An abelian group $F$ is free if and only if $\operatorname{Ext}(F,A)=0$ for
all $A$.

###### Proof.

First suppose that $F$ is free. By applying Lemma 8.4 to the diagram

${F}$${I_{F}}$${F}$$\scriptstyle{s}$$\scriptstyle{1}$$\scriptstyle{q}$

we obtain a homomorphism $s\colon F\to I_{F}$ with $qs=1_{F}$. This gives a
splitting in the usual way, so there is a unique map $r\colon
I_{F}\to\ker(q)=I_{F}^{2}$ with $jr=1-sq$ and we have $rj=1_{I^{2}_{F}}$. Now
for any $A$ we have homomorphisms

$\operatorname{Hom}(I^{2}_{F},A)\xrightarrow{r^{*}}\operatorname{Hom}(I_{F},A)\xrightarrow{j^{*}}\operatorname{Hom}(I^{2}_{F},A)$

with $j^{*}r^{*}=(rj)^{*}=1^{*}=1$. This implies that $j^{*}$ is surjective,
so the cokernel is zero. But $\operatorname{Ext}(F,A)$ is defined to be
$\operatorname{cok}(j^{*})$, so $\operatorname{Ext}(F,A)=0$ as claimed.

Conversely, suppose that $\operatorname{Ext}(F,A)=0$ for all $A$. Let $q\colon
B\to C$ be a surjective homomorphism. If we put $A=\ker(q)$, then Proposition
8.6 gives an exact sequence

$0\to\operatorname{Hom}(U,A)\xrightarrow{j_{*}}\operatorname{Hom}(U,B)\xrightarrow{q_{*}}\operatorname{Hom}(U,C)\xrightarrow{\delta}0\xrightarrow{j_{*}}0\xrightarrow{q_{*}}0\to
0.$

From this we deduce that $q_{*}$ is surjective. By the last part of Lemma 8.4,
this means that $F$ is free. ∎

To complete our study of $\operatorname{Ext}$ groups, we will need to
understand groups $D$ with the “dual” property that
$\operatorname{Ext}(A,D)=0$ 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 $n>0$ and all $v\in V$ there exists $u\in V$ with $nu=v$.
Equivalently, $V$ is divisible iff all the maps $n.1_{V}\colon V\to V$ (for
$n>0$) are surjective.

######  Example 8.9.

The groups ${\mathbb{Q}}$, $\mathbb{R}$ and ${\mathbb{Q}/\mathbb{Z}}$ are all
divisible. The only divisible finite group is the trivial group.

######  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
$d_{n}\colon A\to A$ for $n>0$ such that $n.d_{n}(a)=a$ for all $n$ and $a$
(so in particular $d_{1}(a)=a$), 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 ${\mathbb{Q}}$ or $\mathbb{R}$ we just have
$d_{n}(a)=a/n$. For $A={\mathbb{Q}}/\mathbb{Z}$, every element has a unique
representation as $a=x+\mathbb{Z}$ with $0\leq x<1$, and we put
$d_{n}(a)=x/n+\mathbb{Z}$.

######  Proposition 8.12.

Let $D$ be a divisible group. Then for any injective homomorphism $j\colon
A\to B$, the resulting homomorphism
$j^{*}\colon\operatorname{Hom}(B,D)\to\operatorname{Hom}(A,D)$ is surjective.
Equivalently, given homomorphisms $j$ and $f$ as shown below, there exists
$g\colon B\to D$ with $gj=f$.

${A}$${B}$${D.}$$\scriptstyle{j}$$\scriptstyle{f}$$\scriptstyle{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
$f\in\operatorname{Hom}(A,D)$ lies in the image of
$j^{*}\colon\operatorname{Hom}(B,D)\to\operatorname{Hom}(A,D)$; 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 from the special case of the extension property
where $A=B=\mathbb{Z}$ and $j=n.1_{\mathbb{Z}}$ for some $n>0$.

For the main part of the proof, suppose that $D$ is divisible, and that we are
given $j$ and $f$ as shown. It will be harmless to replace $A$ by the
isomorphic group $j(A)\leq B$ and so assume that $A\leq B$ and that $j$ is
just the inclusion. We then need to find $g\colon B\to D$ with $g|_{A}=f$. For
this we choose a division system $(d_{n})_{n>0}$ for $D$ and a well-ordering
of $B$. For $b\in B$ we let $B_{<b}$ denote the subgroup generated by
$A\cup\\{x\in B\;|\;x<b\\}$, and similarly for $B_{\leq b}$. We then have
$\\{k\in\mathbb{Z}\;|\;kb\in B_{<b}\\}=\mathbb{Z}.n_{b}$ for some $n_{b}\geq
0$. Note that if $n_{b}=0$ then $B_{\leq b}=B_{<b}\oplus\mathbb{Z}$, with the
$\mathbb{Z}$ summand generated by $b$. We say that a homomorphism $g\colon
B_{\leq b}\to D$ is _admissible_ if

  * •

$g|_{A}=f$

  * •

Whenever $x\leq b$ with $n_{x}=0$ we have $g(x)=0$

  * •

Whenever $x\leq b$ with $n_{x}>0$ we have $g(x)=d_{n_{x}}(g(n_{x}x))$.

We claim that for all $b$ there is a unique admissible homomorphism
$g_{b}\colon B_{\leq b}\to D$. If not, let $b$ be the least element for which
this is false (which is meaningful because $B$ is well-ordered). For all
$x<b$, we have a unique admissible map $g_{x}\colon B_{\leq x}\to D$. By
uniqueness, we see that $g_{x}$ agrees with $g_{y}$ on $B_{\leq y}$ whenever
$y\leq x<b$. It follows that the maps $g_{x}$ can be combined to give a map
$g^{\prime}\colon B_{<b}\to D$. If $n_{b}=0$, we find that there is a unique
extension $g_{b}\colon B_{\leq b}\to D$ satisfying $g_{b}(u+kb)=g^{\prime}(u)$
for all $u\in B_{<b}$ and $k\in\mathbb{Z}$. If $n_{b}>0$, then we have an
element $z=d_{n_{b}}(g^{\prime}(n_{b}b))\in D$ with $nz=g^{\prime}(n_{b}b)$,
and we see that there is a unique extension $g_{b}\colon B_{\leq b}\to D$
satisfying $g_{b}(u+kb)=g^{\prime}(u)+kz$ for all $u\in B_{<b}$ and
$k\in\mathbb{Z}$. Either way, we find that $g_{b}$ is the unique admissible
map on $B_{\leq b}$, contrary to assumption. Thus, we have $g_{b}$ for all
$b$, and we again see that $g_{b}$ agrees with $g_{c}$ on $B_{\leq c}$
whenever $c\leq b$, so the maps $g_{b}$ fit together to give a homomorphism
$g\colon B\to D$. This clearly satisfies $g|_{A}=f$ as required. ∎

######  Corollary 8.13.

If $D$ is divisible, then $\operatorname{Ext}(A,D)=0$ for all $A$.

######  Remark 8.14.

The converse statement is also true, but it will be more convenient to prove
that later.

###### Proof.

$\operatorname{Ext}(A,D)$ is by definition the cokernel of
$j^{*}\colon\operatorname{Hom}(I_{A},D)\to\operatorname{Hom}(I^{2}_{A},D)$,
but $j^{*}$ is surjective by the proposition. ∎

We next want to show that every abelian group can be embedded in a divisible
group. This will be proved after some preliminaries.

######  Proposition 8.15.

Let $A$ be an abelian group, and let $a$ be a nontrivial element of $A$. Then
there is a homomorphism $f\colon A\to{\mathbb{Q}/\mathbb{Z}}$ with $f(a)\neq
0$.

###### Proof.

Put $A_{0}=\mathbb{Z}a\leq A$. If $a$ has infinite order then
$A_{0}\simeq\mathbb{Z}$ and we can define $f_{0}\colon
A_{0}\to{\mathbb{Q}/\mathbb{Z}}$ by $f_{0}(ka)=k/2+\mathbb{Z}$. If $a$ has
finite order $n$ we can instead define $f_{0}\colon
A_{0}\to{\mathbb{Q}/\mathbb{Z}}$ by $f_{0}(ka)=k/n+\mathbb{Z}$. Either way we
have $f_{0}(a)\neq 0$. Next, as ${\mathbb{Q}/\mathbb{Z}}$ is divisible,
Proposition 8.12 tells us that the restriction map
$\operatorname{Hom}(A,{\mathbb{Q}/\mathbb{Z}})\to\operatorname{Hom}(A_{0},{\mathbb{Q}/\mathbb{Z}})$
is surjective, so we can choose $f\colon A\to{\mathbb{Q}/\mathbb{Z}}$ with
$f|_{A_{0}}=f_{0}$. In particular, this means that $f(a)\neq 0$, as required.
∎

######  Lemma 8.16.

For any set $I$, the group $\operatorname{Map}(I,{\mathbb{Q}/\mathbb{Z}})$ is
divisible.

###### Proof.

Let $(d_{n})_{n>0}$ be the standard division system for
${\mathbb{Q}/\mathbb{Z}}$ as in Remark 8.11. The maps $u\mapsto d_{n}\circ u$
then give a division system for
$\operatorname{Map}(I,{\mathbb{Q}/\mathbb{Z}})$. ∎

Now suppose we have a family of homomorphisms $f_{i}\colon
A\to{\mathbb{Q}/\mathbb{Z}}$ for $i\in I$. We can combine them to give a
single homomorphism $j\colon
A\to\operatorname{Map}(I,{\mathbb{Q}/\mathbb{Z}})$ by the rule
$j(a)(i)=f_{i}(a)$. Note that the kernel of $j$ is the intersection 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 $f\colon A\to{\mathbb{Q}/\mathbb{Z}}$, and thus to take
$I=\operatorname{Hom}(A,{\mathbb{Q}/\mathbb{Z}})$. This leads us to the
following definitions.

######  Definition 8.17.

We put
$E_{A}=\operatorname{Map}(\operatorname{Hom}(A,{\mathbb{Q}/\mathbb{Z}}),{\mathbb{Q}/\mathbb{Z}})$,
and define $j\colon A\to E_{A}$ by the rather tautological rule
$j(a)(f)=f(a)$. We write $E^{2}_{A}$ for the cokernel of $j$, and $q$ for the
quotient map $E_{A}\to E^{2}_{A}$.

