2. RANK ONE MODULES AND TYPES

E. L. Lady

December 26, 1995

To ask what finite rank torsion free W -modules there are is to ask the question: What are the W -submodules of a finite dimensional Q-vector space? The obvious place to start in investigating this question is with the lowest possible dimension: What are the submodules of Q? At first thought, it may not be obvious that there are any great variety of these. The modules W and Q may be the only ones to come to mind. Indeed we shall see that if W is local, that is to say a discrete valuation ring, then these are essentially the only examples. If W is not local, however, then for each prime ideal p we have the localization W_{p}. Indeed for each multiplicative set S ⊆ Q we have the localization S^{−1}W. These examples are not only submodules but are actually subrings of Q. It turns out that if Spec W is finite then these are essentially the only examples of submodules of Q. However if Spec W is infinite then Q will contain many submodules which are not isomorphic to subrings or to localizations of W.

There is another reason for beginning the study of finite rank torsion free modules with rank-one modules. If one has gone through the calculations in the examples at the end of Chapter 1 carefully, one will have noticed that a lot of what is going on in a finite rank torsion free module G often depends on the rank-one submodules of G, and also on the set of denominators which a particular element g ∈ G allows, which is to say the set of w ∈ W such that g/w ∈ G. In this chapter we will explore these notions and develop tools and vocabulary to keep track of this sort of thing.

Throughout this chapter, the symbols A and B will denote rank-one modules. In particular, A, B ≠ 0. Recall also that throughout this book the term “prime ideal” is used to signify “non-trivial prime ideal.”

examples 2.1. (1) W and Q are non-isomorphic rank-one modules since Q is divisible and W is not.

(2) For every prime ideal p we get rank-one modules W_{p} and p^{−∞}. These are all mutually non-isomorphic, as can be seen from the fact that W_{p} is not p-divisible and is q-divisible for all primes q ≠ p, and p^{−∞} is p-divisible but is not q-divisible for primes q ≠ p.

(3) Lest one think that rank-one modules can always be distinguished by their divisibility properties, recall Example 1.27. If A is the submodule of Q generated by p^{−1} for all p ∈ Spec W, then A has the same divisibility properties as W, viz. it is not p-divisible for any prime p. However if Spec W is infinite then A is not isomorphic to W since it is not finitely generated.

RANK-ONE MODULES. Rank one modules are the atoms, as it were, of the theory of torsion-free modules. The next proposition gives a few obvious instances of this atomicity.

If A and B are rank-one modules with QA = QB, let [B: A] = {w ∈ Q | wA ⊆ B}.

proposition 2.2. Let A and B be rank-one modules.

(1) If QA = QB then A ≈ B ⟺ B = wA for some w ∈ Q.

(2) If H ◁ G and A is a rank-one submodule of G, then either A ⊆ H or A ∩ H = 0.

(3) If A has rank one and φ: A → H, then φ is either monic or trivial.

(4) If A ⊆ Q then for any torsion free module G, the multiplication map A ⊗ G → AG given by a ⊗ g ↦ ag is an isomorphism.

(5) If A ⊆ Q then A is a subring of Q if and only if for all primes p, A_{p} is either W_{p} or Q. Furthermore in this case A = S^{−1}W for some multiplicative set S.

(6) [B: A] ≈ Hom(A, B).

(7) In particular, End A ≈ [A: A] ⊆ Q.

(8) [A: A] = ⋂{W_{p} | A is not p-divisible}.

(9) If wA ⊆ B ⊆ A for some w ≠ 0 ∈ W then A/B ≈ W/I for some ideal I ⊆ W and B = IA.

proof: (3) Ker φ is a pure submodule of A. If Ker φ ≠ 0 then rank Ker φ = 1 = rank A, so Ker φ is also an essential submodule of A, so Ker φ = A and φ = 0.

(2) The quotient map G → G/H restricts to a map φ: A → G/H. If φ is trivial then A ⊆ H, otherwise by (3) φ is monic so A ∩ H = 0.

(4) Recall that AG is the submodule of QG generated by all products ag with a ∈ A, g ∈ G. Thus clearly the multiplication map a ⊗ g ↦ ag is surjective. This map is the restriction of a map Q ⊗ G → QG between the divisible hulls which is known to be an isomorphism. Therefore the multiplication map is also monic.

(5) ( ⇐ ): Clear since A = ⋂ A_{p} and an intersection of subrings is a subring.

( ⇒ ): If A is a subring of Q then 1 ∈ A_{p} so W ⊆ A and W_{p} ⊆ A_{p}. But W_{p} is a discrete valuation ring, and hence by Proposition 1.* a maximal proper subring of Q, so either A_{p} = W_{p} or A_{p} = Q.

(6) Assume wlog that A, B ⊆ Q. As usual we identify Hom(A, B) with the set of φ ∈ Hom(QA, QB) = End Q such that φ(A) ⊆ B. But since Q is a one-dimensional vector space over itself there is a canonical isomorphism End Q ≈ Q given by φ ↦ φ(1). This isomorphism sends Hom(A, B) to [B: A].

Clear from (6).

(1) Clear from (6).

(8) Let R = [A: A]. Then R is a subring of Q, R = ⋂ R_{p}, and for each prime p, R_{p} = W_{p} or R_{p} = Q. Now if p^{−1} ⊆ [A: A] then R_{p} ≠ W_{p} so R_{p} = Q, and if p^{−1} ⊈ [A: A] then R_{p} ≠ Q so R_{p} = W_{p}. But p^{−1} ⊆ [A: A] ⟺ A is p-divisible. Now in the intersection R = ⋂ R_{p} we may omit the factors for which R_{p} = Q. Thus [A: A] = R = ⋂{W_{p} | A is not p-divisible}.

(9) By Proposition 1.22 A/wA has finite length, hence also A/B has finite length and in particular is not faithful. Let I be the annihilator of A/B. Then IA ⊆ B ⊆ A. By Proposition 0.* A/B contains a submodule isomorphic to W/I and in particular length A/B ≥ length W/I. On the other hand by Proposition 1.22 length A/IA ≤ length W/I. Thus length A/B = length W/I and B = IA and

A/B ≈ W/I. ✓

The next Proposition makes the rather strange assertion that addition and multiplication of subrings are equivalent for subrings of Q. This is a consequence of the existence of partial fraction expansions (Lemma 1.28), But here we will give a slick alternative proof using localization.

proposition 2.3. (1) The set of subrings of Q is a complemented lattice isomorphic to the lattice of subsets of the maximal spectrum of W. Specially, if A is a subring of Q then A = ⋂_{Y} W_{p}, where Y is the set of prime ideals p such that A is not p-divisible.

(2) If A and B are subrings of Q then A + B = AB.

proof: (1) ****

(2) It suffices to prove that for each prime ideal p, A_{p} + B_{p} = A_{p}B_{p}. Since A_{p} and B_{p} are either W_{p} or Q, this is easy. ✓

TYPES. We want to associate to each rank-one module A an entity t(A) called the type of A. We want the set of all types to constitute a partially ordered set (in fact, as it turns out, a lattice) in such a way that the following two axioms are satisfied: (1) if A ≈ B then t(A) = t(B); (2) if A ⊆ B then t(A) ≤ t(B).

To accomplish this we introduce an equivalence relation on the class of rank-one modules (or the set of submodules of Q) making A and B equivalent if each is isomorphic to a submodule of the other. When this is the case we will say (anticipating Chapter 3) that A and B are quasi-isomorphic. We then let t(A) be the corresponding equivalence class, and write t(A) ≤ t(B) if A is isomorphic to a submodule of B. We say that t(A) and t(B) are incomparable types if t(A) ≰ t(B) and t(B) ≰ t(A).

It is tautological, then, that the set of types classifies rank-one modules up to quasi-isomorphism.

proposition 2.4. (1) t(A) ≤ t(B) ⟺ Hom(A, B) ≠ 0.

(2) If QA = QB then t(A) ≤ t(B) if and only if there exists w ≠ 0 ∈ W with wA ⊆ B.

(3) If B ⊆ A then t(A) = t(B) if and only if there exists w ≠ 0 ∈ W with wA ⊆ B.

(4) If B ⊆ A then t(A) = t(B) ⟺ A/B has finite length.

(5) If t(A) ≤ t(B) and A is p-divisible for some prime p then B is also p-divisible.

proof: (1) By definition, t(A) ≤ t(B) if and only if there is a monomorphism from A into B. But by Proposition 2.2 all non-trivial maps between rank-one modules are monic. Thus t(A) ≤ t(B) ⟺ Hom(A, B) ≠ 0.

(2) By Proposition 2.2 a submodule of B isomorphic to A has the form wA, for some w ≠ 0 ∈ Q. Replacing w by a multiple, we may assume that w ∈ W.

(3) If B ⊆ A then t(B) ≤ t(A). Thus t(A) = t(B) if and only if B contains a submodule isomorphic to A. As in (2) such a submodule will have the form wA for some w ≠ 0 ∈ W.

(4) By (3) it suffices to prove that there exists w ≠ 0 ∈ W with wA ⊆ B if and only if A/B has finite length. In fact by Proposition 2.2 if wA ⊆ B ⊆ A then A/B ≈ W/I for some non-trivial ideal I, and by Proposition 0.* W/I has finite length. Conversely if A/B has finite length then in particular by Proposition 1.* it is not faithful, so w(A/B) = 0 for some w ≠ 0 ∈ W and so wA ⊆ B.

(5) Since p^{∞}B ◁ B and B has rank one, either p^{∞}B = B or p^{∞}B = 0, depending on whether B is p-divisible or not. If A is p-divisible then p^{∞}A ≠ 0 and if t(A) ≤ t(B) then there is a monomorphism from A to B so p^{∞}B ≠ 0 by Proposition 1.4. Thus p^{∞}B = B so B is p-divisible. ✓

examples 2.5. (1) t(W) ≤ t(A) ≤ t(Q) for every rank-one module A.

(2) t(p^{−∞}) ≤ t(A) ⟺ A is p-divisible and t(A) ≤ t(W_{p}) ⟺ A is not p-divisible.

(3) Let S ⊆ Spec W and let A be the submodule of Q generated by p^{−1} for all p ∈ S. Let B be the submodule of Q generated by p^{−2} for all p ∈ S. If S is infinite then t(W) < t(A) < t(B) < t(W_{p}) and t(A) ≰ t(p^{−∞}) for all primes p. On the other hand, if S is finite then t(A) = t(B) = t(W).

proof: (1) Clear from Lemma 2.4 since Hom(W, A) and Hom(A, Q) are always non-trivial.

