Gamma Invariants and the Torsion-Freeness of Ext

This paper investigates homological properties of the class ∗ B of Abelian groups A such that Ext(A, B) is torsion-free. In particular, our results demonstrate that Gamma-invariants cannot be introduced in a meaningful way for ∗ B.

Ascending chains of subgroups have always been an the essential tool in the study of infinite Abelian groups.This was further emphasized by Shelah's solution of the Whitehead problem and a series of related papers concerning the invariants of the divisible group Ext(A, B) in the case that A and B are torsionfree Abelian groups.In particular, either Ext(A, B) = 0 or r 0 (Ext(A, B)) = 2 ℵ 0 if A and B are countable.On the other hand, the Ext-group may be torsion-free without vanishing.
To simplify our notation, we denote the class of all torsion-free Abelian groups A such that Ext(A, B) = 0 by ⊥ B, while * B consists of the groups A for which Ext(A, B) is torsion-free.Clearly, ⊥ B ⊆ * B.Moreover, since ⊥ B was successfully investigated using Eklof's Γ-invariants [7], it can be expected that these invariants play a similar role in the apparently related discussion of * B. One of the consequences of our discussion is the surprising fact that this is not the case.
To define the Γ-invariant of an Abelian group A of regular cardinality κ with respect to a property P, write A = ∪ α<κ A α where the A α 's form a smooth ascending chain of subgroups of A with A 0 = {0} and |A α | < κ for all α, and consider S = {α < κ|A β /A α does not satisfy P for some β > α}.The Gamma-invariant Γ κ,P (A) of A with respect to P is the equivalence class of S with respect to the intersection with closed and unbounded sets.
Surprisingly, we encounter immediate difficulties when attempting to use these invariants for the property A ∈ * B. Looking at ⊥ B, we realize that its closure with respect to extensions is what enables the use of Γ-invariants.In contrast, * B need not be closed with respect to extensions even in the case B = Z: Proof.Clearly, Hom(A, Z) = 0. Therefore, the induced sequence is exact.If Ext(A, Z) were torsion-free, then Hom(Z, Z) would be isomorphic to a pure subgroup of the torsion-free divisible group Ext(Q, Z), which is impossible.
However, * B is closed with respect to B-cobalanced extensions, where a sequence 0 → U → G → H → 0 is B-cobalanced if B is injective with respect to it, i.e. the induced sequence 0 → H * → G * → U * → 0 is exact.Here, the symbol ( ) * denotes one of the contra-variant functors G * = Hom(G, B) and M * = Hom E (M, B) between the category of Abelian groups and the category of left E-modules.Moreover, ψ G is the induced canonical map G → G * * .For details, see [12].For instance, every sequence 0 Unfortunately, there is no direct connection between the groups in a short exact sequence belonging to * B and the sequence being B-cobalanced as the following example shows:

Example 2. a) There exists group
Proof.For a), let G be an Abelian group such that Ext(G, Z) is non-zero and torsion-free.A free resolution 0 We nevertheless obtain the following Proof.Consider the induced sequence Ext(H, B) Since H is torsion-free, im(β * ) is divisible.Hence a) holds.Moreover, β * is one-to-one if the original sequence is B-cobalanced, from which b) follows directly.
We say that a torsion-free Abelian group A of (regular The category of B-solvable groups consists of all groups G for which θ G is an isomorphism.Moreover, G is B-projective if it isomorphic to a direct summand of ⊕ I B for some index-set I.If B is a torsion-free Abelian group of finite rank, then H B and T B induce an equivalence between the category of B-projective groups and the category of projective right E(B)-modules.Finally, a torsionfree group of finite rank is a finitely faithful S-group if r p (E) = [r p (A)] 2 where r p (G) = dim /p G/pG denotes the p-rank of a torsion-free group G. Arnold showed in [6] Since there exists a closed and unbounded subset E of ω 1 such that C α = A α for all α ∈ E, we may assume that each A α is A-generated.
Since A α is B-generated, it is a B 0 -module where B 0 is the subring of Q generated by { 1 p |B = pB}.Because A is reduced, A α is isomorphic to a subgroups of B Iα for some index-set I α by [3,Proposition 2.2].Since E(B) is right hereditary, it is also right and left Noetherian by [12], and H B (B Iα ) is ℵ 1 -projective.By [2], A α is B-solvable, and where the first two maps are isomorphisms since B is slender.By the Snake Lemma, ψ A α+1 /Aα is a monomorphism.Thus, A α+1 /A α can be embedded into a group of the form B I .Arguing as before, we obtain that A α+1 /A α is Bprojective.Since E(B) is hereditary, the every submodule of the last module in the induced sequence 0 splits.Thus, the bottom row splits too, say Conversely, observe that a B-projective group A is of the form A = ⊕ α<ω 1 B α where each B α is isomorphic to a B-generated subgroup of B since E is right hereditary.Clearly, A α = ⊕ β<α B β is a B-cobalanced filtration.By the well-known Corner-Dugas-Goebel construction, there exists an ℵ 1free left E op -module M such that End (M + ) is E op .Viewing M as a right E-module gives an ℵ 1 -projective E-module which is not projective.By [1], φ M is an isomorphism, and A = T B (M) is an ℵ 1 -B-projective, B-solvable Abelian group which does not have a B-balanced ℵ 1 -filtration.
Furthermore, we obtain Corollary 6.The following statements are undecidable in ZFC: a) If G has cardinality ℵ 1 and Ext(G, Z) is torsion-free, then G has a Zcobalanced filtration of countable subgroups; b) There exists a reduced group H with Ext(H, Z) torsion-free and for which one can find an exact sequence 0 → Z → G → H → 0 with Ext(G, Z) is not torsion-free.
Furthermore, the group A is B-generated if there exists an exact sequence ⊕ I B → A → 0. If A has regular cardinality κ, and |B| < κ, then A is B-generated if and only if has a κ-filtration of B-generated subgroups.Before continuing, we want to remind the reader of some terminology.For every Abelian group B with endomorphism ring E(B), we have a pair (H B , T B ) of adjoint functors between the category of Abelian groups and the category of right E(B)-modules defined by H B (G) = Hom(B, G) and T B (M) = M ⊗ E(B) B for all Abelian groups G and all right E(B)-modules M. Associated with these functors are natural maps θ G : T B H B (G) → G and φ M : M → H B T B (M).

Corollary 5 .
Let B be a finitely faithful S-group.If A is an ℵ 1 -B-projective group of cardinality ℵ 1 , which is not B-projective, then A does not have a B-cobalanced ℵ 1 -filtration with factors in * B.
that a torsion-free group B of finite rank is a finitely faithful S-group if an only if B ∈ * B. Furthermore, every finitely faithful S-group has a hereditary endomorphism ring.Let B a finitely faithful S-group.A reduced B-generated group A of cardinality ℵ 1 has a B-cobalanced filtration of countable subgroups with factors in where the last map is multiplication by p.Therefore, Ext(A, B) is torsion-free, once we have shown that the first map in this sequence is onto.To see this, let f ∈ Hom(A, B/pB).Suppose that f |A α ∈ Hom(A α , B/pB) lifts to a map a) If A is a torsion-free group which has a κ-filtration {A α } α<κ of subgroupsA α such that S = {α < κ | A α+1 /A α / ∈ * B} is not stationary, then A ∈ * B. b) * B if and only if A is B-projective.Proof.a) Since S is not stationary, there is a closed and unbounded subset C of κ with S ∩ C = ∅.Considering the κ-filtration {A α : α ∈ C} of A permits to assume that A α+1 /A α ∈ * B for all α.Observe that multiplication by a prime p induces the exact sequence Hom(A, B) → Hom(A, B/pB) → Ext(A, B) → Ext(A, B)