######  Proposition 8.18.

The groups $E_{A}$ and $E^{2}_{A}$ are divisible, and the sequence

$A\xrightarrow{j}E_{A}\xrightarrow{q}E^{2}_{A}$

is short exact.

###### Proof.

The group $E_{A}$ is divisible by Lemma 8.16, and $E^{2}_{A}$ is a quotient of
$E_{A}$ so it is also divisible. It is clear by construction that $q$ is
surjective with $\operatorname{image}(j)=\ker(q)$. Finally, if $a\in A$ is
nonzero then Proposition 8.15 gives us a homomorphism
$f\in\operatorname{Hom}(A,{\mathbb{Q}/\mathbb{Z}})$ with $f(a)\neq 0$ or
equivalently $j(a)(f)\neq 0$, so $j(a)\neq 0$. This shows that $j$ is
injective, so the sequence is short exact as claimed. ∎

######  Remark 8.19.

Note that for $a\in A$ and
$f,g\in\operatorname{Hom}(A,{\mathbb{Q}/\mathbb{Z}})$ we have

$j(a)(f+g)=(f+g)(a)=f(a)+g(a)=j(a)(f)+j(a)(g),$

so the map
$j(a)\colon\operatorname{Hom}(A,{\mathbb{Q}/\mathbb{Z}})\to{\mathbb{Q}/\mathbb{Z}}$
is a homomorphism. In other words, $j$ can be regarded as an injective
homomorphism

$j\colon
A\to\operatorname{Hom}(\operatorname{Hom}(A,{\mathbb{Q}/\mathbb{Z}}),{\mathbb{Q}/\mathbb{Z}}).$

If we use the briefer notation $A^{*}$ for
$\operatorname{Hom}(A,{\mathbb{Q}/\mathbb{Z}})$, then $j\colon A\to A^{**}\leq
E_{A}$. The group $A^{**}$ is usually not divisible, so $E_{A}$ is more useful
for our immediate applications to $\operatorname{Ext}$ groups. However, the
group $A^{**}$ will reappear later in other contexts.

######  Corollary 8.20.

For all $A$ and $B$, there is a natural exact sequence

$0\to\operatorname{Hom}(A,B)\xrightarrow{j_{*}}\operatorname{Hom}(A,E_{B})\xrightarrow{q_{*}}\operatorname{Hom}(A,E^{2}_{B})\to\operatorname{Ext}(A,B)\to
0.$

###### Proof.

Apply Proposition 8.6 to the sequence $B\to E_{B}\to E^{2}_{B}$, noting that
$\operatorname{Ext}(A,E_{B})=\operatorname{Ext}(A,E^{2}_{B})=0$ by Corollary
8.13. ∎

######  Corollary 8.21.

For any $f_{0},f_{1}\colon A^{\prime}\to A$ we have
$(f_{0}+f_{1})^{*}=f_{0}^{*}+f_{1}^{*}\colon\operatorname{Ext}(A,B)\to\operatorname{Ext}(A^{\prime},B)$.

###### Proof.

The corresponding statement is clearly true for the induced maps on
$\operatorname{Hom}(A,E^{2}_{B})$, and Corollary 8.20 identifies
$\operatorname{Ext}(A,B)$ in a natural way as a quotient group of
$\operatorname{Hom}(A,E^{2}_{B})$. ∎

######  Corollary 8.22.

The natural maps

$\displaystyle\operatorname{Ext}(\bigoplus_{i}A_{i},B)$
$\displaystyle\to\prod_{i}\operatorname{Ext}(A_{i},B)$
$\displaystyle\operatorname{Ext}(A,\prod_{j}B_{j})$
$\displaystyle\to\prod_{j}\operatorname{Ext}(A,B_{j}).$

are isomorphisms.

###### Proof.

The functors $\operatorname{Hom}(-,U)$ (for $U\in\\{B,E_{B},E^{2}_{B}\\}$)
convert coproducts to products, and the cokernel of a product is the product
of the cokernels, so the first statement follows from Corollary 8.20.
Similarly, the functors $\operatorname{Hom}(T,-)$ (for
$T\in\\{A,I_{A},I^{2}_{A}\\}$) preserve products, so the second statement
follows from our original definition of $\operatorname{Ext}$. ∎

######  Proposition 8.23.

Let $A\xrightarrow{i}B\xrightarrow{p}C$ be a short exact sequence of abelian
groups, and let $V$ be an abelian group. Then there is a natural exact
sequence

$0\to\operatorname{Hom}(C,V)\xrightarrow{p^{*}}\operatorname{Hom}(B,V)\xrightarrow{i^{*}}\operatorname{Hom}(A,V)\xrightarrow{\delta}\operatorname{Ext}(C,V)\xrightarrow{p^{*}}\operatorname{Ext}(B,V)\xrightarrow{i^{*}}\operatorname{Ext}(A,V)\to
0.$

###### Proof.

Consider the diagram

${\operatorname{Hom}(C,E_{V})}$${\operatorname{Hom}(B,E_{V})}$${\operatorname{Hom}(A,E_{V})}$${\operatorname{Hom}(C,E^{2}_{V})}$${\operatorname{Hom}(B,E^{2}_{V})}$${\operatorname{Hom}(A,E^{2}_{V})}$$\scriptstyle{q_{*}}$$\scriptstyle{p^{*}}$$\scriptstyle{q_{*}}$$\scriptstyle{i^{*}}$$\scriptstyle{q_{*}}$$\scriptstyle{p^{*}}$$\scriptstyle{i^{*}}$

The rows are short exact by Lemma 7.17 together with Proposition 8.12. The
Snake Lemma therefore gives us a six-term exact sequence involving the kernels
and cokernels of the vertical maps $q_{*}$. Corollary 8.20 identifies these
kernels and cokernels with $\operatorname{Hom}$ and $\operatorname{Ext}$
groups as required. ∎

######  Corollary 8.24.

For any groups $B$ and $V$, and any subgroup $A\leq B$, the restriction
$\operatorname{Ext}(B,V)\to\operatorname{Ext}(A,V)$ is surjective. ∎

######  Corollary 8.25.

There are natural isomorphisms $\operatorname{Ext}(\mathbb{Z}/n,B)\simeq B/nB$
for $n>0$, and $\operatorname{Ext}(\mathbb{Z},B)=0$.

Note that in conjunction with Corollary 8.22 this allows us to calculate
$\operatorname{Ext}(A,B)$ whenever $A$ is finitely generated.

###### Proof.

Corollary 8.7 tells us that $\operatorname{Ext}(\mathbb{Z},B)=0$ for all $B$.
Now consider the short exact sequence

$\mathbb{Z}\xrightarrow{n.1_{\mathbb{Z}}}\mathbb{Z}\xrightarrow{}\mathbb{Z}/n.$

Using Proposition 8.23 we obtain an exact sequence

$0\to\operatorname{Hom}(\mathbb{Z}/n,B)\to\operatorname{Hom}(\mathbb{Z},B)\to\operatorname{Hom}(\mathbb{Z},B)\to\operatorname{Ext}(\mathbb{Z}/n,B)\to\operatorname{Ext}(\mathbb{Z},B)\to\operatorname{Ext}(\mathbb{Z},B)\to
0.$

Now $\operatorname{Ext}(\mathbb{Z},B)=0$, while
$\operatorname{Hom}(\mathbb{Z},B)$ is easily identified with $B$, and
$\operatorname{Hom}(\mathbb{Z}/n,B)$ with $B[n]=\\{b\in B\;|\;nb=0\\}$. We
thus have an exact sequence

$0\to B[n]\to
B\xrightarrow{n.1_{B}}B\xrightarrow{\delta}\operatorname{Ext}(\mathbb{Z}/n,B)\to
0.$

From this it is clear that
$\operatorname{Ext}(\mathbb{Z}/n,B)=\operatorname{cok}(n.1_{B})=B/nB$. ∎

######  Proposition 8.26.

Let $A$ be a torsion group. Then there is a canonical isomorphism

$\operatorname{Ext}(A,\mathbb{Z})=\operatorname{Hom}(A,{\mathbb{Q}/\mathbb{Z}})=\prod_{p}\operatorname{Hom}(\operatorname{tors}_{p}(A),{\mathbb{Q}/\mathbb{Z}}),$

and this maps surjectively to
$\prod_{p}\operatorname{Hom}(A[p],\mathbb{Z}/p)$. In particular, if
$\operatorname{Ext}(A,\mathbb{Z})=0$ then $A=0$.

###### Proof.

As $A$ is torsion, and both $\mathbb{Z}$ and ${\mathbb{Q}}$ are torsion free,
we see that
$\operatorname{Hom}(A,\mathbb{Z})=\operatorname{Hom}(A,{\mathbb{Q}})=0$. As
${\mathbb{Q}}$ and ${\mathbb{Q}/\mathbb{Z}}$ are divisible, we see from
Proposition 8.12 that
$\operatorname{Ext}(A,{\mathbb{Q}})=\operatorname{Ext}(A,{\mathbb{Q}/\mathbb{Z}})=0$.
Thus, if we apply Proposition 8.6 to the short exact sequence
$\mathbb{Z}\to{\mathbb{Q}}\to{\mathbb{Q}/\mathbb{Z}}$ we just get an
isomorphism
$\delta\colon\operatorname{Ext}(A,\mathbb{Z})\to\operatorname{Hom}(A,{\mathbb{Q}/\mathbb{Z}})$.
Next, Proposition 4.9 gives $A=\bigoplus_{p}\operatorname{tors}_{p}(A)$, so
$\operatorname{Hom}(A,{\mathbb{Q}/\mathbb{Z}})=\prod_{p}\operatorname{Hom}(\operatorname{tors}_{p}(A),{\mathbb{Q}/\mathbb{Z}})$.
Now $A[p]$ is a subgroup of $\operatorname{tors}_{p}(A)$ and
${\mathbb{Q}/\mathbb{Z}}$ is divisible so the restriction

$\operatorname{Hom}(\operatorname{tors}_{p}(A),{\mathbb{Q}/\mathbb{Z}})\to\operatorname{Hom}(A[p],{\mathbb{Q}/\mathbb{Z}})$

is surjective. Moreover, any homomorphism from
$A[p]\to{\mathbb{Q}/\mathbb{Z}}$ necessarily lands in
$({\mathbb{Q}/\mathbb{Z}})[p]$, which is a copy of $\mathbb{Z}/p$ generated by
the element $(1/p)+\mathbb{Z}$. By taking the product over all $p$, we get a
surjection

