Mathematical Notes, vol. 75, no. 1, 2004, pp. 93–100. Translated from Matematicheskie Zametki, vol. 75, no. 1, 2004, pp. 100–108. c Original Russian Text Copyright 2004 by P. A. Krylov, E. G. Pakhomova.
When Is the Group Hom(A, B) an Injective E(B)-Module? P. A. Krylov and E. G. Pakhomova Received May 7, 2002
Abstract—Injectivity conditions for the homomorphism group Hom(A, B) regarded as a left module over the endomorphism ring of the group B are found for arbitrary Abelian groups A and B , where B is nonreduced. Key words: Abelian group, injective module, endomorphism ring.
The fact that the circle of problems related to the notion of injectivity is of importance for the theory of Abelian groups is stressed by the setting of Problems 12, 29, and 84D in the books [1, 2] (see also the surveys [3, 4]). The Abelian groups injective over their rings of endomorphisms are described in the paper of Richman and Walker [5]. Ivanov [6] described the cases in which the ring of endomorphisms of an Abelian group is self-injective, i.e., is an injective module over itself. One can pose a more general problem. The homomorphism group Hom(A, B) is a left (right) module over the ring E(B) (the ring E(A)). To be more exact, we have the E(B)- E(A)-bimodule Hom(A, B) . The existence of a natural isomorphism Hom(Z , B) ∼ = B of left E(B)-modules and of the relation Hom(A, A) = E(A) shows that the investigation of the group Hom(A, B) as an E(B)and E(A)-module includes the study of the Abelian group as a module over its endomorphism ring and the consideration of one-sided properties of the endomorphism ring. In [7], the authors characterize Abelian groups A and B such that Hom(A, B) is an injective E(A)-module (and Abelian groups for which Hom(A, B) is an injective E(B)-module if the group B is reduced). In [8], an investigation of general properties of an injective E(B)-module Hom(A, B) was carried out and a series of auxiliary results was obtained. In the present paper, the investigation of the group Hom(A, B) as an injective E(B)-module is generally completed. The rather complicated remaining case of nonreduced group B is considered. The terminology and notation correspond to [1, 2, 9, 10]. All groups in the paper are Abelian; N denotes the set of positive integers, Z the ring of integers, and Q the field of rationals. The letter p stands for some prime number, Kp for the ring of p-adic integers, Ap for the field of padic numbers, and Z(p∞ ) for the quasicyclic p-group. Let G be a group. Then denote by r(G) the rank of the group G , by rp (G) the p-rank of G , i.e., the rank of the quotient group G/pG , by E(G) the endomorphism ring of G , by T (G) the periodic subgroup of G , and by Gp the pcomponent of G , i.e., the maximal subgroup of G which is a p-group. A group G is said to be G and G be the direct sum and the product of m copies of p-divisible if pG = G ; let m m the group G , respectively ( m is a cardinal), and let G[n] = {x ∈ G | nx = 0} , where n ∈ N . For ϕA , where ϕ ranges over a group A , denote by SA (G) the trace of A in G , i.e., SA (G) = all homomorphisms from A into G . Denote by Rn the ring of all matrices of order n over the ring R .
