1. FUNDAMENTALS

E. L. Lady

May 14, 1998

From now on, W will denote a dedekind domain. This means among other things that a W -module is flat if and only if it is torsion free, and is injective if and only if it is divisible. We denote the quotient field of W by Q.

We will henceforth use the term prime ideal to denote a non-trivial prime ideal in W. We will also write Spec W for the maximal spectrum of W, i.e. the set of non-trivial prime ideals. (Since W is a dedekind domain, non-trivial primes are maximal.)

In general, the objective has been to set up the terminology and notation in such a way that to the greatest extent possible the language used will agree with the language used for abelian groups. At the same time, we have included in this Chapter proofs for a number of commonplace facts of abelian group theory, although we have not made an effort to provide proofs that are valid outside the torsion free environment.

CHEAP DEFINITIONS. We can cheaply define a finite rank torsion free W -module as a submodule of a finite dimensional vector space over Q. In fact, if G is a torsion free W -module, then G can be regarded as a submodule of its localization at the zero ideal, which we will denote as QG. QG is a Q-vector space, and the rank of G is the dimension of QG. Thus to say that G has finite rank is to say that QG is finite dimensional. QG is an injective envelope for G and is also sometimes called the divisible hull of G. To say that G is divisible is to say that G = QG.

If S is a multiplicative set in W then we will identify S^{−1}G with the submodule of QG consisting of elements g/s for s ∈ S. We will say that G is p-local for a prime ideal p if G = G_{p}.

In the remainder of this book, in the absence of indication to the contrary all modules under consideration are assumed to be finite rank torsion free. The adjectives “finite rank” and “torsion free” will nonetheless sometimes be inserted redundantly where it seems to improve clarity or diction. In particular, the phrase “If G is a torsion free W -module” is often used as an indicator that the assertion that follows does not really require the hypothesis of finite rank.

The price we pay for the cheap definitions given here is that they do not necessarily apply to general (i.e. not torsion free) W -modules.

One can think of QG as a shorthand for Q ⊗ G. Or if G is explicitly given as a submodule of a Q-vector space V , we can identify QG with the Q-subspace of V generated by G. Thus the notation QG has a certain logic to it.

Thinking of G as a submodule of QG is a convenience in many ways. For instance, if A is a submodule of Q, then the product AG is a well defined submodule of QG, namely the submodule generated by all ag with a ∈ A, g ∈ G. In particular, if p is a prime ideal (or any non-trivial ideal, for that matter), recall that there is a unique submodule of Q which we denote by p^{−1} , determined by the property that p^{−1}p = W. Thus for all n ≥ 0 we get p^{−n}G ⊆ QG. We will define p^{−∞} to be ⋃_{1}^{∞} p^{−n} and p^{−∞}G to be ⋃_{1}^{∞} p^{−n}G. (If p is a principal ideal p = (p_{0}), then p^{−∞}G is simply the localization of G with respect to the multiplicative set S = {p_{0}^{n} | n ≥ 0}.) There is an alternate way of interpreting these notations. In the following proposition we see that the two interpretations agree.

proposition 1.1. (1) p^{−n}G = {x ∈ QG | p^{n}x ⊆ G}.

(2) p^{−∞}G = {x ∈ QG | (∃n) p^{n}x ⊆ G}.

(3) p^{−n}G/G ≈ G/p^{n}G.

proof: (1) If x ∈ QG then p^{n}x ⊆ G ⟺ W x = p^{−n}p^{n}x ⊆ p^{−n}G ⟺ x ∈ p^{−n}G. (2) Clear from (1).

(3) This is clear if W is a principal ideal domain. In fact in that case multiplication by a generator for p^{n} gives an isomorphism p^{−n}G → G which maps G onto p^{n}G. Thus since G_{p} is a module over W_{p}, which is a discrete valuation ring by Proposition 0.*, we get p^{−n}G_{p}/G_{p} ≈ G_{p}/p^{n}G_{p}. But p^{−n}G/G ≈ p^{−n}G_{p}/G_{p} and G/p^{n}G ≈ G_{p}/p^{n}G_{p} by Proposition 0.* since these are all p-primary modules. ✓

Since QG is the localization of G at the zero ideal, we know that a homomorphism φ: G → H extends uniquely to a Q-linear transformation QG → QH. It will be very convenient to in fact identify Hom(G, H) with the set of linear transformations φ ∈ Hom(QG, QH) such that φ(G) ⊆ H. Then we can also identify the divisible hull QHom(G, H) of Hom(G, H) as the subspace of Hom(QG, QH) generated by Hom(G, H). (In most cases, QHom(G, H) will be a proper subspace of Hom(QG, QH).) Mappings in QHom(G, H) are called quasi-homomorphisms and play a vital role in the study of finite rank torsion free modules.

DIVISIBLE AND p-DIVISIBLE MODULES. A non-cheap definition of divisibility is that a W -module G is divisible if 𝔞G = G for every non-trivial ideal 𝔞. It is an easy exercise to see that if G is torsion free, this is equivalent to our cheap definition, viz. that G is a module over Q, or, equivalently, that G = QG. Any torsion free W -module contains a maximal divisible submodule, which we denote d(G). We can obtain d(G) cheaply either as ⋂_{𝔞} 𝔞G, where the intersection is taken over all non-trivial ideals of W, or as the image of the mapping Hom(Q, G) → G given by φ ↦ φ(1).

For any G, we write p^{∞}G = ⋂ p^{n}G. Thus there is a filtration p^{∞}G ⊆ ⋯ ⊆ p^{n}G ⊆ ⋯ ⊆ p^{−n}G ⊆ ⋯ ⊆ p^{−∞}G. proposition 1.2. The following conditions are equivalent:

(1) G = pG.

(2) G = p^{∞}G.

(3) G = p^{−∞}G.

(4) G_{p} is divisible.

proof: 1) ⇒ 2) & 3): Multiplying by p^{−1} shows that G = pG ⟺ p^{−1}G = G. Now from G = pG we get inductively G = p^{n}G for all n, so G = p^{∞}G. And from G = p^{−1}G we get G = p^{−n}G for all n so G = p^{−∞}G.

1) ⟺ 4): G = pG ⟺ G_{q} = pG_{q} for all prime ideals q. But for q ≠ p, qW_{p} = W_{p} so pG_{q} = G_{q} is automatic. Thus G = pG ⟺ G_{p} = pG_{p} ⟺ p^{−∞}G_{p} = G_{p} by the equivalence of 1) and 3). But p^{−∞}G_{p} = p^{−∞}W_{p}G_{p} = QG_{p}, so that G_{p} = pG_{p} if and only if G_{p} is divisible.

(2) ⇒ (1): By Proposition 1.1 p^{−1}p^{n}G = p^{n−1}G, and from this we see that p^{−1}p^{∞}G ⊆ p^{∞}G. Multiplying by p yields p^{∞}G ⊆ pp^{∞}G. Since the opposite inclusion is clear, we see that if G = p^{∞}G then G = pG.

(3) ⇒ (1): We prove similarly that pp^{−∞}G = p^{−∞}G. ✓

We say that a torsion free W -module G is p-divisible if G = pG.

proposition 1.3. (1) G is divisible if and only if G is p-divisible for all primes p. (2) p^{∞}G is the unique maximal p-divisible submodule of G.

proof: (1) If 𝔞 is an ideal in W, then 𝔞 = p_{1} ⋯ p_{n} where the p_{i} are not necessarily distinct prime ideals. If p_{i}G = G for each i, then 𝔞G = G.

(2) We need to see that i.e. that p^{∞}G ⊆ p(p^{∞}G). In fact, if x ∈ p^{∞}G then for all n,

x ∈ p^{n+1}G = p(p^{n}G) so that p^{−1}x ⊆ p^{∞}G, so x ∈ p(p^{∞}G). ✓

note: Part (2) of Proposition 1.3 is not valid without the assumption that G is torsion free.

We say that G is reduced if d(G) = 0 and p-reduced if p^{∞}G = 0.

proposition 1.4. (1) If φ: G → H then φ(d(G)) ⊆ d(H) and φ(p^{∞}G) ⊆ p^{∞}H. (2) A submodule of a reduced module is reduced.

(3) A submodule of a p-reduced module is p-reduced.

proof: (1) We have φ(d(G)) = φ(⋂ 𝔞G) ⊆ ⋂ φ(𝔞G) ⊆ ⋂ 𝔞H = d(H), and φ(p^{∞}G) = φ (⋂ p^{n}G) ⊆ ⋂ p^{n}H = p^{∞}H.

(2) and (3): If H ⊆ G then apply (1) to the inclusion map H ↪ G to see that p^{∞}H ⊆ p^{∞}G and d(H) ⊆ d(G). Thus p^{∞}G = 0 ⇒ p^{∞}H = 0 and

d(G) = 0 ⇒ d(H) = 0. ✓

proposition 1.5. There are natural isomorphisms Hom(Q, G) → d(G) and Hom(p^{−∞}, G) → p^{∞}G given in each case by φ ↦ φ(1). proof: (1) Since d(G) is divisible it is a Q-vector space and so there is a well known natural isomorphism Hom(Q, d(G)) = Hom_{Q}(Q, d(G)) → d(G) given by φ ↦ φ(1). But Hom(Q, d(G)) = Hom(Q, G) since by Proposition 1.4 if φ ∈ Hom(Q, G) then φ(Q) = φ(d(Q)) ⊆ d(G).

(2) Essentially the same. We note that p^{∞}G is a module over the ring p^{−∞}W = p^{−∞}

and so p^{∞}G ≈ Hom(p^{−∞}, p^{∞}G) = Hom(p^{−∞}, G). ✓

PURITY. The concept of purity can also be defined cheaply for torsion free modules. Namely, we can say that a submodule H of a torsion free module G is pure if G/H is torsion free. We will write H ◁ G to indicate that H is a pure submodule of G.

The concept of purity is the most important concept we will use. Pure submodules play the same role in the theory of torsion free modules that normal subgroups play in general group theory: they are precisely the subobjects which are kernels.

The non-cheap definition of purity, which is valid for general W -modules, is given as item (2) of the following proposition.

proposition 1.6. Let G be torsion free and H ⊆ G. The following conditions are equivalent:

(1) H ◁ G.

