Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 43 (2002), No. 1, 111-119.

Ikeda-Nakayama Modules^{∗}

Robert Wisbauer Mohamed F. Yousif Yiqiang Zhou

Heinrich-Heine-University, 40225 Düsseldorf, Germany e-mail: wisbauer@math.uni-duesseldorf.de

The Ohio State University, Lima Campus, Ohio 45804, USA e-mail: yousif.1@osu.edu

Memorial University of Newfoundland, St.John’s, NF A1C 5S7, Canada e-mail: zhou@math.mun.ca

Abstract. Let _{S}M_{R} be an (S, R)-bimodule and denote l_{S}(A) = {s ∈ S : sA = 0} for any submodule A of M_{R}. Extending the notion of an Ikeda-Nakayama ring, we investigate the condition l_{S}(A∩B) = l_{S}(A)+l_{S}(B) for any submodules A, B of M_{R}. Various characterizations and properties are derived for modules with this property. In particular, for S = End(M_{R}), the π-injective modules are those modules M_{R} for which S = l_{S}(A) + l_{S}(B) whenever A ∩ B = 0, and our techniques also lead to some new results on these modules. MSC 2000: 16D50 (primary); 16L60 (secondary)

1. Annihilator conditions

Let R and S be rings and _{S}M_{R} be a bimodule. For any X ⊆ M and any T ⊆ S, denote

l_{S}(X) = {s ∈ S : sX = 0} and r_{M}(T) = {m ∈ M : T m = 0}.

There is a canonical ring homomorphism λ : S ⟶ End(M_{R}) given by λ(s)(x) = sx for x ∈ M and s ∈ S. For any submodules A and B of M_{R} and any t ∈ l_{S}(A ∩ B), define

α_{t} : A + B → M, a + b ↦ ta.

Clearly, α_{t} is a well-defined R-homomorphism.

∗The research was supported in part by NSERC of Canada and a grant from Ohio State University 0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 2002 Heldermann Verlag

Lemma 1. Let _{S}M_{R} be a bimodule and A, B be submodules of M_{R}. The following are equivalent:

(1) l_{S}(A ∩ B) = l_{S}(A) + l_{S}(B).

(2) For any t ∈ l_{S}(A ∩ B), the diagram

0 → A + B → M
↓ α_{t}
M

can be extended commutatively by λ(s), for some s ∈ S.

Proof. (1) ⇒ (2). Suppose (1) holds. For A, B, t given as in (2), write t = u + v where u ∈ l_{S}(A) and v ∈ l_{S}(B). Then, for all a ∈ A and b ∈ B,

α_{t}(a + b) = ta = (u + v)a = va = v(a + b) = λ(v)(a + b).

(2) ⇒ (1). It is clear that l_{S}(A ∩ B) ⊇ l_{S}(A) + l_{S}(B). Let t ∈ l_{S}(A ∩ B). Define α_{t} : A + B ⟶ M as above. By (2), there exists s ∈ S such that λ(s) extends α_{t}.

Thus, for all a ∈ A and b ∈ B, ta = α_{t}(a + b) = λ(s)(a + b) = s(a + b). It follows that (t − s)a + (−s)b = 0 for all a ∈ A and b ∈ B. So, t − s ∈ l_{S}(A) and −s ∈ l_{S}(B), and hence t = (t − s) − (−s) ∈ l_{S}(A) + l_{S}(B). □

Lemma 2. Let _{S}M_{R} be a bimodule and A, B be submodules of M_{R} such that A∩B = 0. The following are equivalent:

(1) S = l_{S}(A) + l_{S}(B).

(2) The diagram

0 → A + B → M
↓ α_{1}
M

can be extended commutatively by λ(s), for some s ∈ S.

Proof. (1) ⇒ (2). Apply Lemma 1 with t = 1.