(2) By Lemma 2.4 t(p^{−∞}) ≰ t(A) if A is not p-divisible and t(A) ≰ t(W_{p}) if A is p-divisible. Thus it remains to see that Hom(p^{−∞}, A) ≠ 0 if A is p-divisible and Hom(A, W_{p}) ≠ 0 if A is not p-divisible. In fact, if A is p-divisible then A = p^{−∞}A and so any non-trivial map W → A induces a non-trivial map p^{−∞} → p^{−∞}A = A. And if A is not p-divisible then W_{p}A ≠ Q and since W_{p} is a discrete valuation ring it follows that W_{p}A ≈ W_{p} (see Proposition 2.6 below). Thus Hom(W_{p}A, W_{p}) ≠ 0 and so a fortiori Hom(A, W_{p}) ≠ 0.

(3) Since W ⊆ A ⊆ B, t(W) ≤ t(A) ≤ t(B). And by (2) t(B) ≤ t(W_{p}). And if W has at least two primes then t(W_{p}) ≰ t(B) because W_{p} is q-divisible for primes q different from p and A is not. Now t(A) ≤ t(W) ⟺ A is isomorphic to an ideal of W, and this is true if and only if A is finitely generated (see Proposition 2.8 below). But A is finitely generated if and only if S is finite, in which case B is also finitely generated so that t(B) = t(W). On the other hand, if S is infinite then for any w ≠ 0 ∈ W there exist primes p in S such that wp^{−2} ⊈ A (namely all those p with w ∉ p) and so wB ⊈ A and thus t(B) ≰ t(A). Now for any prime p, p^{−∞}A is generated over the ring p^{−∞} by the set of all q^{−1} for q ≠ p ∈ Spec W. Thus if W has infinitely many primes then p^{−∞}A is not finitely generated over p^{−∞}, and so t(p^{−∞}A) ≰ t(p^{−∞}). Thus a fortiori t(A) ≰ t(p^{−∞}). ✓

There often seems to be a need for standard notation to denote t(W), t(Q), t(p^{−∞}) and t(W_{p}). Your author would like to suggest somewhat tentatively the following:

1 = t(W),  ∞ = t(Q),  t_{p} = t(p^{−∞}),  w_{p} = t(W_{p}).

However these notations will be rarely used in this book.

The following proposition essentially trivializes any theorem about rank-one modules which can be proved by localization. (I think this was already proved in Chapter 1.)

proposition 2.6. If A is a non-trivial submodule of Q and p a prime ideal, then there exists a unique k with −∞ ≤ k < ∞ and A_{p} = p^{k}W_{p}.

proof: We may as well assume W = W_{p} and A = A_{p}. Then W is a discrete valuation ring and every element of Q can be written in the form u/p^{n} for some n ≥ 0 and some u ∈ W, so that the ideals in W all have the form p^{m} for m ≥ 0. Futhermore p^{−∞} = Q. For any n ≥ 0, p^{n}A ∩ W is an ideal in W so either p^{n}A ∩ W = p^{m} for some m or p^{n}A ∩ W = W in which case W ⊆ p^{n}A. Now if for some n, p^{n}A = p^{m} then A = p^{−n}p^{n}A = p^{m−n}. And if p^{n}A ⊇ W for all n then p^{−n} ⊆ A for all n and

A = Q. ✓

definition 2.7. A non-trivial submodule of Q is called a fractional ideal if it is quasi-isomorphic to W.

proposition 2.8. Let A be a submodule of Q. The following conditions are equivalent: (1) A is a fractional ideal.

(2) (∃w ≠ 0 ∈ W) wA ⊆ W.

(3) A is finitely generated.

(4) There exists B ⊆ Q such that AB = W.

(5) For every rank-one module B, t(A) ≤ t(B).

proof: (1) ⇒ (3): If A is quasi-isomorphic to W then A is isomorphic to a submodule (ideal) of W. Since W is noetherian, A is finitely generated.

(3) ⇒ (2): If A is finitely generated then wA ⊆ W where w is a common denominator for a finite set of generators of A.

(2) ⇒ (1): Given w ≠ 0 with wA ⊆ W, we have A ≈ wA ⊆ W. On the other hand A contains submodules isomorphic to W, namely any cyclic submodule. Thus A and W are isomorphic to submodules of each other, hence are quasi-isomorphic, so A is a fractional ideal.

(4) ⇒ (2): If AB = W then bA ⊆ W for any b ∈ B.

(2) ⇒ (4): If A is a fractional ideal, let B = {b ∈ Q | bA ⊆ W }. By (2) B ≠ 0. But since A is isomorphic to an ideal, A is projective and so AB = W by Proposition 0.*.

(1) ⇒ (5): If A is a fractional ideal then t(A) = t(W). Then t(A) ≤ t(B) for every B by Example 2.5.

(5) ⇒ (2): If t(A) ≤ t(W) then (∃w ≠ 0 ∈ W) wA ⊆ W by Lemma 2.4. ✓

If A is a fractional ideal, then the fractional ideal B such that AB = W is unique, since if AB = W = AB′ then B′ = B′W = B′AB = W B = B. We let A^{−1} denote B. proposition 2.9. Let A and B be non-trivial submodules of Q. The following conditions are equivalent:

(1) t(A) = t(B).

(2) B = IA for some fractional ideal I. Furthermore if B ⊆ A then I is an actual ideal of W.

(3) There exist fractional ideals I and J such that A = (A ∩ B) + I, B = (A ∩ B) + J.

(4) There exist fractional ideals I and J such that A + B = A + I = B + J.

proof: Half of the proof comes down to the following fact: If B ⊆ A and A and B are quasi-isomorphic then there exist fractional ideals I and I′ such that B = IA and A = I′ + B. In fact, by Proposition 2.2 A/B ≈ W/I for some ideal I of W and B = IA. Furthermore if a ∈ A is a pre-image for a generator of the cyclic module A/B then a generates a fractional ideal I′ = W a in Q and A = I′ + B.

This establishes (1) ⇒ (2) since we may assume without loss of generality that B ⊆ A. To prove (1) ⇒ (3) & (4) we first note that if A and B are quasi-isomorphic, then A and B are quasi-isomorphic to A + B and A ∩ B. In fact if A and B are quasi-isomorphic then there exists w ≠ 0 ∈ W with wA ⊆ B. Then w(A + B) ⊆ B ⊆ A + B and wA ⊆ A ∩ B ⊆ A. From this we see that in proving (3) and (4) there is no loss of generality in assuming B ⊆ A, so the result follows from the first paragraph.

(2) ⇒ (1): If B = IA, note that I ∩ W ≠ 0 since W is an essential submodule of Q. And if w ≠ 0 ∈ I then wA ⊆ IA = B. Thus t(A) ≤ t(B). Since A = I^{−1}B it follows similarly that t(B) ≤ t(A).

(3) ⇒ (1): Since I is a fractional ideal there exists w ≠ 0 ∈ W with wI ⊆ A ∩ B. Then wA = w((A ∩ B) + I) ⊆ A ∩ B so t(A) = t(A ∩ B) by Lemma 2.4. Likewise t(B) = t(A ∩ B).

(4) ⇒ (1): Analogously we prove t(A) = t(A + B) = t(B) under the stated hypothesis. ✓

corollary 2.10. Quasi-isomorphism is equivalent to isomorphism for rank-one modules if and only if W is a principal ideal domain. proof: ( ⇐ ): If W is a principal ideal domain and A and B are quasi-isomorphic rank-one modules, then B = IA for some fractional ideal I. Then I is isomorphic to an ideal of W, hence I is principal, say I = (w). Then B = wA so B ≈ A.

( ⇒ ): If all modules quasi-isomorphic to W are isomorphic to W then all ideals are isomorphic to W, hence are principal. ✓

The following proposition contains the basic distributivity rules governing the three binary operations A + B, A ∩ B, and AB defined for non-trivial submodules of Q. The first two of these rules are evident simply from the facts that for any G, AG ≈ A ⊗ G, and that tensoring with a flat module preserves push-outs and pull-backs.

proposition 2.11. Let A, B, C be non-trivial submodules of Q and let G be any finite rank torsion free module. Then

(1) (A + B)G = AG + BG.

(2) (A ∩ B)G = AG ∩ BG.

(3) A ∩ (B + C) = (A ∩ B) + (A ∩ C).

(4) A + (B ∩ C) = (A + B) ∩ (A + C).

(5) If t(B) ≤ t(A) then [A: B]_{p} = [A_{p}: B_{p}].

(6) If [A: B] ≠ 0 then B[A: B] = A.

(7) If t(B) ≤ t(A_{1}) and t(B) ≤ t(A_{2}) then [A_{1} + A_{2} : B] = [A_{1} : B] + [A_{2} : B].

proof: It suffices to prove that first four properties locally, i.e. that (A_{p} + B_{p})G_{p} = A_{p}G_{p} + B_{p}G_{p} etc. But locally the proofs are quite evident from Proposition 2.6 and the facts that p^{k}p^{m} = p^{k+m}, p^{k} + p^{m} = p^{min{k,m}} and p^{k} ∩ p^{m} = p^{max{k,m}}.

(5) This follows from Proposition 1.39 and Proposition 2.2.

✠

(6) Since by (5) [A: B]_{p} = [A_{p}: B_{p}], it suffices to prove the assertion locally, which is easy if we note that [p^{k}: p^{m}] = p^{k−m}. ✓

For the moment let S be the set of all non-trivial submodules of Q. Then Proposition 2.11 implies that S is a lattice-ordered multiplicative semigroup where the lattice operations are distributive. The set F of fractional ideals is a subgroup of S, and in fact is the unique maximal group contained in S.

If A, B ≠ 0 ⊆ Q and s = t(A) and t = t(B), we define st = t(AB), s ∨ t = t(A + B), and s ∧ t = t(A ∩ B). If s ≤ t we define [t : s] = t([B : A]). From what has come above, the following is pretty much straight lattice theory.

proposition 2.12. (1) The types st, s ∧ t, s ∨ t and [t: s] are well defined, independent of the choice of representing modules A and B.

(2) s ∨ t is the least upper bound of s and t and s ∧ t is the greatest lower bound.

(3) For any two rank-one modules A and B such that t(A) ≤ t(B), t(Hom(A, B)) = [t: s], where s = t(A), t = t(B).

proof: (1) In the notation above, we see from Proposition 2.9 that the set of types is simply the quotient semigroup S/F. It then follows immediately that the product st = t(A)t(B) is well defined. Now we need to see that if t(A) = t(A′) then t(A + B) = t(A′ + B) and t(A ∩ B) = t(A′ ∩ B). We can assume wlog that A′ ⊆ A, since otherwise we can compare both A and A′ to A ∩ A′. Then A = A′ + I for some fractional ideal I and so A + B = A′ + B + I so that t(A + B) = t(A′ + B). And A ∩ B = (A′ + I) ∩ B = (A′ ∩ B) + (I ∩ B). But (I ∩ B) is a fractional ideal, since as a submodule of the finitely generated module I it is also finitely generated. Hence t(A ∩ B) = t(A′ ∩ B).

