4. THE t-SOCLE AND t-RADICAL

E. L. Lady

July 10, 1998

Recall Example 1.45. In that example we constructed a rank two module G for which there existed three primes p_{1} , p_{2} , p_{3} such that the three submodules G(p_{i}^{∞}G) had rank one and were distinct. We were then able to argue that if φ ∈ End G then φ must leave the three subspaces Q(p_{i}^{∞}G) distinct and that therefore φ must be given by multiplication by a scalar in Q, so that End G ⊆ Q. In particular, we concluded that End G has no proper itempotents so that G is indecomposable.

There are variations of this example for which the concept p^{∞}G is not adequate.

example 4.1. Suppose that Spec W is infinite and let P_{1} , P_{2} , and P_{3} be three mutually disjoint infinite sets of primes. Let G be the submodule of Q ⊕ Q generated by W ⊕ W together with p_{1}^{−1}(1, 0) for all p_{1} ∈ P_{1} , p_{2}^{−1}(0, 1) for all p_{2} ∈ P_{2} , and p_{3}^{−1}(1, 1) for all p_{3} ∈ P_{3} . Then p^{−∞}G = 0 for all primes p. However we can note that if t_{1} = t(1, 0), t_{2} = t(0, 1), and t_{3} = t(1, 1) then these are three mutually incomparable types. We see that for each i = 1, 2, 3 the set of elements g ∈ G with type t_{i} forms a pure rank-one submodule of G which by Proposition 2.21 must be invariant under any endomorphism of G. Just as in Example 1.45 we can then argue that any φ ∈ End G must have three different eigenvectors, hence must be given by multiplication by a scalar, so that End G is isomorphic to a subring of Q.

In fact, as shown in Example 2.28 we could take any three non-trivial submodules A_{1} , A_{2} , A_{3} of Q such that the three types t_{i} = t(A_{i}) are mutually incomparable and such that A_{i} ∩ A_{j} is the same for all pairs i ≠ j. We can then construct a module G ⊆ Q ⊕ Q by setting G = A_{1}(1, 0) + A_{2}(0, 1) + A_{3}(1, 1). Again, we see that the three rank-one submodules A_{1}(1, 0), A_{2}(0, 1), and A_{3}(1, 1) must be fully invariant, so that End G is isomorphic to a subring of Q.

The relevant concept here is as follows.

definition 4.2. Let A be a rank-one module and t = t(A). For any torsion free module G we define G(t) = {g ∈ G | t(g) ≥ t}. G(t) is sometimes called the t-socle of G. It is a fully invariant pure submodule of G (see Proposition 4.1 and Proposition 4.2 below). In Example 2.28, for instance, as described above, A_{1}(1, 0) = G(t_{1}), A_{2}(0, 1) = G(t_{2}). and A_{3}(1, 1) = G(t_{3}). proposition 4.3. Let t = t(A). Define σ: A ⊗ Hom(A, G) → G to be the evaluation map a ⊗ g ↦ g(a).

(1) σ is a monomorphism with image G(t).

(2) G(t) is a pure submodule of G.

(3) G(t) is generated by the set of rank-one submodules of G isomorphic to A.

proof: (2) If g ∈ G(t) and g′ ∈ G ∩ Qg then by Proposition 2.* t(g′) = t(g) ≥ t so g′ ∈ G(t). Thus G ∩ Q(G(t)) ⊆ G(t) so G(t) is a pure submodule of G.

(3) If g ∈ G then t(g) = t(G ∩ Qg). Thus t(g) ≥ t if and only if t(G ∩ Qg) ≥ t and if this is so then G ∩ Qg ⊆ G(t) because G ∩ Qg has rank-one and by (2) G(t) ◁ G (see Proposition 2.2). Thus every element of G(t) is contained in a pure rank-one submodule B of G with t(B) ≥ t. It thus suffices to see that if B is such a submodule of G then B is generated by rank-one submodules isomorphic to A. At this point we may assume wlog that A and B are submodules of Q. Then in fact by Proposition 2.11 B is generated by all modules of the form wA for w ∈ [B: A]. Since wA ≈ A, this gives the result.

(1) We may suppose A ⊆ Q. As usual, we make the identification Hom(A, G) ⊆ Hom(QA, QG). Then σ is the restriction of the evaluation map σ′: Q ⊗ Hom(Q, QG) → QG. But σ′ is known to be an isomorphism. Thus σ is monic.

On the other hand, part (3) shows that G(t) is generated by all submodules of the form φ(A) for φ ∈ Hom(A, G), and this says that σ is surjective. Therefore σ is an isomorphism. ✓

Analogously to the t-socle we define the t-radical of a torsion free module by setting G[t] = ⋂{ Ker φ | φ ∈ Hom(G, A)}, where A is a rank-one module with t(A) = t. In other words, G[t] consists of those g ∈ G which are sent to 0 by every possible homomorphism of G into a rank-one module of type t. We can see from Proposition 2.* that if 0 ≠ g ∈ G then g ∈ G[t] if and only if whenever H is a pure submodule of G with g ∉ H and rank G/H = 1, then t(G/H) ≰ t.

proposition 4.4. (1) The t-socle and t-radical are subfunctors of the identity, i.e. if φ ∈ Hom(G, H) then φ(G(t)) ⊆ H(t) and φ(G[t]) ⊆ H[t]. Furthermore if φ ∈ QHom(G, H) then φ restricts to quasi-homomorphisms from G(t) to H(t) and from G[t] to H[t].

(2) If G and H are quasi-equal then G(t) and H(t) are quasi-equal, and likewise for G[t] and H[t].

(3) The t-socle and t-radical respect direct sums, i.e. (G_{1} ⊕ G_{2})(t) = G_{1}(t) ⊕ G_{2}(t) and (G_{1} ⊕ G_{2})[t] = G_{1}[t] ⊕ G_{2}[t].

(4) If H ◁ G then H(t) = H ∩ G(t).

(5) If H ⊆ G[t] then G[t]/H = (G/H)[t].

proof: (1) Let φ: G → H. If g ∈ G(t) then t(g) ≥ t and hence by Proposition 2.21 t(φ(g)) ≥ t, so that φ(g) ∈ H(t). Now let g ∈ G[t] and let A be a rank-one module with t(A) = t. If γ ∈ Hom(H, A) then γφ ∈ Hom(G, A) so γφ(g) = 0 since g ∈ G[t], showing that φ(g) ∈ H[t].

(2) As shown in (1), the t-socle and t-radical are functors, and clearly are additive functors. (More specifically, we can think of the t-socle, for instance, as a functor F where F(G) = G(t) and if φ: G → H then F(φ): G(t) → H(t) is simply the restriction of F.) Since additive functors respect direct sums (Proposition 0.*) it follows that the t-socle and t-radical respect direct sums.

(3) By Proposition 2.21 if h ∈ H then t_{H}(h) = t_{G}(h) so that h ∈ H(t) ⟺ t_{H}(h) ≥ t ⟺ t_{G}(h) ≥ t ⟺ h ∈ G(t).

(4) Let K ⊆ G be the inverse image of (G/H)[t] in G and let A be a rank-one module with t(A) = t. If H ⊆ G[t] and φ: G → A then φ(H) = 0, so that φ induces φ̄: G/H → A. Then φ(K) = φ̄((G/H)[t]) = 0. Since this is true for every φ ∈ Hom(G, A), K ⊆ G[t]. On the other hand, if γ: G → G/H is the quotient map then by (1) γ(G[t]) ⊆ (G/H)[t] so that G[t] ⊆ K. Thus K = G[t] and so

(G/H)[t] = K/H = G[t]/H. ✓

proposition 4.5. Let A be a rank-one module and t = t(A). Define ρ: G → Hom(Hom(G, A), A) by ρ(g)(φ) = φ(g).

(1) Ker ρ = G[t].

(2) G[t] is the intersection of all pure submodules H of G such that rank G/H = 1 and t(G/H) ≤ t.

(3) rank G[t] = rank G − rank Hom(G, A).

proof: (1) g ∈ G[t] ⟺ φ(g) = 0 for all φ ∈ Hom(G, A) ⟺ ρ(g)(φ) = 0 for all φ ∈ Hom(G, A) ⟺ ρ(g) = 0.

(2) This just rephrases the definition of G[t]. Explain!

