5. BUTLER MODULES

E. L. Lady

April 17, 1998

Consider again Example 2.28. As described at the beginning of Chapter 4, we can think of the construction there as beginning with three incomparable types t_{1} , t_{2} , and t_{3} satisfying the compatibility condition that t_{i} ∧ t_{j} is the same for all pairs i ≠ j. We then construct a rank-two module G such that the three submodules G(t_{i}) are distinct and G = G(t_{1}) + G(t_{2}) + G(t_{3}). We then observed that if φ ∈ End G then φ must leave these three submodules invariant, and we concluded that QEnd G ≈ Q.

In Example 1.49 similar reasoning was used. Using two incomparable types t_{1} and t_{2} we constructed a module rank-two module H such that H(t_{1}) ⊕ H(t_{2}) is a proper essential submodule of H and is quasi-equal to H. Using the fact that H(t_{1}) and H(t_{2}) must be invariant under any φ ∈ End H we were then able to calculate End H (see Example 2.21.) Furthermore, we also constructed a module H_{2} looking very much like H, but were able to prove that H and H_{2} are not isomorphic by using the fact that an isomorphism between them would have to map H(t_{1}) isomorphically onto H_{2}(t_{1}) and H(t_{2}) isomorphically onto H_{2}(t_{2}). In fact, we could have easily continued by actually calculating Hom(H, H_{2}) in the same way that we calculated End H.

We can easily construct more complicated examples using this same idea. Consider, for instance, the following:

example 5.1. Take the rank-two module G in Example 2.28, given as G = A_{1}(1, 0) + A_{2}(0, 1) + A_{3}(1, 1) ⊆ Q ⊕ Q, where A_{1}, A_{2}, A_{3} are rank-one modules with t_{i} = t(A_{i}) and A_{i} ∩ A_{j} = A_{1} ∩ A_{2} for all pairs i ≠ j. Let A be a rank-one module such that t(A) is not comparable to t_{1} , t_{2} , or t_{3} and such that A∩A_{i} = A_{1} ∩A_{2} for i = 1, 2, 3. Let w_{1}, w_{2} be distinct elements in W, neither of which is either 0 or 1. Construct two rank-two modules G_{1}, G_{2} containing G as follows:

G_{1} = G + A(1, w_{1}) = A_{1}(1, 0) + A_{2}(0, 1) + A_{3}(1, 1) + A(1, w_{1})
G_{2} = G + A(1, w_{2}) = A_{1}(1, 0) + A_{2}(0, 1) + A_{3}(1, 1) + A(1, w_{2}) ⊆ QG.

Let t = t(A). Then as shown in Proposition 2.*

G_{1}(t_{1}) = G_{2}(t_{1}) = A_{1}(1, 0) = G(t_{1})
G_{1}(t_{2}) = G_{2}(t_{2}) = A_{2}(0, 1) = G(t_{2})
G_{1}(t_{3}) = G_{2}(t_{3}) = A_{3}(1, 1) = G(t_{3})
G = G_{i}(t_{1}) + G_{i}(t_{2}) + G_{i}(t_{3})  (i = 1, 2)
G_{1}(t) = A(1, w_{1})
G_{2}(t) = A(1, w_{2}).

It follows that Hom(G_{1}, G_{2}) = 0. proof: If φ: G_{1} → G_{2} then φ_{1}(G(t_{i})) ⊆ G_{2}(t_{i}) for i = 1, 2, 3 and so it follows from the above that φ(G) ⊆ G. As was shown in Example 1.45, it then follows that φ is given by multiplication by some q ∈ Q. (In fact, if we assume further that t_{1} ∧ t_{2} ∧ t_{3} = t(W), as we did in Example 1.45, then q ∈ W.) But now it must also be true that φ(G(t)) ⊆ G(t), i.e. that φ(1, w_{1}) = (q, qw_{1}) is a multiple of (1, w_{2}). This is possible only if qw_{1} = qw_{2} , and since w_{1} ≠ w_{2} it follows that q = 0. Thus

Hom(G_{1}, G_{2}) = 0. ✓

Since there are infinitely many choices for w_{1} and w_{2} , this construction produces an infinite family of rank-two modules, all with the same typesets and the same p-ranks which look very much alike to the naked eye and yet are not quasi-isomorphic. Among other things, this shows what a strong relation quasi-isomorphism is.

In these examples, one is able to compute QEnd G and QHom(G, H) by using the fact that if φ: G → H is a quasi-homomorphism then for all g ∈ G, t(φ(g)) ≥ t(g) (c.f. Proposition 3.7). Before investigating the scope of application of this technique, we will give a final, fairly elaborate but very important example of its use.

lemma 5.2. Let A_{1}, …, A_{n} be rank-one modules with mutually incomparable types such that t_{i} ∧ t_{j} is the same for all pairs i ≠ j. Let V be a finite dimensional Q-space and for each i let V_{i} be a subspace of V . Let t_{0} = t_{1} ∧ t_{2} . Then there exists a module G such that QG = V and for i = 1, …, n, QG(t_{i}) = QV_{i}, and such that G is generated by a finite number of rank-one submodules with types in {t_{0}, …, t_{n}}. proof: Let C_{0} be an essential t_{0}-projective submodule of V and for i = 1, …, n let C_{i} be an essential t_{i}-projective submodule of V_{i}. Let G = ∑_{0}^{n} C_{i}. Then since each C_{i} is a direct sum of rank-one modules of type t_{i}, G is generated by a finite number of rank-one submodules with types in {t_{0}, …, t_{n}}. Clearly QG = V and for i = 1, …, n, C_{i} ⊆ G(t_{i}) so that V_{i} ⊆ QG(t_{i}). We claim that in fact G(t_{i}) = C_{i}_{∗}. Suppose by way of contradiction that there exists g ∈ G with t(g) ≥ t_{i} and g ∉ C_{i}_{∗}. Then there exists H ◁ G with C_{i}_{∗} ⊆ H and g ∉ H and rank G/H = 1, i.e. there exists μ: G → Q such that C_{i} ⊆ Ker μ and μ(g) ≠ 0. Thus t(μ(G)) ≥ t(g) = t_{i}. But since C_{i} ⊆ Ker μ, μ(G) is generated by μ(C_{j}) for j ≠ i, and since C_{j} is t_{j}-projective CT(C_{j}) = {t_{j}} and so t(μ(C_{j})) = t_{j}. Thus t(μ(G)) ≤ sup_{j≠i} t_{j}. But by Proposition 2.* t_{i} ∧sup_{j≠i} t_{j} = sup_{j≠i}{t_{i} ∧t_{j}} = t_{0} < t_{i}, since t_{i} ∧t_{j} = t_{0} for j ≠ i. Thus t_{i} ≰ t(μ(G)), a contradiction. Therefore for all g ∈ G, t(g) ≥ t_{i} ⇒ g ∈ C_{i}_{∗}, so that G(t_{i}) = C_{i}_{∗}, and

so V_{i} = QG(t_{i}). ✓

proposition 5.3. For n ≥ 1 let Λ = Q⟨x_{1}, …, x_{n}⟩ be the ring of polynomials in n non-commuting indeterminates over Q. Suppose that W has at least n + 3 prime ideals. Then there exists a functor B from the category of finite dimensional Λ-modules into the category of finite rank torsion free W -modules such that for every pair of finite dimensional Λ-modules M and N, QHom(B(M), B(N)) ≈ Hom_{Λ}(M, N). In particular, B(M) and B(N) are quasi-isomorphic if and only if M and N are isomorphic Λ-modules. proof: Let t_{1}, …, t_{n+3} be a fixed set of mutually incomparable types such that t_{i} ∧ t_{j} is the same for all pairs i ≠ j. (For instance, choose t_{i} = t(p_{i}^{−∞}), where p_{1}, …, p_{n+3} are distinct prime ideals.) Now let M be a finitely generated Λ-module. By Lemma 5.2, we can define a finite rank torsion free module G such that

QG = M ⊕ M
QG(t_{n+1}) = M ⊕ 0 = {(m, 0) | m ∈ M}
QG(t_{n+2}) = 0 ⊕ M = {(0, m) | m ∈ M}
QG(t_{n+3}) = {(m, m) | m ∈ M}
and QG(t_{i}) = {(m, x_{i}m) | m ∈ M} for i = 1, …, n.

Write G = B(M).

First note that B is a functor on the category of Λ-modules. In fact, if μ: M → N is a homomorphism of Λ-modules then μ induces μ̃ = μ ⊕ μ: M ⊕ M → N ⊕ N. If G = B(M), H = B(N), define B(μ) = μ̃: QG → QH. Note that B(μ)(QG(t_{i})) ⊆ QH(t_{i}) for i = 1, …, n + 3. In fact,

g ∈ QG(t_{n+1}) ⇒ (∃m ∈ M) g = (m, 0) ⇒ μ̃(g) = (μ(m), 0) ∈ QH(t_{n+1}),
g ∈ QG(t_{n+2}) ⇒ (∃m ∈ M) g = (0, m) ⇒ μ̃(g) = (0, μ(m)) ∈ QH(t_{n+2}),
g ∈ QG(t_{n+3}) ⇒ (∃m ∈ M) g = (m, m) ⇒ μ̃(g) = (μ(m), μ(m)) ∈ QH(t_{n+3}),

and for i = 1, …, n,

g ∈ QG(t_{i}) ⇒ (∃m ∈ M) g = (m, x_{i}m) ⇒ μ̃(g) = (μ(m), x_{i}μ(m)) ∈ QH(t_{i})

(using the fact that μ is Λ-linear).

Now by assumption B(M) is generated by a finite number of rank-one submodules with types in {t_{0}, …, t_{n}}. If A is such a rank-one submodule of B(M) and t(A) = t_{i}, then A ⊆ B(M)(t_{i}) and so by the preceding paragraph μ̃(A) ⊆ B(N)(t_{i}). Thus by Proposition 2.* there exists w ≠ 0 ∈ W such that wμ̃(A) ⊆ B(N)(t_{i}) ⊆ B(N). Since B(M) is generated by finitely many such rank-one submodules A, we can multiply together these w to get w ≠ 0 ∈ W such that wμ̃(B(M)) ⊆ B(N). Thus μ̃ is a quasi-homomorphism from B(M) to B(N) and so B is a functor.

To finish the proof, we show that if M and N are Λ-modules then μ ↦ B(μ) is an isomorphism from Hom_{Λ}(M, N) to QHom(B(M), B(N)). Clearly μ ≠ 0 ⇒ B(μ) ≠ 0 so it remains to see that every φ ∈ QHom(B(M), B(N)) has the form B(μ) for some μ: M → N. In fact, let G = B(M) and H = B(N). Then φ: QG = M ⊕ M → N ⊕ N = QH. Then φ(QG(t_{n+1})) ⊆ QH(t_{n+1}), φ(QG(t_{n+2})) ⊆ QH(t_{n+2}), and φ(QG(t_{n+3}) ⊆ QG(t_{n+3}). For m ∈ M, let μ(m) be the first coordinate of φ(m, 0). Then for m ∈ M,

φ(m, 0) = (μ(m), 0)  since φ((M ⊕ 0) ⊆ N ⊕ 0
φ(m, m) = (μ(m), μ(m))  since φ(QG(t_{n+3}) ⊆ QH(t_{n+3})
and φ(0, m) = φ(m, m) − φ(m, 0) = (0, μ(m)).

Thus φ = μ̃. It remains to see that μ is Λ-linear. But for i = 1, … n,

φ(m, x_{i}m) = μ̃(m, x_{i}m) = (μ(m), μ(x_{i}m)),

and since φ(Q(t_{i})) ⊆ H(t_{i}) this must belong to H(t_{i}), i.e. φ(m, x_{i}m) = (μ(m), x_{i}μ(m)), which implies that μ(x_{i}m) = x_{i}μ(m). Since this is true for all i, μ must in fact be Λ-linear. This completes the proof. ✓

It is generally acknowledged for that n ≥ 2 the task of classifying all modules over the ring Λ in Proposition 5.3 is hopeless. Thus, among other things, Proposition 5.3 shows that it is hopeless to expect to every be able to classify finite rank torsion free W -modules even up to quasi-isomorphism.

BUTLER MODULES. Based on the methodology of these examples, it would be easy to conjecture that the existence of homomorphisms, or at least quasi-homomorphisms, between finite rank torsion free modules is determined solely by the types of the elements in these modules. One might indeed look at Proposition 2.37 which states that a map from QG to QH belongs to Hom(G, H) if and only if it increases p-heights, and conjecture the following:

plausible conjecture 5.4. A map φ from QG to QH belongs to QHom(G, H) if and only if t(φ(g)) ≥ t(g) for all g ∈ G, or, equivalently, if and only if for all t ∈ T(G), φ(QG(t)) ⊆ QH(t).

The Pontryagin module (Example 1.47), however, shows that this conjecture is false. In fact, the Pontryagin module G is homogeneous: T(G) = {t(W)}. Thus Conjecture 5.4 would imply that QEnd G = End QG, which by Proposition 4.6 would mean that G dominates itself. It would then follow from Proposition 4.15 that the Pontryagin module is t(W)-projective, i.e. a projective W -module. But this is certainly not the case.

The class of modules for which Conjecture 5.4 is true is very important. In the search for this class, we can begin by noting that it includes completely decomposable modules.

The implications of this are explored in the following proposition: proposition 5.5. (1) If G = α(C) for some α ∈ Hom(C, G) and φ: QG → QH, then φ ∈ QHom(G, H) ⟺ φα ∈ QHom(C, H).

(2) If H is a pure submodule of D and ι: H → D is the inclusion map, and φ: QG → QH, then φ ∈ QHom(G, H) ⟺ ιφ ∈ QHom(G, D).

(3) If A and B are rank-one modules and φ: QA → QB, then φ ∈ QHom(A, B) ⟺ t(A) ≤ t(B).

(4) If A has rank one and t = t(A) and φ: QA → QH, then φ ∈ QHom(A, H) ⟺ φ(QA) ⊆ QH(t).

(5) If B has rank-one and t = t(B) and φ: QG → QB, then φ ∈ QHom(G, B) ⟺ φ(G[t]) = 0.

(6) If G = ⨁_{1}^{n} G_{i} and φ: QG → QH then φ ∈ QHom(G, H) if and only if for all i the restriction of φ to G_{i} is a quasi-homomorphism from G_{i} to H.

(7) If H = ⨁_{1}^{n} H_{i} and φ: QG → QH, then φ ∈ QHom(G, H) if and only if for each i the composition φ_{i} of φ with the projection ⨁ H_{i} → H_{i} belongs to QHom(G, H_{i}).

proof: Each of these is quite obvious.

(1) ( ⇒ ): Clear from general principles.

( ⇐ ): If φα ∈ QHom(C, H) then (∃w ≠ 0) wφα(C) ⊆ H. Since α(C) = G, thus wφ(G) ⊆ H and φ ∈ QHom(G, H).

