1. FUNDAMENTALS. E. L. Lady. May 14, 1998

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1 1 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.

2 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 whichwedenotebyp 1, determined by the property that p 1 p = W.Thusfor all n 0wegetp n G QG. We will define p to be 1 p n and p G to be 1 p n G.(Ifpis a principal ideal p =(p 0 ), then p G is simply the localization of G with respect to the multiplicative set S = {p n 0 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 Wx = 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 givesanisomorphismp n G Gwhich 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.butp 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) ofhom(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 andplayavitalroleinthestudyof 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.itisan 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. 2

3 3 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, sog=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,sothatg 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,thena =p 1 p n where the p i are not necessarily distinct prime ideals. If p i G = G for each i, thenag=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)sothatp 1 x p G,sox p(p G). note: free. Part (2) of Proposition 1.3 is not valid without the assumption that G is torsion We say that G is reduced if d(g) =0 andp-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.

4 proposition 1.5. There are natural isomorphisms Hom(Q, G) d(g) and Hom(p,G) p Ggiven 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) givenbyϕ 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,andsoG/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,sothisisequivalenttoh p G p for all p. (1) (2): Suppose h H ag for some non-trivial ideal a.thena 1 h Gand 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. 4

5 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.thenh Gif 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,sothath 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 )forsomew W and h aw n, so that for all i, ww i aw = a.sincew 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. ThenK 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,thenh 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 5

6 6 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. ThenG QH = G ( QH i )= (G QH i )= Hi =H. proposition If G 1,...,G n are submodules of a finite dimensional Q-space V,then Gi is isomorphic to a pure submodule of G i. proof: **** definition 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 If H G we say that H is p-pure in G if p/ Ass G/H. proposition 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,whichsays that there is no g G with g/ Hand pg H. I.e. no non-trivial element of G/H is annihilated by p. Thusp/ Ass G/H. (1) (4): If p / Ass G/H then p / Ass(G/H) p =AssG 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.butifqis 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 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,sog p S 1 G p.

7 7 proposition (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 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 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 =rankg. (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 =dimqg rank H =rankg. (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 Gthen k has a non-trivial multiple wk in H.Then wk 0 H K.

8 8 (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.ThenrankH=rankH =rankg. proposition 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.ButHis 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 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. Ifwethenletwbe a common denominator for finite set of generators for G, wehavewg W, i.e wg is an ideal. Since G is torsion free, G wg and the result follows. Now if rank G>1letV 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.

9 9 corollary 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 A projective W -module is finitely generated if and only if it has finite rank. proof: **** corollary If W is local and A is a W -module and rank A = 1,thenAis 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. NowifA contains fractions with denominators which are arbitrarily high powers of p, then1 p A A,soA=p Asince 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, thenais finitely generated, hence projective, and by Proposition 1.21 A is free. corollary A W -module G is torsion free if and only if it is flat. proof: If w 0 Wthen the map G w G can be obtained as µ G: W G W G where µ: W W is multiplication by w. SinceWis 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 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,wherer=rankG. proof: We first show that if I is a non-trivial ideal of W then length I/aI =lengthw/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

10 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 rlength W/a. But M is a homomorphic image of F/aF,solengthM rlength 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/ Mand x together with M generates a strictly larger finitely generated submodule of G/aG, contradicting the maximality of M. Hence length G/aG =lengthm rlength W/a. 10 proposition 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.NowifT=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 = lengthg/pg. These two concepts have similar behavior. proposition (1) rank(g H) = rankg +rankh,and p-rank(g H) = p-rank G +p-rankh. (2) If H G then rank G = rankh +rankg/h and p-rank G =p-rankh+p-rankg/h. (3) If H is an essential submodule of G then rank H = rankg and p-rank H p-rank G. (4) p-rank G =p-rankg p. (5) p-rank G rank G. (6) For any torsion free G and H, rank G H = (rankg)(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.

11 (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-rankh+p-rankg/h. (1) A special case of (2). (4) G/pG G p /pg p. (5) By Proposition 1.24 p-rank G =lengthg/pg (rank G)(length W/p) =rankg, 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 =dimqg, rankg=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. 11 proposition The following conditions are equivalent: (1) p-rank G =rankg (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-rankg. Thus if p-rankg =rankgthen rank H p-rank H =p-rankg=rankg rank H.ThusHisan 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,soifg 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-rankg p. If G p is a free W p -module then it is a direct sum of r copies of W p,wherer =rankg. Then p-rank G p = r(p-rank W p )=r=rankg. 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.