✠

(3) If φ ≠ 0 ∈ Hom(G, A) then rank Ker φ = rank G − 1. It follows that in order for G[t] to be the intersection of Ker φ for various φ ∈ Hom(G, A) there must be at least rank G − rank G[t] linearly independant φ of this sort. In other words, rank Hom(G, A) ≥ rank G − rank G[t]. On the other hand, since φ(G[t]) = 0 for all φ ∈ G[t], Hom(G, A) ≈ Hom(G/G[t], A) so rank Hom(G, A) ≤ rank G/G[t] = rank G − rank G[t]. ✓

Assertions (3) and (4) in Proposition 4.4 need not hold without the hypotheses that T(G) and CT(G) are finite respectively. For instance if t = IT(G) then G(s) ⊊ G(t) = G for every s > t. But Example 2.* is an example of a module G with infinite typeset such that IT(G) ∉ T(G).

examples 4.6. (1) Consider the module G = A_{1}(1, 0) + A_{2}(0, 1) + A_{3}(1, 1) in Example 2.28, as described in the beginning of this chapter. OT(G) consists of t_{1} ∨ t_{2} ∨ t_{3} and t_{i} ∨ t_{j} for i ≠ j = 1, 2, 3. Then G[t_{1} ∨ t_{2} ∨ t_{3}] = 0, G[t_{1} ∨ t_{2}] = A_{3} , G[t_{1} ∨ t_{3}] = A_{2} , and G[t_{2} ∨ t_{3}] = A_{1} .

(2) Let G be the Pontryagin module (Example 1.47). If t is p-divisible then G[t] = 0, otherwise G[t] = G. proof: (1) If μ(G) is a rank-one homomorphic image of G let B_{i} = μ(A_{i}(x, y)) for i = 1, 2, 3 and x, y = 0, 1 chosen appropriately so that A_{i}(x, y) ⊆ G. (Here, as previously, A_{i}(x, y) denotes the set of elements (ax, ay) for a ∈ A_{i}.) Then since G = A_{1}(1, 0) + A _{2}(0, 1) + A_{3}(1, 1), μ(G) = B_{1} + B_{2} + B_{3} so that t(μ(G)) is t_{1} ∨ t_{2} , t_{1} ∨ t_{3} , t_{2} ∨ t_{3} or t_{1} ∨ t_{2} ∨ t_{3} , depending on whether one or none of the B_{i} are trivial. Now B_{1} , for instance, is trivial if and only if A_{1}(1, 0) ⊆ Ker μ. Since t_{i} ∨ t_{1} ≰ t_{2} ∨ t_{3} for i = 1, 2, it follows that t(μ(G)) ≤ t_{2} ∨ t_{3} ⟺ A_{1} ⊆ Ker μ so that G[t_{2} ∨ t_{3}] = A_{1}(1, 0). And analogously for the other possibilities.

(2) G is p-reduced and p-rank G = 1 < rank G. Thus if H ◁ G then 0 < p-rank H ≤ p-rank G = 1, so that p-rank H = p-rank G and p-rank G/H = 0, so that G/H is p-divisible. In particular, if A is a rank-one module which is not p-divisible then Hom(G, A) = 0. Thus G[t] = G for all types t which are not p-divisible.

On the other hand, if p′ is a prime different from p then p′-rank G = rank G, so that by Proposition 1.25 every rank-one homomorphic image of G_{p′} is a free W_{p′}-module. It follows that if G/H is a rank-one homomorphic image of G then G/H ≈ p^{−∞} and so t(G/H) ≤ t for any p-divisible type t. It then follows from Proposition 4.5 that G[t] = 0.

FIX THIS! ✓ ✠

proposition 4.7. (1) If s ≤ t then G(s) ⊇ G(t) and G[s] ⊇ G[t].

(2) G(s ∨ t) = G(s)(t) = G(s) ∩ G(t).

(3) G[s](t) = G[s] ∩ G(t).

(4) If T(G) is finite then t ∈ T(G) if and only if G(t) ⊋ G(s) for every s > t.

(5) If CT(G) is finite then t ∈ CT(G) if and only if G[t] ⊊ G[s] for every s < t.

(6) If s ≰ t then G(s) ⊆ G[t].

proof: (1) If s ≤ t and g ∈ G(t) then t(g) ≥ t ≥ s so that g ∈ G(s).

Now let A and B be rank-one modules such that t(A) = s and t(B) = t. If s ≤ t we may suppose wlog that A ⊆ B. Then Hom(G, A) ⊆ Hom(G, B) so if g ∈ G[t] and thus γ(g) = 0 for all γ ∈ Hom(G, B) then a fortiori γ(g) = 0 for all γ ∈ Hom(G, A), so that g ∈ G[s].

(2) By definition of s ∨ t, for g ∈ G, t(g) ≥ s ∨ t ⟺ t(g) ≥ s & t(g) ≥ t. Thus G(s ∨ t) = G(s) ∩ G(t). Furthermore since G(s) ◁ G, G(s)(t) = G(s) ∩ G(t) by Proposition 4.4.

(3) Since G[s] is a pure submodule of G, G[s](t) = G[s] ∩ G(t) by Proposition 4.4.

(4) If t ∈ T(G) then there exists g ∈ G with t(g) = t, and so g ∈ G(t) but g ∉ G(s) for s > t. Thus G(t) ⊊ G(s). Conversely if t ∉ T(G) then no element of G has type t, so if g ∈ G(t) then t(g) > t, which means that QG(t) = ⋃_{s>t} QG(s), where the union can be taken over s with s ∈ T(G) and s > t. If T(G) is finite then since a vector space cannot be a finite union of proper subspaces we conclude that QG(t) = QG(s) for some s > t. Since G(t) and G(s) are pure submodules of G, we conclude that G(t) = G(s).

(5) ( ⇐ ): If G[t] ⊊ G[s] for every s < t then by Proposition 4.5, rank Hom(G, B) < rank Hom(G, A) so QHom(G, B) ⊊ QHom(G, A) whenever A is a rank-one module with t(A) = t and B a rank-one module with t(B) = s < t. Now if CT(G) is finite, then there are only finitely many B with t(B) ∈ CT(G) and t(B) < t (Explain!). Since a vector space over the infinite field Q cannot be a union of a finite number of proper subspaces, thus there exists φ ∈ Hom(G, A) with φ ∉ QHom(G, B) for any B with t(B) < t. Thus t(φ(A)) = t so t ∈ CT(G).

( ⇒ ): If t ∈ CT(G) and s < t and B ⊊ A are rank-one modules with t(A) = t, t(B) = s, then there exists φ ∈ Hom(G, A) with φ(G) = A. Thus φ ∉ Hom(G, B) so that Hom(G, B) ⊊ Hom(G, A). Thus by Proposition 4.6 rank G[t] < rank G[s] so that G[t] ⊊ G[s].

(6) If s ≰ t and A is a rank-one module with t(A) = t and φ ∈ Hom(G, A) then A(s) = 0 and so by Proposition 4.4 φ(G(s)) ⊆ A(s) = 0. Thus G(s) ⊆ G[t]. ✓

DOMINATION. We say that H dominates G if IT(H) ≥ OT(G).

proposition 4.8. (1) H dominates G if and only if QHom(G, H) = Hom(QG, QH).

(2) If H dominates G then G is isomorphic to a submodule of H^{r}, for any r such that r(rank H) ≥ rank G.

(3) Domination is transitive, i.e. if K dominates H and H dominates G, then K dominates G.

proof: (1) Since Hom(QG, QH) is generated by those linear transformations QG → QH with one-dimensional image, QHom(G, H) = Hom(QG, QH) if and only if QHom(G, H) contains every φ: QG → QH with dim φ(QG) = 1, i.e. if and only if for every φ: G → QH with rank φ(G) = 1 there exists w ≠ 0 ∈ W with wφ(G) ⊆ H. The latter condition holds if and only if t(φ(G)) ≤ t(H ∩ φ(QG)). So QHom(G, H) = Hom(QG, QH) ⟺ the type of every rank-one homomorphic image of G is less than or equal to the type of every pure rank-one submodule of H, i.e. s ≤ t for all s ∈ CT(G), t ∈ T(H). But this is equivalent to saying OT(G) ≤ IT(H). (Rewrite this!)

(2) If r(rank H) ≥ rank G then dim QG ≤ dim QH^{r}, so QG is isomorphic to a subspace of QH^{r}. If H dominates G then clearly H^{r} also dominates G and so if φ: QG → QH^{r} is a monomorphism then φ ∈ QHom(G, H^{r}). Then for some w ≠ 0 ∈ W, wφ is a monomorphism from G into H^{r}.

(3) Recall from Proposition 2.33 that IT(H) ≤ OT(H). Thus if K dominates H and H dominates G then OT(G) ≤ IT(H) ≤ OT(H) ≤ IT(K), so that K dominates

G. ✓

proposition 4.9. Let H dominate G and let p be a prime. Then (1) Either H is p-divisible or G_{p} is a free W_{p}-module.

(2) p-rank Hom(G, H) = (p-rank G)(p-rank H).