$\prod_{p}\operatorname{Hom}(\operatorname{tors}_{p}(A),{\mathbb{Q}/\mathbb{Z}})\to\prod_{p}\operatorname{Hom}(A[p],\mathbb{Z}/p)$

as claimed. The last statement follows from the isomorphism
$\operatorname{Ext}(A,\mathbb{Z})=\operatorname{Hom}(A,{\mathbb{Q}/\mathbb{Z}})$
together with Proposition 8.15. ∎

######  Example 8.27.

We can take $A={\mathbb{Q}/\mathbb{Z}}$, and we find that
$\operatorname{Ext}({\mathbb{Q}/\mathbb{Z}},\mathbb{Z})=\operatorname{End}({\mathbb{Q}/\mathbb{Z}})$.
It is easy to see that $({\mathbb{Q}/\mathbb{Z}})[n]$ is a copy of
$\mathbb{Z}/n$, generated by $(1/n)+\mathbb{Z}$. It follows that
$\operatorname{Hom}(({\mathbb{Q}/\mathbb{Z}})[p],\mathbb{Z}/p)\simeq\mathbb{Z}/p$,
and thus that
$\prod_{p}\operatorname{Hom}(({\mathbb{Q}/\mathbb{Z}})[p],\mathbb{Z}/p)$ is
uncountable, and thus that $\operatorname{End}({\mathbb{Q}/\mathbb{Z}})$ is
uncountable. Next, we can apply $\operatorname{Hom}(-,\mathbb{Z})$ and
$\operatorname{Ext}(-,\mathbb{Z})$ to the sequence
$\mathbb{Z}\to{\mathbb{Q}}\to{\mathbb{Q}/\mathbb{Z}}$ to get a six term exact
sequence. As ${\mathbb{Q}}$ and ${\mathbb{Q}/\mathbb{Z}}$ are divisible, we
find that
$\operatorname{Hom}({\mathbb{Q}},\mathbb{Z})=\operatorname{Hom}({\mathbb{Q}/\mathbb{Z}},\mathbb{Z})=0$.
We also have $\operatorname{Hom}(\mathbb{Z},\mathbb{Z})=\mathbb{Z}$ and
$\operatorname{Ext}(\mathbb{Z},\mathbb{Z})=0$. The six term sequence therefore
reduces to a short exact sequence
$\mathbb{Z}\to\operatorname{End}({\mathbb{Q}/\mathbb{Z}})\to\operatorname{Ext}({\mathbb{Q}},\mathbb{Z})$.
As $\mathbb{Z}$ is countable and $\operatorname{End}({\mathbb{Q}/\mathbb{Z}})$
is uncountable, we deduce that $\operatorname{Ext}({\mathbb{Q}},\mathbb{Z})$
is uncountable. In particular, it is nonzero.

######  Definition 8.28.

An _extension of $C$ by $A$_ is a short exact sequence
$A\xrightarrow{i}B\xrightarrow{p}C$. We say that two extensions
$(A\xrightarrow{i_{0}}B_{0}\xrightarrow{p_{0}}C)$ and
$(A\xrightarrow{i_{1}}B_{1}\xrightarrow{p_{1}}C)$ are _equivalent_ if there
exists a map $f\colon B_{0}\to B_{1}$ with $fi_{0}=i_{1}$ and $p_{1}f=p_{0}$.
(Any such $f$ is an isomorphism by a straightforward diagram chase, and using
this we see that this notion of equivalence is reflexive, symmetric and
transitive.)

######  Proposition 8.29.

Let $E=(A\xrightarrow{i}B\xrightarrow{p}C)$ be an extension of $C$ by $A$.

  * (a)

For any homomorphism $h\colon C^{\prime}\to C$, we have an extension
$h^{*}E=(A\xrightarrow{i^{\prime}}B^{\prime}\xrightarrow{p^{\prime}}C^{\prime})$
given by

$\displaystyle B^{\prime}$ $\displaystyle=\\{(b,c^{\prime})\in B\oplus
C^{\prime}\;|\;p(b)=h(c^{\prime})\\}$ $\displaystyle i^{\prime}(a)$
$\displaystyle=(i(a),0)$ $\displaystyle p^{\prime}(b,c^{\prime})$
$\displaystyle=c^{\prime}.$

(We call this the _pullback of $E$ along $h$._)

  * (b)

Suppose we have another extension
$E^{*}=(A\xrightarrow{i^{*}}B^{*}\xrightarrow{p^{*}}C^{\prime})$ and a
commutative diagram

${A}$${B^{*}}$${C^{\prime}}$${A}$${B}$${C}$$\scriptstyle{i^{*}}$$\scriptstyle{p^{*}}$$\scriptstyle{g}$$\scriptstyle{h}$$\scriptstyle{i}$$\scriptstyle{p}$

Then $E^{*}$ is equivalent to $h^{*}E$.

  * (c)

For any map $m\colon C^{\prime}\to B$, the extension $(h+pm)^{*}E$ is
equivalent to $h^{*}E$.

  * (d)

If $E_{0}$ and $E_{1}$ are equivalent extensions of $C$ by $A$, then
$h^{*}E_{0}$ and $h^{*}E_{1}$ are also equivalent.

###### Proof.

  * (a)

We can certainly define a group $B^{\prime}$ and homomorphisms $i^{\prime}$
and $p^{\prime}$ by the given formulae. As $i$ is injective, it is clear that
$i^{\prime}$ is injective. Now suppose that $c^{\prime}\in C^{\prime}$. We
then have $f(c^{\prime})\in C$ and $p\colon B\to C$ is surjective by
assumption, so we can choose $b\in B$ with $p(b)=f(c^{\prime})$. This gives a
point $b^{\prime}=(b,c^{\prime})\in B^{\prime}$ with
$p^{\prime}(b^{\prime})=c^{\prime}$, so we see that $p^{\prime}$ is
surjective. It is immediate that $p^{\prime}i^{\prime}=0$, so
$\operatorname{image}(i^{\prime})\leq\ker(p^{\prime})$. A general element of
$\ker(p^{\prime})$ has the form $b^{\prime}=(b,0)$ with $p(b)=f(0)=0$, so
$b\in\ker(p)=\operatorname{image}(i)$, so $b=i(a)$ for some $a\in A$. This
means that $b^{\prime}=i^{\prime}(a)\in\operatorname{image}(i^{\prime})$. We
conclude that the sequence $h^{*}E$ is indeed short exact, so it gives an
extension of $C^{\prime}$ by $A$.

  * (b)

Now suppose we have a commutative diagram as indicated. As $hp^{*}=pg$ we can
define $g^{\prime}\colon B^{*}\to B^{\prime}$ by
$g^{\prime}(b^{*})=(g(b^{*}),p^{*}(b^{*}))$. We then have
$p^{\prime}g^{\prime}(b^{*})=p^{*}(b^{*})$ and

$g^{\prime}(i^{*}(a))=(g(i^{*}(a)),p^{*}(i^{*}(a)))=(i(a),0)=i^{\prime}(a),$

so $p^{\prime}g^{\prime}=p^{*}$ and $g^{\prime}i^{*}=i^{\prime}$. Thus,
$g^{\prime}$ gives an equivalence between $E^{*}$ and $h^{*}E$.

  * (c)

By the construction in part (a), we have a commutative diagram

${A}$${B^{\prime}}$${C^{\prime}}$${A}$${B}$${C}$$\scriptstyle{i^{\prime}}$$\scriptstyle{p^{\prime}}$$\scriptstyle{g}$$\scriptstyle{h}$$\scriptstyle{i}$$\scriptstyle{p}$

(where $g(b,c^{\prime})=b$). It follows easily that there is also a
commutative diagram

${A}$${B^{\prime}}$${C^{\prime}}$${A}$${B}$${C.}$$\scriptstyle{i^{\prime}}$$\scriptstyle{p^{\prime}}$$\scriptstyle{g+mp^{\prime}}$$\scriptstyle{h+pm}$$\scriptstyle{i}$$\scriptstyle{p}$

Now part (b) tells us that the top row is equivalent to $(h+pm)^{*}$ of the
bottom row, or in other words $h^{*}E\simeq(h+pm)^{*}E$ as claimed.

  * (d)

Suppose we have equivalent extensions
$E_{k}=(A\xrightarrow{i_{k}}B_{k}\xrightarrow{p_{k}}C)$ for $k=0,1$, so there
is an isomorphism $s\colon B_{0}\to B_{1}$ with $si_{0}=i_{1}$ and
$p_{1}s=p_{0}$. We can then define $s^{\prime}\colon B^{\prime}_{0}\to
B^{\prime}_{1}$ by $s^{\prime}(b_{0},c^{\prime})=(s(b_{0}),c^{\prime})$, and
we find that $s^{\prime}i^{\prime}_{0}=i^{\prime}_{1}$ and
$p^{\prime}_{1}s^{\prime}=p^{\prime}_{0}$; this shows that $h^{*}E_{0}$ and
$h^{*}E_{1}$ are equivalent as claimed.

∎

######  Proposition 8.30.

Let $E=(A\xrightarrow{i}B\xrightarrow{p}C)$ be an extension of $C$ by $A$.

  * (a)

For any homomorphism $f\colon A\to A^{\prime}$, we have an extension
$f_{*}E=(A^{\prime}\xrightarrow{i^{\prime}}B^{\prime}\xrightarrow{p^{\prime}}C)$
given by

$\displaystyle R$ $\displaystyle=\\{(f(a),-i(a))\;|\;a\in A\\}\leq
A^{\prime}\oplus B$ $\displaystyle B^{\prime}$
$\displaystyle=(A^{\prime}\oplus B)/R$ $\displaystyle i^{\prime}(a^{\prime})$
$\displaystyle=(a^{\prime},0)+R$ $\displaystyle p^{\prime}((a^{\prime},b)+R)$
$\displaystyle=p(b).$

(We call this the _pushout of $E$ along $f$._)

  * (b)

Suppose we have another extension
$E^{*}=(A\xrightarrow{i^{*}}B^{*}\xrightarrow{p^{*}}C^{\prime})$ and a
commutative diagram

${Aq}$${B}$${C}$${A^{*}}$${B^{*}}$${C}$$\scriptstyle{i}$$\scriptstyle{f}$$\scriptstyle{p}$$\scriptstyle{g}$$\scriptstyle{i^{*}}$$\scriptstyle{p^{*}}$

Then $E^{*}$ is equivalent to $f_{*}E$.