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1. AUXILIARY RESULTS For the convenience of the reader, we cite results frequently used throughout the paper. Theorem 1 [5]. A group B is an injective E(B)-module if and only if B = p∈T Cp ⊕ D , where T is a set of primes, Cp is a finite p-group for any p ∈ T , and D is a divisible group of finite rank, and either D is an unmixed group and Cp = 0 for almost all p or D = 0 . Theorem 2 [7]. Let A and B be groups, let B be reduced, and let S = SA (B) . The group Hom(A, B) is a nonzero injective E(B)-module if and only if there are embeddings Cp ⊆ S ⊆ Cp , p∈T
p∈T
where T is a nonempty set of primes, Cp is a finite p-group for any p ∈ T , and the following conditions hold : (a) SA (P ) = S , where P = p∈T Cp ; (b) the orders of the elements of the group Cp do not exceed the orders of the cyclic direct summands of the p-component of the group A for any p ∈ T ; (c) for any p ∈ T , there is a decomposition B = Hp ⊕ Cp , where Hp is a p-divisible subgroup of B . Let us use the notion of idealization of a bimodule[11, Proposition 7.11]. Let R and Sbe rings, R M r m and let M be an R- S-bimodule. Introduce the ring as the set of all matrices 0 S 0 s (r ∈ R , s ∈ S, m ∈ M ) with the ordinary operations of matrix addition and multiplication. The R M ring is called the idealization of the bimodule M . Denote this ring by the letter K . 0 S In this paper, there are left K-modules having the form of a “column module.” To save room, we write such modules and their elements in the form of rows. Let X be a left R-module, let Y be a left S-module, and let W = (X , Y ) be the set of column vectors (x, y) , where x ∈ X and y ∈ Y . Identify W with the direct sum of the groups X and Y . Suppose that there is an R-homomorphism g : M ⊗S Y → X . We define a “product” of elements in M by elements in Y with values in X by assuming that my = g(m ⊗ y) , where m ∈ M and y ∈ Y . Write r m (x, y) = (rx + my , sy) (1) 0 s for any r ∈ R , s ∈ S , m ∈ M , x ∈ X , and y ∈ Y . A verification shows that W thus becomes a left K-module (in fact, every left K-module can be obtained by the above construction of “column module”). We identify the groups (X , 0) and (0, Y ) with X and Y , respectively. Let us do the same for the elements of W . It is clear that X is a submodule of W . The next assertion can be found in [8]. Proposition 1. (1) The K-module W is injective if and only if X is an injective R-module, Y is an injective S-module, and, for any R-homomorphism ϕ : M → X , there is a y ∈ Y such that ϕ(m) = my for any m ∈ M . (2) The K-module X is injective if and only if it is injective as an R-module and we have HomR (M , X) = 0 . Idealizations of bimodules occur in the following situation. Let B be a group of the form D ⊕ G , where D is a completely characteristic summand, i.e., Hom(D, G) = 0 . Let us write out the idealization of the E(D)- E(G)-bimodule Hom(G, D) , E(D) Hom(G, D) . 0 E(G) MATHEMATICAL NOTES
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The ring E(B) can be identified with the given matrix ring [2, Theorem 106.1], and it can be regarded as the idealization of the bimodule in question. In the above notation we have K = E(B) , R = E(D) , S = E(G) , and M = Hom(G, D) . Let A be another group. We have a left K-module Hom(A, B) . Moreover, Hom(A, B) = Hom(A, D) ⊕ Hom(A, G) , where Hom(A, D) is a left Rmodule and Hom(A, G) is a left S-module. By setting W = Hom(A, B) , X = Hom(A, D) , and Y = Hom(A, G) , we arrive at the “column module” W = (X , Y ) with the module product (1). The product my is the composition of the mappings m and y . If Hom(A, B) is an injective E(B)-module, then it follows from Proposition 1 that Hom(A, D) is an injective E(D)-module, Hom(A, G) is an injective E(G)-module, and the condition concerning the homomorphisms ϕ holds. The injectivity of Hom(A, D) regarded as an E(B)-module is equivalent to the injectivity of Hom(A, D) regarded as an E(D)-module together with the validity of the relation HomE(D) (Hom(G, D), Hom(A, D)) = 0. 