(3) Hom(G, H)_{p} = Hom(G_{p}, H_{p}).

proof: (1) If H is not p-divisible then IT(H) is not p-divisible. If H dominates G, then OT(G) ≤ IT(H), so OT(G) is not p-divisible. Thus by Proposition 2.33 p-rank G = rank G so that G_{p} is a free W_{p}-module by Proposition 1.25.

(3) By Proposition 1.39, Hom(G, H)_{p} ◁ Hom(G_{p}, H_{p}). But since H dominates G, Hom(QG, QH) = QHom(G, H) ⊆ QHom(G_{p}, H_{p}) ⊆ Hom(QG, QH), so that Hom(G, H) is also an essential submodule of Hom(G_{p}, H_{p}). Hence by Proposition 1.17 they are identical.

(2) By Proposition 1.24, p-rank Hom(G, H) = p-rank Hom(G, H)_{p}. By (3) Hom(G, H)_{p} = Hom(G_{p}, H_{p}). Thus by (1) we need consider only the case where G_{p} is a free W_{p}-module or the case where H_{p} is divisible. In the first case Hom(G_{p}, H_{p}) ≈ H_{p}^{r}, with r = rank G = p-rank G, and in the second case by Corollary 1.34 Hom(G_{p}, H_{p}) is also divisible, so that p-rank Hom(G, H) = 0. In either case, the result is clear. ✓

Proposition 4.9 shows that if G and H are p-local modules then H dominates G if and only if either H is divisible or G is a free W_{p}-module. Thus domination is not an interesting concept for p-local modules.

proposition 4.10. Let K dominate G.

(1) If G′ ◁ G then the restriction map Hom(G, K) → Hom(G′, K) is a surjection.

(2) If φ: K → K′ is a surjection, then φ_{∗}: Hom(G, K) → Hom(G, K′) is surjective.

(3) If φ: K → G is a surjection, then φ splits.

(4) If K ◁ G then K is a summand of G.

proof: (1) By Proposition 0.* it suffices to prove that for each prime p the map Hom(G, K)_{p} → Hom(G′, K)_{p} is surjective. But by Proposition 4.9 we can write this as Hom(G_{p}, K_{p}) → Hom(G′_{p}, K_{p}). Furthermore by Proposition 4.9 either K_{p} is divisible, hence injective, in which case clearly this map is a surjection, or G_{p} is projective, in which case G_{p}/G′_{p} is also projective and G′_{p} is a summand of G_{p} and once again the map is a surjection.

(2) Analogous. It suffices to prove that for each p the map Hom(G_{p}, K_{p}) → Hom(G_{p}, K′_{p}) is surjective. This is clear if G_{p} is free. On the other hand, if K_{p} is divisible then so is K′_{p} and the map in question simply becomes Hom(QG, QK) → Hom(QG, QK′), which is a surjection by vector space theory.

(3) Applying (2) with K′ = G we see that the induced map φ_{∗}: Hom(G, K) → Hom(G, G) is surjective. In particular, 1_{G} = φ_{∗}(δ) = φδ for some δ ∈ Hom(G, K), so that φ splits.

(4) Applying (1) with G′ = K we see that the restriction map Hom(G, K) → Hom(K, K) is surjective, so that there exists γ ∈ Hom(G, K) so that the restriction of γ to K is the identity. Thus γ is a projection from G onto

K. ✓

It needs to be noted that parts (3) and (4) of Proposition 4.10 are stated in somewhat fraudulent generality. If G is a homomorphic image of H and H dominates G then G dominates itself, since OT(G) ≤ OT(H) ≤ IT(G). Likewise if H ◁ G and H dominates G then H dominates itself. But we will see shortly that the only modules which dominate themselves are those which are direct sums of mutually quasi-isomorphic rank-one modules (see Corollary 4.15 below). proposition 4.11. If H dominates G then for any finite rank torsion free module K, H ⊗ K dominates G and Hom(G, H ⊗ K) ≈ Hom(G, H) ⊗ K. proof: (1) By Proposition 2.35 IT(H ⊗ K) = IT(H)IT(K) ≥ IT(H) ≥ OT(G).

(2) By Proposition 1.40 the canonical map Hom(G, H) ⊗ K → Hom(G, H ⊗ K) is a monomorphism with pure image. But since H dominates G, rank Hom(G, H) ⊗ K = (rank G)(rank H)(rank K) ≥ rank(Hom(G, H) ⊗ K). Therefore this map must also be surjective. ✓

t-SATURATED AND t-BOUNDED MODULES. We say that G is t-saturated if G(t) = G and t-bounded if G[t] = 0. We say that G is t-projective if it is completely decomposable and homogeneous of type t. In other words, G is t-projective if G = ⨁ A_{i}, where the A_{i} are rank-one modules with t(A_{i}) = t.

proposition 4.12. (1) Pure submodules and homomorphic images of t-saturated modules are t-saturated. Furthermore if G is an essential submodule of H and G is t-saturated then H is t-saturated.

(2) Submodules and homomorphic images of t-bounded modules are t-bounded.

(3) A module quasi-isomorphic to a t-saturated module is t-saturated and a module quasi-isomorphic to a t-bounded module is t-bounded.

(4) Let H be a pure submodule of G. Then G/H is t-bounded ⟺ H ⊇ G[t].

(5) If G_{1}, …, G_{n} are t-saturated submodules of G, then G_{1} + ⋯ + G_{n} is t-saturated.

(6) If G_{1}, …, G_{n} are t-bounded submodules of G, then G_{1} + ⋯ + G_{n} is t-bounded.

proof: (1) If G is t-saturated and φ: G → K is surjective then by Proposition 4.4 ✠

K(t) ⊇ φ(G(t)) = φ(G) = K, so K is t-saturated. And if H ◁ G then by Proposition 4.4 H(t) = H ∩ G(t) = H ∩ G = H, so H is t-saturated. And if G is an essential submodule of H then G = G(t) ⊆ H(t), so that H(t) is an essential submodule of H. Since H(t) ◁ H, H = H(t), so H is t-saturated.

(2) By Proposition 2.33 if H ⊆ G then OT(H) ≤ OT(G) and if H ◁ G then OT(G/H) ≤ OT(G). It then follows from (?next proposition?) that if G is bounded then so are H and G/H.

(3) If G and H are quasi-isomorphic and G is t-saturated then we may assume that G is an essential submodule of H. Therefore by (1) H is t-saturated. Likewise if H is t-bounded then by (5) G is t-bounded.

(4) ( ⇐ ): Let A be a rank-one module with t = t(A). Since Hom(G/G[t], A) ≈ Hom(G, A), by Proposition 4.5 rank Hom(G/G[t], A) = rank Hom(G, A) = rank G/G[t] = rank Hom(G/G[t], Q) = rank Hom(G/G[t], QA), so that by Proposition 4.8 A dominates G/G[t], so OT(G/G[t] ≤ IT(A) = t. But if H ⊇ G[t] then G/H is a homomorphic image of G/G[t] so by Proposition 2.33, OT(G/H) ≤ OT(G/G[t]) ≤ t and so G/H is t-bounded.

( ⇒ ): If G/H is t-bounded let g ∈ G[t]. Let A be a rank-one module with t(A) = t and φ̄: G/H → A. Then φ̄ induces φ: G → A and since g ∈ G[t], then φ(g) = 0. Thus φ̄(g + H) = 0. Since this is true for every φ̄ mapping G into A then g + H ∈ (G/H)[t] = 0, i.e. g ∈ H. Thus G[t] ⊆ H.

(5) & (6) G_{1} + ⋯ + G_{n} is a homomorphic image of G_{1} ⊕ ⋯ ⊕ G_{n}, which is t-saturated or t-bounded if each G_{i} is. Thus the result follows from (1) and (2). ✓

proposition 4.13. (1) G is t-saturated if and only if IT(G) ≥ t.

(2) G is t-bounded if and only if OT(G) ≤ t.

(3) If A is a rank-one module with t = t(A) then G is t-saturated if and only if G dominates A and G is t-bounded if and only if A dominates G.

(4) A module is t-saturated if and only if it is a homomorphic image of a (possibly infinite) direct sum of rank-one modules of type t.

(5) A module is t-bounded if and only if it is isomorphic to a submodule of a t-projective module.

(6) If t is not p-divisible and G is t-bounded then p-rank G = rank G and G_{p} is a free W_{p}-module.

(7) If G is t-bounded and H t-saturated, then H dominates G.

proof: (1) G is t-saturated if and only if G = G(t), i.e. if and only if t(g) ≥ t for all g ∈ G. This is true if and only if IT(G) ≥ t.

(2) ( ⇐ ): Suppose OT(G) ≤ t and let g ≠ 0 ∈ G. Then there exists a map φ: G → Q with φ(g) ≠ 0. Since OT(G) ≤ t, t(φ(G)) ≤ t. Thus there is a non-trivial map from G into the rank-one module φ(G) with t(φ(G)) ≤ t , so g ∉ G[t]. Thus G[t] = 0, so G is t-bounded.

