aa r X i v : . [ m a t h . A T ] J a n ALGEBRAIC THEORY OF ABELIAN GROUPS
N. P. STRICKLAND
Contents
1. Introduction 12. Exactness and splittings 13. Products and coproducts 64. Torsion groups 75. Finitely generated abelian groups 96. Free abelian groups and their subgroups 127. Tensor and torsion products 178. Ext groups 279. Localisation 3710. Colimits of sequences 4011. Limits and derived limits of towers 4612. Completion and derived completion 51References 581.
Introduction
This document aims to give a self-contained account of the parts of abelian group theory that are mostrelevant for algebraic topology. It is almost purely expository, although there are some slightly unusualfeatures in the treatment of tensor products, torsion products and Ext groups. The book [1] is a goodreference for Sections 11 and 12. Earlier sections are more standard and can be found in very many sources.2.
Exactness and splittings
Definition 2.1.
Consider a sequence A f −→ A −→ · · · f r − −−−→ A r of abelian groups and homomorphisms. Wesay that the sequence is exact at A i if image( f i − ) = ker( f i ) ≤ A i (which implies that f i ◦ f i − = 0). Wesay that the whole sequence is exact if it is exact at A i for 0 < i < r .Next, we say that a sequence A f −→ B g −→ C is short exact if it is exact, and also f is injective and g issurjective. Remark 2.2.
One can easily check the following facts.(a) A sequence A f −→ B −→ C is exact iff f is surjective. In particular, a sequence A f −→ B −→ f is surjective.(b) A sequence A −→ B g −→ C is exact iff g is injective. In particular, a sequence 0 −→ B g −→ C is exact iff g is injective.(c) A sequence A −→ B g −→ C −→ D is exact iff g is an isomorphism.(d) A sequence 0 −→ A f −→ B g −→ C −→ A f −→ B g −→ C is short exact.(e) Suppose we have an exact sequence A f −→ B g −→ C h −→ D k −→ E. Date : January 29, 2020. hen g induces a map from cok( f ) = B/f ( A ) to C , and h can be regarded as a map from C toker( k ), and the resulting sequence cok( f ) g −→ C h −→ ker( k )is short exact.(f) If A f −→ B g −→ C is short exact, then f induces an isomorphism A → f ( A ) and g induces anisomorphism B/f ( A ) → C . Thus, if A , B and C are finite we have | B | = | f ( A ) | . | B/f ( A ) | = | A || C | .Similarly, if A and C are free abelian groups of ranks n and m , then B is a free abelian group ofrank n + m . Proposition 2.3 (The five lemma) . Suppose we have a commutative diagram as follows, in which the rowsare exact, and p , p , p and p are isomorphisms: A A A A A B B B B B . f p ≃ f p ≃ f p f p ≃ p ≃ g g g g Then p is also an isomorphism.Proof. First suppose that a ∈ A and p ( a ) = 0. It follows that p f ( a ) = g p ( a ) = g (0) = 0, but p is an isomorphism, so f ( a ) = 0, so a ∈ ker( f ). The top row is exact, so ker( f ) = image( f ), sowe can choose a ∈ A with f ( a ) = a . Put b = p ( a ) ∈ B . We then have g ( b ) = g p ( a ) = p f ( a ) = p ( a ) = 0, so b ∈ ker( g ). The bottom row is exact, so ker( g ) = image( g ), so we can choose b ∈ B with g ( b ) = b . As p is an isomorphism, we can now put a = p − ( b ) ∈ A . We then have p f ( a ) = g p ( a ) = g ( b ) = b = p ( a ). Here p is an isomorphism, so it follows that f ( a ) = a .We now have a = f ( a ) = f f ( a ). However, as the top row is exact we have f f = 0, so a = 0. Weconclude that p is injective.Now suppose instead that we start with an element b ∈ B . Put b = g ( b ) ∈ B and a = p − ( b ) ∈ A .We then have p f ( a ) = g p ( a ) = g ( b ) = g g ( b ) = 0 (because g g = 0). As p is an isomorphism,this means that f ( a ) = 0, so a ∈ ker( f ). As the top row is exact we have ker( f ) = image( f ), so wecan choose a ∈ A with f ( a ) = a . Put b ′ = b − p ( a ) ∈ B . We have g ( b ′ ) = g ( b ) − g p ( a ) = b − p f ( a ) = b − p ( a ) = 0, so b ′ ∈ ker( g ) = image( g ). We can thus choose b ′ ∈ B with g ( b ′ ) = b ′ .Now put a ′ = p − ( b ′ ) ∈ A and a ′ = f ( a ′ ) ∈ A . We find that p ( a ′ ) = p f ( a ′ ) = g p ( a ′ ) = g ( b ′ ) = b ′ = b − p ( a ), so p ( a + a ′ ) = b . This shows that p is also surjective, and so is an isomorphism asclaimed. (cid:3) Proposition 2.4.
Suppose we have a commutative diagram as follows, in which the rows are short exactsequences:
A B CA ′ B ′ C ′ jf qg hj ′ q ′ Then there is a unique homomorphism δ : ker( h ) → cok( f ) such that δ ( q ( b )) = a ′ + f ( A ) whenever g ( b ) = j ′ ( a ′ ) . Moreover, this fits into an exact sequence → ker( f ) j −→ ker( g ) q −→ ker( h ) δ −→ cok( f ) j −→ cok( g ) q −→ cok( h ) → . Proof. A snake for the above diagram is a list ( c, b, a ′ , a ) such that(1) c ∈ ker( h ) ≤ C (2) b ∈ B with qb = c (3) a ′ ∈ A ′ with j ′ a ′ = gb ∈ B ′ (4) a is the image of a ′ in cok( f ).It is easy to see that the snakes form a subgroup of ker( h ) × B × A ′ × cok( f ). We claim that for all c ∈ ker( h ),there exists a snake starting with c . Indeed, as q is surjective, we can choose b ∈ B satisfying (2). Then ′ g ( b ) = hq ( b ) = h ( c ) = 0, so g ( b ) ∈ ker( q ′ ) = img( j ′ ), so we can choose a ′ ∈ A ′ satisfying (3). Finally, wecan define a to be the image of a ′ in cok( f ), so that (4) is satisfied: this gives a snake as required. Next, weclaim that any two snakes starting with c have the same endpoint. By subtraction we reduce to the followingclaim: if (0 , b, a ′ , a ) is a snake, then a = 0, or equivalently a ′ ∈ img( f ). Indeed, condition (2) says that b ∈ ker( q ) = img( j ), so we can find a ∈ A with b = ja . Now j ′ ( f a − a ′ ) = j ′ f a − gb = gja − gb = gb − gb = 0,and j ′ is injective, so f a = a ′ as required. This allows us to construct a map δ : ker( h ) → cok( f ) as follows:we define δ ( c ) to be the endpoint of any snake starting with c .We now need to check exactness of the resulting sequence.(1) As j : A → B is injective, it is clear that the restricted map j : ker( f ) → ker( g ) is also injective.(2) As the composite A j −→ B q −→ C is zero, the same is true of the restricted composite ker( f ) j −→ ker( g ) q −→ ker( h ). Moreover, suppose we have b ∈ ker( g ) with qb = 0. By the original exactness assumption wecan find a ∈ A with ja = b . Now j ′ f a = gja = gb = 0 but j ′ is injective so f a = 0 so a ∈ ker( f ).Thus, b is in the image of the map j : ker( f ) → ker( g ).(3) Suppose we have b ∈ ker( g ). Then ( qb, b, ,
0) is a snake starting with qb , showing that δqb = 0.Conversely, suppose that c ∈ ker( h ) with δc = 0, so there exists a snake ( c, b, a ′ , a ′ ∈ img( f ), say a ′ = f a for some a ∈ A . Put b ′ = b − ja ∈ B . Snakecondition (2) gives qb = c but also qj = 0 so qb ′ = c . On the other hand, snake condition (3) gives gb = j ′ a ′ = j ′ f a = gja so gb ′ = 0. This means that c is in the image of the map q : ker( g ) → ker( h ).(4) Suppose we have c ∈ ker( h ) with δ ( c ) = a . This means that there is a snake ( c, b, a ′ , a ). We claimthat the induced map j ′ : cok( f ) → cok( g ) sends a to 0, or equivalently that j ′ a ′ ∈ img( g ). This isclear because j ′ a ′ = gb by the snake axioms. Conversely, suppose that a ∈ cok( f ) and that a mapsto 0 in cok( g ). This means that we can find a ′ ∈ A representing a and that ja ′ lies in the image of g , say ja ′ = gb for some b ∈ B . If we put c = qb ∈ C we find that hc = hqb = q ′ gb = q ′ j ′ a ′ = 0, so c ∈ ker( h ). By construction we see that ( c, b, a ′ , a ) is a snake so a ∈ img( δ ).(5) As the composite A ′ j ′ −→ B ′ q ′ −→ C ′ is zero, the same is clearly true for the induced maps cok( f ) → cok( g ) → cok( h ). Conversely, suppose we have an element b ∈ cok( g ) that maps to zero in cok( h ).We can choose b ′ ∈ B ′ representing b , and then q ′ b ′ must lie in img( h ), say q ′ b ′ = hc . As q issurjective we can choose b ∈ B with qb = c . This gives q ′ b ′ = hqb = q ′ gb , so the element b ′ − gb liesin ker( q ′ ), which is the same as img( j ′ ). We can therefore choose a ′ ∈ A ′ with b ′ = gb + j ′ a ′ . If welet a denote the image of a ′ in cok( f ), we find that b = j ′ a in cok( g ).(6) Finally, suppose we have c ∈ cok( h ). We can then choose a representing element c ′ ∈ C ′ . As q ′ issurjective we can choose b ′ ∈ B ′ with q ′ b ′ = b , then we can put b = [ b ′ ] ∈ cok( g ). We find that q ′ b = c . This shows that q ′ : cok( g ) → cok( h ) is surjective. (cid:3) Definition 2.5. A split short exact sequence is a diagram A B C i pr s where pi = 0 rs = 0 ri = 1 A ps = 1 C ir + sp = 1 B . This can also be displayed as
A CBA C i pr s Example 2.6.
Given abelian groups A and C , there is a split short exact sequence A ⊕ C C i ′ p ′ r ′ s ′ given by i ′ ( a ) = ( a, p ′ ( a, c ) = cs ′ ( c ) = (0 , c ) r ′ ( a, c ) = a. The above example is essentially the only example, as we see from the following result:
Proposition 2.7.
Suppose we have a split short exact sequence
A B C i pr s
Then there is an isomorphism f : B → A ⊕ C given by f ( b ) = ( r ( b ) , p ( b )) with inverse f − ( a, c ) = i ( a ) + s ( c ) .Moreover, the diagram A B CA A ⊕ C C i pr f ≃ si ′ p ′ r ′ s ′ commutes in the sense that f i = i ′ f s = s ′ pf = p ′ rf = f ′ . Proof.
We can certainly define homomorphisms B f −→ A ⊕ C g −→ B by f ( b ) = ( r ( b ) , p ( b )) and g ( a, c ) = i ( a ) + s ( c ). We then have gf ( b ) = ( ir + sp )( b ) = b and f g ( a, c ) = ( ri ( a ) + rs ( c ) , pi ( a ) + ps ( c )) = ( a, c ) so f and g are mutually inverse isomorphisms. We also have f i ( a ) = ( ri ( a ) , pi ( a )) = ( a,
0) = i ′ ( a ), and theequations f s = s ′ , pf = p ′ and rf = f ′ can be verified equally easily. (cid:3) Our terminology is justified by the following observation:
Lemma 2.8. If A B C i pr s is a split short exact sequence, then A i −→ B p −→ C is a short exact sequence.Proof. Suppose that i ( a ) = 0. As ri = 1 A we have a = r ( i ( a )) = r (0) = 0. This shows that ker( i ) = 0, so i isinjective. Next, we have ps = 1 C , so for all c ∈ C we have c = p ( s ( c )) ∈ image( p ); so p is surjective. We alsohave pi = 0, so image( i ) ≤ ker( p ). Finally, we have ir + sp = 1 B , so for b ∈ B we have b = i ( r ( b )) + s ( p ( b )).If b ∈ ker( p ) this reduces to b = i ( r ( b )) ∈ image( i ), so ker( p ) ≤ image( i ) as required. (cid:3) Proposition 2.9.
Let A i −→ B p −→ C be a short exact sequence. (a) For any map r : B → A with ri = 1 A , there is a unique map s : C → B such that ( i, p, r, s ) gives asplit short exact sequence. (b) For any map s : C → B with ps = 1 C , there is a unique map r : B → A such that ( i, p, r, s ) gives asplit short exact sequence.Proof. We will prove (a) and leave the similar proof of (b) to the reader. Define f = 1 − ir : B → B . As ri = 1 we have f i = i − i ( ri ) = 0, so f vanishes on image( i ), which is the same as ker( p ). We therefore have awell-defined map s : C → B given by s ( c ) = f ( b ) for any b with p ( b ) = c . This means that sp = f = 1 − ir , orin other words 1 B = ir + sp . We also have pi = 0 so psp = p (1 − ir ) = p , so ( ps − p = 0. As p is surjective his implies that ps − ps = 1 C . Finally, we have ri = 1 so rsp = r (1 − ir ) = r − ( ri ) r = 0 but p is surjective so rs = 0. Thus, all the conditions for a split short exact sequence are verified. If s ′ : C → B is another map giving a split short exact sequence then we can subtract the equations ir + sp = 1 and ir + s ′ p = 1 to get ( s − s ′ ) p = 0 but p is surjective so s = s ′ ; this shows that s is unique. (cid:3) Proposition 2.10.
Let B be an abelian group, and let A and C be subgroups such that B = A + C and A ∩ C = 0 . Let i : A → B and s : C → B be the inclusion maps. The there is a unique pair of homomorphisms A r ←− B p −→ C giving a split short exact sequence.Proof. Consider an element b ∈ B . As B = A + C we can find ( a, c ) ∈ A ⊕ C such that b = a + c . Supposewe have another pair ( a ′ , c ′ ) ∈ A ⊕ C with b = a ′ + c ′ . We put x = a − a ′ , and by rearranging the equation a + c = a ′ + c ′ we see that x = c ′ − c . The first of these expressions shows that x ∈ A , and the second that x ∈ C . As A ∩ C = 0 this means that x = 0, so a = a ′ and c = c ′ . Thus, the pair ( a, c ) is unique, so wecan define maps A r ←− B p −→ C by r ( b ) = a and p ( b ) = c . It is straightforward to check that these give a splitshort exact sequence. (cid:3) Proposition 2.11.
Let B be an abelian group, and let e : B → B be a homomorphism with e = e . Then image( e ) = ker(1 − e ) and ker( e ) = image(1 − e ) and B = image( e ) ⊕ image(1 − e ) .Proof. First, if b ∈ image( e ) then b = e ( a ) for some a , so (1 − e )( b ) = e ( a ) − e ( a ) = 0, so b ∈ ker(1 − e ).Conversely, if b ∈ ker(1 − e ) then b − e ( b ) = 0 so b = e ( b ) ∈ image( e ). This shows that image( e ) = ker(1 − e )as claimed. Now put f = 1 − e . We then have f = 1 − e + e = 1 − e + e = f , so f is another idempotentendomorphism of B . We can thus apply the same logic to see that image( f ) = ker(1 − f ), or in other wordsimage(1 − e ) = ker( e ).Now consider an arbitary element b ∈ B . We can write b as e ( b )+(1 − e )( b ), so b ∈ image( e )+image(1 − e );this shows that B = image( e )+image(1 − e ). Now suppose that b ∈ image( e ) ∩ image(1 − e ) = ker(1 − e ) ∩ ker( e ).This means that (1 − e )( b ) = e ( b ) = 0, so b = e ( b ) = 0. This means that image( e ) ∩ image(1 − e ) = 0, so thesum is direct. (cid:3) We now give a useful application of Proposition 2.7 to the theory of additive functors. We recall thedefinition:
Definition 2.12. A covariant functor from abelian groups to abelian groups is a construction that gives anabelian group F ( A ) for each abelian group A , and a homomorphism f ∗ : F ( A ) → F ( B ) for each homomor-phism f : A → B , in such a way that:(a) For identity maps we have (1 A ) ∗ = 1 F ( A ) for all A .(b) For homomorphisms A f −→ B g −→ C we have ( gf ) ∗ = g ∗ f ∗ : F ( A ) → F ( C ).We say that F is a additive if ( f + f ) ∗ = ( f ) ∗ + ( f ) ∗ for all f , f : A → B . Example 2.13.
Fix an integer n >
0. We can then define an additive functor F by F ( A ) = A [ n ] = { a ∈ A | na = 0 } , and another additive functor G by G ( A ) = A/nA . In both cases the homomorphisms f ∗ arejust the obvious ones induced by f . Proposition 2.14.
Let F be an additive covariant functor as above. Then for any abelian groups A and C we have an natural isomorphism f : F ( A ⊕ C ) → F ( A ) ⊕ F ( C ) given by f ( b ) = ( r ′∗ ( b ) , p ′∗ ( b )) with inverse f − ( a, c ) = i ′∗ ( a ) + s ′∗ ( c ) .Proof. Put B = A ⊕ C , and recall that 1 B = i ′ r ′ + s ′ p ′ . As F is an additive functor we have1 F ( B ) = (1 B ) ∗ = ( i ′ r ′ ) ∗ + ( s ′ p ′ ) ∗ = i ′∗ r ′∗ + s ′∗ p ′∗ . In the same way the equations r ′ i ′ = 1, p ′ s ′ = 1, p ′ i ′ = 0 and r ′ s ′ = 0 give r ′∗ i ′∗ = 1, p ′∗ s ′∗ = 1, p ′∗ i ′∗ = 0 and r ′∗ s ′∗ = 0, so we have a split short exact sequence F ( A ) F ( B ) F ( C ) i ′∗ p ′∗ r ′∗ s ′∗ hus, Proposition 2.7 gives us an isomorphism F ( A ⊕ C ) = F ( B ) → F ( A ) ⊕ F ( C ), and by unwinding thedefinitions we see that this is given by the stated formulae. (cid:3) There is a similar statement for contravariant functors as follows.
Definition 2.15. A contravariant functor from abelian groups to abelian groups is a construction thatgives an abelian group F ( A ) for each abelian group A , and a homomorphism f ∗ : F ( B ) → F ( A ) for eachhomomorphism f : A → B , in such a way that:(a) For identity maps we have (1 A ) ∗ = 1 F ( A ) for all A .(b) For homomorphisms A f −→ B g −→ C we have ( gf ) ∗ = f ∗ g ∗ : F ( C ) → F ( A ).We say that F is additive if ( f + f ) ∗ = f ∗ + f ∗ for all f , f : A → B . Proposition 2.16.
Let F be an additive contravariant functor as above. Then for any abelian groups A and C we have an natural isomorphism f : F ( A ⊕ C ) → F ( A ) ⊕ F ( C ) given by f ( b ) = (( i ′ ) ∗ ( b ) , ( s ′ ) ∗ ( b )) withinverse f − ( a, c ) = ( r ′ ) ∗ ( a ) + ( p ′ ) ∗ ( c ) .Proof. Essentially the same as Proposition 2.14. (cid:3) Products and coproducts
If we have a finite list of abelian groups A , . . . , A n , we can form the product group Q ni =1 A i = A ×· · ·× A n ,which is also denoted by L ni =1 A i = A ⊕ · · · ⊕ A n . This should be familiar. These constructions can begeneralised to cover families of abelian groups A i indexed by a set I that may be infinite, and need not beordered in any natural way. This is a little more subtle, and in particular L i A i is not the same as Q i A i inthis context. In this section we will briefly outline the relevant definitions and properties. Definition 3.1.
Let I be a set, and let ( A i ) i ∈ I be a family of abelian groups indexed by I . The productgroup Q i ∈ I A i is the set of all systems a = ( a i ) i ∈ I consisting of an element a i ∈ A i for each i ∈ I . We makethis into an abelian group by the obvious rule( a i ) i ∈ I ± ( b i ) i ∈ I = ( a i ± b i ) i ∈ I . For each k ∈ I we define π k : Q i ∈ I A i → A k by π k (( a i ) i ∈ I ) = a k . This is clearly a homomorphism. We alsodefine ι k : A k → Q i ∈ I A i by ι k ( a ) i = ( a ∈ A k if i = k ∈ A i if i = k. Example 3.2. If I = { , , . . . , n } , then Q i ∈ I A i is just the set of n -tuples ( a , . . . , a n ) with a i ∈ A i , asbefore. Example 3.3.
Suppose we have a fixed group U , and we take A i = U for all i . Then Q i ∈ I A i is just theset Map( I, U ) of all functions from I to U , considered as a group under pointwise addition. Remark 3.4.
It is easy to see that a homomorphism f : U → Q i ∈ I A i is essentially the same thing as a familyof homomorphisms f i : U → A i , one for each i ∈ I . Indeed, given such a family we define f : U → Q i ∈ I A i by f ( u ) = ( f i ( u )) i ∈ I , and we can then recover the original homomorphisms f i as the composites π i ◦ f . Thismeans that Q i ∈ I A i is a product for the groups A i in the general sense considered in category theory. Definition 3.5.
Given an element a = ( a i ) i ∈ I ∈ Q i ∈ I A i , the support of a is the setsupp( a ) = { i ∈ I | a i = 0 } ⊆ I. We put M i ∈ I A i = { a ∈ Y i ∈ I A i | supp( a ) is a finite set } . It is easy to see that supp( a ± b ) ⊆ supp( a ) ∪ supp( b ), and thus that L i ∈ I A i is a subgroup of Q i ∈ I A i . Wecall it the coproduct of the family ( A i ) i ∈ I . We also note that supp( ι k ( a )) ⊆ { k } , so ι k can be regarded as ahomomorphism A k → L i ∈ I A i . emark 3.6. If the index set I is finite then all supports are automatically finite and so the coproduct isthe same as the product. In fact, we only need the set I ′ = { i | A i = 0 } to be finite for this to hold.Definition 3.5 is again compatible with the more general definition coming from category theory, as wesee from the following result: Proposition 3.7.
Suppose we have an abelian group V , and a system of homomorphisms g i : A i → V forall i ∈ I . Then there is a unique homomorphism g : L i ∈ I A i → V such that g ◦ ι k = g k for all k ∈ I .Proof. Given a point a = ( a i ) i ∈ I ∈ L i ∈ I A i , we define g ( a ) = X i ∈ supp( a ) g i ( a i ) ∈ V. The terms in the sum are meaningful because a i ∈ A i and g i : A i → V , and supp( a ) is finite so there onlyfinitely many terms so it is not a problem to add them up. If we replace supp( a ) by some larger finite set J ⊆ I then this gives us some extra terms but they are all zero so the sum is unchanged. After taking J = supp( a ) ∪ supp( b ) it becomes easy to see that g ( a + b ) = g ( a ) + g ( b ), so g is a homomorphism. Usingsupp( ι k ( a )) ⊆ { k } we see that g ◦ ι k = g k , as required. Let g ′ : L i ∈ I A i → V be another homomorphismwith g ′ ◦ ι k = g k for all k . If we have an element a as before, we observe that a = P i ∈ supp( a ) ι i ( a i ), and byapplying g ′ to this we get g ′ ( a ) = X i ∈ supp( a ) g ′ ( ι i ( a i )) = X i ∈ supp( a ) g i ( a i ) = g ( a ) , so g is unique as claimed. (cid:3) Remark 3.8.
It would at worst be a tiny abuse of notation to say that g ( a ) = P i ∈ I g i ( a i ). This is a sumwith infinitely many terms, which would not normally be meaningful, but only finitely many of the termsare nonzero, so the rest can be ignored. Remark 3.9.
Suppose we have an abelian group A , and a family of subgroups ( A i ) i ∈ I . There is then aunique homomorphism σ : L i ∈ I A i → A such that σ ◦ ι k : A k → A is just the inclusion for all k . Moreexplicitly, we just have σ ( a ) = P i ∈ supp( a ) a i . If this map σ is an isomorphism, we will say (with anotherslight abuse of notation) that A = L i ∈ I A i .4. Torsion groups
Definition 4.1.
Let A be an abelian group.(a) We say that an element a ∈ A is a torsion element if na = 0 for some integer n > A ) for the set of torsion elements of A . This is easily seen to be a subgroup, becauseif na = 0 and mb = 0 then nm ( a ± b ) = 0.(c) We say that A is a torsion group if every element is torsion, or equivalently tors( A ) = A . At theother extreme, we say that A is torsion-free if tors( A ) = 0.(d) Now fix a prime p . We say that a is a p -torsion element if p k a = 0 for some k ≥
0. We write tors p ( A )for the set of p -torsion elements, which is again a subgroup.(e) We say that A is a p -torsion group if every element is p -torsion, or equivalently tors p ( A ) = A . Atthe other extreme, we say that A is p -torsion free if tors p ( A ) = 0. Remark 4.2.
We write n. A for the endomorphism of A given by a na . Then tors( A ) = S n> ker( n. A ),and A is torsion-free if and only if the maps n. A (for n >
0) are all injective.
Example 4.3. If A is a finite abelian group with | A | = n then Lagrange’s Theorem tells us that na = 0for all a ∈ A , so A is a torsion group. For another instructive proof of the same fact, consider the element z = P x ∈ A x . As x runs over A , the elements a + x also run over A , so z = P x ∈ A ( a + x ) = na + z , so na = 0. It is a curious fact, which we leave to the reader, that z itself is actually zero in all cases exceptwhen | A | = 2. Example 4.4.