2. THE CASE OF A DIVISIBLE GROUP B Theorem 3. Let A be an arbitrary group, and let D be a divisible group. The E(D)-module Hom(A, D) is nonzero and injective if and only if A is an aperiodic group, D an unmixed group a finite rank, and, if Dp = 0 , then Ap = 0 for any p . Proof. Necessity. Suppose that A is a periodic group. Then there is a p-component Dp such that Hom(A, Dp ) = 0 . By Proposition 1, Hom(A, Dp ) is an injective E(Dp )-module. The multiplication of the group Dp by p is not a zero divisor in E(Dp ) . Hence Hom(A, Dp ) is divisible by p [10, Proposition 2.16], which contradicts [1, Theorem 47.1]. Thus, the group A has an element a of infinite order. ∞ Let us prove that ∞r(D) < ∞ . Suppose the contrary. Write D = i=1 Di , where Di = 0 for any i . Set W = i=1 Di and choose an element d ∈ W − D . Since the group W is injective, there is a homomorphism ϕ : A → W taking a to d . Let πi : D → Di be the projection, and let ϕi be the composition of ϕ with the projection W → Di , i ≥ 1 , where ϕi ∈ πi Hom(A, D) . By [5, Lemma 6], there is a ψ : A → D such that πi ψ = ϕi for any i . It is clear that ϕ = ψ , which is impossible. Thus, r(D) < ∞ . Suppose that D is a mixed group, D0 is the torsion-free part of D , and Dp is a p-component in D . By Proposition 1, Hom(A, Dp ⊕ D0 ) is an injective module over the ring E(Dp ⊕ D0 ) . Assume that D = Dp ⊕ D0 . Let us lead this assumption to a contradiction. Choose a p-basis subgroup V of the group A . We have the induced sequence of modules 0 → Hom(A/V , D) → Hom(A, D) → Hom(V , D) → 0 (by the term “module,” etc., we mean here the term “ E(D)-module”). The group Hom(A/V , D) is a p-adic algebraically compact group [1, Theorem 47.7]. Thus, this is a p-servant injective group which is a p-servant subgroup of Hom(A, D) [1, Proposition 44.7]. Hence the sequence is split, and the outermost modules in this sequence are injective. Take a cyclic direct summand C of the group V . Then Hom(C , D) is an injective module. Let C ∼ = Z(pk ) , k ∈ N . Since k k ∼ Hom(C , D) = D[p ] , it follows that D[p ] is a direct summand in D , which is impossible. If C ∼ = Z , then Hom(C , D) ∼ = D is an injective module. However, this contradicts Theorem 1. Hence V = 0 , and A is a p-divisible group. Set K = ∩ ker α , where α ranges over all homomorphisms A → D , and write A1 = A/K . Here A1 is the direct sum of a nonzero p-divisible torsion-free group and of a divisible p-group. One can assume that the group A by itself is a p-divisible torsion-free group. Take the exact sequence of modules 0 → Hom(U/A, D) → Hom(U , D) → Hom(A, D) → 0 MATHEMATICAL NOTES
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in which U stands for the divisible hull of A . The periodic group U/A has zero p-component. Thus, Hom(U/A, D) = 0 and Hom(U , D) ∼ = Hom(A, D) . The group Q is a direct summand of U , and therefore Hom(Q, D) is an injective module. One can apply 1 to this module (see also the paragraph after the proposition). Let Proposition ∞ us write Dp = m Z(p ) and D0 = n Q , where m, n ∈ N . We have R = E(Dp ) = (Kp )m , and Hom(Q, Dp ) = Ap , M = Hom(D0 , Dp ) = X = Hom(Q, Dp ) =
n
m
n
m
Ap , Y = Hom(Q, D) = D , and HomR (M , X) =
n
HomR
m
Ap ,
Ap .
m