  * (c)

For any map $n\colon B\to A^{\prime}$, the extension $(f+ni)_{*}E$ is
equivalent to $f_{*}E$.

  * (d)

If $E_{0}$ and $E_{1}$ are equivalent extensions of $C$ by $A$, then
$f_{*}E_{0}$ and $f_{*}E_{1}$ are also equivalent.

###### Proof.

  * (a)

We can certainly define groups $R$ and $B^{\prime}$, and a homomorphism
$i^{\prime}$, by the given formulae. If $(a^{\prime},b)\in R$ then there
exists $a\in A$ with $a^{\prime}=f(a)$ and $b=-i(a)$, so $p(b)=-p(i(a))=0$.
Given this, we see that the formula $p^{\prime}((a^{\prime},b)+R)=p(b)$ also
gives a well-defined map $B^{\prime}\to C$.

If $i^{\prime}(a^{\prime})=0$ we must have $(a^{\prime},0)\in R$, so there
exists $a\in A$ with $f(a)=a^{\prime}$ and $i(a)=0$. As $i$ is injective this
gives $a=0$ and then $a^{\prime}=f(0)=0$. This proves that $i^{\prime}$ is
injective. As $p$ is surjective, it is immediate that $p^{\prime}$ is also
surjective. Next, we have
$p^{\prime}i^{\prime}(a^{\prime})=p^{\prime}((a^{\prime},0)+R)=p(0)=0$, so
$\operatorname{image}(i^{\prime})\leq\ker(p^{\prime})$. Conversely, suppose we
have an element $b^{\prime}=(a^{\prime},b)+R\in B^{\prime}$ with
$p^{\prime}(b^{\prime})=0$. This means that $p(b)=0$, so $b=i(a)$ for some
$a\in A$. We then find that $(f(a),-i(a))\in R$

$b^{\prime}=(a^{\prime},b)+R=(a^{\prime},b)+(f(a),-i(a))+R=(a^{\prime}+f(a),0)+R=i^{\prime}(a^{\prime}+f(a))\in\operatorname{image}(i^{\prime}).$

We conclude that the sequence $f_{*}E$ is indeed short exact, so it gives an
extension of $C$ by $A^{\prime}$.

  * (b)

Now suppose we have a commutative diagram as indicated. Define
$g^{\prime\prime}\colon A^{\prime}\oplus B\to B^{*}$ by
$g^{\prime\prime}(a^{\prime},b)=i^{*}(a^{\prime})+g(b)$. We then have
$g^{\prime\prime}(f(a),-i(a))=(i^{*}f-gi)(a)=0$, so $g^{\prime\prime}(R)=0$,
so there is a unique homomorphism $g^{\prime}\colon B^{\prime}\to B^{*}$ given
by $g^{\prime}(x+R)=g^{\prime\prime}(x)$. This means that
$g^{\prime}i^{\prime}(a)=g^{\prime}((a^{\prime},0)+R)=g^{\prime\prime}(a^{\prime},0)=i^{*}(a^{\prime})$,
so $g^{\prime}i^{\prime}=i^{*}$. We also have

$p^{*}g^{\prime}((a^{\prime},b)+R)=p^{*}i^{*}(a^{\prime})+p^{*}g(b)=0+p(b)=p(b),$

so $p^{*}g^{\prime}=p$. Thus, $g^{\prime}$ gives the required equivalence from
$f_{*}E$ to $E^{*}$.

  * (c)

By the construction in part (a), we have a commutative diagram

${A}$${B}$${C}$${A^{\prime}}$${B^{\prime}}$${C}$$\scriptstyle{i}$$\scriptstyle{f}$$\scriptstyle{p}$$\scriptstyle{g}$$\scriptstyle{i^{\prime}}$$\scriptstyle{p^{\prime}}$

(where $g(b)=(0,b)+R$). It follows easily that there is also a commutative
diagram

${A}$${B}$${C}$${A^{\prime}}$${B^{\prime}}$${C}$$\scriptstyle{i}$$\scriptstyle{f+ni}$$\scriptstyle{p}$$\scriptstyle{g+i^{\prime}n}$$\scriptstyle{i^{\prime}}$$\scriptstyle{p^{\prime}}$

Now part (b) tells us that the bottom row is equivalent to $(f+ni)_{*}$ of the
top row, or in other words $f_{*}E\simeq(f+ni)_{*}E$ as claimed.

  * (d)

Suppose we have equivalent extensions
$E_{k}=(A\xrightarrow{i_{k}}B_{k}\xrightarrow{p_{k}}C)$ for $k=0,1$, so there
is an isomorphism $s\colon B_{0}\to B_{1}$ with $si_{0}=i_{1}$ and
$p_{1}s=p_{0}$. We can then define $s^{\prime}\colon B^{\prime}_{0}\to
B^{\prime}_{1}$ by

$s^{\prime}((a^{\prime},b_{0})+R_{0})=(a^{\prime},s(b_{0}))+R_{1}.$

We find that $s^{\prime}i^{\prime}_{0}=i^{\prime}_{1}$ and
$p^{\prime}_{1}s^{\prime}=p^{\prime}_{0}$; this shows that $h^{*}E_{0}$ and
$h^{*}E_{1}$ are equivalent as claimed.

∎

######  Proposition 8.31.

Let $A$ and $C$ be abelian groups, and let $\operatorname{Ext}^{\prime}(C,A)$
denote the set of equivalence classes of extensions of $C$ by $A$. Let $Q$
denote the extension $(I_{C}^{2}\xrightarrow{j}I_{C}\xrightarrow{q}C)$. Then
there is a well-defined bijection

$\zeta\colon\operatorname{Ext}(C,A)=\frac{\operatorname{Hom}(I_{C}^{2},A)}{j^{*}(\operatorname{Hom}(I_{C},A))}\to\operatorname{Ext}^{\prime}(C,A)$

given by $\zeta(\alpha+\operatorname{image}(j^{*}))=[\alpha_{*}(Q)]$.

###### Proof.

Using Proposition 8.30 (especially part (c)) we see that there is a well-
defined map $\zeta$ as described. We must show that it is a bijection.
Consider an arbitrary extension $E=(A\xrightarrow{i}B\xrightarrow{p}C)$. As
$I_{C}$ is free and $p$ is surjective, Lemma 8.4 gives us a homomorphism
$\beta\colon I_{C}\to B$ with $p\beta=q$. This means that $p\beta j=qj=0$, so
$\operatorname{image}(\beta j)\leq\ker(p)=\operatorname{image}(i)$, so there
is a unique map $\alpha\colon I^{2}_{C}\to A$ with $i\alpha=\beta j$. We now
have a commutative diagram

${I^{2}_{C}}$${I_{C}}$${C}$${A}$${B}$${C,}$$\scriptstyle{j}$$\scriptstyle{\alpha}$$\scriptstyle{q}$$\scriptstyle{\beta}$$\scriptstyle{i}$$\scriptstyle{p}$

so Proposition 8.30(b) tells us that $E\simeq\alpha_{*}Q$, or in other words
$[E]=\zeta(\alpha+\operatorname{image}(j^{*}))$. This proves that $\zeta$ is
surjective. Suppose we also have
$[E]=\zeta(\alpha^{\prime}+\operatorname{image}(j^{*}))$. There is then
another commutative diagram

${I^{2}_{C}}$${I_{C}}$${C}$${A}$${B}$${C.}$$\scriptstyle{j}$$\scriptstyle{\alpha^{\prime}}$$\scriptstyle{q}$$\scriptstyle{\beta^{\prime}}$$\scriptstyle{i}$$\scriptstyle{p}$

In particular we have $p\beta^{\prime}=q=p\beta$ so
$p(\beta^{\prime}-\beta)=0$, so $\beta^{\prime}-\beta$ factors through
$\ker(p)=\operatorname{image}(i)$, so there is a unique homomorphism
$\gamma\colon I_{C}\to A$ with $\beta^{\prime}=\beta+i\gamma$. Now
$\beta^{\prime}j=i\alpha^{\prime}$ and $\beta j=i\alpha$ so the equation
$\beta^{\prime}=\beta+i\gamma$ yields $i\alpha^{\prime}=i\alpha+i\gamma j$, or
$i(\alpha^{\prime}-\alpha-\gamma j)=0$. As $i$ is injective we conclude that
$\alpha^{\prime}=\alpha+j^{*}(\gamma)$, so $\alpha^{\prime}$ and $\alpha$ have
the same image in $\operatorname{cok}(j^{*})=\operatorname{Ext}(C,A)$. This
proves that $\zeta$ is also injective. ∎

The above proposition gives a bijection from the group
$\operatorname{Ext}(C,A)$ to the set $\operatorname{Ext}^{\prime}(C,A)$. There
is thus a unique group structure on $\operatorname{Ext}^{\prime}(C,A)$ for
which this bijection is a homomorphism. We would like to understand this more
intrinsically.

######  Definition 8.32.

Suppose we have two extensions
$E_{k}=(A\xrightarrow{i_{k}}B_{k}\xrightarrow{p_{k}}C)$ for $k=0,1$. The _Baer
sum_ of $E_{0}$ and $E_{1}$ is the sequence
$E_{2}=(A\xrightarrow{i_{2}}B_{2}\xrightarrow{p_{2}}C)$ where

$\displaystyle U$ $\displaystyle=\\{(b_{0},b_{1})\in B_{0}\oplus
B_{1}\;|\;p_{0}(b_{0})=p_{1}(b_{1})\\}$ $\displaystyle V$
$\displaystyle=\\{(i_{0}(a),-i_{1}(a))\;|\;a\in A\\}$ $\displaystyle B_{2}$
$\displaystyle=U/V$ $\displaystyle i_{2}(a)$
$\displaystyle=(i_{0}(a),0)+V=(0,i_{1}(a))+V$ $\displaystyle
p_{2}((b_{0},b_{1})+V)$ $\displaystyle=p_{0}(b_{0})=p_{1}(b_{1}).$

######  Proposition 8.33.

In the above context, the sequence $E_{2}$ is an extension of $C$ by $A$, with
$[E_{2}]=[E_{0}]+[E_{1}]$ in $\operatorname{Ext}^{\prime}(C,A)$. Moreover, the
zero element in $\operatorname{Ext}^{\prime}(C,A)$ is the equivalence class
consisting of all split extensions.