Finally, suppose that A′ ⊆ A and B ⊆ B′ and and s = t(A) = t(A′) and t = t(B) = t(B′) and Hom(A, B) ≠ 0. Then [B: A] ⊆ [B′: A′] since if wA ⊆ B then a fortiori wA′ ⊆ B′. Thus t[B: A] ≤ t[B′: A′]. But by Lemma 2.4 there exist w, w′ ≠ 0 ∈ W with wA ⊆ A′ and w′B′ ⊆ B. Then ww′[B′: A′] ⊆ [B: A], so that t[B′: A′] = t(ww′[B′: A′]) ≤ t[B: A].

(2) First, we claim that for any s and t, s ≤ t ⟺ s ∧ t = s ⟺ s ∨ t = t. In fact, let s = t(A) and t = t(B). Then s ∧ t = s if and only if (∃w ≠ 0 ∈ W) wA ⊆ A ∩ B and this is equivalent to (∃w ≠ 0 ∈ W) wA ⊆ B, i.e. s = t(A) ≤ t(B) = t. And likewise s ∨ t = t ⟺ (∃w ≠ 0 ∈ W) w(A + B) ⊆ B ⟺ (∃w ≠ 0 ∈ W) wA ⊆ B ⟺ s ≤ t.

Now let r be any type. Then

r ≤ s & r ≤ t ⟺ r ∧ s = r & r ∧ t = r.

If this is the case then r ∧ s ∧ t = r so that r ≤ s ∧ t. On the other hand if r ≤ s ∧ t then r ∧ s ∧ t = r and so r ∧ s = (r ∧ s ∧ t) ∧ s = r ∧ s ∧ t = r so that r ≤ s and likewise r ≤ t. Thus for any type r, r ≤ s ∧ t ⟺ r ≤ s & r ≤ t, i.e. s ∧ t is a greatest lower bound for s and t. The proof that s ∨ t is a least upper bound is entirely analogous.

(3) We may assume wlog that A, B ⊆ Q. The by Proposition 2.2

t(Hom(A, B)) = t([B: A]) = [t: s]. ✓

corollary 2.13. The set of types is a distributive lattice and furthermore for any types s, t_{1} , t_{2} we have s(t_{1} ∨ t_{2}) = st_{1} ∨ st_{2} and s(t_{1} ∧ t_{2}) = st_{1} ∧ st_{2} .

proof: Immediate from Proposition 2.12 and Proposition 2.11. ✓

example 2.14. t(p^{−∞}) ∧ t(W_{p}) = t(W) and t(p^{−∞}) ∨ t(W_{p}) = t(Q).

Unlike the set of submodules of Q, however, the set of types is not a complete lattice if Spec W is infinite, which is to say that for an arbitrary set S of types, sup S and inf S need not exist. We will postpone giving an example until some new concepts are available (Example 2.20 below).

IDEMPOTENT TYPES. We say that a type t is idempotent if t^{2} = t. (In earlier, less enlightened times, the term non-nil was frequently used for this concept. In this respect, as well as in using the notation t_{1}t_{2} rather than t_{1} + t_{2} , your author has preferred to follow [Fuchs].)

proposition 2.15. Let A, B ⊆ Q and t_{1} = t(A), t_{2} = t(B).

(1) If t_{1} ≤ t_{2} then t_{1}[t_{2} : t_{1}] = t_{2} .

(2) If s and t are any types, then s ≤ t ⟺ t = ss′ for some type s′.

(3) A type t is idempotent if and only if t = t(A) for some subring A of Q.

(4) For any type t, [t: t] is idempotent.

(5) The type t is idempotent ⟺ [t: t] = t.

(6) If t_{1}t_{2} = t_{1} then t_{2} ≤ [t_{1} : t_{1}].

(7) If A is a subring of Q and B a non-trivial submodule of Q then t(B) = t(A) ⟺ B is a fractional A-ideal in Q.

(8) In particular if A and B are subrings of Q with t(A) = t(B) then A = B.

(9) The set of itempotent types is a complemented lattice isomorphic to the set of subsets of Spec W. proof: (1) Clear from Proposition 2.11.

(2) ( ⇐ ): Writing 1 = t(W) we have 1 ≤ s′ and so s = s1 ≤ ss′.

( ⇒ ): If s ≤ t then t = s[t: s].

(3) ( ⇐ ): If A is a subring of Q then by Proposition 2.3, AA = A + A = A. Thus if t = t(A) then t^{2} = t(AA) = t(A) = t so t is idempotent.

(4) If t = t(A) then [t: t] is the type for the ring End A so [t: t] is idempotent by (3).

(5) ( ⇐ ): Immediate from (4).

( ⇒ ): If t^{2} = t then [t : t] = [t^{2} : t] and clearly by definition [t^{2} : t] ≥ t. But [t: t] ≤ t for any type t by (1) and (2) since t = t[t: t].

(3) ( ⇒ ): If t is idempotent then t = [t: t] by (5) so t is the type of the ring Hom(A, A), which can be identified with a subring of Q since Hom(A, A) ⊆ Hom(Q, Q) ≈ Q.

(6) **** ✠

(7) If B is a fractional A-ideal in Q then B and A are isomorphic to A-submodules of each other, hence a fortiori isomorphic to W -submodules of each other. Hence t(A) = t(B). Conversely if t(A) = t(B) then B = IA for some fractional ideal I. But IA is a finitely generated A-module, hence is a fractional A-ideal.

(8) By (6) if A and B are subrings of Q and t(A) = t(B) then B is a fractional A-ideal. Since 1 ∈ B it follows that A ⊆ B. Likewise B ⊆ A.

(9) Immediate from (3) and Proposition 2.3. ✓

proposition 2.16. If A ⊆ Q and A is p-divisible for all but finitely many p, then t(A) is idempotent. proof: If we let P be the finite set of primes for which A is not p-divisible and if we assume wlog that 1 ∈ A then ⋂_{P} W_{p} ⊆ ⋂_{P} A_{p} = A. By Proposition 2.6 for each prime p ∈ P there exists a positive integer n_{p} such that p^{n_{p}}A ⊆ W_{p}. If we choose w ≠ 0 ∈ ⋂_{P} p^{n_{p}} then wA ⊆ ⋂_{P} W_{p} ⊆ A, so that A is quasi-isomorphic to the ring ⋂_{P} W_{p}. Then t(A) is idempotent by Proposition 2.15. ✓

corollary 2.17. If W is semi-local then the rank-one W -modules are precisely the subrings A of Q and the fractional A-ideals for these subrings, so that all types are idempotent. proof: By Proposition 2.16 all types are idempotent. Thus if B is a non-trivial submodule of Q then t(B) = t(A) for some subring A of Q. Therefore B = IA for some fractional ideal I. Thus B is the A-fractional ideal generated by I. ✓

proposition 2.18. If t, t′ are types and t is idempotent, then tt′ = t ∨ t′. proof: wlog we may assume that t = t(A), t′ = t(B), where A is a subring of Q and W ⊆ B. As in the proof of Proposition 2.3 we will prove that for each prime p, A_{p}B_{p} = A_{p} + B_{p}. Since either A_{p} = W_{p} or A_{p} = Q, and since W_{p} ⊆ B_{p} the conclusion is clear in either case. ✓

If t = t(A), then we say that t is p-divisible if A is p-divisible. We say that a type t is locally trivial (or locally free) if t is not p-divisible for any prime p. This is equivalent to saying that t = t(A) where A is a rank-one module such that A_{p} ≈ W_{p} for every prime p.

proposition 2.19. (1) A type t is locally trivial if and only if [t: t] = t(W). (2) If t(A) is locally trivial then End A ≈ W.

(3) If t is locally trivial then for any s, [st: t] = s.

(4) If t is locally trivial and ts_{1} = ts_{2} then s_{1} = s_{2} .

(5) If t is locally trivial and s ≠ t(W) then st > t.

(6) A type t can be written as t = t_{1}t_{2} = t_{1} ∨ t_{2} where t_{2} is locally trivial and t_{1} = [t: t], so that t_{1} is idempotent, and t_{1} ∧ t_{2} = t(W). Furthermore this representation of t as the product of an idempotent type and a locally trivial type whose greatest lower bound is t(W) is unique.

proof: (1) By Proposition 2.15 t = t[t : t], hence by Proposition 2.15 [t: t] ≤ t. Hence if [t: t] = t(R) where R is a subring of Q and t is locally trivial then R_{p} ≠ Q for all p. Since W_{p} is a maximal proper subring of Q, R_{p} = W_{p} for all p. Then R = ⋂ R_{p} = W. Conversely if t = t(A) is not locally trivial then A is p-divisible for at least one prime p and so p^{−∞} ⊆ End A, so [t: t] > t(W).

(2) Immediate from (1).

(3) If t = t(A), s = t(B) then [st: t] = t(Hom(A, A ⊗ B)) = t(Hom(Hom(A, A), B)) = t(Hom(W, B)) = t(B) = s by (2).

(4) By (3) s_{1} = [s_{1}t: t] = [s_{2}t: t] = s_{2} .

(5) By Proposition 2.15 st ≥ t. If st = t then st = t(W)t, so s = t(W) by (4).

(6) Let t_{1} = [t: t]. Then t_{1} is idempotent. By Proposition 2.15 and Proposition 2.3 the set of idempotent types is a complemented lattice, so there exists a unique idempotent type r such that t_{1} ∧ r = t(W) and t_{1}r = t_{1} ∨ r = t(Q). Let t_{2} = r ∧ t. Then t(W) ≤ t_{1} ∧ t_{2} = t_{1} ∧ t ∧ r ≤ t_{1} ∧ r = t(W). Since r is p-divisible only for those primes p such that t is not p-divisible, we see that t_{2} = r ∧ t is not p-divisible for any p, so that t_{2} is locally trivial. And t_{1}t_{2} = t_{1}(r ∧ t) = t_{1}r ∧ ([t: t]t) = t(Q) ∧ t = t. ✓

We can now give the promised example of a set of types having no least upper bound, and a set with no greatest lower bound.

example 2.20. Suppose that Spec W is infinite. Let S be the set of all locally trivial types and T the set of all types of the form t(B) where B is p-divisible for all but finitely many primes p. Then s < t for all s ∈ S and t ∈ T. Furthermore S has no least upper bound and T has no greatest lower bound. proof: Let s = t(A) ∈ S and t = t(B) ∈ T. By Proposition 2.16 t is idempotent, so by Proposition 2.15 we may suppose that B is a subring of Q. Then by Proposition 2.17 the B-module AB is quasi-isomorphic to a subring of Q. Since A is locally trivial, AB is p-divisible by only those primes p for which B is p-divisible, and so the subring of Q quasi-isomorphic to AB must in fact be B. Thus s ≤ t(AB) = t(B) = t. But clearly s ≠ t because s is locally trivial and t is not.