(2) For every ideal 𝔞 of W, H ∩ 𝔞G = 𝔞H.

(3) H = G ∩ QH

(4) For every prime p the induced mapping H/pH → G/pG is a monomorphism.

(5) For every prime p, H_{p} is a pure submodule of G_{p}.

proof: (1) ⇒ (2) ⇒ (4) ⇒ (5) ⇒ (3) ⇒ (1).

(1) ⟺ (3): A torsion element of G/H is represented by g ∈ G such that wg ∈ H for some w ≠ 0 ∈ W, which is the same as saying that g ∈ QH. Thus the torsion submodule of G/H equals (G ∩ QH)/H, and so G/H is torsion free if and only if G ∩ QH = H.

(1) ⟺ (5): G/H is torsion free if and only if (G/H)_{p} is torsion free for all non-trivial primes p. But (G/H)_{p} ≈ G_{p}/H_{p}, so this is equivalent to H_{p} ◁ G_{p} for all p.

(1) ⇒ (2): Suppose h ∈ H ∩ 𝔞G for some non-trivial ideal 𝔞. Then 𝔞^{−1}h ⊆ G and every x ∈ 𝔞^{−1}h yields a coset x + H ∈ G/H annihilated by 𝔞. Thus if G/H is torsion free the coset x + H must be trivial, so that 𝔞^{−1}h ⊆ H. Thus h ∈ 𝔞H, showing that H ∩ 𝔞G ⊆ 𝔞H. The opposite inclusion is trivial.

(2) ⟺ (4): H/pH → G/pG is a monomorphism if and only if H ∩ pG = pH, since H ∩ pG/pH is the kernel of H/pH → G/pG.

(2) ⇒ (1): G/H is torsion free if and only if p ∉ Ass G/H for all non-trivial primes p, i.e. if and only if there does not exist a non-trivial coset g + H ∈ G/H annihilated by p. Now if p(g + H) = 0 ∈ G/H then pg ⊆ H so pg ⊆ H ∩ pG = pH by (2). Then g ∈ p^{−1}pH = H by Proposition 1.1, so g + H = 0. It follows that p ∉ Ass G/H for all primes p and so G/H is torsion free. ✓ example 1.7. Let G = W^{n} and let H be the cyclic submodule of G generated by g = (w_{1}, …, w_{n}) ∈ W^{n}. Then H ◁ G if and only if w_{1}, …, w_{n} generate the unit ideal in W (i.e. w_{1}, …, w_{n} are relatively prime). proof: ( ⇒ ): Suppose w_{1}, …, w_{n} generate the ideal 𝔞 in W. Then g = (w_{1}, …, w_{n}) ∈ 𝔞G. If H ◁ G this means g ∈ 𝔞H, so that H ⊆ 𝔞H. Since H ≈ W it follows that W ⊆ 𝔞W = 𝔞. Therefore 𝔞 = W.

( ⇐ ): Suppose that w_{1}, …, w_{n} generate the unit ideal. For any ideal 𝔞 of W let h ∈ H ∩ 𝔞G. Then h = (ww_{1}, …, ww_{n}) for some w ∈ W and h ∈ 𝔞W^{n}, so that for all i, ww_{i} ∈ 𝔞W = 𝔞. Since w_{1}, …, w_{n} generate W this means that wW ⊆ 𝔞, i.e. w ∈ 𝔞. Then h = w(w_{1}, …, w_{n}) ∈ 𝔞H. This shows that H is pure in G. ✓

The following are well known properties of purity:

proposition 1.8. Let H ⊆ K ⊆ G.

(1) If H ◁ G then H ◁ K.

(2) Purity is transitive, i.e. if H ◁ K and K ◁ G then H ◁ G.

(3) Suppose H ◁ G. Then K ◁ G if and only if K/H ◁ G/H.

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

(5) If H ◁ G then the image of H in G/d(G) is pure in G/d(G).

(6) A pure submodule of a divisible module is divisible.

(7) If G is the direct union of a family of submodules {G_{i}}_{i∈I} and H is a pure submodule of each G_{i}, then H ◁ G.

(8) If H is the direct union of a family of pure submodules H_{i} of G then H ◁ G.

proof: (1) Clear.

(2) Since QH ⊆ QK, G ∩ QH = G ∩ QK ∩ QH = K ∩ QH = H.

(3) (G/H)/(K/H) ≈ G/K.

(4) If G = H ⊕ L then G/H ≈ L.

(5) We need to show that (H + d(G))/d(G) is pure in G/d(G). Since d(G) ◁ G, by (3) this is equivalent to showing that H + d(G) is pure in G. Now in the short exact sequence

0 → (H + d(G))/H → G/H → G/(H + d(G)) → 0

the left end is divisible, hence is a summand. Therefore the right end is isomorphic to a summand of G/H, hence is torsion free. Thus H + d(G) is a pure submodule of G.

(6) If G = QG and H ◁ G then H = G ∩ QH = QG ∩ QH = QH.

(7) G/H is the direct union of the family of torsion free modules G_{i}/H, hence is torsion free.

(8) For any ideal 𝔞, 𝔞G ∩ H = 𝔞G ∩ ⋃ H_{i} = ⋃(𝔞G ∩ H_{i}) = ⋃ 𝔞H_{i} ⊆ 𝔞H since H_{i} ◁ G. Since 𝔞H ⊆ 𝔞G ∩ H in any case, this shows that H ◁ G. ✓

Condition (8) in Proposition 1.7 was included for the sake of completeness. It has no relevance to what happens in this book, though, since a finite rank module cannot be the union of an infinite ascending chain of pure submodules. In fact finite rank torsion free modules have both the ascending and descending chain conditions on pure submodules since the length of a chain of pure submodules cannot be greater than the rank of the entire module.

proposition 1.9. If G is torsion free, then the intersection of any family of pure submodules of G is a pure submodule of G. proof: Let H = ⋂ H_{i} where H_{i} ◁ G. Then G ∩ QH = G ∩ (⋂ QH_{i}) = ⋂(G ∩ QH_{i}) =

⋂ H_{i} = H. ✓

proposition 1.10. If G_{1}, …, G_{n} are submodules of a finite dimensional Q-space V , then ⋂ G_{i} is isomorphic to a pure submodule of ⨁ G_{i}. proof: ****

definition 1.11. If H ⊆ G we write H_{∗} = G ∩ QH. H_{∗} is a pure submodule of G and is called the pure submodule of G generated by H or the purification of H in G. H_{∗} can also be obtained as the intersection of all pure submodules of G containing H.

definition 1.12. If H ⊆ G we say that H is p-pure in G if p ∉ Ass G/H.

proposition 1.13. The following conditions are equivalent:

(1) H is p-pure in G.

(2) H ∩ pG = pH.

(3) For all n ≥ 1, H ∩ p^{n}G = p^{n}H.

(4) H_{p} is a pure submodule of G_{p}.

proof: (2) ⇒ (1): If H ∩ pG = pH then p^{−1}H ∩ G = p^{−1}(H ∩ pG) = H, which says that there is no g ∈ G with g ∉ H and pg ⊆ H. I.e. no non-trivial element of G/H is annihilated by p. Thus p ∉ Ass G/H.

(1) ⇒ (4): If p ∉ Ass G/H then p ∉ Ass(G/H)_{p} = Ass G_{p}/H_{p}. But then if we consider G_{p}/H_{p} as a W_{p}-module, the only possible associated prime is the zero ideal. It then follows from Proposition 0.* that Ass G_{p}/H_{p} ⊆ {0}, i.e. G_{p}/H_{p} is torsion free, so that H_{p} is a pure submodule of G_{p}.

(4) ⇒ (3): If H_{p} ◁ G_{p} then H_{p} ∩ p^{n}G_{p} = p^{n}H_{p}. But if q is a prime different than p then p^{n}H_{q} = H_{q} and so H_{q} ∩ p^{n}G_{q} = p^{n}H_{q}. Thus by Proposition 0.9, H ∩ p^{n}G = p^{n}H.

(3) ⇒ (2): Trivial. ✓

The equivalence of conditions (2) and (3) in Proposition 1.13 is not valid without the assumption that G is torsion free.

proposition 1.14. If S is a multiplicative set in W then G is p-pure in S^{−1}G for all primes p with p ∩ S = ∅.

proof: If p ∩ S = ∅ then S^{−1}G_{p} = G_{p}, so G_{p} ◁ S^{−1}G_{p}. ✓

proposition 1.15. (1) For any G, p^{∞}G and d(G) are pure submodules of G. (2) If H ◁ G then d(H) = H ∩ d(G).

(3) If H is p-pure in G then p^{∞}H = H ∩ p^{∞}G.

proof: (1) ***

(2) If H ◁ G then for every ideal 𝔞, 𝔞H = H ∩ 𝔞G. Thus d(H) = ⋂_{𝔞 ≠0} 𝔞H = ⋂(H ∩ 𝔞G) = H ∩ d(G).

(3) Analogous, using the fact that if H is p-pure in G then for every n ≥ 1,

p^{n}H = H ∩ p^{n}G. ✓

proposition 1.16. Let R be a torsion free W -algebra. Then W is a p-pure submodule of R if and only if R_{p} is not divisible.

proof: W is p-pure in R if and only if Q ∩ R_{p} = W_{p}. But as a discrete valuation ring, W_{p} is a maximal proper subring of Q, so either Q ∩ R_{p} = W_{p} or Q ∩ R_{p} = Q. In the first case, R_{p} cannot be divisible by Proposition 1.15, since Q ∩ R_{p} ◁ R_{p}. In the second case R_{p} is in fact a Q-algebra and hence is divisible. ✓

ESSENTIAL SUBMODULES. We say that H is an essential submodule of G if H ⊆ G and every element of G has a non-trivial multiple in H. This is equivalent to, but more convenient than, the usual definition, namely that every non-trivial submodule of G intersects H non-trivially. It is usually more convenient to say that H is an essential submodule of G than to say that G is an essential extension of H, although we will say that too sometimes.

The traditional terminology in the theory of torsion free abelian groups has been to say full subgroup rather than essential subgroup.

proposition 1.17. Let H ⊆ G. The following conditions are equivalent:

(1) H is an essential submodule of G.

(2) QH = QG.

(3) G/H is torsion.

(4) rank H = rank G.