(2) Obvious.

(3) QHom(A, B) is a subspace of the one-dimensional vector space Hom(QA, QB). Hence either QHom(A, B) = Hom(QA, QB) or QHom(A, B) = 0, depending on whether or not t(A) ≤ t(B).

(4) Assuming φ ≠ 0, let B = H ∩ φ(QA). Then B is a pure rank-one submodule of H and φ ∈ QHom(A, H) ⟺ φ ∈ QHom(A, B). By (1) this is true if and only if t(B) ≥ t(A) = t. But the latter is equivalent to saying that B ⊆ QG(t), i.e. φ(A) ⊆ QG(t).

(5) Let K = G ∩ Ker φ. Then rank G/K = 1 and φ ∈ QHom(G, H) ⟺ the induced map φ̄: G/K → QB belongs to QHom(G/K, B), i.e. if and only if t(G/K) ≤ t(B) = t. By Proposition 4.10 this is true if and only if K ⊆ G[t], i.e. if and only if φ(G[t]) = 0.

(6) ( ⇐ ): Clear on general principles.

( ⇒ ): If φ ∈ QHom(G_{i}, H) then (∃w_{i} ≠ 0) w_{i}φ(G_{i}) ⊆ H. Then wφ(G) ⊆ H, where w = w_{1} ⋯ w_{n}, so φ ∈ QHom(G, H).

(7) Analogous. ✓

One obvious moral that can be drawn from the sequence of observations in Proposition 5.5 is that homomorphic images of completely decomposable modules satisfy Conjecture 5.4.

definition 5.6. We say that a finite rank torsion free module G is a Butler module if G is a homomorphic image of a completely decomposable module. proposition 5.7. (1) G is a Butler module if and only if G is generated by a finite number of its rank-one submodules.

(2) A module quasi-isomorphic to a Butler module is a Butler module.

proof: (1) If G is a Butler module then it is a homomorphic image of a completely decomposable module C, and is generated by the images of the rank-one summands of C. Conversely, if G is generated by a finite set of rank-one submodules A_{i} then G is a homomorphic image of ⨁ A_{i}.

(2) Suppose G is a Butler module and H ∼ G. We may suppose G ⊆ H. Now G is generated by a finite set of rank-one submodules and H is generated by G together with a finite set of elements h_{i} ∈ H. Then H is generated by the rank-one submodules which generate G plus the finite set of cyclic submodules generated by the h_{i}, hence H is a Butler module by (1). ✓

Now putting several items of Proposition 5.5 together yields

proposition 5.8. Let G be a Butler module and H be a torsion free module and φ: QG → QH. The following conditions are equivalent:

(1) φ ∈ QHom(G, H)

(2) φ(QG(t)) ⊆ QH(t) for all t ∈ T(G).

(3) For all g ∈ G, t_{H}(φ(g)) ≥ t_{G}(g). Furthermore, if S is a subset of T(G) such that G is quasi-equal to ∑_{S} G(t), then for the implication (2) ⇒ (1) it suffices to verify (2) for those t ∈ S. proof: (1) ⇒ (2): By Proposition 3.7.

(2) ⇒ (3): Assume (2) and let g ∈ G and t = t(g). Then g ∈ G(t) so by (2) φ(g) ∈ H(t), i.e. t_{H}(φ(g)) ≥ t = t_{G}(g).

(3) ⇒ (1): Assume that t(φ(g)) ≥ t(g) for all g ∈ G. Since G is a Butler module, then there is a completely decomposable module C = ⨁ A_{i} and a surjection α: C → G. By Proposition 5.5, in order to prove that φ ∈ QHom(G, H) it suffices to show that φα ∈ QHom(C, H). Furthermore, for all c ∈ C, t_{H}(φα(c)) ≥ t_{G}(α(c)) ≥ t_{C}(c). Then by Proposition 5.5 (4), the restriction of φα to each rank-one summand A_{i} of C is a quasi-homomorphism. Therefore by Proposition 5.5 (6), φα ∈ QHom(C, H). Thus

φ ∈ QHom(G, H). ✓

corollary 5.9. Let G and H be Butler modules. If QG = QH then G and H are quasi-equal if and only if QG(t) = QH(t) for all t ∈ T(G) ∪ T(H). (If this is so, then by Proposition 3.8 T(G) = T(H).) proof: G and H are quasi-equal if and only if the identity map on QG is both a quasi-homomorphism from G to H and from H to G. By Proposition 5.8 this is equivalent to saying that QG(t) ⊆ QH(t) and QH(t) ⊆ QG(t) for all t ∈ T(G) and

t ∈ T(H). ✓

BUTLER’S THEOREM. In [Butler] Butler proved the following: theorem 5.10. The class of Butler modules is the same as the class of pure submodules of completely decomposable modules.

example 5.11. Consider again the module G in Example 1.45: G is a rank-two module and G = A_{1}(1, 0) + A_{2}(0, 1) + A_{3}(1, 1), where the A_{i} are submodules of Q with 1 ∈ A_{i} for all i and A_{i} ∩ A_{j} is the same for all pairs i ≠ j. Write A_{0} = A_{1} ∩ A_{2} . Now let B_{1} = A_{2} + A_{3} , B_{2} = A_{1} + A_{3} , and B_{3} = A_{1} + A_{2} . By Proposition 2.*

B_{1} ∩ B_{2} = A_{2} ∩ A_{1} + A_{2} ∩ A_{3} + A_{3} ∩ A_{1} + A_{3} ∩ A_{3} = A_{0} + A_{0} + A_{0} + A_{3} = A_{3}.

Likewise

B_{2} ∩ B_{3} = A_{1} and B_{1} ∩ B_{3} = A_{2}.

Then G is isomorphic to the pure closure of the submodule of the completely decomposable module B_{1} ⊕ B_{2} ⊕ B_{3} generated by a_{1} = (0, 1, −1), a_{2} = (−1, 0, 1) and a_{3} = (−1, 1, 0).

(For simplicity, one may want to suppose, as in the original construction, that A_{i} = p_{i}^{−∞}, where p_{1}, p_{2}, p_{3} are distinct primes. Then G = (p_{1}^{−∞}, 0) + (0, p_{2}^{−∞}) + p_{3}^{−∞}(1, 1), and, for instance, B_{1} = p_{2}^{−∞}p_{3}^{−∞}.) proof: Let H be the pure closure of the submodule of B_{1} ⊕ B_{2} ⊕ B_{3} generated by a_{1} , a_{2} , amnd a_{3} . For i = 1, 2, 3, let A′_{i} be the pure closure of the submodule of B_{1} ⊕ B_{2} ⊕ B_{3} generated by a_{i}. The map (0, x, −x) ↦ x is an isomorphism from A′_{1} to B_{2} ∩ B_{3} , so that A′_{1} ≈ B_{2} ∩ B_{3} = A_{1} . Likewise A′_{2} ≈ A_{2} and A′_{3} ≈ A_{3} . Note that since a_{3} = a_{1} + a_{2} , QH is generated over Q by a_{1} and a_{2} . Thus rank H = 2. Furthermore Q(A′_{1} + A′_{2} + A′_{3}) = QH and A′_{1} + A′_{2} + A′_{3} ◁ B_{1} ⊕ B_{2} ⊕ B_{3} (why?) so that A′_{1} + A′_{2} + A′_{3} = H. Finally this shows that there is an isomorphism φ from G to H given by (1, 0) ↦ a_{1} , (0, 1) ↦ a_{2} and (1, 1) ↦ a_{3} . In fact, φ is monic and φ(A_{i}) = ⟨a_{i}⟩_{∗} for i = 1, 2, 3, and thus since G = A_{1} + A_{2} + A_{3} , φ(G) = H. ✓

After contemplating a few examples such as this, it is easy to become completely convinced of the validity of Butler’s Theorem. In fact, it is not that hard to devise specific procedures for representing a Butler module as a pure submodule of a completely decomposable module, and vice versa. However the problem of actually proving that such procedures work is another matter.

BALANCED AND COBALANCED SUBMODULES. Before embarking on the proof of Butler’s Theorem, we need to note that as far as representing G as a homomorphic image of a completely decomposable module or a pure submodule of a completely decomposable module, there are good surjections and good pure embeddings, and others which are less desirable. For instance, let G ⊆ Q be such that G = A + B where A, B ⊆ Q with t(A), t(B) < t(G). Then G is trivially a Butler module, and G is a homomorphic image of A ⊕ B. However representing the rank-one module as a homomorphic image of a rank-two module A ⊕ B is clearly uneconomical. Furthermore T(A ⊕ B) contains types which are not in T(G), so that this way of representing G as a homomorphic image of a completely decomposable module casts less enlightenment on the properties of G than one might wish. Similarly if we have H = A ∩ B with A, B ⊆ Q and t(H) < t(A), t(B), then H is isomorphic to a pure submodule of A ⊕ B, but once again this is not the most desirable way of representing H as a pure submodule of a completely decomposable module.

The concept we need is the following:

definition 5.12. Let H ◁ G and let φ: G → G/H be the quotient map. We say that H of G is a balanced submodule of G if for every type t, φ(G)(t) = φ(G(t)), i.e. (G/H)(t) = (G(t) + H)/H. We also say that φ: G → G/H is a balanced surjection. If φ ∈ QHom(G, K) is a quasi-surjection, we say that φ is balanced if for every type t, φ(G)(t) is quasi-equal to φ(G(t)).

We say that H is a co-balanced submodule of G if for every type t, H[t] = H ∩ G[t]. This is equivalent to the assertion that for every type t, the map H/H[t] → G/G[t] induced by the inclusion H ↪ G is monic.

proposition 5.13. (1) A surjection φ: G → K is balanced if and only if every k ∈ K has a pre-image g ∈ G with t(g) = t(k).

(2) A pure submodule H of G is cobalanced if and only if every μ ∈ Hom(H, Q) is the restriction of a map μ′ ∈ Hom(G, Q) such that t(μ′(G)) = t(μ(H)).

proof: (1) ( ⇒ ): Let φ be a balanced surjection of G onto K and let k ∈ K and t = t(k). Then k ∈ K(t) and since φ is balanced there exists g ∈ G(t) with φ(g) = k. Then t(g) ≥ t. But by Proposition 2.21 t(g) ≤ t, so t(g) = t = t(k).

( ⇐ ): Let k ∈ K(t) for some type t. If there exists g ∈ G with φ(g) = k and t(g) = t(k), then t(g) ≥ t so that g ∈ G(t) and thus k ∈ φ(G(t)). Thus if every k ∈ K has a pre-image of the same type, then for every type t, K(t) = φ(G(t)), i.e. φ is balanced.

(2) ( ⇒ ): By Proposition 1.* H/H[t] is isomorphic to a pure submodule of G/H[t]. And by Proposition 4.2 (G/H)[t] = G[t]/H[t]. Now if H is cobalanced in G then H[t] = H ∩ G[t], so that (H/H[t]) ∩ (G/H[t])[t] = (H/H[t]) ∩ (G[t]/H[t]) = (H ∩ G[t])/H[t] = 0. Thus by Proposition 4.21 H/H[t] is a quasi-summand of G/H[t]. Now let μ: H → Q and let A = μ(H) and t = t(A). Then μ(H[t]) = 0 so μ induces μ̄: H/H[t] → A, and since H/H[t] is a quasi-summand of G/H[t] this extends to μ̄′ ∈ QHom(G/H[t], A). But then μ̄′ induces μ′ ∈ QHom(G, A) extending μ. Since μ′(G) is quasi-equal to A, t(μ′(G)) = t(A) = t.

( ⇐ ): By Proposition 4.2 H[t] ⊆ G[t], so H[t] ⊆ H ∩ G[t]. Now let A be a rank-one module with t(A) = t and let μ: H → A. If μ extends to a map μ′: G → Q with t(μ′(G)) = t, then μ′(G[t]) = 0, so μ(H ∩ G[t]) = 0. Thus if every μ ∈ Hom(H/Q) extends to μ′ ∈ Hom(G, Q) with t(μ′(G)) = t(μ(G)), then H ∩ G[t] ⊆ H[t], so H ∩ G[t] = H[t]. Since this holds for every type t, H is a cobalanced submodule of

G. ✓

corollary 5.14. (1) If H is a balanced pure submodule of G then T(G/H) ⊆ T(G). (2) If H is a cobalanced pure submodule of G then CT(H) ⊆ CT(G). proof: (1) If H is a balanced pure submodule of G and t ∈ T(G/H) then t = t(ḡ) for some ḡ ∈ G/H. By Proposition 5.13 ḡ has a pre-image g ∈ G such that t(g) = t. Thus t ∈ T(G).

(2) If H is a cobalanced pure submodule of G and t ∈ CT(H) then there exists μ ∈ Hom(H, Q) with t(μ(H)) = t. By Proposition 5.13 μ extends to μ′ ∈ Hom(G, Q) with t(μ′(G)) = t. Thus t ∈ CT(G). ✓

Using the concept of balanced surjection, we can give a generalized version of Baer’s Lemma.

proposition 5.15. If K is completely decomposable and φ: G → K is a balanced surjection, then φ splits. proof: Let K = ⨁_{t} K_{t} , where each K_{t} is t-projective. It suffices to see that for each t, the restriction of φ to a map φ^{−1}(K_{t}) → K_{t} is a split surjection. In fact, if for each t, σ_{t} : K_{t} → φ^{−1}(K_{t}) ⊆ G is such that φσ_{t} = 1_{K_{t}} , then the family {σ_{t}}_{t} combine to give a map K = ⨁ K_{t} → G splitting φ.

Now K_{t} ⊆ K(t) = φ(G(t)) since φ is balanced. Thus since φ is surjective, K_{t} = φ( G(t) ∩ φ^{−1}(K_{t})) and it suffices to see that the surjection G(t) ∩ φ^{−1}(K_{t}) → K_{t} splits. But φ^{−1}(K_{t}) ◁ G and so G(t) ∩ φ^{−1}(K_{t}) is a pure submodule of the t-saturated module G(t), and hence by Proposition 4.* is t-saturated. Since K_{t} is t-bounded, it follows from Baer’s Lemma (Theorem 4.13) that G(t) ∩ φ^{−1}(K_{t}) → K_{t} splits. ✓

PROOF OF BUTLER’S THEOREM. We now begin proving the results which will result in a proof of Butler’s Theorem.

definition 5.16. If G ◁ ⨁ A_{i}, where the A_{i} are rank-one modules, and if B is a pure rank-one submodule of G, we define the support of B to be Supp B = {i ∈ I | b_{i} ≠ 0} where b ≠ 0 ∈ B. (It is easy to see that this is independant of the choice of b.) We say that B is a distinguished rank-one submodule of G if there does not exist a rank-one submodule B′ ◁ G with Supp(B′) ⊊ Supp(B). If b ≠ 0 ∈ B, we also sometimes write Supp(b) = Supp(B).

proposition 5.17. If G is a pure submodule of a completely decomposable module ⨁ A_{i}, then there are only finitely many distinguished pure rank-one submodules of G, and G is generated by these distinguished submodules. Furthermore, if {B_{j}}_{j=1}^{m} are these distinguished submodules, the map ⨁_{1}^{m} B_{j} → G induced by the inclusion maps B_{j} ↪ G is a balanced surjection, so that G is a homomorphic image of a completely decomposable module. proof: First note that if B and B′ are distinguished rank-one submodules of G and Supp B = Supp B′, then B = B′. For if Supp B = Supp B′ and B ≠ B′ then one can find b ∈ B, b′ ∈ B′ with b ≠ b′ but where b and b′ agree in at least one coordinate. Then the pure submodule generated by b − b′ is non-trivial and has strictly smaller support than B, a contradiction.