This shows that for any prime p, t(W_{p}) is an upper bound for S. Therefore if S has a least upper bound t then t ≤ t(W_{p}), so t is not p-divisible for any prime p. Therefore a least upper bound for S must be locally trivial. But no locally trivial type s is an upper bound for S, since t(W) is not an upper bound and if s ≠ t(W) then s^{2} > s and s^{2} ∈ S.

Essentially the same argument shows that T cannot have a greatest lower bound. If t is a lower bound for T then t ≤ t(W_{p}) so t is not p-divisible for any prime p. Thus t must be locally trivial. But all locally trivial types are lower bounds for t, so a locally trivial type t cannot be the greatest lower bound because t^{2} > t. (Unless t = t(W), but clearly t(W) is not the greatest lower bound.) ✓

THE TYPE OF AN ELEMENT. We have already disposed of the naive notion that every finite rank torsion free module is a direct sum of rank-one modules. However what is true is that every torsion free module is a union of rank-one modules, in fact of its pure rank-one submodules. Namely, every non-trivial element g of a finite rank torsion free module G belongs to a unique pure rank-one submodule of G, namely ⟨g⟩_{∗} = G ∩ Qg. The nature of this pure rank-one submodule provides vital information about how the element g sits inside the module G.

If 0 ≠ g ∈ QG, we write t(g) = t(G ∩ Qg) and call this the type of g (in G). (Most authors define t(g) only for g ∈ G, but it is sometimes useful to think of t( ) as a generalized valuation defined on the whole vector space QG.) As a useful although illogical convention we set t(0) equal to t(Q).

proposition 2.21. (1) If g ∈ G and w ≠ 0 ∈ W then t(g) = t(wg).

(2) If H ⊆ G and h ∈ H, then t_{H}(h) ≤ t_{G}(h).

(3) If H ◁ G and h ∈ H, then t_{H}(h) = t_{G}(h).

(4) If φ: G → H and g ∈ G then t(φ(g)) ≥ t(g).

(5) If g, h ∈ G then t(g + h) ≥ t(g) ∧ t(h).

(6) If G = H ⊕ K and h ∈ H, k ∈ K, then t(h + k) = t(h) ∧ t(k).

proof: (1) Clear since g and wg generate the same pure rank-one submodule of G. (3) Since H ◁ G, H ∩ Qg = G ∩ Qg.

(4) The map φ restricts to a map ⟨g⟩_{∗} → ⟨φ(g)⟩_{∗}, so t(g) ≤ t(φ(g)) by Proposition 2.4.

(2) Apply (4) to the inclusion map H ↪ G.

(6) Let A = ⟨h⟩_{∗}, B = ⟨k⟩_{∗}, and let α: A → Q, β: B → Q be the unique embeddings such that α(h) = β(k) = 1. (We assume here wlog that h, k ≠ 0.) Then there is a short exact sequence 0 → C ↪ A ⊕ B ⟶^{φ} Q where φ(h, k) = α(h) − β(k) ∈ Q. Then C = Ker φ is a rank-one pure submodule of A ⊕ B and hence a pure submodule of H ⊕ K, and h + k ∈ Ker φ. Thus C = ⟨h + k⟩_{∗}. But C ≈ α(A) ∩ β(B). It follows that t(h + k) = t(C) = t(A) ∧ t(B) = t(h) ∧ t(k).

(5) If ⟨g⟩_{∗} = ⟨h⟩_{∗} then this is clear. Otherwise by Proposition 2.2 ⟨g⟩_{∗} ∩ ⟨h⟩_{∗} = 0 and so ⟨g⟩_{∗} ⊕ ⟨h⟩_{∗} ⊆ G. By (2) t(g + h) is greater than or equal to the type of g + h in ⟨g⟩_{∗} ⊕ ⟨h⟩_{∗}, which by (6) is t(g) ∧ t(h). ✓

With the help of Proposition 2.21 we can look back at Example 1.49 in a more conceptual way. example 2.22. Let W be ℤ or any ring such that 5 is neither trivial nor invertible. (There is nothing special about 5, but using a specific number seems to make the example easier to follow.) Take two rank-one modules A and B such that t(A) and t(B) are locally trivial and incomparable, and t(A) ∧ t(B) = t(W) and choose elements a ∈ A, b ∈ B with a ∉ 5A, b ∉ 5B. (Such elements exist because of the local triviality of t(A) and t(B).) Let H be the submodule of QA ⊕ QB generated by A ⊕ B together with (a + b)/5. Then End H is isomorphic to the subring of W ⊕ W generated by 5W ⊕ 5W together with (1, 1). In particular, End H has no proper idempotents, so that H is indecomposable. proof: Write s = t(A), t = t(B). Note that t(a) = s, t(b) = t, and if x ∉ A ∪ B then by Proposition 2.21 t(x) = t(W). Now if φ ∈ End H then t(φ(a)) ≥ s and since s and t are incomparable and strictly greater than t(W) this is only possible if φ(a) ∈ A. Likewise φ(b) ∈ B. In other words, φ restricts to endomorphisms of A and B. Since t(A) and t(B) are locally trivial, by Proposition 2.19 End A ≈ End B ≈ W, so that if φ ∈ End H then φ(a) = wa and φ(b) = w′b for w, w′ ∈ W. Then

φ (\frac{a + b}{5}) = \frac{wa}{5} + \frac{w′b}{5}.

But (wa + w′b)/5 = w(a + b)/5 + (w′ − w)b/5 ∈ H ⟺ (w′ − w)b/5 ∈ H ⟺ w ≡ w′ (mod 5), i.e. if and only if (w, w′) = (w, w)+5(0, x) for some x ∈ W. Thus the mapping φ ↦ (w, w′) is an isomorphic from End H onto the subring of W ⊕ W generated by 5W ⊕ 5W together with (1, 1).

Now if φ is to be idempotent then w = 1, 0 and w′ = 1, 0. Since w ≡ w′ (mod 5) this forces either w = w′ = 1, in which case φ = 1_{H}, or w = w′ = 0, in which case φ = 0. Thus End H has no non-trivial idempotents. ✓

THE TYPESET AND COTYPESET. We define the typeset of G, written T(G), to be the set of types of all non-trivial elements of G, or equivalently the set of types of all pure rank-one submodules of G. Dually, we define the cotype set of G, CT(G) to be the set of types of all rank-one homomorphic images of G.

We say that G is homogeneous (or t-homogeneous) if T(G) = {t} for some type t. We say that G is cohomogeneous (or t-cohomogeneous) if CT(G) = {t}.

proposition 2.23. If G = H ⊕ K then

(1) T(G) consists of T(H) ∪ T(K) together with all types s ∧ t with s ∈ T(H) and t ∈ T(K).

(2) CT(H ⊕ K) consists of CT(H) ∪ CT(K) together with all types s ∨ t with s ∈ CT(H) and t ∈ CT(K).

proof: (1) If g = h + k ∈ H ⊕ K then by Proposition 2.21 t(g) = t(h) ∧ t(k). (This includes the cases h = 0 or k = 0 since t(0) = t(Q) and t ∧ t(Q) = t.)

(2) If μ: H⊕K → Q then μ(H⊕K) = μ(H)+μ(K) so t(μ(H⊕K)) = t(μ(H))∨t(μ(K)) unless μ(H) = 0 or μ(K) = 0 in which case t(H ⊕ K) = t(μ(H)) or

t(μ(H ⊕ K)) = t(μ(K)). ✓

A direct sum of rank-one modules is called completely decomposable. As the following corollary shows, the typeset of a completely decomposable module cannot be very complicated.

corollary 2.24. If G = A_{1} ⊕ ⋯ ⊕ A_{n} where the A_{i} are rank-one modules and t_{i} = t(A_{i}), then

(1) T(G) consists of {t_{1}, …, t_{n}} and all types obtained from this set by taking greatest lower bounds.

(2) CT(G) consists of {t_{1}, …, t_{n}} and all types obtained from this set by taking least upper bounds. ✓

examples 2.25. (1) Let G be the submodule of Q ⊕ Q generated by p_{1}^{−∞}(1, 0), p_{2}^{−∞}(0, 1), and p_{3}^{−∞}(1, 1), where p_{1} , p_{2} , and p_{3} are distinct primes. Then T(G) consists of t(W) and t(p_{i}^{−∞}) for i = 1, 2, 3. CT(G) = {t((p_{1}p_{2})^{−∞}), t((p_{1}p_{3})^{−∞}), t((p_{2}p_{3})^{−∞}), t((p_{1}p_{2}p_{3})^{−∞})}.

(2) Pontryagin module. Let G be the submodule of Q ⊕ Q ⊕ Q generated by (1, 0, 0), (0, 1, 0), (0, 0, 1) and the sets {(1, v_{n}, 0)/p^{n} | n = 1, 2, … } and and {(1, 0, w_{n})/p^{n} | n = 1, 2, … }, where {v_{n}}_{1}^{∞} and {w_{n}}_{1}^{∞} are sequences in W such that (∀n) v_{n+1} ≡ v_{n} (mod p^{n}) and (∀n) w_{n+1} ≡ w_{n} (mod p^{n}) and there do not exist u, v, w ∈ W such that (∀n) u + vv_{n} + ww_{n} ≡ 0 (mod p^{n}). Then T(G) = {t(W)} and CT(G) = {t(p^{−∞})} so G is both homogeneous and cohomogeneous. proof: (2) By Example 1.47 every rank-one submodule of G is finitely generated, hence quasi-isomorphic to W. Thus T(G) = {t(W)}. It was also proved that G is p-reduced. Now let μ: G → Q and let H = Ker μ. Since G, and therefore also H, is p-reduced, 1 ≤ p-rank H ≤ p-rank G = 1. Thus p-rank G/H = 0 and we can prove t(μ(G)) = t(G/H) = t(p^{−∞}). Thus CT(G) = {t(p^{−∞})}. ✓

Corollary 2.23 gives another proof that the Pontryagin module is indecomposable. Namely, if G = H ⊕ K with H, K non-trivial, then one of H or K must have rank one. Since T(G) = {t(W)}, this means that G has a summand isomorphic to W. But this is inconsistent with the fact that CT(G) = {t(p^{−∞})}.