The last sum is isomorphic to n HomKp (Ap , Ap ) = n Ap [12, Theorem 1.2]. Since the set HomR (M , X) is of continuum cardinality and Y is a countable set, it follows that the condition on ϕ in Proposition 1 fails. A contradiction. Thus, D is an unmixed group. Let us prove for the periodic group D that Ap = 0 for Dp = 0 . Suppose that C is an indecomposable direct summand of the group Ap . Then Hom(C , D) is an injective module. As above, it turns out that the case C ∼ = Z(pk ) is impossible. Let C ∼ = Z(p∞ ) . With regard to the ∞ structure of the ring E(D) and the group Hom(Z(p ), D) , we see the following. An element p1 , which is not a zero divisor in E(D) , must divide the group Hom(C , D) [10, Proposition 2.16], and this is impossible. Hence Ap = 0 . Sufficiency. If D is a divisible torsion-free group of finite rank, then E(D) is the full ring of matrices over Q . Every module over this ring is injective. For the periodic group D , we have the ∼ module decomposition Hom(A, D) = p Hom(A, Dp ) . Therefore, assume that D is a divisible pgroup of finite rank and Ap = 0 . Let us show that the module Hom(A, D) is injective. Denote by V one of the p-basis subgroups of A and consider the exact sequence of modules 0 → Hom(A/V , D) → Hom(A, D) → Hom(V , D) → 0. Let us prove that the outermost modules are injective. Since V = s Z , where s is a cardinal number, it follows that Hom(V , D) ∼ = s D . It remains to note that D is = s Hom(Z, D) ∼ an injective module (Theorem 1). Take the module Hom(A/V , D) . Write G = A/V . Since Ap = 0 , it follows that Gp = 0 . Hence Hom(T (G), D) = 0 and Hom(G, D) ∼ = Hom(G/T (G), D) . We assume that the group G itself is p-divisible and torsion-free. Let U be the divisible hull of G ; the p-component of the periodic group U/G is equal to zero, Hom(U/G, D) = 0 , and Hom(G, D) ∼ = Hom(U , D) ∼ = t Hom(Q, D) for some cardinal number t . It suffices to prove that the module Hom(Q, D) is injective. However, this fact is known because E(D) ∼ = (Kp )m , where m ∈ N , and Hom(Q, Z(p∞ )) ∼ Ap . Hom(Q, D) ∼ = = m
m
3. THE CASE OF NONREDUCED AND INDIVISIBLE GROUP B Let A and B be groups, where B is nonreduced and indivisible. Write B = D ⊕ G , where D is the divisible part and G the reduced part of B . By assumption, D, G = 0 . The letters A , B , D , and G preserve their meaning till the end of the paper. It is more convenient to state the result for aperiodic and periodic groups A separately. MATHEMATICAL NOTES
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Theorem 4. If A is an aperiodic group and B is a nonreduced indivisible group, then Hom(A, B) is an injective E(B)-module if and only if SA (G) = p∈T Cp , where T is a set of primes, Cp is a finite p-group, and the following conditions hold : (a) SA (P ) = SA (G) , where P = p∈T Cp ; (b) the orders of the elements of the group Cp do not exceed the orders of the cyclic direct summands of the group Ap for any p ∈ T ; (c) D is an unmixed group of finite rank ; (d) if D is torsion-free, then B = D ⊕ SA (G) ⊕ H , where H is a periodic group; (e) if D is a periodic group, then Ap = 0 , Gp is a bounded group, and rp (G/T (G)) = r(G/T (G)) < ∞ for any p with Dp = 0 ; moreover, if pA = A , then B = D ⊕ SA (G) ⊕ Y , where Y is a periodic group. Proof. The word “module,” etc., will mean “ E(B)-module.” The group A has an element a of infinite order. Every homomorphism a → D can be extended to a homomorphism A → D . This implies that D ⊆ SA (B) and SA (B) = D ⊕ SA (G) . Let Hom(A, B) be an injective module. By Proposition 1, Hom(A, D) is an injective E(D)module, and Hom(A, G) is an injective E(G)-module. Thus, the trace SA (G) has the structure indicated in Theorem 2, and conditions (a) and (b) hold. As far as the trace is concerned, it remains to show that it is equal to the sum of the groups Cp . By Theorem 3, the group D has finite rank, it is an unmixed group, and Ap = 0 if Dp = 0 . Therefore, assertion (c) holds. Let us prove assertion (d). Let W be the injective hull