It is clear that any free abelian group is torsion-free. The groups Q and R are torsion-freebut not free. xample 4.5. Consider the quotient group A = Q / Z . The subset A n = { Z = n + Z , n + Z , . . . , n − n + Z } is a cyclic subgroup of order n . Any element a ∈ Q / Z can be written as a = m/n + Z for some m, n ∈ Z with n >
0. We can also write m as qn + r for some q, r ∈ Z with 0 ≤ r < n and observe that a = m/n + Z = r/n + q + Z = r/n + Z ∈ A n . This proves that A is the union of the subgroups A n . As na = 0 for all a ∈ A n ,we see that A is a torsion group. One can also check that A n ≤ A m if and only if n divides m . In particular,for each prime p we have a chain of subgroups A p ≤ A p ≤ A p ≤ · · · ≤ [ k A p k = tors p ( A ) . Example 4.6.
Now consider instead the group R / Z . Suppose we have a torsion element a = t + Z . Thismeans that for some integer n > nt ∈ Z , which implies that t is rational. It follows thattors( R / Z ) = Q / Z . A similar argument shows that R / Q is torsion-free.The following result is known as the Chinese Remainder Theorem. Proposition 4.7.
Suppose we have positive integers n , . . . , n r any two of which are coprime, and we put n = Q i n i . Define φ : Z /n → ( Z /n ) × · · · × ( Z /n r ) by φ ( k + n Z ) = ( k + n Z , . . . , k + n r Z ) . Then (a)
There exist integers e , . . . , e r such that P i e i = 1 and e i = 1 (mod n i ) and e i = 0 (mod n/n i ) . (b) The map φ is an isomorphism.Proof. For i = j we know that n i and n j are coprime, so we can choose integers a ij and b ij with a ij n i + b ij n j =1. We then put f ij = b ij n j = 1 − a ij n i , so f ij = 1 (mod n i ) and f ij = 0 (mod n j ). Now fix i , and let g i be the product of the numbers f ij as j runs over the remaining indices. We find that g i = 1 (mod n i ),but g i is divisible by the product of all the n j , or equivalently by n/n i . Thus, the numbers g i almost haveproperty (a), but we will need a slight adjustment to make the sum equal to one. However, we are now readyto prove (b). Given any integers m , . . . , m r , we have φ ( X i m i g i + n Z ) = ( m + n Z , . . . , m r + n r Z ) . This proves that φ is surjective, and the domain and codomain of φ both have order n , so φ must actually bean isomorphism. By construction we have φ ( P i g i + n Z ) = φ (1 + n Z ), and φ is injective, so P i g i = 1 + nk for some k . We define e i = g i for i < r , and e r = 1 − P i Corollary 4.8. Suppose that the prime factorisation of n is n = p v · · · p v r r , where the primes p i are all dis-tinct. Then there are integers e , . . . , e r such that P i e i = 1 and e i = 1 (mod p v i i ) and e i = 0 (mod n/p v i i ) .Moreover, the natural map Z /n → ( Z /p v ) × · · · × ( Z /p v r r ) is an isomorphism. (cid:3) Proposition 4.9. For any abelian group A we have tors( A ) = L p tors p ( A ) .Proof. Suppose we have a torsion element a ∈ A , so na = 0 for some n > 0. We can factor this as Q ri =1 p v i i and then choose integers e i as in Corollary 4.8. Now e i = 0 (mod n/p v i i ) so p v i i e i is divisible by n , so p v i i e i a = 0, so e i a ∈ tors p i ( A ). We also have P i e i = 1, so a = P i e i a ∈ P i tors p i ( A ). This shows thattors( A ) = P p tors p ( A ).To show that the sum is direct, suppose we have a finite list of distinct primes p , . . . , p r , and elements a i ∈ tors p i ( A ) with P i a i = 0; we must show that a i = 0 for all i . As a i ∈ tors p i ( A ) we have p v i i a i = 0 forsome v i ≥ 0. We again choose numbers e i as in Corollary 4.8. As p v i i a i = 0 and e i = 1 (mod p v i i ) we have i a i = a i . On the other hand, for j = i we have e i = 0 (mod p v j j ) and so e i a j = 0. We can thus multiplythe relation P j a j = 0 by e i to get a i = 0 as required. (cid:3) Lemma 4.10. The quotient group A/ tors( A ) is always torsion-free.Proof. Suppose we have a torsion element a = x + tors( A ) in A/ tors( A ). This means that for some n > na = 0 or equivalentlt nx ∈ tors( A ). This in turn means that for some m > mnx = 0,which shows that x itself is a torsion element in A . This means that the coset a = x + tors( A ) is zero, asrequired. (cid:3) Finitely generated abelian groups Let A be an abelian group, and let a , . . . , a r be elements of A . We then have a homomorphism f : Z r → A given by f ( n , . . . , n r ) = n a + · · · + n r a r . In particular, if e i is the i ’th standard basis vector in Z r then f ( e i ) = a i . Conversely, if we start with ahomomorphism f : Z r → A we can put a i = f ( e i ) ∈ A and we find that f ( n , . . . , n r ) = f ( X i n i e i ) = X i n i a i , so everything fits together as before. The image of f is the smallest subgroup of A containing all the elements a i , or in other words the subgroup generated by { a , . . . , a r } . This justifies the following definition: Definition 5.1. We say that an abelian group A is finitely generated if there exists a surjective homomor-phism f : Z r → A for some r . Example 5.2. Suppose that A is actually finite, so we can choose a list a , . . . , a r that contains all theelements of A . The corresponding map Z r → A is certainly surjective, so A is finitely generated.Our main aim in this section is to prove the following classification theorem: Theorem 5.3. Let A be a finitely generated abelian group. Then A can be decomposed the direct sum of afinite list of subgroups, each of which is isomorphic either to Z , or to Z /p v for some prime p and some v > .The number of subgroups of each type in the decomposition is uniquely determined, although the precise listof subgroups is not. Remark 5.4. Note that Proposition 4.7 gives a decomposition of the stated type for the cyclic group Z /n .The groups Z r themselves are of course finitely generated. It is convenient to observe that no two of themare isomorphic: Lemma 5.5. If Z r is isomorphic to Z s , then r = s .Proof. Any isomorphism f : A → B induces an isomorphism A/ A → B/ B , so in particular | A/ A | = | B/ B | . We have Z r / Z r = ( Z / r , which has order 2 r , and the claim follows easily. (cid:3) This means that the term ’rank’ in the following definition is well-defined: Definition 5.6. We say that an abelian group A is free of rank r if it is isomorphic to Z r . Lemma 5.7. Let A be a subgroup of Z . Then either A = 0 ≃ Z or A = d Z ≃ Z for some (unique) d > .Proof. The case where A = 0 is trivial, so suppose that A = 0. As A = − A we see that A must contain atleast one strictly positive integer. Let d be the smallest strictly positive integer in A . It is than clear that d Z ⊆ A . Conversely, suppose that n ∈ A . As d > n must lie between id and ( i + 1) d for some i ∈ Z , say n = id + j with 0 ≤ j < d . Now j = n − id ∈ A and 0 ≤ j < d , which contradicts the definingproperty of d unless j = 0. We thus have n = di , showing that A = d Z as claimed. (cid:3) Proposition 5.8. Let A be a subgroup of Z r ; then A is free of rank at most r . roof. For i ≤ r we put F i = { x ∈ Z r | x i +1 = · · · = x r = 0 } ≃ Z i . We let π s : Z r → Z be the projection map x x s , and put J = { j | π j ( A ∩ F j ) = 0 } = { j | A ∩ F j > A ∩ F j − } . We can list the elements of this set as j < · · · < j s for some s ≤ r (possibly s = 0). We see from Lemma 5.7that π j p ( A ∩ F j p ) must have the form d p Z for some d p > a p ∈ A ∩ F j p such that π j p ( a p ) = d p for all p . We claim that the list a , . . . , a s is a basis for A over Z , so that A is a free abelian groupas claimed. More precisely, we claim that a , . . . , a p is always a basis for A ∩ F j p . In the case p = 0 we havethe empty list and the zero group so the claim is clear. When p > p − 1. Consider an arbitrary element u ∈ A ∩ F j p . By the definition of d p , we have π j p ( u ) = m p d p for some m p ∈ Z . The element u ′ = u − m p a p then lies in A ∩ F j p and satisfies π j p ( u ′ ) = 0 so in fact u ′ ∈ A ∩ F j p − .By the induction hypothesis there are unique integers m , . . . , m p − with u ′ = m a + · · · + m p − a p − , andit follows that u = u ′ + m p a p = m a + · · · + m p a p . This shows that u can be expressed as an integercombination of a , . . . , a p , and a similar argument shows that the expression is unique. This completes theinduction step, and after s steps we see that A itself has a basis as claimed. (cid:3) Remark 5.9. In the above proof, we can alter our choice of a p by subtracting off suitable multiples of a p − , a p − and so on in turn to ensure that 0 ≤ π j q ( a p ) < d q for 1 ≤ q < p . One can then check that the resultingbasis satisfying this auxiliary condition is in fact unique. Corollary 5.10. If A is finitely generated and B is a subgroup of A then B and A/B are also finitelygenerated.Proof. Choose a surjective homomorphism f : Z r → A . The composite Z r f −→ A π −→ A/B is again surjective,so A/B is finitely generated. Now put F = { x ∈ Z r | f ( x ) ∈ B } , and let g : F → B be the restriction of f .For b ∈ B ≤ A we can choose x ∈ Z r with f ( x ) = b (because f is surjective). Then x ∈ F be the definition of f , and g ( x ) = f ( x ) = b ; this proves that g is surjective. Moreover, F is a subgroup of Z r , so it is isomorphicto Z s for some s ≤ r by the proposition. It now follows that B is finitely generated. (cid:3) Proposition 5.11. Suppose that F is finitely generated and torsion free; then F is free.Proof. Choose a surjective homomorphism f : Z r → F with r as small as possible. Put A = ker( f ); itwill suffice to show that A = 0. If not, choose some nonzero element a = ( a , . . . , a r ) ∈ ker( f ). Let d be the greatest common divisor of a , . . . , a r (or equivalently, the number d > P i a i Z = d Z ,which exists by Lemma 5.7). We find that a/d ∈ Z r and d f ( a/d ) = f ( a ) = 0 in F , so f ( a/d ) is a torsionelement, but F is assumed torsion free, so f ( a/d ) = 0. We may thus replace a by a/d and assume that d = 1, so P i a i Z = Z . We can thus choose integers b , . . . , b r with P i a i b i = 1. Now define β : Z r → Z by β ( x ) = P i x i b i and put U = ker( β ). As β ( a ) = 1 we find that x − β ( x ) a ∈ U for all x , and it follows that Z r = U ⊕ Z a . Proposition 5.8 tells us that U is free, and using the splitting Z r = U ⊕ Z a we see that U hasrank r − 1. As f ( a ) = 0 we also see that f ( U ) = f ( U ⊕ Z a ) = f ( Z r ) = F , so f restricts to give a surjectivehomomorphism U → F . This contradicts the assumed minimality of r , so we must have A = 0 after all, so f : Z r → F is an isomorphism. (cid:3) Corollary 5.12. Let A be a finitely generated abelian group. Then tors( A ) is finite, and there exists afinitely generated free subgroup F ≤ A such that A = tors( A ) ⊕ F ≃ tors( A ) ⊕ Z s for some s .Proof. Firstly, Corollary 5.10 tells us that tors( A ) is finitely generated, so we can choose a finite list ofgenerators, say a , . . . , a r . These must be torsion elements, so we can choose n i > n i a i = 0.This means that the corresponding surjection f : Z r → tors( A ) factors through the finite quotient group Q ri =1 ( Z /n i ), so tors( A ) is finite as claimed. Next, the quotient group A/ tors( A ) is finitely generated (byCorollary 5.10) and torsion-free (by Lemma 4.10) so it is free of finite rank by Proposition 5.11. We canthus choose an isomorphism g : Z s → A/ tors( A ). Now choose an element a i ∈ A representing the coset g ( e i ) (for i = 1 , . . . , s ) and define g : Z s → A by g ( x ) = P i x i a i , and put F = g ( Z s ). If we let q denote the uotient map A → A/ tors( A ) we have qg = g , which is an isomorphism. It follows that g : Z s → F is anisomorphism, so F is free as claimed. Next, let h : A → F be the composite A q −→ A/ tors( A ) g − −−→ Z s g −→ F. We find that qh = q , so q ( a − h ( a )) = 0, so a − h ( a ) ∈ tors( A ) for all a . This implies that a = ( a − h ( a ))+ h ( a ) ∈ tors( A ) + F , so A = tors( A ) + F . Moreover, the intersection tors( A ) ∩ F is both torsion and torsion-free, soit must be zero, so the sum is direct. (cid:3) This corollary allows us to focus on tors( A ), which is a finite group, of order n say. Proposition 4.9 givesa splitting tors( A ) = L p tors p ( A ), and it is clear that tors p ( A ) can only be nonzero if p divides n . In thatcase, tors p ( A ) will be a finite abelian group whose order is a power of p . Lemma 5.13. Let A be an abelian group of order p v . Suppose we have an element c of order p w , and thatevery other element has order dividing p w , and that the subgroup C = Z c has nontrivial intersection withevery nontrivial subgroup. Then A = C .Proof. Consider a nontrivial element a ∈ A . The order of a + C in A/C will then be p i for some i ≤ w . Wethen have p i a = mc for some m ∈ Z , and the assumption ( Z a ) ∩ C = 0 means that mc = 0. We can thuswrite mc = up j c for some j < w and some u such that u = 0 (mod p ). It follows that the order of mc in A is p w − j , and thus that the order of a in A is p w − j + i . By assumption, this is at most p w , so i ≤ j . We canthus put b = a − up j − i c , and observe that p i b = 0. Now b is congruent to a mod C , so it again has order p i in A/C , but p i b is already zero in A , so ( Z b ) ∩ C = 0. As C meets every nontrivial subgroup, we must have Z b = 0, so a = up j − i c ∈ C . This means that A = C as claimed. (cid:3) Corollary 5.14. Let A be an abelian group of order p v , and suppose that the largest order of any elementof A is p w . Then A ≃ B ⊕ ( Z /p w ) for some subgroup B of order p v − w .Proof. Choose an element c of order p w , and let C be the subgroup that it generates, so C ≃ Z /p w . Amongthe subgroups B with B ∩ C = 0, choose one of maximal order. Then put A = A/B , and let C be theimage of C in A , which is isomorphic to C because B ∩ C = 0. It will suffice to prove that A = B + C , orequivalently that A = C . By the lemma, we need only check that C has nontrivial intersection with everynontrivial subgroup of A . Consider a nonzero element a ∈ A , and choose a representing element a ∈ A \ B .Then Z a + B is strictly larger than B and must meet C nontrivially, so there exists k ∈ Z and b ∈ B with ka + b ∈ C \ { } . If ka + b were in B it would give a nontrivial element of B ∩ C , contrary to assumption.It follows that ka is nontrivial in A and lies in C , as required. (cid:3) Corollary 5.15. Let A be an abelian group of order p v . Then A is isomorphic to L ri =1 Z /p w i for some list w , . . . , w r of positive integers with P i w i = v .Proof. This follows by an evident induction from Corollary 5.14. (cid:3) Definition 5.16. Let A be a finite abelian group. For any prime p and positive integer k , we put F kp ( A ) = { a ∈ p k − A | pa = 0 } . This is a finite abelian group of exponent p , so it has order p v for some v . We define f kp ( A ) to be this v , andwe also put g kp ( A ) = f kp ( A ) − f k +1 p ( A ). Proposition 5.17. Let A be a finite abelian group. (a) If A ≃ A ′ , then F kp ( A ) ≃ F kp ( A ′ ) for all p and k , so f kp ( A ) = f kp ( A ′ ) and g kp ( A ) = g kp ( A ′ ) . (b) If A = B ⊕ C then F kp ( A ) = F kp ( B ) ⊕ F kp ( C ) , so f kp ( A ) = f kp ( B )+ f kp ( C ) and g kp ( A ) = g kp ( B )+ g kp ( C ) . (c) If A has order not divisible by p , then F kp ( A ) = 0 and so f kp ( A ) = g kp ( A ) = 0 . (d) Suppose that A has a decomposition as a sum of subgroups Z /p v i i (with v i > ). Then g kp ( A ) is thenumber of times that Z /p k occurs in the decomposition.Proof. Parts (a) to (c) are straightforward and are left to the reader. Given these, part (d) reduces to theclaim that g kp ( Z /p j ) is one when j = k , and zero otherwise. One can see from the definitions that f kp ( Z /p j )is one when k ≤ j , and zero when k > j ; the claim follows easily from this. (cid:3) roof of Theorem 5.3. Let A be a finitely generated abelian group. Corollary 5.12 and subsequent remarksshow that A ≃ Z s ⊕ L mi =1 tors p i ( A ) for some finite list of primes p i . After applying Corollary 5.15 to eachof the groups tors p i ( A ), we get the claimed splitting of A as a sum of copies of Z and Z /p w j j . The number s is the rank of the quotient group A/ tors( A ), which is well-defined by Lemma 5.5. We can also apply thelast part of Proposition 5.17 to tors( A ) to see that the number of summands of each type is independent ofthe choice of splitting. (cid:3) Free abelian groups and their subgroups In various places we have already used the free abelian group Z [ I ] generated by a set I . We start with amore careful formulation of this construction. One approach is to define Z [ I ] = L i ∈ I Z as in Definition 3.5.That is essentially what we will do, but we will spell out some details. Definition 6.1. Let I be any set. We write Map( I, Z ) for the set of all maps u : I → Z . These form anabelian group under pointwise addition. For any map u : I → Z , the support is the setsupp( u ) = { i ∈ I | u ( i ) = 0 } ⊆ I. We put Map ( I, Z ) = { u : I → Z | supp( u ) is finite } . It is easy to see that supp( u ± v ) ⊆ supp( u ) ∪ supp( v ), and thus that Map ( I, Z ) is a subgroup of Map( I, Z ).Next, for any i ∈ I we define δ i : I → Z by δ i ( j ) = ( j = i j = i. Note that supp( δ i ) = { i } so δ i ∈ Map ( I, Z ). Remark 6.2. If I is a finite set with n elements then we see that Map ( I, Z ) = Map( I, Z ) ≃ Z n . Thesituation is a little more subtle when I is infinite.The following lemma shows that Map ( I, Z ) is generated freely, in a certain sense, by the elements δ i . Lemma 6.3. Let A be an abelian group. Then for any function f : I → A there is a unique homomorphism f : Map ( I, Z ) → A such that f ( δ i ) = f ( i ) for all i ∈ I .Proof. We put f ( u ) = X i ∈ supp( u ) u ( i ) f ( i ) . The terms are meaningful, because each u ( i ) is in Z and each f ( i ) is in A so we can multiply to get anelement of A . The sum is meaningful because u ∈ Map ( I, Z ), so supp( u ) is finite, so there are only finitelymany terms to add. More explicitly, if supp( u ) = { i , . . . , i r } and u ( i t ) = n t ∈ Z for all t then f ( u ) = n f ( i ) + · · · + n r f ( i r ) ∈ A. Note that it would be harmless to replace supp( u ) by any finite set J with supp( u ) ⊆ J ⊆ I ; this wouldintroduce some extra terms, but they would all be zero. After taking J = supp( u ) ∪ supp( v ) we can checkthat f ( u + v ) = f ( u ) + f ( v ), so f is a homomorphism. From the definitions it is clear that f ( δ i ) = f ( i ).Now let α : Map ( I, Z ) → A be another homomorphism with α ( δ i ) = f ( i ). Put β = α − f , so β ( δ i ) = 0 forall i . It is not hard to see that a general element u as above can be expressed in the form u = n δ i + · · · + n r δ i r . It follows that β ( u ) = n β ( δ i ) + · · · + n r β ( δ i r ) = n . · · · + n r . . This shows that β = 0, so α = f as required. (cid:3) The notation used so far is convenient for giving the definition, and the proof of the above freenessproperty, but not for the applications. We thus introduce the following alternative: efinition 6.4. We write Z [ I ] for Map ( I, Z ), and [ i ] for δ i . We say that an abelian group A is free if it isisomorphic to Z [ I ] for some I .The freeness property now takes the following form: Lemma 6.5. Let A be an abelian group. Then for any function f : I → A there is a unique homomorphism f : Z [ I ] → A such that f ([ i ]) = f ( i ) for all i ∈ I . One key result is as follows: Theorem 6.6. If A is a free abelian group, then every subgroup of A is also free. This is a generalisation of Proposition 5.8, which covered the case where A is finitely generated. Belowwe will state and prove some more refined statements. To prove Theorem 6.6 itself, we can use Remark 6.14and the case k = ⊤ of Lemma 6.16 (in the notation of Definition 6.13).To extend the proof of Proposition 5.8 to cover infinitely generated groups, we need two ingredients.Firstly, we need to modify the inductive argument so that it works for a suitable class of infinite orderedsets. Next, we need to show that any set can be ordered in the required way. The precise structure that weneed is as follows: Definition 6.7. A well-ordering on a set I is a relation on I (denoted by i ≤ j ) such that(a) For all i ∈ I we have i ≤ i .(b) For all i, j ∈ I we have either i ≤ j or j ≤ i , and if both hold then i = j .(c) For all i, j, k ∈ I , if i ≤ j and j ≤ k then i ≤ k .