singular. In addition, the module M is not singular, by assumption. Therefore, M is not an essential extension of a singular module. Then there exist two closed submodules X and Y in M such that X is a singular essential submodule in X, Y is a nonsingular nonzero module, X ∩ Y = 0, and X ⊕ Y is an essential submodule in M . (1). Let y be a nonzero element of the nonsingular module Y . Then the right annihilator r(y) of the element y is not an essential right ideal of the ring A. Therefore, B ∩ r(y) = 0 for some nonzero right ideal B. Since the ring A is prime, the right annihilator r(B) is equal to the zero. By Remark 1.12, B contains a finite subset {b1 , . . . , bn } such that r(B) = r({b1 , . . . , bn }) = 0. Then r(b1 ) ∩ · · · ∩ r(bn )} = r(B) = 0. A module B n contains the submodule X = b1 A ⊕ · · · ⊕ bn A, which is isomorphic to the module AA /r(b1 ) ⊕ · · · ⊕ AA /r(bn ). Therefore, there exists a monomorphism AA → X ⊆ B n ⊆ M n and M is a finitely faithful module. (2). It is known that for any ring R, every R-module that is injective with respect to some finitely faithful module is injective (see, e.g., [15, Theorem 2.15]). Therefore, (2) follows from (1). (3). There exists a submodule X1 in M such that X ⊆ X1 and X1 /X = Sing(M/X). We assume that X1 properly contains the closed submodule X of the module M . Then X2 = X1 ∩ Y is a nonzero nonsingular submodule in X1 , X2 ∩ X = 0. Therefore, the singular module X1 /X contains a submodule that is isomorphic to the nonzero nonsingular module X2 . This is a contradiction. Therefore, X = X1 and the module M/X is nonsingular. By Lemma 1.10(3), the module X is injective with respect to the nonzero nonsingular module Y . By (2), the module X is injective. Therefore, there exists a direct decomposition M = X ⊕ U , where X is an injective module that is an essential extension of the singular module Sing(M ) of the module M , and U is an automorphism-extendable nonsingular nonzero module. By Remark 1.11, the module X is singular. If the module U is uniform, then there is nothing to prove. It remains to consider the case where the module U is not uniform. Then there exist two nonzero closed submodules U1 and U2 of the nonsingular module such that U1 ∩ U2 = 0 and U1 ⊕ U2 is an essential submodule in U . The modules U/U1 and U/U2 are nonsingular. By Lemma 1.10(3), the nonzero nonsingular modules U1 and U2 are injective with respect to each other. By (2), the modules U1 and U2 are injective. Then U1 ⊕ U2 is an injective essential submodule in U . Therefore, U = U1 ⊕ U2 is an injective module. (4). By Remarks 2 and 4, the automorphism-invariant module M is a pseudo-injective automorphism-extendable module. By (3), we can assume that M = X ⊕ U , where X is an injective module and U is a pseudo-injective nonsingular uniform module. By [1, Lemma 3.5], the pseudo-injective uniform module U is quasi-injective. By (2), the quasi-injective nonsingular A-module U is injective. Since M = X ⊕ U , the module M is injective. Lemma 1.14. Let M be a module, and let E be the injective hull of M . (1) The module M is quasi-injective if and only if M is a fully invariant submodule in E. (2) If M is a quasi-injective module, then every automorphism of an arbitrary submodule of M can be extended to an automorphism of the module M . Proof. (1). The assertion is known; see [7]. (2). The assertion follows from (1) and Proposition 1.3.

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Lemma 1.15. Let M be a module, and let M =

 i∈I

Mi , where all Mi are essential quasi-injective sub-

modules in M . Then M is a quasi-injective module.

Proof. Let E be the injective hull of the module M . Since all Mi are essential submodules in M , all Mi are essential submodules of the injective module E. Therefore, E is the injective hull of the module Mi for every i ∈ I. Let f be an endomorphism of the module E. Since all modules Mi are quasi-injective, it follows from Lemma 1.14(1) that f (Mi ) ⊆ Mi for every i ∈ I. Then     Mi = f (Mi ) ⊆ Mi = M. f (M ) = f i∈I

i∈I

i∈I

By Lemma 1.14(1), the module M is quasi-injective. Lemma 1.16. Let M be a module M that is the direct sum of quasi-injective modules Mi , i ∈ I.  Mj is Mi -injective, then M is a quasi-injective module. (1) If for every i ∈ I the module j=i