of the module Hom(A, D) . Suppose that ϕ ∈ W and ϕ ∈ / Hom(A, D) . There is a p such that ϕA has nonzero projection to Cp (by Theorem 2, every component Cp is a direct summand of B). Denoting the projection B → Cp by π , we obtain πϕ : A → Cp and 0 = πϕ ∈ W . Since Hom(A, D) is an essential submodule of W , there is an α ∈ E(B) such that 0 = απϕ ∈ Hom(A, D) , which is impossible. Thus, Hom(A, D) = W , and Hom(A, D) is an injective module. Take the divisible hull U of the group A . Then Hom(U/A, D) = 0 and Hom(U , D) ∼ = Hom(A, D) . The group Q is contained in U . Hence Hom(Q, D) is an injective module. The relation Hom(Q, D) ∼ = D implies that D is injective. Thus, the decomposition B = D ⊕ G is a module decomposition, i.e., G is a completely characteristic subgroup and Hom(G, D) = 0 . This implies that the group G is periodic and SA (G) = p∈T Cp . Since G = Hp ⊕ Cp and pHp = Hp (Theorem 2), it follows that Cp = Gp for any p ∈ T . One can now write B = D ⊕ SA (G) ⊕ H , where H is a periodic group. Let us pass to (e). Fix a number p with Dp = 0 . Set Ep = Dp ⊕ Cp and Ep = Dp if Cp = 0 . By Theorem 2, B = Ep ⊕ H for some group H . Further, Hom(A, T (H)) is a submodule of Hom(A, B) . Since the group Dp is injective, it follows that the module Hom(A, Ep ⊕ T (H)) is essential in Hom(A, B) . Write Hom(A, B) = W1 ⊕ W2 , where W1 (W2 ) is the injective hull of the module Hom(A, Ep ) (Hom(A, T (H))) . We also have the group decomposition Hom(A, B) = Hom(A, Ep ) ⊕ Hom(A, H). As in (d), one can prove the relation Hom(A, Ep ) = W1 . Let us prove that Hom(A, H) = W2 . For ϕ ∈ W2 , write ϕ = α + β , where α ∈ Hom(A, Ep ) and β ∈ Hom(A, H) . Let π : B → Ep be a projection. Then πϕ = πα+πβ = πα = 0 for α = 0 . On the other hand, πϕ = πα ∈ W1 ∩W2 = 0 ; a contradiction. Thus, α = 0 and ϕ = β ∈ Hom(A, H) . We have the inclusion W2 ⊆ Hom(A, H) , and hence these sets are equal. The decomposition Hom(A, B) = Hom(A, Ep ) ⊕ Hom(A, H) is a module decomposition. Hence Hom(SA (H), Ep ) = 0 . Thus, SA (H) and SA (G) are periodic groups, and SA (G) = p∈T Cp . One can assume that pA = A . As above, we write B = Dp ⊕ H ( Cp = 0 and Ep = Dp ). Set R = E(Dp ) , M = Hom(H , Dp ) , and X = Hom(A, Dp ) . Since Hom(A, Dp ) is an injective module (it is equal to W1 ), it follows from Proposition 1 that HomR (M , X) = 0 , and hence HomKp (Hom(H , Z(p∞ )), Hom(A, Z(p∞ ))) = 0 MATHEMATICAL NOTES
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(see [12, Theorem 1.2] and a similar argument in the proof of Theorem 3). Further, Hom(A, Z(p∞ )) ∼ = Hom(U , Z(p∞ )), where U is the divisible hull of the group A/T (A) (with regard to the relations Ap = 0 and pA = A). Hence Hom(A, Z(p )) ∼ = Hom ∞
t
Q, Z(p ) ∼ Hom(Q, Z(p∞ )) ∼ Ap , = = ∞
t
t
where t = r(A/T (A)) . It now follows from (2) that Hom(H , Z(p∞ )) is a periodic group. Using the structure of this group [1, Theorem 47.1], we see that the component Hp is bounded. Let us now note that Gp = Hp . Let F be a p-basis subgroup of L , where L = H/T (H) . The group Hom(L, Z(p∞ )) is periodic. Using the structure of this group again, we see that L/F is a periodic group, r(F ) < ∞ , and rp (L) = r(L) < ∞ . This was to be proved because G/T (G) ∼ = L. Let us now assume that Dp = 0 and pA = A for some p . Choose a p-basis subgroup V of A . We have the p-servant exact sequence of modules 0 → Hom(A/V , Ep ) → Hom(A, Ep ) → Hom(V , Ep ) → 0 (where Ep is the group defined above). This sequence terminates at the zero group on the right by the p-servant injectivity of the group Ep . One can prove that the sequence is split as in the proof of Theorem 3. Therefore, Hom(V , Ep ) and Hom(Z, Ep ) are injective modules (the group V is free because Ap = 0). It follows from Ep ∼ = Hom(Z, Ep ) that the module