To see that G is generated by the B_{j}, it suffices to see that for each prime ideal p, G_{p} is generated by the (B_{j})_{p}. But now the (B_{j})_{p} are precisely the distinguished rank-one submodules of G_{p}. Therefore there is no loss of generality in supposing that W is a local ring. Under this assumption, there is also no loss of generality in assuming that for all i, A_{i} = W or A_{i} = Q. Now if W is local then a rank-one submodule B ⊆ ⨁ A_{i} is pure if and only if either B = pB or B ⊈ ⨁ pA_{i}, and in either of these cases there must exist i such that B contains an element whose ith coordinate is 1. Now given g ∈ G, we want to show that g ∈ ∑ B_{j}, where the sum extends over the distinguished rank-one pure submodules of G. The idea is to use induction on the size of Supp g. If g ∈ B_{j} for some j, we are done. Otherwise, Supp g is not minimal and we choose B_{j} with Supp B_{j} ⊊ Supp g. Since B_{j} contains an element with ith coordinate 1 for some i ∈ Supp B_{j} ⊆ Supp g, B_{j} contains an element b whose ith coordinate agrees with the ith coordinate of g. It suffices to see that g − b is in the submodule generated by the B_{j}. But this follows by induction, since Supp(b − g) ⊊ Supp g.

Finally, the assertion that ⨁_{1}^{m} B_{j} → G is a balanced surjection is simply the claim that for each type t, G(t) is generated by those B_{j} with B_{j} ⊆ G(t) and t(B_{j}) ≥ t. In fact, applying the proof above to G(t) shows that G(t) is generated by those B_{j} with B_{j} ⊆ G(t). Furthermore t(B_{j}) ≥ t for all such B_{j} since they are pure submodules of

G. ✓

The proof of the other direction in Butler’s Theorem is dual to the above. Let G be a Butler module and {A_{i}}_{1}^{n} be a finite family of rank-one submodules generating G. By a linear functional on G we will mean a homomorphism μ: G → Q. For a linear functional μ we define Supp(μ) = {i | μ(A_{i}) ≠ 0}. We say that μ is a distinguished linear functional if μ is non-trivial and there does not exist a non-trivial linear functional μ′: G → Q with Supp(μ′) ⊊ Supp(μ).

Notice that if μ and μ′ are linear functionals on G and μ′ = qμ for some q ≠ 0 ∈ Q, then μ and μ′ have the same support. We will think of two linear functionals as equivalent if they are scalar multiples of each other, which is the same as saying that they have the same kernel.

Notice also that the property of being distinguished is not intrinsic to μ and G but depends on the family {A_{i}} of rank-one modules generating G.

proposition 5.18. If G is a Butler module then there are only finitely many distinguished linear functionals on G (up to scalar multiples) and the product map γ: G → ⨁ μ(G), where the μ range over the distinguished linear functionals on G, is a cobalanced pure embedding of G into the completely decomposable module ⨁ μ(G). proof: First note that if μ is a distinguished linear functional, then Supp μ uniquely determines μ up to a scalar multiple. In fact, suppose Supp μ = Supp μ′. Replacing μ and μ′ by suitable multiples, we can arrange that μ and μ′ agree on some A_{j} with j ∈ Supp μ (since rank Hom(A_{j}, Q) = 1). Then j ∉ Supp(μ − μ′), so that Supp(μ − μ′) ⊊ Supp μ. Since μ is distinguished, it follows that μ − μ′ = 0, so that μ = μ′.

It now follows that there are only finitely many distinguished linear functionals, up to scalar multiples. Let μ_{1}, … μ_{r} be these distinguished functionals. To see that the mapping γ: g ↦ (μ_{1}(g), …, μ_{r}(g)) embeds G as a pure submodule of ⨁_{1}^{r} μ_{i}(G), it suffices to see that γ embeds G_{p} as a pure submodule of ⨁_{1}^{r} μ_{i}(G)_{p} for every prime p. We then easily see that there is no loss of generality in supposing that W is local. In this case, each A_{i} is isomorphic to either W or Q and G/d(G) is finitely generated, so G is the direct sum of d(G) and a free W -module. Furthermore if μ is a linear functional on G then either μ(G) = Q or μ(G) ≈ W. We will normalize the distinguished linear functionals so that if μ_{i}(G) ≠ Q then μ_{i}(G) = W. We now claim that the distinguished linear functionals μ_{1}…. , μ_{r} generate Hom(G, Q) as a Q-space, and that the distinguished functionals with μ_{i}(G) = W generate Hom(G, W) over W. To prove the second asserion, we will prove by induction on the size of Supp μ that for μ ∈ Hom(G, W), μ is a linear combination with coefficients in W of μ_{i} with μ_{i}(G) = W. If μ is not distinguished, then Supp μ is not minimal and there exists a distinguished functional μ_{i} with Supp μ_{i} ⊆ Supp μ, and μ_{i}(G) = W since clearly μ_{i}(G) ≠ Q. Now since μ_{i}(G) = W, it follows that μ_{i}(A_{j}) ⊈ p for some j, so μ_{i}(A_{j}) = W since W is local. Now Hom(A_{j}, W) is cyclic (since rank A_{j} = 1) and the restriction of μ_{i} to A_{j} is a generator since μ_{i}(A_{j}) = W. Thus for some w ∈ W, wμ_{i} and μ have the same restriction of A_{j}, so that j ∉ Supp(μ − wμ_{i}). The claim then follows by induction.

The proof that the complete set of linear functionals μ_{1}, …, μ_{r} generates Hom(G, Q) as a Q-space is analogous but easier. It then follows that for g ≠ 0 ∈ G there exists a distinguished linear functional μ_{i} with μ_{i}(g) ≠ 0, from which it follows that γ: g ↦ (μ_{1}(g), …, μ_{r}(g)) is a monomorphism.

To prove purity, we need to show that if g ∉ pG then there exists a distinguished functional μ_{i} with μ_{i}(g) ∉ pμ_{i}(G) = p. Since we have seen that the set of distinguished functionals with μ_{i}(G) = W generate Hom(G, W), it suffices to prove that there exists μ ∈ Hom(G, W) with μ(g) ∉ p. But since g ∉ pG and in particular g ∉ d(G), and since G/d(G) is free, this is clear. ✓

Between Proposition 5.17 and Proposition 5.18 the proof of Butler’s Theorem is complete.

corollary 5.19. Let H be a Butler module and φ: QG → QH for some G. The following conditions are equivalent:

(1) φ ∈ QHom(G, H)

(2) φ(QG[t]) ⊆ QH[t] for every t ∈ CT(G).

(3) For all μ: H → Q, t(μφ(G)) ≤ t(μ(H)).

proof: (1) ⟺ (2): Since H is a pure submodule of a direct sum of rank-one modules, by parts (2) and (7) of Proposition 5.5, it suffices to suppose that rank H = 1, in which case the result follows from Proposition 5.5 (5). (Note that if rank H = 1 then by Lemma 2.4 H[t] = 0 if t ≥ t(H), otherwise H[t] = H.)

(1) ⇒ (3): We may assume that in fact φ ∈ Hom(G, H). Let μ: H → Q and let t = t(μ(H)). Then by Proposition 4.4 φ(G[t]) ⊆ H[t] ⊆ Ker μ, i.e. μφ(G[t]) = 0. Therefore by Proposition 4.4 t(μφ(G)) ≤ t = t(μ(H)).

(3) ⇒ (2): Let g ∈ G[t] for some type t and let A be a rank-one module with t(A) = t. Let μ ∈ Hom(H, A). Then t(μ(H)) ≤ t and if t(μφ(G)) ≤ t(μ(H)) ≤ t then μφ(g) = 0 since g ∈ G[t]. If this is true for every μ ∈ Hom(H, A) it follows that φ(g) ∈ H[t]. Thus if t(μφ(G)) ≤ t(μ(H)) for every μ ∈ Hom(H, A) then

φ(QG[t]) ⊆ QH[t]. ✓

corollary 5.20. Quasi-pure submodules and homomorphic images of Butler modules are Butler modules. proof: The assertion about homomorphic images is clear, since if G is a homomorphic image of a completely decomposable module C and H is a homomorphic image of G, then H is also a homomorphic image of C. Now if H is a quasi-pure submodule of G, then H is quasi-equal to G ∩ QH, and by Proposition 5.7 H is a Butler module if and only if G ∩ QH is a Butler module. Thus it suffices to suppose that H ◁ G. Now if G is a Butler module then G is a pure submodule of a completely decomposable module C, and H is also a pure submodule of C, so by the Theorem 5.10 H is a Butler module. ✓

Arbitrary submodules of Butler modules need not be Butler modules. For instance, divisible modules are Butler modules (since they are completely decomposable) and any G is a submodule of the divisible module QG. But there exist finite rank torsion free modules with are not Butler modules, for instance the Pontryagin module (Example 1.47).

corollary 5.21. Let G be a Butler module and H ⊆ QG. Then H is a quasi-pure submodule of G if and only if H is a Butler module and for all h ∈ H, t_{H}(h) = t_{G}(h). proof: Let H′ = H ∩ QG. Then H′ ◁ G so by Corollary 5.20 H′ is a Butler module. And H is quasi-pure in G if and only if H is quasi-equal to H′.

( ⇒ ): If H is quasi-equal to H′ then H is a Butler module by Proposition 5.7. Furthermore if h ∈ H then by Proposition 3.7 the type of h in H is the same as its type in H′, and thus the same as its type in G.

( ⇐ ): If every element h ∈ H has the same type in G as in H then since H′ ◁ G, by Proposition 2.* it has the same type in H′ as in H. Thus by Proposition 5.9 H is quasi-equal to H′, hence is a quasi-pure submodule of G. ✓

TYPESET AND COTYPESET.

corollary 5.22. If G is a Butler module then T(G) and CT(G) are finite and generate the same lattice. proof: Suppose that G is generated by pure rank-one submodules A_{1}, …, A_{n} and let t_{i} = t(A_{i}) ∈ T(G). If t ∈ CT(G) then there exists a surjection φ: G → B where B is a rank-one module with t(B) = t. Then B = φ(G) = ∑ φ(A_{i}) and either φ(A_{i}) ≈ A_{i} or φ(A_{i}) = 0, so that t = sup{t_{i} | φ(A_{i}) ≠ 0}. Thus CT(G) is contained in the lattice generated by T(G) and in particular is finite.

Conversely, G is isomorphic to a pure submodule of ⨁ B_{j} where rank B_{j} = 1 for all i and we may assume wlog that π_{j}(G) = B_{j}, where π_{j} is the projection onto B_{j}. Thus t(B_{j}) ∈ CT(G). Now if t ∈ T(G) then t = t(g) for some g ∈ G. Since G ◁ ⨁ B_{j}, by Proposition 2.21 we can compute t(g) in ⨁ B_{j}. Then by Proposition 2.* t(g) = inf {t(B_{j}) | π_{j}(g) ≠ 0}. Thus T(G) is finite and contained in the lattice generated

by CT(G). ✓

corollary 5.23. Let G be a Butler module and H a quasi-pure submodule of G. Then T(H), T(G/H), and CT(H), and CT(G/H) are contained in the lattice generated by T(G). proof: By the previous Corollary, it suffices to prove that T(H) and CT(G/H) are contained in the lattice generated by T(G) or by CT(G). But T(H) ⊆ T(G) and CT(G/H) ⊆ CT(G) even without assuming that G is a Butler module. ✓

corollary 5.24. If G is a Butler module and T(G) contains a unique maximal type t, then G can be written as G_{t} ⊕ G′ where G_{t} is t-projective and t ∉ T(G′). proof: If t is the unique maximal type in T(G) then by Corollary 5.23 t = OT(G) and by by Proposition 4.22 G(t) is t-projective and G = G(t) ⊕ G′ for some G′. Since G′ ◁ G, by Proposition 4.2 G′(t) = G′ ∩ G(t) = 0. Thus t ∉ T(G′). ✓

corollary 5.25. If G is a Butler module and T(G) is totally ordered, then G is completely decomposable. In particular, a homogeneous Butler module is t-projective for some t. proof: By induction on the size of T(G), which is finite by Proposition 5.22. If T(G) = {t} then by Proposition 5.22 CT(G) = {t}, so that IT(G) = OT(G) = t and G is both t-saturated and t-bounded. Therefore by Proposition 4.14 G is t-projective.

In the general case, if T(G) is totally ordered then it contains a unique maximal type t. Then by Proposition 5.24 G = G_{t} ⊕ G′ where G_{t} is t-projective and t ∉ T(G′). Thus T(G′) ⊊ T(G) (since G′ ◁ G), so by induction G′ is completely decomposable. Thus G is completely decomposable. ✓

example 5.26. In Example 1.49 we started with two rank-one modules A and B such that t(A) and t(B) are incomparable, and constructed an indecomposable module H as the submodule of QA ⊕ QB generated by A ⊕ B together with (a + b)/w for a certain a ∈ A and b ∈ B and w ≠ 0 ∈ W. This construction will not produce an indecomposable module if t(A) and t(B) are not incomparable. In fact, since G is then quasi-equal to the completely decomposable module A ⊕ B it is a Butler module, and T(G) = {t(A), t(B), t(A) ∧ t(B)}. if then t(A) ≤ t(B) or t(B) ≤ t(A) then by Corollary 5.25 G must be completely decomposable and thus not indecomposable.

More specifically, consider the example where W is a principal ideal domain and p and q are distinct primes and w ∈ W, w ∉ p ∪ q. Let A and B be the submodules of Q ⊕ Q defined by A = (p^{−∞}, 0) and B = (0, p^{−∞}q^{−∞}). Let G be the submodule of Q ⊕ Q generated by A ⊕ B together with (1, 1)/w.