(5) Every non-trivial submodule of G intersects H non-trivially.

(6) For every torsion free K the restriction map Hom(G, K) → Hom(H, K) is monic.

proof: (1) ⟺ (2): Restated, the definition of essentiality says that H is an essential submodule of G if and only if G ⊆ QH. Clearly this is equivalent to QG = QH.

(2) ⟺ (4): QH = QG ⟺ dim QH = dim QG ⟺ rank H = rank G.

(1) ⟺ (3): G/H is torsion if and only if for all g ∈ G there exists w ≠ 0 ∈ W such that wg ∈ H, which by definition means that H is an essential submodule of G.

(1) ⇒ (5): Suppose that H is an essential submodule of G. If K is a non-trivial submodule of G and k ≠ 0 ∈ G then k has a non-trivial multiple wk in H. Then wk ≠ 0 ∈ H ∩ K.

(5) ⇒ (1): If g ≠ 0 ∈ G, then to say that the cyclic submodule of G generated by g intersects H non-trivially is equivalent to saying that g has a non-trivial multiple in H.

(3) ⇒ (6): 0 → Hom(G/H, K) → Hom(G, K) → Hom(H, K). But since G/H is torsion and K is torsion free, Hom(G/H, K) = 0. Thus Hom(G, K) → Hom(H, K) is monic.

(6) ⇒ (4): Let K = G/H_{∗} and let φ: G → K be the quotient map. Then the restriction of φ to H is trivial since H ⊆ H_{∗}. Thus if Hom(G, K) → Hom(H, K) is monic then φ is trivial, which means G = H_{∗}. Then rank H = rank H_{∗} = rank G. ✓

proposition 1.18. Let H be a submodule of G.

(1) H is a pure submodule of G if and only if there is no proper essential extension of H in G.

(2) H is an essential submodule of G if and only if H_{∗} = G.

(3) If H is both pure and essential in G then H = G.

proof: (1) Let H ⊆ K ⊆ G. Then H is an essential submodule of K if and only if K ⊆ QH. Thus the maximal essential extension of H in G is G ∩ QH. But H is pure if and only if G ∩ QH = H.

(2) Let H ⊆ K ⊆ G. Then K is pure in G if and only if G/K is torsion free. Hence the minimal pure submodule of G containing H is the inverse image of the torsion submodule of G/H. On the other hand H is an essential submodule of G if and only G/H is torsion, i.e. if and only if the minimal pure submodule of G containing H is G itself.

(3) Clear from (2). ✓

FINITELY GENERATED MODULES. A lot of the preceding is actually valid over an arbitrary integral domain W, especially if one replaces the term “torsion free” by “flat.” The thing that makes dedekind domains different, and that makes most of the rest of this book work, is the following Proposition.

theorem 1.19. A finitely generated torsion free W -module is projective, and in fact is isomorphic to a direct sum of ideals. proof: It suffices to prove the second assertion, since ideals in a Dedekind domain are projective. We use induction on rank G. If rank G = 1 then QG ≈ Q and we may assume wlog that G ⊆ Q. If we then let w be a common denominator for finite set of generators for G, we have wG ⊆ W, i.e wG is an ideal. Since G is torsion free, G ≈ wG and the result follows.

Now if rank G > 1 let V be a subspace of QG with codimension 1 and let H = V ∩ G. Then G/H is isomorphic to a submodule of the one-dimensional space QG/V , hence rank G/H = 1 and by the first part G/H is isomorphic to an ideal. In practicular G/H is projective, hence H is a summand of G. By induction H is a direct sum of ideals and since G ≈ H ⊕ G/H the result follows. ✓ corollary 1.20. If W is a principal ideal domain, then finitely generated torsion free W -modules are free. proof: If W is a principal ideal domain then all non-trivial ideals are isomorphic

to W. ✓

proposition 1.21. A projective W -module is finitely generated if and only if it has finite rank. proof: ****

corollary 1.22. If W is local and A is a W -module and rank A = 1, then A is either free or divisible. proof: Since rank A = 1, QA ≈ Q. Therefore we may assume wlog that A ⊆ Q. Now if W is local then it is a discrete valuation ring, and thus is a unique factorization domain with only one prime p. Now if A contains fractions with denominators which are arbitrarily high powers of p, then 1 ∈ p^{∞}A ◁ A, so A = p^{∞}A since rank A = 1, and thus A is p-divisible, hence divisible. On the other hand, if there is a highest power of p that occurs in the denominators of fractions belonging to A, then A is finitely generated, hence projective, and by Proposition 1.21 A is free. ✓

corollary 1.23. A W -module G is torsion free if and only if it is flat. proof: If w ≠ 0 ∈ W then the map G ⟶^{w} G can be obtained as μ ⊗ G: W ⊗ G → W ⊗ G where μ: W → W is multiplication by w. Since W is an integral domain, μ is monic, so if G is flat then μ ⊗ G is also monic, showing that G is torsion free.

On the other hand, any module G is a direct union of its finitely generated submodules. By Theorem 1.19, if G is torsion free then these finitely generated submodule are projective, hence flat. Then G is flat, since a direct union of flat modules is flat. (This is because the tensor product commutes with direct limits.) ✓

As an immediate application of Theorem 1.19, we have the following very important fact:

proposition 1.24. If G is a finite rank torsion free module and 𝔞 a non-trivial ideal, then G/𝔞G has finite length. In fact, length G/𝔞G ≤ r length W/𝔞, where r = rank G. proof: We first show that if I is a non-trivial ideal of W then length I/𝔞I = length W/𝔞. Since I/𝔞I is a torsion module, by Proposition 0.* it is a direct sum of its localizations (I/𝔞I)_{p}, and it then suffices to prove that length(I_{p}/𝔞I_{p}) = length(W_{p}/𝔞W_{p}). But since W_{p} is a principal ideal domain, I_{p} ≈ W_{p} so the assertion in this case is clear.

Now let G be any finite rank torsion free module. Let M be a finitely generated submodule of G/aG and let F be the submodule of G generated by representatives for a finite set of generators of M. Thus there is a surjection F → M which induces a surjection F/𝔞F → M since 𝔞M = 0. By the Theorem, since F is finitely generated it is a direct sum of at most r ideals, so by the paragraph above length F/𝔞F ≤ r length W/𝔞. But M is a homomorphic image of F/𝔞F, so length M ≤ r length W/𝔞.

Thus r length W/𝔞 is a bound on the lengths of the finitely generated submodules of G/𝔞G. It follows that there is a maximal such finitely generated submodule M. Then G/𝔞G = M, otherwise there exists x ∈ G/𝔞G with x ∉ M and x together with M generates a strictly larger finitely generated submodule of G/𝔞G, contradicting the maximality of M. Hence length G/𝔞G = length M ≤ r length W/𝔞. ✓

proposition 1.25. Let H be an essential submodule of G. Then for each prime p, G_{p}/H_{p} is the direct sum of a divisible p-primary module and a finite length p-primary module.

(?? This is only true if G is finitely generated.) proof: Since H_{p} is essential in G_{p}, G_{p}/H_{p} is torsion. Thus 0 ∉ Ass G_{p}/H_{p}, and since this is a module over W_{p} it follows that Ass G_{p}/H_{p} = {p}, i.e. it is p-primary. Now let H′ ⊆ G_{p} be such that H_{p} ⊆ H′ and H′/H_{p} = d(G_{p}/H_{p}). Then G_{p}/H_{p} = d(G_{p}/H_{p}) ⊕ G_{p}/H′. Now if T = G_{p}/H′ then T is a reduced p-primary module and T/pT ≈ G_{p}/pG_{p}. But by Proposition 1.24 G_{p}/pG_{p} has finite length. Thus by Proposition 0.* G_{p}/H′ has finite length. ✓

RANK AND p-RANK. We have defined rank G as dim QG. For any prime ideal p we now define p-rank G = length G/pG. These two concepts have similar behavior.

proposition 1.26. (1) rank(G ⊕ H) = rank G + rank H, and p-rank(G ⊕ H) = p-rank G + p-rank H.

(2) If H ◁ G then rank G = rank H + rank G/H and p-rank G = p-rank H + p-rank G/H.

(3) If H is an essential submodule of G then rank H = rank G and p-rank H ≥ p-rank G.

(4) p-rank G = p-rank G_{p}.

(5) p-rank G ≤ rank G.

(6) For any torsion free G and H, rank G ⊗ H = (rank G)(rank H) and p-rank G ⊗ H = (p-rank G)(p-rank H).

(7) rank G = 0 if and only if G = 0.

(8) p-rank G = 0 if and only if G is p-divisible.

(9) p-rank G equals the number of summands of QG/G isomorphic to W (p^{∞}).

proof: (2) The assertion about ranks is clear. For the first assertion about p-ranks we need to note that by Proposition 1.6 if H ◁ G then H/pH can be identified as a submodule of G/pG, and in fact we get a short exact sequence

0 → H/pH → G/pG → (G/H)/p(G/H) → 0.

(In other words, tensoring the pure exact sequence 0 → H → G → G/H → 0 with W/p preserves exactness.) Since the length of the middle term of this sequence must equal the sum of the lengths of the two ends, we have p-rank G = p-rank H + p-rank G/H.

(1) A special case of (2).

(4) G/pG ≈ G_{p}/pG_{p}.

(5) By Proposition 1.24 p-rank G = length G/pG ≤ (rank G)(length W/p) = rank G, since p is a maximal ideal and so W/p is a simple module.

(6) Immediate since Q(G ⊗ H) ≈ QG ⊗ QH and (G ⊗ H)/p(G ⊗ H) ≈ G/pG ⊗ H/pH.

(7) Since rank G = dim QG, rank G = 0 ⟺ QG = 0 ⟺ G = 0.

(8) By Proposition 1.2 G is p-divisible if and only if G = pG.

(9) Since QG/G is a torsion divisible module, the claim is that p-rank G = length (QG/G)[p]. But (QG/G)[p] = p^{−1}G/G ≈ G/pG by Proposition 1.1.