###### Proof.

First, if $i_{2}(a)=0$ we must have $(i_{0}(a),0)\in V$, so
$(i_{0}(a),0)=(i_{0}(a^{\prime}),-i_{1}(a^{\prime}))$ for some $a^{\prime}\in
A$. As $i_{1}$ is injective and $i_{1}(a^{\prime})=0$ we have $a^{\prime}=0$,
so the equation $i_{0}(a)=i_{0}(a^{\prime})$ gives $i_{0}(a)=0$ and then
$a=0$. This shows that $i_{2}$ is injective. Next, suppose we have $c\in C$.
As both $p_{0}$ and $p_{1}$ are surjective we can choose $b_{0}\in B_{0}$ and
$b_{1}\in B_{1}$ with $p_{0}(b_{0})=p_{1}(b_{1})=c$. The element
$b_{2}=(b_{0},b_{1})+V\in B_{2}$ then satisfies $p_{2}(b_{2})=c$, so $p_{2}$
is surjective. Next, as $p_{0}i_{0}=0=p_{1}i_{1}$ we see from the definitions
that $p_{2}i_{2}=0$, so $\operatorname{image}(i_{2})\leq\ker(p_{2})$. Now
suppose we have an element $b_{2}=(b_{0},b_{1})+V\in B_{2}$ with
$p_{2}(b_{2})=0$. This means that $p_{0}(b_{0})=0=p_{1}(b_{1})$, so there is a
unique element $a_{0}\in A$ with $b_{0}=i_{0}(a_{0})$, and also a unique
element $a_{1}\in A$ with $b_{1}=i_{1}(a_{1})$. Put $a_{2}=a_{0}+a_{1}$ and
note that

$i_{2}(a_{2})=i_{2}(a_{0})+i_{2}(a_{1})=(i_{0}(a_{0}),0)+(0,i_{1}(a_{1}))+V=(b_{0},b_{1})+V=b_{2}.$

This proves that $\ker(p_{2})=\operatorname{image}(i_{2})$, so we have an
extension as claimed. Now suppose that
$[E_{k}]=\zeta(\alpha_{k}+\operatorname{image}(j^{*}))$ for $k=0,1$, so there
are commutative diagrams

${I^{2}_{C}}$${I_{C}}$${C}$${A}$${B_{k}}$${C}$$\scriptstyle{j}$$\scriptstyle{\alpha_{k}}$$\scriptstyle{q}$$\scriptstyle{\beta_{k}}$$\scriptstyle{i_{k}}$$\scriptstyle{p_{k}}$

for $k=0,1$. We define $\alpha_{2}\colon I^{2}_{C}\to A$ by
$\alpha_{2}(y)=\alpha_{0}(y)+\alpha_{1}(y)$, and we define $\beta_{2}\colon
I_{C}\to B_{2}$ by $\beta_{2}(x)=(\beta_{0}(x),\beta_{1}(x))+V$. It is
straightforward to check that this gives a commutative diagrams as above,
showing that

$[E_{2}]=\zeta(\alpha_{2}+\operatorname{image}(j^{*}))=\zeta(\alpha_{0}+\operatorname{image}(j^{*}))+\zeta(\alpha_{1}+\operatorname{image}(j^{*}))=[E_{0}]+[E_{1}].$

Thus, the sum in $\operatorname{Ext}^{\prime}(C,A)$ is the Baer sum, as
claimed. The zero element is $\zeta(0)$, which is the pushout of the extension
$Q=(I_{C}^{2}\xrightarrow{j}I_{C}\xrightarrow{q}C)$ along the map $0\colon
I^{2}_{C}\to A$. If we use the notation of Proposition 8.30 in this context we
have

$\displaystyle R$ $\displaystyle=\\{(0,-j(y))\;|\;y\in I_{C}^{2}\\}=0\oplus
I_{C}^{2}\leq A\oplus I_{C}$ $\displaystyle B^{\prime}$
$\displaystyle=\frac{A\oplus I_{C}}{R}=\frac{A\oplus I_{C}}{0\oplus
I_{C}^{2}}=A\oplus(I_{C}/I^{2}_{C})\simeq A\oplus C$ $\displaystyle
i^{\prime}(a)$ $\displaystyle=(a,0)$ $\displaystyle p^{\prime}(a,c)$
$\displaystyle=c.$

Thus, $0_{*}Q$ is just the obvious split extension of $C$ by $A$. ∎

We now present a result that will help us relate homology groups to cohomology
groups. There are very standard theorems that deduce information about
cohomology from information about homology. To go in the opposite direction we
need the following proposition, which is less well-known.

######  Proposition 8.34.

Suppose that $\operatorname{Hom}(A,\mathbb{Z})$ and
$\operatorname{Ext}(A,\mathbb{Z})$ are finitely generated. Then $A$ is
finitely generated.

The proof will follow after some lemmas.

######  Lemma 8.35.

Suppose that $\operatorname{Hom}(A,\mathbb{Z})$ is finitely generated. Then
$A=B\oplus F$ for some subgroups $B$ and $F$ such that $F$ is free and
finitely generated, and $\operatorname{Hom}(B,\mathbb{Z})=0$.

###### Proof.

Choose maps $f_{1},\dotsc,f_{r}\colon A\to\mathbb{Z}$ that generate
$\operatorname{Hom}(A,\mathbb{Z})$, and define $f\colon A\to\mathbb{Z}^{r}$ by
$f(a)=(f_{1}(a),\dotsc,f_{r}(a))$. Now $f(A)$ is a subgroup of
$\mathbb{Z}^{r}$, so it is free, with basis $f(a_{1}),\dotsc,f(a_{s})$ say.
Put $B=\ker(f)$, and let $F\leq A$ be the subgroup generated by
$a_{1},\dotsc,a_{s}$. We find that $f\colon F\to f(A)\simeq\mathbb{Z}^{s}$ is
an isomorphism, and thus that $A=B\oplus F$. Consider a homomorphism $g\colon
B\to\mathbb{Z}$. Then the composite

$A=B\oplus F\xrightarrow{\text{proj}}B\xrightarrow{g}\mathbb{Z}$

is an element of the group $\operatorname{Hom}(A,\mathbb{Z})$, which is
generated by the maps $f_{i}$, but $f_{i}(B)=0$, so we see that $g=0$. This
proves that $\operatorname{Hom}(B,\mathbb{Z})=0$. ∎

######  Lemma 8.36.

Suppose that
$\operatorname{Hom}(A,\mathbb{Z})=\operatorname{Ext}(A,\mathbb{Z})=0$. Then
$A=0$.

###### Proof.

Corollary 8.24 implies that
$\operatorname{Ext}(\operatorname{tors}(A),\mathbb{Z})=0$, so Proposition 8.26
gives $\operatorname{tors}(A)=0$, so $A$ is torsion free. We thus have short
exact sequences $A\xrightarrow{n}A\to A/n$ for all $n>0$, and using the
resulting six term sequences we deduce that
$\operatorname{Ext}(A/n,\mathbb{Z})=0$. As $A/n$ is torsion we can use
Proposition 8.26 again to see that $A/n=0$, so $n.1_{A}$ is surjective. It is
also injective because $\operatorname{tors}(A)=0$, so it is an isomorphism. We
can thus make $A$ into a vector space over ${\mathbb{Q}}$ by the rule
$(m/n).a=(n.1_{A})^{-1}(ma)$. Linear algebra therefore tells us that either
$A$ is zero, or it has ${\mathbb{Q}}$ as a summand. In the latter case
$\operatorname{Ext}(A,\mathbb{Z})$ would contain the uncountable group
$\operatorname{Ext}({\mathbb{Q}},\mathbb{Z})$ as a summand, which is
impossible as $\operatorname{Ext}(A,\mathbb{Z})=0$. We therefore have $A=0$ as
claimed. ∎

###### Proof of Proposition 8.34.

Using Lemma 8.35 we reduce to the case where
$\operatorname{Hom}(A,\mathbb{Z})=0$. We next claim that there are only
finitely many primes $p$ for which $A[p]\neq 0$. Indeed, for any such $p$ we
see (by linear algebra over the field $\mathbb{Z}/p$) that
$\operatorname{Hom}(A[p],\mathbb{Z}/p)\neq 0$. If there are infinitely many
such primes, we deduce that the group
$P=\prod_{p}\operatorname{Hom}(A[p],\mathbb{Z}/p)$ is uncountable, which is
impossible as Proposition 8.26 tells us that $P$ is a quotient of the finitely
generated group $\operatorname{Ext}(A,\mathbb{Z})$. We can thus choose $p$
such that $A[p]=0$, so we have a short exact sequence
$A\xrightarrow{p}A\xrightarrow{}A/p$. Proposition 8.23 then tells us that
multiplication by $p$ is surjective on $\operatorname{Ext}(A,\mathbb{Z})$. By
the structure theory of finitely generated groups, we see that
$\operatorname{Ext}(A,\mathbb{Z})$ must be finite, of order $n$ say, and that
$n$ must be coprime to $p$.