( ⇒ ): Let G be t-bounded and let A be a rank-one module such that t(A) = t. Then by Proposition 4.5 rank Hom(G, A) = rank G/G[t] = rank G = rank Hom(G, Q) = rank Hom(G, QA). Thus QHom(G, A) = Hom(G, QA) so that by Proposition 4.8, A dominates G, i.e. OT(G) ≤ IT(A) = t(A).

(3) This is just a restatement of (1) and (2).

(4) Let G be t-saturated. Then G is a homomorphic image of ⨁{A ⊆ G | rank A = 1 and t(A) = t}. The converse follows from Proposition 4.12 (1).

✠ ????

(5) ( ⇒ ): By (3) G is t-bounded if and only if A dominates G, where t = t(A). Thus by Proposition 4.8 G is isomorphic to a submodule of A^{r}, i.e. to a submodule of a t-projective module of rank r.

( ⇐ ): It A is a rank-one module with t(A) = t then A is t-bounded by (1). It follows that t-projective modules are t-bounded, hence submodules of t-projective modules are t-bounded by Proposition 4.12 (2).

(6) If G is t-bounded and t is not p-divisible, then OT(G) is not p-divisible since OT(G) ≤ t. Thus by Proposition 2.33 p-rank G = rank G. Then by Proposition 1.25, G_{p} is a free W_{p}-module.

(7) If G is t-bounded and H is t-saturated, then OT(G) ≤ t ≤ IT(H). Hence H dominates G. ✓

The following Proposition will partially justify the use of the term “t-projective.” proposition 4.14. Let A be a subring of Q, let t = t(A), and let G is any finite rank torsion free module.

(1) G is t-saturated if and only if G is isomorphic to an A-module.

(2) G is t-projective if and only if G is isomorphic to a projective A-module.

proof: (1) If G is an A-module and g ≠ 0 ∈ G then Ag ⊆ G ∩ Qg and since Ag ≈ A we see that t(A) ≤ t(G ∩ Qg) = t(g), so g ∈ G(t). Thus G is t-saturated. Conversely if G is t-saturated then we will show that G is an A-submodule of QG, i.e that AG ⊆ G. If g ≠ 0 ∈ G then Ag ⊆ QG and Ag ≈ A. Then t(G ∩ Ag) = t(G ∩ Qg ∩ Ag) = t(g) ∧ t(A) = t(A) since t(g) ≥ t(A). Thus by Proposition 2.15, G ∩ Ag is a finitely generated A-submodule of Qg, and hence of QG. Thus Ag ⊆ A(G ∩ Ag) ⊆ G ∩ Ag ⊆ G. Since this holds for all g ∈ G, G is an A-submodule of QG.

(2) By Proposition 0.* A is a dedekind domain. (Recall that by convention the word “subring” means a W -subalgebra, i.e. A is a subring of Q containing W.) Thus an A-module is projective if and only if it is a direct sum of A-ideals. On the other hand, G is t-projective if and only if G is a direct sum of rank-one modules of type t. But by Proposition 2.15 a rank-one module has type t if and only if it is isomorphic to a fractional ideal in A, and hence to a projective A-module. Hence G is t-projective if and only if G is isomorphic to a projective A-module. ✓

examples 4.15. (1) It t = t(W) then for any G, G(t) = G and G[t] is a summand of G and is the smallest pure submodule H of G such that G/H is projective. Every G is t-saturated and G is t-bounded if and only if it is projective.

(2) If t = t(Q) then for any G, G[t] = 0 and G(t) = d(G). Every G is t-bounded and G is t-saturated if and only if it is divisible.

proof: By Proposition 2.32 IT(G) ≥ t(W) and OT(G) ≤ t(Q) for every G, and OT(G) = t(W) if and only if G is projective and IT(G) = t(Q) if and only if G is divisible. Thus since by Proposition 4.12 (6) G[t(W)] is the maximal pure submodule H of G such that G/H is t-bounded, it follows that it is the maximal pure submodule such that G/H is projective. Since G/G[t] is projective, G[t] is a summand.

Likewise G(t(Q)) is the maximal t(Q)-saturated submodule of G, i.e. the maximal divisible submodule of G. ✓

BAER’S LEMMA. What makes everything work is the following theorem, due to Reinhold Baer. It is essentially just a special case of Proposition 4.10, however it seems worthwhile to also show how the traditional proof can be adapted to the dedekind domain case.

theorem 4.16. If H ◁ G and G/H is t-bounded and G = H + G(t), then G = H ⊕ L for some L ⊆ G(t). first proof. G(t)/(G(t) ∩ H) ≈ (G(t) + H)/H = G/H. Now since G(t) is t-saturated and G/H is t-bounded, G(t) dominates G/H. Thus by Proposition 4.10 G(t) ∩ H is a summand of G(t), say G(t) = (G(t) ∩ H) ⊕ L. Then since L ⊆ G(t), H ∩ L = G(t) ∩ H ∩ L = 0 and H ⊕ L = H + (G(t) ∩ H) + L = H + G(t) = G.

a more traditional proof: For this proof, we need to assume that G/H is t-projective. It turns out that this involves no loss of generality, since the result in this case is adequate to prove Corollary 4.14 below, which implies that if G/H is t-bounded then it must be t-projective. Assume then that G/H = ⨁ A_{i}, where theA_{i} are rank-one modules of type t. Let φ: G → G/H be the quotient map. We need to find σ: G/H → G splitting φ. It suffices to find σ_{i}: A_{i} → G for each i such that φσ_{i} = 1_{A_{i}}.

✠

Now fix i. Since G = G(t) + Ker φ, there exists a pure rank-one submodule B of G(t) with 0 ≠ φ(B) ⊆ A_{i}. Then t(B) ≥ t and also since Hom(B, A_{i}) ≠ 0, t(B) ≤ t. So t(φ(B)) = t(B) = t = t(A_{i}). Thus by Proposition 2.2, A_{i}/φ(B) is cyclic. Let a ∈ A_{i} be a pre-image for a generator of A_{i}/φ(B). Since A ⊆ φ(G(t)) there exists a pure rank-one submodule C of G(t) with a ∈ φ(C). Thus φ(B) + φ(C) = A_{i}. The restrictions of φ to B and C are not trivial, hence are monic by Proposition 2.*, so there exist σ_{B}: φ(B) → B and σ_{C}: φ(C) → C such that φσ_{B} = 1_{B} and φσ_{C} = 1_{C}. Now by Proposition 2.* [A_{i}: A_{i}] = [φ(B) + φ(C): A_{i}] = [φ(B): A_{i}] + [φ(C): A_{i}], so there exist in particular u ∈ [φ(B): A_{i}] and v ∈ [φ(C): A_{i}] such that u + v = 1. Now define σ: A_{i} → G by σ_{i}(x) = σ_{B}(ux) + σ_{C}(vx). Note that by construction ux ∈ φ(B) and vx ∈ φ(C), and furthermore φσ_{i}(x) = φσ_{B}(ux) + φσ_{C}(vx) = ux + vx = x. I.e. φσ_{i} = 1_{A_{i}}.

Now combining these maps σ_{i} yields a map σ: G/H = ⨁ A_{i} → G such that φσ = 1_{G/H}. Thus φ splits and H is a summand of G. ✓

proposition 4.17. A finite rank torsion free module G is both t-saturated and t-bounded if and only if G is t-projective. proof: ( ⇐ ): Clear.

( ⇒ ): By induction on rank G. Suppose that G is both t-saturated and t-bounded. If G has rank-one we see that t(G) = t and G is trivially t-projective. If rank G > 1, let H ◁ G be such that rank G/H = 1. Then by Proposition 4.12 G/H is t-bounded and so by Baer’s Lemma, G ≈ H ⊕ G/H. By induction, H is t-projective. Therefore G is t-projective. ✓

corollary 4.18. (1) If G and H are finite rank torsion free modules then G and H mutually dominate each other if and only if there exists a type t such that G and H are each t-projective.

(2) If G and H are t-projective modules and QG = QH then G and H are quasi-equal.

(3) A finite rank torsion free module G dominates itself if and only if G is t-projective for some type t.

proof: (1) G and H mutually dominate each other if and only if IT(G) ≤ OT(G) ≤ IT(H) ≤ OT(H) ≤ IT(G) (c.f. Proposition 2.33), i.e. if and only if G and H are both t-saturated and t-bounded, where t = IT(G) = IT(H) = OT(G) = OT(H). By Proposition 4.17 this is equivalent to G and H both being t-projective.

(2) If G and H are t-projective and QG = QH, then G and H dominate each other by (1). Thus 1_{QG} belongs to both QHom(G, H) and QHom(H, G). Thus G and H are quasi-equal.