(3) The assertion about ranks is clear since QG = QH. And QG/G is a homomorphic image of QG/H = QH/H since H ⊆ G, so by (8) p-rank G ≤ p-rank H. ✓

warning: One should note that the usage here of the term “p-rank” is not consistent with the way it is used for torsion modules. For this reason many authors (perhaps wiser than yours truly) avoid using this word in the present context and leave p-rank as a concept for which there is notation but no name.

proposition 1.27. The following conditions are equivalent:

(1) p-rank G = rank G

(2) No non-trivial torsion free homomorphic image of G is p-divisible.

(3) G_{p} is a free W_{p}-module.

proof: (1) ⇒ (2): Let H ◁ G be such that G/H is p-divisible. Then p-rank G/H = 0 so p-rank H = p-rank G. Thus if p-rank G = rank G then rank H ≥ p-rank H = p-rank G = rank G ≥ rank H. Thus H is an essential submodule of G so by Proposition 1.18 H = G and G/H = 0.

(2) ⇒ (3): Choose H ◁ G such that rank G/H = 1. By Proposition 1.22 if G/H is not p-divisible then G_{p}/H_{p} is a free W_{p}-module, so that H_{p} is a direct summand of G_{p}. Thus H_{p} is a homomorphic image of G_{p}, so if G_{p} has no p-divisible homomorphic images neither can H_{p}. By induction we conclude that H_{p} is a free W_{p}-module, so that G_{p} ≈ H_{p} ⊕ (G_{p}/H_{p}) and thus G_{p} is free over W_{p}.

(3) ⇒ (1): By Proposition 1.26 p-rank G = p-rank G_{p}. If G_{p} is a free W_{p}-module then it is a direct sum of r copies of W_{p}, where r = rank G. Then p-rank G_{p} = r(p-rank W_{p}) = r = rank G. ✓

One of the things that makes linear algebra work as well as it does is the theorem that a linear transformation from a finite dimensional vector space into itself is a monomorphism if and only if it is surjective. For finite rank torsion free modules only one direction of this theorem is valid. proposition 1.28. Let φ be an endomorphism of G.

(1) φ is monic if and only if φ(G) is an essential submodule of G.

(2) If φ(G) = G, then φ is an automorphism of G.

proof: (1) Ker φ = 0 ⟺ rank Ker φ = 0 ⟺ rank φ(G) = rank G ⟺ φ(G) is an essential submodule of G.

(2) If φ(G) = G then by (1) φ is monic as well as surjective, hence is an automorphism. ✓

lemma 1.29. [partial fractions expansion] Let 𝔞_{1}, …, 𝔞_{n} be ideals such that 𝔞_{i} + 𝔞_{j} = W (i.e. 𝔞_{i} and 𝔞_{j} are relatively prime) for all pairs i ≠ j. Then 𝔞_{1}^{−1} + ⋯ + 𝔞_{n}^{−1} = 𝔞_{1}^{−1} ⋯ 𝔞_{n}^{−1} . proof: Note that if 𝔞_{1} + 𝔞_{2} = W and 𝔞_{1} + 𝔞_{3} = W then 𝔞_{1} + 𝔞_{2}𝔞_{3} = W, since W = (𝔞_{1} + 𝔞_{2})(𝔞_{1} + 𝔞_{3}) = 𝔞_{1}(𝔞_{1} + 𝔞_{2} + 𝔞_{3}) + 𝔞_{2}𝔞_{3} ⊆ 𝔞_{1} + 𝔞_{2}𝔞_{3} . It then follows by induction that it suffices to prove the result for the case n = 2. But if 𝔞_{1} + 𝔞_{2} = W then

𝔞_{1}^{−1}𝔞_{2}^{−1} = 𝔞_{1}^{−1}𝔞_{2}^{−1}(𝔞_{2} + 𝔞_{1}) = 𝔞_{1}^{−1} + 𝔞_{2}^{−1} . ✓

For instance, if W = ℤ then the submodule of Q generated by 1/3 and 1/5 is the same as the submodule generated by 1/15. In fact, \frac{1}{3} = \frac{5}{15} , \frac{1}{5} = \frac{3}{15} , and \frac{1}{15} = \frac{2}{5} − \frac{1}{3} .

example 1.30. Let A be the submodule of Q generated by the set of p^{−1} for all p ∈ Spec W. A is the direct union of all products p_{1}^{−1} ⋯ p_{n}^{−1} such that p_{1}, …, p_{n} are distinct primes. (If W is a principal ideal domain, then A consists of all fractions with “square-free denominators,” i.e. all u/v with u, v ∈ W and (∀p) v ∉ p^{2} .) Then rank A = 1 and p-rank A = 1 for all p. Thus one cannot distinguish A and W on the basis of rank and p-rank, and yet if Spec W is infinite A and W are not isomorphic. proof: (1) Obviously A is the direct union of those submodules of Q of the form p_{1}^{−1} + ⋯ + p_{n}^{−1} , where p_{1}, …, p_{n} are distinct primes. But by Lemma 1.29 p_{1}^{−1} + ⋯ + p_{n}^{−1} = p_{1}^{−1} ⋯ p_{n}^{−1} .

(2) If Spec W is infinite, there is an infinite strictly ascending chain W ⊂ p_{1}^{−1} ⊂ p_{1}^{−1}p_{2}^{−1} ⊂ ⋯ of submodules of A. Therefore A is not noetherian, and hence not finitely generated. Thus A ≉ W.

(3) Clearly rank A = 1 and so p-rank A ≤ 1. Now if p-rank A = 0 then A would be p-divisible, i.e. A = p^{−∞}A. Since W ⊆ A this would imply p^{−∞}W = p^{−∞} ⊆ A. But in fact even p^{−2} ⊈ A since if p^{−2} ⊆ 𝔞^{−1} where 𝔞 is a product of distinct primes then cross multiplying yields 𝔞 ⊆ p^{2} , a contradiction. Thus p-rank A = 1. ✓

FINITE RANK TORSION FREE RINGS. Except in Chapter 11, we will use the word ring to mean W -algebra. By a torsion free ring, or finite rank torsion free ring, we shall mean a W -algebra that has the indicated property as a W -module.

We will be working a lot more with non-commutative rings than most abelian group theorists or commutative ring theorists are used to. Most of the ring theory we will use is not terribly sophisticated, although a knowledge of the Wedderburn Theorem will be very useful. The most commonly used concept from ring theory will be that of the nil radical. definition 1.31. If L is a left ideal in a finite rank torsion free ring R, we say that L is nilpotent if L^{r} = 0 for some r, i.e. whenever ℓ_{1}, …, ℓ_{r} ∈ L then ℓ_{1} ⋯ ℓ_{r} = 0. (This is stronger than merely requiring that all the elements in L are nilpotent.) We define nil rad R, the nil radical of R, to be the left ideal generated by the set of all nilpotent left ideals.

In general ring theory, the term “prime radical” is more customary for this concept. However the term nil radical is well established in the theory of finite rank torsion free modules and is in fact appropriate from ring theoretic considerations in this context.

The point is that the theory of finite rank torsion free rings in several respects closely resembles the theory of finite dimensional algebras over a field, since a lot of things happening with a finite rank torsion free ring R are really happening in QR.

proposition 1.32. Let R be a finite rank torsion free ring and N = nil rad R. Then

(1) N is a two-sided ideal in R.

(2) If r = rank R then N^{r} = 0.

(3) All elements in N are nilpotent.

(4) N is a pure submodule of R.

(5) QR is a finite dimensional algebra over Q and QN is the radical of QR.

proof: (1) If L is a nilpotent left ideal and r ∈ R, then it is easy to see that Lr is also nilpotent, so that Lr ⊆ N. Since N is generated by the set of all such L, it follows that Nr ⊆ N. Thus N is a two-sided ideal.

(5) Let Ñ denote the radical of QR, by which we mean the nil radical, which is the same as the Jacobson radical since QR is a finite dimensional algebra. Since r = dim QR, it is known from the theory of finite dimensional algebras that Ñ^{r} = 0. Thus (R ∩ Ñ)^{r} = 0, so R ∩ Ñ is a nilpotent ideal, so R ∩ Ñ ⊆ N = nil rad R, and so Ñ ⊆ QN. Conversely it is easy to see that if L is a nilpotent left ideal in R then QL is a nilpotent left ideal in QR. Thus QL ⊆ nil rad QR = Ñ, and we thus conclude that N ⊆ Ñ. It follows that N = R ∩ Ñ and QN = Ñ.

(4) Since N = R ∩ Ñ = R ∩ QN, N is a pure submodule of R.

(2) Since QN^{r} = 0 then N^{r} = 0.

(3) If n ∈ N then n^{r} ∈ N^{r} = 0. ✓

We say that a left (or right) ideal in a ring R is nil is it consists of nilpotent elements. Clearly nilpotent left ideals are nil, but for rings in general nil left ideals need not be nilpotent. For finite rank torsion free rings, however, just as for finite dimensional algebras over a field, the two concepts agree.

proposition 1.33. Let R be a finite rank torsion free ring.

(1) Every nil left ideal is nilpotent.

(2) If r ∈ R and sr is nilpotent for every s ∈ R, then r ∈ nil rad R. proof: (1) Let L be a nil left ideal. Then it is easy to see that QL is a nil left ideal in QR. Furthermore clearly if QL is nilpotent then so is L. Thus wlog we may assume that R = QR, so that result is standard from the theory of finite dimensional algebras. (In outline, the proof is as follows: Replacing L by L^{r} for large r, one sees that there is no loss of generality in supposing that L^{2} = L. There is then a standard lemma that states that either L = 0 or L contains an idempotent. But the second alternative is inconsistent with the fact that L is nil.)

(2) To say that every sr is nilpotent is to say that the left ideal Rr is nil. Then by (1) Rr is a nilpotent left ideal and it follows that r ∈ Rr ⊆ nil rad R. ✓

For any W -module G, the set of endomorphisms of G forms a ring, which we denote End G. The properties of the ring End G are very important, and will be developed in the chapters that follow, especially Chapter 3, Chapter 11, and Chapter 7. For the moment, we note only the following facts:

proposition 1.34. If rank G = r, then End G is a finite rank torsion free ring with rank at most r^{2} .

proof: According to our conventions, endomorphisms of G are linear transformations on QG, so that End G is a subring of the r^{2}-dimensional Q-algebra End_{Q} QG (which is isomorphic to the ring of r × r-matrices over Q.) Thus End G is torsion free with rank at

most r^{2} . ✓

It is fairly uncommon for the rank of End G to be r^{2} , except in the obvious case r = 1. Exactly when this happens will be discussed in Chapter 4.