Ep is injective. Therefore, the decomposition B = Ep ⊕ H is a module decomposition, i.e., H is a completely characteristic subgroup. As in (d), we see that B = D ⊕ SA (G) ⊕ Y , where Y is a periodic group. Sufficiency. Let D be torsion-free. All summands of the decomposition B = D ⊕ S ⊕ H ( S = SA (G)) are completely characteristic. Hence we have the product of rings E(B) = E(D) × E(S) × E(H) and the module decomposition Hom(A, B) = Hom(A, D) ⊕ Hom(A, S). Let us show that both summands are injective. By [7, Lemma 2.2], this is equivalent to the condition that Hom(A, D) is an injective E(D)-module and Hom(A, S) is an injective E(S)module. Let D be a periodic group. Then Hom(A, B) is a product of modules of the following three forms: Hom(A, Dp ) , Hom(A, Dp ⊕ Cp ) , and Hom(A, Cp ) . Let us show that each of them is injective. Take Hom(A, Dp ) . By Theorem 3, Hom(A, Dp ) is an injective E(Dp )-module (it is assumed that Cp = 0). Write B = Dp ⊕ H and set M = Hom(H , Dp ) and X = Hom(A, Dp ) . Let us show that Hom(M , X) = 0 . We first prove that M is a periodic group. The component Hp is bounded, and hence M∼ = Hom(Hp , Dp ) ⊕ Hom(H/Hp , Dp ), where Hom(Hp , Dp ) is a bounded group. The group Hom(H/Hp , Dp ) coincides with the group Hom(L, Dp ) , where L = H/T (H) . Further, Hom(L, Dp ) = Hom(F , Dp ) ⊕ Hom(L/F , Dp ) , where F is a p-basis subgroup of L . Since rp (L) = r(L) < ∞ , it follows that the group Hom(F , Dp ) is periodic. We now note that L/F is a periodic group, and hence Hom(L/F , Dp ) = Hom(E , Dp ) , where E is the p-component of the group L/F . Since rp (L) = r(L) < ∞ , it follows MATHEMATICAL NOTES
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that E is the direct sum of cyclic groups [13, Theorem 0.2]. Thus, Hom(L/F , Dp ) is a finite group, and the group Hom(L, Dp ) is periodic together with M . Let us make some comments concerning X . If pA = A , then B is a periodic group and M = 0 . If pA = A , then the group X is torsion-free. In any case, Hom(M , X) = 0 . By Proposition 1, Hom(A, Dp ) is an injective module. Let us consider the module Hom(A, Dp ⊕ Cp ) . Since pA = A , it follows that B is a periodic group. Both summands in the decomposition B = (Dp ⊕ Cp ) ⊕ H are completely characteristic. It suffices to prove that Hom(A, Dp ⊕ Cp ) is an injective module over the endomorphism ring K of the group Ep , where Ep = Dp ⊕ Cp [7, Lemma 2.2]. Since Ap = 0 and pA = A , it follows that the p-basis subgroup V of the group A is free. The induced sequence of K-modules 0 → Hom(A/V , Ep ) → Hom(A, Ep ) → Hom(V , Ep ) → 0 is split (see above). It suffices to establish that the outermost K-modules are injective. We have Hom(V , Ep ) ∼ =
Hom(Z , Ep ) ∼ =
s
Ep
s
( s is a cardinal number), where Ep is an injective K-module (Theorem 1). We set M = Hom(Cp , Dp ) and X = Hom(A/V , Ep ) . Then Hom(M , X) = 0 because M is a finite group and X is torsion-free. By Theorem 3, X is an injective E(Dp )-module. Hence X is injective as a K-module (Proposition 1). Let us consider the module Hom(A, Cp ) (it is assumed that Dp = 0). It follows from pA = A that B is periodic and the summands in the decomposition B = Cp ⊕ H are completely characteristic. It suffices to show that Hom(A, Cp ) is injective as an E(Cp )-module. However, this is known by Theorem 2. Theorem 5. If A is a periodic group and B is an indivisible nonreduced group, then Hom(A, B) is a nonzero injective E(B)-module if and only if SA (B) = p∈T Cp , where T is a nonempty set of primes, Cp is a finite p-group for any p ∈ T , and the conditions (b) and (c) of Theorem 2 are satisfied. Proof. Necessity. Let us write B = D ⊕ G (as we wrote before Theorem 4). By Proposition 1, Hom(A, D) is an injective E(D)-module and Hom(A, G) is an injective