(d) For any nonempty subset J ⊆ I there is an element j ∈ J such that j ≤ j for all j ∈ J . (In otherwords, j is smallest in J .) Remark 6.8. We have stated the axioms in a form that is conceptually natural but inefficient. Axiom (a)follows from (d) by taking J = { i } , the first half of (b) follows from (d) by taking J = { i, j } , and with a littlemore argument one can deduce (c) by taking J = { i, j, k } and appealing to the second half of (b). Thus, wereally only need (d) together with the second half of (b). Example 6.9. The obvious ordering of N is a well-ordering, as is the obvious ordering on N ∪ {∞} . Theobvious ordering on Z is not a well-ordering, because the whole set does not have a smallest element. Wecan choose a bijection f : N → Z (for example, by setting f (2 n ) = n and f (2 n + 1) = − n − 1) and use thisto transfer the standard ordering of N to a nonstandard ordering of Z that is a well-ordering. Alternatively,we can specify a well-ordering on Z by the rules0 < < < < < · · · < − < − < − < · · · . By constructions such as these, one can give explicit well-orderings of most naturally occurring countablesets. Remark 6.10. Let I be a well-ordered set. If I is nonempty then it must have a smallest element, whichwe denote by ⊥ I or just ⊥ . Now suppose that i ∈ I and i is not maximal, so the set I >i = { j ∈ I | j > i } is nonempty. Then I >i must have a smallest element. We denote this by s ( i ), and call it the successor of i .We say that an element j ∈ I is a successor if j = s ( i ) for some i . For example, in N ∪ {∞} the elements 0and ∞ are not successors, but all other elements are successors. Theorem 6.11. Every set admits a well-ordering. The proof will be given after some preliminaries.There is no known well-ordering of R , and indeed it is probably not possible to specify a well-orderingconcretely, although the author does not know of any precise theorems to that effect. Similarly, there is noknown well-ordering of the set of subsets of N , or of most other naturally occurring uncountable sets. Theproblem is that one needs to make an infinite number of arbitrary choices, which cannot be done explicitly.However, we shall assume the Axiom of Choice, a standard principle of Set Theory, which says that suchchoices are nonetheless possible. More precisely, we shall assume that every set has a choice function, in thefollowing sense: efinition 6.12. Let I be a set, and let P ′ ( I ) denote the set of nonempty subsets of I . A choice function for I is a function c : P ′ ( I ) → I such that c ( J ) ∈ J for all J ∈ P ′ ( I ). (In other words, c ( J ) is a “chosen”element of J .)If I is well-ordered, we can define a choice function by taking c ( J ) to be the smallest element of J .Conversely, if we are given a choice function then we can use it to construct a well-ordering, as we nowexplain. In the literature it is more common to do this by proving Zorn’s Lemma as an intermediate step,but here we have chosen to bypass that. Proof of Theorem 6.11. Let I be a set, and let c be a choice function for I . Let P ( I ) be the set of all subsetsof I , and put P ∗ ( I ) = P ( I ) \ { I } . Define d : P ∗ ( I ) → I by d ( J ) = c ( I \ J ), and then define e : P ( I ) → P ( I )by e ( J ) = ( J ∪ { d ( J ) } if J = II if J = I. We call this the expander function. Clearly we have J ⊆ e ( J ) for all J , with equality iff J = I . Now saythat a subset A ⊆ P ( I ) is saturated if(a) Whenever J ∈ A , we also have e ( J ) ∈ A .(b) For any family of sets in A , the union of that family is also in A .We say that a set J is compulsory if it lies in every saturated family. For example, by applying (b) tothe empty family we see that the set J = ∅ is compulsory. It follows using (a) that the sets J n = e n ( ∅ )are compulsory for all n . Axiom (b) then tells us that the set J ω = S ∞ n =0 J n is compulsory, as is the set J ω +1 = e ( J ω ). If we had developed the theory of infinite ordinals, we could use transfinite recursion to definecompulsory sets J α for all ordinals α . As we have not discussed that theory, we will instead use an approachthat avoids it. We let C denote the family of all compulsory sets. This is clearly itself a saturated family.We say that a set J ∈ C is comparable if for all other K ∈ C we have either J ⊆ K or K ⊆ J . Let D ⊆ C bethe set of all comparable sets; we will show that this is saturated, and thus equal to C . Consider a familyof comparable sets J α , with union J say, and another set K ∈ C . For each α we have J α ⊆ K or K ⊆ J α ,because J α is comparable. If J α ⊆ K for all K then clearly J ⊆ K . Otherwise we must have K ⊆ J α forsome α but J α ⊆ J so K ⊆ J . This shows that J is comparable, so D is closed under unions. Now considera comparable set J ; we claim that e ( J ) is also comparable. To see this, put E J = { K | e ( J ) ⊆ K or K ⊆ J } . By a similar argument to the previous paragraph, this is closed under unions. Suppose that K ∈ E J .(a) If e ( J ) ⊆ K then clearly e ( J ) ⊂ e ( K ), so e ( K ) ∈ E J .(b) If K = J then e ( J ) = e ( K ), so e ( K ) ∈ E J .(c) Suppose instead that K ⊂ J . As K ∈ C we also have e ( K ) ∈ C , and J is comparable so either e ( K ) ⊆ J or J ⊂ e ( K ). In the latter case we have K ⊂ J ⊂ e ( K ), so | e ( K ) \ K | ≥ 2, but | e ( K ) \ K | ≤ e ( K ) ⊆ J , so again e ( K ) ∈ J .We now see that E J is a saturated subset of C , so it must be all of C . It follows easily from this that e ( J ) iscomparable. This means that D is a saturated subset of C , so it must be all of C , so all compulsory sets arecomparable, or in other words C is totally ordered by inclusion.We now claim that C is in fact well-ordered by inclusion. To see this, consider a nonempty family ofcompulsory sets K α . Let J be the union of all compulsory sets that are contained in T α K α . As C is closedunder unions, we see that J is actually the largest compulsory set that is contained in T α K α . In particular,the larger set e ( J ) cannot be contained in T α J α , so for some α we have e ( J ) K α . Now K α ∈ C and wehave seen that C = E J and using this we see that K α ⊆ J . From this it follows easily that K α is the smallestset in the family, as required.Next, put C ∗ = C \ { I } , which is again well-ordered by inclusion. For i ∈ I we let p ( i ) be the union of allcompulsory sets that do not contain i . As C is closed under unions this defines a map p : I → C ∗ . We canalso restrict d to get a map d : C ∗ → I in the opposite direction. Note that e ( p ( i )) = p ( i ) ∪ { d ( p ( i )) } is acompulsory set not contained in p ( i ), so we must have i ∈ e ( p ( i )), but i p ( i ) by construction, so we musthave d ( p ( i )) = i . In the opposite direction, suppose we start with a compulsory set J ∈ C ∗ , and put i = d ( J ), o e ( J ) = J ∐ { i } . Now p ( i ) ∈ C and we have seen that C = E J so either e ( J ) ⊆ p ( i ) or p ( i ) ⊆ J . The first ofthese would imply that i ∈ p ( i ), contradicting the definition of p ( i ), so we must instead have p ( i ) ⊆ J . Onthe other hand, J is one of the sets in the union that defines p ( i ), so J ⊆ p ( i ), so J = p ( i ) = p ( d ( J )). Thisproves that the maps I p −→ C ∗ d −→ I are mutually inverse bijections. We can thus introduce a well-ordering of I by declaring that i ≤ j iff p ( i ) ⊆ p ( j ). (cid:3) We can now start to prove as promised that subgroups of free abelian groups are free. It will be enoughto prove that every subgroup of Z [ I ] is free, and by Theorem 6.11 we may assume that I is well-ordered. Wewill need some auxiliary definitions. Definition 6.13. Let I be a well-ordered set, and let A be a subgroup of Z [ I ]. We put I ⊤ = I ∐{⊤} , orderedso that i ≤ ⊤ for all i ∈ I (which is again a well-ordering). Put I As each B ( j ) is nonempty, we see that there exist adapted bases. Implicitly we are usingthe Axiom of Choice here: there exists a choice function c for A , and then we can take a ( j ) = c ( B ( j )) forall j . However, we will prove as Proposition 6.20 that there is a unique adapted basis satisfying a certainnormalisation condition, which enables us to avoid this use of choice. Remark 6.15. Let a be an adapted basis for A Let a be an adapted basis for A Lemma 6.17. Let a be an adapted basis for A We actually claim that more generally, the restricted maps f n : Z [ J The method that we used to deduce Lemma 6.16 from 6.17 is called transfinite induction ;it is evidently an extension of the usual kind of induction over the natural numbers. We will use transfiniteinduction again below without spelling it out so explicitly.As we remarked previously, Theorem 6.6 follows from Theorem 6.11, Lemma 6.16 and Remark 6.14.Because we need Theorem 6.11, the proof is unavoidably nonconstructive. Nonetheless, we can remove oneset of arbitrary choices by pinning down a specific adapted basis, as we now explain. Definition 6.19. Let a be an adapted basis for A There is a unique normalised adapted basis for A . This follows by transfinite induction from the following lemma: Lemma 6.21. Suppose that for all m < ⊤ there is a unique normalised basis for A Definition 7.1. Let A be an abelian group. We make the free abelian group Z [ A ] into a commutative ringby the rule ( X i n i [ a i ]) . ( X j m j [ b j ]) = X i,j n i m j [ a i + b j ](so in particular [ a ][ b ] = [ a + b ]). We define a ring homomorphism ǫ : Z [ A ] → Z by ǫ ( P i n i [ a i ]) = P i n i , andwe define I A to be the kernel of ǫ . We write I A for the ideal generated by all products xy with x, y ∈ I A .We also define a group homomorphism q : Z [ A ] → A by q ( P i n i [ a i ]) = P i n i a i . Proposition 7.2. (a) The abelian groups I A and I A are both free. (b) More specifically, the elements h a i = [ a ] − [0] for a ∈ A \ form a basis for I A . (c) Put h a, b i = h a ih b i = [ a + b ] − [ a ] − [ b ] + [0] = h a + b i − h a i − h b i . Then the set of all elements of this form generates I A as an abelian group. (d) There is a natural short exact sequence I A j −→ I A q −→ A (where j is just the inclusion).Proof. (a) Both I A and I A are subgroups of Z [ A ], so they are free by Theorem 6.6.(b) In the case of I A it is easy to be more concrete. Suppose we have an element x = P i n i [ a i ] ∈ I A .Then P i n i = 0, so x can also be written as P i n i ([ a i ] − [0]), and it is clearly harmless to omit anyterms where a i = 0, so we see that x is in the subgroup generated by the elements [ a ] − [0] with a = 0. It is easy to see that all such elements lie in I A and that they are independent over Z , sothey form a basis for I A as claimed.(c) Let M be the subgroup of I A generated by all elements of the form h a, b i . As the elements [ a ] − [0]generate I A as an ideal, it is clear from the description h a, b i = h a ih b i that M generates I A as anideal, so it will be enough to check that M itself is already an ideal. This follows easily from theidentity [ x ] h a, b i = h a, b + x i − h a, x i , which can be verified directly from the definitions.(d) It is clear from the definitions that q ( h a, b i ) = a + b − a − b + 0 = 0, so qj = 0 by part (c), so wehave an induced map q : I A /I A → A . In the opposite direction, we can define s : A → I A /I A by s ( a ) = [ a ] − [0] + I A . We have s ( a + b ) − s ( a ) − s ( b ) = [ a + b ] − [ a ] − [ b ] + [0] + I A = h a, b i + I A = I A , which means that s is a homomorphism. It is visible that qs = 1 A , so s is injective and q is surjective.It is also clear from (b) that s ( A ) generates I A /I A but s is a homomorphism so s ( A ) is already asubgroup of I A /I A , so s is surjective. This means that s is an isomorphism, with inverse q . As q isan isomorphism we see that I A j −→ I A q −→ A is exact. (cid:3) Definition 7.3. Let A and B be abelian groups. We regard A and B as subgroups of A × B in the obviousway. In Z [ A × B ] we let J be the ideal generated by all elements h a i with a ∈ A , and we let K be the idealgenerated by all elements h b i with b ∈ B . We then put A ⊗ B = JK/ ( J K + JK ), and write a ⊗ b for thecoset h a, b i + J K + JK ∈ A ⊗ B . We also write Tor( A, B ) = ( J K ∩ JK ) / ( J K ). Remark 7.4. Note that A ⊗ B is generated by elements of the form a ⊗ b . Moreover, because h a + a ′ , b i − h a, b i − h a ′ , b i = h a ih a ′ ih b i ∈ J K h a, b + b ′ i − h a, b i − h a, b ′ i = h a ih b ih b ′ i ∈ JK we see that these satisfy a ⊗ ( b + b ′ ) = ( a ⊗ b ) + ( a ⊗ b ′ )( a + a ′ ) ⊗ b = ( a ⊗ b ) + ( a ′ ⊗ b ) . It follows easily that a ⊗ a ∈ A , and 0 ⊗ b = 0 for all b ∈ B , and ( na ) ⊗ ( mb ) = nm ( a ⊗ b ) forall n, m ∈ Z . In fact, A ⊗ B can be defined more loosely as the abelian group generated by symbols a ⊗ b subject only to the relations a ⊗ ( b + b ′ ) = ( a ⊗ b ) + ( a ⊗ b ′ ) and ( a + a ′ ) ⊗ b = ( a ⊗ b ) + ( a ′ ⊗ b ). emark 7.5. Suppose we have homomorphisms f : A → A ′ and g : B → B ′ . These give a homomorphism f × g : A × B → A ′ × B ′ , which induces a ring map ( f × g ) • : Z [ A × B ] → Z [ A ′ × B ′ ]. This sends the ideals J and K to the coresponding ideals in Z [ A ′ × B ′ ] and so induces a homomorphism A ⊗ B → A ′ ⊗ B ′ , whichwe denote by f ⊗ g . By construction we have ( f ⊗ g )( a ⊗ b ) = f ( a ) ⊗ g ( b ). It is not hard to see that thisconstruction is functorial, in the sense that 1 A ⊗ B = 1 A ⊗ B and that ( f ′ ⊗ g ′ )( f ⊗ g ) = ( f ′ f ) ⊗ ( g ′ g ) for all f ′ : A ′ → A ′′ and g ′ : B ′ → B ′′ . It is also bilinear in the following sense: if f , f : A → A ′ and g , g : B → B ′ then ( f + f ) ⊗ ( g + g ) = ( f ⊗ g ) + ( f ⊗ g ) + ( f ⊗ g ) + ( f ⊗ g )as homomorphisms from A ⊗ B to A ′ ⊗ B ′ . Remark 7.6. As in Proposition 7.2, one can check that(a) J is freely generated as an abelian group by the elements h a i [ b ] = [ a + b ] − [ b ] with a ∈ A \ { } and b ∈ B .(b) K is freely generated as an abelian group by the elements [ a ] h b i = [ a + b ] − [ a ] with a ∈ A and b ∈ B \ JK is freely generated as an abelian group by the elements h a, b i with a ∈ A \ b ∈ B \ J K is generated as an abelian group by the elements h a ih a ′ ih b i with a, a ′ ∈ A and b ∈ B .(e) JK is generated as an abelian group by the elements h a ih b ih b ′ i with a ∈ A and b, b ′ ∈ B .We can generalise the identities in Remark 7.4 as follows: Definition 7.7. Let A , B and V be abelian groups. We say that a function f : A × B → V is bilinear if forall a, a ′ ∈ A and all b, b ′ ∈ B we have f ( a, b + b ′ ) = f ( a, b ) + f ( a, b ′ ) f ( a + a ′ , b ) = f ( a, b ) + f ( a ′ , b ) . More generally, we say that a map g : A × · · · × A n → V is multilinear (or more specifically n -linear ) if foreach k and each a , . . . , a k − , a k +1 , . . . , a n , the map x g ( a , . . . , a k − , x, a k +1 , . . . , a n )is a homomorphism from A k to V . Example 7.8. (a) Matrix multiplication defines a bilinear map µ : M n ( Z ) × M n ( Z ) → M n ( Z ) by µ ( M, N ) = M N .(b) The dot product defines a bilinear map R × R → R , the cross product defines a bilinear map R × R → R , and the triple product ( u, v, w ) u. ( v × w ) defines a trilinear map R × R × R → R .(c) By construction, we have a bilinear map ω : A × B → A ⊗ B defined by ω ( a, b ) = a ⊗ b .Example (c) above is in a sense the universal example, as explained by the following result: Proposition 7.9. Let f : A × B → V be a bilinear map. Then there is a unique homomorphism f : A ⊗ B → V such that f ( a ⊗ b ) = f ( a, b ) for all a ∈ A and b ∈ B (or equivalently f ◦ ω = f ).Proof. As A ⊗ B is generated by the elements a ⊗ b , it is clear that f will be unique if it exists.By Lemma 6.5, there is a unique homomorphism f : Z [ A × B ] → V such that f ([ a, b ]) = f ( a, b ) for all a and b . Note that for a ∈ A we have f ([ a ]) = f ([ a, f ( a, 0) = 0, and similarly f ([ b ]) = 0 for b ∈ B , so f ( h a, b i ) = f ([ a, b ]) − f ([ a, − f ([0 , b ]) + f ([0 , f ( a, b ) . Note also that f ( h a ih a ′ ih b i ) = f ( h a + a ′ , b i ) − f ( h a, b i ) − f ( h a ′ , b i ) = f ( a + a ′ , b ) − f ( a, b ) − f ( a ′ , b ) = 0 , so (using Remark 7.6(d)) we see that f ( J K ) = 0. Similarly, we have f ( JK ) = 0, so f induces ahomomorphism f : A ⊗ B = JKJ K + JK → V with f ( a ⊗ b ) = f ( h a, b i + J K + JK ) = f ( h a, b i ) = f ( a, b )as required. (cid:3) emark 7.10. For example, we have a bilinear map µ : M n ( Z ) × M n ( Z ) → M n ( Z ) given by µ ( M, N ) = M N ,so there is a unique homomorphism µ : M n ( Z ) ⊗ M n ( Z ) → M n ( Z ) such that µ ( M ⊗ N ) = M N . Ratherthan spelling this out explicitly, we will usually just say that µ : M n ( Z ) ⊗ M n ( Z ) → M n ( Z ) is defined by µ ( M ⊗ N ) = M N .It will be convenient to reformulate Proposition 7.9 in a different way. We write Hom( A, B ) for theset of homomorphisms from A to B , considered as a group under pointwise addition. Similarly, we writeBilin( A, B ; V ) for the group of bilinear maps from A × B to V . Proposition 7.11. For any abelian groups A , B and V , there are natural isomorphisms Hom( A ⊗ B, V ) ≃ Bilin( A, B ; V ) ≃ Hom( A, Hom( B, V )) ≃ Hom( B, Hom( A, V )) . More specifically, if we have elements f ∈ Hom( A ⊗ B, V ) f ∈ Bilin( A, B ; V ) f ∈ Hom( A, Hom( B, V )) f ∈ Hom( B, Hom( A, V )) then they are related by the above isomorphisms if and only if for all a ∈ A and b ∈ B we have f ( a ⊗ b ) = f ( a, b ) = f ( a )( b ) = f ( b )( a ) . Proof. This is mostly trivial. For any map f : A × B → V , we can define a map f : A → Map( B, V )by f ( a )( b ) = f ( a, b ). If f satisfies the right-linearity condition f ( a, b + b ′ ) we see that f ( a )( b + b ′ ) = f ( a )( b ) + f ( a )( b ′ ), so f ( a ) is a homomorphism from B to V , or in other words f is a map from A toHom( B, V ). If f also satisfies the left linearity condition f ( a + a ′ , b ) = f ( a, b ) + f ( a ′ , b ) then we seethat f ( a + a ′ ) is the sum of the homomorphisms f ( a ) and f ( a ′ ), so f itself is a homomorphism, or inother words f ∈ Hom( A, Hom( B, V )). All of this is reversible, so we have an isomorphism Bilin( A, B ; V ) ≃ Hom( A, Hom( B, V )). We can define f ( b )( a ) = f ( a, b ) to obtain a similar isomorphism Bilin( A, B ; V ) ≃ Hom( B, Hom( A, V )). Finally, Proposition 7.9 gives an isomorphism Bilin( A, B ; V ) ≃ Hom( A ⊗ B, V ). (cid:3) Proposition 7.12. There are natural isomorphisms as follows: η A : Z ⊗ A → A η A ( n ⊗ a ) = naτ AB : A ⊗ B → B ⊗ A τ AB ( a ⊗ b ) = b ⊗ aα ABC : A ⊗ ( B ⊗ C ) → ( A ⊗ B ) ⊗ C α ABC ( a ⊗ ( b ⊗ c )) = ( a ⊗ b ) ⊗ c. In other words, the operation ( − ) ⊗ ( − ) is commutative, associative and unital up to natural isomorphism.Proof. First, we certainly have a bilinear map η ′ A : Z × A → A given by η ′ A ( n, a ) = na , and by Proposition 7.9this gives a homomorphism η A : Z ⊗ A → A as indicated. We can also define a homomorphism ζ A : A → Z ⊗ A by ζ A ( a ) = 1 ⊗ a , and it is clear that η A ζ A = 1 A . In the opposite direction, we must show that the map ξ = 1 Z ⊗ A − ζ A η A : Z ⊗ A → Z ⊗ A is zero. By Proposition 7.11, it will suffice to show that the corresponding homomorphism ξ ′ : Z → Hom( A, Z ⊗ A ) is zero. This is given by ξ ′ ( n )( a ) = ( n ⊗ a ) − (1 ⊗ na )so visibly ξ ′ (1) = 0 but ξ ′ is a homomorphism so ξ ′ ( n ) = n.ξ ′ (1) = 0 for all n as required.Next, Lemma 6.5 tells us that there is a unique homomorphism τ ′ AB : Z [ A × B ] → Z [ B × A ] with τ ′ AB [ a, b ] =[ b, a ]. It is visible that this is an isomorphism (with inverse τ ′ BA ) and that τ ′ AB ( R AB ) = R BA so there is aninduced isomorphism τ AB : A ⊗ B → B ⊗ A , with inverse τ BA .Now fix a ∈ A , and define α ′′ ( a ) : B × C → ( A ⊗ B ) ⊗ C by α ′′ ( a )( b, c ) = ( a ⊗ b ) ⊗ c . This is bilinear, andmoreover α ′′ ( a + a ′ ) = α ′′ ( a ) + α ′′ ( a ′ ), so we have a homomorphism α ′′ : A → Bilin( B, C ; ( A ⊗ B ) ⊗ C ) . We also have an isomorphismBilin( B, C ; ( A ⊗ B ) ⊗ C ) ≃ Hom( B ⊗ C, ( A ⊗ B ) ⊗ C )and using that we obtain a homomorphism α ′ : A → Hom( B ⊗ C ; ( A ⊗ B ) ⊗ C ) . haracterised by α ′ ( a )( b ⊗ c ) = ( a ⊗ b ) ⊗ c . As in Proposition 7.11 this corresponds to a homomorphism α : A ⊗ ( B ⊗ C ) → ( A ⊗ B ) ⊗ C , given by α ( a ⊗ ( b ⊗ c )) = α ′ ( a )( b ⊗ c ) = ( a ⊗ b ) ⊗ c. In the same way, we can construct β : ( A ⊗ B ) ⊗ C → A ⊗ ( B ⊗ C ) with β (( a ⊗ b ) ⊗ c ) = a ⊗ ( b ⊗ c ). Thismeans that βα ( x ) = x whenever x has the form a ⊗ ( b ⊗ c ), but elements of that form generate A ⊗ ( B ⊗ C ),so βα = 1. A similar argument shows that αβ = 1, so α is an isomorphism as required. (cid:3) Proposition 7.13. For any families of abelian groups ( A i ) i ∈ I and ( B j ) j ∈ J there is a natural isomorphism M i ∈ I A i ! ⊗ M j ∈ J B j ≃ M ( i,j ) ∈ I × J ( A i ⊗ B j ) . In particular, for any A , B and C we have A ⊗ ( B ⊕ C ) ≃ ( A ⊗ B ) ⊕ ( A ⊗ C ) .Proof. For brevity, let L and R be the left and right hand sides of the claimed isomorphism. For each i and j we can define a bilinear map g ′ ij : A i × B j → L by g ′ ij ( a, b ) = ι i ( a ) ⊗ ι j ( b ). This gives a homomorphism g ij : A i ⊗ B j → L , and Proposition 3.7 tells us that there is a unique g : R → L with g ◦ ι ij = g ij for all i and j . In the opposite direction, we can define a bilinear map f ′ : ( L A i ) × ( L B j ) → R by f ′ ( a, b ) = X i ∈ supp( a ) X j ∈ supp( b ) ι ij ( a i ⊗ b j ) . This corresponds in the usual way to a homomorphism f : L → R . We leave it to the reader to check that f g = 1 R and gf = 1 L . (cid:3) Corollary 7.14. Given sets I and J and an abelian group B , we have natural isomorphisms Z [ I ] ⊗ B ≃ L i ∈ I B and Z [ I ] ⊗ Z [ J ] ≃ Z [ I × J ] . In particular, this gives Z r ⊗ B ≃ B r and Z r ⊗ Z s ≃ Z rs .Proof. By applying the Proposition to the family { Z } i ∈ I and the family consisting of the single group B ,we obtain Z [ I ] ⊗ B ≃ L i ∈ I B . If instead we use the family { Z } j ∈ J on the right hand side, we obtain Z [ I ] ⊗ Z [ J ] ≃ Z [ I × J ]. (cid:3) Another straightforward example is as follows: Proposition 7.15. For any integer n and any abelian group A there is a natural isomorphism ( Z /n ) ⊗ A ≃ A/nA . In particular, we have ( Z /n ) ⊗ ( Z /m ) = Z / ( n, m ) , where ( n, m ) denotes the greatest common divisorof n and m .Proof. We can define a bilinear map f ′ : ( Z /n ) ⊗ A → A/nA by f ′ ( k + n Z , a ) = ka + nA , and this inducesa homomorphism f : ( Z /n ) ⊗ A → A/nA . In the opposite direction, we can define g : A/nA → ( Z /n ) ⊗ A by g ( a + nA ) = (1 + n Z ) ⊗ a . We leave it to the reader to check that these are well-defined and that f g and gf are identity maps. In particular, this gives ( Z /n ) ⊗ ( Z /m ) = Z / ( n Z + m Z ), but it is standard that n Z + m Z = ( n, m ) Z . (cid:3) We saw in Section 5 that every finitely generated abelian group is a direct sum of groups of the form Z or Z /p v . We can thus use