As a very rough rule of thumb, the more complicated G is the smaller one expects rank End G to be. In a certain not very accurate sense, the rank of End G is a measure of the amount of symmetry in G. In particular, rank End G may be either larger or smaller than rank G.

By Proposition 1.32 if s = rank End G and N = nil rad End G, then N^{s} = 0. We now see rhat furthermore N^{r} = 0, where r = rank G. Since r ≤ s in many cases (although r ≥ s in many others) this is sometimes a stronger result.

Recall that a submodule H of G is called fully invariant if φ(H) ⊆ H for all φ ∈ End G.

proposition 1.35. Let G be a non-trivial finite rank torsion free module and let N = nil rad End G. Let r = rank G. Then

(1) rank NG < rank G.

(2) N^{r} = 0.

(3) NG is a fully invariant submodule of G.

(4) If H is a fully invariant submodule of G and G = H + NG then G = H. proof: By NG is meant, of course, the submodule of G generated by all elements φ(g) with g ∈ G, φ ∈ N.

(1) Note that End G acts on QG, so that N acts on QG. Now if rank NG = rank G then NQG = QG, and so it follows inductively that N^{k}QG = QG for all k. Since N^{k} = 0 for large k and QG ≠ 0, this is a contradiction.

(2) Continuing the reasoning in (1), we get rank G > rank NG > rank N^{2}G > …, where the chain terminates only when it reaches the point that N^{k}G = 0. But the chain can have length at most r = rank G. Thus N^{r}G = 0. But clearly this implies N^{r} = 0.

(3) If φ ∈ End G, then φN ⊆ N because N is an ideal in End G. Thus φ(NG) ⊆ NG, showing that NG is fully invariant.

(4) Since H is fully invariant, NH ⊆ H. Thus if G = H + NG then G = H + N(H + NG) = H + NH + N^{2}G = H + N^{2}G = ⋯ = H + N^{k}G for all k. But

by (1) N^{r} = 0 so G = H + N^{r}G = H. ✓

A very important classical property of the nil radical is that units or idempotents modulo the nil radical lift to the whole ring, as stated in the following lemma. This result is general ring theory, and does not require that R be a W -algebra or be torsion free. We will see the importance of this lemma in Chapter 3 and especially in Chapter 11.

lemma 1.36. Let N be an ideal of a ring R containing only nilpotent elements.

(1) Let r ∈ R be such that the image of r in R/N is invertible. Then r is invertible in R.

(2) Let r ∈ R be such that the image of r in R/N is idempotent. Then there exists an idempotent e ∈ R such that e ≡ r (mod N).

proof: (1) ****

(2) If e ≡ r (mod N) then the image of e in R/N is idempotent, so e^{2} − e ∈ N and since all elements of N are nilpotent, (e^{2} − e)^{k} = 0 for some k. Choose e so that k is as small as possible. We claim that k = 1. If not, note that (2e − 1)^{2} = 1 + 4(e^{2} − e) is invertible since e^{2} − e is nilpotent and thus 2e − 1 is also invertible. Now e(2e − 1) = (2e − 1)e and so (2e − 1)^{−1}e = e(2e − 1)^{−1} . Let n = −(e^{2} − e)(2e − 1)^{−1} . Then n commutes with e. It now follows that n^{k} = (−1)^{k}(e^{2} − e)^{k}(2e − 1)^{−k} = 0. Since n commutes with e, (e + n)^{2} − (e + n) = e^{2} − e + n(2e − 1) + n^{2} = e^{2} − e − (e^{2} − e)(2e − 1)^{−1}(2e − 1) + n^{2} = n^{2} . Now n^{k} = 0 and if k > 1 then 2(k − 1) = 2k − 2 ≥ 2k − k = k so that (n^{2})^{k−1} = 0. In other words, if k > 1 then [(e + n)^{2} − (e + n)]^{k−1} = 0. Since e + n ≡ e ≡ r (mod N) this contradicts the minimality of k. Thus k = 1 so that e^{2} − e = 0 and e is an idempotent, as required. ✓

TENSOR PRODUCTS AND HOM. The tensor product and Hom functors play a much more crucial role in the theory of finite rank torsion free modules than they do in the theory of torsion modules. If G and H are finite rank torsion free modules, then so are G ⊗ H and Hom(G, H), and these functors provide a convenient way to get at many important constructions. At the same time, it must be noted that the actual computation of G ⊗ H and Hom(G, H) is generally a formidable problem. lemma 1.37. For any finitely generated submodule 𝔞 of Q,

(1) Hom(𝔞G, 𝔞H) = Hom(G, H).

(2) Hom(G, 𝔞H) = 𝔞 Hom(G, H).

(3) Hom(𝔞G, H) = 𝔞^{−1} Hom(G, H).

proof: Since 𝔞 is finitely generated it is isomorphic to an ideal, hence is projective, hence by Proposition 0.* there exists a submodule 𝔞^{−1} of Q such that 𝔞𝔞^{−1} = W.

(1) Clearly Hom(𝔞G, 𝔞H) ⊆ Hom(G, H) since if φ: QG → QH and φ(G) ⊆ H then φ(𝔞G) ⊆ 𝔞H. Conversely, if φ(𝔞G) ⊆ 𝔞H then φ(G) = φ(𝔞^{−1}𝔞G) ⊆ 𝔞^{−1}𝔞H = H. Thus Hom(𝔞G, 𝔞H) ⊆ Hom(G, H).

(2) Clearly 𝔞 Hom(G, H) ⊆ Hom(G, 𝔞H) since if φ ∈ Hom(G, H) then 𝔞φ(G) ⊆ 𝔞H. Now applying this to Hom(G, 𝔞H) and 𝔞^{−1} yields Hom(G, 𝔞H) = 𝔞𝔞^{−1} Hom(G, 𝔞H) ⊆ 𝔞 Hom(G, 𝔞^{−1}𝔞H) = 𝔞 Hom(G, H).

(3) Analogous. This also follows from (1) and (2) since Hom(𝔞G, H) =

Hom(𝔞^{−1}𝔞G, 𝔞^{−1}H) = Hom(G, 𝔞^{−1}H) = 𝔞^{−1} Hom(G, H). ✓

corollary 1.38.

(1) p^{∞} Hom(G, H) = Hom(G, p^{∞}H).

(2) d(Hom(G, H)) = Hom(G, d(H)).

proof: (1) By Lemma 1.37 p^{n} Hom(G, H) = Hom(G, p^{n}H) for all n. Thus the Corollary follows once we see that Hom(G, p^{∞}H) = ⋂_{n} Hom(G, p^{n}H). But this simply says that if φ ∈ Hom(G, H) then φ(G) ⊆ p^{∞}H ⟺ (∀n) φ(G) ⊆ p^{n}H, and this assertion is clearly true.

(2) By Lemma 1.37 d(Hom(G, H)) = ⋂_{a} a Hom(G, H) = ⋂ Hom(G, aH) =

Hom(G, ⋂ aH) = Hom(G, d(H)). ✓

proposition 1.39. (1) rank G ⊗ H = (rank G)(rank H).

(2) p-rank(G ⊗ H) = (p-rank G)(p-rank H).

(3) rank Hom(G, H) ≤ (rank G)(rank H).

(4) p-rank Hom(G, H) ≤ (p-rank G)(p-rank H).

proof: (1) Clear since Q(G ⊗ H) ≈ Q ⊗ G ⊗ H ≈ Q ⊗ G ⊗ Q ⊗ H ≈ QG ⊗ QH. (2) (G ⊗ H)/p(G ⊗ H) ≈ G/pG ⊗ H/pH.

(3) Hom(G, H) ⊆ Hom(QG, QH) and rank Hom(QG, QH) = (rank G)(rank H).

(4) By Lemma 1.37 p Hom(G, H) = Hom(G, pH). Thus from the exact sequence

0 → Hom(G, pH) → Hom(G, H) → Hom(G, H/pH)

we deduce that Hom(G, H)/p Hom(G, H) is isomorphic to a submodule of Hom(G, H/pH). Thus p-rank Hom(G, H) ≤ length Hom(G, H/pH). But Hom(G, H/pH) is isomorphic to Hom(G/pG, H/pH) since any homomorphism φ ∈ Hom(G, H/pH) annihilates pG. And since G/pG and H/pH are vector spaces over W/p,

p-rank Hom(G, H) ≤ length Hom(G/pG, H/pH) = dim Hom(G/pG, H/pH)
= (dim G/pG)(dim H/pH) = (p-rank G)(p-rank H). ✓

We have noted that torsion free W -modules are flat. Thus if H ⊆ G then for any torsion free K we can (and will) identify H ⊗ K as a submodule of G ⊗ K.

proposition 1.40. If H ◁ L then for any G, Hom(G, H) ◁ Hom(G, L) and G ⊗ H ◁ G ⊗ L. proof: Hom(G, L)/ Hom(G, H) is isomorphic to a submodule of Hom(G, L/H). Since H ◁ L, L/H is torsion free and so Hom(G, L/H) is torsion free.

The proof for the tensor product is analogous. ✓

proposition 1.41. If H is a p-pure submodule of L, then for any G, Hom(G, H) is p-pure in Hom(G, L) and G ⊗ H is p-pure in G ⊗ L. proof: There is an exact sequence

0 → Hom(G, H) → Hom(G, L) → Hom(G, L/H).

It suffices to prove that p ∉ Ass Hom(G, L/H). In fact, since p ∉ Ass L/H and p is maximal, there is no non-trivial submodule M of L/H such that pM = 0. In particular, if φ ≠ 0 ∈ Hom(G, L/H) then pφ(G) ≠ 0. Thus pφ ≠ 0, showing that p ∉ Ass Hom(G, L/H).