(2) If M is an automorphism-extendable module, then M is a quasi-injective module. Proof. (1). The assertion is known (see, e.g., [11, Proposition 1.18]).  Mj for every i ∈ I, it follows (2). Since M is an automorphism-extendable module and M = Mi ⊕ j = i  from Lemma 1.10(2) that for every i ∈ I, the module Mj is Mi -injective. By (1), M is a quasi-injective j=i module. Lemma 1.17. Let A be a serial Artinian ring. (1) Every A-module is a direct sum of uniserial modules of finite length. (2) Every automorphism-extendable A-module is quasi-injective. Proof. (1). The assertion is known (see, e.g., [18, 55.16]).  Mi , where all Mi (2). Let M be an automorphism-extendable A-module. By (1), the module M = i∈I

are uniserial modules of finite length. We fix i ∈ I. By Lemma 1.16(2), it is sufficient to prove that the module Mi is quasi-injective. Let Ei be the injective hull of the uniform module Mi . Then Ei is an indecomposable module. By (1), E is a uniserial module of finite length. Therefore, all submodules of the module Ei form a finite chain. Then Mi is a fully invariant submodule in Ei . By Lemma 1.14(1), the module Mi is quasi-injective. Lemma 1.18. Let A be a hereditary Noetherian prime ring, and let M be an automorphism-extendable A-module. (1) For every nonzero ideal B of the ring A, the factor ring A/B is a serial Artinian ring. (2) If r(M ) = 0, then the module M is quasi-injective. (3) If B is a nonzero ideal of the ring A and X = {m ∈ M | mB = 0}, then X is a quasi-injective module. (4) Let {Bi } i∈I be some set of nonzero ideals of the ring A, and let Xi = {m ∈ M | mBi = 0}, i ∈ I. Xi and all Xi are essential submodules in M , then M is a quasi-injective module. If M = i∈I

(5) If A is a bounded ring, then every singular right A-module T is semi-Artinian. Proof. (1). The assertion is known; see [2]. (2). We denote by R the proper factor ring A/r(M ) of the ring A. By (1), R is a serial Artinian ring. Since M is an automorphism-extendable A-module, M is an automorphism-extendable R-module. By Lemma 1.17(2), M is a quasi-injective R-module. Therefore, M is a quasi-injective A-module. (3). Let f be an endomorphism of the module M . Then f (X)B = f (XB) = f (0) = 0,

f (X) ⊆ X. 591

Therefore, X is a fully invariant submodule of the automorphism-extendable module M . Therefore, X is an automorphism-extendable A-module with nonzero annihilator. By (2), X is a quasi-injective module. (4). By (3), all the modules Mi are quasi-injective. By Lemma 1.15, the module M is quasi-injective. (5). Since every factor module of a singular module is singular, it is sufficient to prove that for every nonzero element t ∈ T , the cyclic module tA contains a simple submodule. The essential right ideal r(t) of the bounded ring A contains some nonzero ideal B. By (1), the factor ring A/B is Artinian. Then tA is a homomorphic image of the Artinian A-module A/B. Therefore, tA is a nonzero Artinian module and tA contains a simple submodule. Lemma 1.19. Let A be a hereditary Noetherian prime ring, {Bj }j∈J be the set of all proper invertible ideals of the ring A, {Pi }i∈I be the set of all maximal elements of the set {Bj }j∈J , and let M be a torsion right A-module. For every i ∈ I, we denote by Mi the submodule in M consisting of all elements in M that are annihilated by some power of the ideal Pi .  Xi for every submodule X in M , where Xi = X ∩Mi , i ∈ I. In addition, Hom(Xi , Xj ) = 0 (1) X = i∈I

for any distinct subscripts i, j ∈ I. (2) The module M is automorphism-extendable (quasi-injective) if and only if all modules Mi are automorphism-extendable (respectively, quasi-injective). (3) The module M is automorphism-extendable if and only if M is quasi-injective. In this case, every automorphism of an arbitrary submodule of M can be extended to an automorphism of M . Proof. (1). The assertion is known; see [12]. (2). The assertion is directly verified with the use of (1). (3). It follows from Lemma 1.14(2) that, in any quasi-injective module M , every automorphism of an arbitrary submodule can be extended to an automorphism of M . Conversely, we assume that M is an automorphism-extendable module. It follows from (2) that, without loss of generality, it is sufficient to consider the case in which M = Mi for some i ∈ I. We set ∞ ∞  P = Pi and Xn = {m ∈ M | xP n = 0}, n = 1, 2, . . . . Then Xn ⊆ Xn+1 for every n, M = Xn = Xn , n=1