(2) ⇒ (1). It suffices to show that 1 ∈ l_{S}(A) + l_{S}(B). Note that α_{1} : A + B ⟶ M is given by α_{1}(a + b) = a (a ∈ A and b ∈ B). By (2), there exists s ∈ S such that λ(s) extends α_{1} . Arguing as in the proof of ‘(2) ⇒ (1)’ of Lemma 1, we have 1 = (1 − s) − (−s) ∈ l_{S}(A) + l_{S}(B). □

Lemma 3. Let _{S}M_{R} be a bimodule such that _{S}M is faithful and A, B be complements of each other in M_{R}. The following are equivalent:

(1) S = l_{S}(A) + l_{S}(B).

(2) S = l_{S}(A) ⊕ l_{S}(B).

(3) M = A ⊕ B and, for the projection f of M onto A along B, f = λ(s) for some s ∈ S.

Proof. (1) ⇒ (3). By (1), we have S = l_{S}(A) + l_{S}(B). Write 1_{S} = u + v where u ∈ l_{S}(A) and v ∈ l_{S}(B). It follows that a = va for all a ∈ A, b = ub for all b ∈ B and vB = uA = 0. Thus, B ⊆ r_{M}(v) ⊆ r_{M}(v^{2}) and r_{M}(v^{2}) ∩ A = 0. Since B is complement of A in M_{R}, we have B = r_{M}(v) = r_{M}(v^{2}). Similarly, A = r_{M}(u) = r_{M}(u^{2}). Next we show that (vu)M ∩ (A + B) = 0. For any z ∈ (vu)M ∩ (A + B), write z = vux = a + b, where x ∈ M, a ∈ A and b ∈ B. Noting that vu = uv, we have that (v^{2}u^{2})x = (vu)(a + b) = 0. So, u^{2}x ∈ r_{M}(v^{2}) = r_{M}(v), and this gives that u^{2}vx = vu^{2}x = 0. So, vx ∈ r_{M}(u^{2}) = r_{M}(u). Thus, z = vux = uvx = 0. So, (vu)M ∩ (A + B) = 0. Since A + B is essential in M_{R}, (vu)M = 0, and hence vu = 0 since _{S}M is faithful. So, uM ⊆ r_{M}(v) = B and vM ⊆ r_{M}(u) = A, and hence M = vM + uM = A + B = A ⊕ B.

Let f be the projection of M onto A along B. Then f(M) = A and (1 − f)(M) = B. Noting that _{S}M is faithful, we have l_{S}(A) = l_{S}(f(M)) = {s ∈ S : λ(s)f(M) = 0} = {s ∈ S : λ(s)f = 0} and l_{S}(B) = l_{S}((1 − f)(M)) = {s ∈ S : λ(s)(1 − f) = 0}. Thus, λ(u)f = 0 and λ(v)(1 − f) = 0. It follows that

0 = λ(v)(1 − f) = λ(1 − u)(1 − f) = (1 − λ(u))(1 − f) = 1 − f − λ(u),

and thus f = 1 − λ(u) = λ(1 − u) = λ(v).

(3) ⇒ (2). By (3), M = A ⊕ B. Let f be the projection of M onto A along B. Then f^{2} = f ∈ End(M_{R}), A = f(M) and B = (1 − f)(M). By (3), f = λ(s) for some s ∈ S. It follows that (s^{2} − s)M = λ(s^{2} − s)(M) = (f^{2} − f)(M) = 0. So, s^{2} = s, since _{S}M is faithful. And so,

l_{S}(A) = l_{S}(f(M)) = l_{S}(sM) = l_{S}(s) = S(1 − s),

and, similarly, l_{S}(B) = Ss. Thus, S = l_{S}(A) ⊕ l_{S}(B).

(2) ⇒ (1). Obvious. □

A module M_{R} is called π-injective (or quasi-continuous) if every submodule is essential in a direct summand (C1) and, for any two direct summands M_{1}, M_{2} with M_{1} ∩M_{2} = 0, M_{1} ⊕M_{2} is also a direct summand (C3) (see [8]). It is known that M_{R} is π-injective if and only if M = A ⊕ B whenever A and B are complements of each other in M_{R} (see [8, Theorem 2.8]).

Corollary 4. Let _{S}M_{R} be a bimodule such that _{S}M is faithful. The following are equivalent:

(1) For any submodules A and B of M_{R} with A ∩ B = 0, S = l_{S}(A) + l_{S}(B).