For the tensor product, simply note that if H_{p} ◁ L_{p} then G ⊗ H_{p} ◁ G ⊗ L_{p}, since G ⊗ (L_{p}/H_{p}) is torsion free. ✓

The major obstacle that prevents localization from being a really powerful tool for studying finite rank torsion free modules is that localization does not commute with Hom, i.e. Hom(G, H)_{p} is not necessarily isomorphic to Hom(G_{p}, H_{p}). For instance if Spec W is infinite and B is the submodule of Q generated by all p^{−1} (Example 1.30), then Hom(B, W) = 0 but for each prime ideal p, Hom(B_{p}, W_{p}) ≈ W_{p}. We do however have the following theorem.

proposition 1.42. For any multiplicative set S ⊂ W, S^{−1} Hom(G, H) ◁ Hom(S^{−1}G, S^{−1}H). proof: By Proposition 0.* Hom(S^{−1}G, S^{−1}H) = Hom(G, S^{−1}H). We will show that S^{−1} Hom(G, H) is p-pure in Hom(G, S^{−1}H) for all primes p. If p ∩ S ≠ ∅ then this is clear by Proposition 1.13 since S^{−1} Hom(G, H)_{p} = QHom(G, H). On the other hand for p with p ∩ S = ∅ H is p-pure in S^{−1}H by Proposition 1.14 and so by Proposition 1.41 Hom(G, H) is p-pure in Hom(G, S^{−1}H). Thus S^{−1} Hom(G, H) is p-pure in Hom(G, S^{−1}H) for these p as well. Thus S^{−1} Hom(G, H) is pure in

Hom(S^{−1}G, S^{−1}H). ✓

proposition 1.43. For any finite rank torsion free modules G, H, and K, Hom(G, H) ⊗ K is embedded as a pure submodule in Hom(G, H ⊗ K) by the mapping that takes φ ⊗ k to the homomorphism g ↦ φ(g) ⊗ k. proof: We will show that for every prime ideal p the induced map

Γ: (Hom(G, H) ⊗ K)/p(Hom(G, H) ⊗ K) → Hom(G, H ⊗ K)/p Hom(G, H ⊗ K)

is monic. As seen in the proof of Proposition 1.39, Hom(G, H)/p Hom(G, H) is naturally isomorphic to a submodule of Hom(G/pG, H/pH), and likewise the codomain of Γ is isomorphic to a submodule of Hom(G/pG, H/pH ⊗ K/pK). Furthermore, with these identifications Γ is the restriction of the canonical map

Γ′: Hom(G/pG, H/pH) ⊗ K/pK → Hom(G/pG, H/pH ⊗ K/pK).

But Γ′ is a map of vector spaces over W/p and as such is well known to be an isomorphism. Hence Γ is monic. ✓

It is useful to define the reduced tensor product of two finite rank torsion free W -modules G and H as G ⋆ H = (G ⊗ H)/d(G ⊗ H). The reduced tensor product has the properties of a tensor product with respect to the category of reduced torsion free W -modules, as shown by the following proposition.

proposition 1.44. (1) If K is a reduced W -module then Hom(G ⋆ H, K) ≈ Hom(G, Hom(H, K)).

(2) If R is a torsion free W -algebra and M is a reduced R-module then Hom(G, M) ≈ Hom_{R}(R ⋆ G, M).

(3) The reduced tensor product is neither left nor right exact, however if H ◁ G then H ⋆ K is canonically isomorphic to a pure submodule of G ⋆ K and the map G ⋆ K → (G/H) ⋆ K is surjective.

proof: (1) In fact, since K is reduced Hom(G ⋆ H, K) = Hom((G ⊗ H)/d(G ⊗ H), K) ≈ Hom(G ⊗ H, K) ≈ Hom(G, Hom(H, K)).

(2) Similarly Hom_{R}(R ⋆ G, M) ≈ Hom_{R}(R ⊗ G, M) ≈ Hom(G, M) where the second isomorphism is standard module theory and the first holds because if φ: R ⊗ G → M is R-linear then it is also W -linear, so φ(d(R ⊗ M)) = 0.

(3) If H ◁ G then H ⊗ K ◁ G ⊗ K (since by assumption K is torsion free). Then by Proposition 1.8 (H ⊗ K) + d(G ⊗ K) is pure in G ⊗ K. It follows that H ⋆ K is embedded as a pure submodule of G ⋆ K. Furthermore since G ⊗ K → G/H ⊗ K is surjective, the induced map G ⋆ K → G/H ⋆ K must also be surjective. ✓

INDECOMPOSABLE MODULES. We say that a torsion free W -module G is indecomposable if it is not possible to write G as a direct sum H ⊕ K with both H and K non-trivial.

If G has finite rank r then it is clear that G can be written as a direct sum of at most r indecomposable modules. One might suspect, in fact, that there would always be exactly r terms in a complete breakdown of G as a direct sum, i.e. that every indecomposable module has rank one. This is not the case, however, and except for some very special rings W (most notably complete valuation rings) there are indecomposable W -modules of every finite rank. (There also exist indecomposable W -modules of infinite rank. See [Fuchs].)

We start by indicating the most obvious cases:

proposition 1.45. (1) Rank-one modules are indecomposable.

(2) If G is a p-reduced module with p-rank one then G is indecomposable.

proof: (1) Clear.

(2) If p-rank G = 1 and G = H ⊕ K then p-rank H + p-rank K = 1 so either H or K has p-rank zero, i.e. either H or K is p-divisible. But by Proposition 1.4 H and K are p-reduced. Thus H = 0 or K = 0. ✓

One may notice that the hypothesis in (2) above can be weakened to “if G has p-rank one and has no p-divisible summand … ” As a practical matter, though, if one is trying to prove that a module G is indecomposable it is usually hard to establish that G has no p-divisible summand except by showing that G is p-reduced.

The question that arises is: How does one produce modules with p-rank one? From Proposition 1.26 we see that if G ⊆ G′ ⊆ QG then p-rank G′ ≤ p-rank G. Since p-rank QG = 0, it seems plausible that if G is not p-divisible then somewhere between G and QG there are modules G′ with p-rank one. In fact, a cheap way of obtaining such a G′ is to note that if p-rank G > 0 then the divisible torsion module QG/G must have a summand isomorphic to W (p^{∞}). One can then choose G′ with G′ ⊇ G such that QG/G′ ≈ W (p^{∞}). We will omit the proof that this works, since this construction has the severe drawback that one will probably have no idea what the resulting module G′ looks like.

Instead, we will start with G and adjoin various elements of QG until the p-rank has been reduced as desired. The way to accomplish the desired reduction of p-rank is to adjoin elements having, as it were, denominators which are powers of p. In the simplest case, we can take an element g ∈ G and adjoin to G all the elements of p^{−n}g, n = 1, 2, …. The following example uses this approach:

example 1.46. Let p_{1} , p_{2} , and p_{3} be distinct prime ideals in W. Let G be the free module W ⊕ W and let G′ be the submodule of QG generated by G together with p_{1}^{−∞}(1, 0), p_{2}^{−∞}(0, 1) and p_{3}^{−∞}(1, 1), where p_{1}^{−∞}(1, 0), for instance, denotes the set of all elements (x, 0) with x ∈ p_{1}^{−∞}. Then G′ is indecomposable. proof: For p = p_{1}, p_{2}, p_{3} , G′ has p-rank one since, for instance, there is a short exact sequence 0 → ⟨(1, 0)⟩_{∗} → G′ → G′/⟨(1, 0)⟩_{∗} → 0 and the left end is p-divisible and thus has p_{1}-rank zero and the right end has p_{1}-rank at most one since its rank is one. However Proposition 1.45 does not apply, since G′ is not p-reduced for p = p_{1}, p_{2}, p_{3} . However we can still use underlying idea of Proposition 1.45. If G′ = H ⊕ K and p = p_{1}, p_{2}, p_{3} , then since p-rank G′ = 1 either p-rank H = 0 or p-rank K = 0. Since there are three choices for p and only two summands H and K, we see that one of the summands must have p-rank zero for two different p, say p_{1}-rank H = p_{2}-rank H = 0.

Then H ⊆ p_{1}^{∞}G′ ∩ p_{2}^{∞}G′ = 0. ✓

To be able to use Proposition 1.45 (2) as given, we need a more subtle construction than that of the previous example. To enlarge G to get G′ with p-rank G′ = 1 and yet still have p^{∞}G = 0, we cannot simply put larger and larger denominators under a fixed numerator g. Instead, we let the numerator move as the denominators increase.

example 1.47. Pontryagin module. This construction can be carried out to produce p-reduced modules of any rank with p-rank one, but to simplify notation we will only consider rank three. Start with W ⊕ W ⊕ W and let {v_{i}}_{1}^{∞}, {w_{i}}_{1}^{∞} be sequences of elements from W such that v_{n+1} ≡ v_{n} (mod p^{n}) and w_{n+1} ≡ w_{n} (mod p^{n}) for all n. To obtain G, adjoin to W ⊕ W ⊕ W all the elements of p^{−n}(v_{n}, 1, 0) and p^{−n}(w_{n}, 0, 1) for n = 1, 2, … (Here by p^{−n}(v_{n}, 1, 0) is meant the set of all elements (xv_{n}, x, 0) for x ∈ p^{−n}.)

Let A be the cyclic submodule of G generated by (1, 0, 0). Let G_{0} = W ⊕ W ⊕ W and for n = 1, 2, … let G_{n} = A + p^{−n}(v_{n}, 1, 0) + p^{−n}(w_{n}, 0, 1). We make the following claims:

(1) For all n, G_{n} is a projective W -module (and free if p is a principal ideal).

(2) (∀n) G_{n} ⊆ G_{n+1} .

(3) (∀n) G_{n} ⊆ pG_{n+1} + A.

(4) G is the direct union of the G_{n}.

(5) A ◁ G.

(6) p-rank G = 1 unless u_{1} ≡ v_{1} ≡ 0 (mod p).

(7) p^{∞}G ≠ 0 ⟺ there exist u, v, w ∈ W such that u + vv_{n} + ww_{n} ≡ 0 (mod p^{n}) for all n.

(8) If p^{∞}G = 0 then every rank-one submodule of G is finitely generated.

proof: (1) It is easy to see that in fact G_{n} = A ⊕ p^{−n}(v_{n}, 1, 0) ⊕ p^{−n}(w_{n}, 0, 1) ≈ W ⊕ p^{−n} ⊕ p^{−n}.