Now let A′ be the submodule of G generated by p^{−∞}(1, 1) together with w^{−1} . (By the existence of partial fraction expansions (Lemma 1.28) A′ = p^{−1}(1, 1)/w.) Then A′ ∩ B = 0 and since A ⊆ p^{−∞}(1, 1) ⊕ B, A ⊆ A′ ⊕ B ⊆ G. Since (1, 1)/w ∈ A′ ⊕ B it follows that G = A′ ⊕ B, confirming that G is completely decomposable. ✓

We can now generalize Lemma 5.2 to a general existence theorem for Butler modules. proposition 5.27. Let T be a finite set of types such that for each t ∈ T, t is not the least upper bound of the set of types in T strictly less than t. Let U be a finite dimensional Q-space and for each t ∈ T let U_{t} be a subspace of U such that if s ≤ t then U_{s} ⊇ U_{t} . Then there exists a Butler module G such that QG = U and for each t ∈ T, QG(t) = U_{t} . Furthermore G is unique up to quasi-equality. FIX THIS! proof: It follows from Proposition 5.9 that if such a Butler module G exists then it is unique up to quasi-equality. Now let t_{0} = inf T and let C_{0} be an essential t_{0}-projective submodule of U. And for each t ∈ T let C_{t} be an essential t-projective submodule of U_{t} . By Proposition 4.15 each C_{t} is well defined up to quasi-equality. Now let G be the module generated by C_{0} and all the C_{t} . Then G is generated by a finite set of rank-one modules since this is true of each C_{t} , so by Proposition 5.7 G is a Butler module. Clearly QG = U. We claim that for all t ∈ T, C_{t} is a quasi-pure submodule of G and QG(t) = QC_{t} .

**** Thus wlog we may assume that C_{t} ◁ G. Certainly it is clear that C_{t} ⊆ G(t). Conversely, consider the quotient map γ: G → G/C_{t} . By Proposition 4.2 γ(G(t)) ⊆ (G/C_{t})(t). Thus to show that G(t) ⊆ C_{t} it suffices to show that (G/C_{t})(t) = 0. Now by hypothesis QC_{s} ⊆ QC_{t} whenever s ≥ t. Thus (G/C_{t}) is a homomorphic image of ⨁_{s≱t} C_{s} , and so by Proposition 5.22 T(G/C_{t}) is contained in the lattice generated by {s ∈ T | s ≱ t}. ****

CRITICAL TYPES. According to Corollary 5.9, if G is a Butler module then we know G up to quasi-equality once we know the finite family of subspaces QG(t) of QG, for t ∈ T(G). And in fact a closer examination of the proof of Corollary 5.9 shows that we don’t even need to know all the QG(t), but only certain “critical” ones (see Proposition 5.* below).

definition 5.28. A critical type for a Butler module G is a type t ∈ T(G) such that G^{∗}(t) is not an essential submodule of G(t). We write T′(G) to denote the set of critical types of G.

lemma 5.29. If G is a Butler module and t = IT(G) then G^{∗}(t)_{∗} = G[t]. proof: By Proposition 4.27 G^{∗}(t)_{∗} ⊆ G[t] (since G[t] ◁ G). On the other hand, since G is a Butler module it is generated by a finite set of rank-one pure submodules. If t = IT(G) then all these rank-one submodules have type at least t, and the ones with types greater than t belong to G^{∗}(t)_{∗}. Thus G/G^{∗}(t)_{∗} is generated by a finite set of rank-one modules with type t and thus is a homomorphic image of the direct sum of these rank-one modules, hence is t-projective by Proposition 4.16, and in particular is t-bounded. It then follows from Proposition 4.10 that G[t] ⊆ G^{∗}(t)_{∗}. ✓

proposition 5.30. Let G be a Butler module.

(1) For every type t, G^{∗}(t) is a quasi-pure submodule of G and there exists a t-projective pure submodule G_{t} of G such that G(t) = G_{t} ⊕ G^{∗}(t)_{∗}.

(2) t is a critical type of G if and only if G_{t} ≠ 0.

(3) If t ∈ T(G) and t < inf {s ∈ T(G) | s > t}, then t is critical. In particular, all maximal elements of T(G) are critical. proof: (1) G^{∗}(t) is a homomorphic image of ⨁{G(s) | s ∈ T(G) and s > t}, where the direct sum is finite since T(G) is finite. Since IT(G(s)) ≥ s > t, IT(G^{∗}(t)) ≥ t. Now each G(s) is a Butler module by Proposition 5.20 since G(s) ◁ G. Therefore G^{∗}(t) is a Butler module by Proposition 5.20. Thus by Proposition 5.21 G^{∗}(t) is quasi-pure in G if and only if for all g ∈ G^{∗}(t), the type t′ of g in G^{∗}(t) is t_{G}(g). Now if t_{G}(g) = s > t then g ∈ G(s) ◁ G and G(s) ◁ G^{∗}(t). Thus by Proposition 2.21 both t′ and t_{G}(g) are the same as the type of g in G(s) so t′ = t_{G}(g). Thus we need only consider g ∈ G^{∗}(t) with t_{G}(g) = t. But then t′ ≥ IT(G^{∗}(t)) ≥ t and t′ ≤ t_{G}(g) = t by Proposition 2.* since G^{∗}(t) ⊆ G. Thus t′ = t_{G}(g) in this case as well. It follows that G^{∗}(t) is a quasi-pure submodule of G.

Now since G^{∗}(t)_{∗} ⊆ G(t) there is no loss of generality in supposing that G = G(t), i.e. that t = IT(G). Then by Proposition 4.23 that G = G_{t} ⊕ G[t] for some t-projective module G_{t} . Since G[t] = G^{∗}(t)_{∗} by Lemma 5.29, the result follows.

(2) By definition, t ∈ T′(G) if and only if G^{∗}(t)_{∗} ≠ G(t), i.e. if and only if G_{t} ≠ 0.

(3) Suppose that t < inf {s ∈ T(G) | s > t} and let g ∈ G with t(g) = t. We claim that g ∉ G^{∗}(t)_{∗}, which will prove that G(t) ≠ G^{∗}(t)_{∗}, so that t ∈ T′(G). In fact, if g ∈ G^{∗}(t)_{∗} then (∃w ≠ 0 ∈ W) wg = ∑_{I} g_{i} with t(g_{i}) > t. Then by Proposition 2.* t(g) = t(wg) ≥ inf {t(g_{i}) | i ∈ I}, contradicting the hypothesis on t. ✓

proposition 5.31. (1) If G and H are Butler modules and G ∼ H, then T′(G) = T′(H).

(2) If G = ⨁_{1}^{n} A_{i} where the A_{i} are rank-one modules, then T′(G) = {t(A_{i}) | i = 1, …, n}.

proof: (2) If G = ⨁_{1}^{n} A_{i} then by Proposition 4.* G(t) = ⨁_{t(A_{i})≥t} A_{i} and G^{∗}(t) = ⨁_{t(A_{i})>t} A_{i}. Thus G^{∗}(t) is an essential submodule of G(t) if and only if G^{∗}(t) = G(t), and thus if and only if there exists A_{i} with t(Ai) ≥ t but t(Ai) ≯ t, i.e. if and only if t = t(A_{i}) for some i. Thus T′(G) = {t(A_{1}), …, t(A_{n})}.

(1) If G ∼ H then from Proposition 4.2 it follows that G(t) ∼ H(t) and G^{∗}(t) ∼ H^{∗}(t). Thus if t ∈ T′(G), so that rank G^{∗}(t) < rank G(t), then rank H^{∗}(t) < rank H(t), so t ∈ T′(H). ✓

theorem 5.32. If for each t ∈ T′(G) a t-projective submodule G_{t} of G is chosen as in Proposition 5.30 then G is quasi-equal to ∑_{T′(G)} G_{t} . Furthermore the map ⨁_{T′(G)} G_{t} → G induced by the inclusions G_{t} ⊆ G is a balanced quasi-surjection.

proof: We will prove by downward induction on t ∈ T(G) that ∑{G_{s} | s ∈ T′(G) and s ≥ t} for is quasi-equal to G(t).

If t is maximal in T(G) then by Proposition 5.30 t ∈ T′(G). Furthermore there are no g ≠ 0 ∈ G with t(g) > t so G^{∗}(t) = 0, and so by Proposition 5.30 G(t) = G_{t} ⊕ G^{∗}(t)_{∗} = G_{t} , establishing the claim in this case.

Now suppose that for all t′ ∈ T(G) with t′ > t it has been proved that G(t′) is quasi-equal to ∑_{s≥t′} G_{s} . It follows that G^{∗}(t) = ∑_{t′>t} G(t′) is quasi-equal to ∑_{s>t} G_{s} (using the fact that by Corollary 5.22 there are only finitely many t′ ∈ T(G).) Since by Proposition 5.30 G^{∗}(t) is a quasi-pure submodule of G, it follows that G^{∗}(t)_{∗} is quasi-equal to ∑_{s>t} G_{s} . Thus G(t) = G_{t} ⊕ G^{∗}(t)_{∗} is quasi-equal to ∑_{s≥t} G_{s} .

Since T(G) is finite, we finally conclude that for t = IT(G), G = G(t) is quasi-equal to the submodule of G generated by all G_{t} for t ∈ T′(G).

Now let φ: ⨁ G_{t} → G be the map induced by the inclusions G_{t} ⊆ G. Since for any type t, (⨁ G_{t′})(t) = ⨁_{t′≥t} G_{t′}, the assertion that φ is a balanced quasi-surjection is simply the claim that for each t, G(t) is quasi-equal to the submodule generated by all G_{t′} for t′ ≥ t. But this has been shown already. ✓

proposition 5.33. If G is a Butler module and t a type, then G[t] is equal to the pure closure of the submodule generated by the set of G_{s} such that s ∈ T′(G) and s ≰ t. proof: Let G⟨t⟩ be the pure submodule generated by the set of G_{s} such that s ∈ T′(G) and s ≰ t. By Proposition 4.4 s ≰ t ⇒ G(s) ⊆ G[t], so a fortiori G_{s} ⊆ G[t]. Thus G⟨t⟩ ⊆ G[t]. To see that G[t] ⊆ G⟨t⟩ it suffices by Proposition 4.10 to see that G/G⟨t⟩ is t-bounded. But by Proposition 5.32 G is quasi-equal to the submodule generated by all G_{s} , and so G/G⟨t⟩ is quasi-generated by the images of these submodules, and clearly we need use only with s ≤ t, since all other G_{s} are contained in G⟨t⟩. Since each such G_{s} is s-projective and thus t bounded, G/G⟨t⟩ is t-bounded by Proposition 4.10. Thus G[t] ⊆ G⟨t⟩ and so G[t] = G⟨t⟩. ✓

corollary 5.34. If t is minimal in T′(G) then (G(t) + G[t])_{∗} = G. proof: If s ≠ t ∈ T′(G) then since t is minimal in T′(G), s ≰ t, so by Proposition 5.33 G_{s} ⊆ G[t]. Since G_{t} ⊆ G(t), G(t) + G[t] thus contains G_{s} for all s ∈ T′(G). Thus by Proposition 5.32 G(t) + G[t] is quasi-equal to G. ✓

Recall that if t is a type then t- rank G = rank G/G[t].

corollary 5.35. Let G be a Butler module and t a type.

(1) G has no non-trivial t-projective quasi-summand if and only if the set all G_{s} for s ∈ T′(G) and s ≠ t generates an essential submodule of G.

(2) If t is minimal in T′(G) then t- rank G is the rank of a maximal t-projective quasi-summand of G.

proof: (2) By Proposition 4.23 the rank of a maximal t-projective quasi-summand of G equals rank(G(t) + G[t])/G[t]. But if t is minimal in T′(G) then by Proposition 4.29 G(t) + G[t] is quasi-equal to G. Thus the rank of a maximal t-projective quasi-summand of G equals rank G/G[t] = t- rank G.

(1) ( ⇐ ): If G is quasi-equal to C ⊕ K, where C is t-projective, then for all s ≠ t, G_{s} can be chosen so that G_{s} ⊆ K (why?). Thus if C ≠ 0 then the submodule generated by all G_{s} is not essential in G.

( ⇒ ): By Proposition 4.23 G has no quasi-summand of type t if and only if G(t) ⊆ G[t]. But if H is the pure closure of the submodule generated by all G_{s} with s ≠ t then by Proposition 5.33 G[t] ⊆ H. It follows that if G has no quasi-summand of type t then G(t) ⊆ H, in which case G_{t} ⊆ H and so by the above H is quasi-equal to G, and so H = G since H ◁ G. ✓ COCRITICAL TYPES. The dual of Proposition 5.30 is equally valid, although less used.

definition 5.36. A cocritical type for a Butler module G is a type t ∈ CT(G) such that G^{∗}[t] ≠ G[t]. We write CT′(G) for the set of cocritical types of G.

proposition 5.37. Let G be a Butler module.

(1) If t = OT(G) then G(t) is a t-projective direct summand of G, and G(t) = G^{∗}[t].

(2) If t ∈ CT(G) and t > sup{s ∈ CT(G) | s < t}, then t is cocritical. In particular, if t is minimal in CT(G) then t is cocritical.

proof: (1) If t = OT(G) then G is t-bounded, and G(t) is by definition t-saturated. Thus by Proposition 4.21 G(t) is a direct summand of the t-bounded module G, say G = G(t) ⊕ H. By Proposition 4.27 G(t) ⊆ G^{∗}[t]. Now by Theorem 5.10 G is a pure submodule of a completely decomposable module ⨁ A_{i}. And if g ∈ G and g ∉ G(t) then by Proposition 2.32 t(g) < OT(G) = t, and hence by Proposition 2.21 g must have a non-trivial coordinate in some A_{i} with t(A_{i}) < t. If μ_{i} is the restriction to G of the projection ⨁ A_{i} → A_{i}, then μ_{i}(g) ≠ 0, so that g ∉ G[t(A_{i})] and so g ∉ G^{∗}[t]. Thus G^{∗}[t] ⊆ G(t).

(2) Let μ: G → A with A = μ(G) and t(A) = t. Then μ(G[t]) = 0. Suppose G^{∗}[t] = G[t]. Then for g ∉ G[t], g ∉ G^{∗}[t] so that there exists s < t with g ∉ G[s], and thus there exists A′ ⊂ A with t(A′) = s < t(A) and μ′: G → A′ with μ′(g) ≠ 0. Then A = μ(G) is generated by such A′, and in fact by finitely many of them (since G is generated by finitely many of its rank-one submodules). Thus t = t(A) is the least upper bound of such t(A′). This contradicts the assumption on t, so G^{∗}[t] ≠ G[t] and t is cocritical. ✓

theorem 5.38. For each t ∈ CT′(G), G^{∗}[t]/G[t] is a direct summand of G/G[t]. Let β_{t} : G → G^{∗}[t]/G[t] be the composition of the quotient mapping G → G/G[t] with a projection of G/G[t] onto G^{∗}[t]/G[t]. Let

β: G → ⨁_{CT′(G)} G^{∗}[t]/G[t]

be given by the product of the maps β_{t} . Then β is a monomorphism mapping G onto a cobalanced pure submodule of ⨁ G^{∗}[t]/G[t]. proof: Let t ∈ CT′(G). Then OT(G/G[t]) ≤ t since G/G[t] is t-bounded and on the other hand t ∈ CT(G/G[t]) since t ∈ CT(G). Thus t = OT(G/G[t]) and it follows from Proposition 4.2 that G^{∗}[t]/G[t] = (G/G[t])^{∗}[t]. Thus by Proposition 5.37 G^{∗}[t]/G[t] is a summand of G/G[t].