Next we have short exact sequences $A[n]\xrightarrow{i}A\xrightarrow{f}nA$ and
$nA\xrightarrow{j}A\xrightarrow{g}A/n$, where $f(a)=na$ and $g$ is the
quotient map. By assumption we have $\operatorname{Hom}(A,\mathbb{Z})=0$, and
also
$\operatorname{Hom}(A[n],\mathbb{Z})=\operatorname{Hom}(A/n,\mathbb{Z})=0$
because $\mathbb{Z}$ is torsion free. From the six term sequences we find that
$\operatorname{Hom}(nA,\mathbb{Z})=0$. We also find that
$j^{*}\colon\operatorname{Ext}(A,\mathbb{Z})\to\operatorname{Ext}(nA,\mathbb{Z})$
is surjective and
$f^{*}\colon\operatorname{Ext}(nA,\mathbb{Z})\to\operatorname{Ext}(A,\mathbb{Z})$
is injective, but the composite $f^{*}j^{*}=(jf)^{*}$ is just multiplication
by $n$. As $n$ was defined to be the order of
$\operatorname{Ext}(A,\mathbb{Z})$ we deduce that $f^{*}j^{*}=0$, which
implies that $\operatorname{Ext}(nA,\mathbb{Z})=0$. Lemma 8.36 therefore tells
us that $nA=0$, so $A$ is torsion and
$\operatorname{Ext}(A,\mathbb{Z})=\operatorname{Hom}(A,{\mathbb{Q}/\mathbb{Z}})=A^{*}$.
This is finitely generated by assumption. Moreover, we have $nA=0$ and
therefore $nA^{*}=0$, so $A^{*}$ is a finite group. It follows that $A^{**}$
is finite but Remark 8.19 gives an embedding $A\to A^{**}$ so $A$ is finite,
as required. ∎

##  9\. Localisation

######  Definition 9.1.

A _multiplicative set_ is a set $S$ of positive integers that contains $1$ and
is closed under multiplication.

######  Definition 9.2.

Let $A$ be an abelian group, and let $S$ be a multiplicative set. We introduce
an equivalence relation on the set $A\times S$ by declaring that
$(a,s)\sim(b,t)$ iff $bsx=atx$ for some $x\in S$. We write $a/s$ for the
equivalence class of the pair $(a,s)$, and we write $A[S^{-1}]$ for the set of
equivalence classes. We make this into an 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 show that the given relation really is an equivalence relation.
It is clearly reflexive (as we can take $x=1$) and symmetric. Suppose that
$(a,s)\sim(b,t)\sim(c,u)$, so there are elements $x,y\in S$ with $atx=bsx$ and
$buy=cty$. Put $z=txy\in S$ and note that

$auz=(au)(txy)=(atx)(uy)=(bsx)(uy)=(buy)(sx)=(cty)(sx)=(cs)(txy)=csz,$

so $(a,s)\sim(c,u)$, as required.

Next, we should check that addition is well-defined. More specifically,
suppose that $(a_{0},s_{0})\sim(a_{1},s_{1})$ and
$(b_{0},t_{0})\sim(b_{1},t_{1})$. Put
$(c_{k},u_{k})=(a_{k}t_{k}+b_{k}s_{k},s_{k}t_{k})$; we must show that
$(c_{0},u_{0})\sim(c_{1},u_{1})$. By hypothesis there are elements $x,y\in S$
such that $a_{0}s_{1}x=a_{1}s_{0}x$ and $b_{0}t_{1}y=b_{1}t_{0}y$. We multiply
these two equations by $t_{0}t_{1}y$ and $s_{0}s_{1}x$ respectively, and then
add them together to get

$a_{0}s_{1}xt_{0}t_{1}y+b_{0}t_{1}ys_{0}s_{1}x=a_{1}s_{0}xt_{0}t_{1}y+b_{1}t_{0}ys_{0}s_{1}x,$

or equivalently

$(a_{0}t_{0}+b_{0}s_{0})(s_{1}t_{1})xy=(a_{1}t_{1}+b_{1}s_{1})(s_{0}t_{0})xy.$

If we put $z=xy\in S$ this can be rewritten as $c_{0}u_{1}z=c_{1}u_{0}z$, so
$(c_{0},u_{0})\sim(c_{1},u_{1})$ as required.

Finally, we should show that addition is commutative and associative, that the
element $0/1$ is an additive identity, and that $(-a)/s$ 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 $A[S^{-1}]$ if and only if there
exists $t\in S$ with $at=0$.

######  Remark 9.5.

In the case $A=\mathbb{Z}$ it is not hard to see that $\mathbb{Z}[S^{-1}]$ can
be identified with the set $\\{n/s\in{\mathbb{Q}}\;|\;n\in\mathbb{Z},\;s\in
S\\}$, which is a subring of ${\mathbb{Q}}$.

######  Remark 9.6.

Consider the case where $A$ is finite, so $A$ is the direct sum of its Sylow
subgroups, say $A=A_{1}\oplus\dotsb\oplus A_{r}$ with $|A_{i}|=p_{i}^{v_{i}}$
for some primes $p_{1},\dotsc,p_{r}$ and integers $v_{i}>0$. We then find that
$A[S^{-1}]=\bigoplus_{k}A_{k}[S^{-1}]$. Suppose that there exists $n\in S$
that is divisible by $p_{k}$. In $A_{k}[S^{-1}]$ we then have
$a/s=(an^{v_{k}})/(sn^{v_{k}})=0/(sn^{v_{k}})=0$, so $A_{k}[S^{-1}]=0$. On the
other hand, if there is no such $n$ then for each $s\in S$ we see that
$s.1_{A_{k}}$ is invertible for all $s\in S$, and using this we will see later
that $A_{k}[S^{-1}]=A_{k}$. Thus, $A[S^{-1}]$ 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 $S=\\{n^{k}\;|\;k\geq 0\\}$ we write $A[n^{-1}]$ or $A[1/n]$ for
$A[S^{-1}]$.

  * (b)

If $p$ is prime and $S=\\{n>0\;|\;n\neq 0\pmod{p}\\}=\mathbb{N}\setminus
p\mathbb{N}$ then we write $A_{(p)}$ for $A[S^{-1}]$. This is called the _$p$
-localisation_ of $A$.

  * (c)

If $S=\\{n\in\mathbb{N}\;|\;n>0\\}$ then we write $A{\mathbb{Q}}$ or $A_{(0)}$
for $A[S^{-1}]$. This is called the _rationalisation_ of $A$.

######  Proposition 9.8.

Let $S$ be a multiplicative set. Then any homomorphism $f\colon A\to B$ gives
a homomorphism $f[S^{-1}]\colon A[S^{-1}]\to B[S^{-1}]$ by the rule
$f[S^{-1}](a/s)=f(a)/s$. This construction gives an additive functor, and
there is a natural map $\eta\colon A\to A[S^{-1}]$ given by $\eta(a)=a/1$.

###### Proof.

First, we see from the definitions that if $(a_{0},s_{0})\sim(a_{1},s_{1})$
then $(f(a_{0}),s_{0})\sim(f(a_{1}),s_{1})$. This shows that $f[S^{-1}]$ is
well-defined. It also follows directly from the definitions that it is a
homomorphism. Next, if we have maps $A\xrightarrow{f}B\xrightarrow{g}C$ then

$(gf)[S^{-1}](a/s)=gf(a)/s=g[S^{-1}](f(a)/s)=g[S^{-1}](f[S^{-1}](a/s)),$

which shows that our construction is functorial. We also claim that $\eta$ is
natural, which means that for any $f\colon A\to B$ the square

${A}$${B}$${A[S^{-1}]}$${B[S^{-1}]}$$\scriptstyle{f}$$\scriptstyle{\eta}$$\scriptstyle{\eta}$$\scriptstyle{f[S^{-1}]}$

commutes. This is again straightforward. ∎

######  Remark 9.9.

When there is no danger of confusion, we will just write $f$ rather than
$f[S^{-1}]$ for the induced map $A[S^{-1}]\to B[S^{-1}]$.

######  Proposition 9.10.

If $A\xrightarrow{f}B\xrightarrow{g}C$ is exact (or short exact), then so is
the localised sequence

$A[S^{-1}]\xrightarrow{f[S^{-1}]}B[S^{-1}]\xrightarrow{g[S^{-1}]}C[S^{-1}].$

###### Proof.

First, as $gf=0$ and localisation is functorial we see that
$g[S^{-1}]f[S^{-1}]=(gf)[S^{-1}]=0$, so
$\operatorname{image}(f[S^{-1}])\leq\ker(g[S^{-1}])$. Now consider an element
$b/s\in\ker(g[S^{-1}])$. We then have $g(b)/s=0/1$ in $B[S^{-1}]$, or
equivalently $g(b)\,x=0$ for some $x\in S$. This means that $g(xb)=0$ so
$xb\in\ker(g)=\operatorname{image}(f)$, so there exists $a\in A$ with
$f(a)=xb$. This implies that $f[S^{-1}](a/(xs))=f(a)/(xs)=xb/xs=b/s$, so
$b/s\in\operatorname{image}(f[S^{-1}])$. We now see that the sequence
$A[S^{-1}]\xrightarrow{f[S^{-1}]}B[S^{-1}]\xrightarrow{g[S^{-1}]}C[S^{-1}]$ is
exact as claimed. Now suppose that the original sequence is short exact. It is
equivalent to say that the sequences $0\to A\to B$ and $A\to B\to C$ and $B\to
C\to 0$ are all exact, and it follows from this that the sequences $0\to
A[S^{-1}]\to B[S^{-1}]$ and $A[S^{-1}]\to B[S^{-1}]\to C[S^{-1}]$ and
$B[S^{-1}]\to C[S^{-1}]\to 0$ are also exact. We can then reassemble these
pieces to see that the sequence
$A[S^{-1}]\xrightarrow{f[S^{-1}]}B[S^{-1}]\xrightarrow{g[S^{-1}]}C[S^{-1}]$ is
again short exact. ∎

######  Proposition 9.11.

There is a natural isomorphism $\mu\colon\mathbb{Z}[S^{-1}]\otimes A\to
A[S^{-1}]$ given by $\mu((n/s)\otimes a)=(na)/s$.

###### Proof.

First, it is straightforward to check that there is a well-defined bilinear
map $\mu_{0}\colon\mathbb{Z}[S^{-1}]\times A\to A[S^{-1}]$ given by
$\mu_{0}(n/s,a)=(na)/s$. By the universal property of tensor products, this
gives a homomorphism $\mu\colon\mathbb{Z}[S^{-1}]\otimes A\to A[S^{-1}]$ with
$\mu((n/s)\otimes a)=(na)/s$. In the opposite direction, we would like to
define $\nu\colon A[S^{-1}]\to\mathbb{Z}[S^{-1}]\otimes A$ by
$\nu(a/s)=(1/s)\otimes a$. To see that this is well-defined, suppose that
$a/s=b/t$, so $xta=xsb$ for some $x\in S$. From the definition of tensor
products, we have $mu\otimes v=u\otimes mv$ in $U\otimes V$ for all $u\in U$,
$v\in V$ and $m\in\mathbb{Z}$. We can apply this with
$u=1/(stx)\in\mathbb{Z}[S^{-1}]$ and $v=a$ and $m=tx$ to get $(1/s)\otimes
a=(1/(stx))\otimes xta$. By a symmetrical argument, we have $(1/t)\otimes
b=(1/(stx))\otimes xsb$, but $xta=xsb$ so we find that $(1/s)\otimes
a=(1/t)\otimes b$, as required. It is clear that $\mu\nu=1_{A[S^{-1}]}$. The
other way around, we have

$\nu\mu((n/s)\otimes a)=\nu((na)/s)=(1/s)\otimes na=(n/s)\otimes a.$

Thus, $\nu$ is inverse to $\mu$. ∎

######  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 $a\in A$ there exists $s\in S$ with $sa=0$.
We say that $A$ is _$S$ -local_ if for each $s\in S$, the endomorphism
$s.1_{A}\colon A\to A$ is invertible.