n=1

and Xn is an essential submodule in M for every n. By Lemma 1.18(4), M is a quasi-injective module. Lemma 1.20. Let A be a semiprime right Goldie ring, and let M be a right A-module. (1) Sing(M ) = t(M ). In particular, M is nonsingular (singular ) if and only if M is a torsion-free (respectively, torsion) module. (2) Any essential extension of each singular right A-module is a singular torsion module. (3) If there exists a submodule X in M such that the modules X and M/X are singular, then the module M is singular. (4) Sing(M ) is a closed submodule in M and the module M/ Sing(M ) is nonsingular. Proof. (1). The assertion is proved in [5, Proposition 6.9]. (2). The assertion follows from (1) and Remark 1.11. (3). The assertion is proved in [5, Proposition 6.10]. (4). The assertion follows from (1) and (2). Proposition 1.21. Let A be a bounded hereditary Noetherian prime ring, and let M be a right A-module. The following conditions are equivalent: (1) M is an automorphism-extendable module; (2) every automorphism of an arbitrary submodule of the module M can be extended to an automorphism of the module M ; (3) either M is a quasi-injective torsion module, or M is an injective nontorsion module, or M = X ⊕ U , where X is an injective torsion module and U is an automorphism-extendable uniform nonzero torsion-free module. 592

Proof. By Lemma 1.20, the module M is nonsingular (singular) if and only if M is a torsion-free (respectively, torsion) module, and any essential extension of any singular right A-module is a singular torsion module. If M is a torsion module, then the equivalences (1) ⇐⇒ (2) and (2) ⇐⇒ (3) follow from Lemma 1.19(3). We assume that M is a nontorsion module. By Lemma 1.20, the module M is not an essential extension of a singular module. Then the equivalences (1) ⇐⇒ (2) and (2) ⇐⇒ (3) follow from Lemma 1.13(3), Corollary 1.7, and Lemma 1.20. Remark 1.22. Let A be a right Ore domain, Q be the classical right division ring of fractions of A, and let U be a nonzero right A-module. It is well known that QA is the injective hull of the module AA . In addition, the module U is a uniform torsion-free module if and only if U is isomorphic to a submodule of the module QA . In the last case, QA is the injective hull of the module U . Lemma 1.23. Let A be a right Ore domain, Q be the right classical division ring of fractions of A, and let M be a nontorsion right A-module. The following conditions are equivalent: (1) M is an automorphism-extendable module; (2) every automorphism of an arbitrary submodule of M can be extended to an automorphism of the module M ; (3) either M is an injective nontorsion module, or M = X ⊕U , where X is an injective torsion module and U is an automorphism-extendable module that is isomorphic to the nonzero submodule in QA . Proof. Lemma 1.23 follows from Lemma 1.13(3), Corollary 1.7, and Remark 1.22. Proposition 1.24. Let A be an invariant hereditary domain, and let M be a right A-module. The following conditions are equivalent: (1) M is an automorphism-extendable module; (2) every automorphism of an arbitrary submodule of the module M can be extended to an automorphism of the module M ; (3) M is a skew-injective module; (4) M is a completely integrally closed module; (5) either M is a quasi-injective torsion module, or M is an injective nontorsion module, or M = X ⊕ U , where X is an injective torsion module and U is a uniform nonzero torsion-free module. Proof. Since A is a hereditary uniform domain, A is a Noetherian domain (see, e.g., [15, Proposition 4.34]). The implication (4) =⇒ (3) follows from Lemma 1.8(1). The implication (3) =⇒ (1) is true in the general case. The equivalence (1) ⇐⇒ (2) and the implication (1) =⇒ (5) follow from Proposition 1.21. (5) =⇒ (4). Let Q be the classical division ring of fractions of the invariant hereditary Noetherian domain A. The module QA is distributive by [15, Theorem 11.17]. By Proposition 1.9, Lemma 1.20, and Remark 1.22, it is sufficient to prove that an arbitrary nonzero submodule U of the module QA is skew-injective. By Remark 1.22, QA is the injective hull of the module U . Let X be an essential submodule in U , and let f be an endomorphism of the module with f (X) ⊆ X. It is sufficient to prove that f (U ) ⊆ U . Let h : Q → Q/X be the natural epimorphism. Since f (X) ⊆ X, we have that f induces the endomorphism f¯ of the module h(Q). By Lemma 1.18(5), the singular module h(Q) is semi-Artinian. Since the module QA is distributive, the module h(Q) is distributive. Since h(Q) is a distributive semi-Artinian module, all submodules of h(Q) are fully invariant in h(Q) (see, e.g., [14, Corollary 2 to Theorem 4.1]).   Therefore, f¯ h(U ) ⊆ h(U ). Therefore, f (U ) ⊆ U + X = U , which is required. Remark 1.25 (the completion of the proof of Theorems 1–4). Theorem 1 follows from Lemma 1.13(3) and Corollary 1.7. Theorem 2 follows from Lemma 1.20 and Proposition 1.21. Theorem 3 follows from Lemma 1.20 and Proposition 1.24. Theorem 4 follows from Lemma 1.13(4). 593