(2) If A and B are complements of each other in M_{R}, then S = l_{S}(A) + l_{S}(B).

(3) If A and B are complements of each other in M_{R}, then S = l_{S}(A) ⊕ l_{S}(B).

(4) M is π-injective and, for any f^{2} = f ∈ End(M_{R}), f = λ(s) for some s ∈ S.

Proof. (1) ⇔ (2) is obvious, and (2) ⇔ (3) ⇔ (4) is by Lemma 3. □

For submodules A, B of M_{R}, let

π : M/(A ∩ B) → M/A ⊕ M/B, m + (A ∩ B) ↦ (m + A, m + B)

be the canonical R-homomorphism. The next lemma can easily be verified.

Lemma 5. Let M_{R} be an R-module with S = End(M_{R}) and A, B be submodules of M_{R}. The following are equivalent:

(1) l_{S}(A ∩ B) = l_{S}(A) + l_{S}(B).

(2) For any R-homomorphism f : M/(A ∩ B) ⟶ M, the diagram

0 → M/(A ∩ B) →^{π} M/A ⊕ M/B
↓ f
M

can be extended commutatively by some g : M/A ⊕ M/B ⟶ M.

2. Ikeda-Nakayama modules

A well known result of Ikeda and Nakayama [6] says that every right self-injective ring R satisfies the so called Ikeda-Nakayama annihilator condition, i.e., l_{R}(A ∩ B) = l_{R}(A) + l_{R}(B) for all right ideals A, B of R. Rings with the Ikeda-Nakayama annihilator condition, called right Ikeda-Nakayama rings, were studied in [2]. Extending this notion we call M_{R} an Ikeda-Nakayama module (IN-module) if

l_{S}(A ∩ B) = l_{S}(A) + l_{S}(B)

for any submodules A and B of M_{R} where S = End(M_{R}). Clearly, every quasi-injective module is an IN-module (Lemma 1) and every IN-module is π-injective (Corollary 4).

Proposition 6. The following are equivalent for a module M_{R} with S = End(M_{R}):

(1) M_{R} is an IN-module.

(2) For any finite set {A_{i} : i = 1, …, n} of submodules of M_{R},

l_{S}(A_{1} ∩ ⋯ ∩ A_{n}) = l_{S}(A_{1}) + ⋯ + l_{S}(A_{n}).

(3) For any submodules A, B of M_{R} and any f ∈ S with f(A ∩ B) = 0, the diagram

0 → A + B → M
↓ α_{f}
M

can be extended commutatively by some g : M ⟶ M.

(4) For any submodules A, B of M_{R} and any R-homomorphism f : M/(A ∩ B) ⟶ M, the diagram

0 → M/(A ∩ B) →^{π} M/A ⊕ M/B
↓ f
M

can be extended commutatively by some g : M/A ⊕ M/B ⟶ M.

Proof. (1) ⇒ (2) can be easily proved by using induction on n; (2) ⇒ (1) is obvious;

(1) ⇔ (3) is by Lemma 1; and (1) ⇔ (4) is by Lemma 5. □

Remark 7. The equivalences (1) ⇔ (2) ⇔ (3) in Proposition 6 can be proved to hold for an arbitrary bimodule _{S}M_{R}.

Many characterizations of π-injective modules are given in [13, 41.21 & 41.23]. In particular, the equivalence “(1) ⇔ (2)” of the next theorem is contained in [13, 41.21].

Theorem 8. The following are equivalent for a module M_{R} with S = End(M_{R}):

(1) M is π-injective.

(2) For any submodules A and B of M_{R} with A ∩ B = 0 , S = l_{S}(A) + l_{S}(B).

(3) For any submodules A and B of M_{R} with A ∩ B = 0 and any f ∈ S, the diagram

0 → A + B → M
↓ α_{f}
M

can be extended commutatively by some g : M ⟶ M.

(4) For any submodules A, B of M_{R} with A ∩ B = 0, the diagram

0 → A + B → M
↓ α_{1}
M

can be extended commutatively by some g : M ⟶ M.