Proposition 7.13, Corollary 7.14 and Proposition 7.15 to understand the tensorproduct of any two finitely generated abelian groups.We next consider the interaction between tensor products and exactness. Proposition 7.16. Let A , B , C and U be abelian groups. (a) If we have an exact sequence A j −→ B q −→ C → , then the resulting sequence U ⊗ A ⊗ j −−→ U ⊗ B ⊗ q −−→ U ⊗ C → is also exact. (In other words, tensoring is right exact .) (b) If we have an injective map j : A → B and U is torsion-free, then ⊗ j : U ⊗ A → U ⊗ B is alsoinjective. c) If we have a short exact sequence → A j −→ B q −→ C → , and U is torsion-free, then the resulting sequence → U ⊗ A ⊗ j −−→ U ⊗ B ⊗ q −−→ U ⊗ C → is also short exact. For the first part, we will use the following criterion: Lemma 7.17. A sequence A j −→ B q −→ C → is exact iff for every abelian group V , the resulting sequence → Hom( C, V ) q ∗ −→ Hom( B, V ) j ∗ −→ Hom( A, V ) is exact. Remark 7.18. The evident analogous statement with short exact sequences is not valid. We will investigatethis in more detail later. Proof. Let S denote the first sequence, and write Hom( S , V ) for the second one.Suppose that S is exact, so q is surjective and ker( q ) = image( j ). Suppose that f ∈ ker( q ∗ ), so f : C → V and f q = 0 : B → V . This means that f ( q ( b )) = 0 for all b ∈ B , but q is surjective, so f ( c ) = 0 for all c ∈ C ,so f = 0. This proves that ker( q ∗ ) = 0, so q ∗ is injective. Now suppose that g ∈ ker( j ∗ ), so g : B → V and gj = 0, or equivalently g (image( j )) = 0, or equivalently g (ker( q )) = 0. We thus have a well-definedmap f : C → V given by f ( c ) = g ( b ) for any b with q ( b ) = c . Now f ∈ Hom( C, V ) and q ∗ ( f ) = f q = g , so g ∈ image( q ∗ ). This proves that ker( j ∗ ) = image( q ∗ ), so Hom( S , V ) is exact.Conversely, suppose that Hom( S , V ) is exact for all V . Take V = cok( q ) = C/q ( B ) and let f : C → V be the evident projection (which is surjective). By construction we have q ∗ ( f ) = 0 but q ∗ is assumed to beinjective so f is zero as well as being surjective. This implies that C/q ( B ) = 0 so q is surjective.Now instead take V = C . As Hom( S , C ) is exact, we certainly have j ∗ q ∗ = 0 : Hom( C, C ) → Hom( A, C ).In particular, we see that j ∗ q ∗ (1) = 0 in Hom( A, C ), or in other words that qj = 0 : A → C . This impliesthat image( j ) ≤ ker( q ).Finally, take V = cok( j ) = B/j ( A ), and let g : B → V be the projection. Then j ∗ ( g ) = gj = 0, so g ∈ ker( j ∗ ) = image( q ∗ ), so there exists f : C → B/j ( A ) with f q = g : B → B/j ( A ). Now if q ( b ) = 0 then b + image( j ) = g ( b ) = f ( q ( b )) = 0, so b ∈ image( j ). This proves that ker( q ) = image( j ), so S is exact asclaimed. (cid:3) Proof of Proposition 7.16. (a) By Lemma 7.17, the sequence0 → Hom( C, V ) q ∗ −→ Hom( B, V ) j ∗ −→ Hom( A, V )is exact for all V . As V is arbitrary we can replace it by Hom( U, V ), where U and V are botharbitrary. This gives an exact sequence0 → Hom( C, Hom( U, V )) q ∗ −→ Hom( B, Hom( U, V )) j ∗ −→ Hom( A, Hom( U, V )) , and we can use Proposition 7.11 to rewrite it as0 → Hom( U ⊗ C, V ) (1 ⊗ q ) ∗ −−−−→ Hom( U ⊗ B, V ) (1 ⊗ j ) ∗ −−−−→ Hom( U ⊗ A, V ) . Finally we apply Lemma 7.17 in the opposite direction to see that the sequence U ⊗ A → U ⊗ B → U ⊗ C → j : A → B , and that U is torsion-free. We mustshow that (1 ⊗ j ) : U ⊗ A → U ⊗ B is injective. If U is actually free then we may assume that U = Z [ I ] for some set I . In this case Corollary 7.14 tells us that 1 ⊗ j is just a direct sum of copies of j and the claim is clear. In particular, this holds whenever U is finitely generated and torsion-free,as we see from Proposition 5.11. The real issue is to deduce the infinitely generated case from thefinitely generated case. Suppose we have an element x ∈ U ⊗ A with (1 ⊗ j )( x ) = 0. We can write x in the form x = P ni =1 u i ⊗ a i say. We then have P ni =1 u i ⊗ j ( a i ) = 0. Going back to Definition 7.3,we deduce that P i [ u i , j ( a i )] can be expressed in Z [ U × B ] as a finite Z -linear combination of terms f the form [ u + u ′ , b ] − [ u, b ] − [ u ′ , b ] or [ u, b + b ′ ] − [ u, b ] − [ u, b ′ ]. Choose such an expression, and let S be the (finite) set of elements of U that are involved in that expression, together with the elements u , . . . , u n occuring in our expression for x . Let U be the subgroup of U generated by S , which isfinitely generated and torsion-free. We now have an element x = P i u i ⊗ a i in U ⊗ A and we findthat (1 ⊗ j )( x ) = 0 in U ⊗ B . By the finitely generated case we see that x = 0, but x is the imageof x under the evident homomorphism A ⊗ U → A ⊗ U , so x = 0 as required.(c) This is just a straightforward combination of (a) and (b). (cid:3) Note that in part (b) of the Proposition, it is definitely necessary to assume that U is torsion-free. Indeed,we can take j : A → B to be n. Z : Z → Z , and then 1 ⊗ j is just n. U , and these maps are injective for all n > U is torsion free.For various purposes it is important to understand the kernel of 1 ⊗ j in more detail. We will first discussthe case U = Z /n , which is quite straightforward. Definition 7.19. We write A [ n ] for { a ∈ A | na = 0 } , which can also be identified with Hom( Z /n, A ). (Ahomomorphism u : Z /n → A corresponds to the element u (1 + n Z ) ∈ A [ n ].) Proposition 7.20. Fix an integer n > . For any short exact sequence A j −→ B q −→ C , there is a uniquehomomorphism δ : C [ n ] → A/nA = ( Z /n ) ⊗ A such that δ ( q ( b )) = a + nA whenever nb = j ( a ) . Moreover,this fits into an exact sequence → A [ n ] j −→ B [ n ] q −→ C [ n ] δ −→ A/n j −→ B/n q −→ C/n → . Proof. This is just the Snake Lemma (Proposition 2.4) applied to the diagram A B CA B C jn. A qn. B n. C j q (cid:3) It turns out that there are similar six-term exact sequences in much greater generality, involving thegroups Tor( A, B ) introduced in Definition 7.3. We start by recording an obvious fact: Lemma 7.21. There is an isomorphism τ AB : Z [ A × B ] → Z [ B × A ] given by τ AB ([ a, b ]) = [ b, a ] , and thisinduces an isomorphism τ AB : Tor( A, B ) → Tor( B, A ) with inverse τ BA . (cid:3) Lemma 7.22. Let A and B be abelian groups, and let J and K be ideals in Z [ A × B ] as in Definition 7.3.Then there are natural exact sequences → Tor( A, B ) → A ⊗ I B ⊗ j B −−−→ A ⊗ I B ⊗ q B −−−→ A ⊗ B → and → Tor( A, B ) → I A ⊗ B j A ⊗ −−−→ I A ⊗ B q A ⊗ −−−→ A ⊗ B → . Proof. Let J and K be as in Definition 7.3. As JK ≤ JK and J K ≤ J K we have a map f : ( JK ) / ( J K ) → ( JK ) / ( J K ) given by f ( x + J K ) = x + J K . The cokernel is JK/ ( J K + JK ), which is A ⊗ B bydefinition. The kernel is ( JK ∩ J K ) / ( J K ), which is Tor( A, B ) by definition. In other words, we havean exact sequence 0 → Tor( A, B ) → JK J K f −→ JKJ K → A ⊗ B → . Next, we have Z [ A × B ] = Z [ A ] ⊗ Z [ B ] by Corollary 7.14. For p, q ≥ I pA ≤ Z [ A ] and I qB ≤ Z [ B ]. These are free abelian groups by Theorem 6.6, so using Proposition 7.16(b) we see that theevident homomorphisms pA ⊗ I qB I pA ⊗ Z [ B ] Z [ A ] ⊗ I qB Z [ A ] ⊗ Z [ B ]are all injective. This means that I pA ⊗ I qB can be identified with its image in Z [ A × B ], which is just J p K q . Our exact sequence now takes the form0 → Tor( A, B ) → I A ⊗ I B I A ⊗ I B f −→ I A ⊗ I B I A ⊗ I B → A ⊗ B → . Next, as tensoring is right exact (Proposition 7.16(a)) we can identify ( I A ⊗ I B ) / ( I A ⊗ I B ) with ( I A /I A ) ⊗ I B ,which is A ⊗ I B by Proposition 7.2. Similarly, we can identify ( I A ⊗ I B ) / ( I A ⊗ I B ) with A ⊗ I B , so our exactsequence becomes 0 → Tor( A, B ) → A ⊗ I B −→ A ⊗ I B −→ A ⊗ B → (cid:3) Corollary 7.23. If A or B is torsion-free then Tor( A, B ) = 0 . In particular, this holds if A or B is free.Proof. If A is torsion-free then the homomorphism A ⊗ I B → A ⊗ I B is injective by Proposition 7.16(b), but Tor( A, B ) is the kernel so Tor( A, B ) = 0. The other case followssymmetrically. (cid:3) We next discuss the functorial properties of Tor groups. Suppose we have homomorphisms f : A → A ′ and g : B → B ′ . As discussed in Remark 7.5, these give a homomorphism f × g : A × B → A ′ × B ′ , which inducesa ring map ( f × g ) • : Z [ A × B ] → Z [ A ′ × B ′ ], sending J and K to the coresponding ideals in Z [ A ′ × B ′ ].It therefore induces a homomorphism Tor( A, B ) → Tor( A ′ , B ′ ), which we denote by Tor( f, g ). This makesTor( A, B ) a functor of the pair ( A, B ).Now suppose we have two homomorphisms f , f : A → A ′ . We then have ring maps ( f ) • , ( f ) • and( f + f ) • from Z [ A ] to Z [ A ′ ], and it is not true that ( f + f ) • = ( f ) • + ( f ) • . Because of this, it is notobvious that Tor( f + f , g ) = Tor( f , g ) + Tor( f , g ). However, this does turn out to be true, as we nowprove. Proposition 7.24. (a) Given homomorphisms f , f : A → A ′ and g : B → B ′ we have Tor( f + f , g ) = Tor( f , g ) +Tor( f , g ) . (b) Given homomorphisms f : A → A ′ and g , g : B → B ′ we have Tor( f, g + g ) = Tor( f, g ) +Tor( f, g ) . (c) Given families of groups { A i } i ∈ I and { B j } j ∈ J there is a natural isomorphism M i,j Tor( A i , B j ) → Tor M i A i , M j B j . Proof. Lemma 7.22 allows us to describe Tor( A, B ) as the kernel of the map 1 ⊗ j : A ⊗ I B → A ⊗ I B , andclaim (a) follows easily from this. Claim (b) is proved similarly.For (c), let us put A ∗ = L i A i and B ∗ = L j B j and T = L i,j Tor( A i , A j ). We have inclusions ι i : A i → A ∗ and ι j : B j → B ∗ , which give homomorphisms Tor( ι i , ι j ) : Tor( A i , B j ) → Tor( A ∗ , B ∗ ). Wealso have inclusions ι ij : Tor( A i , B j ) → T . By the universal property of coproducts (Proposition 3.7) thereis a unique homomorphism φ : T → Tor( A ∗ , B ∗ ) with φ ◦ ι ij = Tor( ι i , ι j ) for all i and j . It is this map thatwe claim is an isomorphism.For fixed j , we can identify Tor( A ∗ , B j ) with the kernel of the evident map A ∗ ⊗ I B j → A ∗ ⊗ I B j . Fromthis it follows easily that Tor( A ∗ , B j ) = L i Tor( A i , B j ). A similar argument shows that Tor( A ∗ , B ∗ ) = L j Tor( A ∗ , B j ), and by putting these together we obtain Tor( A ∗ , B ∗ ) ≃ T as claimed. We leave it to thereader to check that the isomorphism arising from this argument is the same as φ . (cid:3) roposition 7.25. For any abelian group U and any short exact sequence A → B → C there is a naturalexact sequence → Tor( U, A ) → Tor( U, B ) → Tor( U, C ) → U ⊗ A → U ⊗ B → U ⊗ C → . Proof. Apply the Snake Lemma (Proposition 2.4) to the diagram I U ⊗ A I U ⊗ B I U ⊗ CI U ⊗ A I U ⊗ B I U ⊗ C, noting that the rows are short exact because I U and I U are free. (cid:3) Remark 7.26. Let A and B be abelian groups. We can always choose a free abelian group F and asurjective homomorphism p : F → B . Indeed, one possibility is to use the natural map q : I B → B fromProposition 7.2, but we can also make a less natural choice that is typically much smaller. We then let F ′ denote the kernel of p , and note that this is again free by Theorem 6.6. Now the Proposition gives an exactsequence 0 → Tor( A, F ′ ) → Tor( A, F ) → Tor( A, B ) → A ⊗ F ′ ⊗ i −−→ A ⊗ F → A ⊗ B → . As F ′ and F are free we have Tor( A, F ′ ) = Tor( A, F ) = 0, so we conclude that Tor( A, B ) is isomorphic tothe kernel of 1 ⊗ i .A more traditional approach to Tor groups is to define Tor( A, B ) as the kernel of 1 ⊗ i . In this ap-proach one has to work to prove that the resulting group is well-defined up to canonical isomorphism, andthat Tor( A, B ) ≃ Tor( B, A ) and so on. However, one makes more contact with the general techniques ofhomological algebra, which are important for other reasons. Remark 7.27. Suppose we have subgroups A ′ ≤ A and B ′ ≤ B . Using Proposition 7.25 we see that themaps Tor( A ′ , B ′ ) Tor( A ′ , B )Tor( A, B ′ ) Tor( A, B )are all injective, so we can regard Tor( A ′ , B ′ ) as a subgroup of Tor( A, B ). Proposition 7.28. Tor( A, B ) is the union of the subgroups Tor( A ′ , B ′ ) for finite subgroups A ′ ≤ A and B ′ ≤ B .Proof. Any element of x ∈ Tor( A, B ) has the form x = y + z + J K for some y ∈ J K and z ∈ JK . Wecan write y as P ri =1 n i h a i ih a ′ i ih b i i for some n i ∈ Z and a i , a ′ i ∈ A and b i ∈ B . Similarly, we can write z as P sj =1 m j h c j ih d j ih d ′ j i for some m j ∈ Z and c j ∈ A and d j , d ′ j ∈ B . Let A be the subgroup of A generatedby the elements a i , a ′ i and c j , and let B be the subgroup of B generated by the elements b i , d j and d ′ j . It isthen clear that x ∈ Tor( A , B ) ≤ Tor( A, B ). Moreover, A and B are finitely generated, so the subgroups A ′ = tors( A ) and B ′ = tors( B ) are finite. It follows from Theorem 5.3 that A = A ′ ⊕ P and B = B ′ ⊕ Q ,where P and Q are free. This means that Tor( P, B ′ ) = Tor( P, Q ) = Tor( A ′ , Q ) = 0 by Corollary 7.23, soTor( A , B ) = Tor( A ′ , B ′ ). The claim follows. (cid:3) Corollary 7.29. Tor( A, B ) is always a torsion group.Proof. In view of the proposition, it will suffice to prove this when A is finite. In that case there exists n such that n. A = 0, and so Tor( n. A , B ) = 0 : Tor( A, B ) → Tor( A, B ). However, using Proposition 7.24we see that Tor( n. A , B ) = n. Tor(1 A , B ) = n. Tor( A,B ) . This means that nx = 0 for all x ∈ Tor( A, B ), soTor( A, B ) is a torsion group. (cid:3) We next show how to construct elements of Tor( A, B ) more explicitly. Proposition 7.30. a) For n > and a ∈ A [ n ] and b ∈ B [ n ] there is an element e n ( a, b ) ∈ Tor( A, B ) given by e n ( a, b ) = n h a ih b i + J K ∈ J K ∩ JK J K = Tor( A, B ) . (b) The map e n : A [ n ] × B [ n ] → Tor( A, B ) is bilinear, so it induces a homomorphism A [ n ] ⊗ B [ n ] → Tor( A, B ) . (c) Suppose we have elements a ∈ A [ n ] and b ∈ B [ m ] . Let d be any common divisor of n and m . Then ( n/d ) a ∈ A [ m ] and ( m/d ) b ∈ A [ n ] and we have e n ( a, ( m/d ) b ) = e nm/d ( a, b ) = e m (( n/d ) a, b ) . (d) Suppose we have a short exact sequence → A j −→ B q −→ C → giving rise to an exact sequence → Tor( U, A ) → Tor( U, B ) → Tor( U, C ) δ −→ U ⊗ A → U ⊗ B → U ⊗ C → as in Proposition 7.25. Suppose that u ∈ U [ n ] and c ∈ C [ n ] . Then δ ( e n ( u, c )) can be describedas follows: we choose b ∈ B with q ( b ) = c , then there is a unique a ∈ A with j ( a ) = nb , and δ ( e n ( u, c )) = u ⊗ a .Proof. (a) First, as na = 0 in A ≃ I A /I A we see that n h a i ∈ I A and so n h a ih b i ∈ I A ⊗ I B = J K . Similarly wehave n h b i ∈ I B so n h a ih b i = h a i . ( n h b i ) ∈ I A ⊗ I B = JK . Thus, we have n h a ih b i ∈ J K ∩ JK andthe definition of e n is meaningful.(b) Next, recall that h a + a ′ i − h a i − h a ′ i = h a ih a ′ i ∈ I A , so n h a + a ′ ih b i − n h a ih b i − n h a ′ ih b i = ( h a ih a ′ i ) . ( n h b i ) ∈ I A ⊗ I B = J K , so e n ( a + a ′ , b ) = e n ( a, b ) + e n ( a ′ , b ). By a symmetrical argument we see that e n ( a, b ) is also linearin b , as required.(c) Suppose we have elements a ∈ A [ n ] and b ∈ B [ m ]. Let d be any common divisor of n and m , so n = pd and m = qd for some p, q > 0. Now m. ( n/d ) a = pqda = q.na = 0, so ( n/d ) a = pa ∈ A [ m ],and similarly ( m/d ) b = qb ∈ B [ m ]. Next note that pd h a ih qb i − pqd h a ih b i = ( n h a i )( h qb i − q h b i ) ∈ I A ⊗ I B = J K , so e pd ( a, qb ) = e pqd ( a, b ), or e n ( a, ( m/d ) b ) = e nm/d ( a, b ). By a symmetrical argument, we also have e nm/d ( a, b ) = e m (( n/d ) a, b ).(d) For u ∈ U [ n ] and c ∈ C [ n ] we note that n h u i ∈ I U so we can put e ′ n ( u, c ) = ( n h u i ) ⊗ c ∈ I U ⊗ C . Notethat it is not legitimate to rewrite this as h u i ⊗ nc because h u i 6∈ I U . However this rewriting becomeslegitimate if we work in I U ⊗ C , and the result is zero because nc = 0. In other words, e ′ n ( u, c ) liesin the kernel of the map I U ⊗ A → I U ⊗ A , which was identified with Tor( U, A ) in Lemma 7.22. Itis not hard to check that e ′ n ( u, c ) corresponds to e n ( u, c ) under this identification. Now consider thediagram I U ⊗ A I U ⊗ B I U ⊗ CI U ⊗ A I U ⊗ B I U ⊗ C. ⊗ ji ⊗ ⊗ qi ⊗ i ⊗ ⊗ j ⊗ q By inspecting the proof of Proposition 2.4 we find the following prescription for the connectingmap δ : Tor( U, C ) → A ⊗ C . Given z ∈ I U ⊗ C with ( i ⊗ z ) = 0 we choose y ∈ I U ⊗ B with(1 ⊗ q )( y ) = z , then we check that ( i ⊗ y ) ∈ ker(1 ⊗ q ) = image(1 ⊗ j ) so there exists x ∈ I U ⊗ A with (1 ⊗ j )( x ) = ( i ⊗ y ), and the image of x in U ⊗ A = ( I U /I U ) ⊗ A is by definition δ ( z ). Nowsuppose again that we are given u ∈ U [ n ] and c ∈ C [ n ] and take z = e ′ n ( u, c ). As q is surjective wecan choose b ∈ B with q ( b ) = a . We then have q ( nb ) = na = 0 so nb ∈ ker( q ) = image( j ), so thereexists a ∈ A with j ( a ) = nb . We can thus put y = ( n h u i ) ⊗ b ∈ I U ⊗ B and x = h u i ⊗ a ∈ I U ⊗ A , and e find that (1 ⊗ q )( y ) = z and ( i ⊗ y ) = h u i ⊗ nb = (1 ⊗ j )( h u i ⊗ a ). This means that δ ( e n ( u, a ))is the image of h u i ⊗ a in U ⊗ A , which is just u ⊗ a as claimed. (cid:3) Corollary 7.31. For any abelian group U and n > , the map u e n ( u, n Z ) gives an isomorphism U [ n ] Tor( U, Z /n ) .Proof. Consider the short exact sequence Z n −→ Z → Z /n . As Tor( U, Z ) = 0 and U ⊗ Z = Z the six-termexact sequence reduces to 0 → Tor( U, Z ) δ −→ U n −→ U −→ U/n −→ , which means that δ gives an isomorphism δ : Tor( U, Z ) → U [ n ]. Part (d) of the proposition tells us that δ ( e n ( u, n Z )) = u for all u ∈ U [ n ], and the claim follows from this. (cid:3) Corollary 7.32. For any abelian groups A and B , the group Tor( A, B ) is generated by all the elements e n ( a, b ) for n > and a ∈ A [ n ] and b ∈ B [ n ] .Proof. In view of Proposition 7.28, it is enough to prove this when A and B are finite. Now A and B canbe written as direct sums of finite cyclic groups, and using Proposition 7.24(c) we reduce to the case where A and B are themselves cyclic. That case is immediate from Corollary 7.31. (cid:3) Corollary 7.33. For any abelian group B , there is a natural isomorphism Tor( Q / Z , B ) ≃ tors( B ) .Proof. Let A m be the subgroup of Q / Z generated by 1 /m + Z , and define ǫ m : B [ m ] → Tor( A m , B ) ≤ Tor( Q / Z , B ) by ǫ m ( b ) = e n (1 /m + Z , b ). Note that A m is cyclic of order m , so the previous result tellsus that ǫ m is an isomorphism. Now suppose that m divides n , so B [ m ] ≤ B [ n ]. We can take d = m inProposition 7.30(c) to see that e n ( a, b ) = e m (( n/m ) a, b ) whenever na = 0 and mb = 0. Taking a = 1 /n + Z we deduce that ǫ n ( b ) = ǫ m ( b ) whenever mb = 0, so ǫ n | B [ m ] = ǫ m . It follows that there is a uniquehomomorphism ǫ : tors( B ) = [ n> B [ n ] → Tor( Q / Z , B ) . Every finite subgroup of Q / Z has the form A n for some n , and it follows from this using Proposition 7.28that ǫ is an isomorphism.There is an explicit construction of the inverse which is quite instructive. An element x ∈ Q / Z is a cosetof Z in Q , and any such coset intersects the interval [0 , 1) in a single point, which we call λ ( x ). The function λ : Q / Z → Q is not a homomorphism, but λ ( x + x ′ ) and λ ( x ) + λ ( x ′ ) are both representatives of the samecoset x + x ′ , so we at least have λ ( x + x ′ ) − λ ( x ) − λ ( x ′ ) ∈ Z . We can extend λ to give a group homomorphism Z [ Q / Z ] → Q by λ ( P i n i [ x i ]) = P i n i λ ( x i ). We then findthat λ ( h x ih x ′ i ) = λ ([ x + x ′ ] − [ x ] − [ x ′ ] + [0]) = λ ( x + x ′ ) − λ ( x ) − λ ( x ′ ) ∈ Z , and thus that λ ( I Q / Z ) ≤ Z . We therefore have a homomorphism λ ⊗ I Q / Z ⊗ B → Z ⊗ B = B . Lemma 7.22allows us to regard Tor( Q / Z , B ) as a subgroup of I Q / Z ⊗ B , so we can restrict to this subgroup to get ahomomorphism µ : Tor( Q / Z , B ) → B . We leave it to the reader to check that the image of µ is tors( B ), andthat µ : Tor( Q / Z , B ) → tors( B ) is inverse to ǫ . (cid:3) Proposition 7.34. Define maps L n,m,d ( A [ nd ] ⊗ B [ md ]) L p ( A [ p ] ⊗ B [ p ]) Tor( A, B ) λρ ǫ by λ ( i n,m,d ( a ⊗ b )) = i nd ( a ⊗ mb ) ρ ( i n,m,d ( a ⊗ b )) = i md ( na ⊗ b ) ǫ ( i p ( a ⊗ b )) = e p ( a, b ) . Then the sequence n,m,d ( A [ nd ] ⊗ B [ md ]) L p ( A [ p ] ⊗ B [ p ]) Tor( A, B ) 0 λ − ρ ǫ is exact, so Tor( A, B ) is the cokernel of λ − ρ .Proof. Let T ( A, B ) be the cokernel of λ − ρ , and let e ′ p ( a, b ) be the image of i p ( a ⊗ b ) in T ( A, B ), so byconstruction we have e ′ nd ( a, mb ) = e ′ md ( na, b ) whenever nda = 0 and mdb = 0. Part (c) of Proposition 7.30shows that ǫλ = ǫρ , so img( λ − ρ ) ≤ ker( ǫ ), so there is a unique homomorphism ǫ : T ( A, B ) → Tor( A, B )such that ǫ ( e ′ p ( a, b )) = e p ( a, b ) for all p ≥ a ∈ A [ p ] and b ∈ B [ p ]. This is surjective by Corollary 7.32.We must show that this is also injective.Consider the special case where A and B are finitely generated. It is easy to see that there are naturalsplittings T ( A ⊕ A ′ , B ⊕ B ′ ) ≃ T ( A, B ) ⊕ T ( A, B ′ ) ⊕ T ( A ′ , B ) ⊕ T ( A ′ , B ′ )Tor( A ⊕ A ′ , B ⊕ B ′ ) ≃ Tor( A, B ) ⊕ Tor( A, B ′ ) ⊕ Tor( A ′ , B ) ⊕ Tor( A ′ , B ′ )so we can reduce to the case where A and B are cyclic. If A = Z or B = Z it is clear that T ( A, B ) = 0 =Tor( A, B ), so we may assume that A = Z /r and B = Z /s say. Let t be the least common multiple of r and s , so we have an element u = e ′ t (1 + r Z , s Z ) ∈ T ( A, B ). Note that T ( A, B ) is generated by elements ofthe form v = e p ( a + r Z , b + s Z ) with pa ∈ r Z and pb ∈ s Z , which implies that pab = tc for some c ∈ Z . Wethus have an element x = i c, ,t (1 + r Z , s Z ) + i ,b,ap (1 + r Z , s Z ) − i a, ,p (1 + r Z , b + s Z ) ∈ M n,m,d ( A [ nd ] ⊗ B [ md ]) , and we find that( λ − ρ )( x ) = i p (( a + r Z ) ⊗ ( b + s Z )) − i t ( c + r Z , s Z ) = i p (( a + r Z ) ⊗ ( b + s Z )) − c i t (1 + r Z , s Z )so v = cu . This proves that u generates T ( A, B ). It is also clear that ru = su = 0, so if we put d = ( r, s ) wehave du = 0. This means that T ( A, B ) is cyclic of order dividing d but the map T ( A, B ) → Tor( A, B ) ≃ Z /d is surjective so it must be an isomorphism.We now revert to the general case, where A and B may be infinitely generated. Consider an element w ∈ ker( ǫ ) ≤ T ( A, B ). We can write w as P Ni =1 e ′ p i ( a i , b i ) say, and then let A ′ be the subgroup of A generated by a , . . . , a N , and let B ′ be the subgroup of B generated by b , . . . , b N . There is then an evidentelement w ′ ∈ T ( A ′ , B ′ ) that maps to w in T ( A, B ). Now consider the diagram T ( A ′ , B ′ ) Tor( A ′ , B ′ ) T ( A, B ) Tor( A, B ) . ǫ ≃ ǫ The top map is an isomorphism by the special case considered above, and the right hand map is injective byRemark 7.27. By chasing w ′ around the diagram we see that w = 0. Thus, the map ǫ : T ( A, B ) → Tor( A, B )is injective as required. (cid:3) Ext groups Definition 8.1. Let A and B be abelian groups, and let j be the inclusion I A → I A . We define Ext( A, B ) tobe the kernel of the map j ∗ : Hom( I A , B ) → Hom( I A , B ). Now suppose we have homomorphisms f : A ′ → A and g : B → B ′ . We then have compatible homomorphisms f • : I A ′ → I A and f • : I A ′ → I A , and we can usethese in an evident way to construct a commutative square of mapsExt( A, B ) Ext( A, B ′ )Ext( A ′ , B ) Ext( A ′ , B ′ ) g ∗ f ∗ f ∗ g ∗ emark 8.2. Let q be the usual map I A → A , with kernel I A . Consider the sequence0 → Hom( A, B ) q ∗ −→ Hom( I A , B ) j ∗ −→ Hom( I A , B ) −→ Ext( A, B ) → . By combining Lemma 7.17 with Definition 8.1, we see that this is exact. Remark 8.3. If we have two homomorphisms g , g : B → B ′ , it is clear that( g + g ) ∗ = g ∗ + g ∗ : Ext( A, B ) → Ext( A, B ′ ) . If we have two homomorphisms f , f : A ′ → A , it is not obvious from the definitions that ( f + f ) ∗ = f ∗ + f ∗ ,but later we will give a different description of the Ext groups that makes this fact clear. Lemma 8.4. Let F be a free abelian group. Then for every surjective homomorphism q : B → C , the resultingmap q ∗ : Hom( F, B ) → Hom( F, C ) is also surjective. Equivalently, for every pair of homomorphisms f and q as shown with q surjective, there exists g with qg = f . FB C. g fq Conversely, every group F with this property is free.Proof. Firstly, there exists a map g as shown if and only if the element f ∈ Hom( F, C ) lies in the image of q ∗ : Hom( F, B ) → Hom( F, C ); so the two versions of the statement are indeed equivalent. To prove them,we may assume that F = Z [ I ] for some I . We then have elements f ([ i ]) ∈ C and q : B → C is surjective sowe can choose b i ∈ B with q ( b i ) = f ([ i ]). Now there is a unique homomorphism g : Z [ I ] → B with g ([ i ]) = b i for all i , and qg = f as required.Conversely, let F be any abelian group that has the property under consideration. Take C = F and B = I F , let q : B → C be the usual surjection I F → F , and let f : F → C be the identity. Then there mustexist g : F → I F with qg = 1 F . This means that g is injective, so F is isomorphic to a subgroup of the freegroup I F , so F is free by Theorem 6.6. (cid:3) Remark 8.5. In the more general context of modules over an arbitrary ring, the property in the lemma iscalled projectivity . Thus, we have shown that an abelian group is projective if and only if it is free. The sameargument shows that an R -module is projective if and only if it is a direct summand in a free R -module.The analogue of Theorem 6.6 is not valid for modules over a general ring, so projective modules need not befree. Proposition 8.6. Let U be an abelian group, and let A j −→ B q −→ C be a short exact sequence of abeliangroups. Then there is a natural exact sequence → Hom( U, A ) j ∗ −→ Hom( U, B ) q ∗ −→ Hom( U, C ) δ −→ Ext( U, A ) j ∗ −→ Ext( U, B ) q ∗ −→ Ext( U, C ) → . Proof. We first claim that the sequence0 → Hom( U, A ) j ∗ −→ Hom( U, B ) q ∗ −→ Hom( U, C )is exact. Indeed, if j ∗ ( α ) = 0 then j ( α ( u )) = 0 for all u , but j is injective so α ( u ) = 0 for all u , so α = 0; this proves that j ∗ is injective. Next, suppose that q ∗ ( β ) = 0, so q ( β ( u )) = 0 for all u ∈ U , so β ( u ) ∈ ker( q ) = image( j ), so there exists α ( u ) ∈ A with β ( u ) = j ( α ( u )). This element α ( u ) is in fact unique,because j is injective. We also have j ( α ( u + u ′ ) − α ( u ) − α ( u ′ )) = j ( α ( u + u ′ )) − j ( α ( u )) − j ( α ( u ′ )) = β ( u + u ′ ) − β ( u ) − β ( u ′ ) = 0 , but j is injective so α ( u + u ′ ) − α ( u ) − α ( u ′ ) = 0, so the map α : U → A is a homomorphism. Clearly j ∗ ( α ) = β , so we see that ker( q ∗ ) = image( j ∗ ) as required.Now consider the sequence0 → Hom( I U , A ) j ∗ −→ Hom( I U , B ) q ∗ −→ Hom( I U , C ) → . s U was arbitrary we can replace it by I U to see that j ∗ is injective and image( j ∗ ) = ker( q ∗ ). As I U is freewe also see from Lemma 8.4 that q ∗ is surjective, so the sequence is short exact. The same argument applieswith I U replaced by I U .Now consider the diagramHom( I U , A ) Hom( I U , B ) Hom( I U , C )Hom( I U , A ) Hom( I U , B ) Hom( I U , C ) . j ∗ q ∗ j ∗ q ∗ We have just seen that the rows are short exact, so we can apply the Snake Lemma (Proposition 2.4) to geta six-term exact sequence involving the kernels and cokernels of the vertical maps. Remark 8.2 identifiesthese kernels and cokernels as Hom and Ext groups, as required. (cid:3) Corollary 8.7. An abelian group F is free if and only if Ext( F, A ) = 0 for all A .Proof. First suppose that F is free. By applying Lemma 8.4 to the diagram FI F F s q we obtain a homomorphism s : F → I F with qs = 1 F . This gives a splitting in the usual way, so thereis a unique map r : I F → ker( q ) = I F with jr = 1 − sq and we have rj = 1 I F . Now for any A we havehomomorphisms Hom( I F , A ) r ∗ −→ Hom( I F , A ) j ∗ −→ Hom( I F , A )with j ∗ r ∗ = ( rj ) ∗ = 1 ∗ = 1. This implies that j ∗ is surjective, so the cokernel is zero. But Ext( F, A ) isdefined to be cok( j ∗ ), so Ext( F, A ) = 0 as claimed.Conversely, suppose that Ext( F, A ) = 0 for all A . Let q : B → C be a surjective homomorphism. If weput A = ker( q ), then Proposition 8.6 gives an exact sequence0 → Hom( U, A ) j ∗ −→ Hom( U, B ) q ∗ −→ Hom( U, C ) δ −→ j ∗ −→ q ∗ −→ → . From this we deduce that q ∗ is surjective. By the last part of Lemma 8.4, this means that F is free. (cid:3) To complete our study of Ext groups, we will need to understand groups D with the “dual” property thatExt( A, D ) = 0 for all A . These will turn out to be the divisible groups, as defined below. Definition 8.8. Let V be an abelian group. We say that V is divisible if for all integers n > v ∈ V there exists u ∈ V with nu = v . Equivalently, V is divisible iff all the maps n. V : V → V (for n > 0) aresurjective. Example 8.9. The groups Q , R and Q / Z are all divisible. The only divisible finite group is the trivialgroup. Remark 8.10. It is clear that any quotient of a divisible group is divisible. Remark 8.11. If A is divisible, then (using the Axiom of Choice) we can choose functions d n : A → A for n > n.d n ( a ) = a for all n and a (so in particular d ( a ) = a ), and we can also ensure that d n (0) = 0 for all n . Of course, d n need not be a homomorphism. We will call such a system of maps a division system for A . In many cases one can make an explicit choice for d n . For Q or R we just have d n ( a ) = a/n . For A = Q / Z , every element has a unique representation as a = x + Z with 0 ≤ x < 1, and weput d n ( a ) = x/n + Z . Proposition 8.12. Let D be a divisible group. Then for any injective homomorphism j : A → B , theresulting homomorphism j ∗ : Hom( B, D ) → Hom( A, D ) is surjective. Equivalently, given homomorphisms j and f as shown below, there exists g : B → D with gj = f . BD. jf g Conversely, any group D that has this property is divisible.Proof. Firstly, there exists a map g as shown if and only if the element f ∈ Hom( A, D ) lies in the image of j ∗ : Hom( B, D ) → Hom( A, D ); so the two versions of the statement are indeed equivalent.Next, we claim that if D has the above extension property then it is divisible. This is immediate fromthe special case of the extension property where A = B = Z and j = n. Z for some n > D is divisible, and that we are given j and f as shown. Itwill be harmless to replace A by the isomorphic group j ( A ) ≤ B and so assume that A ≤ B and that j is justthe inclusion. We then need to find g : B → D with g | A = f . For this we choose a division system ( d n ) n> for D and a well-ordering of B . For b ∈ B we let B
0. Note that if n b = 0then B ≤ b = B g ( x ) = d n x ( g ( n x x )).We claim that for all b there is a unique admissible homomorphism g b : B ≤ b → D . If not, let b be the leastelement for which this is false (which is meaningful because B is well-ordered). For all x < b , we havea unique admissible map g x : B ≤ x → D . By uniqueness, we see that g x agrees with g y on B ≤ y whenever y ≤ x < b . It follows that the maps g x can be combined to give a map g ′ : B 0, then we have an element z = d n b ( g ′ ( n b b )) ∈ D with nz = g ′ ( n b b ), and we see that there is a uniqueextension g b : B ≤ b → D satisfying g b ( u + kb ) = g ′ ( u ) + kz for all u ∈ B
A, D ) = 0 for all A . Remark 8.14. The converse statement is also true, but it will be more convenient to prove that later. Proof. Ext( A, D ) is by definition the cokernel of j ∗ : Hom( I A , D ) → Hom( I A , D ), but j ∗ is surjective by theproposition. (cid:3) We next want to show that every abelian group can be embedded in a divisible group. This will be provedafter some preliminaries. Proposition 8.15. Let A be an abelian group, and let a be a nontrivial element of A . Then there is ahomomorphism f : A → Q / Z with f ( a ) = 0 .Proof. Put A = Z a ≤ A . If a has infinite order then A ≃ Z and we can define f : A → Q / Z by f ( ka ) = k/ Z . If a has finite order n we can instead define f : A → Q / Z by f ( ka ) = k/n + Z .Either way we have f ( a ) = 0. Next, as Q / Z is divisible, Proposition 8.12 tells us that the restriction mapHom( A, Q / Z ) → Hom( A , Q / Z ) is surjective, so we can choose f : A → Q / Z with f | A = f . In particular,this means that f ( a ) = 0, as required. (cid:3) Lemma 8.16. For any set I , the group Map( I, Q / Z ) is divisible.Proof. Let ( d n ) n> be the standard division system for Q / Z as in Remark 8.11. The maps u d n ◦ u thengive a division system for Map( I, Q / Z ). (cid:3) Now suppose we have a family of homomorphisms f i : A → Q / Z for i ∈ I . We can combine them to givea single homomorphism j : A → Map( I, Q / Z ) by the rule j ( a )( i ) = f i ( a ). Note that the kernel of j is the ntersection of the kernels of all the homomorphisms f i . Thus, if the family is large enough we can hope that j will be injective.The most canonical thing to do is to consider the family of all homomorphisms f : A → Q / Z , and thusto take I = Hom( A, Q / Z ). This leads us to the following definitions. Definition 8.17. We put E A = Map(Hom( A, Q / Z ) , Q / Z ), and define j : A → E A by the rather tautologicalrule j ( a )( f ) = f ( a ). We write E A for the cokernel of j , and q for the quotient map E A → E A . Proposition 8.18. The groups E A and E A are divisible, and the sequence A j −→ E A q −→ E A is short exact.Proof. The group E A is divisible by Lemma 8.16, and E A is a quotient of E A so it is also divisible. It is clearby construction that q is surjective with image( j ) = ker( q ). Finally, if a ∈ A is nonzero then Proposition 8.15gives us a homomorphism f ∈ Hom( A, Q / Z ) with f ( a ) = 0 or equivalently j ( a )( f ) = 0, so j ( a ) = 0. Thisshows that j is injective, so the sequence is short exact as claimed. (cid:3) Remark 8.19. Note that for a ∈ A and f, g ∈ Hom( A, Q / Z ) we have j ( a )( f + g ) = ( f + g )( a ) = f ( a ) + g ( a ) = j ( a )( f ) + j ( a )( g ) , so the map j ( a ) : Hom( A, Q / Z ) → Q / Z is a homomorphism. In other words, j can be regarded as aninjective homomorphism j : A → Hom(Hom( A, Q / Z ) , Q / Z ) . If we use the briefer notation A ∗ for Hom( A, Q / Z ), then j : A → A ∗∗ ≤ E A . The group A ∗∗ is usually notdivisible, so E A is more useful for our immediate applications to Ext groups. However, the group A ∗∗ willreappear later in other contexts. Corollary 8.20. For all A and B , there is a natural exact sequence → Hom( A, B ) j ∗ −→ Hom( A, E B ) q ∗ −→ Hom( A, E B ) → Ext( A, B ) → . Proof. Apply Proposition 8.6 to the sequence B → E B → E B , noting that Ext( A, E B ) = Ext( A, E B ) = 0by Corollary 8.13. (cid:3) Corollary 8.21. For any f , f : A ′ → A we have ( f + f ) ∗ = f ∗ + f ∗ : Ext( A, B ) → Ext( A ′ , B ) .Proof. The corresponding statement is clearly true for the induced maps on Hom( A, E B ), and Corollary 8.20identifies Ext( A, B ) in a natural way as a quotient group of Hom( A, E B ). (cid:3) Corollary 8.22. The natural maps Ext( M i A i , B ) → Y i Ext( A i , B )Ext( A, Y j B j ) → Y j Ext( A, B j ) . are isomorphisms.Proof. The functors Hom( − , U ) (for U ∈ { B, E B , E B } ) convert coproducts to products, and the cokernelof a product is the product of the cokernels, so the first statement follows from Corollary 8.20. Similarly,the functors Hom( T, − ) (for T ∈ { A, I A , I A } ) preserve products, so the second statement follows from ouroriginal definition of Ext. (cid:3) Proposition 8.23. Let A i −→ B p −→ C be a short exact sequence of abelian groups, and let V be an abeliangroup. Then there is a natural exact sequence → Hom( C, V ) p ∗ −→ Hom( B, V ) i ∗ −→ Hom( A, V ) δ −→ Ext( C, V ) p ∗ −→ Ext( B, V ) i ∗ −→ Ext( A, V ) → . Proof. Consider the diagram om( C, E V ) Hom( B, E V ) Hom( A, E V )Hom( C, E V ) Hom( B, E V ) Hom( A, E V ) q ∗ p ∗ q ∗ i ∗ q ∗ p ∗ i ∗ The rows are short exact by Lemma 7.17 together with Proposition 8.12. The Snake Lemma thereforegives us a six-term exact sequence involving the kernels and cokernels of the vertical maps q ∗ . Corollary 8.20identifies these kernels and cokernels with Hom and Ext groups as required. (cid:3) Corollary 8.24. For any groups B and V , and any subgroup A ≤ B , the restriction Ext( B, V ) → Ext( A, V ) is surjective. (cid:3) Corollary 8.25. There are natural isomorphisms Ext( Z /n, B ) ≃ B/nB for n > , and Ext( Z , B ) = 0 . Note that in conjunction with Corollary 8.22 this allows us to calculate Ext( A, B ) whenever A is finitelygenerated. Proof. Corollary 8.7 tells us that Ext( Z , B ) = 0 for all B . Now consider the short exact sequence Z n. Z −−−→ Z −→ Z /n. Using Proposition 8.23 we obtain an exact sequence0 → Hom( Z /n, B ) → Hom( Z , B ) → Hom( Z , B ) → Ext( Z /n, B ) → Ext( Z , B ) → Ext( Z , B ) → . Now Ext( Z , B ) = 0, while Hom( Z , B ) is easily identified with B , and Hom( Z /n, B ) with B [ n ] = { b ∈ B | nb = 0 } . We thus have an exact sequence0 → B [ n ] → B n. B −−−→ B δ −→ Ext( Z /n, B ) → . From this it is clear that Ext( Z /n, B ) = cok( n. B ) = B/nB . (cid:3) Proposition 8.26. Let A be a torsion group. Then there is a canonical isomorphism Ext( A, Z ) = Hom( A, Q / Z ) = Y p Hom(tors p ( A ) , Q / Z ) , and this maps surjectively to Q p Hom( A [ p ] , Z /p ) . In particular, if Ext( A, Z ) = 0 then A = 0 .Proof. As A is torsion, and both Z and Q are torsion free, we see that Hom( A, Z ) = Hom( A, Q ) = 0. As Q and Q / Z are divisible, we see from Proposition 8.12 that Ext( A, Q ) = Ext( A, Q / Z ) = 0. Thus, if we applyProposition 8.6 to the short exact sequence Z → Q → Q / Z we just get an isomorphism δ : Ext( A, Z ) → Hom( A, Q / Z ). Next, Proposition 4.9 gives A = L p tors p ( A ), so Hom( A, Q / Z ) = Q p Hom(tors p ( A ) , Q / Z ).Now A [ p ] is a subgroup of tors p ( A ) and Q / Z is divisible so the restrictionHom(tors p ( A ) , Q / Z ) → Hom( A [ p ] , Q / Z )is surjective. Moreover, any homomorphism from A [ p ] → Q / Z necessarily lands in ( Q / Z )[ p ], which is a copyof Z /p generated by the element (1 /p ) + Z . By taking the product over all p , we get a surjection Y p Hom(tors p ( A ) , Q / Z ) → Y p Hom( A [ p ] , Z /p )as claimed. The last statement follows from the isomorphism Ext( A, Z ) = Hom( A, Q / Z ) together withProposition 8.15. (cid:3) Example 8.27. We can take A = Q / Z , and we find that Ext( Q / Z , Z ) = End( Q / Z ). It is easy to see that( Q / Z )[ n ] is a copy of Z /n , generated by (1 /n ) + Z . It follows that Hom(( Q / Z )[ p ] , Z /p ) ≃ Z /p , and thusthat Q p Hom(( Q / Z )[ p ] , Z /p ) is uncountable, and thus that End( Q / Z ) is uncountable. Next, we can applyHom( − , Z ) and Ext( − , Z ) to the sequence Z → Q → Q / Z to get a six term exact sequence. As Q and Q / Z are divisible, we find that Hom( Q , Z ) = Hom( Q / Z , Z ) = 0. We also have Hom( Z , Z ) = Z and Ext( Z , Z ) = 0.The six term sequence therefore reduces to a short exact sequence Z → End( Q / Z ) → Ext( Q , Z ). As Z is countable and End( Q / Z ) is uncountable, we deduce that Ext( Q , Z ) is uncountable. In particular, it isnonzero. efinition 8.28. An extension of C by A is a short exact sequence A i −→ B p −→ C . We say that two extensions( A i −→ B p −→ C ) and ( A i −→ B p −→ C ) are equivalent if there exists a map f : B → B with f i = i and p f = p . (Any such f is an isomorphism by a straightforward diagram chase, and using this we see thatthis notion of equivalence is reflexive, symmetric and transitive.) Proposition 8.29. Let E = ( A i −→ B p −→ C ) be an extension of C by A . (a) For any homomorphism h : C ′ → C , we have an extension h ∗ E = ( A i ′ −→ B ′ p ′ −→ C ′ ) given by B ′ = { ( b, c ′ ) ∈ B ⊕ C ′ | p ( b ) = h ( c ′ ) } i ′ ( a ) = ( i ( a ) , p ′ ( b, c ′ ) = c ′ . (We call this the pullback of E along h . ) (b) Suppose we have another extension E ∗ = ( A i ∗ −→ B ∗ p ∗ −→ C ′ ) and a commutative diagram A B ∗ C ′ A B C i ∗ p ∗ g hi p Then E ∗ is equivalent to h ∗ E . (c) For any map m : C ′ → B , the extension ( h + pm ) ∗ E is equivalent to h ∗ E . (d) If E and E are equivalent extensions of C by A , then h ∗ E and h ∗ E are also equivalent.Proof. (a) We can certainly define a group B ′ and homomorphisms i ′ and p ′ by the given formulae. As i isinjective, it is clear that i ′ is injective. Now suppose that c ′ ∈ C ′ . We then have f ( c ′ ) ∈ C and p : B → C is surjective by assumption, so we can choose b ∈ B with p ( b ) = f ( c ′ ). This gives a point b ′ = ( b, c ′ ) ∈ B ′ with p ′ ( b ′ ) = c ′ , so we see that p ′ is surjective. It is immediate that p ′ i ′ = 0, soimage( i ′ ) ≤ ker( p ′ ). A general element of ker( p ′ ) has the form b ′ = ( b, 0) with p ( b ) = f (0) = 0, so b ∈ ker( p ) = image( i ), so b = i ( a ) for some a ∈ A . This means that b ′ = i ′ ( a ) ∈ image( i ′ ). Weconclude that the sequence h ∗ E is indeed short exact, so it gives an extension of C ′ by A .(b) Now suppose we have a commutative diagram as indicated. As hp ∗ = pg we can define g ′ : B ∗ → B ′ by g ′ ( b ∗ ) = ( g ( b ∗ ) , p ∗ ( b ∗ )). We then have p ′ g ′ ( b ∗ ) = p ∗ ( b ∗ ) and g ′ ( i ∗ ( a )) = ( g ( i ∗ ( a )) , p ∗ ( i ∗ ( a ))) = ( i ( a ) , 0) = i ′ ( a ) , so p ′ g ′ = p ∗ and g ′ i ∗ = i ′ . Thus, g ′ gives an equivalence between E ∗ and h ∗ E .(c) By the construction in part (a), we have a commutative diagram A B ′ C ′ A B C i ′ p ′ g hi p (where g ( b, c ′ ) = b ). It follows easily that there is also a commutative diagram A B ′ C ′ A B C. i ′ p ′ g + mp ′ h + pmi p Now part (b) tells us that the top row is equivalent to ( h + pm ) ∗ of the bottom row, or in otherwords h ∗ E ≃ ( h + pm ) ∗ E as claimed.(d) Suppose we have equivalent extensions E k = ( A i k −→ B k p k −→ C ) for k = 0 , 1, so there is an iso-morphism s : B → B with si = i and p s = p . We can then define s ′ : B ′ → B ′ by s ′ ( b , c ′ ) = ( s ( b ) , c ′ ), and we find that s ′ i ′ = i ′ and p ′ s ′ = p ′ ; this shows that h ∗ E and h ∗ E are equivalent as claimed. Proposition 8.30. Let E = ( A i −→ B p −→ C ) be an extension of C by A . (a) For any homomorphism f : A → A ′ , we have an extension f ∗ E = ( A ′ i ′ −→ B ′ p ′ −→ C ) given by R = { ( f ( a ) , − i ( a )) | a ∈ A } ≤ A ′ ⊕ BB ′ = ( A ′ ⊕ B ) /Ri ′ ( a ′ ) = ( a ′ , 0) + Rp ′ (( a ′ , b ) + R ) = p ( b ) . (We call this the pushout of E along f . ) (b) Suppose we have another extension E ∗ = ( A i ∗ −→ B ∗ p ∗ −→ C ′ ) and a commutative diagram Aq B CA ∗ B ∗ C if pgi ∗ p ∗ Then E ∗ is equivalent to f ∗ E . (c) For any map n : B → A ′ , the extension ( f + ni ) ∗ E is equivalent to f ∗ E . (d) If E and E are equivalent extensions of C by A , then f ∗ E and f ∗ E are also equivalent.Proof. (a) We can certainly define groups R and B ′ , and a homomorphism i ′ , by the given formulae. If( a ′ , b ) ∈ R then there exists a ∈ A with a ′ = f ( a ) and b = − i ( a ), so p ( b ) = − p ( i ( a )) = 0. Giventhis, we see that the formula p ′ (( a ′ , b ) + R ) = p ( b ) also gives a well-defined map B ′ → C .If i ′ ( a ′ ) = 0 we must have ( a ′ , ∈ R , so there exists a ∈ A with f ( a ) = a ′ and i ( a ) = 0. As i isinjective this gives a = 0 and then a ′ = f (0) = 0. This proves that i ′ is injective. As p is surjective,it is immediate that p ′ is also surjective. Next, we have p ′ i ′ ( a ′ ) = p ′ (( a ′ , 0) + R ) = p (0) = 0, soimage( i ′ ) ≤ ker( p ′ ). Conversely, suppose we have an element b ′ = ( a ′ , b ) + R ∈ B ′ with p ′ ( b ′ ) = 0.This means that p ( b ) = 0, so b = i ( a ) for some a ∈ A . We then find that ( f ( a ) , − i ( a )) ∈ Rb ′ = ( a ′ , b ) + R = ( a ′ , b ) + ( f ( a ) , − i ( a )) + R = ( a ′ + f ( a ) , 0) + R = i ′ ( a ′ + f ( a )) ∈ image( i ′ ) . We conclude that the sequence f ∗ E is indeed short exact, so it gives an extension of C by A ′ .