2. Examples and Remarks Example 2.1. Let F be a field, and let A be the algebra over F with two generators x and y and one defining relation xy − yx = 1. We prove that AA and A A are the automorphism-extendable modules, and the modules AA and A A are not skew-injective. It is well known that A is a simple principal right (left) ideal domain, A is not a division ring, and the group U (A) of invertible elements of the domain A coincides with F \ 0. In particular, U (A) is contained in the center of the domain A and A is a hereditary Noetherian prime ring. Let a be a nonzero noninvertible element of the domain A. Since the definition of the ring A is left-right symmetrical, it is sufficient to prove the following two assertions. (∗) For every automorphism α of the module aAA , there exists an invertible element u ∈ U (A) such that α(ab) = uab for all b ∈ A. (∗∗) There exists an endomorphism f of the module aA that cannot be extended to an endomorphism of the module AA . (∗). Since α(aA) = aA, we have that α(a) = au and a = auv for some elements u, v ∈ A. Then uv = 1. Since A is a domain, we have that vu = 1 and u ∈ U (A) ⊂ F . Then uab = aub = α(ab) for all b ∈ A. (∗∗). Since A is a simple domain and a is a nonzero noninvertible element, we have AaA = A = Aa. Therefore, ab ⊆ Aa for some element b ∈ A. Since A is a Noetherian domain, A has the classical division ring of fractions, which contains the element a−1 . Then aba−1 aA ⊆ aA and the relation f (ac) = aba−1 c defines an endomorphism f of the submodule aAA in AA . We assume that f can be extended to an endomorphism ϕ of the module AA . We set d = ϕ(a). Then ab = aba−1 a = f (a)a = ϕ(a)a = da ∈ Aa. This is a contradiction. Remark 2.2. Let A be a principal right ideal domain, and let U be the group of invertible elements of A. The following conditions are equivalent: (1) AA is an automorphism-extendable module; (2) every automorphism of an arbitrary submodule of the module AA can be extended to an automorphism of the module AA ; (3) aU ⊆ U a for every element a ∈ A. Proof. The implication (2) =⇒ (1) is directly verified. (1) =⇒ (3). We have to prove that au ∈ U a for any two elements a ∈ A and u ∈ U . Without loss of generality, we can assume that a = 0. We denote by ϕ the mapping from aA into A such that ϕ(ab) = aub for every element b ∈ A. Since aA = auA and A is a domain, ϕ is an automorphism of the module aA. Since AA is an automorphism-extendable module, the automorphism ϕ can be extended to some endomorphism f of the module AA . We set v = f (1) ∈ A. Since vA = A and A is a domain, we have v ∈ U . Then va = f (a) = au and aU ⊆ U a. (3) =⇒ (2). Let X be a submodule in AA , and let ϕ be an automorphism of X. Since A is a principal right ideal domain, X = aA for some element a ∈ X. Without loss of generality, we can assume that a = 0. Since aA = ϕ(a)A, there exist elements u, w ∈ A such that ϕ(a) = au and a = auw. Since A is a domain, 1 = uw and u ∈ U . By assumption, aU ⊆ U a. Therefore, au = va for some v ∈ U . We denote by f automorphism of the module AA such that f (b) = vb for all b ∈ A. Then f (a) = va = au = ϕ(a). Therefore, the automorphism f is an extension of the automorphism ϕ. Remark 2.3. Let D be a noncommutative division ring. Then D[x] is a principal right (left) ideal domain that is not an automorphism-extendable right or left D[x]-module. In addition, if the division ring D is finite-dimensional over its center F , then D[x] is a bounded hereditary Noetherian prime ring. Proof. It is well known that D[x] is a principal right (left) ideal domain, and the group U of invertible elements of D[x] coincides with multiplicative group of the division ring D. In particular, D[x] is a bounded 594