(5) For any submodules A, B of M_{R} with A ∩ B = 0 and any f ∈ S, the diagram

0 → M →^{π} M/A ⊕ M/B
↓ f
M

can be extended commutatively by some g : M/A ⊕ M/B ⟶ M.

(6) For any submodules A and B of M_{R} with A ∩ B = 0, S_{0} = l_{S_{0}}(A) + l_{S_{0}}(B) where S_{0} is the subring of S generated by all idempotents of S.

(7) If A and B are complements of each other in M_{R}, then S = l_{S}(A) ⊕ l_{S}(B). In each of the conditions (2)–(6), the pair A, B of submodules with A ∩ B = 0 can be replaced by a pair A, B of submodules such that they are complements of each other in M_{R}.

Proof. (2) ⇔ (3) ⇔ (4) ⇔ (5): By Lemmas 1, 2 and 5.

(1) ⇔ (2) ⇔ (7): By Corollary 4.

(1) ⇔ (6): Apply Corollary 4 to the bimodule _{S_{0}}M_{R}. □

One condition in the equivalence list of Theorem 8 says that, if A, B are complements of each other in M_{R}, then the map α_{1} : A ⊕ B ⟶ M given by α_{1}(a + b) = a extends to M. This is an improvement of a result of Smith and Tercan [11, Thm.4] where it was proved that M_{R} is π-injective if and only if M satisfies (P_{2}), i.e., if A and B are complement submodules of M with A ∩ B = 0, then every map from A ⊕ B to M extends to M.

Remark 9. Two modules X and Y are said to be orthogonal and written X ⊥ Y if they have no nonzero isomorphic submodules. A submodule N of the module M is called a type submodule if, whenever N ⊂ P ⊆ M, there exists 0 ≠ X ⊆ P such that N ⊥ X. Two submodules X and Y of M are said to be type complements of each other in M if they are complements of each other in M such that X ⊥ Y. The module M is called TS if each of its type submodules is a direct summand of M. The module M is said to satisfy (T_{3}) if, whenever X and Y are type submodules as well as direct summands such that X ⊕ Y is essential in M, X ⊕ Y = M. As shown in [14], a module M satisfies both TS and (T_{3}) if and only if, whenever A, B are type complements of each other in M, M = A ⊕ B. The module satisfying TS and (T_{3}) can be regarded as the ‘type’ analogue of the notion of π-injective modules. Several characterizations of this ‘type’ analogue of π-injective modules have been obtained in [14]. Some new characterizations of this notion can be obtained by restating Theorem 8 with ‘A ∩ B = 0’ being replaced by ‘A ⊥ B’, ‘A, B are complements of each other in M’ replaced by ‘A, B are type complements of each other in M’, and “all idempotents of S” by “all idempotents f with f(M) ⊥ Ker(f)”.

Proposition 10. Let C be the center of End(M_{R}). The following are equivalent:

(1) For any submodules A, B of M_{R} with A ∩ B = 0, C = l_{C}(A) + l_{C}(B).

(2) M_{R} is π-injective and every idempotent of End(M_{R}) is central.

(3) M_{R} is π-injective and every direct summand of M_{R} is fully invariant.

Proof. (1) ⇔ (2). Apply Corollary 4 to the bimodule _{C}M_{R}.

(2) ⇒ (3). Let X be a direct summand of M_{R}. Then X = f(M) for some f^{2} = f ∈ End(M_{R}). For any g ∈ End(M_{R}), since f is central by (2), g(X) = g(f(M)) = f(g(M)) ⊆ f(M) = X. This shows that X is a fully invariant submodule of M_{R}.