(3) The hypothesis that v_{n+1} ≡ v_{n} (mod p^{n}) can be restated as v_{n} ∈ v_{n+1} + p^{n}, so that (v_{n}, 1, 0) ∈ (v_{n+1}, 1, 0) + p^{n}(1, 0, 0) = (v_{n+1}, 1, 0) + p^{n}A. Thus p^{−n}(v_{n}, 1, 0) ⊆ pp^{−(n+1)}(v_{n+1}, 1, 0)+A ⊆ pG_{n+1} +A. Likewise p^{−n}(w_{n}, 1, 0) ⊆ pG_{n+1} +A. Since G_{n} = A ⊕ p^{−n}(v_{n}, 1, 0) ⊕ p^{−n}(w_{n}, 0, 1) it follows that G_{n} ⊆ pG_{n+1} + A.

(2) Immediate from (3) since by construction A ⊆ G_{n+1} .

(4) Clear.

(5) For all n, A ◁ G_{n} since A is a direct summand of G_{n}. It therefore follows from (4) that A ◁ G.

(6) By (2) and (3) and (4), G = ⋃ G_{n} ⊆ A + p ⋃ G_{n+1} = A + pG. Thus G/pG ≈ A/(A ∩ pG), so G/pG is cyclic. Thus p-rank G ≤ 1. It follows from (5) that G is not p-divisible, therefore p-rank G = 1.

(7) ( ⇒ ): First one should note that since W ⊕ W ⊕ W ⊆ G_{n}, if x ≡ x′, y ≡ y′ and z ≡ z′ (mod p^{n}) then (x, y, z) ∈ p^{n}G ⟺ (x′, y′, z′) ∈ p^{n}G.

Now suppose that there exist u, v, w ∈ W with u + vv_{n} + ww_{n} ∈ p^{n} for all n. Then we claim that (−u, v, w) ∈ p^{n}G for all n, Since −u ≡ vv_{n} + ww_{n} (mod p^{n}) for all n, by the preceding remark it suffices to prove that (vv_{n} + ww_{n}, v, w) ∈ p^{n}G. But in fact (vv_{n} + ww_{n}, v, w) = v(v_{n}, 1, 0) + w(w_{n}, 0, 1) ∈ p^{n}G for all n, so that (−u, v, w) ∈ p^{∞}G.

( ⇐ ): If p^{∞}G ≠ 0 then G_{0} ∩ p^{∞}G ≠ 0, since G_{0} is an essential submodule of G. Thus there exist u, v, w ∈ W and not all in pW such that (u, −v, −w) ∈ p^{n}G for all n. Now for any n, (v_{n}, 1, 0), (w_{n}, 0, 1) ∈ p^{n}G. Thus (u + vv_{n} + ww_{n}, 0, 0) = (u, −v, −w) + v(v_{n}, 1, 0) + w(w_{n}, 0, 1) ∈ p^{n}G ∩ A = p^{n}A, since A ◁ G. Thus u + vv_{n} + ww_{n} ≡ 0 (mod p^{n}) for all n.

(8) Let B be a rank-one submodule of G and let B_{0} = B ∩ G_{0} . Then B_{0} is finitely generated since G_{0} is. Since G is p-reduced, so is B and so B_{p} ≠ QB. Since rank B = 1, QB ≈ Q so QB = ⋃ p^{−n}B_{0p}. Thus there exists a maximum n with p^{−n}B_{0p} ⊆ B_{p}, and for this n, p^{n}B_{p} ⊆ B_{0p}. On the other hand for primes q ≠ p, G_{q} = G_{0q}, and therefore B_{q} = B_{0,q}. Then by Proposition 0.9, p^{n}B ⊆ B_{0} , so p^{n}B is finitely generated and thus B = p^{−n}p^{n}B is finitely generated. ✓

The Pontryagin module will be explored in more depth in Chapter 6.

An alternate (and in fact equivalent) way of producing p-reduced modules with p-rank one is to simply take G = V ∩ Ŵ_{p}, where V is any finite dimensional Q-subspace of QŴ_{p}. In other words, G is simply a pure submodule of Ŵ_{p}. Then G must be p-reduced since Ŵ_{p} is p-reduced, so that 0 < p-rank G ≤ p-rank Ŵ_{p} = 1.

If we look at the Example 1.47, we can note that the conditions imposed on v_{n} and w_{n} ensure that sequences {v_{n}}_{1}^{∞} and {w_{n}}_{1}^{∞} converge to elements v̂ and ŵ in Ŵ_{p}. If we let G be the module constructed in Example 1.47 and let V be the Q-subspace of Ŵ_{p} generated by 1, v̂ and ŵ, then one sees that V ∩ Ŵ_{p} = G_{p}. proof: Taking coordinates with respect to the basis 1, −v̂, −ŵ for V we see that the element with coordinates (v_{n}, 1, 0) is simply (v_{n} − v̂). Then p^{−n}(v_{n}, 1, 0) = p^{−n}(v_{n} − v̂) ⊆ V ∩ Ŵ_{p} since v_{n} ≡ v̂ (mod p^{n}). Likewise the coordinates (w_{n}, 0, 1) correspond to the element w_{n} − ŵ and p^{−n}(w_{n} − ŵ) ⊆ V ∩ Ŵ_{p}. Thus G ⊆ V ∩ Ŵ_{p} and since V ∩ Ŵ_{p} is clearly p-local, G_{p} ⊆ V ∩ Ŵ_{p}. Furthermore, since rank G_{p} = 3, G_{p} is an essential submodule of V ∩ Ŵ_{p}. Therefore by Proposition 1.18 it suffices to prove that G_{p} ◁ V ∩ Ŵ_{p}.

As before, let A be the cyclic submodule of G generated by (1, 0, 0). Then according to our identification, A_{p} = W_{p} ◁ Ŵ_{p}. Thus by Proposition 1.8 to show that G_{p} ◁ V ∩ Ŵ_{p} it suffices to see that G_{p}/A_{p} ◁ (V ∩ Ŵ_{p})/A_{p}. But rank G_{p}/A_{p} = 2 and the construction of G shows that the cosets (0, 1, 0) + A_{p} = (v_{n}, 1, 0) + A_{p} and (0, 0, 1) + A_{p} = (w_{n}, 0, 1) + A_{p} belong to p^{∞}(G_{p}/A_{p}). Thus G_{p}/A_{p} is p-divisible and hence divisible since it is p-local. Thus it is a summand of (V ∩ Ŵ_{p})/A_{p} and a fortiori pure in (V ∩ Ŵ_{p})/A_{p}. This completes the proof that G_{p} = V ∩ Ŵ_{p}. ✓

The ideas behind Example 1.47 and the remarks in the above paragraph will be explored more thoroughly in Chapter 6.

From general module theory we know that the ultimate criterion for indecomposability is the following:

proposition 1.48. G is indecomposable if and only if End G has no idempotents other than 0 and 1.

It might well seem that this proposition would be of limited practical value, since in most cases it would be more difficult to compute End G than to determine whether G has any non-trivial summands. In fact the most common way of spotting idempotents in End G is by locating summands of G. However there are cases where it is easy to see that End G has no proper idempotents. Consider again Example 1.46.

Example 1.46 [continued]. If G is 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 End G ≈ W. In particular, End G has no proper idempotents. proof: Let φ: G → G, and as usual think of φ as a linear transformation QG → QG such that φ(G) ⊆ G. Now if p = p_{1}, p_{2}, p_{3} , then φ(p^{∞}G) ⊆ p^{∞}G. This means that φ must leave invariant the subspaces of QG = Q ⊕ Q generated by (1, 0), (0, 1), and (1, 1). In other words, (1, 0), (0, 1), and (1, 1) are eigenvectors of φ. Now eigenvectors corresponding to distinct eigenvalues are linearly independent, so since these three eigenvectors are not linearly independent we see that they correspond to a single eigenvalue w ∈ Q. It then follows that φ(g) = wg for all g ∈ G, i.e. φ is simply multiplication by w. Furthermore since (w, 0) = w(1, 0) ∈ G it follows that w ∈ p_{1}^{−∞}. Similarly one concludes that w ∈ p_{2}^{−∞} and w ∈ p_{3}^{−∞}. since p_{1}^{−∞} ∩ p_{2}^{−∞} ∩ p_{3}^{−∞} = W, it follows that End G = W. ✓

Since all finite rank torsion free W -modules are direct sums of indecomposable ones, it might seem to follow that the study of finite rank torsion free modules reduces to the problem of understanding all indecomposable ones. This would be true if the direct sum decomposition of a module were unique (up to isomorphism). However this is not the case, so that there are sometimes, as it were, more than one set of building blocks one can put together to get the same end result. There are many entertaining elementary examples of this. Unfortunately, at this point in the book we have a scarcity of theoretical tools, so that constructing such examples involves a fair amount of not totally transparent computation. Therefore at this point we will be content with a single example. To simplify notation, we suppose that W is the ring of integers. However it should be apparent that any dedekind domain with an infinite number of prime ideals would work just as well, after making very minor modifications to the example.

example 1.49. Let W be the ring of integers. Let P_{1} and P_{2} be disjoint infinite sets of prime ideals not containing, say, 5. (For instance one can let P_{1} consist of those p with p ≡ 1 (mod 10) and P_{2} consist of those with p ≡ −1 (mod 10).) Let H_{1} be the submodule of Q ⊕ Q generated by all (p^{−1}, 0) for p ∈ P_{1} and (0, p^{−1}) for p ∈ P_{2} together with (1, 1)/5. Let H_{2} be generated by the same set of elements except with (1, 1)/5 replaced with (1, 2)/5. Let A be the submodule of Q generated by all p^{−1} for p ∈ P_{2} . Then

(1) H_{1} and H_{2} are not isomorphic.