(b) Now suppose we have a commutative diagram as indicated. Define g ′′ : A ′ ⊕ B → B ∗ by g ′′ ( a ′ , b ) = i ∗ ( a ′ ) + g ( b ). We then have g ′′ ( f ( a ) , − i ( a )) = ( i ∗ f − gi )( a ) = 0, so g ′′ ( R ) = 0, so there is a uniquehomomorphism g ′ : B ′ → B ∗ given by g ′ ( x + R ) = g ′′ ( x ). This means that g ′ i ′ ( a ) = g ′ (( a ′ , 0) + R ) = g ′′ ( a ′ , 0) = i ∗ ( a ′ ), so g ′ i ′ = i ∗ . We also have p ∗ g ′ (( a ′ , b ) + R ) = p ∗ i ∗ ( a ′ ) + p ∗ g ( b ) = 0 + p ( b ) = p ( b ) , so p ∗ g ′ = p . Thus, g ′ gives the required equivalence from f ∗ E to E ∗ .(c) By the construction in part (a), we have a commutative diagram A B CA ′ B ′ C if pgi ′ p ′ (where g ( b ) = (0 , b ) + R ). It follows easily that there is also a commutative diagram A B CA ′ B ′ C if + ni pg + i ′ ni ′ p ′ Now part (b) tells us that the bottom row is equivalent to ( f + ni ) ∗ of the top row, or in other words f ∗ E ≃ ( f + ni ) ∗ E as claimed. d) Suppose we have equivalent extensions E k = ( A i k −→ B k p k −→ C ) for k = 0 , 1, so there is an isomor-phism s : B → B with si = i and p s = p . We can then define s ′ : B ′ → B ′ by s ′ (( a ′ , b ) + R ) = ( a ′ , s ( b )) + R . We find that s ′ i ′ = i ′ and p ′ s ′ = p ′ ; this shows that h ∗ E and h ∗ E are equivalent as claimed. (cid:3) Proposition 8.31. Let A and C be abelian groups, and let Ext ′ ( C, A ) denote the set of equivalence classesof extensions of C by A . Let Q denote the extension ( I C j −→ I C q −→ C ) . Then there is a well-defined bijection ζ : Ext( C, A ) = Hom( I C , A ) j ∗ (Hom( I C , A )) → Ext ′ ( C, A ) given by ζ ( α + image( j ∗ )) = [ α ∗ ( Q )] .Proof. Using Proposition 8.30 (especially part (c)) we see that there is a well-defined map ζ as described.We must show that it is a bijection. Consider an arbitrary extension E = ( A i −→ B p −→ C ). As I C is free and p is surjective, Lemma 8.4 gives us a homomorphism β : I C → B with pβ = q . This means that pβj = qj = 0,so image( βj ) ≤ ker( p ) = image( i ), so there is a unique map α : I C → A with iα = βj . We now have acommutative diagram I C I C CA B C, jα qβi p so Proposition 8.30(b) tells us that E ≃ α ∗ Q , or in other words [ E ] = ζ ( α + image( j ∗ )). This proves that ζ is surjective. Suppose we also have [ E ] = ζ ( α ′ + image( j ∗ )). There is then another commutative diagram I C I C CA B C. jα ′ qβ ′ i p In particular we have pβ ′ = q = pβ so p ( β ′ − β ) = 0, so β ′ − β factors through ker( p ) = image( i ), so there is aunique homomorphism γ : I C → A with β ′ = β + iγ . Now β ′ j = iα ′ and βj = iα so the equation β ′ = β + iγ yields iα ′ = iα + iγj , or i ( α ′ − α − γj ) = 0. As i is injective we conclude that α ′ = α + j ∗ ( γ ), so α ′ and α have the same image in cok( j ∗ ) = Ext( C, A ). This proves that ζ is also injective. (cid:3) The above proposition gives a bijection from the group Ext( C, A ) to the set Ext ′ ( C, A ). There is thusa unique group structure on Ext ′ ( C, A ) for which this bijection is a homomorphism. We would like tounderstand this more intrinsically. Definition 8.32. Suppose we have two extensions E k = ( A i k −→ B k p k −→ C ) for k = 0 , 1. The Baer sum of E and E is the sequence E = ( A i −→ B p −→ C ) where U = { ( b , b ) ∈ B ⊕ B | p ( b ) = p ( b ) } V = { ( i ( a ) , − i ( a )) | a ∈ A } B = U/Vi ( a ) = ( i ( a ) , 0) + V = (0 , i ( a )) + Vp (( b , b ) + V ) = p ( b ) = p ( b ) . Proposition 8.33. In the above context, the sequence E is an extension of C by A , with [ E ] = [ E ]+[ E ] in Ext ′ ( C, A ) . Moreover, the zero element in Ext ′ ( C, A ) is the equivalence class consisting of all split extensions. roof. First, if i ( a ) = 0 we must have ( i ( a ) , ∈ V , so ( i ( a ) , 0) = ( i ( a ′ ) , − i ( a ′ )) for some a ′ ∈ A . As i is injective and i ( a ′ ) = 0 we have a ′ = 0, so the equation i ( a ) = i ( a ′ ) gives i ( a ) = 0 and then a = 0.This shows that i is injective. Next, suppose we have c ∈ C . As both p and p are surjective we canchoose b ∈ B and b ∈ B with p ( b ) = p ( b ) = c . The element b = ( b , b ) + V ∈ B then satisfies p ( b ) = c , so p is surjective. Next, as p i = 0 = p i we see from the definitions that p i = 0, soimage( i ) ≤ ker( p ). Now suppose we have an element b = ( b , b ) + V ∈ B with p ( b ) = 0. This meansthat p ( b ) = 0 = p ( b ), so there is a unique element a ∈ A with b = i ( a ), and also a unique element a ∈ A with b = i ( a ). Put a = a + a and note that i ( a ) = i ( a ) + i ( a ) = ( i ( a ) , 0) + (0 , i ( a )) + V = ( b , b ) + V = b . This proves that ker( p ) = image( i ), so we have an extension as claimed. Now suppose that [ E k ] = ζ ( α k + image( j ∗ )) for k = 0 , 1, so there are commutative diagrams I C I C CA B k C jα k qβ k i k p k for k = 0 , 1. We define α : I C → A by α ( y ) = α ( y ) + α ( y ), and we define β : I C → B by β ( x ) =( β ( x ) , β ( x )) + V . It is straightforward to check that this gives a commutative diagrams as above, showingthat [ E ] = ζ ( α + image( j ∗ )) = ζ ( α + image( j ∗ )) + ζ ( α + image( j ∗ )) = [ E ] + [ E ] . Thus, the sum in Ext ′ ( C, A ) is the Baer sum, as claimed. The zero element is ζ (0), which is the pushout ofthe extension Q = ( I C j −→ I C q −→ C ) along the map 0 : I C → A . If we use the notation of Proposition 8.30 inthis context we have R = { (0 , − j ( y )) | y ∈ I C } = 0 ⊕ I C ≤ A ⊕ I C B ′ = A ⊕ I C R = A ⊕ I C ⊕ I C = A ⊕ ( I C /I C ) ≃ A ⊕ Ci ′ ( a ) = ( a, p ′ ( a, c ) = c. Thus, 0 ∗ Q is just the obvious split extension of C by A . (cid:3) We now present a result that will help us relate homology groups to cohomology groups. There are verystandard theorems that deduce information about cohomology from information about homology. To go inthe opposite direction we need the following proposition, which is less well-known. Proposition 8.34. Suppose that Hom( A, Z ) and Ext( A, Z ) are finitely generated. Then A is finitely gener-ated. The proof will follow after some lemmas. Lemma 8.35. Suppose that Hom( A, Z ) is finitely generated. Then A = B ⊕ F for some subgroups B and F such that F is free and finitely generated, and Hom( B, Z ) = 0 .Proof. Choose maps f , . . . , f r : A → Z that generate Hom( A, Z ), and define f : A → Z r by f ( a ) =( f ( a ) , . . . , f r ( a )). Now f ( A ) is a subgroup of Z r , so it is free, with basis f ( a ) , . . . , f ( a s ) say. Put B = ker( f ),and let F ≤ A be the subgroup generated by a , . . . , a s . We find that f : F → f ( A ) ≃ Z s is an isomorphism,and thus that A = B ⊕ F . Consider a homomorphism g : B → Z . Then the composite A = B ⊕ F proj −−→ B g −→ Z is an element of the group Hom( A, Z ), which is generated by the maps f i , but f i ( B ) = 0, so we see that g = 0. This proves that Hom( B, Z ) = 0. (cid:3) Lemma 8.36. Suppose that Hom( A, Z ) = Ext( A, Z ) = 0 . Then A = 0 . roof. Corollary 8.24 implies that Ext(tors( A ) , Z ) = 0, so Proposition 8.26 gives tors( A ) = 0, so A is torsionfree. We thus have short exact sequences A n −→ A → A/n for all n > 0, and using the resulting six termsequences we deduce that Ext( A/n, Z ) = 0. As A/n is torsion we can use Proposition 8.26 again to seethat A/n = 0, so n. A is surjective. It is also injective because tors( A ) = 0, so it is an isomorphism. Wecan thus make A into a vector space over Q by the rule ( m/n ) .a = ( n. A ) − ( ma ). Linear algebra thereforetells us that either A is zero, or it has Q as a summand. In the latter case Ext( A, Z ) would contain theuncountable group Ext( Q , Z ) as a summand, which is impossible as Ext( A, Z ) = 0. We therefore have A = 0as claimed. (cid:3) Proof of Proposition 8.34. Using Lemma 8.35 we reduce to the case where Hom( A, Z ) = 0. We next claimthat there are only finitely many primes p for which A [ p ] = 0. Indeed, for any such p we see (by linearalgebra over the field Z /p ) that Hom( A [ p ] , Z /p ) = 0. If there are infinitely many such primes, we deducethat the group P = Q p Hom( A [ p ] , Z /p ) is uncountable, which is impossible as Proposition 8.26 tells us that P is a quotient of the finitely generated group Ext( A, Z ). We can thus choose p such that A [ p ] = 0, sowe have a short exact sequence A p −→ A −→ A/p . Proposition 8.23 then tells us that multiplication by p issurjective on Ext( A, Z ). By the structure theory of finitely generated groups, we see that Ext( A, Z ) must befinite, of order n say, and that n must be coprime to p .Next we have short exact sequences A [ n ] i −→ A f −→ nA and nA j −→ A g −→ A/n , where f ( a ) = na and g is the quotient map. By assumption we have Hom( A, Z ) = 0, and also Hom( A [ n ] , Z ) = Hom( A/n, Z ) = 0because Z is torsion free. From the six term sequences we find that Hom( nA, Z ) = 0. We also find that j ∗ : Ext( A, Z ) → Ext( nA, Z ) is surjective and f ∗ : Ext( nA, Z ) → Ext( A, Z ) is injective, but the composite f ∗ j ∗ = ( jf ) ∗ is just multiplication by n . As n was defined to be the order of Ext( A, Z ) we deduce that f ∗ j ∗ = 0, which implies that Ext( nA, Z ) = 0. Lemma 8.36 therefore tells us that nA = 0, so A is torsion andExt( A, Z ) = Hom( A, Q / Z ) = A ∗ . This is finitely generated by assumption. Moreover, we have nA = 0 andtherefore nA ∗ = 0, so A ∗ is a finite group. It follows that A ∗∗ is finite but Remark 8.19 gives an embedding A → A ∗∗ so A is finite, as required. (cid:3) Localisation Definition 9.1. A multiplicative set is a set S of positive integers that contains 1 and is closed undermultiplication. Definition 9.2. Let A be an abelian group, and let S be a multiplicative set. We introduce an equivalencerelation on the set A × S by declaring that ( a, s ) ∼ ( b, t ) iff bsx = atx for some x ∈ S . We write a/s for theequivalence class of the pair ( a, s ), and we write A [ S − ] for the set of equivalence classes. We make this intoan abelian group by the rule a/s + b/t = ( at + bs ) /st. Remark 9.3. Various checks are required to ensure that this definition is meaningful. First, we must showthat the given relation really is an equivalence relation. It is clearly reflexive (as we can take x = 1) andsymmetric. Suppose that ( a, s ) ∼ ( b, t ) ∼ ( c, u ), so there are elements x, y ∈ S with atx = bsx and buy = cty .Put z = txy ∈ S and note that auz = ( au )( txy ) = ( atx )( uy ) = ( bsx )( uy ) = ( buy )( sx ) = ( cty )( sx ) = ( cs )( txy ) = csz, so ( a, s ) ∼ ( c, u ), as required.Next, we should check that addition is well-defined. More specifically, suppose that ( a , s ) ∼ ( a , s ) and( b , t ) ∼ ( b , t ). Put ( c k , u k ) = ( a k t k + b k s k , s k t k ); we must show that ( c , u ) ∼ ( c , u ). By hypothesisthere are elements x, y ∈ S such that a s x = a s x and b t y = b t y . We multiply these two equations by t t y and s s x respectively, and then add them together to get a s xt t y + b t ys s x = a s xt t y + b t ys s x, or equivalently ( a t + b s )( s t ) xy = ( a t + b s )( s t ) xy. If we put z = xy ∈ S this can be rewritten as c u z = c u z , so ( c , u ) ∼ ( c , u ) as required. inally, we should show that addition is commutative and associative, that the element 0 / − a ) /s is an additive inverse for a/s . All this is left to the reader. Remark 9.4. From the definitions we see that a/s = 0 in A [ S − ] if and only if there exists t ∈ S with at = 0. Remark 9.5. In the case A = Z it is not hard to see that Z [ S − ] can be identified with the set { n/s ∈ Q | n ∈ Z , s ∈ S } , which is a subring of Q . Remark 9.6. Consider the case where A is finite, so A is the direct sum of its Sylow subgroups, say A = A ⊕ · · · ⊕ A r with | A i | = p v i i for some primes p , . . . , p r and integers v i > 0. We then find that A [ S − ] = L k A k [ S − ]. Suppose that there exists n ∈ S that is divisible by p k . In A k [ S − ] we then have a/s = ( an v k ) / ( sn v k ) = 0 / ( sn v k ) = 0, so A k [ S − ] = 0. On the other hand, if there is no such n then for each s ∈ S we see that s. A k is invertible for all s ∈ S , and using this we will see later that A k [ S − ] = A k . Thus, A [ S − ] is just the direct sum of some subset of the Sylow subgroups. Definition 9.7. We use different notation for the most popular cases, as follows:(a) If S = { n k | k ≥ } we write A [ n − ] or A [1 /n ] for A [ S − ].(b) If p is prime and S = { n > | n = 0 (mod p ) } = N \ p N then we write A ( p ) for A [ S − ]. This iscalled the p -localisation of A .(c) If S = { n ∈ N | n > } then we write A Q or A (0) for A [ S − ]. This is called the rationalisation of A . Proposition 9.8. Let S be a multiplicative set. Then any homomorphism f : A → B gives a homomorphism f [ S − ] : A [ S − ] → B [ S − ] by the rule f [ S − ]( a/s ) = f ( a ) /s . This construction gives an additive functor,and there is a natural map η : A → A [ S − ] given by η ( a ) = a/ .Proof. First, we see from the definitions that if ( a , s ) ∼ ( a , s ) then ( f ( a ) , s ) ∼ ( f ( a ) , s ). This showsthat f [ S − ] is well-defined. It also follows directly from the definitions that it is a homomorphism. Next, ifwe have maps A f −→ B g −→ C then( gf )[ S − ]( a/s ) = gf ( a ) /s = g [ S − ]( f ( a ) /s ) = g [ S − ]( f [ S − ]( a/s )) , which shows that our construction is functorial. We also claim that η is natural, which means that for any f : A → B the square A BA [ S − ] B [ S − ] fη ηf [ S − ] commutes. This is again straightforward. (cid:3) Remark 9.9. When there is no danger of confusion, we will just write f rather than f [ S − ] for the inducedmap A [ S − ] → B [ S − ]. Proposition 9.10. If A f −→ B g −→ C is exact (or short exact), then so is the localised sequence A [ S − ] f [ S − ] −−−−→ B [ S − ] g [ S − ] −−−−→ C [ S − ] . Proof. First, as gf = 0 and localisation is functorial we see that g [ S − ] f [ S − ] = ( gf )[ S − ] = 0, soimage( f [ S − ]) ≤ ker( g [ S − ]). Now consider an element b/s ∈ ker( g [ S − ]). We then have g ( b ) /s = 0 / B [ S − ], or equivalently g ( b ) x = 0 for some x ∈ S . This means that g ( xb ) = 0 so xb ∈ ker( g ) = image( f ),so there exists a ∈ A with f ( a ) = xb . This implies that f [ S − ]( a/ ( xs )) = f ( a ) / ( xs ) = xb/xs = b/s , so b/s ∈ image( f [ S − ]). We now see that the sequence A [ S − ] f [ S − ] −−−−→ B [ S − ] g [ S − ] −−−−→ C [ S − ] is exact asclaimed. Now suppose that the original sequence is short exact. It is equivalent to say that the sequences0 → A → B and A → B → C and B → C → → A [ S − ] → B [ S − ] and A [ S − ] → B [ S − ] → C [ S − ] and B [ S − ] → C [ S − ] → A [ S − ] f [ S − ] −−−−→ B [ S − ] g [ S − ] −−−−→ C [ S − ] is again shortexact. (cid:3) roposition 9.11. There is a natural isomorphism µ : Z [ S − ] ⊗ A → A [ S − ] given by µ (( n/s ) ⊗ a ) = ( na ) /s .Proof. First, it is straightforward to check that there is a well-defined bilinear map µ : Z [ S − ] × A → A [ S − ]given by µ ( n/s, a ) = ( na ) /s . By the universal property of tensor products, this gives a homomorphism µ : Z [ S − ] ⊗ A → A [ S − ] with µ (( n/s ) ⊗ a ) = ( na ) /s . In the opposite direction, we would like to define ν : A [ S − ] → Z [ S − ] ⊗ A by ν ( a/s ) = (1 /s ) ⊗ a . To see that this is well-defined, suppose that a/s = b/t , so xta = xsb for some x ∈ S . From the definition of tensor products, we have mu ⊗ v = u ⊗ mv in U ⊗ V forall u ∈ U , v ∈ V and m ∈ Z . We can apply this with u = 1 / ( stx ) ∈ Z [ S − ] and v = a and m = tx to get(1 /s ) ⊗ a = (1 / ( stx )) ⊗ xta . By a symmetrical argument, we have (1 /t ) ⊗ b = (1 / ( stx )) ⊗ xsb , but xta = xsb so we find that (1 /s ) ⊗ a = (1 /t ) ⊗ b , as required. It is clear that µν = 1 A [ S − ] . The other way around, wehave νµ (( n/s ) ⊗ a ) = ν (( na ) /s ) = (1 /s ) ⊗ na = ( n/s ) ⊗ a. Thus, ν is inverse to µ . (cid:3) Definition 9.12. Let S be a multiplicative set, and let A be an abelian group. We say that A is S -torsion if for all a ∈ A there exists s ∈ S with sa = 0. We say that A is S -local if for each s ∈ S , the endomorphism s. A : A → A is invertible.Some care is needed in relating the above definition to the traditional terminology in the most popularcases: Definition 9.13. (a) Definition 4.1(c) is equivalent to the following: we say that A is a torsion group if it is S -torsion,where S = { n ∈ N | n > } .(b) We say that A is rational if it is S -local. (We will see that in this case, A can be regarded as avector space over Q .)(c) Now let p be a prime number. We say that A is p -torsion if it is p N -torsion, where p N = { p n | n ∈ N } .(d) However, we say that A is p -local if it is S p -local, where S p = N \ p N . Proposition 9.14. (a) The group A [ S − ] is always S -local. (b) The map η : A → A [ S − ] is an isomorphism if and only if A is S -local. (c) If A is S -local then it can be regarded as a module over the ring Z [ S − ] ≤ Q by the rule ( n/s ) .a =( s. A ) − ( na ) . (d) Suppose that f : A → B is a homomorphism, and that B is S -local. Then there is a unique homo-morphism f ′ : A [ S − ] → B such that f ′ ◦ η = f : A → B .Proof. (a) One can check that for each s ∈ S there is a well-defined map d s : A [ S − ] → A [ S − ] given by d s ( a/t ) = a/ ( st ). This is inverse to s. A [ S − ] .(b) If η is an isomorphism then it follows from (a) that A is S -local. Conversely, if A is S -local one cancheck that the formula ζ ( a/s ) = ( s. A ) − ( a ) gives a well-defined map ζ : A [ S − ] → A , and that thisis inverse to η .(c) First we must check that the multiplication rule is well-defined. Suppose that n/s = m/t , so ntx = msx for some x ∈ S . As this is an equation in Z and x > tn = ms . If wewrite n A for n. A and so on, we deduce that t A n A = m A s A : A → A . We can compose on the leftby t − A and on the right by s − A to get n A s − A = t − A m A . As s − A is a homomorphism, it commuteswith multiplication by n , so s − A n A = t − A m A . This means that the definition of multiplicationis consistent. We will leave it to the reader that it has the usual associativity and distributivityproperties.(d) We define f ′ : A [ S − ] → B by f ′ ( a/s ) = ( s. B ) − ( f ( a )). We leave it to the reader to show thatthis is well-defined and is a homomorphism. Note that f ′ ( a/ 1) = f ( a ), so f ′ η = f . If f ′′ is anotherhomomorphism with f ′′ η = f , we have s A ( f ′′ ( a/s )) = s.f ′′ ( a/s ) = f ′′ ( s. ( a/s )) = f ′′ ( a/ 1) = f ′′ ( η ( a )) = f ( a ) . s s A is invertible we can rewrite this as f ′′ ( a/s ) = s − A ( f ( a )) = f ′ ( a/s ). As a/s was arbitrary thismeans that f ′′ = f ′ , which gives that claimed uniqueness statement. (cid:3) Remark 9.15. As a consequence of (b), we can identify A [ S − ] with A [ S − ][ S − ]. There is a slight subtletyhere: there are two apparently different isomorphisms A [ S − ] → A [ S − ][ S − ], and to keep everything straightit is necessary to prove that they are the same. Indeed, for any B , we have a map η B : B → B [ S − ]. Wecan specialise to the case B = A [ S − ] to get a map η A [ S − ] : A [ S − ] → A [ S − ][ S − ], given by η A [ S − ] ( a/s ) =( a/s ) / 1. Alternatively, we can apply Proposition 9.8 to the map η A : A → A [ S − ] to get another map η A [ S − ] : A [ S − ] → A [ S − ][ S − ], given by ( η A [ S − ])( a/s ) = ( a/ /s . It is clear that s.η A [ S − ] ( a/s ) = ( a/ / s. ( η A [ S − ])( a/s ) , and multiplication by s is an isomorphism on A [ S − ][ S − ], so η A [ S − ] = η A [ S − ]. Proposition 9.16. (a) If we have a short exact sequence A → B → C in which two of the three termsare S -local, then so is the third. (b) Direct sums, products and retracts of S -local groups are S -local. (c) The kernel, cokernel and image of any homomorphism between S -local groups are S -local. (d) p -torsion groups are p -local.Proof. (a) Note that U is S -local iff for each n ∈ S we have U [ n ] = 0 and U/n = 0. Recall also thatProposition 7.20 gives exact sequences0 → A [ n ] j −→ B [ n ] q −→ C [ n ] δ −→ A/n j −→ B/n q −→ C/n → . The claim follows by diagram chasing.(b) This is clear, because direct sums, products and retracts of isomorphisms are isomorphisms.(c) Let f : A → B be a homomorphism between S -local groups. Then img( f ) is a subgroup of B so for n ∈ S we have img( f )[ n ] ≤ B [ n ] = 0. Similarly, img( f ) is a quotient of A so img( f ) /n is a quotientof A/n and so is zero. It follows that img( f ) is S -local. We can therefore apply (a) to the shortexact sequences ker( f ) → A f −→ img( f ) and img( f ) → B → cok( f ) to see that ker( f ) and cok( f ) arealso S -local.