Not every set of types can be the typeset of some finite rank torsion free module. The following proposition gives some necessary conditions for a set of types to have the form T(G).

proposition 2.26. (1) The length of any strictly ascending (or descending) chain in T(G) is at most rank G.

(2) If S ⊆ T(G) then inf S exists.

(3) If T(G) is finite and S ⊆ T(G), then inf S ∈ T(G).

proof: For any type t we define G(t) = {g ∈ G | t(g) ≥ t}. We will study the properties of G(t) in Chapter 4, but for the moment we need to note that G(t) is a pure submodule of G and that s ≤ t ⇒ G(s) ⊇ G(t).

(1): A strictly ascending chain t_{1} < ⋯ < t_{n} in T(G) yields a descending chain of pure subgroups of G: G(t_{1}) ▷ ⋯ ▷ G(t_{n}). Now note that if all t_{i} ∈ T(G) then these inclusions must be strict since for each t_{i} there exists g ∈ G with t(g) = t_{i} and so g ∈ G(t_{i}) but g ∉ G(t_{i+1}). Hence such a chain cannot be longer than rank G.

(2): If S′′ ⊆ S′ are finite subsets of S then inf S′ and inf S′′ exist and inf S′ ≤ inf S′′ so that G(inf S′′) ⊆ G(inf S′). Since these are pure submodules of G, we see that among such finite subsets of S there exists an S′ such that G(inf S′) is maximal of this sort. We now claim that inf S′ is the greatest lower bound of S. It suffices to prove that inf S′ is a lower bound, since clearly no type strictly larger than inf S′ could be a lower bound. Let t ∈ S. There exists g ∈ G with t(g) = t. Then t(g) ≥ inf (S′ ∪ {t}). so g ∈ G(inf (S′ ∪ {t})) ⊇ G(inf S′). By the choice of S′, the inclusion here must be an equality, so g ∈ G(inf S′). Hence t = t(g) ≥ inf S′. Thus we see that inf S′ is a lower bound for S, and consequently is the greatest lower bound.

(3): If T(G) is finite, consider the set of types in t ∈ T(G) such that t ≥ inf S and let t_{1}, …, t_{n} ∈ T(G) be the types which are minimal in this set. Then if t(g) ≥ inf S then t(g) ≥ t_{i} for some i. Translated, this says that G(inf S) = G(t_{1}) ∪ ⋯ ∪ G(t_{n}) and so QG(inf S) = QG(t_{1}) ∪ … QG(t_{n}). Since a vector space over an infinite field cannot be the union of a finite number of proper subspaces, it follows that QG(inf S) = QG(t_{i}) for some i. Since G(t_{i}) is a pure submodule of G, it follows that G(inf S) = G(t_{i}). As in the the proof of (2) we conclude that inf S = t_{i} ∈ T(G). ✓

corollary 2.27. If rank G = 2 then T(G) either consists of (1) a single type, or (2) an infinite set of incomparable types {t_{i}}_{i∈I} such that t_{i} ∧ t_{j} is the same for all choices of i ≠ j, or (3) a finite or infinite set of types {t_{i}}_{i∈I} including a type t_{0} such that t_{i} ∧ t_{j} = t_{0} for all choices of i ≠ j. proof: By the proof of Proposition 2.26, if t_{1} and t_{2} are distinct types in T(G) then t_{1} ∧ t_{2} = inf T(G). Thus if inf T(G) ∈ T(G) we must have either case (1) or case (3), and if inf T(G) ∉ T(G) then case (2) must hold, since by Proposition 2.26 in this case T(G) must be infinite. ✓

example 2.28. In Example 1.45 we constructed a module G with rank two using three distinct primes p_{1}, p_{2} , and p_{3} . If we let A, B, and C denote the three rank-one modules p_{i}^{−∞}, then G was constructed as as the submodule of Q ⊕ Q generated by A(1, 0), B(0, 1), and C(1, 1). (Here C(1, 1), for instance, is the set of elements (c, c) with c ∈ C.) Then T(G) consists of t_{1}, t_{2}, t_{3} , and t(W). (Note that t(W) = t_{i} ∧ t_{j} if i ≠ j = 1, 2, 3.)

From this we easily see that G is indecomposable. In fact, since G has rank two, if G is not indecomposable then G = H ⊕ K where H and K are rank-one modules. Then by Corollary 2.27 T(G) would consist of t(H), t(K), and t(H) ∧ t(K), which is only three types.

Clearly one can generalize this example in many ways. One can use three rank-one modules which do not have the form p_{i}^{−∞}. It is essential, however, the types of these three rank-one modules conform to Corollary 2.27.

For instance, suppose for convenience that W has only three primes, p_{1}, p_{2} , and p_{3} , and that we let A, B, and C be the three rank-one modules W_{p_{i}}, i = 1, 2, 3. But for i ≠ j, t(W_{i}) ∧ t(W_{k}) = p_{k}^{−∞}, where {i, j, k} = {1, 2, 3}. This does not conform to Corollary 2.27, so if we construct a module G = W_{p_{1}}(1, 0) + W_{p_{2}}(0, 1) + W_{p_{3}}(1, 1) then T(G) will not be as desired. In fact, this construction will product a divisible module G. For any (x, y) ∈ G ⊆ Q ⊕ Q, (x, y) = (x − y, 0) + (y, y) ∈ p_{1}^{−∞}G, and likewise (x, y) ∈ p_{2}^{−∞}G, p_{3}^{−∞}G. Thus G is p-divisible for all primes p, hence is divisible.

For a finite set T satisfying conditions (1) or (2) of Corollary 2.27, there always exists a rank-two module with T(G) = T. In case (1) we can simply choose G = A ⊕ A, where A is a rank-one module with t(A) = t. For case (3) the following Proposition applies.

proposition 2.29. Let n ≥ 2 A_{1}, …, A_{n} be submodules of Q such that A_{i} ∩ A_{j} is the same for all pairs i ≠ j, and let t_{i} = t(A_{i}). Let v_{1}, …, v_{n}, w_{1}, …, w_{n} ∈ W be such that for each i v_{i} and w_{i} are relatively prime (i.e. the ideal generated by v_{i} and w_{i} is W), and such that the n ordered pairs (v_{i}, w_{i}) generate distinct subspaces of Q ⊕ Q. Let

G = A_{1}(v_{1}, w_{1}) + ⋯ + A_{n}(v_{n}, w_{n}) ⊆ Q ⊕ Q.

Then each A_{i}(v_{i}, w_{i}) is a pure rank-one submodule of G with type t_{i}, and t(G) = {t_{1}, …, t_{n}, t_{1} ∧ t_{2}}. proof: Let A′_{i} = A_{i}(v_{i}, w_{i}) ⊆ G. To see that A′_{i} ◁ G it suffices by Proposition 1.*

✠ to see that for each prime p, (A′_{i})_{p} ◁ G_{p}. Thus no generality is lost in supposing that W is local. Then by Proposition 2.6 A_{i} = p^{k_{i}} for −∞ ≤ k_{i} < ∞, and so for all i, j, A_{i} ⊆ A_{j} or A_{j} ⊆ A_{i}. Suppose, say, A_{2} ⊆ A_{1} . Then by assumption for i ≠ 1, A_{2} = A_{1} ∩ A_{2} = A_{1} ∩ A_{i} = A_{i} since A_{2} ≠ A_{1} , and so A_{i} ⊆ A_{2} . Likewise since A_{i} ∩ A_{2} = A_{1} ∩ A_{2} = A_{2} for all i ≠ 2, it follows that A_{2} ⊆ A_{i} for all i. Thus A_{i} = A_{2} for all i ≥ 2. Now if A_{i} = Q for all i, then A′_{i} ◁ G by Proposition 1.*. Otherwise A_{i} = A_{2} ≠ Q for i ≠ 1 and to see that A′_{i} ◁ G it suffices to prove that A_{2}^{−1}(A′_{i}) ◁ A_{2}^{−1}G. In other words, no generality is lost in assuming that W ⊆ A_{1} and A_{i} = W for i ≠ 1.

Now drop the assumption that W is local. Since there is an isomorphism A_{i} ≈ A′_{i} given by a ↦ a(v_{i}, w_{i}), it follows that for all i, t(A_{i}) ∈ T(G). Note that if B is any pure rank-one submodule of T(G) then t(B) ≥ t(A_{1}) ∧ ⋯ ∧ t(A_{n}) since ****. Also note that for any i, G/A′_{i} is a rank-one module generated by the (non-trivial) images of A′_{j} for j ≠ i′, and so t(G/A′_{i}) = sup{t(A_{j} | j ≠ i}. If now B is a rank-one module with B ≠ A′_{i} for all i then for all i the mapping G → G/A′_{i} maps B nontrivially into G/A′_{i}. Thus by Proposition 2.21 t(B) ≤ t(G/A′_{i}) = sup_{j≠i} t(A_{j}) so that t(B) ≤ inf_{i} sup_{j≠i} t(A_{j}) = t(A_{1}) ∧ ⋯ ∧ t(A_{n}) ≤ t(B). Thus

t(A_{1}) ∧ ⋯ ∧ t(A_{n}) ∈ T(G). ✓

For an infinite set T, things are not so straightfoward. One needs to be use meticulous care to construct an example with a prescribed typeset, if exists at all. The following example will illustrate the fact that if T(G) is infinite, we cannot expect T(G) to be closed under greatest lower bounds. example 2.30. Suppose that W = ℤ or any other ring having at least as many prime ideals as the cardinality of its quotient field Q. Let p_{w} for w ∈ Q be distinct prime ideals and let p_{∞} be an additional prime. For w ∈ Q let

A_{w} = {(x, wx)/p_{w}^{n} | x, wx ∈ W, n ≥ 0}

and let A_{∞} = {(0, 1)/p_{∞}^{n} | n ≥ 0}. Let G be the submodule of Q ⊕ Q generated by all the A_{w} and A_{∞}. Then T(G) = {t(p_{x}^{−∞}) | x = w ∈ Q or x = ∞}. If S is any subset of T(G) containing at least two types, then inf S = t(W) ∉ T(G).

✠ FIX THIS!

For the cotypeset, we have a dual result to Proposition 2.26:

proposition 2.31. (1) The length of any strictly ascending chain in CT(G) is at most rank G.

(2) If S ⊆ CT(G) then sup S exists.