(3) This follows by applying (1) to the case H = G. ✓

proposition 4.19. (1) Pure submodules and torsion free homomorphic images of t-projective modules are t-projective.

(2) A pure submodule of a t-projective module is a direct summand.

(3) A module quasi-isomorphic to a t-projective module is t-projective.

(4) A t-projective submodule H of a t-projective module G is a quasi-summand of G.

proof: (1) By Proposition 4.12 a pure submodule or homomorphic image of a t-projective module is both t-saturated and t-bounded. Thus the result follows from Proposition 4.17.

(2) If G is t-projective and H ◁ G, then by Proposition 4.17 G is t-saturated and by Proposition 4.12 G/H is t-bounded. Thus H is a summand of G by Baer’s Lemma (Theorem 4.16).

(3) If G is t-projective and H ∼ G, the H is t-bounded and t-saturated, by Proposition 4.17 H is t-projective.

(4) By (2) H_{∗} is a summand of H and hence by (1) is t-projective. Since H and H_{∗} have the same rank, it is thus clear that they are quasi-isomorphic. Thus by Proposition 3.9 the inclusion map H ↪ H_{∗} is a quasi-isomorphism. Hence since H_{∗} is a summand of G, H is a quasi-summand. ✓

The following proposition is extremely useful for avoiding calculations when constructing examples:

proposition 4.20. If H ◁ G and G = H + K where K is t-projective for some type t, then there exists a pure submodule K′ of K such that G = H ⊕ K′.

proof: We can apply Baer’s Lemma. In fact, since K ⊆ G(t), clearly G = H + G(t) and G/H is t-projective since it is a homomorphic image of K. Thus by Baer’s Lemma H is a summand of G. Now we need to see that the complementary summand can be chosen to be a submodule of K. One can verify this rather tediously by retracing the proof of Baer’s Lemma. A faster way is to note that by Baer’s Lemma K = (H ∩ K) ⊕ K′ for some K′ ⊆ K. As in the proof of Theorem 4.16 one then sees that G = H ⊕ K′. ✓ example 4.21. Consider again Example 1.49. In this example H is generated by B ⊕ C together with (b + c)/w where B and C are rank-one modules with incomparable types, and b ∈ B, c ∈ C are such that b ∉ wB, c ∉ wC. We let A be a module isomorphic to B and let a ∈ A correspond to b ∈ B under some isomorphism. Now let B_{2} be the pure rank-one submodule of A ⊕ H generated by a + 2b, and let H_{2} be the purification of B_{2} + C. Then there is an isomorphism θ from H_{2} to the submodule of QB ⊕ QC generated by B ⊕ 2C together with (b + 2c)/w, given by θ(a + 2b) = b and θ(2c) = c. We have seen that under suitable hypotheses on A and B, H and H_{2} are not isomorphic. But there exists a rank-one submodule A_{2} of A ⊕ H such that A_{2} ⊕ H_{2} = A ⊕ H, and if W is a principal ideal domain then A_{2} ≈ A. proof: Let G = A ⊕ H. Clearly G = H_{1} + (A ⊕ B). Since A ⊕ B is t(A)-projective, by Proposition 4.20 A ⊕ H = A_{2} ⊕ H_{2} for some rank-one submodule A_{2} of A ⊕ B. By Proposition 4.19 A_{2} is t-projective, i.e. t(A_{2}) = t(A). Thus if W is a principal ideal domain then A_{2} ≈ A by Corollary 2.10. ✓

In contrast to most of the theorems in this chapter and the next, the dual theorem to Baer’s Lemma is not valid. The dual theorem would assert that if H is a t-projective pure submodule of G such that H ∩ G[t] = 0 then H is a direct summand. We have the following simple counter-example:

example 4.22. Consider again Example 1.49. In this example H is generated by B ⊕ C together with (b + c)/w where B and C are rank-one modules with incomparable types, and b ∈ B, c ∈ C are such that b ∉ wB, c ∉ wC. Now let t = t(B). Then H[t] = C and B ∩ H[t] = 0. But B is not a direct summand of H. proof: If φ ∈ Hom(H, B) then φ(C) = 0 since t(C) and t are incomparable. Thus C ⊆ H[t]. On the other hand, since H/C ≈ B, H[t] ⊆ C by Proposition 4.12. We have seen earlier that H is indecomposable, so B is not a summand of H. ✓

However we do have the following two propositions, which are slightly weaker.

proposition 4.23. Let G be t-bounded and H a pure submodule of G. Let K be a t-saturated module. Then every homomorphism φ: H → K extends to a homomorphism G → K. proof: If G is t-bounded and K t-saturated then K dominates G. Hence this is a special case of Proposition 4.10. ✓

proposition 4.24. (1) If H is a t-saturated submodule of G and H ∩ G[t] = 0 then H is t-projective and G is quasi-equal to H ⊕ L for some submodule L with G[t] ⊆ L and H ⊕ G[t] is a quasi-pure submodule of G.

(2) If in addition H + G[t] is a pure submodule of G then G = H ⊕ L for some L with G[t] ⊆ L.

(3) If G is t-bounded and H is a pure t-saturated submodule of G, then H is a t-projective direct summand of G. proof: (3) If G is t-bounded and H t-saturated, then Proposition 4.23 applied to 1_{H} shows that H is a summand of G. Since H is both t-saturated and t-bounded, it is t-projective by Proposition 4.17.

(2) If H + G[t] = H ⊕ G[t] is a pure submodule of G then (H ⊕ G[t])/G[t] is a pure t-saturated submodule of the t-bounded module G/G[t], and hence is a summand of G/G[t] by (3). Thus there is a projection G/G[t] → (H ⊕ G[t])/G[t] which leads to a sequence of maps

G → G/G[t] → (H ⊕ G[t])/G[t] ⟶^{≈} H.

Let θ be the composition of these maps. Then θ restricts to the identity on H, showing that G = H ⊕ Ker θ. Note that G[t] ⊆ Ker θ.

(1) (H ⊕ G[t])/G[t] is t-bounded since it is a submodule of G/G[t] and is t-saturated since it is isomorphic to H. And (H ⊕ G[t])_{∗}/G[t] is also t-bounded and t-saturated. Hence these are both t-projective and hence quasi-isomorphic, since they have the same rank. Thus by Proposition 3.9 they are quasi-equal, so (H ⊕ G[t])/G[t] is a quasi-pure submodule of G/G[t]. Thus by Proposition 3.14 H ⊕ G[t] is a quasi-pure submodule of G. Now the reasoning in (2) applies, except that in this case we can only conclude that θ is a quasi-projection of G “quasi-onto” H. Thus G is quasi-equal to H ⊕ L for some L

with G[t] ⊆ L. ✓

proposition 4.25. If t = OT(G) then G(t) is t-projective and is a summand of G. proof: Since t = OT(G), G is t-bounded. Since G(t) ◁ G, by Proposition 4.12 G(t) is t-bounded. Since it is t-saturated, it is t-projective by Proposition 4.17 and is a summand of G by Proposition 4.24. ✓

proposition 4.26. Let G be a finite rank torsion free module.

(1) G(t) + G[t] = G_{t} ⊕ G[t], where G_{t} is a t-projective module.

(2) G(t) + G[t] is a quasi-pure submodule of G.

(3) G_{t} is a maximal t-projective quasi-summand of G.

(4) (G(t) + G[t])_{∗}/G[t] is a t-projective direct summand of G/G[t].

(5) G_{t} ∼ (G(t) + G[t])_{∗}/G[t].

proof: (1) (G(t) + G[t])/G[t] is t-bounded since it is a submodule of the t-bounded module G/G[t]. Hence by Theorem 4.16 G[t] is a summand of G(t) + G[t] and the complementary summand is t-projective.

(2) By Proposition 4.24 G_{t} ⊕ G[t] = G(t) + G[t] is quasi-pure in G.

(3) Since G_{t} is t-saturated and G_{t} ∩ G[t] = 0, by Proposition 4.24 G_{t} is a quasisummand of G. Clearly it is a maximal t-projective quasi-summand, since if H is a t-projective quasi-summand of G then H ⊆ G(t) and if π: G → H is a quasi-projection then π(G[t]) ⊆ H[t] = 0, so that H ∩ G[t] = 0.

(4) (G(t) + G[t])_{∗}/G[t] is a t-saturated submodule of the t-bounded module G/G[t]. Therefore it is a t-projective direct summand of G/G[t] by Proposition 4.24.

(5) G(t) + G[t] is quasi-equal to (G(t) + G[t])_{∗} by (2), so that

G_{t} ≈ (G(t) + G[t])/G[t] ∼ (G(t) + G[t])_{∗}/G[t]. ✓

COMPLETELY DECOMPOSABLE MODULES. Recall that a module is called completely decomposable if it is a direct sum of rank-one modules. We can now rephrase this by saying that a module is completely decomposable if it is a direct sum of t-projective modules for various t.