(d) Let A be a p -torsion group. Consider m ∈ Z \ p Z ; we must show that m. A is an isomorphism. Forany a ∈ A we can choose k ≥ p k a = 0, and p k is coprime with m so we can choose r, s ∈ Z with p k r + ms = 1. It follows that a = msa = ( m. A )( sa ). Using this we see that m. A issurjective. Also, if ma = 0 then a = sma = 0, so A [ m ] = 0, so m. A is also injective. (cid:3) Proposition 9.17. If A is a torsion group, then A (0) = 0 and A ( p ) = tors p ( A ) .Proof. First suppose that A is a p -torsion group, so A = S k A [ p k ]. If m is not divisible by p then we can find n > mn = 1 (mod p k ), so n. A [ p k ] is inverse to m. A [ p k ] . It follows that m. A is also an isomorphism.This holds for all m ∈ Z \ p Z , so A is p -local, so A = A ( p ) . Now suppose instead that A is a q -torsiongroup for some prime q = p . For any element a ∈ A we have q v a = 0 for some v ≥ 0, so in A ( p ) we have a/m = ( q v a ) / ( q v m ) = 0 / ( q v m ) = 0. This shows that A ( p ) = 0. Finally, for a general torsion group A wehave A = L q tors q ( A ) by Proposition 4.9, so A ( p ) = L q tors q ( A ) ( p ) . By the special cases that we have justdiscussed, this sum contains only the single factor tors p ( A ) ( p ) = tors p ( A ), as claimed. It is also clear fromRemark 9.4 that A (0) = 0. (cid:3) Colimits of sequences Definition 10.1. By a sequence we mean a diagram of the form A f −→ A f −→ A f −→ A f −→ · · · . iven such a sequence and natural numbers i ≤ j , we write f ij for the composite A i f i −→ A i +1 −→ · · · f j − −−−→ A j . In particular, f ii is the identity map of A i , and f i,i +1 = f i . Definition 10.2. Given such a sequence, we consider the group A + = L i A i . For each k we have aninclusion ι k : A k → A + and also a homomorphism ι k +1 ◦ f k : A k → A + . We let R k denote the image of( ι k − ι k +1 f k ) : A k → A + , and put R + = P k R k ≤ A + and lim −→ i A i = A + /R + . This group is called the colimit of the sequence.Next, we write ı k for the composite A k ι k −→ A + −→ A + /R + = lim −→ i A i . By construction we have ı k = ı k +1 f k , so the following diagram commutes: A A A A · · · lim −→ i A i lim −→ i A i lim −→ i A i lim −→ i A i · · · ı f ı f ı f ı f This implies that ı k = ı m f k,m whenever k ≤ m . Remark 10.3. The definition can be reformulated slightly as follows. We can define an endomorphism S of A + by S ( a , a , a , a , . . . ) = (0 , f ( a ) , f ( a ) , f ( a ) , f ( a ) , · · · ) . Equivalently, S is the unique map such that the following diagram commutes for all k : A k A k +1 A + A + ι k f k ι k +1 S We can then say that lim −→ i A i is the cokernel of 1 − S : A + → A + . Note that if a i is the first nonzero entryin a then the i ’th entry in (1 − S )( a ) is again a i , so (1 − S )( a ) = 0. This shows that 1 − S is injective, so weactually have a short exact sequence A + A + lim −→ i A i . − S Remark 10.4. The colimit can also be characterised by a universal property, as follows. A cone for thesequence is a group B with a collection of maps u k : A k → B such that u k +1 f k = u k for all k ≥ 0. Byconstruction, the maps ı k : A k → lim −→ i A i form a cone. We claim that for any cone { A k u k −→ B } k ∈ N there is aunique homomorphism u ∞ : lim −→ i A i → B such that u ∞ ı k = u k for all k . Indeed, Proposition 3.7 gives us aunique map v : A + → B with vi k = u k for all k , and the cone property tells us that v ( R k ) = 0 for all k , so v ( R + ) = 0, so v induces a map u ∞ : lim −→ i A i = A + /R + → B . This is easily seen to be the unique map suchthat u ∞ ı k = u k for all k .In many cases colimits are just unions, as we now explain. Proposition 10.5. We have ı ( A ) ≤ ı ( A ) ≤ ı ( A ) ≤ · · · ≤ lim −→ i A i . Moreover, lim −→ i A i is the union of the groups ı k ( A k ) , and we have ı k ( a ) = 0 iff f k,m ( a ) = 0 for some m ≥ k . roof. First, when k ≤ m we have ı k = ı m f k,m , and this implies that ı k ( A k ) ≤ ı m ( A m ). Next, as lim −→ i A i isa quotient of A + , we see that every element a ∈ lim −→ i A i can be written as a = P Nk =0 ı k ( a k ) for some N ≥ a k ∈ A k . Now ı k ( a k ) ∈ ı k ( A k ) ≤ ı N ( A N ) for all k , so a ∈ ı N ( A N ). Thus lim −→ i A i = S N ı N ( A N ) asclaimed. Now suppose we have a ∈ A k and that f km ( a ) = 0 for some m ≥ k . Using ı k = ı m f km we deducethat ı k ( a ) = 0. Conversely, suppose that ı k ( a ) = 0, so i k ( a ) ∈ R + , so i k ( a ) = N − X m =0 ( i m ( b m ) − i m +1 ( f m ( b m )))for some N > k and some b , . . . , b N − with b i ∈ A i . Now let h : L Nm =0 A m → A N be the map given by f mN on A m , or more formally the unique map with hi m = f mN . We note that h ( i m ( b m ) − i m +1 ( f m ( b m ))) = f m,N ( b m ) − f m +1 ,N ( f m ( b m )) = 0 , so we can apply h to the above equation for i k ( a ) to get f kN ( a ) = 0 as required. (cid:3) Corollary 10.6. Suppose we have a sequence { A i } i ∈ N and a cone { u i : A i → B } i ∈ N giving rise to a homo-morphism u ∞ : lim −→ i A i → B . Then (a) The image of u ∞ is the union of the subgroups u k ( A k ) . (b) The map u ∞ is injective iff whenever u k ( a ) = 0 , there exists m ≥ k with f km ( a ) = 0 .Proof. This is clear from the Proposition. (cid:3) Example 10.7. Let A be any abelian group. Suppose we have a chain of subgroups A ≤ A ≤ A ≤ A ≤ · · · ≤ A. Let f k : A k → A k +1 be the inclusion. Then the colimit of the resulting sequence is just S i A i . Indeed, theinclusions A n → S i A i form a cone, which clearly has both the properties in Corollary 10.6.The following examples are instructive as well as useful. Proposition 10.8. Let A be an arbitrary abelian group. Then the colimit of the sequence A n −→ A n −→ A n −→ A n −→ A −→ · · · is A [1 /n ] , whereas the colimit of the sequence A −→ A −→ A −→ A −→ A −→ · · · is the rationalisation A (0) .Proof. We will prove the second statement; the first is similar but easier. Let C denote the colimit, sowe have maps ı k : A → C with ı k ( a ) = ( k + 1) ı k +1 ( a ). Define u n : A → A (0) by u n ( a ) = a/n !. As(( n + 1) a ) / ( n + 1)! = a/n ! we see that these maps form a cone, so there is a unique map u ∞ : C → A (0) with u ∞ ı k = u k for all k . Any element a ∈ A (0) can be written as a = a ′ /n for some a ′ ∈ A and n > 0, so a = (( n − a ′ ) /n ! = u n (( n − a ′ ), so A (0) is the union of the images of the maps u n . Now suppose that u n ( a ′ ) = 0. By the definition of A (0) this just means that ma ′ = 0 for some m > 0. The map f n,n + m in oursequence is multiplication by the integer p = ( n + 1)( n + 2) · · · ( n + m ) = ( n + m )! n ! = m ! (cid:18) n + mn (cid:19) , which is divisible by m , so f n,n + m ( a ′ ) = 0. The claim follows by Corollary 10.6. (cid:3) Proposition 10.9. Suppose we have a commutative diagram as shown A A A A · · · B B B B · · · f p f p f p f p g g g g et ı k be the canonical map A k → lim −→ i A i , and let k be the canonical map B k → lim −→ i B i . Then there is aunique map p ∞ such that the diagram A k lim −→ i A i B k lim −→ i B ip k ı k p ∞ k commutes for all k . Moreover, if all the maps p k are injective, or surjective, or bijective, then p ∞ has thesame property.Proof. The maps k p k : A k → lim −→ i B i satisfy k p k = k +1 g k p k = k +1 p k +1 f k , so they form a cone for the sequence { A i } . There is thus a unique map p ∞ : lim −→ i A i → lim −→ i B i with p ∞ ı k = k p k for all k , as claimed.(a) Now suppose that all the maps p k are injective. Consider an element a ∈ ker( p ∞ ). By Proposi-tion 10.5 we have a = ı k ( a ′ ) for some k and some a ′ ∈ A k . We then have k ( p k ( a ′ )) = p ∞ ( ı k ( a ′ )) = p ∞ ( a ) = 0. The same proposition therefore tells us that g km ( p k ( a ′ )) = 0 for some m ≥ k . Now g km p k = p m f km and p m is injective, so f km ( a ′ ) = 0. This means that a = i k ( a ′ ) = i m ( f km ( a ′ )) = 0.Thus, p ∞ is injective as claimed.(b) Suppose instead that all the maps p k are surjective. Consider an element b ∈ lim −→ i B i . By Proposi-tion 10.5 we have b = k ( b ′ ) for some k and some b ′ ∈ B k . As p k is surjective, we can choose a ′ ∈ A k with p k ( a ′ ) = b ′ , and then put b = ı k ( a ); we find that p ∞ ( a ) = b . Thus, p ∞ is also surjective.(c) If the maps p k are all isomorphisms, then (a) and (b) together imply that p ∞ is an isomorphism. (cid:3) Proposition 10.10. Suppose we have a commutative diagram as shown, in which all the columns are exact: A A A A · · · B B B B · · · C C C C · · · f p f p f p f p g q g q g q g q h h h h Then the resulting sequence lim −→ i A i p ∞ −−→ lim −→ i B i q ∞ −−→ lim −→ i C i is also exact.Proof. First, any element a ∈ lim −→ i A i has the form a = ı n ( a ′ ) for some n and a ′ . We can then chase a ′ around the diagram A n lim −→ i A i B n lim −→ i A i C n lim −→ i A iı n p n p ∞ n q n q ∞ k n o see that q ∞ ( p ∞ ( a )) = 0. Conversely, suppose we have an element b ∈ ker( q ∞ ). We then have b = n ( b ′ )for some n and some b ′ ∈ B n . We then have k n ( q n ( b ′ )) = q ∞ ( n ( b ′ )) = q ∞ ( b ) = 0, so h n,m ( q n ( b ′ )) = 0 forsome m ≥ n . Now h n,m ( q n ( b ′ )) = q m ( g n,m ( b ′ )), so g n,m ( b ′ ) ∈ ker( q m ) = image( p m ), so we can find a ′ ∈ A m with p m ( a ′ ) = g n,m ( b ′ ). Now put a = ı m ( a ′ ). We find that p ∞ ( a ) = m ( p m ( a ′ )) = m ( g n,m ( b ′ )) = n ( b ′ ) = b, so b ∈ image( p ∞ ). The claim follows. (cid:3) Proposition 10.11. Suppose we have a sequence A f −→ A f −→ A f −→ A f −→ · · · and a nondecreasing function u : N → N such that u ( i ) → ∞ as i → ∞ . Put g i = f u ( i ) ,u ( i +1) = ( A u ( i ) f u ( i ) −−−→ A u ( i )+1 f u ( i )+1 −−−−→ · · · f u ( i +1) − −−−−−−→ A u ( i +1) ) , so we have a sequence A u (0) g −→ A u (1) g −→ A u (2) g −→ A u (3) g −→ · · · Then there is a canonical isomorphism lim −→ j A u ( j ) = lim −→ i A i .Proof. Let ı n : A n → lim −→ i A i and n : A u ( n ) → lim −→ j A u ( j ) be the usual maps. As ı n = ı n +1 f n for all n wefind by induction that ı n = ı m f n,m for all n ≤ m . By applying this to the pair u ( k ) ≤ u ( k + 1), we seethat ı u ( k ) = ı u ( k +1) g k , so the maps ı u ( k ) form a cone for the sequence { A u ( j ) } j ∈ N , so there is a unique map p : lim −→ j A u ( j ) → lim −→ i A i with p k = ı u ( k ) for all k . In the opposite direction, suppose we have n ∈ N . As u ( j ) → ∞ as j → ∞ , we can choose k such that n ≤ u ( k ), and form the composite q nk = ( A n f n,u ( k ) −−−−→ A u ( k ) k −→ lim −→ j A u ( j ) ) . As k = k +1 g k = k +1 f u ( k ) ,u ( k +1) and f u ( k ) ,u ( k +1) f n,u ( k ) = f n,u ( k +1) we see that q n,k = q n,k +1 . Thus q nk isindependent of k , so we can denote it by q n . We also find that q n = q n +1 f n , so the maps q n form a cone forthe sequence { A i } i ∈ N , so there is a unique map q : lim −→ i A i → lim −→ j A u ( j ) with qı n = q n for all n .Now note that any element a ∈ lim −→ i A i has the form a = ı n ( a ′ ) for some n and a ′ ∈ A n . If we choose k with u ( k ) ≥ n we have pq ( a ) = pq n ( a ′ ) = p k f n,u ( k ) ( a ′ ) = ı u ( k ) f n,u ( k ) ( a ′ ) = ı n ( a ′ ) = a, so pq is the identity. A similar argument shows that qp is the identity. (cid:3) Proposition 10.12. Suppose we have a commutative diagram A A A A · · · A A A A · · · A A A A · · · A A A A · · ·· · · · · · · · · · · · f g f g f g f g f g f g f g f g f g f g f g f g f g f g f g f g nd thus sequences lim −→ j A j g ∞ −−→ lim −→ j A j g ∞ −−→ lim −→ j A j g ∞ −−→ lim −→ j A j g ∞ −−→ · · · and lim −→ i A i f ∞ −−→ lim −→ i A i f ∞ −−→ lim −→ i A i f ∞ −−→ lim −→ i A i f ∞ −−→ · · · Suppose we also put h i = g i,j +1 f ij = f i +1 ,j g ij : A ii → A i +1 ,i +1 , giving a third sequence A h −→ A h −→ A h −→ A h −→ · · · Then there are canonical isomorphisms lim −→ i lim −→ j A ij ≃ lim −→ i A ii ≃ lim −→ j lim −→ i A ij . Example 10.13. In conjunction with Proposition 10.8, this will give A [ 1 n ][ 1 m ] = A [ 1 nm ] = A [ 1 m ][ 1 n ] Proof. Put A ++ = L n,m A nm , and let i nm : A nm → A ++ be the canonical inclusion. Put P nm = image( i nm − i n,m +1 f nm : A nm → A ++ ) Q nm = image( i nm − i n +1 ,m g nm : A nm → A ++ ) . From the definitions we have M n lim −→ j A nj = A ++ / X n,m P nm M m lim −→ i A im = A ++ / X n,m Q nm . It follows easily that lim −→ i lim −→ j A ij = A ++ / X n,m P nm + X n,m Q nm ! = lim −→ j lim −→ i A ij . We write A ∞∞ for this group, and we write ı nm for the obvious map A nm → A ∞∞ . By construction, thefollowing diagram commutes: A nm A n,m +1 A n +1 ,m A ∞∞ . f nm g nm ı nm ı n,m +1 ı n +1 ,m It follows that ı kk = ı k +1 ,k +1 h k , so the maps ı kk form a cone for the sequence { A ii } i ∈ N . Thus, if wewrite k for the usual map A kk → lim −→ i A ii , we find that there is a unique map p : lim −→ i A ii → A ∞∞ with p k = ı kk for all k . In the opposite direction, suppose we have n, m, k ∈ N with n, m ≤ k . By composing f ’sand g ’s in various orders we can form a number of maps A nm → A kk but they are all the same because theoriginal diagram is commutative. We write u nmk for this map, and put q nmk = k u nmk : A nm → lim −→ i A ii .Now k = k +1 h k and h k u nmk = u n,m,k +1 so q n,m,k = q n,m,k +1 . Thus q nmk is independent of k (providedthat k ≥ max( n, m )) so we can denote it by q nm . We can now put the maps q nm together to give a map q ′ : A ++ → lim −→ i A ii with q ′ i nm = q nm . When k > max( n, m ) we have u nmk = u n,m +1 ,k f nm = u n +1 ,m,k g nm ,and using this we see that q ′ ( P nm ) = 0 = q ′ ( Q nm ). There is thus an induced map A ∞∞ → lim −→ i A ii with qı nm = q nm . We leave it to the reader to check that q is inverse to p . (cid:3) Limits and derived limits of towers Definition 11.1. A tower is a diagram of the form B f ←− B f ←− B f ←− B f ←− · · · Given such a tower an integers i ≥ j , we write f ij for the composite B i f i − −−−→ B i − f i − −−−→ · · · f j −→ B j . Note that f ii is the identity map, and f i +1 ,i = f i , and f jk f ij = f ik whenever i ≥ j ≥ k . Definition 11.2. Suppose we have a tower as above. The limit (or inverse limit ) of the tower is the grouplim ←− i B i = { a ∈ Y i B i | a i = f i ( a i +1 ) for all i } . Equivalently, if we define D : Q i B i → Q i B i by D ( a ) i = a i − f i ( a i +1 ), then lim ←− i B i = ker( D ). We also writelim ←− i B i for the cokernel of D . We define p n : lim ←− i B i → B n by p n ( a ) = a n , so p n = f n p n +1 . Remark 11.3. The limit can also be characterised by a universal property, as follows. A cone for the toweris a group A with a collection of maps u k : A → B k such that u k = f k u k +1 for all k . Tautologically, the maps p k : lim ←− i B i → B k form a cone. Moreover, for any cone { A u k −→ B k } k ∈ N we can define u ∞ : A → lim ←− i B i by u ∞ ( a ) = ( u ( a ) , u ( a ) , u ( a ) , . . . ) , and this is the unique map with p k u ∞ = u k for all k . Example 11.4. Suppose we have a chain of subgroups B ≥ B ≥ B ≥ · · · and we take the maps f i to be the inclusion maps. Then we see from the definitions thatlim ←− i B i = { ( b, b, b, · · · ) | b ∈ \ i B i } ≃ \ i B i . Example 11.5. Suppose we have a system of groups C i (for i ∈ N ). Put B k = Q ki =0 C i and let f k : B k +1 → B k be the obvious projection map. If b ∈ lim ←− k B k then b k ∈ Q ki =0 C i so b kk ∈ C k . We can thus define d : lim ←− k B k → Y k C k by d ( b ) = ( b , b , b , . . . ) . By the definition of lim ←− k B k , we have b kj = b jj for j ≤ k . Using this, we see that d is an isomorphism. Example 11.6. Suppose we have a tower in which the groups are arbitrary but the maps are all zero. Thenthe map D is the identity, so both lim ←− and lim ←− are zero. Example 11.7. Suppose we have a sequence A f −→ A f −→ A f −→ A f −→ · · · , and another group B . This gives us a towerHom( A , B ) f ∗ ←− Hom( A , B ) f ∗ ←− Hom( A , B ) f ∗ ←− Hom( A , B ) f ∗ ←− · · · The elements of lim ←− i Hom( A i , B ) are precisely the cones from the sequence { A i } i ∈ N to B , which biject withhomomorphisms from lim −→ i A i to B . In other words, we havelim ←− i Hom( A i , B ) = Hom(lim −→ i A i , B ) . xample 11.8. Fix a prime p . We then have a tower Z /p ←− Z /p ←− Z /p ←− Z /p ←− · · · . The inverse limit is called the ring of p -adic integers , and is denoted by Z p . We will investigate it in moredetail in Section 12. We can also form a tower Z / ←− Z / ←− Z / ←− Z / ←− · · · The inverse limit is called the profinite completion of Z , and is denoted by b Z . Using the Chinese RemainderTheorem (Proposition 4.7) one can show that b Z = Q p Z p . For yet another description, recall that Q / Z = [ n ( Q / Z )[ n !] = lim −→ n ( Q / Z )[ n !] , so End( Q / Z ) = lim ←− n Hom(( Q / Z )[ n !] , Q / Z ) . Now we have a map m : Z → End( Q / Z ) defined by m ( k ) = k. Q / Z , and this fits in a commutative diagram Z End( Q / Z ) Z /n ! Hom(( Q / Z )[ n !] , Q / Z ) . m restrict m n Using the fact that ( Q / Z )[ n !] is generated by (1 /n !) + Z we see that m n is an isomorphism. By passingto inverse limits, we obtain an isomorphism m ∞ : b Z → End( Q / Z ). Proposition 11.9. Suppose we have a commutative diagram as shown A A A A · · · B B B B · · · p p f p f p f f g g g g and we define p = Q i p i : Q i A i → Q i B i . Then the central square below commutes, so there are inducedmaps p ∞ and p ∞ as shown. lim ←− i A i Q i A i Q i A i lim ←− i A i lim ←− i B i Q i B i Q i B i lim ←− i B i . p ∞ p D p p ∞ D Proof. Clear from the definitions. (cid:3) Proposition 11.10. Suppose we have a commutative diagram as shown, in which all the columns are shortexact: A A A A · · · B B B B · · · C C C C · · · p p f p f p f f q q g q g q g g h h h h Then there is an associated exact sequence lim ←− i A i lim ←− i B i lim ←− i C i lim ←− i A i lim ←− i B i lim ←− i C ip ∞ q ∞ δ p ∞ q ∞ Proof. Apply the Snake Lemma to the diagram i A i Q i B i Q i C i Q i A i Q i B i Q i C i Q i p i D Q i q i D D Q i p i Q i q i in which the rows are easily seen to be short exact. (cid:3) In practice the groups lim ←− i A i are usually either zero, or enormous and untractable. We will thus be veryinterested in results that force them to be zero. Proposition 11.11. Suppose we have a tower in which the maps f i : A i +1 → A i are all surjective. Then lim ←− i A i = 0 , and the projection maps p k : lim ←− i A i → A k are all surjective.Proof. Consider an element a ∈ Q i A i . We will choose elements b k ∈ A k recursively as follows: we startwith b = 0, and then take b n to be any element with f n − ( b n ) = b n − − a n − . These elements b k give anelement b ∈ Q i A i with D ( b ) = a , so D is surjective and lim ←− i A i = cok( D ) = 0.Now suppose we have an element a ∈ A k . Define c i = f k,i ( a ) for all i ≤ k . Then define c i ∈ A i recursivelyfor i > k by choosing c i to be any element with f i − ( c i ) = c i − . This gives an element c ∈ lim ←− i A i with p k ( c ) = a . (cid:3) Definition 11.12. We say that a tower A f ←− A f ←− · · · is nilpotent if for all i there exists j > i such that f ji = 0 : A j → A i . Proposition 11.13. For a nilpotent tower as above, we have lim ←− i A i = lim ←− i A i = 0 .Proof. Define E : Q i A i → Q i A i by E ( a ) i = ∞ X j = i f j,i ( a j ) . Although the sum is formally infinite, the nilpotence hypothesis means that there are only finitely manynonzero terms, so the expression is meaningful. It is then not hard to check that DE = ED = 1, so thekernel and cokernel of D are zero. (cid:3) Definition 11.14. Consider a tower A f ←− A f ←− · · · , so for each i we have a descending chain of subgroups A i ≥ f i +1 ,i ( A i +1 ) ≥ f i +2 ,i ( A i +2 ) ≥ f i +3 ,i ( A i +3 ) ≥ · · · We say that the tower is Mittag-Leffler if for each i there exists j ≥ i such that f ki ( A k ) = f ji ( A j ) for all k ≥ j (so the above chain is eventually constant). Example 11.15. Towers of surjections are Mittag-Leffler, as are nilpotent towers. Proposition 11.16. If all the groups A i are finite, then the tower is Mittag-Leffler. Similarly, if the groups A i are finite-dimensional vector spaces over a field K , and the maps f i are all K -linear, then the tower isMittag-Leffler.Proof. In the first case, we just choose j ≥ i such that the order | f ji ( A j ) | is as small as possible; it thenfollows that f ki ( A k ) = f ji ( A j ) for k ≥ i . In the second case, use dimensions instead of orders. (cid:3) Proposition 11.17. If A is a Mittag-Leffler tower, we have lim ←− i A i = 0 .Proof. By the Mittag-Leffler condition, there is a subgroup A ′ i ≤ A i such that f ji ( A j ) = A ′ i for all sufficientlylarge j . Thus, for j very large we have both A ′ i +1 = f j,i +1 ( A j ) and A ′ i = f ji ( A j ) = f i ( f j,i +1 ( A j )) = f i ( A ′ i +1 ).Thus, the groups A ′ i form a subtower of A , with surjective maps f ′ i : A ′ i +1 → A ′ i . Now put A ′′ i = A i /A ′ i , sothere are induced maps f ′′ i : A ′′ i +1 → A ′′ i , giving a third tower. If j is much larger than i we have f ji ( A j ) = A ′ i and so f ′′ ji = 0 : A ′′ j → A ′′ i ; this shows that the tower A ′′ is nilpotent. Now apply Proposition 11.10 to theshort exact sequence A ′ → A → A ′′ to give an exact sequence im ←− i A ′ i lim ←− i A i lim ←− i A ′′ i lim ←− i A ′ i lim ←− i A i lim ←− i A ′′ iδ Here lim ←− i A ′ i = 0 because the maps in A ′ are surjective, and lim ←− i A ′′ i = 0 because A ′′ is nilpotent, solim ←− i A i = 0 as claimed. (We also have lim ←− i A ′′ i = 0 and so lim ←− i A ′ i = lim ←− i A i .) (cid:3) We also have the following result analogous to Proposition 10.11, which again indicates that lim ←− i A i andlim ←− i A i only depend on the asymptotic behaviour of the tower. Proposition 11.18. Suppose we have a tower A f ←− A f ←− A f ←− A f ←− · · · and a nondecreasing function u : N → N such that u ( i ) → ∞ as i → ∞ . Put g i = f u ( i +1) ,u ( i ) = ( A u ( i +1) f u ( i +1) − −−−−−−→ A u ( i +1) − · · · f u ( i ) −−−→ A u ( i ) ) , so we have a tower A u (0) g ←− A u (1) g ←− A u (2) g ←− A u (3) g ←− · · · Then there are canonical isomorphisms lim ←− j A u ( j ) = lim ←− i A i and lim ←− j A u ( j ) = lim ←− i A i .Proof. Define v : N → N by v ( i ) = min { j | u ( j ) ≥ i } . We will construct a diagram as follows: Q j A u ( j ) Q i A i Q j A u ( j ) Q j A u ( j ) Q i A i Q j A u ( j ) φD ′ ψD D ′ λ µ The maps are: D ′ ( b ) j = b j − f u ( j +1) ,u ( j ) ( b j +1 ) D ( a ) i = a i − f i +1 ,i ( a i +1 ) φ ( b ) i = f uv ( i ) ,i ( b v ( i ) ) ψ ( a ) j = a u ( j ) λ ( b ) i = X u ( j )= i b j µ ( a ) j = X u ( j ) ≤ i