(3) If CT(G) is finite and S ⊆ CT(G) then sup S ∈ CT(G).

proof: (1) If t_{1} < ⋯ < t_{n} is a strictly ascending chain in CT(G), with t_{i} = t(A_{i}), then rank Hom(G, A_{1}) < ⋯ < rank Hom(G, A_{n}) < rank Hom(G, Q) = rank G.

(2) Among those rank-one modules A with t(A) = sup S′ for some finite subset S′ ⊆ S, choose one such that Hom(G, A) has maximum possible rank. (Note that in any case rank Hom(G, A) ≤ rank G.) We claim that sup S′ is a least upper bound for S. It suffices to prove that it is an upper bound, since clearly no smaller type could be an upper bound. Let t ∈ S and let S′′ = S ∪ {t}. Let A′′ be a rank-one module with t(A′′) = sup S′′. Then by choice of A, rank Hom(G, A) = rank Hom(G, A′′), in other words, Hom(G, A) is an essential submodule of Hom(G, A′′). Thus if μ: G → Q with t(μ(G)) = t then since t ≤ sup S′′ we may assume (replace μ with a scalar multiple) that μ ∈ Hom(G, A′′), and then in fact there exists w ≠ 0 ∈ W such that wμ ∈ Hom(G, A). It follows that t = t(μ(G)) ≤ t(A) = sup S′. Thus sup S′ is in fact a least upper bound for S.

(3) If CT(G) is finite, let S = {t_{1}, …, t_{n}}. We may suppose that S contains all types t ∈ CT(G) such that t ≤ sup S. Choose A_{i} ⊆ Q such that t(A_{i}) = t_{i}. Let A ⊆ Q with t(A) = sup S. We may in fact suppose that A = A_{1} + ⋯ + A_{n}. As usual, look at Hom(G, A_{i}) and Hom(G, A) as submodules of Hom(G, Q). Now if μ ∈ Hom(G, Q) and μ(G) ⊆ A then t(μ(G)) ≤ sup S so t(μ(G)) = t_{i} for some i. Then for some w ≠ 0 ∈ W, wμ ∈ Hom(G, A_{i}). Thus QHom(G, A) = QHom(G, A_{1}) ∪ ⋯ ∪ QHom(G, A_{n}). Since a vector space over an infinite field cannot be a union of a finite number of proper subspaces, we conclude that QHom(G, A) = QHom(G, A_{i}) for some i. Thus if μ: G → Q and t(μ(G)) ≤ sup S, then (∃w ≠ 0 ∈ W) wμ ∈ Hom(G, A_{i}), so that t(μ(G)) ≤ t_{i}. Thus t_{i} is an upper bound for S, and clearly must be the least upper bound. ✓

definition 2.32. We define the inner type of G, denoted IT(G), to be inf T(G), and the outer type OT(G) as sup CT(G). We define the exterior type ET(G) to be t(⋀^{r} G), where r = rank G. (Note that ⋀^{r} G is a rank-one module since ⋀^{r} G ⊆ ⋀^{r} QG ≈ Q.) proposition 2.33. (1) G is divisible if and only if IT(G) = t(Q).

(2) G is projective if and only if OT(G) = t(W).

proof: (1) In other words, G is divisible if and only if t(g) = t(Q) for all g ∈ G. (Note that by Example 2.5 t = t(Q) is equivalent to t ≥ t(Q).)

2) Note that t ≤ t(W) is equivalent to t = t(W). By Corollary 2.24 if G is projective then CT(G) = t(W), so OT(G) = t(W). Conversely if OT(G) = t(W) then CT(G) = t(W) so every rank-one homomorphic image of G is projective. By induction on rank G we conclude that G is projective. ✓

proposition 2.34. (1) For any t ∈ T(G), t ≤ OT(G).

(2) For any t ∈ CT(G), IT(G) ≤ t.

(3) IT(G) ≤ OT(G) ≤ ET(G).

(4) If H ⊆ G then OT(H) ≤ OT(G) and ET(H) ≤ ET(G).

(5) If H ◁ G then IT(H) ≥ IT(G).

(6) If H is an essential submodule of G then IT(H) ≤ IT(G).

(7) If H ◁ G then IT(G/H) ≥ IT(G) and OT(G/H) ≤ OT(G).

(8) OT(G) is p-divisible if and only if p-rank G < rank G.

proof: (4) If φ maps H onto a rank-one module A, then φ extends to a map φ′: G → QA and t(A) = t(φ(H)) ≤ t(φ′(G)) ∈ CT(G). Thus every type in CT(H) is less than or equal to a type in CT(G), so OT(H) ≤ OT(G). Now if r = rank G and s = rank H and K is a submodule of G maximally disjoint from H then QG = QH ⊕ QK and there is a non-trivial map ⋀^{s} H ⊗ ⋀^{r−s} K → ⋀^{r} G given by x ⊗ y ↦ x ∧ y. We then conclude that ET(H) ≤ ET(H)ET(K) ≤ ET(G).

(1) If A is a rank-one pure submodule of G then by (4) t(A) = OT(A) ≤ OT(G).

(5) If H ◁ G then T(H) ⊆ T(G) since any pure rank-one submodule of H is also a pure submodule of G. Thus IT(G) = inf T(G) ≤ inf T(H) = IT(H).

(6) If H is an essential submodule of G then any pure rank-one submodule of G has the form G ∩ QB, where B is a pure rank-one submodule of H. Furthermore t(B) ≤ t(G ∩ QB) since B ⊆ G ∩ QB. Thus for every s ∈ T(H) there exists t ∈ T(G) with s ≤ t and conversely. Thus IT(H) ≤ IT(G).

(7) For any coset g + H ∈ G/H, t(g + H) ≥ t(g) ∈ T(G) by Proposition 2.21. Thus IT(G/H) ≥ IT(G). Furthermore, if A is a homomorphic image of G/H, then A is also a homomorphic image of G, so CT(G/H) ⊆ CT(G), so OT(G/H) = sup(CT(G/H)) ≤ sup(CT(G)) = OT(G).

(2) If A is a rank-one homomorphic image of G then by (6) IT(G) ≤ IT(A) = t(A).

(3) It is clear from (1) and (2) that IT(G) ≤ OT(G). Now let μ: G → Q and let H = Ker μ. Choose a maximal linearly independent set h_{1}, …, h_{r−1} in H. Then we get ν: G → ⋀^{r} G by ν(g) = h_{1} ∧ ⋯ ∧ h_{r−1} ∧ g. Now μ and ν have the same kernel, so if we identify ⋀^{r} G as a submodule of Q it follows that μ and ν are scalar multiples of each other. Thus t(μ(G)) = t(ν(G)) ≤ t(⋀^{r} G)) = ET(G). Since this is true for all μ ∈ Hom(G, Q) it follows that OT(G) ≤ ET(G).

(8) By Proposition 1.25 p-rank G < rank G if and only if G has a homomorphic image which is p-divisible. This will be true if and only if G has a rank-one p-divisible homomorphic image (since the homomorphic image of a p-divisible module is p-divisible), i.e. if and only if CT(G) contains a p-divisible type. By the proof of Proposition 2.31 OT(G) is the least upper bound of a finite subset of CT(G), hence OT(G) is p-divisible if and only if some element of CT(G) is p-divisible. Thus p-rank G < rank G if and only if OT(G) is p-divisible. ✓

It is often the case that T(G) and CT(G) are infinite. Nonetheless there always exist finite subsets S ⊆ T(G) and S′ ⊆ CT(G) such that IT(G) = inf S and OT(G) = sup S′. This was seen in the proofs of Proposition 2.26 and Proposition 2.31, but the following proposition is more precise.

proposition 2.35. Let r = rank G.

(1) Let g_{1}, …, g_{r} be a maximal linearly independent set in G. Then inf {t(g_{1}), …, t(g_{r})} = IT(G).

(2) Let μ_{1}, …, μ_{r} be a maximal linearly independent set in Hom(G, Q). Then OT(G) = sup{t(μ_{1}(G)), …, t(μ_{r}(G))}.

proof: (1) Since g_{1}, …, g_{r} generate an essential submodule of G, if g ∈ G then (∃w ≠ 0 ∈ W) wg = ∑ w_{i}g_{i} with w_{i} ∈ W. Then by Proposition 2.21 t(g) = t(wg) ≥ inf {t(w_{i}g_{i}) | w_{i} ≠ 0} ≥ inf {t(g_{i}) | i = 1, …, r}. Thus IT(G) = inf T(G) ≥ inf {t(g_{i}) | i = 1, … r}. Since the opposite inequality is clear, the result follows.

(2) Since μ_{1}, …, μ_{r} form a basis for Hom(G, Q), for any μ ∈ Hom(G, Q) there exist w, w_{1}, …, w_{r} ∈ W such wμ = ∑ w_{i}μ_{i}. Then wμ(G) is contained in the submodule of Q generated by μ_{1}(G), …, μ_{r}(G), so t(μ(G)) = t(wμ(G)) ≤ sup{t(μ_{i}(G)) | i = 1, …, r}. Thus OT(G) ≤ sup{t(μ_{1}(G)), …, t(μ_{r}(G))}. Since the opposite inequality is clear, the result follows. ✓

proposition 2.36. (1) T(G ⊗ H) contains all types of the form st for s ∈ T(G), t ∈ T(H).

(2) CT(G ⊗ H) contains all types of the form st for s ∈ CT(G), t ∈ CT(H).

(3) IT(G ⊗ H) = IT(G)IT(H).

(4) OT(G ⊗ H) = OT(G)OT(H).

(5) ET(G ⊕ H) = ET(G)ET(H).

proof: (1) If A and B are pure submodules of G ⊗ H with t(A) = s and t(B) = t, then A ⊗ B is a pure submodule of G ⊗ H with t(A ⊗ B) = st.

(2) Let α: G → Q and β: H → Q with t(α(G)) = s ∈ CT(G), t(β(H)) = t ∈ CT(H). Then α ⊗ β: G ⊗ H → Q with t((α ⊗ β)(G ⊗ H)) = t(α(G)β(H)) = st. Thus st ∈ CT(G ⊗ H).

(3) It is clear from (1) and the fact that (by Proposition 2.35) IT(G ⊗ H) is actually the greatest lower bound of a finite subset of T(G ⊗ H) that IT(G ⊗ H) ≤ IT(G)IT(H). Now let x = ∑ g_{i} ⊗ h_{i} ∈ G ⊗ H. Then by Proposition 2.21 t(x) ≥ inf {t(g_{i} ⊗ h_{i}} ≥ IT(G)IT(H), so that IT(G ⊗ H) ≥ IT(G)IT(H).