By Jónsson’s Theorem (Theorem 3.24) a maximal completely decomposable quasisummand of a module G is determined uniquely up to quasi-isomorphism. It is somewhat surprising, although quite elementary, that the same is true of a maximal completely decomposable actual summand of G.

proposition 4.27. (1) A maximal t-projective summand of a finite rank torsion free module G is unique up to quasi-isomorphism.

(2) A maximal completely decomposable summand of G is unique up to quasi-isomorphism.

proof: (1) Let G = C ⊕ H = D ⊕ K, where C and D are t-projective and H and K have no non-trivial t-projective summand. Then H ∩ D ◁ D so by Proposition 4.19 H ∩ D is a summand of D, and hence a summand of G, and hence a summand of H. But by assumption H has no non-trivial t-projective summand. Thus H ∩ D = 0, so the projection of G onto C restricts to a monomorphism from D into C. Likewise there is a monomorphism from C into D, so that by Proposition 3.1 C and D are quasi-isomorphic.

(2) Let G = H ⊕ ⨁ C_{t} where each C_{t} is t-projective and H has no non-trivial completely decomposable summand. Then in particular for each t, H has no t-projective summand so by (1) each C_{t} is unique up to quasi-isomorphism. Thus ⨁ C_{t} is unique up to quasi-isomorphism. ✓

Note that if W is a principal ideal domain then …

✠

In light of Proposition 4.27 it is very plausible to infer think that if C is a maximal completely decomposable summand of G and D a maximal completely decomposable summand of H, then C ⊕ D is a maximal completely decomposable summand of G ⊕ H. However, this need not be the case. In fact, as the following example shows, it is possible for G and H to both have no non-trivial completely decomposable summand at all and yet for G ⊕ H to have a very large one.

example 4.28. Let A_{1}, …, A_{n} be rank-one modules with mutually incomparable types, and for each i let B_{i} be a rank-one module with B_{i} ≈ A_{i}. Suppose further that there exist distinct primes p and q such that none of the A_{i} is p-divisible or q-divisible, and for each i choose a_{i} ∈ A_{i} with a_{i} ∉ pA_{i} and b_{i} ∈ B_{i} with b_{i} ∉ qB_{i}. Let G be the submodule of QA_{1} ⊕ ⋯ ⊕ QA_{n} generated by A_{1} ⊕ ⋯ ⊕ A_{n} together with p^{−1}(a_{1} + ⋯ + a_{n}) and let H be the submodule of QB_{1} ⊕ ⋯ ⊕ QB_{n} generated by B_{1} ⊕ ⋯ ⊕ B_{n} together with q^{−1}(b_{1} + ⋯ + b_{n}). Then G and H are indecomposable, and in particular have no non-trivial completely decomposable summands, but G ⊕ H has a completely decomposable summand of rank n. proof: ****

definition 4.29. We define G^{∗}(t) as the submodule generated by {g ∈ G | t(g) > t} and G^{∗}[t] as ⋂_{s<t} G[s]. (In other words, g ∈ G^{∗}[t] if and only if μ(g) = 0 whenever μ: G → A where A is a rank-one module with t(A) < t.)

proposition 4.30. (1) G^{∗}(t) ⊆ G(t) ∩ G[t].

(2) G(t) + G[t] ⊆ G^{∗}[t]

proof: (1) Clearly G^{∗}(t) ⊆ G(t). And for all g with t(g) > t, if φ: G → A with t(A) = t, then φ(g) = 0 by Proposition 2.21, so that g ∈ G[t]. Thus G^{∗}(t) ⊆ G[t].

(2) Since by Proposition 4.12 s < t ⇒ G[s] ⊇ G[t], it follows that G[t] ⊆ G^{∗}[t]. And if g ∈ G(t) then g ∈ G[s] for all s < t since t(g) ≥ t > s and so if φ: G → A with t(A) = s then φ(g) = 0. Thus g ∈ G^{∗}[t], and so G(t) ⊆ G^{∗}[t]. ✓

proposition 4.31. Let G be completely decomposable and G = ⨁ G_{t} , where for each t, G_{t} is t-projective. If t is any type then

(1) G(t) = ⨁_{s≥t} G_{s} .

(2) G^{∗}(t) = ⨁_{s>t} G_{s} .

(3) G[t] = ⨁_{s≰t} G_{s} .

(4) G^{∗}[t] = ⨁_{s≮t} G_{s} .

proof: Since G_{s} is t-projective it is t-saturated and t-bounded.

(1) G_{s}(t) = G_{s} if s ≥ t and G_{s}(t) = 0 if s ≱ t. Thus by Proposition 4.4 G[t] = ⨁ G_{s}[t] = ⨁_{s≥t} G_{s} .

(2) G_{s}^{∗}(t) = G_{s} if s > t and G_{s}^{∗}(t) = 0 otherwise. Thus G^{∗}(t) = ⨁ G_{s}^{∗}(t) = ⨁_{s>t} G_{s} .

(3) G_{s}[t] = 0 if s ≤ t and G_{s}[t] = G_{s} otherwise. Thus G[t] = ⨁ G_{s}[t] = ⨁_{s≰t} G_{s} .

(4) G_{s}^{∗}[t] = 0 if and only if G_{s}[t′] = 0 for some t′ < t and this is true if and only if s < t. Otherwise G_{s}^{∗}[t] = G_{s} . Thus G^{∗}[t] = ⨁ G_{s}^{∗}[t] = ⨁_{s≮t} G_{s} . (why?) ✓

If g ∈ G then to say that g ∈ G^{∗}[t] is to say that φ(g) = 0 for every φ: G → Q such that t(φ(g)) < t. Clearly this is true if either g ∈ G[t] or t(g) = t. We have already taken note of this above as the formula G(t) + G[t] ⊆ G^{∗}[t] in Proposition 4.30. It seems naively plausible that G[t] and G(t) would account for all of G^{∗}[t], i.e. that G^{∗}[t] = G(t) + G[t]. Surprisingly enough, though, this plausible equality is valid only if G is completely decomposable.

proposition 4.32. (1) A finite rank torsion free module G is completely decomposable if and only if if and only if G^{∗}[t] = G(t) + G[t] for all t ∈ CT(G).

(2) G is quasi-isomorphic to a completely decomposable module if and only if G^{∗}[t] = (G(t) + G[t])_{∗} for all t ∈ CT(G). proof: By Proposition 4.26 G(t) + G[t] is a quasi-pure submodule of G, so G^{∗}[t] = (G(t) + G[t])_{∗} if and only if G^{∗}[t] is quasi-equal to G(t) + G[t]. For every type t, define F_{t}(G) = G^{∗}[t]/(G(t) + G[t]). (Note that F_{t}(G) need not be torsion free.) Then F_{t} is a functor in the obvious way, and respects direct sums since it is additive. The assertion then is that G is completely decomposable ⟺ F_{t}(G) = 0 for all t ∈ CT(G) and G is quasi-isomorphic to a completely decomposable module if and only if F_{t}(G) has finite length for all t.

( ⇒ ): Since F_{t} respects direct sums, to see that F_{t}(G) = 0 for all types t when G is completely decomposable it suffices to prove that F_{t}(G) = 0 for modules G with rank G = 1. Now if rank G = 1 and t ≤ t(G) then G(t) = G^{∗}[t] = G, so G = G(t) + G[t] = G^{∗}[t] and F_{t}(G) = 0. On the other hand, if t > t(G) then G^{∗}[t] = 0 and again F_{t}(G) = 0. Finally, if rank G = 1 and t and t(G) are incomparable, then G[t] = G^{∗}[t] = G so that G = G[t] + G(t) = G^{∗}[t] and F_{t}(G) = 0 in this case as well.

Now if G is quasi-equal to a completely decomposable module G′ then G^{∗}[t] is quasi-equal to G^{′∗}[t] and G(t) + G[t] is quasi-equal to G′(t) + G′[t]. Thus if G^{∗}[t] = G(t) + G[t] then G^{′∗}[t] is quasi-equal to G′(t) + G′[t]. (???)