Some care is needed in relating the above definition to the traditional
terminology in the most popular cases:

######  Definition 9.13.

  * (a)

Definition 4.1(c) is equivalent to the following: we say that $A$ is a
_torsion group_ if it is $S_{0}$-torsion, where
$S_{0}=\\{n\in\mathbb{N}\;|\;n>0\\}$.

  * (b)

We say that $A$ is _rational_ if it is $S_{0}$-local. (We will see that in
this case, $A$ can be regarded as a vector space over ${\mathbb{Q}}$.)

  * (c)

Now let $p$ be a prime number. We say that $A$ is _$p$ -torsion_ if it is
$p^{\mathbb{N}}$-torsion, where
$p^{\mathbb{N}}=\\{p^{n}\;|\;n\in\mathbb{N}\\}$.

  * (d)

However, we say that $A$ is _$p$ -local_ if it is $S_{p}$-local, where
$S_{p}=\mathbb{N}\setminus p\mathbb{N}$.

######  Proposition 9.14.

  * (a)

The group $A[S^{-1}]$ is always $S$-local.

  * (b)

The map $\eta\colon A\to A[S^{-1}]$ 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
$\mathbb{Z}[S^{-1}]\leq{\mathbb{Q}}$ by the rule $(n/s).a=(s.1_{A})^{-1}(na)$.

  * (d)

Suppose that $f\colon A\to B$ is a homomorphism, and that $B$ is $S$-local.
Then there is a unique homomorphism $f^{\prime}\colon A[S^{-1}]\to B$ such
that $f^{\prime}\circ\eta=f\colon A\to B$.

###### Proof.

  * (a)

One can check that for each $s\in S$ there is a well-defined map $d_{s}\colon
A[S^{-1}]\to A[S^{-1}]$ given by $d_{s}(a/t)=a/(st)$. This is inverse to
$s.1_{A[S^{-1}]}$.

  * (b)

If $\eta$ is an isomorphism then it follows from (a) that $A$ is $S$-local.
Conversely, if $A$ is $S$-local one can check that the formula
$\zeta(a/s)=(s.1_{A})^{-1}(a)$ gives a well-defined map $\zeta\colon
A[S^{-1}]\to A$, and that this is inverse to $\eta$.

  * (c)

First we must check that the multiplication rule is well-defined. Suppose that
$n/s=m/t$, so $ntx=msx$ for some $x\in S$. As this is an equation in
$\mathbb{Z}$ and $x>0$ it reduces to $tn=ms$. If we write $n_{A}$ for
$n.1_{A}$ and so on, we deduce that $t_{A}n_{A}=m_{A}s_{A}\colon A\to A$. We
can compose on the left by $t_{A}^{-1}$ and on the right by $s_{A}^{-1}$ to
get $n_{A}s_{A}^{-1}=t_{A}^{-1}m_{A}$. As $s_{A}^{-1}$ is a homomorphism, it
commutes with multiplication by $n$, so $s_{A}^{-1}n_{A}=t_{A}^{-1}m_{A}$.
This means that the definition of multiplication is consistent. We will leave
it to the reader that it has the usual associativity and distributivity
properties.

  * (d)

We define $f^{\prime}\colon A[S^{-1}]\to B$ by
$f^{\prime}(a/s)=(s.1_{B})^{-1}(f(a))$. We leave it to the reader to show that
this is well-defined and is a homomorphism. Note that $f^{\prime}(a/1)=f(a)$,
so $f^{\prime}\eta=f$. If $f^{\prime\prime}$ is another homomorphism with
$f^{\prime\prime}\eta=f$, we have

$s_{A}(f^{\prime\prime}(a/s))=s.f^{\prime\prime}(a/s)=f^{\prime\prime}(s.(a/s))=f^{\prime\prime}(a/1)=f^{\prime\prime}(\eta(a))=f(a).$

As $s_{A}$ is invertible we can rewrite this as
$f^{\prime\prime}(a/s)=s_{A}^{-1}(f(a))=f^{\prime}(a/s)$. As $a/s$ was
arbitrary this means that $f^{\prime\prime}=f^{\prime}$, which gives that
claimed uniqueness statement.

∎

######  Remark 9.15.

As a consequence of (b), we can identify $A[S^{-1}]$ with $A[S^{-1}][S^{-1}]$.
There is a slight subtlety here: there are two apparently different
isomorphisms $A[S^{-1}]\to A[S^{-1}][S^{-1}]$, and to keep everything straight
it is necessary to prove that they are the same. Indeed, for any $B$, we have
a map $\eta_{B}\colon B\to B[S^{-1}]$. We can specialise to the case
$B=A[S^{-1}]$ to get a map $\eta_{A[S^{-1}]}\colon A[S^{-1}]\to
A[S^{-1}][S^{-1}]$, given by $\eta_{A[S^{-1}]}(a/s)=(a/s)/1$. Alternatively,
we can apply Proposition 9.8 to the map $\eta_{A}\colon A\to A[S^{-1}]$ to get
another map $\eta_{A}[S^{-1}]\colon A[S^{-1}]\to A[S^{-1}][S^{-1}]$, given by
$(\eta_{A}[S^{-1}])(a/s)=(a/1)/s$. It is clear that

$s.\eta_{A[S^{-1}]}(a/s)=(a/1)/1=s.(\eta_{A}[S^{-1}])(a/s),$

and multiplication by $s$ is an isomorphism on $A[S^{-1}][S^{-1}]$, so
$\eta_{A[S^{-1}]}=\eta_{A}[S^{-1}]$.

######  Proposition 9.16.

  * (a)

If we have a short exact sequence $A\to B\to C$ in which two of the three
terms are $S$-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 $n\in S$ we have $U[n]=0$ and $U/n=0$.
Recall also that Proposition 7.20 gives exact sequences

$0\to
A[n]\xrightarrow{j}B[n]\xrightarrow{q}C[n]\xrightarrow{\delta}A/n\xrightarrow{j}B/n\xrightarrow{q}C/n\to
0.$

The claim follows by diagram chasing.

  * (b)

This is clear, because direct sums, products and retracts of isomorphisms are
isomorphisms.

  * (c)

Let $f\colon A\to B$ be a homomorphism between $S$-local groups. Then
$\operatorname{img}(f)$ is a subgroup of $B$ so for $n\in S$ we have
$\operatorname{img}(f)[n]\leq B[n]=0$. Similarly, $\operatorname{img}(f)$ is a
quotient of $A$ so $\operatorname{img}(f)/n$ is a quotient of $A/n$ and so is
zero. It follows that $\operatorname{img}(f)$ is $S$-local. We can therefore
apply (a) to the short exact sequences $\ker(f)\to
A\xrightarrow{f}\operatorname{img}(f)$ and $\operatorname{img}(f)\to
B\to\operatorname{cok}(f)$ to see that $\ker(f)$ and $\operatorname{cok}(f)$
are also $S$-local.

  * (d)

Let $A$ be a $p$-torsion group. Consider $m\in\mathbb{Z}\setminus
p\mathbb{Z}$; we must show that $m.1_{A}$ is an isomorphism. For any $a\in A$
we can choose $k\geq 0$ such that $p^{k}a=0$, and $p^{k}$ is coprime with $m$
so we can choose $r,s\in\mathbb{Z}$ with $p^{k}r+ms=1$. It follows that
$a=msa=(m.1_{A})(sa)$. Using this we see that $m.1_{A}$ is surjective. Also,
if $ma=0$ then $a=sma=0$, so $A[m]=0$, so $m.1_{A}$ is also injective.

∎

######  Proposition 9.17.

If $A$ is a torsion group, then $A_{(0)}=0$ and
$A_{(p)}=\operatorname{tors}_{p}(A)$.

###### Proof.

First suppose that $A$ is a $p$-torsion group, so $A=\bigcup_{k}A[p^{k}]$. If
$m$ is not divisible by $p$ then we can find $n>0$ with $mn=1\pmod{p^{k}}$, so
$n.1_{A[p^{k}]}$ is inverse to $m.1_{A[p^{k}]}$. It follows that $m.1_{A}$ is
also an isomorphism. This holds for all $m\in\mathbb{Z}\setminus p\mathbb{Z}$,
so $A$ is $p$-local, so $A=A_{(p)}$. Now suppose instead that $A$ is a
$q$-torsion group for some prime $q\neq p$. For any element $a\in A$ we have
$q^{v}a=0$ for some $v\geq 0$, 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 $A$ we have
$A=\bigoplus_{q}\operatorname{tors}_{q}(A)$ by Proposition 4.9, so
$A_{(p)}=\bigoplus_{q}\operatorname{tors}_{q}(A)_{(p)}$. By the special cases
that we have just discussed, this sum contains only the single factor
$\operatorname{tors}_{p}(A)_{(p)}=\operatorname{tors}_{p}(A)$, as claimed. It
is also clear from Remark 9.4 that $A_{(0)}=0$. ∎

##  10\. Colimits of sequences

######  Definition 10.1.

By a _sequence_ we mean a diagram of the form

$A_{0}\xrightarrow{f_{0}}A_{1}\xrightarrow{f_{1}}A_{2}\xrightarrow{f_{2}}A_{3}\xrightarrow{f_{3}}\dotsb.$

Given such a sequence and natural numbers $i\leq j$, we write $f_{ij}$ for the
composite

$A_{i}\xrightarrow{f_{i}}A_{i+1}\xrightarrow{}\dotsb\xrightarrow{f_{j-1}}A_{j}.$

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 $A_{+}=\bigoplus_{i}A_{i}$. For
each $k$ we have an inclusion $\iota_{k}\colon A_{k}\to A_{+}$ and also a
homomorphism $\iota_{k+1}\circ f_{k}\colon A_{k}\to A_{+}$. We let $R_{k}$
denote the image of $(\iota_{k}-\iota_{k+1}f_{k})\colon A_{k}\to A_{+}$, and
put $R_{+}=\sum_{k}R_{k}\leq A_{+}$ and
$\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}=A_{+}/R_{+}$.
This group is called the _colimit_ of the sequence.