(4) As seen in the proof of (2), if s ∈ OT(G) and t ∈ OT(H) then st ∈ OT(G ⊗ H) so that st ≤ OT(G ⊗ H). Now by Proposition 2.35 OT(G) and OT(H) are the least upper bounds of finite subsets of CT(G) and CT(H). It thus follows that OT(G)OT(H) ≤ OT(G ⊗ H). Now consider maximal linearly independent sets α_{1}, …, α_{r} ∈ Hom(G, Q) and β_{1}, …, β_{s} ∈ Hom(H, Q). We claim that {α_{i} ⊗ β_{j} | 1 ≤ i ≤ r, 1 ≤ j ≤ s} is a maximal linearly independent set in Hom(G ⊗ H, Q). In fact, for any i, j, ⋂_{i′≠i} Ker α_{i′} and ⋂_{j′≠j} Ker β_{j′} must be non-trivial, so there exist g ≠ 0 ∈ G and h ≠ 0 ∈ H with α_{i′}(g) = 0 = β_{j′} for all i′ ≠ i, j′ ≠ j. Then necessarily (α_{i} ⊗ β_{j})(g ⊗ h) ≠ 0. Then given any linear relation ∑_{i,j} w_{ij}α_{i} ⊗ β_{j} = 0 we substitute g ⊗ h and see that w_{ij} = 0, showing the linear independence of α_{1} ⊗ β_{1}, …, α_{r} ⊗ β_{s}. Furthermore t((α_{i} ⊗ β_{j})(G ⊗ H)) = t(α_{i}(G)t(β_{j}(H)) ≤ OT(G)OT(H). But by Proposition 2.35 OT(G ⊗ H) is the least upper bound of the types t((α ⊗ β)(G ⊗ H)). Thus OT(G ⊗ H) = OT(G)OT(H).

(5) It is elementary that if r = rank G and s = rank H then ⋀^{r+s}(G ⊕ H) ≈ (⋀^{r} G) ⊗ (⋀^{s} H). In fact, an element of ⋀^{r+s} G ⊕ H has the form (g_{1} + h_{1}) ∧ ⋯ ∧ (g_{r+s} + h_{r+s}) Now using the multilinearity of the exterior product, we see that ⋀^{r+s} G ⊕ H is generated by elements of the form x_{1} ∧ ⋯ ∧ x_{r+s} where for each k, x_{k} ∈ G or x_{k} ∈ H. Furthermore r of these x_{k} must belong to G and s to H, otherwise x_{1}, …, x_{r+s} are linearly dependent and x_{1} ∧ ⋯ ∧ x_{r+s} = 0. Furthermore we can bring all the G factors to the front and push all the H factors to the back, causing at worst a change in sign. Thus we see that ⋀^{r+s} G ⊕ H is generated by elements of the form g_{1} ∧ ⋯ ∧ g_{r} ∧ h_{1} ∧ ⋯ ∧ h_{s}. From this it is clear that there is a surjection (⋀^{r} G) ⊗ (⋀^{s} H) → ⋀^{r+s}(G ⊕ H). Since both domain and codomain have rank one, this is an isomorphism. ✓

It is a natural mistake to assume that T(G ⊗ H) is actually equal to the set of all st for s ∈ T(G), t ∈ T(H). A counterexample is as follows:

example 2.37. Let R be a pure subring of Ŵ_{p} with rank R > 1. Define μ: R ⊗ R → R by μ(r ⊗ s) = rs. Then Ker μ is a non-trivial divisible submodule of R ⊗ R, so that t(Q) ∈ R ⊗ R. However T(R) = {t(W_{p})}. first proof: Note that μ is surjective since if r ∈ R then r = μ(1 ⊗ r). Since p-rank R ⊗ R = 1 = p-rank R we see that p-rank Ker μ = 0 and since Ker μ is p-local it follows that Ker μ is divisible. Since rank R ⊗ R > rank R, Ker μ ≠ 0. Thus R ⊗ R is not reduced, so t(Q) ∈ T(R ⊗ R). second proof: The following more computational proof that Ker μ is divisible is also enlightening: We can note that Ker μ consists of those ∑ r_{i} ⊗ r′_{i} ∈ R ⊗ R such that ∑ r_{i}r′_{i} = 0. Now R/W_{p} is divisible since it is p-local and has trivial p-rank, so that for every non-trivial ideal 𝔞 of W, R/W_{p} = 𝔞(R/W_{p}) = (𝔞R + W_{p})/W_{p}, i.e. R = W_{p} + 𝔞R. Thus there exist w_{i}, w′_{i} ∈ W_{p} such that r_{i} ≡ w_{i}, r′_{i} ≡ w′_{i} (mod 𝔞R). Then ∑ r_{i} ⊗ r′_{i} ≡ ∑ w_{i} ⊗ w′_{i} = ∑ w_{i}w′_{i} ⊗ 1 ≡ ∑ r_{i}r′_{i} ⊗ 1 = 0 (mod 𝔞(R ⊗ R)). Thus if

∑ r_{i} ⊗ r′_{i} ∈ Ker μ then ∑ r_{i} ⊗ r′_{i} ∈ ⋂ 𝔞(R ⊗ R) = d(R ⊗ R). ✓

As a general rule, determining T(G ⊗ H) is a difficult problem.

CHARACTERISTICS. We now show how to classify rank-one modules up to quasi-isomorphism, i.e. how to classify types.

Recall from Proposition 2.6 that if A is a non-trivial submodule of Q and p a prime ideal, then A_{p} = p^{k_{p}}W_{p} for a unique k_{p} with −∞ ≤ k_{p} < ∞. We now write χ(p) = −k_{p}. Then since we always have A = ⋂ A_{p} = ⋂ p^{−χ(p)}W_{p}, there is a one-to-one correspondence between the submodules of Q and the functions χ: Spec W → ℤ ∪ {∞} with all but finitely many χ(p) ≥ 0. Such a function is called a characteristic (or height sequence), and so a type can be identified with an equivalence class of characteristics (which is, in fact, the traditional way of defining types). We can formalize this as follows:

theorem 2.38. Let ℤ^{+} be the additive semi-group of non-negative integers, and let ℤ^{++} be ℤ^{+} together with ∞. Then the lattice-ordered multiplicative semi-group of types is isomorphic to ∏_{p} ℤ^{++} / ⨁_{p} ℤ^{+} , where the product and direct sum are indexed by the set of (non-trivial) prime ideals in W.

proof: If A ≠ 0 ⊆ Q and χ_{A} is the associated characteristic, then χ(p) ≥ 0 for all p if and only if W = ⋂ W_{p} ⊆ A. Thus the mapping A ↦ (χ_{A}(p))_{p∈Spec W} sends the set of rank-one submodules of Q containing W into ∏ ℤ^{++} . It is apparent that this function is a homomorphism (taking multiplication to addition). Furthermore it is a surjection since if for every p ∈ Spec W we are given an non-negative “extended integer” χ(p) ∈ ℤ^{++} , then A = ⋂_{p} p^{−χ(p)}W_{p} ≠ 0 (in fact W ⊆ A) and the characteristic corresponding to A is χ.

Now if I is an ideal and χ is the characteristic corresponding to I, then χ(p) < ∞ for all p, i.e. χ(p) ∈ ℤ^{+} , and furthermore χ(p) = 0 except for the finitely many p containing I, so that χ represents an element of ⨁ ℤ^{+} . Conversely if χ is associated with a rank-one module A and χ(p) ∈ ℤ^{+} for all p and χ(p) = 0 for all but finitely many p, then A is the ideal ∏ p^{−χ(p)} .

Now if A and B are non-trivial submodules of Q, then t(A) = t(B) if and only if there exist non-trivial ideals 𝔞 and 𝔟 such that 𝔞A = 𝔟B. By the above, this is true if and only if the corresponding elements of ∏_{p} ℤ^{++} are congruent modulo ⨁_{p} ℤ^{+} . Therefore the map that sends each rank-one module A such that W ⊆ A to its characteristic is a isomorphism from the set of types of submodules of Q containing W onto ∏ ℤ^{++}/ ⨁ ℤ^{+} . But in fact if A is a non-trivial submodule of Q, then A ∩ W is a non-trivial ideal 𝔞 ⊆ W, and thus 𝔞W ⊆ A and W ⊆ 𝔞^{−1}A. Since t(A) = t(𝔞^{−1}A) we see that every type is represented by a submodule of Q containing W so that the function described is in fact a isomorphism from the set of all types into

∏ ℤ^{++}/ ⨁ ℤ^{+} . ✓

proposition 2.39. If t has the characteristic χ, then t is idempotent if and only if χ(p) is either 0 or ∞ for all but finitely many prime ideals p.

proof: By Proposition 2.15 t is idempotent if and only if t = t(A) where A is a subring of Q. But by Proposition 2.2 A is a subring of Q if and only if for all p, A_{p} = W_{p} or A_{p} = Q, i.e. for all p, χ(p) = 0 or χ(p) = ∞. ✓

In general, if G is a W -module and g ∈ QG, we write ht_{p}(g) = sup{n | g ∈ p^{n}G}, and call this the p-height of g in G. Note that ht_{p}(g) ≥ 0 ⟺ g ∈ G_{p} and that if w ≠ 0 ∈ W then ht_{p}(wg) = ht_{p}(g) except for finitely many primes p, namely those p which contain w. Note also that if A ⊆ Q and A_{p} = p^{−χ(p)}W_{p}, then χ(p) = ht_{p}(1). Thus for a submodule A of Q we get the characteristic χ_{A} defined by χ_{A}(p) = ht_{p}(1) for each prime ideal p, and we see that this is precisely the characteristic associated with A as described in the preceding paragraphs.

Note also the following:

proposition 2.40. Let φ: QG → QH. Then φ ∈ Hom(G, H) ⟺ φ increases p-heights, i.e. (∀p)(∀g ∈ QG) ht_{p}(φ(g)) ≥ ht_{p}(g). proof: ( ⇐ ): We have φ ∈ Hom(G, H) ⟺ φ(G) ⊆ H ⟺ (∀p) φ(G_{p}) ⊆ H_{p} ⟺ (∀p)(∀g ∈ QG) ht_{p}(g) ≥ 0 ⇒ ht_{p}(φ(g)) ≥ 0. Certainly this is the case of φ increases heights.

( ⇒ ): If φ(G) ⊆ H then for all integers n, φ(p^{n}G) ⊆ p^{n}H. Thus

g ∈ p^{n}G ⇒ φ(g) ∈ p^{n}H. ✓

Proposition 2.40 is important in that it shows that the category of finite rank torsion free W -modules can be identified with a category of multi-valuated finite dimensional vector spaces [Rotman].