E(G)-module. In fact, Hom(A, D) = 0 because Ap = 0 for Dp = 0 by Theorem 3. Hence SA (B) = SA (G) = p∈T Cp , where T is nonempty. Conditions (b) and (c) in Theorem 2 are also satisfied. Sufficiency. For a given p ∈ T , write B = Hp ⊕ Cp , where pHp = Hp . Suppose that Dp = 0 . With regard to the structure of the trace SA (B) , one can see that Ap = 0 and Cp = 0 ; a contradiction. Thus, Dp = 0 . This implies that the summands Hp and Cp are completely characteristic. By Theorem 2 and by [7, Lemma 2.2], Hom(A, Cp ) is an injective E(B)-module. We now have Cp = Hom A, Cp ∼ Hom(A, Cp ), Hom(A, B) = Hom A, = p∈T
p∈T
p∈T
which proves that Hom(A, B) is injective and completes the proof of the theorem. Theorems 2–5 imply injectivity conditions for the E(B)-module Hom(A, B) in some principal situations. One can also find the structure of the injective E(B)-module Hom(A, B) . Let Cp be a finite p-group, and let Dp ( D0 ) be a divisible p-group (torsion-free group, respectively) of finite rank. The left modules Hom(Z, Cp ) , Hom(Z, Dp ⊕ Cp ) , Hom(Z, Dp ) , Hom(Q, Dp ⊕ Cp ) , Hom(Q, Dp ) , and Hom(Q, D0 ) are injective over the corresponding endomorphism rings. The MATHEMATICAL NOTES
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injective E(B)-module Hom(A, B) is the product of some numberof copies of each of these modules for different primes p , i.e., of modules Cp , Dp ⊕ Cp , Dp , np Ap ( np ∈ N), and D0 . Condition (a) in Theorems 2 and 4 can be replaced by some more obvious group-theoretic condition. In Theorem 2, the trace S is a servant completely characteristic subgroup of the product p∈T Cp . Condition (a) generalizes the property of a subgroup S to be a servant completely characteristic subgroup of p∈T Cp (as used by Ivanov in [6]). For these reasons, one can derive Ivanov’s results [6] on self-injective endomorphism rings from Theorems 2–5 and from [7]. ACKNOWLEDGMENTS This research was financially supported by the Russian Foundation for Basic Research under grant no. 00-01-00876. REFERENCES 1. L. Fuchs, Infinite Abelian Groups. vol. 1 Academic Press, New York–London, 1970. 2. L. Fuchs, Infinite Abelian Groups. vol. 1 Academic Press, New York–London, 1973. 3. A. P. Mishina and A. V. Mikhalev, “Infinite Abelian groups: methods and results,” Fundam. Prikl. Mat., 1 (1995), no. 2, 319–375. 4. A. P. Mishina, “Abelian groups,” J. Math. Sci., 76 (1995), no. 6, 2721–2792. 5. F. Richman and E. Walker, “Modules over PIDs that are injective over their endomorphism rings,” Ring theory, Academic Press, New York, 1972, pp. 363–372. 6. A. V. Ivanov, “Abelian groups with self-injective rings of endomorphisms and with rings of endomorphisms with the annihilator condition,” in: Abelian Groups and Modules [in Russian], Tomsk Gos. Univ., Tomsk, 1982, pp. 93–109. 7. P. A. Krylov and E. G. Pakhomova, “Abelian groups as injective modules over endomorphism rings,” Fundam. Prikl. Mat., 4 (1998), no. 4, 1365–1384. 8. P. A. Krylov and E. G. Pakhomova, “Investigation of the group Hom(A, B) as an injective E(B)module,” Abelian Groups and Modules [in Russian], Tomsk Gos. Univ., Tomsk, 1996, pp. 132–169. 9. J. Lambek, Lectures on Rings and Modules, Second ed., Chelsea Publishing Co., New York, 1976. 10. G. E. Puninskii and A. A. Tuganbaev, Rings and Modules [in Russian], SOYuZ, Moscow, 1998. 11. C. Faith, Algebra: Rings, Modules and Categories. I, Springer-Verlag, New York-Heidelberg, 1973. 12. P. M. Cohn, Free Rings and Their Relations, Academic Press, London-New York, 1971. 13. D. M. Arnold, “Finite rank torsion-free abelian groups and rings,” Lecture Notes in Math., 931 (1982). Tomsk State University E-mail : (P. A. Krylov)
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MATHEMATICAL NOTES
Vol. 75
No. 1
2004