(3) ⇒ (2). Let f, g ∈ End(M_{R}) with f^{2} = f. By (3), g(f(M)) ⊆ f(M) and g((1 − f)(M)) ⊆ (1 − f)(M). It follows that fgf = gf and (1 − f)g(1 − f) = g(1 − f). Thus, g − gf = g(1 − f) = (1 − f)g(1 − f) = g − gf − fg + fgf = g − gf − fg + gf = g − fg. This shows that fg = gf. □

3. Applications

In the rest of the paper, we discuss some applications of Theorem 8. Recall that a module M is called continuous if (C1) holds and every submodule isomorphic to a direct summand is itself a direct summand of M (C2). As a generalization of (C2)-condition, a module M_{R} is called GC2 if, for any submodule N of M_{R} with N ≅ M, N is a summand of M. Note that if R is the 2 × 2 upper triangular matrix ring over a field, then R_{R} satisfies both (C1) and (GC2) but it does not satisfy (C3).

Proposition 11. Let M_{R} be a module with S = End(M_{R}). The following are equivalent:

(1) For any family {A_{i} : i ∈ I} of submodules of M_{R} with ∩_{i∈I}A_{i} = 0, S = Σ_{i∈I}l_{S}(A_{i}).

(2) M_{R} is finitely cogenerated and, for any finite family {A_{i} : i = 1, …, n} of submodules of M_{R} with ∩_{i=1}^{n}A_{i} = 0, the map

M →^{h} ⊕_{i=1}^{n}M/A_{i}, m ↦ (m + A_{1}, …, m + A_{n}),

splits.

(3) M_{R} is finitely cogenerated and, for any finite family {A_{i} : i = 1, …, n} of submodules of M_{R} with ∩_{i=1}^{n}A_{i} = 0, S = Σ_{i=1}^{n}l_{S}(A_{i}).

If M_{R} satisfies both (1) and (GC2), then M_{R} is continuous and S is semiperfect.

Proof. It is straightforward to verify the equivalences (1) ⇔ (2) ⇔ (3).

Suppose that M_{R} satisfies both (1) and (GC2). By Theorem 8, M_{R} is π-injective. Thus, by [8, Lemma 3.14], M is continuous. To show that S is semilocal, let σ : M ⟶ M be a monomorphism. Then M = σ(M) ⊕ N for some N ⊆ M (by the GC2-condition). It must be that N = 0 since M is finite dimensional (indeed, finitely cogenerated). So, σ is an isomorphism. Therefore, M satisfies the assumptions in Camps-Dicks [3, Thm.5], and so End(M) is semilocal. But, by [8, Prop.3.5 & Lemma 3.7], idempotents of S/J(S) lift to idempotents of S, and thus S is semiperfect. □

A ring R is called right Kasch if every simple right R-module embeds in R_{R}, or equivalently if l(I) ≠ 0 for any maximal right ideal I of R.

Corollary 12. If R satisfies the condition that, for any set {A_{i} : i ∈ I} of right ideals such that ∩_{i∈I}A_{i} = 0, R = Σ_{i∈I}l_{R}(A_{i}) and R_{R} satisfies (GC2), then R is a semiperfect right continuous ring with a finitely generated essential right socle. In particular, R is left and right Kasch.

Proof. The first part follows from Theorem 11. The second part is by [9, Lemma 4.16]. □

A ring R is called strongly right IN if, for any set {A_{i} : i ∈ I} of right ideals, l_{R}(∩_{i∈I}A_{i}) = Σ_{i∈I}l_{R}(A_{i}). The ring R is called right dual if every right ideal of R is a right annihilator. It is well-known that every two-sided dual ring is strongly left and right IN.

Corollary 13. The following are equivalent for a ring R:

(1) R is a two-sided dual ring.

(2) R is strongly left and right IN, and left (or right) GC2.

(3) R is left and right finitely cogenerated, left and right IN, and left (or right) GC2.

Proof. (1) ⇒ (2): Obvious.

(2) ⇒ (3): It is clear by Corollary 12.