( ⇐ ): Now suppose F_{t}(G) has finite length for all t ∈ CT(G). We will prove by induction on rank that G is quasi-isomorphic to a completely decomposable module, and that if in fact F_{t}(G) = 0 for all t then G is completely decomposable. If rank G = 1 or, more generally, if CT(G) = {t} for some t, then G[t] = 0 and G^{∗}[t] = G, so that F_{t}(G) = 0 implies that G = G(t), so G is both t-saturated and t-bounded, hence is t-projective and a fortiori completely decomposable. Now in general note that by Proposition 2.26 CT(G) must contain minimal elements. If CT(G) contains more than one element, let s be minimal in CT(G). Then G[s′] = 0 for all s′ < s, so G^{∗}[s] = G. The hypothesis applied for t = s together with Proposition 4.26 yields that G = G^{∗}[s] is quasi-equal to G(s) + G[s] = G_{s} ⊕ G[s] for some s-projective module G_{s} . Then for any t the fact that F_{t} is an additive functor yields F_{t}(G) = F_{t}(G_{s}) ⊕ F_{t}(G[s]). Then by induction on rank we conclude that G[s] is quasi-equal to a completely decomposable module, so that G_{s} ⊕ G[s] is quasi-equal to q completely decomposable module, and thus G is quasi-equal to a completely decomposable module and is completely decomposable if F_{t}(G) = 0 for all t. ✓

In Proposition 4.19 we saw that a module quasi-isomorphic to a t-projective module is also t-projective. It is not in general true, however, that a module quasi-isomorphic to a completely decomposable module is completely decomposable. For instance in Example 1.49 the module H formed from the completely decomposable module B ⊕ C by adjoining an element (b + c)/w is quasi-isomorphic to B ⊕ C but is not itself completely decomposable: in fact, it is indecomposable and has rank two. H was obtained by filling in glue in the space between the two incomparable modules B and C. If we had tried to add glue on top of B or C, by adjoining to B ⊕ C an element b/w, for instance, with b ∈ B, the construction would not have worked. It would also not have worked if if t(B) = t(C) since in that case by Proposition 4.19 H would be t-projective. In fact, if t(B) = t(C) and H = (B ⊕ C) + W (b + c)/w, then let B′ be the rank-one pure submodule of H containing (b + c)/w. Then B′ ∼ B ∼ C and H = B′ ⊕ C. We will also see in Chapter 5 that H is necessarily completely decomposable if t(B) ≤ t(C) (Proposition 5.*).

Modules quasi-isomorphic to completely decomposable modules are called almost completely decomposable and will be studied in Chapter 5.

The idea that one can test for complete decomposability of a module G by seeing whether a family of functors F_{t} vanish on G is quite fascinating. As an immediate consequence, we get the first part of the following classical theorem.

theorem 4.33. (1) A direct summand of a completely decomposable module is completely decomposable.

(2) If G is completely decomposable, and G = ⨁ G_{t} where the G_{t} are t-projective modules, then the G_{t} are uniquely determined up to isomorphism. In fact, G_{t} ≈ (G(t) + G[t])/G[t] ≈ G(t)/G^{∗}(t). proof: (1) For any type t, define the functor F_{t} as in the proof of the preceding Proposition. Then G is completely decomposable if and only if F_{t}(G) = 0 for all t ∈ CT(G). But since F_{t} is an additive functor, F_{t}(H ⊕ K) = F_{t}(H) ⊕ F_{t}(K). Hence if H ⊕ K is completely decomposable, then so are H and K.

(2) Let G = ⨁ G_{t} where each G_{t} is t-projective. By Proposition 4.31 G(t) = ⨁_{s≥t} G_{s} and G[t] = ⨁_{s≰t} G_{s} ⊇ ⨁_{s>t} G_{s} , so that G(t) + G[t] = G_{t} ⊕ G[t]. Thus G_{t} is the same G_{t} as occurred in Proposition 4.32 and G_{t} ≈ (G(t) + G[t])/G[t]. Furthermore since G^{∗}(t) = ⨁_{s>t} G_{s} , G_{t} ≈ G(t)/G^{∗}(t). ✓

corollary 4.34. If W is a principal ideal domain and G and H are quasi-isomorphic completely decomposable modules, then G and H are in fact isomorphic. proof: If G = ⨁ G_{t} ∼ H = ⨁ H_{t} , where for each t the G_{t} and H_{t} are t-projective, then by Theorem 3.24 G_{t} ∼ H_{t} for each t. Thus it suffices to suppose that G and H are t-projective for some type t. By Corollary 2.10 if W is a principal ideal domain then the rank-one summands of G are isomorphic to the rank-one summands of H. Thus

G ≈ H. ✓

WARFIELD ISOMORPHISM THEOREMS. In the results that follow it is worth remembering that by Proposition 2.2, if A is a subring of W then A = S^{−1}W for some multiplicative set S and that for any G, A ⊗ G is naturally isomorphic to S^{−1}G.

proposition 4.35. Let A have rank one and let A_{0} = End A. Recall that t(A_{0}) = [t(A) : t(A)]. Let G and H be finite rank torsion free modules.

(1) G(t) ≈ A_{0} ⊗ G(t).

(2) G(t) ≈ A ⊗ Hom(A, G).

(3) Hom(A, A ⊗ G) ≈ A_{0} ⊗ G.

(4) Hom(A ⊗ G, A ⊗ H) ≈ Hom(G, A_{0} ⊗ H) ≈ Hom(A_{0} ⊗ G, A_{0} ⊗ H). proof: (2) Already proved in Proposition 4.3.

(3) Since t(A ⊗ G) ≥ t = OT(A), A ⊗ G dominates A. Thus by Proposition 4.11 Hom(A, A ⊗ G) ≈ Hom(A, A) ⊗ G = A_{0} ⊗ G.

(4) By (3) and the adjointness between Hom and the tensor product, Hom(A ⊗ G, A ⊗ H) ≈ Hom(G, Hom(A, A ⊗ H)) ≈ Hom(G, A_{0} ⊗ H) ≈ Hom(A_{0} ⊗ G, A_{0} ⊗ H) where the last isomorphism holds by Proposition 0.*, since A_{0} is a subring of Q. ✓

proposition 4.36. If A is a subring of Q and t = t(A) then (1) G(t) ≈ Hom(A, G).

(2) For any H, Hom(G, H)(t) = Hom(G, H(t)).

(3) For any G, Hom(G, A) is a t-projective module.

proof: (1) Since A is a subring of Q then Hom(A, G) is an A module. Furthermore A = S^{−1}W for some multiplicative set S and so by Proposition 4.7 and Proposition 0.*, G(t) ≈ A ⊗ Hom(A, G) = A ⊗_{A} Hom(A, G) ≈ Hom(A, G).

(2) By (1), Hom(G, H)(t) ≈ Hom(A, Hom(G, H)) ≈ Hom(A ⊗ G, H) ≈ Hom(G, Hom(A, H)) ≈ Hom(G, H(t)).

(3) **** ✓

corollary 4.37. If R is any dedekind domain over W (not necessarily with finite rank) then for any finite rank torsion free W -module G, Hom(G, R) is a projective R-module. proof: By Proposition 0.*, Hom(G, R) ≈ Hom_{R}(R ⊗ G, R). But R ⊗ G is a finite rank torsion free R-module. Therefore we may as well replace the ground ring by R, in other words no generality is lost in supposing R = W so that the result reduces to proposition 4.36. ✓

The following corollary is essentially just an elaboration of what was proved in Proposition 4.14. We take the opportunity, though, to give a different proof.

corollary 4.38. Let A be a subring of Q and t = t(A). Then for any G the following conditions are equivalent:

(1) G is t-saturated.

(2) G is an A-submodule of QG.

(3) A ⊗ G ≈ G.

(4) Hom(A, G) ≈ G.

proof: (1) ⇒ (4): By Corollary 4.37.

(4) ⇒ (3): As noted in the proof of Corollary 4.37, A ⊗ Hom(A, G) ≈ Hom(A, G).

(3) ⇒ (2): Clear since A ⊗ G is an A-submodule of QA ⊗ G.

(2) ⇒ (1): If G is an A-module then by Proposition 0.* Hom(A, G) ≈ Hom_{A}(A, G) ≈ G. Thus by Corollary 4.37 G(t) ≈ G. Thus rank G(t) = rank G. Since G(t) is a pure submodule of G, we see that G(t) = G, i.e. G is t-saturated. ✓ corollary 4.39. Let A be a rank-one module and let G and H be t(A)-saturated modules. Then

(1) The map Hom(G, H) → Hom(A ⊗ G, A ⊗ H) given by φ ↦ 1_{A} ⊗ φ is an isomorphism.

(2) The map Hom(G, H) → Hom(Hom(A, G), Hom(A, H)) given by φ ↦ φ_{∗} is an isomorphism.

proof: (1) If A_{0} = [A: A] then by Corollary 4.38 and Proposition 4.35 Hom(G, H) ≈ Hom(A_{0} ⊗ G, A_{0} ⊗ H) ≈ Hom(A ⊗ G, A ⊗ H).