RICHMAN TYPE. Another approach to the concept of types is as follows: We define an equivalence relation on torsion W -modules by making two torsion modules T_{1} and T_{2} equivalent if they differ by a finite length summand (i.e. if there exist finite length modules M_{1} and M_{2} such that T_{1} ⊕ M_{1} ≈ T_{2} ⊕ M_{2} .) Then for any finite rank torsion free module G we define the Richman type of G to be the equivalence class of G/F, where F is any finitely generated essential submodule of G. We will show below that the Richman type is well defined.

The equivalence relation on torsion modules introduced here can be clarified by the following Lemma:

lemma 2.41. Let T and T′ be torsion modules with length T [p] < ∞ and length T′[p] < ∞ for all p.

(1) If T ⊆ T′ then T and T′ are equivalent if and only if T′/T has finite length.

(2) If T′ ≈ T/K then T and T′ are equivalent if and only if K has finite length.

proof: (1) ( ⇒ ): Easy.

( ⇐ ): T and T′ are the direct sums of their p-primary components, and the condition that T′[p] < ∞ implies that each p-primary component is the direct sum of a divisible module and a finite length module. Now if T′/T has finite length then only a finite number of p-primary components can differ between T and T′ (namely for those p ∈ Ass T′/T), and even there the divisible submodules must agree. Thus if M is the direct sum of the reduced parts of the p-primary components of T for p ∈ Ass T′/T, and M′ is the analogous component for T′, then T ⊕ M′ ≈ T′ ⊕ M.

(2) Essentially the same. One wants to let M be the sum of the reduced parts of the p-primary components of T for p ∈ Ass K and analogously for M′. ✓

If T is a torsion module such that length T [p] ≤ 1 for all p, then by Proposition 0.* T ≈ ⨁_{p} W (p^{k_{p}}), where the direct sum is taken over all prime ideals and 0 ≤ k_{p} ≤ ∞, with the convention that W (p^{0}) = 0. We can thus associate a characteristic χ by setting χ(p) = k_{p} for all p. This gives a one-to-one correspondence between the set of all characteristics and the set of isomorphism classes of torsion modules such that T [p] ≤ 1 for all p.

proposition 2.42. (1) The Richman type of a finite rank torsion free module G is well defined.

(2) Two rank-one modules have the same type if and only if they have the same Richman type.

(3) Let W ⊆ A ⊆ Q and let T = A/W. If χ is the characteristic corresponding to T as indicated above, then for every prime p, χ(p) = ht_{p}(1), so that t(A) is the type corresponding to the characteristic χ.

(4) If T is a torsion module then there exists a rank-one module whose Richman type is the class of T if and only if T [p] is simple for all but finitely many primes p and T [p] has finite length for all p.

proof: (1) Let F and F′ be finitely generated essential submodules of G. To see that G/F and G/F′ differ by a finite length summand, we may compare both quotients to G/(F ∩ F′). Thus there is no loss of generality in supposing F ⊆ F′. Then F′/F is a finitely generated torsion module over a dedekind domain, hence has finite length. Also note that G/F ⊆ QG/F ≈ (Q/W)^{r} and so for each prime p, length G[p] ≤ r < ∞. Thus G/F′ is a homomorphic image of G/F with finite length kernel F′/F, so by the preceding Lemma G/F and G/F′ are equivalent.

(3) Since A_{p} = p^{− ht_{p}(1)}W_{p} and (A/W)[p^{∞}] = (A_{p}/W_{p})[p^{∞}], this is clear.

(2) Clear from (3).

(4) The stated condition amounts to saying that T is equivalent to a torsion module whose p-socle is simple for all p. But to say that T [p] is simple for all p is the same as saying that for all p the p-primary component of T is either cyclic or is W (p^{∞}), i.e. T [p^{∞}] = W (p^{χ(p)}) for some χ(p) with 1 ≤ χ(p) ≤ ∞. Thus the result follows

from (3). ✓

The Richman type has not proved to be highly useful in the theory of finite rank torsion free modules, however it seems to hold out some ineffable promise of eventual value that prevents it from falling into total obscurity. It does contain information a little sharper than that given by the inner type, outer type, and exterior type.

proposition 2.43. Let rank G = r and let F be a finitely generated essential submodule of G. For each prime ideal p of G we can write (G/F)_{p} ≈ W (p^{i_{p}}) ⊕ ⋯ ⊕ W (p^{o_{p}}) with 0 ≤ i_{p} ≤ ⋯ ≤ o_{p} ≤ ∞, where the direct sum includes precisely r summands, some of which may be trivial. (We adopt the convention that W (p^{0}) = 0.) Let e_{p} = i_{p} + ⋯ + o_{p}. Let χ_{I}, χ_{O}, and χ_{E} be the characteristics given by χ_{I}(p) = i_{p}, χ_{O}(p) = o_{p}, and χ_{E}(p) = e_{p} for p ∈ Spec W. Then χ_{I}, χ_{O}, and χ_{E} are the characteristics corresponding to IT(G), OT(G), and ET(G). proof: (1) Since F is projective, F = F_{1} ⊕ ⋯ ⊕ F_{r}, where each F_{i} is isomorphic to an ideal of W. For each i let A_{i} = F_{i ∗}. Then A_{1} ⊕ ⋯ ⊕ A_{r} ⊆ G and by Proposition 2.35 IT(G) = inf {t(A_{1}), …, t(A_{r})}. Furthermore if the p-primary component of A_{j}/F_{j} is isomorphic to W (p^{k_{j}}), then i_{p} = min{k_{1}, …, k_{r}}. Therefore it suffices to prove the assertion about inner types in the rank-one case. But this is just Proposition 2.42.

(2) Continuing the same notation, let H_{i} be the pure submodule of G generated by all F_{j} for j ≠ i. Write B_{i} = G/H_{i} and let μ_{i}: G → B_{i} be the quotient map. Then μ_{i}(F) is a finitely generated essential finitely generated submodule of B_{i}. Let χ_{i} be the characteristic corresponding to B_{i}/μ_{i}(F) as in the proof of Proposition 2.42, i.e. the p-primary component of B_{i}/μ_{i}(F) is W (p^{χ_{i(p)}}). By Proposition 2.42 t(B_{i}) is the class of the characteristic χ_{i}. Since there is a surjection G/F → B_{i}/μ_{i}(F), it follows that χ_{i}(p) ≤ o_{p} = χ_{O}(p). Thus if t_{O} is the type associated with χ_{O} then t_{O} ≤ t(B_{i}), i.e. χ_{i} ≤ χ_{O}. By Proposition 2.35 we know that OT(G) = sup{t(B_{1}), …, t(B_{r})}. Thus OT(G) ≤ t_{O}. On the other hand, since ⋂ H_{i} = 0, G is isomorphic to a submodule of ⨁ μ_{i}(G) and so G/F is isomorphic to a submodule of ⨁ μ_{i}(G)/μ_{i}(F) ≈ ⨁ W (p^{χ_{i}(p)}) and so o_{p} ≤ max{χ_{1}(p), …, χ_{r}(p)} and we conclude that t_{O} ≤ sup{t(B_{1}), …, t(B_{r})} = OT(G) ≤ t_{O}.

(3) ⋀^{r} F is a finitely generated essential submodule of ⋀^{r} G. Thus ET(G) is the class of the characteristic associated with the torsion module ⋀^{r} G / ⋀^{r} F. Now the elements of this torsion module have representatives of the form g_{1} ∧ ⋯ ∧ g_{r} where g_{1}, …, g_{r} are linearly independent in G. Furthermore p^{m}(g_{1} ∧ ⋯ ∧ g_{r} + ⋀^{r} F) = 0 ⟺ m can be written as m_{1} + ⋯ + m_{r} with p^{m_{i}}g_{i} ∈ F for all i (so that, in particular, all the cosets g_{i} + F generate p-primary submodules of G/F.) From this it is clear that if χ is the characteristic associated with ⋀^{r} G / ⋀^{r} F then for each p, e_{p} ≤ χ(p). Conversely, if we write (G/F)_{p} as a direct sum of r p-primary cyclic modules (with some possibly trivial) and let g_{1}, …, g_{r} be representatives for generators of these cyclic summands, then g_{1} ∧ ⋯ ∧ g_{r} generates a submodule of ⋀^{r} G/ ⋀^{r} F isomorphic to W (p^{e_{p}}) so that

e_{p} ≤ χ(p). ✓

If G is a finite rank torsion free module and to each g ∈ QG one associates the family of extended integers {ht_{p}(g)}_{p∈Spec W} , then QG becomes a multiply-valuated vector space. The import of Proposition 2.40 is that in this way the category of finite rank torsion free modules becomes essentially the same as the category of such multiply-valuated vector spaces.

This point of view has not proved to be highly fruitful, however, precisely because it is not any real simplification: it is simply a restatement of the problem.

The concept of the type of an element provides an apparently minimal simplification of the height point of view. In essence this amounts to defining an equivalence relation on QG, whereby for w ≠ 0 ∈ W, g and wg are considered equivalent. This small simplification makes types a much more useful tool than heights for torsion free modules.

Types – and for that matter heights – work well because of three facts:

(1) If G is any torsion free module then G = ⋂_{Spec W} G_{p}.

(2) For every (non-trivial) prime p, W_{p} is a discrete valuation ring.

(3) The rank-one modules over a discrete valuation ring are as described in Proposition 2.6.

In order to get conditions (1) and (2), we are restricted to the environment of torsion free modules over dedekind domains. However it is possible to get most of the theorems of the present chapter with less stringent restrictions on W if we impose further restrictions on the modules G under consideration. Namely, we can look at integral domains W such that W_{p} is a discrete valuation ring for all height-one primes p. We need to then restrict the theory to those modules G such that G = ⋂ G_{p} where the intersection is taken over the set of height-one prime ideals. It turns out that, if we assume W noetherian, the class of rings in question is the class of noetherian Krull domains. And if we assume in addition that W_{m} is a unique factorization domain for each maximal ideal m – a not unreasonable hypothesis – then the rank-one modules A such that A = ⋂_{ht p=1} A_{p} are precisely the flat rank-one modules. Thus we get a theory of types for rank-one flat modules over locally factorial noetherian integral domains [Lady].

Unfortunately, once one tries to go beyond the realm of this chapter into the more general theory of finite rank torsion free modules, nothing works very well unless one has a dedekind domain. For flat modules, one has a notion of purity, which we can define cheaply by saying that H is a pure submodule of a flat module G is G/H is flat. One of the main problems that arises, though, is that if φ ∈ Hom(G, K), φ(G) need not be flat.