(3) ⇒ (1): Suppose ∩_{i∈I}A_{i} = 0 where all A_{i} are right ideals R. Since R is right finitely cogenerated, ∩_{i∈F} A_{i} = 0 where F is a finite subset of I. Thus, R = l_{R}(∩_{i∈F} A_{i}) = Σ_{i∈F} l_{R}(A_{i}) because of the IN-condition, and hence R = Σ_{i∈I}l_{R}(A_{i}). By Corollary 12, R is left and right Kasch. Since R is left and right IN, it follows from [2, Lemma 9] that R is a two-sided dual ring. □

The GC2-condition in Corollary 12 and in Corollary 13(3) can not be removed. To see this, let R be the trivial extension of ℤ and the ℤ-module ℤ_{2^{∞}}. Then R has an essential minimal ideal, so R is finitely cogenerated and, for any set {A_{i} : i ∈ I} of right ideals of R, R = Σ_{i∈I}l_{R}(A_{i}). Moreover, R is IN. But R contains non-zero divisors which are not invertible, so R is not GC2. Clearly, R is not Kasch, so it is not semiperfect by Corollary 12. We do not know if the GC2-condition can be removed in Corollary 13(2).

Proposition 14. Suppose every finitely generated left ideal of R is a left annihilator. Then the following are equivalent:

(1) Every closed right ideal of R is a right annihilator of a finite subset of R.

(2) R_{R} satisfies (C1).

(3) R is right continuous.

Proof. (3) ⇒ (2): Obvious.

(2) ⇒ (1): If I_{R} is closed in R_{R}, then I = eR for some e^{2} = e ∈ R. Hence I = r(1 − e).

(1) ⇒ (2): Let I_{R} and K_{R} be complements of each other in R_{R}. Then, by (1), I = r_{R}(a_{1}, …, a_{n}) and K = r_{R}(b_{1}, …, b_{m}) where a_{i}, b_{j} ∈ R. Thus,

R = l_{R}(I ∩ K) = l_{R}[r_{R}(a_{1}, …, a_{n}) ∩ r_{R}(b_{1}, …, b_{m})]
= l_{R}(r_{R}(Σ_{i=1}^{n}Ra_{i} + Σ_{j=1}^{m}Rb_{j})) = Σ_{i=1}^{n}Ra_{i} + Σ_{j=1}^{m}Rb_{j}
= l_{R}(I) + l_{R}(K).

Thus, by Theorem 8, R_{R} is π-injective, and in particular R_{R} satisfies (C1).

(2) ⇒ (3): Since r_{R}(l_{R}(F)) = F for all finitely generated left ideals F of R, R is right P-injective, and hence satisfies the right C2-condition. Thus, R is right continuous. □

A ring R is called a right CF-ring (resp. right FGF-ring) if every cyclic (resp. finitely generated) right R-module embeds in a free module. The ring R is called right FP-injective if every R-homomorphism from a finitely generated submodule of a free right R-module F into R extends to F. Note that every right self-injective ring is right F P -injective, but not conversely. Also every finitely generated left ideal of a right F P -injective ring is a left annihilator (see [7]). The well known FGF problem asks whether every right FGF-ring is QF. It is known that every right self-injective, right FGF-ring is QF. In fact, Björk [1] and Tolskaya [12] independently proved that every right self-injective, right CF-ring is QF. On the other hand, Nicholson-Yousif [10, Theorem 4.3] shows that every right FP-injective ring for which every 2-generated right module embeds in a free module is QF. Our next corollary extends the two results.

Corollary 15. Suppose R is a right CF-ring such that every finitely generated left ideal is a left annihilator. Then R is a QF-ring.

Proof. Since R is right CF, every right ideal is a right annihilator of a finite subset of R. By Proposition 14, R_{R} is π-injective. Then, by [5, Corollary 2.9], R is right artinian. Clearly, R is two-sided mininjective. So, R is QF by [9, Cor.4.8]. □

Corollary 16. Every right CF, right FP-injective ring is QF. In particular, every right FGF, right FP-injective ring is QF.

A ring R is called right FPF-ring if every finitely generated faithful right R-module is a generator of Mod-R, the category of all right R-modules. A ring is left (resp. right) duo if every left (resp. right) ideal is two sided. We conclude by noticing that every right FPF-ring which is left or right duo is π-injective. The next corollary follows from Theorem 8 and the proof of [4, 3.1A2, p.3.2].

Corollary 17. Let R be a right FPF-ring. If R is a left or right duo ring, then R_{R} is π-injective. In particular, every commutative FPF-ring is π-injective.

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Received November 12, 2000