1. The expression therefore cancels down to b j − f u ( j ) ,i ( b j ). On theother hand, we also find that v ( i ) = j and v ( i + 1) = j , so D ( φ ( b )) i = f ii ( b j ) − f u ( j ) ,i ( b j ) = λ ( D ′ ( b )) i . hus, the left square commutes. For the right square, consider an element a ∈ Q i A i . We have µ ( D ( a )) j = X u ( j ) ≤ i i . From thedefinitions we have λ ( µ ( a )) i = X u ( j )= i X u ( j ) ≤ h u ( j ). In this case we have τ ( b ) j +1 = 0, so b j + D ( τ ( b )) j = b j + τ ( b ) j = b j + X k Definition 12.1. Let p be a prime. For any abelian group A we have a tower of surjections0 = A/p ←− A/p ←− A/p ←− A/p ←− · · · We write A p for the inverse limit of this tower, and call this the p -completion of A . In particular, we have agroup Z p , whose elements are called p -adic integers . For any A we have a homomorphism η : A → A p givenby η ( a ) = ( a + p A, a + p A, a + p A, a + p A, . . . ) . We say that A is p -complete if η is an isomorphism. Example 12.2. Let A be a finite abelian group, so A splits as B ⊕ C say, where | B | is a power of p and | C | is coprime to p . For all k we have p k C = C , and for large k we have p k B = 0. It follows that A p = B . Inparticular, we see from this that finite abelian p -groups are p -complete. Example 12.3. Suppose that A is divisible. Then for all k we have p k A = A , so A/p k = 0; it follows that A p = 0. In particular, we have Q p = 0 and ( Q / Z ) p = 0. Remark 12.4. The symbol Q p is often used for Q ⊗ Z p , which is not zero; it is known as the field of p -adicrationals. However, this is different from the group that we have called Q p , which is trivial. Proposition 12.5. For all A and k ≥ the projection π k : A → A/p k A induces an isomorphism A p /p k A p = A/p k A .Proof. For notational simplicity, we will treat only the case k = 1. The general case is essentially the same,after we have used Proposition 11.18 to identify A p with the inverse limit of the sequence A/p k ←− A/p k ←− A/p k ←− · · · . First, the projection π : Q i A/p i → A/p restricts to give a homomorphism φ : A p = lim ←− i A/p i → A/p .Proposition 11.11 tells us that this is surjective. As A/p has exponent p , the subgroup pA p is contained inthe kernel. We need to prove that the kernel is precisely pA p . Suppose that a ∈ ker( φ ). Choose a i ∈ A representing the component of a in A/p i A . As A/p A = 0 and φ ( a ) = 0 we can take a = a = 0. By thedefinition of the inverse limit we see that a i is the image of a i +1 in A/p i A , so a i +1 = a i + p i b i for some b i ∈ A (and we may take b = 0). Now put c n = P ni =1 p i − b i ∈ A . It is visible that c n +1 = c n + p n b n +1 = c n (mod p n A n ), so the cosets c n + p n A define an element c ∈ A p . We also see by induction that pc k = a k +1 = a k (mod p k A k ), so pc = a . Thus a ∈ pA p as claimed. (cid:3) Remark 12.6. We see from the proposition that A p = ( A p ) p , so A p is p -complete, as one would expect.There is a subtlety analogous to Remark 9.15 here; we leave it to the reader to check that the two naturalmaps A p → ( A p ) p are the same, and they are both isomorphisms.We next examine the structure of Z p in more detail. First, as the groups Z /p k have canonical ringstructures, and the maps Z /p k +1 → Z /p k are ring maps, we see that Z p is a subring of Q k Z /p k . We willwrite π k for the projection Z p → Z /p k , which is a surjective ring homomorphism. Note also that for n ∈ Z we have η ( n ) = 0 iff π k η ( n ) = 0 for all k iff n is divisible by p k for all k iff n = 0. Thus, η gives an injectivering map Z → Z p . We will usually suppress notation for this and regard Z as a subring of Z p . Definition 12.7. For a ∈ Z p we define v ( a ) = min { k | π k ( a ) = 0 } (or v ( a ) = ∞ if π k ( a ) = 0 for all k , whichmeans that a = 0). We also define d ( a, b ) = p − v ( a − b ) , with the convention p −∞ = 0. Proposition 12.8. The function d defines a metric on Z p (called the p -adic metric ), with respect to whichit is complete and compact. Moreover, the subspace Z is dense.Proof. It is clear that d ( a, b ) = d ( b, a ), and that this is nonnegative and vanishes if and only if a = b . Thisjust leaves the triangle inequality d ( a, c ) ≤ d ( a, b ) + d ( b, c ). This is clear if a = b or b = c , so suppose that a = b = c . Put m = min( v ( a − b ) , v ( b − c )). For k < m we have π k ( a ) = π k ( b ) = π k ( c ). It follows that v ( c − a ) ≥ m , so d ( a, c ) ≤ p − m = max( d ( a, b ) , d ( b, c )) ≤ d ( a, b ) + d ( b, c ) s required.Now consider a Cauchy sequence ( a , a , a , . . . ) in Z p . Given k ∈ N we can choose m such that d ( a i , a j ) 0, and thata complete metric space is compact if and only if it is totally bounded. The set T k is a 2 − k -net, so Z p istotally bounded, so it is compact. (cid:3) Proposition 12.9. Put D = { , , , . . . , p − } (the set of p -adic digits). Then there is a bijection σ : ∞ Y i =0 D → Z p given by σ ( u ) = P i u i p i (a convergent sum with respect to the p -adic metric). In particular, Z p is uncount-able. We call σ − ( a ) the base p expansion of a .Proof. It is elementary that the corresponding map Q k − i =0 D → Z /p k is a bijection, and the claim follows bypassing to inverse limits. (cid:3) Proposition 12.10. The ring Z p is torsion free and is an integral domain. It is also a local ring, with p. Z p being the unique maximal ideal. The group of units is Z × p = { a ∈ Z p | π ( a ) = 0 } = Z p \ p Z p , and every nonzero element is a unit times p k for some k .Proof. First, consider an element a ∈ Z p with π ( a ) = 1. We see from Proposition 12.5 that a = 1 − px forsome x ∈ Z p . It follows easily that the series P i ( px ) i is Cauchy, so it converges to some b ∈ Z p , and we findthat ab = 1. More generally, suppose merely that π ( a ) = 0 in Z /p . As Z /p is a field, we can find b ∈ Z suchthat π ( b ) is inverse to π ( a ). We then find that ab is invertible by the previous case, and it follows that a is invertible. Conversely, as π is a ring map it certainly sends units to nonzero elements. We therefore seethat Z × p = { a ∈ Z p | π ( a ) = 0 } = Z p \ p Z p as claimed. Note also that Z p /p Z p is the field Z /p , so p Z p is a maximal ideal. If m is any maximal ideal wemust have m ∩ Z × p = ∅ , so m ≤ p Z p , so m = p Z p by maximality. Thus, Z p is a local ring.Next, using base p expansions we see easily that multiplication by p is injective, and that every nonzeroelement a ∈ Z p can be written as a = p k b for some k ≥ b with π ( b ) = 0, so b ∈ Z × p . As multiplicationby p k is injective and b is invertible we see that multiplication by a is injective. This means that Z p is anintegral domain. By considering a ∈ Z ⊂ Z p we also see that Z p is torsion free. (cid:3) We can now understand the completion of free modules. Definition 12.11. Let I be a set, and let f be a function from I to Z p . We say that f is asymptoticallyzero if for all k , the set { i | v ( f ( i )) < k } is finite. We write AZ ( I ) for the set of asymptotically zero maps,which is a group under addition. Proposition 12.12. The completion Z [ I ] p is naturally isomorphic to AZ ( I ) .Proof. First, we put ( Z /p k )[ I ] = { f : I → Z /p k | { i | f ( i ) = 0 } is finite } . We write π k for the projection Z → Z /p k , or the projection Z p → Z /p k . We then write π ′ k ( f ) = π k ◦ f ; thisdefines a map AZ ( I ) → ( Z /p k )[ I ], which we can restrict to Z [ I ] ≤ AZ ( I ). Note that π ′ k ( f ) = 0 iff f ( i ) ∈ p k Z p for all i , in which case we can define g = f /p k : I → Z p and we find that g is again asymptotically zero, o f ∈ p k AZ ( I ). This shows that π ′ k induces an isomorphism AZ ( I ) /p k AZ ( I ) → ( Z /p k )[ I ]. By a similarargument, it also induces an isomorphism Z [ I ] /p k Z [ I ] → ( Z /p k )[ I ]. Thus, we have Z [ I ] p = lim ←− k ( Z /p k )[ I ].The maps π ′ k : AZ ( I ) → ( Z /p k )[ I ] therefore assemble to give a homomorphism π ′ : AZ ( I ) → Z [ I ] p . In theopposite direction, suppose we have g ∈ Z [ I ] p . For each k ≥ π k ( g ) ∈ Z [ I ] /p k Z [ I ] =( Z /p k )[ I ], so there is a unique map g k : I → { , , . . . , p k − } with π k ( g )( i ) = g k ( i ) (mod p k ) for all k . Note that the set { i | g k ( i ) = 0 } is finite for all k . If we fix i ,we find that the sequence { g k ( i ) } k ≥ is Cauchy, converging to some element g ∞ ( i ) ∈ Z p say, and we have g ∞ ( i ) = g k ( i ) (mod p k ) for all k . This means that g ∞ ∈ AZ ( I ) and π ′ ( g ∞ ) = g , so π ′ is surjective. We alsohave π ′ ( h ) = 0 iff h ( i ) is divisible by p k for all i and k , which implies that h = 0. Thus, the map π ′ is anisomorphism. (cid:3) Proposition 12.13. For any abelian groups A and B there is a natural map µ : A p ⊗ B p → ( A ⊗ B ) p , whichinduces an isomorphism ( A p ⊗ B p ) p → ( A ⊗ B ) p .Proof. Note that p k . A ⊗ B = ( p k . A ) ⊗ B = 1 A ⊗ ( p k . B ). Using the right exactness of tensor products wesee that ( A ⊗ B ) /p k = ( A/p k ) ⊗ B = A ⊗ ( B/p k ) = ( A/p k ) ⊗ ( B/p k ) . Combining this with Proposition 12.5 gives( A p ⊗ B p ) /p k = ( A p /p k ) ⊗ ( B p /p k ) = ( A/p k ) ⊗ ( B/p k ) = ( A ⊗ B ) /p k . Passing to inverse limits gives an isomorphism ( A p ⊗ B p ) p = ( A ⊗ B ) p . We can compose this with the map η : A p ⊗ B p → ( A p ⊗ B p ) p to get a map µ : A p ⊗ B p → ( A ⊗ B ) p , which is what we most often need forapplications. (cid:3) Proposition 12.14. There is a natural map Z p ⊗ A → A p , which is an isomorphism when A is finitelygenerated.Proof. The map is just the composite Z p ⊗ A ⊗ η −−→ Z p ⊗ A p µ −→ ( Z ⊗ A ) p = A p (or it can be defined more directly by the method used for µ ). By the classification of finitely generatedabelian groups, it will suffice to prove that we have an isomorphism when A = Z or A = Z /p k or A = Z /q k for some prime q = p . The case A = Z is clear. When A = Z /p k we have A p = A as in Example 12.2.We also Z p ⊗ A = Z p /p k Z p by the right exactness of tensoring, and this is the same as Z /p k = A byProposition 12.5, so Z p ⊗ A = A p as claimed. Finally, if q is different from p then q k is invertible in Z p so Z p ⊗ Z /q k = Z p /q k Z p = 0, and similarly ( Z /q k ) p = 0 as in Example 12.2 again. (cid:3) Corollary 12.15. If A → B → C is an exact sequence of finitely generated abelian groups, then the resultingsequence A p → B p → C p is also exact.Proof. As Z p is torsion free, Proposition 7.16 tells us that the sequence Z p ⊗ A → Z p ⊗ B → Z p ⊗ C isexact. (cid:3) We can now assemble our results to prove something closely analogous to Proposition 9.14: Proposition 12.16. (a) The group A p is always p -complete. (b) The map η : A → A p is an isomorphism if and only if A is p -complete. (c) If A is p -complete then it can be regarded as a module over Z p . (d) Suppose that f : A → B is a homomorphism, and that B is p -complete. Then there is a uniquehomomorphism f ′ : A p → B such that f ′ ◦ η = f : A → B .Proof. (a) As mentioned previously, this follows from Proposition 12.5 by taking inverse limits.(b) This is true by definition, and is only mentioned to complete the correspondence with Proposi-tion 9.14. c) Proposition 12.13 gives a map Z p ⊗ A = Z p ⊗ A p → ( Z ⊗ A ) p = A p = A. For r ∈ Z p and a ∈ A we can thus define ra = µ ( r ⊗ a ). Equivalently, this is characterised by thefact that π k ( ra ) = π k ( r ) π k ( a ) in A/p k A (where we have used the obvious structure of A/p k A as amodule over Z /p k ). From this description it is clear that our multiplication rule is associative, unitaland distributive, so it makes A into a module over Z p .(d) The map f : A → B induces an isomorphism f p : A p → B p , and we have an isomorphism η : B → B p .We can and must take f ′ = η − ◦ f p . (cid:3) While our definition of completion is quite natural and straightforward, its exactness properties for infin-itely generated groups are very delicate, and they do not relate well to topological constructions. We willtherefore introduce a different definition that often agrees with completion, but has better formal properties. Definition 12.17. For any abelian group A , we let A [[ x ]] denote the group of formal power series v ( x ) = P ∞ i =0 a i x i with a i ∈ A for all i . This is a module over Z [[ x ]] by the obvious rule X i n i x i ! X j a j x j = X k k X i =0 n i a k − i ! x k . We define L A = A [[ x ]] / (( x − p ) .A [[ x ]]) L A = { v ( x ) ∈ A [[ x ]] | ( x − p ) v ( x ) = 0 } . We can identify A with the set of constant series in A [[ x ]], and then restrict the quotient map A [[ x ]] → L A to A to get a natural map η : A → L A . We say that A is Ext- p -complete if η : A → L A is an isomorphismand L A = 0. We also call L A the derived completion of A . Remark 12.18. Readers familiar with the general theory of derived functors should consult Corollary 12.27to see why the term is appropriate here. Remark 12.19. It will follow from Proposition 12.28 that the condition L A = 0 is actually automaticwhen η : A → L A is an isomorphism. However, it is easier to develop the theory if we have both conditionsin the initial definition. Remark 12.20. There is an evident product map µ : A [[ x ]] ⊗ B [[ x ]] → ( A ⊗ B )[[ x ]]given by µ X i a i x i ! ⊗ X j b j x j = X k X k = i + j a i ⊗ b j x k . This induces a map µ : ( L A ) ⊗ ( L B ) → L ( A ⊗ B ), which fits in a commutative diagram A ⊗ B ( L A ) ⊗ ( L B ) L ( A ⊗ B ) . η ⊗ η ηµ Remark 12.21. If we have a short exact sequence A −→ B −→ C , we can apply the Snake Lemma to thediagram A [[ x ]] B [[ x ]] C [[ x ]] A [[ x ]] B [[ x ]] C [[ x ]] x − p x − p x − p o obtain an exact sequence L A L B → L C → L A → L B ։ L C. Proposition 12.22. (a) If we have a short exact sequence A → B → C in which two of the three termsare Ext- p -complete, then so is the third. (b) Finite sums and retracts of Ext- p -complete groups are Ext- p -complete. (c) The kernel, cokernel and image of any homomorphism between Ext- p -complete groups are Ext- p -complete. (d) The product of any (possibly infinite) family of Ext- p -complete groups is Ext- p -complete. (e) If p k . A = 0 for some k then A is Ext- p -complete. (f) If A is p -complete then it is Ext- p -complete.Proof. (a) Chase the diagram A B CL A L B L C L A L B L C (b) Clear.(c) Consider a homomorphism f : A → B between Ext- p -complete groups, and the resulting short exactsequences img( f ) → B → cok( f ) and ker( f ) → A → img( f ). These give diagramsimg( f ) B cok( f ) L img( f ) 0 L cok( f ) L img( f ) L B L cok( f ) ≃ and ker( f ) A img( f ) L ker( f ) 0 L img( f ) L ker( f ) L A L img( f ) . ≃ From the first diagram we see that L img( f ) = 0 and that the map img( f ) → L img( f ) is injective,and from the second we see that the map img( f ) → L img( f ) is surjective; thus img( f ) is Ext- p -complete. Given this, it is a special case of (a) that ker( f ) and cok( f ) are also Ext- p -complete asclaimed.(d) This is clear, because ( Q i A i )[[ x ]] = Q i A i [[ x ]].(e) If k = 1 the definitions give L A = A [[ x ]] / ( x.A [[ x ]]) = A and L A = { v ( x ) ∈ A [[ x ]] | xv ( x ) = 0 } = 0 asrequired. The general case follows by induction using (a) and the short exact sequence pA → A → A/pA .(f) As the tower { A/p k } consists of surjections, the lim ←− term is zero. As A is p -complete, we thereforehave a short exact sequence A Y k A/p k ։ Y k A/p k . The second and third terms are Ext- p -complete by parts (e) and (d), so A is Ext- p -complete bypart (a). (cid:3) Proposition 12.23. Let A be any abelian group. Then L A = lim ←− k A [ p k ] and there is a natural short exactsequence lim ←− k A [ p k ] L A A p . ξ ζ he limit symbols here refer to the tower A [ p ] p ←− A [ p ] p ←− A [ p ] p ←− A [ p ] p ←− · · · Proof. First, an element of L A is a series v ( x ) = P i a i x i with ( x − p ) v ( x ) = 0, which means that pa = 0and pa i +1 = a i for all i ≥ 0. It follows inductively that p i +1 a i = 0 for all i , so the sequence (0 , a , a , . . . ) isan element of lim ←− i A [ p i ]. All steps here can be reversed so L A = lim ←− i A [ p i ]. Next, define ζ ′ i : A [[ x ]] → A/p i A by ζ ′ i ( X j a j x j ) = X j
0. Form this it follows easily that p i +1 b i = 0 for all i ≥ 0, so we have an element b ′ = (0 , b , b , . . . ) ∈ Q i A [ p i ]. We now see that c = D ( b ′ ), so c represents the zero element of lim ←− i A [ p i ], thus, the map ξ : lim ←− i A [ p i ] → ker( ζ ) is injective. (cid:3) Remark 12.24. It is clear from the definitions that the diagram AL A A pη ηζ commutes. Definition 12.25. We say that an abelian group A has bounded p -torsion if there exists k ≥ p k . tors p ( A ) = 0. Corollary 12.26. If A has bounded p -torsion (in particular, if A is a free abelian group) then L A = 0 and L A = A p .Proof. The tower { A [ p i ] } is nilpotent, so lim ←− i A [ p i ] = lim ←− i A [ p i ] = 0 by Proposition 11.13. (cid:3) Corollary 12.27. Suppose we have a short exact sequence P f −→ Q → A where P and Q are free abeliangroups. Then L A and L A are the cokernel and kernel of the induced map P p → Q p .Proof. This is immediate from Corollary 12.26 and Remark 12.21. (cid:3) Proposition 12.28. For any abelian group A , the groups L A , L A , A p and lim ←− k A [ p k ] are all Ext- p -complete. roof. The groups A [ p k ] and A/p k are Ext- p -complete by part (e) of Proposition 12.22. It follows by part (d)that Q k A [ p k ] and Q k A/p k are Ext- p -complete, and then by part (c) that the groups lim ←− k A [ p k ] = L A ,lim ←− k A [ p k ] and lim ←− k A/p k = A p are Ext- p -complete. We can thus apply part (a) to the short exact sequencelim ←− k A [ p k ] L A A pξ ζ to deduce that L A is Ext- p -complete. (cid:3) Proposition 12.29. For any abelian group A , we have L A = 0 iff A p = 0 iff A/p = 0 .Proof. Proposition 12.23 shows that A p is a quotient of L A , and Proposition 12.5 shows that A/p is aquotient of A p . Conversely, if A/p = 0 then p. A is surjective, so A/p k = 0 for all k and the maps in thetower { A [ p k ] } are all surjective. It follows that A p = lim ←− k A/p k = 0 and (using Proposition 11.17) thatlim ←− k A [ p k ] = 0, so the short exact sequence in Proposition 12.23 shows that L A = 0. (cid:3) We next explain a more traditional construction of the functors L and L . This involves a group knownas Z /p ∞ . Definition 12.30. Define f k : Z /p k → Z /p k +1 by f k ( a + p k Z ) = pa + p k +1 Z , so we have a sequence0 = Z /p Z /p Z /p Z /p Z /p · · · f f f f f We define Z /p ∞ to be the colimit of this sequence. Proposition 12.31. There are canonical isomorphisms Z /p ∞ = Z [1 /p ] / Z = tors p ( Q / Z ) = ( Q / Z ) ( p ) = Q / Z ( p ) . Proof. First, consider the diagram Z Z Z Z Z · · · Z Z Z Z Z · · · Z /p Z /p Z /p Z /p Z /p · · · p p p p p p p p pf f f f f Using Propositions 10.10 and 10.8 we obtain a short exact sequence Z → Z [1 /p ] → Z /p ∞ , so Z /p ∞ = Z [1 /p ] / Z . Next, for a ∈ Q we note that a + Z is a p -torsion element in Q / Z iff p k a ∈ Z for some k , iff a ∈ Z [1 /p ]. It follows that tors p ( Q / Z ) = Z [1 /p ] / Z . We also know from Proposition 9.17 that tors p ( Q / Z ) =( Q / Z ) ( p ) . It is clear that Q is p -local, so Q ( p ) = Q . We can thus apply Proposition 9.10 to the sequence Z Q ։ Q / Z to see that ( Q / Z ) ( p ) = Q / Z ( p ) . (cid:3) Proposition 12.32. There are natural isomorphisms L A = Ext( Z /p ∞ , A ) and L A = Hom( Z /p ∞ , A ) .Proof. In this proof we will identify Z /p ∞ with Z [ p ] / Z . Put F = L ∞ i =0 Z , and let e i be the i ’th basis vectorin F . Define maps F φ −→ F ψ −→ Z /p ∞ by φ ( e i ) = e i − − pe i ψ ( e i ) = p − i − + Z (where e − is interpreted as 0), or equivalently φ ( n , n , n , . . . ) = ( pn − n , pn − n , pn − n , . . . ) φ ( m , m , m , . . . ) = X i m i p − i − + Z . By considering the first nonzero entry in n = ( n , n , . . . ), we see that φ is injective. Any element of Z /p ∞ can be written as k/p i +1 + Z for some i ≥ k ∈ Z , and this is the same as ψ ( ke i ), so ψ is surjective. t is clear from the definitions that ψφ = 0, so img( φ ) ≤ ker( ψ ). Conversely, suppose we have n ∈ F with ψ ( n ) = 0, so the number q = P i n i p − − i actually lies in Z . Put m i = P j
0, so m ∈ F . We find that φ ( m ) = n , so the sequence F φ −→ F ψ −→ Z /p ∞ is short exact. As F is free, this gives us an exact sequenceHom( Z /p ∞ , A ) Hom( F, A ) Hom( F, A ) Ext( Z /p ∞ , A ) . ψ ∗ φ ∗ Next, for any series v ( x ) = P i a i x i ∈ A [[ x ]] we have a homomorphism α ( v ( x )) : F → A given by α ( v ( x ))( e i ) = a i for all i ≥ 0. This construction gives an isomorphism A [[ x ]] → Hom( F, A ). If we make the conventions a − = 0 and e − = 0 we also have α (( x − p ) v ( x ))( e j ) = α X i ( a i − − pa i ) x i ! ( e j ) = a j − − pa j = α ( v ( x ))( e j − − pe j ) = α ( v ( x ))( φ ( e j ))Thus, multiplication by x − p on A [[ x ]] corresponds to φ ∗ on Hom( F, A ), and the claim follows from this. (cid:3) References [1] A. K. Bousfield and Daniel M. Kan, Homotopy limits, completions and localizations , Lecture notes in Mathematics, vol. 304,Springer–Verlag, 1972., Lecture notes in Mathematics, vol. 304,Springer–Verlag, 1972.