Now for each t ∈ CT′(G) let C_{t} = G^{∗}[t]/G[t] and let C = ⨁_{CT′(G)} C_{t} . For each t ∈ CT(G), let

β̃_{t} : G/G[t] → ⨁_{s≤t} C_{s}

be the map induced by β_{s} for s ≤ t. We will show by induction on t that β̃ is monic with pure image. If t is minimal in CT(G), then β̃_{t} = β_{t} and G^{∗}[t] = G and C_{t} = G^{∗}[t]/G[t] = G/G[t]. Thus β̃_{t} is the identity, hence is a pure monomorphism. Now suppose that it has been shown that β̃_{s} is a pure monomorphism for all s ∈ CT(G) with s < t. Let p be a prime and let g ∈ G with g + G[t] ∉ p(G/G[t]). If g ∈ G[s] for all s with s < t then g ∈ G^{∗}[t] and so β_{t}(g) = g + G[t] ∉ pC_{t} and so β̃(g) ∉ ⨁_{s≤t} pC_{s} . On the other hand, if g ∉ G[s] for some s < t, choose s to be minimal with this property. Then g ∈ G^{∗}[s] and so β_{s}(g) = g + G[s] ∈ G^{∗}[s]/G[s]. Now **** In both cases it follows that β(g) ∉ pC.

The fact that all the induced maps G/G[t] → C/C[t] are monic shows that β is cobalanced, provided that it is a pure monomorphism. But this in turn follows by applying the above for t = CT(G), since in that case G/G[t] = G and C/C[t] = C. ✓

proposition 5.39. If G is a Butler module then T(G) and CT(G) are contained in the lattice generated by either T′(G) or CT′(G). proof: By Proposition 5.32 G is a balanced homomorphic image of C = ⨁ G_{t} , where t ranges over T′(G) and G_{t} is t-projective. Then by Proposition 2.* T(C) is contained in the lattice generated by T′(G). But by Corollary 5.23 and Corollary 5.14 T′(G) ⊆ T(G) ⊆ T(C). This shows that T(G) and T′(G) generated the same lattice.

Analogously since G is a balanced pure submodule of D = ⨁ Ḡ_{t} , where t ranges over CT′(G) and Ḡ_{t} is t-projective, it follows that CT′(G) ⊆ CT(G) ⊆ CT(D) and that CT(G) and CT′(G) generate the same lattice. But by Proposition 5.23 T(G) is contained in the lattice generated by CT(G) and vice-versa. This establishes the Proposition. ✓

THE t-SOCLE AND t-RADICAL.

theorem 5.40. Let G be a finite rank torsion free W -module. The following conditions are equivalent:

(1) G is a Butler module.

(2) T(G) is finite and for all t ∈ T(G), G^{∗}(t) is a quasi-pure submodule of G and there exists a pure t-projective submodule G_{t} ⊆ G such that G(t) = G_{t} ⊕ G^{∗}(t)_{∗}.

(3) CT(G) is finite and for all t ∈ CT(G), the obvious embedding δ: G/G^{∗}[t] → ⨁{G/G[s] | s ∈ CT(G), s < t} has a quasi-pure image, and G^{∗}[t]/G[t] is a [maximal] t-projective direct summand of G/G[t].

proof: (1) ⇒ (2): If G is a Butler module, then G is a pure submodule of a completely decomposable module C and by Proposition 2.33 T(G) ⊆ T(C), so T(G) is finite. And by Proposition 5.30 for each t, G(t) = G_{t} ⊕G^{∗}(t)_{∗} for some t-projective G_{t} . Furthermore G^{∗}(t) is a quasi-pure submodule of G by Proposition 5.30.

(2) ⇒ (1): By induction on the size of T(G). If T(G) = {t} then G^{∗}(t) = 0 and by so hypothesis G = G_{t} where G_{t} is t-projective and hence a Butler module.

Now note that if t ∈ T(G) then T(G(t)) = {s ∈ T(G) | s ≥ t} and that if s ≥ t then by Proposition 4.* G(t)(s) = G(s ∧ t) = G(s). It follows that if G satisfies the stated hypothesis then so does G(t). Furthermore since G(t) ◁ G, rank G(t) < rank G unless G(t) = G, i.e. unless t = IT(G). Thus by induction on rank G(t) is a Butler module except possibly for t = IT(G). But by assumption if t = IT(G) then G = G(t) = G_{t} ⊕ G^{∗}(t)_{∗}, where G_{t} is t-projective (since IT(G) ∈ T(G) by Proposition 2.25). Thus if suffices to prove that for t = IT(G), G^{∗}(t)_{∗} is a Butler module. But by hypothesis G^{∗}(t)_{∗} is quasi-equal to G^{∗}(t) and G^{∗}(t) is generated by G(s) for a finite number of s > t, and hence is a homomorphic image of ⨁_{s} G(s). As we have seen, each G(s) is a Butler module by induction on rank, so G is a Butler module since homomorphic images of Butler modules are Butler modules by Corollary 5.20.

(1) ⇒ (3): If G is a Butler module, then G is a homomorphic image of a completely decomposable module C and CT(G) ⊆ CT(C), therefore CT(G) is finite. For the other two claims, we may assume wlog that G[t] = 0. It then follows from Proposition 5.37 that G^{∗}[t] = G(t) is t-projective and a summand of G. Clearly it is maximal as such, since any t-projective submodule of G must be contained in G(t).

Now for s ∈ CT(G) and s < t write C_{s} = G/G[s]. In considering the embedding G/G^{∗}[t] → ⨁ C_{s} , we may assume wlog that G^{∗}[t] = 0. Thus t ∉ CT′(G) and since G[t] = 0, G is t-bounded. Thus s < t for every s ∈ CT′(G). Now by Proposition 5.* G can be embedded as a cobalanced pure submodule of a completely decomposable module ⨁ A_{i}. Furthermore since **** we may assume that for all the rank-one modules A_{i}, t(A_{i}) < t. Now let p be a prime and g ∈ G with g ∉ pG. Then since G ◁ ⨁ A_{i}, there is at least one i such that the coordinate of g in A_{i} is not in pA_{i}. Thus if μ_{i}: G → A_{i} is the restriction to G of the projection onto A_{i} then μ_{i}(g) ∉ pA_{i}. But if s = t(A_{i}) then μ_{i} induces a map G/G[s] → A_{i}. It follows that the image of g in G/G[s] is not in p(G/G[s]) = pC_{s} . Thus δ(g) ∉ p(⨁ C_{s}). This shows that δ(G) ◁ ⨁ C_{s} .

(3) ⇒ (1): Note that if t ∈ CT(G) then CT(G/G[t]) = {s ∈ CT(G) | s ≤ t} and that if s ≤ t then by Proposition 4.* (G/G[t])([s]) = G[s]/G[t] and thus (G/G[t])^{∗}[s] = G^{∗}[s]/G[t]. It follows that if G satisfies the stated conditions then so does G/G[t] and it follows that G/G[t] is a Butler module by induction on rank except possibly in the case G[t] = 0, i.e. where t = OT(G).

Now by assumption, if t = OT(G) then G = G/G[t] ≈ G^{∗}[t] ⊕ (G/G^{∗}[t]), where G^{∗}[t] = G^{∗}[t]/G[t] is t-projective (since by Proposition 2.29 OT(G) ∈ CT(G)). It thus suffices to prove that if t = OT(G) then G/G^{∗}[t] is a Butler module. But by hypothesis G/G^{∗}[t] is isomorphic to a quasi-pure submodule of ⨁_{s<t} G/G[s], and as we have seen each G/G[s] is a Butler module by induction. Thus the result follows from the fact that quasi-pure submodules of Butler modules are Butler modules (Proposition 5.20). ✓

It is customary in presenting the following result to mention that it corrects a misprint in [Lady]. As a measure of how confusing the result is, your author feels compelled to confess that he actually stated it correctly in the original manuscript of and then erroneously “corrected” it while reading the galley proofs.

proposition 5.41. Let G be a Butler module and t a type. Then G(t) is the intersection of all G[s] with s ∈ CT(G) and s ≱ t. proof: Let G⟨t⟩ = ⋂_{s≱t} G[s]. By Proposition 4.4 if s ≱ t then G(t) ⊆ G[s]. Thus G(t) ⊆ G⟨t⟩. Conversely, to see that G⟨t⟩ ⊆ G(t) it suffices to see that G⟨t⟩ is t-saturated. Now by Proposition 5.* G is a pure cobalanced submodule of a completely decomposable module C and so if we define C⟨t⟩ analogously to G⟨t⟩ then the fact that G is cobalanced in C implies that G⟨t⟩ = G ∩ C⟨t⟩ and by Proposition 4.2 the fact that G is pure in C implies that G(t) = G ∩ C(t). Thus it suffices to see that C(t) = C⟨t⟩. Now if C = ⨁ A_{i} where the A_{i} are rank-one modules, then by Lemma 4.28 C(t) = ⨁{A_{i} | t(A_{i}) ≥ t} = ⨁ A_{i}(t). And for s ≱ t, C[s] = ⨁ A_{i}[s] = ⨁{A_{i} | t(A_{i}) ≰ s}. Now A_{i} ⊈ A[s] = 0 for s = t(A_{i}) so if t(A_{i}) ≱ t then A ⊈ C⟨t⟩. Conversely, if t(A_{i}) ≥ t and s ≱ t then s ≱ t(A_{i}) so A_{i}[s] = A_{i}, and since this is true for all s with s ≱ t it follows that A_{i} ⊆ C⟨t⟩. Thus A_{i} ⊆ C⟨t⟩ ⟺ t(A_{i}) ≥ t, and so C(t) = C⟨t⟩. ✓

lemma 5.42. Let G be a Butler module and t a type, let G_{t} be such that G(t) = G_{t} ⊕ G^{∗}(t)_{∗} (c.f. Proposition 5.*). Let Ḡ_{t} = G^{∗}[t]/G[t] and Ḡ′_{t} = (G(t) + G[t])_{∗}/G[t].

(1) G^{∗}(t)_{∗} = G(t)[t].

(2) G_{t} = G′_{t} ⊕ (G_{t} ∩ G[t]) for some G′_{t} .

(3) G^{∗}(t)_{∗} = G(t) ∩ G[t] if and only if G_{t} ∩ G[t] = 0. In particular this is true if t ∉ T′(G).

(4) Ḡ′_{t} is a summand of Ḡ_{t} .

(5) G′_{t} and Ḡ′_{t} are t-projective modules.

(6) G′_{t} ∼ Ḡ′_{t} .

proof: (1) We may as well assume that G = G(t). Then by Proposition 5.* G = G_{t} ⊕ G^{∗}(t)_{∗} for some G_{t} and so G^{∗}(t)_{∗} ⊆ G[t] = G_{t}[t] ⊕ G^{∗}(t)_{∗}[t] = G^{∗}(t)_{∗}[t]. Thus G(t)[t] = G[t] = G^{∗}(t)_{∗}.

(2) This is just a refinement of Proposition 4.23. By Proposition 4.27 G^{∗}(t)_{∗} ⊆ G[t], so that G(t) + G[t] = G_{t} + G^{∗}(t)_{∗} + G[t] = G_{t} + G[t]. Since G_{t} is t-projective by Proposition 4.17 there exists a summand G′_{t} of G_{t} such that G(t) + G[t] = G′_{t} ⊕ G[t]. Then G_{t} = G′_{t} ⊕ (G_{t} ∩ G[t]).

(3) Since G^{∗}(t)_{∗} ⊆ G[t], G(t) ∩ G[t] = (G_{t} ⊕ G^{∗}(t)_{∗}) ∩ G[t] = (G_{t} ∩ G[t]) ⊕ G^{∗}(t)_{∗}. Thus G(t) ∩ G[t] = G^{∗}(t)_{∗} if and only if G_{t} ∩ G[t] = 0. In particular this is true if t ∉ T′(G), since in that case G_{t} = 0.

(4) By Proposition 4.23 (G(t) + G[t])_{∗} ◁ G^{∗}[t]. Therefore Ḡ′_{t} ◁ Ḡ_{t} . Since Ḡ_{t} is t-projective by Theorem 5.40, by Proposition 4.16 Ḡ′_{t} is a summand of Ḡ_{t} .

(5) By Theorem 5.40 G_{t} and Ḡ_{t} are t-projective modules. Since G′_{t} is a summand of G_{t} , and Ḡ′_{t} a summand of Ḡ_{t} , these are also t-projective.

(6) This was proved in Proposition 4.23. ✓

proposition 5.43. Let G be a Butler module, t a type, and let G_{t} be such that G(t) = G_{t} ⊕ G^{∗}(t)_{∗} and let Ḡ_{t} = G^{∗}[t]/G[t].

(1) There is a quasi-exact sequence

0 ⟶ (G(t) ∩ G[t])/G^{∗}(t)_{∗} ⟶ G(t)/G^{∗}(t)_{∗} ⟶^{ζ} G^{∗}[t]/G[t] ⟶ G^{∗}[t]/(G(t) + G[t])_{∗} ⟶ 0 which is quasi-split in the sense that the image of each map is a quasi-summand of the module that contains it.

(2) G(t) ∩ G[t] = G^{∗}(t)_{∗} if and only if G_{t} is a quasi-summand of G, and in this case it is a maximal quasi-summand of G.

(3) G(t) + G[t] = G_{t} ⊕ G[t] if and only if G_{t} is a quasi-summand of G.

(4) G(t) + G[t] is quasi-equal to G^{∗}[t] if and only if Ḡ_{t} is isomorphic to a quasi-summand of G.

proof: (1) Let G′_{t} be as in Lemma 5.42. Since G(t)+G[t] = G′_{t}⊕G[t] and G′_{t} ⊆ G_{t} , the projection of G(t) + G[t] onto G′_{t} restricts to a projection θ′ of G_{t} onto G′_{t} with kernel G_{t} ∩ G[t]. On the other hand, G′_{t} ≈ (G(t) + G[t])/G[t] ⊆ G^{∗}[t]/G[t], so from θ′ we get ζ′: G_{t} → G^{∗}[t]/G[t] with Ker ζ′ = Ker θ′ = G_{t} ∩ G[t] and ζ′(G_{t}) = (G(t) + G[t])/G[t]. Since G_{t} ≈ G(t)/G^{∗}(t)_{∗}, this induces a map ζ: G(t)/G^{∗}(t)_{∗} → G^{∗}[t]/G[t] with Ker ζ = (G(t) ∩ G[t])/G^{∗}(t)_{∗}. Now the image of ζ is (G(t) + G[t])/G[t] and by Proposition 4.23 this is a quasi-pure submodule of G, hence is quasi-equal to (G(t) + G[t])_{∗}/G[t]. Thus the sequence as shown is quasi-exact.