(2) Since by Proposition 4.35 the natural maps σ_{G}: A ⊗ Hom(A, G) → A and σ_{H}: A ⊗ Hom(A, H) → H are isomorphisms, it follows from (1) that

Hom(Hom(A, G), Hom(A, H)) ≈ Hom(A ⊗ Hom(A, G), A ⊗ Hom(A, H)) ≈ Hom(G, H).

Let η denote the composition of these isomorphisms and let θ: Hom(G, H) → Hom(Hom(A, G), Hom(A, H)) be given by θ(φ) = φ_{∗}.

Now let φ ∈ Hom(G, H) and g ∈ G. Then ηθ(φ)(g) = η(φ_{∗})(g) = σ_{H}(φ_{∗}σ_{G}^{−1}(g)) = σ_{H}σ_{H}^{−1}(φ(g)) = φ(g) by the naturality of σ. Thus ηθ(φ) = φ so that ηθ = 1.

On the other hand, let ψ ∈ Hom(Hom(A, G), Hom(A, H)) and μ ∈ Hom(A, G). Then

θη(ψ)(μ) = θ(σ_{H}(1 ⊗ ψ)σ_{G}^{−1})(μ)
= (σ_{H}(1 ⊗ ψ)σ_{G}^{−1})_{∗}(μ)
= σ_{H}(1 ⊗ ψ)σ_{G}^{−1}μ
= (1 ⊗ ψ)σ_{G}σ_{G}^{−1}μ
= ψ(μ)

so that θη(ψ) = ψ. Thus η is an isomorphism with inverse θ. ✓

corollary 4.40. If [t(A): t(A)] ≤ IT(G) then Hom(A, A ⊗ G) ≈ G and Hom(A ⊗ G, A ⊗ H) ≈ Hom(G, H).

proof: If A_{0} = End A then by hypothesis G and H are t(A_{0})-saturated. Thus by Proposition 4.* A_{0} ⊗ G ≈ G and A_{0} ⊗ H ≈ H, so the result follows from Proposition 4.35. ✓

corollary 4.41. Let A be a subring of Q and t ≥ t(A). Then for any G, (1) A ⊗ G[t] = (A ⊗ G)[t].

(2) A ⊗ G[t] is a summand of A ⊗ G.

(3) A ⊗ (G/G[t]) is t-projective. proof: (1) We can identify G with the set of elements 1 ⊗ g ∈ A ⊗ G. Thus if x ∈ G[t] then 1 ⊗ x ∈ (W ⊗ G)[t] and if φ: A ⊗ G → A then since by Proposition 0.* φ is A-linear, φ(a ⊗ x) = aφ(1 ⊗ x) = 0. Thus A ⊗ G[t] ⊆ (A ⊗ G)[t] without any assumptions on t. Therefore by Proposition 4.12 it suffices to prove that (A ⊗ G)/(A ⊗ G[t]) is t-bounded. In fact, (A ⊗ G)/(A ⊗ G[t]) ≈ A ⊗ G/G[t] and OT(A ⊗ G/G[t]) = OT(A)OT(G/G[t]) = t(A) OT(G/G[t]) ≤ (t(A))^{2} = t(A) ≤ t since t(A) is idempotent by Proposition2.*.

(3) By (1), A⊗(G/G[t]) ≈ (A⊗G)/(A⊗G[t]), and this is t-bounded. By Corollary 4.40 A ⊗ (G/G[t]) is also t-saturated. Thus by Proposition 4.17 it is t-projective.

(2) By Corollary 4.40 A ⊗ G is t-saturated. Since A ⊗ (G/G[t]) is t-projective by (3), A ⊗ G[t] is a summand of A ⊗ G by Baer’s Lemma (Theorem 4.16). ✓

proposition 4.42. (1) If K is a t-bounded module, then for any G, Hom(G, K) is t-bounded.

(2) If t is idempotent and K is t-saturated, then Hom(G, K) is t-saturated.

proof: (1) Let K be t-bounded. Let δ: G → Hom(Hom(G, K), K) be defined by δ(g)(φ) = φ(g). Then by Proposition 4.4, for any g ∈ G, δ(g) maps Hom(G, K)[t] into K[t] = 0. Thus if φ ∈ Hom(G, K)[t] and g ∈ G then φ(g) = δ(g)(φ) = 0. Since this is true for all g ∈ G, thus φ = 0. Hence Hom(G, K)[t] = 0 so Hom(G, K) is t-bounded.

(2) If t is idempotent then by Proposition 2.15 t = t(A) for some subring A of Q. By Proposition 4.14 if K is saturated then K is an A-module. But then Hom(G, K) is also an A-module, hence is also t-saturated. ✓

proposition 4.43. Let A be a rank-one module and t = t(A). Then for any G, IT(Hom(A, G)) = [IT(G(t)): t]. proof: First note that Hom(A, G) = Hom(A, G(t)). If B is a pure rank-one submodule of G(t), then by Proposition 1.37 Hom(A, B) is a pure rank-one submodule of Hom(A, G). Furthermore these are all the pure rank-one submodules of Hom(A, G) since if φ ∈ Hom(A, G) then φ ∈ Hom(A, φ(A)_{∗}). But by Proposition 2.*, t(Hom(A, B)) = [t(B): t]. Taking the greatest lower bound over all B ◁ G(t) yields IT(Hom(A, G)) = [IT(G(t)): t]. ✓

corollary 4.44. For any W -module G, there exists a rank-one module A with locally trivial type and a module G_{0} with IT(G_{0}) = [IT(G): IT(G)] (so that, in particular, IT(G_{0}) is idempotent) such that G ≈ A ⊗ G_{0} . proof: Let t′ = IT(G), let t_{0} = [t′: t′], and let A′ be a rank-one module with t′ = t(A′). Let G_{0} = Hom(A′, G). By Proposition 4.43, IT(G_{0}) = [IT(G(t)): t′] = [IT(G): t′] = [t′: t′] = t_{0} . By Proposition 4.35, G = G(t′) ≈ A′ ⊗ Hom(A′, G) = A′ ⊗ G_{0} . Now by Proposition 2.19 t′ can be written t′ = t[t′: t′] = tt_{0} , where t is locally trivial. Thus if A is a rank-one module with t(A) = t and if A_{0} = [A′: A′] (so that t(A_{0}) = IT(G_{0})) then

G ≈ A′ ⊗ G_{0} ≈ A ⊗ A_{0} ⊗ G_{0} ≈ A ⊗ G_{0} , since A ⊗ A_{0} ≈ A_{0}A = A. (why?) ✓

We define the t-rank of G as rank Hom(G, A) where A is a rank-one module with t(A) = t.

proposition 4.45. Let A be a rank-one module with type t.

(1) t- rank G = rank G/G[t].

(2) Hom(G, A) is t-bounded.

(3) If t is idempotent, then Hom(G, A) is t-projective.

(4) If t is idempotent then t- rank G is the rank of a maximal t-projective quasi-direct summand of A ⊗ G.

(5) If t is idempotent then t- rank(G ⊗ H) = (t- rank G)(t- rank H).

proof: (1) This was proved in Proposition 4.5.

(2) This is a special case of Proposition 4.42.

(3) If t is idempotent, then by Proposition 4.42 Hom(G, A) is t-saturated. Since it is t-bounded by (2), by Proposition 4.17 it is t-projective.

(4) If t is idempotent then Hom(G, A) ≈ Hom_{A}(A ⊗ G, A) so that t- rank G = t- rank(A ⊗ G) and in we lose no generality in thinking of A as the ground ring. In other words, wlog we may suppose A = W. But then t- rank G = rank Hom(G, W) = rank G/G[t]. But by Proposition 4.26 G/G[t] = G(t)/G[t] is isomorphic to a maximal t-projective summand of G.

(5) Hom(G ⊗ H, A) ≈ Hom(G, Hom(H, A)). Now Hom(H, A) is an A-module, hence is t-projective by (2), i.e. is a direct sum of s rank-one modules with type t, where s = t- rank H. Thus t- rank(G⊗H) = rank Hom(G⊗H, A) = rank Hom(G, Hom(H, A)) = (t- rank H)(rank Hom(G, A)) = (t- rank G)(t- rank H). ✓

proposition 4.46. Let t = IT(G). Then (1) G = G_{t} ⊕ G[t], for some t-projective module G_{t} .

(2) t- rank G is the rank of a maximal t-projective quasi-summand of G. And, in fact, G has a t-projective actual summand of this rank.

proof: (1) Since t = IT(G), G = G(t) = G(t) + G[t]. Thus by Proposition 4.26 G = G_{t} ⊕ G[t], where G_{t} is t-projective.

(2) By Proposition 4.26 G_{t} is a maximal t-projective quasi-summand of G. Since by Proposition 4.45 t- rank G = rank G/G[t] = rank G_{t} , the result follows. And by (1), G actually has a t-projective summand of this rank. ✓