hereditary Noetherian prime ring. We assume that D[x]D[x] is an automorphism-extendable module. By assumption, dd1 = d1 d for some nonzero elements d and d1 of the division ring D. By Remark 2.2, (d + x)d1 ⊆ U d. In addition, U = D \ 0. Therefore, (d + x)d1 = d2 (d + x) for some element d2 of the division ring D. Then d1 x = d2 x and dd1 = d2 d. Therefore, dd1 = d1 d. This is a contradiction. Similarly, we obtain that the module D[x] D[x] is also not automorphism-extendable. We assume that D is finite-dimensional over its center F . It is well known that for every polynomial f ∈ D[x], there exists a polynomial g ∈ D[x] such that f g is a nonzero polynomial in F [x]; see, e.g., [8, 16.9]. Then f g is a nonzero central element of the domain D[x], which is contained in the principal right (left) ideal domain f D[x]. Therefore, D[x] is a bounded hereditary Noetherian prime ring. Example 2.4. Let H be the division ring of Hamiltonian quaternions, and let R be the field of real numbers. Since the noncommutative division ring H is finite-dimensional over its center R, it follows from Remark 2.3 that H[x] is a bounded principal right (left) ideal domain that is not an automorphism-extendable right or left D[x]-module. In particular, H[x] is a bounded hereditary Noetherian prime ring that is not an automorphism-extendable right or left D[x]-module. Remark 2.5. Let A be a regular ring. If the module AA is skew-injective, then the module AA is injective by [15, Theorem 10.47]. ∞ } be a countable set of copies of the field Z/2Z, and let A be the subring of Example 2.6. Let {Fi=1 the direct product of all Fi consisting of all sequences that stabilize at finite step. Then A is a commutative regular ring. In [3, Example 9], it is proved that A is an automorphism-invariant module that is not quasi-injective. By Remarks 2 and 2.5, AA is an automorphism-extendable module that is not skew-injective.

Remark 2.7. By [15, Proposition 10.22(1)], every semi-Artinian skew-injective module is quasi-injective. In particular, all finite skew-injective modules are quasi-injective. Example 2.8. Let F be a field of order 2, and let A be the 5-dimensional F -algebra over the field F consisting of all 3 × 3 matrices of the form ⎞ ⎛ f11 f12 f13 ⎝ 0 f22 0 ⎠ , 0 0 f33 where fij ∈ F . In [13], it is proved that e11 A = e11 F +e12 F +e13 F is a finite cyclic automorphism-invariant projective module that is not quasi-injective. By Remark 2 at the beginning of the paper and Remark 2.7, e11 A is an automorphism-extendable module that is not skew-injective. Open question 2.9. In Theorems 1–3, each considered automorphism-extendable module M satisfies the property that all automorphisms of submodules in M can be extended to automorphisms of M . Is this true for every automorphism-extendable module M ? The author is supported by the Russian Foundation for Basic Research, project 08-01-00693-a. REFERENCES 1. A. Alahmadi, N. Er, and S. K. Jain, “Modules which are invariant under monomorphisms of their injective hulls,” J. Aust. Math. Soc., 79, No. 3, 2265–2271 (2005). 2. D. Eisenbud and P. Griffith, “Serial rings,” J. Algebra, 17, 389–400 (1971). 3. N. Er, S. Singh, and A. K. Srivastava, “Rings and modules which are stable under automorphisms of their injective hulls,” J. Algebra, 379, 223–229 (2013). 4. K. R. Goodearl, Ring Theory, Marcel Dekker, New York (1976). 5. K. R. Goodearl and R. B. Warfield, An Introduction to Noncommutative Noetherian Rings, Cambridge Univ. Press, Cambridge (1989). 595

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