(2) By Lemma 5.42 G(t) ∩ G[t] = G^{∗}(t)_{∗} if and only if G′_{t} = G_{t} . Now G′_{t} ⊆ G_{t} and by Proposition 4.23 G′_{t} is a maximal quasi-summand of G. Thus G′_{t} = G_{t} if and only if G_{t} is a quasi-summand of G, and so G(t) ∩ G[t] = G^{∗}(t)_{∗} if and only if G_{t} is a quasi-summand of G, and in this case it is a maximal quasi-summand of G.

(3) This follows since G(t) + G[t] = G′_{t} ⊕ G[t] and G′_{t} is a summand of G_{t} and by (2) G′_{t} = G_{t} if and only if G_{t} is a quasi-summand of G.

(4) Let ζ′: G_{t} → G^{∗}[t]/G[t] = Ḡ_{t} be as in (1). Since G_{t} = G′_{t} ⊕ Ker ζ, ζ′(G′_{t}) = ζ′(G_{t}) = (G(t) + G[t])/G[t], which as noted in (3) is quasi-equal to (G(t) + G[t])_{∗}/G[t] = Ḡ′_{t} . On the other hand, by Lemma 5.42 Ḡ′_{t} ≈ G′_{t} which is a maximal quasi-summand of G. It then follows that Ḡ_{t} = Ḡ′_{t} if and only if Ḡ_{t} is the image of a quasi-summand of G. Thus G(t) + G[t] is quasi-equal to G^{∗}[t] if and only if Ḡ_{t} is isomorphic to a quasi-summand of G. ✓

HOM AND TENSOR PRODUCT.

proposition 5.44. If G is a s-projective and H t-projective then G ⊗ H is st-projective. proof: If G is s-projective and H t-projective then IT(G) = OT(G) = s and IT(H) = OT(H) = t. By Proposition 2.35 IT(G ⊗ H) = OT(G ⊗ H) = st. Thus G ⊗ H is both st-saturated and st-bounded, hence is st-projective by Proposition 4.*. ✓

proposition 5.45. (1) If G and H are Butler modules, then so are G ⊗ H and Hom(G, H).

(2) If H is a Butler module such that T(H) consists solely of idempotent types, then for any finite rank torsion free module G, Hom(G, H) is a Butler module whose typeset contains only idempotent types.

(3) If G is a Butler module such that T(G) contains only idempotent types, then for any H, Hom(G, H) is is isomorphic to a quasi-pure submodule of ⨁_{T′(G)} H(t)^{k_{t}} , where k_{t} = rank G_{t} . proof: (1) If G is a homomorphic image of ⨁ A_{i}, and H a homomorphic image of ⨁ B_{j}, where the A_{i} and B_{j} are rank-one modules, then G ⊗ H is a homomorphic image of ⨁ A_{i} ⊗ B_{j}, and each A_{i} ⊗ B_{j} is a rank-one module. Therefore G ⊗ H is a Butler module. Furthermore if H is a pure submodule of ⨁ C_{k}, where the C_{k} are rank-one modules, then Hom(G, H) ◁ ⨁ Hom(G, C_{k}) ◁ ⨁_{i,k} Hom(A_{i}, C_{k}). Since each Hom(C_{k}, A_{i}) is a rank-one module, Hom(G, H) is a Butler module by Theorem 5.10.

(2) If T(H) contains only idempotent types then by Proposition 5.22 the same is true of CT(H). Therefore the rank-one modules C_{k} in the previous paragraph all have idempotent types. Thus by Proposition 4.39 each Hom(G, C_{k}) is t_{k}-projective, where t_{k} = t(C_{k}). Thus Hom(G, H) is isomorphic to a pure submodule of the completely decomposable module ⨁ Hom(G, C_{k}) and therefore is a Butler module by Theorem 5.10.

(3) If A is a rank-one module and t = t(A) is idempotent, then by Proposition 4.36 Hom(A, H) ≈ H(t). Thus if t ∈ T′(G) and G_{t} is a t-projective module such that G(t) = G_{t} ⊕ G^{∗}(t)_{∗}, then Hom(G_{t}, H) ≈ H(t)^{k_{t}} , where k_{t} = rank G_{t} . But by Proposition 5.32 G is a quasi-homomorphic image of ⨁_{T′(G)} G_{t} , so that Hom(G, H) is quasi-isomorphic to a pure submodule of ⨁_{T′(G)} Hom(G_{t}, H) ≈ ⨁ H(t)^{k_{t}} . ✓

proposition 5.46. Let t be an idempotent type. Let G and H be Butler modules such that G has no non-trivial t-projective quasi-summand. For each s ∈ T′(G) ∪ T′(H), choose G_{s} and H_{s} such that G(s) = G_{s} ⊕ G^{∗}(s) and H(s) = H_{s} ⊕ H^{∗}(s) (c.f. Proposition 5.30).

(1) Let G′ and H′ be the submodules of G and H generated by G_{s} and H_{s} for all s ≠ t. Then G ⊗ H is quasi-equal to (G′ ⊗ H_{t}) + (G ⊗ H′).

(2) Every element of T′(G ⊗ H) has the form st for s ∈ T′(G), t ∈ T′(H).

(3) If t is minimal in T′(H) then t ∉ T′(G ⊗ H).

proof: (1) By Proposition 5.30 H is quasi-equal to H_{t} +H′, and so G⊗H is quasi-equal to (G ⊗ H_{t}) + (G ⊗ H′). Thus it suffices to see that G ⊗ H_{t} is quasi-equal to G′ ⊗ H_{t} . By Proposition 5.30 G′ is an essential submodule of G and G′ + G_{t} is quasi-equal to G, so we may assume wlog that G = G′ + G_{t} . Thus G/G′ is a torsion module which is a homomorphic image of G_{t} , so there is a short exact sequence 0 → K → G_{t} → G/G′ → 0 where rank K = rank G. Now G_{t} and H_{t} are t-projective and since t is idempotent it follows from Proposition 5.44 that G_{t} ⊗ H_{t} is t-projective. Then K ⊗ H_{t} is t-saturated and also t-bounded since it is a submodule of G_{t} ⊗ H_{t} . Thus K ⊗ H_{t} is t-projective and quasi-equal to G_{t} ⊗ H_{t} by Proposition 3.9. Therefore (G/G′) ⊗ H_{t} has finite length, so G′ ⊗ G_{t} is quasi-equal to G ⊗ H_{t} . This completes the proof that G ⊗ H is quasi-equal to (G ⊗ H_{t}) + (G ⊗ H′).

(2) ****

(3) By (1) G is quasi-equal to the submodule generated by G_{s} ⊗ H_{t} and G_{t} ⊗ H_{s′} and G_{s} ⊗ H_{s′} for all s ≠ t ∈ T′(G) and s′ ≠ t ∈ T′(H). Thus if t is minimal in T′(G) ∪ T′(H) it follows that G ⊗ H is generated by rank-one modules all having types ss′ ≰ t. From this one sees that G(t) ⊆ G^{∗}(t)_{∗}, so that t ∉ T′(G ⊗ H). ✓

Consider a Butler module G such that T′(G) contains only idempotent types. Form the sequence T′(G), T′(G ⊗ G), T′(G ⊗ G ⊗ G), …. Recall from Proposition 2.* that for idempotent types s and t, st = s ∨ t. It thus follows from (2) of Proposition 5.46 that T′(G^{⊗n}) ⊆ T′(G) for all n. On the other hand, by (3) of Proposition 5.46 as long as no rank-one quasi-summands occur there is an upward pressure on the critical typesets T′(G^{⊗n}), in the sense that the minimal elements of T′(G^{⊗n}) cannot belong to the critical typeset of G^{⊗n+1} . If this were continue indefinitely, we would eventually derive the absurdity that for large enough n, T′(G^{⊗n}) = ∅. Since this cannot happen, it follows that for some n G^{⊗n} must have a rank-one quasi-summand.

This illustrates the idea that when one tensors Butler modules whose typesets are idempotent there is a tendancy for rank-one quasi-summands to appear, and the more one tensors the stronger this tendancy becomes. One can contrast this with what happens with Butler modules whose typesets contains only locally trivial types. We will see in Chapter 9 that if G and H are Butler modules of this sort then it is extremely difficult for G ⊗ H to have a rank-one quasi-summand.

ALMOST COMPLETELY DECOMPOSABLE MODULES. Consider again Example 4.25. In this example we constructed two indecomposable modules G and H, with completely decomposable submodules C = A_{1} ⊕ ⋯ ⊕ A_{n} ≈ D = B_{1} ⊕ ⋯ ⊕ B_{n} such that G/C ≈ W/p and H/D ≈ W/q. We then saw that G ⊕ H = K ⊕ L where L is completely decomposable. Thus G ⊕ H, a direct sum of 2 indecomposable modules, can also be written as a direct sum of n + 1 indecomposable modules (since L is the direct sum of n rank-one modules).

Heuristically, we can explain Example 4.25, at least after a fashion, as follows: Since G/C is cyclic, the module G consists of the completely decomposable module C glued together by a single element in p^{−1}(a_{1} + ⋯ + a_{n}). Likewise H consists of D glued together by a single element belonging to q^{−1}(b_{1} + ⋯ + b_{n}). But since G/C ⊕ H/D ≈ W/p ⊕ W/q ≈ W/pq, a cyclic module, we can find a single element g ∈ G ⊕ H such that the image of g is a generator of G ⊕ H / C ⊕ D. If we can then find a pure submodule K of G ⊕ H of rank n containing g, then K contains all the glue for G ⊕ H and consequently G ⊕ H must be the direct sum of K and a direct sum of rank-one modules (since there is no glue left to stick the remaining rank-one quasi-summands together).

The key point here seems to be that modules which are obtained from completely decomposable modules by using a finite number of pieces of “glue” are likely to be easy to manipulate in terms of obtaining various direct sum decompositions. And that if G is such a module, and C the underlying completely decomposable module, then the quotient G/C is a good measure of how far G is from being completely decomposable and how G will behave in direct sum decompositions.

To formalize this, we introduce the concept of almost completely decomposable modules.

We say that G is almost completely decomposable if G is quasi-isomorphic to a completely decomposable module. Since completely decomposable modules are Butler modules, by Proposition 5.7 so are almost completely decomposable modules. proposition 5.47. Let G be almost completely decomposable.

(1) T′(G) consists of those types t such that G has a quasi-summand of type t.

(2) T′(G) = CT′(G).

(3) T(G) consists of all types of the form inf S, for S ⊆ T′(G).

(4) CT(G) consists of all types of the form sup S for S ⊆ T′(G).

(5) If s, t ∈ T′(G) and G(t) = G_{t} ⊕ G^{∗}(t)_{∗} and t ≤ s then G_{t} ∩ G[s] = 0.

proof: Since G is quasi-equal to a completely decomposable module, by Proposition 3.8 and Proposition 5.30 it suffices to prove the theorem for completely decomposable modules. But in this case (1) was proved in Proposition 5.30, (2) was proved in Proposition 5.31, and (3) and (4) in Corollary 2.24. And (5) is clear for completely decomposable modules since if G = ⨁ G_{t} then G[s] = ⨁_{s′≰s} G_{s′} so if t ≤ s then

G_{t} ∩ G[s] = 0. ✓

proposition 5.48. Let G be a Butler module Then G is completely decomposable if and only if for each t ∈ CT(G), G[t] = ∑{G(s) | s ≰ t}. proof: ( ⇒ ): Easy.

( ⇐ ): For each t ∈ T′(G) let G_{t} ◁ G be such that G(t) = G_{t} ⊕ G^{∗}(t)_{∗}. We will prove that G = ⨁ G_{t} . Suppose first, by way of contradiction, that for some t ∈ T′(G), G_{t} ∩ ∑{G_{s} | s ≠ t} ≠ 0. ????

Let t = IT(G). Then by Proposition 4.40 G = G_{t} ⊕ G[t] for some t-projective module G_{t} . It then suffices to prove that G[t] is completely decomposable. By assumption, G[t] = ∑{G(s) | s ≰ t} = ∑{G(s) | s ∈ T(G) | s ≠ t}, since t = IT(G). But this simply says that G = G^{∗}(t) and so G^{∗}(t)_{∗} = G^{∗}(t). ????

By Proposition 4.29 G is completely decomposable if and only if for all t ∈ CT(G), G^{∗}[t] = G(t) + G[t]. Now if G[t] = ∑{G(s) | s ≰ t} then G(t) + G[t] = ∑{G(s) | s ≮ t}. Now by Proposition 4.27 G(t) + G[t] ⊆ G^{∗}[t]. ****

proposition 5.49. Let G be a Butler module and for every t ∈ T′(G) let G_{t} ◁ G(t) be such that G(t) = G_{t} ⊕ G^{∗}(t)_{∗}. The following conditions are equivalent:

(1) G is almost completely decomposable.

(2) For every type t, G_{t} is a quasi-summand of G.

(3) The sum ∑_{T′(G)} G_{t} is direct.

(4) For every type t, G^{∗}[t]/G[t] is isomorphic to a quasi-summand of G.

(5) For every type t, G^{∗}(t)_{∗} = G(t) ∩ G[t].

(6) For every t, G^{∗}[t] = (G(t) + G[t])_{∗}.

(7) For every t ∈ CT(G), G[t] is quasi-equal to ∑{G(s) | s ≰ t}.

(8) For every t ∈ T′(G), G(t) + G[t] = G_{t} ⊕ G[t].

proof: (1) ⇒ (2): This just comes down to seeing that (2) is true for completely decomposable modules. If G is quasi-equal to a completely decomposable module C = ⨁ C_{t} , then by Proposition 4.2 G(t) is quasi-equal to C(t) and G^{∗}(t)_{∗} is quasi-equal to C^{∗}(t)_{∗}. Furthermore C(t) = C_{t} ⊕ C^{∗}(t), so that G(t) is quasi-equal to C_{t} ⊕ G^{∗}(t)_{∗}. It follows that we may choose G_{t} = (C_{t})_{∗}. Since C_{t} is a summand of C and G is quasi-equal to C, it follows that G_{t} is a quasi-summand of G. And in fact this will be true no matter what choice of G_{t} we make, as can be seen, for instance, from the equivalence of (2) and (3).