(2) A ⊕ H_{1} ≈ A ⊕ H_{2} .

proof: (1) Let B = {(x, 0) | x ∈ Q, (x, 0) ∈ H_{1}}. More briefly, B = H_{1} ∩ (Q ⊕ 0). Then B is generated by {(p^{−1}, 0) | p ∈ P_{1}}. From this we see easily that B = H_{2} ∩ (Q ⊕ 0) as well. We claim that if φ is an isomorphism from H_{1} to H_{2} then φ must restrict to an automorphism of B. In fact, since (p^{−1}, 0) ∈ p′H_{1} for all p′ ≠ p ∈ P_{1} it follows that φ(p^{−1}, 0) ∈ p′H_{2} for all p′ ∈ P_{1} . Since P_{1} is infinite this cannot be the case if φ(p^{−1}, 0) = (y, z) with z ≠ 0, because (y, z) ∈ p′H_{2} for p′ ∈ P_{1} implies z ∈ p′W_{p′}, in other words of p′ divides the numberator of z (since P_{2} ∩ P_{1} = ∅). Since there are an infinite number of such p′ this forces z = 0. Therefore φ(1, 0) = (x, 0) for some x ≠ 0 ∈ Q, so φ(B) ⊆ (Q ⊕ 0) ∩ H_{2} = B. Applying the same consideration to φ^{−1} shows that if φ is to be an automorphism then necessarily φ(B) = B. But if pursuing this same reasoning even further shows that in fact φ(1, 0) = (u, 0) where u is invertible in W. (I.e. u = ±1. Recall that W = ℤ.) In fact, W ⊕ 0 can be characterized as the submodule of B consisting of those (x, 0) ∈ B such that (x, 0) ∈ pB for all p ∈ P_{1} . From this it is clear that if φ is an automorphism of B then φ(W ⊕ 0) = W ⊕ 0, which is the same as saying that φ(1, 0) = (u, 0) where u is invertible in W. Likewise we get φ(0, 1) = (0, v) for some invertible v ∈ W.

Thus φ(1, 1)/5 = (u, v)/5, where u, v = ±1. But if (u, v)/5 ∈ H_{2} then 2v^{−1}(u, v)/5 = (2v^{−1}u, 2)/5 ∈ H_{2} and so (2v^{−1}u, 2) − (1, 2) = (2v^{−1}u − 1, 0) ∈ 5H_{2} ∩ (W ⊕ 0) = 5B ∩ (W ⊕ 0) = 5W ⊕ 0 since 5 ∉ P_{1} . But since u, v = ±1, 2v^{−1}u − 1 = −3, 1 ∉ 5ℤ, a contradiction. Thus there cannot be an isomorphism φ from H_{1} to H_{2} .

(2) Let us change notation at this point, identifying H_{1} ⊕ A as a submodule of Q ⊕ Q ⊕ Q, with H_{1} ⊆ Q ⊕ Q ⊕ 0. Thus H_{1} ⊕ A is generated by all (p^{−1}, 0, 0) for p ∈ P_{1} , (0, p^{−1}, 0) and (0, 0, p^{−1}) for p ∈ P_{2} together with (1, 1, 0)/5. Now let H′ be the submodule of Q ⊕ Q ⊕ Q generated by all (p^{−1}, 0, 0) for p ∈ P_{1} and p^{−1}(0, 3, 5) for p ∈ P_{2} together with (3, 3, 5)/5. Since (3, 3, 5)/5 = (3, 3, 0)/5 + (0, 0, 1) ∈ H_{1} ⊕ A, clearly H′ ⊆ H_{1} ⊕ A. Furthermore H′ ≈ H_{2} where the isomorphism is given by a map (1, 0, 0) ↦ (1, 0, 0), (0, 3, 5) ↦ (0, 1, 0), the crucial point being that (3, 3, 5)/5 ↦ (3, 1, 5)/5 = 3(1, 2, 0)/5 + (0, −1, 1) ∈ H_{1} ⊕ A. Now let A′ be the submodule of Q ⊕ Q ⊕ Q generated by p^{−1}(0, 1, 2) for all p ∈ P_{2} . It is fairly clear that A′ ∩ H′ = 0. We claim that H′ ⊕ A′ = H_{1} ⊕ A. It suffices to see that H_{1}, A ⊆ H′ ⊕ A′. In fact, for p ∈ P_{1} , p^{−1}(1, 0, 0) ∈ H′, and for p ∈ P_{2} , p^{−1}(0, 1, 0) = p^{−1}[2(0, 3, 5)−5(0, 1, 2)] ∈ H′ ⊕A′ and p^{−1}(0, 0, 1) = p^{−1}[−(0, 3, 5) + 3(0, 1, 2)] ∈ H′ ⊕ A′. And finally, (1, 1, 0)/5 = 2(3, 3, 5)/5 − (1, 0, 0) − (0, 1, 2). Thus H_{1} ⊕ A = H′ ⊕ A′ ≈ H_{2} ⊕ A.

It seems only fair to point out here that calculations of this sort can usually be avoided by use of the tools that will be developed in the following three chapters, especially Chapter 4. Nonetheless it is worth studying this example carefully before acquiring such tools. A thorough understanding of calculations such as these provides (in your author’s opinion) insights that slick proofs using sophisticated theorems do not.

The comment preceding Example 1.49 that this construction requires a dedekind domain W with an infinite number of primes is very significant. We will see in Chapter 11, in fact, that if Spec W is finite then every finite rank torsion free W -module A has the cancellation property: If A ⊕ H_{1} ≈ A ⊕ H_{2} then H_{1} ≈ H_{2} .

Without going into more calculations, I would simply like to point out now that there are a plethora of such examples, violating every obvious conjecture about the uniqueness of direct sum decompositions. There exist modules G and H for instance such that G ⊕ G ≈ H ⊕ H but G ≉ H. For any r > 2 one can find G and H such that G and H are indecomposable but G ⊕ H can be broken up into the direct sum of r non-trivial modules. Such examples can be constructed over any dedekind domain with an infinite number of prime ideals. Even over a local ring pathological direct sum decompositions usually exist, although they are much more complicated than the example above.

COMPLETE MODULES. Recall that for a given prime p, the p-adic completion Ĝ_{p} of G was defined in Chapter 0 as a certain p-adically closed submodule of ∏ G/p^{n}G.

lemma 1.50. If p is a (non-trivial) prime and H is a submodule of a W -module G, then H is p-adically dense in G if and only if G/H is p-divisible.

proof:

proposition 1.51. If θ: G → Ĝ_{p} is the canonical map, then Ker θ = p^{∞}G, θ(G) is p-pure in Ĝ_{p}, and Ĝ_{p}/θ(G) is divisible.

proof:

proposition 1.52. Ŵ_{p} is a discrete valuation ring with maximal ideal generated by p and W_{p} ◁ Ŵ_{p}.

proof:

proposition 1.53. For any G, Ĝ_{p} is reduced but is q-divisible for every prime q ≠ p.

proof:

proposition 1.54. If G is a finite rank torsion free W -module then Ĝ_{p} ≈ Ŵ_{p} ⋆ G.

proof:

If W is the ring of integers and G has finite rank, then all G/p^{n}G are finite and so Ĝ_{p} is compact by Tychonof’s Theorem. In general, we can only say that the G/p^{n}G have finite length. But it still seems intuitively plausible that Ĝ_{p} would have all the algebraic properties that one would expect of a compact module. In fact, abelian group theorists will recognize that Ĝ_{p} belongs to the class of “algebraically compact” modules, which are characterized by the following proposition.

proposition 1.55. If Ĝ_{p} is a pure submodule of any (not necessarily torsion free) W -module X, then Ĝ_{p} is a summand of X. proof: ****

One should not expect that Ĝ_{p} is a finite rank W -module, except in the trivial case where W = Ŵ_{p}. However we have the following:

proposition 1.56. Let G be a finite rank torsion free W -module.

(1) Ĝ_{p} is torsion free as a W -module.

(2) Ĝ_{p} is a finite rank torsion free Ŵ_{p}-module.

(3) The rank of Ĝ_{p} over Ŵ_{p} equals rank G.

(4) Ĝ_{p}/d(Ĝ_{p}) is a free Ŵ_{p}-module with rank (over Ŵ_{p}) equal to p-rank G.

proof: ****

At one time, as reflected in [Fuchs], pathological direct sum decompositions such as Example 1.49 were the mainstay of the theory of finite rank torsion free abelian groups. Now their main interest (in your author’s humble opinion) lies in their entertainment value. However one should be aware that a large part of the theory to which this book is devoted was originally developed as a way of understanding such pathological direct sum decompositions or of finding contexts where they can be avoided.

In Math Reviews abelian group theory sits ensconced among all the other varieties of group theory, having no more in common with its neighbors than a handful of words such as “group,” “subgroup,” and “homomorphism.”

Kaplansky, in his “little red book” [Kapl], asserted that abelian group theory is really the study of modules over principal ideal domains, and since then most abelian group theorists tend to feel more at home with commutative ring theory than with group theory in general.

On the other hand, if abelian group theory is a branch of commutative ring theory it is certainly a highly specialized branch. In the study of torsion groups the ring theorist finds more concepts that are strange to him than concepts which are familiar, and this cannot be remedied by a mere change in terminology.

The downside of the influence of Kaplanksy’s statement was that it promoted the idea that it was unnecessary and perhaps even pretentious to state results for modules over principal ideal domains, since results for abelian groups would automatically carry over. To some extent this attitude was justified as long as abelian group theory was largely the study of torsion modules, although even in [Kapl] it was not totally clear how to translate the proof of Ulm’s Theorem, using Kaplanksy’s back-and-forth method, into the context of modules over an uncountable ring.

In the realm of torsion free modules, however, the idea that results for abelian groups will automatically hold for modules over arbitrary principal ideal domains is simply not tenable. The theory of finite rank torsion free abelian groups is full of results that depend on countability, or on having characteristic zero, or working over a ring whose quotient field is a perfect field, as well as proofs using quite specialized results from number theory. Among these is your author’s theorem that for a finite rank torsion free abelian group G there are, up to isomorphism, only finitely many groups isomorphic to summands of G (see Chapter 11.) This is not true for modules over an arbitrary principal ideal domain.

The transition from modules over principal ideal domains to modules over dedekind domains is actually a much smaller leap than that from the integers to an arbitrary PID. Moreover in the process of reformulating theorems and proofs to be valid over dedekind domains, one sees these results in a new and – in your author’s opinion – “more correct” way.

And one becomes more aware of the fact that finite rank torsion free abelian group theory is moving away from abelian group theory in general in much the same fashion that abelian group theory has moved away from general group theory. Unlike the theory of torsion groups, the theory of finite rank torsion free modules is becoming something that fits in fairly well with the mainstream of commutative ring theory.

The only thing that has prevented this from being apparent has been terminology. In this book, I have used for the most part terminology familiar to commutative ring theorists. Abelian groups theorists will have to make some minor adjustments, which may be facilitated by the brief table below.