12 proposition 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,thenϕis an automorphism of G. proof: (1) Ker ϕ = 0 rank Ker ϕ = 0 rank ϕ(g) = rankg 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 [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 a 1 n =a 1 1 a 1 n. 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. Butif a 1 +a 2 =W then a 1 1 a 1 2 = a 1 1 a 1 2 (a 2 + a 1 )=a 1 1 +a 1 2. For instance, if W = Z then the submodule of Q generated by 1/3 and1/5isthe same as the submodule generated by 1/15. In fact, 1 3 = 5 15, 1 5 = 3 15,and 1 15 = example 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 1 n 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 =1forallp. Thus one cannot distinguish A and W on the basis of rank and p-rank, and yet if SpecW is infinite A and W are not isomorphic. proof: (1) Obviously A is the direct union of those submodules of Q of the n,wherep 1,...,p n are distinct primes. But by Lemma 1.29 form p p 1 p p 1 n =p 1 1 p 1 n. (2) If Spec W is infinite, there is an infinite strictly ascending chain W p 1 1 p 1 1 p 1 2 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 Awould be p-divisible, i.e. A = p A. Since W A this would imply p W = p A. Butin 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. 12

13 definition If L is a left ideal in a finite rank torsion free ring R, wesaythatlis nilpotent if L r =0forsomer, 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, thenil 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 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 =rankr 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.SinceNis 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 =dimqr, it is known from the theory of finite dimensional algebras that Ñ r =0. Thus (R Ñ)r =0,soR Ñ is a nilpotent ideal, so R Ñ N = nil rad R, andso 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. ThusQL 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. 13 proposition 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, thenr nil rad R.

14 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. Thuswlog 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 = 0orLcontains 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 If rank G = r,thenend 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 =rankendg and N =nilradendg,thenn s =0. Wenow see rhat furthermore N r =0,wherer=rankG. 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 Let G be a non-trivial finite rank torsion free module and let N =nilradendg. Let r =rankg. Then (1) rankng < 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. 14

15 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, sothatn acts on QG. NowifrankNG =rankg 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 =rankg. Thus N r G = 0. But clearly this implies N r =0. (3) If ϕ EndG, thenϕn N because N is an ideal in EndG. 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 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 =0forsomek.Choose eso 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.Thenncommutes 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 andifk>1then2(k 1) = 2k 2 2k k = k so that (n 2 ) k 1 =0. In other words, if k>1then[(e+n) 2 (e+n)] k 1 =0. Since e+n e r (mod N) this contradicts the minimality of k. Thus k =1 sothat e 2 e=0 ande 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. 15

16 16 lemma For any finitely generated submodule a of Q, (1) Hom(aG, ah) =Hom(G, H). (2) Hom(G, ah) =ahom(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) anda 1 yields Hom(G, ah) =aa 1 Hom(G, ah) a Hom(G, a 1 ah) =ahom(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) 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, pn H). But this simply says that if ϕ Hom(G, H) thenϕ(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) rank G H =(rankg)(rank H). (2) p-rank(g H) =(p-rankg)(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) =(rankg)(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)

17 17 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) lengthhom(g/pg, H/pH) = dimhom(g/pg, H/pH) =(dimg/pg)(dim H/pH) =(p-rankg)(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 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 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) thenpϕ(g) 0.Thuspϕ 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 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) isp-pure in Hom(G, S 1 H) for all primes p. Ifp 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) isp-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).

18 18 proposition For any finite rank torsion free modules G, H,andK, 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 andassuchiswellknowntobean 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) 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

19 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) 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-rankk = 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 = 0orK = 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. Inthe 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 Let p 1, p 2,andp 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 If G = H K and 19

20 20 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, sayp 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 =1andyet 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 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 )andw 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(andfreeifpis 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.

21 (7) ( ): First one should note that since W W W G n,ifx 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, sothat( 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. Nowforanyn,(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.ThenB 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,sop 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 Ŵp,whereV is any finite dimensional Q-subspace of QŴp. In other words, G is simply a pure submodule of Ŵ p. Then G must be p-reduced since Ŵp is p-reduced, so that 0 < p-rank G p-rank Ŵp =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 ˆv and ŵ in Ŵp.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, ˆv and ŵ, then one sees that V Ŵp = G p. proof: Taking coordinates with respect to the basis 1, ˆv, ŵ for V we see that the element with coordinates (v n, 1, 0) is simply (v n ˆv). Then p n (v n,1,0) = p n (v n ˆv) V Ŵp since v n ˆv (mod p n ). Likewise the coordinates (w n, 0, 1) correspond to the element w n ŵ and p n (w n ŵ) V Ŵp. Thus G V Ŵp and since V Ŵp is clearly p-local, G p V Ŵp. Furthermore, since rank G p =3,G p is an essential submodule of V Ŵp. Therefore by Proposition 1.18 it suffices to prove that G p V Ŵp. As before, let A be the cyclic submodule of G generated by (1, 0, 0). Then according to our identification, A p = W p Ŵp. Thus by Proposition 1.8 to show that G p V Ŵp it suffices to see that G p /A p (V Ŵp)/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 Ŵp)/A p and a fortiori pure in (V Ŵp)/A p. This completes the proof that G p = V Ŵp. 21