(2) ⇒ (3): Suppose that each G_{t} is a quasi-summand of G. Note that no quasisummand of G_{t} is quasi-isomorphic to any quasi-summand of G_{t′} for t′ ≠ t. Thus by Theorem 3.24 ∑ G_{t} is direct, and generates a quasi-summand of G.

(3) ⇒ (1): By Proposition 5.32 G is quasi-equal to ∑_{T′(G)} G_{t} . Thus if ∑_{T′(G)} G_{t} = ⨁_{T′(G)} G_{t} then G is quasi-equal to a completely decomposable module, hence is almost completely decomposable.

(1) ⟺ (6): Proved in Proposition 4.29.

(4) ⟺ (6): Proved in Proposition 5.43.

(2) ⟺ (5): Proved in Proposition 5.43.

(1) ⇒ (7): By analogous reasoning, it suffices to prove that C[t] = ∑_{s≰t} C(s), and this is clear from the formulas above, since ⨁_{s≰t} C_{s} = ∑_{s≰t} C(s).

(7) ⇒ (1): ****

(2) ⟺ (8): By Proposition 5.43. ✓

proposition 5.50. Let G be an almost completely decomposable module and s, t types. Then

(1) G[s][t] = G[s] ∩ G[t].

(2) G(s)[t] = G(s) ∩ G[t].

proof: (1) It suffices to prove this for completely decomposable modules, since if G is quasi-equal to a completely decomposable module C then G[s], G[t], G[s][t], and G[s] ∩ G[t] are all pure submodules of G and are quasi-equal to C[s], C[t], C[s][t], and C[s] ∩ C[t]. But if G = ⨁ A_{i}, where the A_{i} are rank-one modules, then C[s] = ⨁{A_{i} | t(A_{i}) ≰ s}, and C[s][t] = ⨁{A_{i} | t(A_{i}) ≰ s and t(A_{i}) ≰ t} = C[s] ∩ C[t].

(2) Analogous. ✓

REGULATING SUBMODULES. Since an almost completely decomposable module G is quasi-isomorphic to a completely decomposable module, by Proposition 3.* there exist completely decomposable submodules C ⊆ G such that G/C has finite length. We want to single out certain of these completely decomposable submodules, called regulating submodules, which have special importance. In order to describe their properties, we first need to define the module-theoretic analogue of the index of a subgroup in a group.

notation 5.51. If H is a submodule of G such that G/H has finite length, we define an ideal [G : H] such that the following properties hold:

(1) If G = H then [G : H] = W.

(2) If G/H is simple, then [G : H] = p where p is the unique prime ideal such that G/H ≈ W/p.

(3) If H ⊆ K ⊆ G, then [G : H] = [G : K][K : H].

It follows from the Jordan-Hölder Theorem that these three properties recursively determine [G : H] for all H ⊆ G such that G/H has finite length, and that [W : 𝔞] = 𝔞 for any non-trivial ideal 𝔞.

In other words, if G/H has finite length consider a composition series H = G_{0} ⊊ G_{1} ⊊ … ⊊ G_{n} = G. Then each G_{i}/G_{i−1} is a simple module, so G_{i}/G_{i−1} ≈ W/p_{i} for some prime (maximal) ideal p_{i}. Then [G: H] = p_{1} ⋯ p_{n}.

(Unfortunately, the notation [G: H] is used in some books on commutative ring theory in a slightly different sense. This is one case where we have decided to be consistent with group theoretic language instead of with ring theory. Since our use of this notation does not extend beyond the present chapter, your author hopes that any confusion will be minimal and brief.)

lemma 5.52. (1) Let C ⊆ G be such that G/C has finite length and let I = [G: C]. Then IG ⊆ C.

(2) If C ⊆ G and D ⊆ H are such that G/C and H/D have finite length, then [G ⊕ H: C ⊕ D] = [G: C][H: D].

(3) If H ⊊ K ⊊ G then [G: H] ⊊ [G: K] and [G: H] ⊊ [K: H].

proof: (1) By an easy induction on length G/C one sees that I(G/C) = 0 so that IG ⊆ C.

(2) C ⊕ D ⊆ C ⊕ H ⊆ G ⊕ H. Thus [G ⊕ H: C ⊕ D] = [G ⊕ H: C ⊕ H][C ⊕ H: C ⊕ D]. But it is easy to see that [G ⊕ H: C ⊕ H] = [G: C] and [C ⊕ H: C ⊕ D] = [H: D].

(3) In fact, by the defining property of the index function, [G: H] = [G: K][K: H] and [K: H] ≠ W, [G: K] ≠ W. Thus [G: H] is a proper subideal of both [G: K] and

[K: H]. ✓

Note that (3) of Lemma 5.52 says that the index function [G: C] is, as it were, covariant in the second variable C. Your author tends to find this sometimes confusing.

definition 5.53. Let G be almost completely decomposable and for each t ∈ T′(G) let G_{t} be a pure completely decomposable submodule of G such that G(t) = G_{t} ⊕ G^{∗}(t)_{∗} (see Proposition 5.40). Recall from Proposition 5.49 that G is quasi-equal to ⨁ G_{t} . We say that ⨁ G_{t} is a regulating submodule of G.

Regulating submodules are not unique, except in certain special cases. In fact, it can be shown that in general not even the quotient G/ ⨁ G_{t} is uniquely determined independently of the choice of G_{t} . However we now see that the index [G: C] is the same for all regulating submodules C of G.

lemma 5.54. Let G be an almost completely decomposable module and let C = ⨁ C_{t} be a submodule of G quasi-equal to G, where each C_{t} is t-projective. Let t be minimal in T′(G) = CT′(G).

(1) C is a regulating submodule of G if and only G(t) + G[t] = C_{t} ⊕ G[t] and C[t] is a regulating submodule of G[t].

(2) If C is a regulating submodule of G then [G : C] = [G: G(t) + G[t]] [G[t]: C[t]].

(3) G(t) + G[t] = G^{∗}[t] = G. proof: (1) C is a regulating submodule of G if and only if for each s ∈ T′(G), G(s) = C_{s} ⊕ G^{∗}(s)_{∗}. By Proposition 5.49 this will be true if and only if G(s) + G[s] = C_{s} ⊕ G[s]. Furthermore since t is minimal IN t′(g), C[t] = ⨁_{s≠t} C_{t} and by Proposition 4.4 G(s) ⊆ G[t] for s ≠ t, so that G[t](s) = G(s) and G[t]^{∗}(s) = G^{∗}(s)_{∗}. It follows that G(s) = C_{s} ⊕ G^{∗}(s) for all s ≠ t ∈ T′(G) if and only if C[t] is a regulating submodule of G[t]. Thus C is a regulating submodule of G if and only if C[t] is a regulating submodule of G[t] and G(t) + G[t] = C_{t} ⊕ G[t].

(2) Since C ⊆ G(t)+G[t] ⊆ G, [G: C] = [G: G(t)+G[t]] [G(t)+G[t]: C]. Furthermore if C is a regulating submodule of G then G(t) + G[t] = C_{t} ⊕ G[t] and C = C_{t} ⊕ C[t], so that [G(t) + G[t]: C] = [G[t]: C[t]]. Thus [G: C] = [G: G(t) + G[t]] [G[t]: C[t]].

(3) By Proposition 5.49, (G(t) + G[t])_{∗} = G^{∗}[t]. Now by Proposition 5.47, CT(G) is contained in the lattice generated by T′(G). Thus if t is minimal in T′(G) and φ ∈ Hom(G, Q) then t(φ(G)) ≮ t since t(φ(G)) ∈ CT(G). Thus G[s] = G for all s < t,

so G^{∗}[t] = G. ✓

proposition 5.55. Let G be an almost completely decomposable module, and let C = ⨁ G_{t} be a regulating submodule of G.

(1) If C′ is a completely decomposable submodule of G which is quasi-equal to G, then [G : C′] ⊆ [G: C].

(2) If C′ is a completely decomposable submodule of G which is quasi-equal to G then C′ is a regulating submodule of G if and only if [G : C′] = [G: C].

proof: (1) Let C′ be a completely decomposable submodule of G which is quasi-equal to G, and write C′ = ⨁_{T′(G)} C′_{t} , where C′_{t} is t-projective. We use induction on the size of T′(G), the result being clear if G is homogeneous since in that case G is quasi-isomorphic to a t-projective module for some t and so by Proposition 4.16 G itself is t-projective and thus G = G_{t} so G = C and [G: C] = W.

If G is not homogeneous, choose t minimal in T′(G) = CT′(G). Then by Lemma 5.54 (G(t) + G[t])_{∗} = G^{∗}[t] = G. Then by Lemma 5.54 C[t] is a regulating submodule of G[t]. Now C′_{t} ⊕ G^{∗}(t)_{∗} ⊆ G(t) and so C′_{t} ⊕ G[t] ⊆ G(t) + G[t]. Thus, recalling that the index operation is covariant in the second variable,

[G: C′] = [G: C′_{t} ⊕ G[t]] [C′_{t} ⊕ G[t]: C′] ⊆ [G: G(t) + G[t]] [G[t]: C′[t]],

since C′ = C′_{t} ⊕ C′[t]. By induction [G[t]: C′[t]] ⊆ [G[t]: C[t]] so that

[G: C′] ⊆ [G: G(t) + G[t]] [G[t]: C[t]] = [G: G(t) + G[t]] [G_{t} ⊕ G[t]: G_{t} ⊕ C[t]]

= [G: G(t) + G[t]] [G(t) + G[t]: C] = [G: C],

since G(t) + G[t] = G_{t} ⊕ G[t] by Proposition 5.49 and G_{t} ⊕ C[t] = G_{t} ⊕ ⨁_{s≠t} G_{s} = ⨁ G_{t} = C.

(2) It follows from (1) that if C and C′ are regulating submodules of G then [G: C′] ⊆ [G: C] and [G: C] ⊆ [G: C′] so that C = C′.

Conversely, suppose that C′ is a completely decomposable submodule module of G quasi-equal to G and that [G: C′] = [G: C], where C is a regulating submodule. Then the inclusions in the displayed formulas above must be equalities. It that follows for t minimal in T′(G), [G: C′_{t} ⊕ G[t]] = [G: G(t) + G[t]], so that G(t) + G[t] = C′_{t} ⊕ G[t]. Likewise it must be true that [G[t]: C′[t]] = [G[t]: C[t]]. By induction on the size of T′(G) it then follows that C′[t] is a regulating submodule of G[t], and thus by Lemma 5.54 C′ is a regulating submodule of C. ✓

definition 5.56. If G is an almost completely decomposable module and C a regulating submodule, then we call [G: C] the regulating index of G and write i(G) = [G: C].

Proposition 5.55 says that among the ideals of the form [G: C], where C ranges over the completely decomposable submodules quasi-equal to G, there is a unique largest one, namely i(G), and that a completely decomposable submodule C of G which is quasi-equal to G is a regulating submodule if and only if [G: C] = i(G). Thus i(G) is in some sense a measure of how far G is from being completely decomposable. The larger the idea i(G) is, the closer G is to being completely decomposable.

Those who do not like the usage of the group-theoretic word “index” in this context might prefer to call i(G) the regulating defect. Or perhaps regulating excess might be more appropriate.

Lemma 5.54 asserts that if t is minimal in T′(G) then i(G) = i(G[t])[G: G(t) + G[t]]. This gives a recursive procedure for computing i(G) without the need to actually find a regulating submodule of G. In applying this it is useful to remember that by Proposition 5.50 and Proposition 4.4 for any type s, G[t][s] = G[t] ∩ G[s] and G[t](s) = G[t] ∩ G(s).

example 5.57. Let A_{1}, A_{2}, A_{3} , and B be non-trivial submodules of Q containing 1 such that t(A_{1}), t(A_{2}), t(A_{3}) are mutually incomparable and t(B) < t(A_{1}), t(A_{2}) and t(B) ≰ t(A_{3}) (and necessarily t(A_{i}) ≰ t(B) for any i, otherwise t(A_{i}) ≤ t(B) < t(A_{1})). Suppose also that there exists a prime p such that 1 ∉ pA_{1}, pA_{2}, pA_{3}, pB. Let p′ be an element of p with p′ ∉ p^{2} . Let G = (A_{1} ⊕ A_{2} ⊕ A_{3} ⊕ B) + p^{−2}(1, 1, −p′, −p′) Then i(G) = p^{2} . proof: Let C = A_{1} ⊕ A_{2} ⊕ A_{3} ⊕ B, so that G = C + p^{−2}(1, 1, −p′, −p′). It is easy to se that C is a regulating submodule of G (although not the only one). However instead of using this fact, we can use Lemma 5.54, as indicated above. If s = t(B) and t_{i} = t(A_{i}), then the minimal elements in T′(G) are s and t_{3} . Now since p^{−1}p′ ⊆ p^{−1}p = W ⊆ A_{3}, B, then p^{−1}(1, 1, p′, p′) ⊆ p^{−1}(1, 1, 0, 0) + C. It follows that explain!

G[t_{1}] = (0 ⊕ A_{2} ⊕ A_{3} ⊕ 0).

Likewise

G(s) = C(s)_{∗} = (A_{1} ⊕ A_{2} ⊕ 0 ⊕ B) + p^{−1}(1, 1, 0, 0)

and

G[s] = (A_{1} ⊕ A_{2} ⊕ A_{3} ⊕ 0) + p^{−1}(1, 1, 0, 0).

Thus G(s)+G[s] = C+p^{−1}(1, 1, 0, 0), and pG ⊆ (G(s)+G[s]) so that [G: G(s)+G[s]] = p. Now T′(G[s]) = {t_{1}, t_{2}, t_{3}} and all these are minimal critical types for G[s]. Now since G(t_{1}) = A_{1} ⊕ 0 ⊕ 0 ⊕ 0 ◁ G[s] ◁ G, then G[s](t_{1}) = G(t_{1}) = A_{1} ⊕ 0 ⊕ 0 ⊕ 0. And G[s][t_{1}] = (0 ⊕ A_{2} ⊕ A_{3} ⊕ 0). Thus G[s] /(G[s](t_{1}) + G[s][t_{1}]) is generated by the image of p^{−1}(1, 1, 0, 0), so that [G[s]: (G[s](t_{1}) + G[s][t_{1}])] = p. Therefore

i(G) = i(G[s])[G: G(s) + G[s]] = pi(G[s]) = pi(G[s][t_{1}])[G[s]: G[s](t_{1}) + G[s][t_{1}]]
= p^{2}i(G[s][t_{1}]) = p^{2}

since G[s][t_{1}] = 0 ⊕ A_{2} ⊕ A_{2} ⊕ 0, a completely decomposable module. ✓

proposition 5.58. (1) If G and H are almost completely decomposable modules then i(G ⊕ H) = i(G)i(H).

(2) An almost completely decomposable module G is completely decomposable if and only if i(G) = W.

(3) If G and H are almost completely decomposable modules and G ⊕ L ≈ H ⊕ L for some finite rank torsion free module L, then i(G) = i(H).

proof: (1) If C and D are regulating submodules of G and H then it is immediately apparent that C ⊕ D is a regulating submodule of G ⊕ H. Thus i(G ⊕ H) = [G ⊕ H: C ⊕ D] = [G: C][H: D] = i(G)i(H).

(2) G is completely decomposable if and only if G is a regulating submodule for itself, i.e. if and only if i(G) = [G: G] = W.

(3) We use induction on the size of T′(G). If T′(G) = {t}, then G is t-projective and so if G⊕L ≈ H ⊕L for some L then H ∼ G by Theorem 3.24, and so by Proposition 4.16 H must also be t-projective. Thus i(G) = i(H) = W. If T′(G) has more than one type, let t be minimal in T′(G). By Lemma 5.54 i(G) = i(G[t])[G: G(t) + G[t]]. Furthermore by Lemma 5.54 G = (G(t) + G[t])_{∗}, and likewise H = (H(t) + H[t])_{∗}. Now let F be the additive functor given by F(X) = (X(t) + X[t])_{∗}/(X(t) + X[t]). By Proposition 4.23 F(L), F(G), and F(H) all have finite length. Since G ⊕ L ≈ H ⊕ L, F(G) ⊕ F(L) ≈ F(H) ⊕ F(L), and so F(G) ≈ F(H) by the Krull-Schmidt Theorem for finite length modules, so that [G: G(t) + G[t]] = [(G(t) + G[t])_{∗}: G(t) + G[t]] = [(H(t) + H[t])_{∗}: H(t) + H[t]] = [H: H(t) + H[t]]. And by induction i(G[t]) = i(H[t]).

Thus i(G) = i(H). ✓

We can now make explicit the assertion in the remark preceding Proposition 5.47 that in Example 4.25 the submodule K of G ⊕ H “uses up all the glue” that holds G and H together, so that if G ⊕ H = K ⊕ L then L must be completely decomposable. In fact, we can now see that this simply means that i(K) = i(G ⊕ H) so that i(L) = W and therefore L is completely decomposable.

We now get a Theorem, which anticipates the ideas of Chapter 10. We first need the following lemma, which is vaguely analogous to (but much weaker than) Nakayama’s Lemma. lemma 5.59. If G is a reduced finite rank torsion free module and I is a non-trivial ideal, then there exists 0 ≠ J ⊆ I such that JM ⊊ M for every non-trivial submodule M of G. proof: If IM = M and IN = N then I(M + N) = M + N. Thus there exists a unique largest submodule M of G such that IM = M. Now if IM = M then IM_{∗} = M_{∗}, so it follows that M ◁ G. If M ≠ 0 then M ⊈ p^{∞}G for some prime p since G is reduced, so M is not p-divisible. Thus M ⊋ pM = pIM and pI ⊆ I. Thus there exist non-trivial ideals I′ ⊆ I such that I′M ≠ M. Now if M′ ⊆ G is such that I′M′ = M′ then M′ ⊇ IM′ ⊇ I′M′ = M′ so that M′ ⊆ M by the maximality of M, and again if we choose M′ maximal with this property then M′ ◁ G and M′ ⊊ M. Since G has the descending chain condition on pure submodules, eventually we find an ideal J ⊆ I such that JM = 0 only for M = 0. ✓

proposition 5.60. If G, H, and L are finite rank torsion free modules such that G⊕L ≈ H ⊕L, and let I be a non-trivial ideal of W. Then there exists a monomorphism φ: G → H such that H/φ(G) has finite length and [H: φ(G)] is comaximal with I. proof: It suffices to prove the Proposition for G/d(G) and H/d(H) since ****. Thus we may suppose wlog that G is reduced. Thus it follows from Lemma 5.59 that, replacing I by a still smaller non-trivial ideal, there is no loss of generality in suppose that IM ⊊ M for every non-trivial submodule M of G.

Now let R = End G and define a function H from finite rank torsion free modules to right R-modules by H(X) = Hom(G, X)/I Hom(G, X). Then H(X) has finite length (as a W -module and a fortiori as an R-module) and so from the isomorphism H(G) ⊕ H(L) ≈ H(H) ⊕ H(L) we deduce that H(G) ≈ H(L) by the Krull-Schmidt Theorem for finite length modules. In particular, H(G) is a cyclic R-module.

Now define a functor F from finite rank torsion free modules to finite length W -modules by letting F(X) be the image of the composition G ⊗ Hom(G, X) → X → X/IX, where the first map is simply given by g ⊗ φ ↦ φ(g). Since every g ∈ G has the form φ(g) where φ = 1 ∈ Hom(G, G), it follows that F(G) = G/IG. As with the functor H in the previous paragraph we conclude from the classical Krull-Schmidt Theorem that F(H) ≈ F(G) = G/IG. But now H/IH ⊕ L/IL ≈ G/IG ⊕ L/IL, and since this is an isomorphism of finite length modules we conclude that H/IH ≈ G/IG. Thus F(H) ≈ F(G) = G/IG ≈ H/IH, and since F(H) ⊆ H/IH and these are finite length modules we conclude that F(H) = H/IH.

Now let γ ∈ Hom(G, H) be a generator for the right R-module H(H). Then F(H) is generated by elements of the form γφ(g) + IH ∈ H/IH, so that H/IH = F(H) = (γ(H) + IH)/IH. This shows that γ induces a surjection γ̄: G/IG → H/IH. Since G/IG and H/IH are isomorphic finite length modules it follows that γ̄ is an isomorphism. Thus if K = Ker γ then K ⊆ IG, but K ◁ G so K ⊆ IK ⊆ K. But by assumption, IK = K implies K = 0. Thus γ is monic.

Now 0 = Ker γ̄ = (γ^{−1}(IH) + IG)/IG and it follows that I(H/γ(G)) = (γ(G) + IH)/γ(G) = H/γ(G), so that if I ⊆ p for some prime p then p ∉ Ass(H/γ(G)) and thus W/p does not appear in a composition series for H/γ(G), so that [H: γ(G)] ⊈ p. Thus no prime ideal contains both I and [H: γ(G)], so [H: γ(G)] + I = W, as required. ✓

We will postpone looking at all the fascinating consequences of Proposition 5.60 for finite rank torsion free modules in general until Chapter 10. For almost completely decomposable modules, however, we have a very specific result, as follows:

theorem 5.61. Suppose that W is a principal ideal domain and let G and H be almost completely decomposable W -modules. The following conditions are equivalent:

(1) There exists a finite rank torsion free module L such that G ⊕ L ≈ H ⊕ L.

(2) G ⊕ C ≈ H ⊕ C, where C is a regulating submodule of G.

(3) i(G) = i(H) and G contains a submodule G′ such that G′ is quasi-equal to G, G′ ≈ H and the ideals [G : G′] and i(G) are comaximal.

proof: (1) ⇒ (3): Suppose that G ⊕ L ≈ H ⊕ L for some L. Then by Proposition 5.58 i(G) = i(H). Furthermore by Proposition 5.60 there exists γ: G → H such that γ(G) ≈ G and γ(G) is quasi-equal to H and [H: γ(G)] + i(G) = W.

(2) ⇒ (1): Clear.

(3) ⇒ (2): Suppose that i(G) = i(H) and there there exists φ ∈ Hom(H, G) such that φ is a quasi-isomorphism and i(G) + [G: φ(H)] = W. Choose u ∈ [G: φ(H)], v ∈ i(G), such that u + v = 1. Let C be a regulating submodule of G. Then [G: C] = i(G). Thus by Lemma 5.52 vG ⊆ i(G)G ⊆ C and uG ⊆ φ(H). Let α: G → H ⊕ C be defined by α(g) = (φ^{−1}(ug), vg), and let β: H ⊕ C → G by β(h, c) = φ(h) + c. Then βα(g) = φ(φ^{−1}(ug)) + vg = (u + v)g = g so that α is a split monomorphism. Thus H ⊕ C ≈ G ⊕ D for some D. Since G ∼ H, by Theorem 3.24 D ∼ C. Furthermore by Proposition 5.58 i(G)i(D) = i(H)i(C) and since i(G) = i(H) it follows that i(D) = i(C) = W, since C is completely decomposable. Thus by Proposition 5.58 D is completely decomposable. Since W is a principal ideal domain and C and D are quasi-isomorphic completely decomposable modules, D ≈ C by Proposition 4.28. Thus

G ⊕ C ≈ H ⊕ C. ✓

example 5.62. Example 1.49 is one good example for Theorem 5.62. We will use the same idea, but make it a little more complicated. Let A_{1}, A_{2}, A_{3} be rank-one modules with mutually incomparable types, such that A_{i} ∩ A_{j} is the same for all pairs i ≠ j and such that there exists a prime p such that none of the A_{i} is p-divisible. We may assume wlog that 1 ∈ A_{i} and 1 ∉ pA_{i} for all i. Let u, v ∉ p. Let

C = A_{1} ⊕ A_{2} ⊕ A_{3}
G =(A_{1} ⊕ A_{2} ⊕ A_{3}) + p^{−1}(1, 1, 1)
H =(A_{1} ⊕ A_{2} ⊕ A_{3}) + p^{−1}(1, u, v) ⊆ Q ⊕ Q ⊕ Q.

Then C is a regulating submodule of both G and H, i(G) = i(H) = p, and G ⊕ C ≈ H ⊕ C′, for some completely decomposable module C′ quasi-isomorphic to C. proof: Let t_{i} = t(A_{i}). By the assumption on the A_{i}, t_{i} ∧ t_{j} = IT(G) for all pairs i ≠ j. Then T(G) = T(H) = {t_{1}, t_{2}, t_{3}, IT(G)}, T′(G) = T′(H) = {t_{1}, t_{2}, t_{3}}. For i = 1, 2, 3, G(t_{i}) = A_{i} ⊆ A_{1} ⊕ A_{2} ⊕ A_{3} and G^{∗}(t_{i}) = 0, so that G_{t_{i}} = A_{i}. Thus C = A_{1} ⊕ A_{2} ⊕ A_{3} = G_{t_{1}} ⊕ G_{t_{2}} ⊕ G_{t_{3}} , so C is a regulating submodule of G and i(G) = i(H) = [G: C] = [H: D] = p. Now let G′ be the submodule of G generated by uvA_{1} ⊕ vA_{2} ⊕ uA_{3} together with uvp^{−1}(1, 1, 1) and let φ: QH → QG be defined by φ(x, y, z) = (uvx, vy, uz). Then φ(A_{1}) = uvA_{1} , φ(A_{2}) = vA_{2} , φ(A_{3}) = uA_{3} , and φ(p^{−1}(1, u, v)) = uvp^{−1}(1, 1, 1) so that in fact φ(H) = G′ and G′ ≈ H. Furthermore G/G′ ≈ W/(uv) ⊕ W/(u) ⊕ W/(v) so that [G: G′] is the principal ideal generated by u^{2}v^{2} , which by assumption is relatively prime to p. Therefore by (the proof of) Theorem 5.61 G ⊕ C ≈ H ⊕ C′ for some C′ ∼ C.

In fact, this can be seen more explicitly from Proposition 5.15, and one can even see that we can choose C′ = C. Since p is maximal and u, v ∉ p there exist r, s ∈ W and p_{1}, p_{2} ∈ p with ru + p_{1} = sv + p_{2} = 1. Let G̃ be the pure closure of the submodule of H ⊕ A_{2} ⊕ A_{3} generated by (1, 0, 0, 0, 0), (0, u, 0, p_{1}, 0), and (0, 0, v, 0, p_{2}). Then G ≈ G̃ by the mapping

ψ: (1, 0, 0) ↦ (1, 0, 0, 0, 0)
(0, 1, 0) ↦ (0, u, 0, p_{1}, 0)
(0, 0, 1) ↦ (0, 0, v, 0, p_{2}).

Note that ψ(p^{−1}(1, 1, 1)) = p^{−1}(1, u, v, p_{1}, p_{2}) ⊆ Qψ(G)∩(H ⊕W ⊕W) ⊆ G̃. Furthermore, with C ⊆ H as in the previous paragraph, G̃ + (C ⊕ A_{2} ⊕ A_{3}) = H ⊕ A_{2} ⊕ A_{3} since

p^{−1}(1, u, v, 0, 0) ⊆ p^{−1}[(1, 0, 0, 0, 0) + (0, u, 0, p_{1}, 0) + (0, 0, v, 0, p_{2})] + p^{−1}(0, 0, 0, p_{1}, p_{2})
⊆ H ⊕ A_{2} ⊕ A_{3}

because p_{i}p^{−1} ⊆ pp^{−1} = W ⊆ A_{2}, A_{3} . Now we claim that G̃ is a balanced submodule of H ⊕A_{2} ⊕A_{3} . Define γ: H ⊕A_{2} ⊕A_{3} → A_{2} ⊕A_{3} by γ(x, y, z, y′, z′) = (p_{1}y −uy′, p_{2}z −vz′). (Note that γ(p^{−1}(1, u, v, 0, 0)) ∈ p^{−1}(p_{1}u, p_{2}v) ∈ A_{2} ⊕ A_{3} .) Since γ(0, p_{1}, 0, −r, 0) = (p_{1} + ru, 0) = (1, 0) and γ(0, 0, 1, 0, −s) = (0, p_{2} + sv) = (0, 1) it follows that A_{2} = γ((H ⊕ A_{2} ⊕ A_{3})(t_{2})) and A_{3} = γ((H ⊕ A_{2} ⊕ A_{3})(t_{3})), so that γ is a balanced surjection. Clearly G̃ ⊆ Ker γ and since the image of γ has rank 2 and G̃ ◁ H ⊕ A_{2} ⊕ A_{3} , G̃ = Ker γ. By Proposition 5.15 γ is split, so that

H ⊕ A_{2} ⊕ A_{3} ≈ G̃ ⊕ A_{2} ⊕ A_{3} ≈ G ⊕ A_{2} ⊕ A_{3}.

Thus a fortiori H ⊕ C ≈ G ⊕ C. ✓

A module G which is known to be completely decomposable is determined up to quasi-isomorphism once we know that rank of G(t) for each t ∈ T(G). Thus completely decomposable modules are a model in the search for classes of modules which can be determined by simple invariants.

Unfortunately there are very few classes of modules which can be completely described by a finite family of integers. However Butler modules are a natural generalization of completely decomposable modules in that a Butler module is determined up to quasi-isomorphism by a finite family of subspace of QG, namely the subspaces QG(t) for t ∈ T(G).

The category whose objects consist of finite dimensional Q-vector spaces together with a finite family of subspaces (and whose morphisms consist of those linear transformations which respect those subspaces) seems (and is) far simpler than the category of finite rank torsion free modules under quasi-homomorphisms. However understanding it is by no means child’s play. One advantage of this category – the category of representations of quivers – is that a lot of algebraists have studied it recently.