Next, we write $\overline{\imath}_{k}$ for the composite

$A_{k}\xrightarrow{\iota_{k}}A_{+}\xrightarrow{}A_{+}/R_{+}=\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}.$

By construction we have $\overline{\imath}_{k}=\overline{\imath}_{k+1}f_{k}$,
so the following diagram commutes:

${A_{0}}$${A_{1}}$${A_{2}}$${A_{3}}$${\dotsb}$${\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}}$${\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}}$${\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}}$${\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}}$${\dotsb}$$\scriptstyle{\overline{\imath}_{0}}$$\scriptstyle{f_{0}}$$\scriptstyle{\overline{\imath}_{1}}$$\scriptstyle{f_{1}}$$\scriptstyle{\overline{\imath}_{2}}$$\scriptstyle{f_{2}}$$\scriptstyle{\overline{\imath}_{3}}$$\scriptstyle{f_{3}}$

This implies that $\overline{\imath}_{k}=\overline{\imath}_{m}f_{k,m}$
whenever $k\leq m$.

######  Remark 10.3.

The definition can be reformulated slightly as follows. We can define an
endomorphism $S$ of $A_{+}$ by

$S(a_{0},a_{1},a_{2},a_{3},\dotsc)=(0,f_{0}(a_{0}),f_{1}(a_{1}),f_{2}(a_{2}),f_{3}(a_{3}),\dotsb).$

Equivalently, $S$ is the unique map such that the following diagram commutes
for all $k$:

${A_{k}}$${A_{k+1}}$${A_{+}}$${A_{+}}$$\scriptstyle{\iota_{k}}$$\scriptstyle{f_{k}}$$\scriptstyle{\iota_{k+1}}$$\scriptstyle{S}$

We can then say that
$\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}$ is the
cokernel of $1-S\colon A_{+}\to A_{+}$. Note that if $a_{i}$ is the first
nonzero entry in $a$ then the $i$’th entry in $(1-S)(a)$ is again $a_{i}$, so
$(1-S)(a)\neq 0$. This shows that $1-S$ is injective, so we actually have a
short exact sequence

${A_{+}}$${A_{+}}$${\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}.}$$\scriptstyle{1-S}$

######  Remark 10.4.

The colimit can also be characterised by a universal property, as follows. A
_cone_ for the sequence is a group $B$ with a collection of maps $u_{k}\colon
A_{k}\to B$ such that $u_{k+1}f_{k}=u_{k}$ for all $k\geq 0$. By construction,
the maps $\overline{\imath}_{k}\colon
A_{k}\to\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}$
form a cone. We claim that for any cone
$\\{A_{k}\xrightarrow{u_{k}}B\\}_{k\in\mathbb{N}}$ there is a unique
homomorphism
$u_{\infty}\colon\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}\to
B$ such that $u_{\infty}\overline{\imath}_{k}=u_{k}$ for all $k$. Indeed,
Proposition 3.7 gives us a unique map $v\colon A_{+}\to B$ 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
$u_{\infty}\colon\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}=A_{+}/R_{+}\to
B$. This is easily seen to be the unique map such that
$u_{\infty}\overline{\imath}_{k}=u_{k}$ for all $k$.

In many cases colimits are just unions, as we now explain.

######  Proposition 10.5.

We have

$\overline{\imath}_{0}(A_{0})\leq\overline{\imath}_{1}(A_{1})\leq\overline{\imath}_{2}(A_{2})\leq\dotsb\leq\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}.$

Moreover, $\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}$
is the union of the groups $\overline{\imath}_{k}(A_{k})$, and we have
$\overline{\imath}_{k}(a)=0$ iff $f_{k,m}(a)=0$ for some $m\geq k$.

###### Proof.

First, when $k\leq m$ we have
$\overline{\imath}_{k}=\overline{\imath}_{m}f_{k,m}$, and this implies that
$\overline{\imath}_{k}(A_{k})\leq\overline{\imath}_{m}(A_{m})$. Next, as
$\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}$ is a
quotient of $A_{+}$, we see that every element
$a\in\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}$ can be
written as $a=\sum_{k=0}^{N}\overline{\imath}_{k}(a_{k})$ for some $N\geq 0$
and $a_{k}\in A_{k}$. Now
$\overline{\imath}_{k}(a_{k})\in\overline{\imath}_{k}(A_{k})\leq\overline{\imath}_{N}(A_{N})$
for all $k$, so $a\in\overline{\imath}_{N}(A_{N})$. Thus
$\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}=\bigcup_{N}\overline{\imath}_{N}(A_{N})$
as claimed. Now suppose we have $a\in A_{k}$ and that $f_{km}(a)=0$ for some
$m\geq k$. Using $\overline{\imath}_{k}=\overline{\imath}_{m}f_{km}$ we deduce
that $\overline{\imath}_{k}(a)=0$. Conversely, suppose that
$\overline{\imath}_{k}(a)=0$, so $i_{k}(a)\in R_{+}$, so

$i_{k}(a)=\sum_{m=0}^{N-1}(i_{m}(b_{m})-i_{m+1}(f_{m}(b_{m})))$

for some $N>k$ and some $b_{0},\dotsc,b_{N-1}$ with $b_{i}\in A_{i}$. Now let
$h\colon\bigoplus_{m=0}^{N}A_{m}\to A_{N}$ 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. ∎

######  Corollary 10.6.

Suppose we have a sequence $\\{A_{i}\\}_{i\in\mathbb{N}}$ and a cone
$\\{u_{i}\colon A_{i}\to B\\}_{i\in\mathbb{N}}$ giving rise to a homomorphism
$u_{\infty}\colon\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}\to
B$. Then

  * (a)

The image of $u_{\infty}$ is the union of the subgroups $u_{k}(A_{k})$.

  * (b)

The map $u_{\infty}$ is injective iff whenever $u_{k}(a)=0$, there exists
$m\geq k$ with $f_{km}(a)=0$.

###### Proof.

This is clear from the Proposition. ∎

######  Example 10.7.

Let $A$ be any abelian group. Suppose we have a chain of subgroups

$A_{0}\leq A_{1}\leq A_{2}\leq A_{3}\leq\dotsb\leq A.$

Let $f_{k}\colon A_{k}\to A_{k+1}$ be the inclusion. Then the colimit of the
resulting sequence is just $\bigcup_{i}A_{i}$. Indeed, the inclusions
$A_{n}\to\bigcup_{i}A_{i}$ 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

$A\xrightarrow{n}A\xrightarrow{n}A\xrightarrow{n}A\xrightarrow{n}A\xrightarrow{}\dotsb$

is $A[1/n]$, whereas the colimit of the sequence

$A\xrightarrow{1}A\xrightarrow{2}A\xrightarrow{3}A\xrightarrow{4}A\xrightarrow{}\dotsb$

is the rationalisation $A_{(0)}$.

###### Proof.

We will prove the second statement; the first is similar but easier. Let $C$
denote the colimit, so we have maps $\overline{\imath}_{k}\colon A\to C$ with
$\overline{\imath}_{k}(a)=(k+1)\overline{\imath}_{k+1}(a)$. Define
$u_{n}\colon A\to A_{(0)}$ 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 $u_{\infty}\colon
C\to A_{(0)}$ with $u_{\infty}\overline{\imath}_{k}=u_{k}$ for all $k$. Any
element $a\in A_{(0)}$ can be written as $a=a^{\prime}/n$ for some
$a^{\prime}\in A$ and $n>0$, so
$a=((n-1)!a^{\prime})/n!=u_{n}((n-1)!a^{\prime})$, so $A_{(0)}$ is the union
of the images of the maps $u_{n}$. Now suppose that $u_{n}(a^{\prime})=0$. By
the definition of $A_{(0)}$ this just means that $ma^{\prime}=0$ for some
$m>0$. The map $f_{n,n+m}$ in our sequence is multiplication by the integer

$p=(n+1)(n+2)\dotsb(n+m)=\frac{(n+m)!}{n!}=m!\left(\begin{array}[]{c}{n+m}\\\
{n}\end{array}\right),$

which is divisible by $m$, so $f_{n,n+m}(a^{\prime})=0$. The claim follows by
Corollary 10.6. ∎

######  Proposition 10.9.

Suppose we have a commutative diagram as shown

${A_{0}}$${A_{1}}$${A_{2}}$${A_{3}}$${\dotsb}$${B_{0}}$${B_{1}}$${B_{2}}$${B_{3}}$${\dotsb}$$\scriptstyle{f_{0}}$$\scriptstyle{p_{0}}$$\scriptstyle{f_{1}}$$\scriptstyle{p_{1}}$$\scriptstyle{f_{2}}$$\scriptstyle{p_{2}}$$\scriptstyle{f_{3}}$$\scriptstyle{p_{3}}$$\scriptstyle{g_{0}}$$\scriptstyle{g_{1}}$$\scriptstyle{g_{2}}$$\scriptstyle{g_{3}}$

Let $\overline{\imath}_{k}$ be the canonical map
$A_{k}\to\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}$,
and let $\overline{\jmath}_{k}$ be the canonical map
$B_{k}\to\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}B_{i}$.
Then there is a unique map $p_{\infty}$ such that the diagram

${A_{k}}$${\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}}$${B_{k}}$${\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}B_{i}}$$\scriptstyle{p_{k}}$$\scriptstyle{\overline{\imath}_{k}}$$\scriptstyle{p_{\infty}}$$\scriptstyle{\overline{\jmath}_{k}}$

commutes for all $k$. Moreover, if all the maps $p_{k}$ are injective, or
surjective, or bijective, then $p_{\infty}$ has the same property.

###### Proof.

The maps $\overline{\jmath}_{k}p_{k}\colon
A_{k}\to\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}B_{i}$
satisfy

$\overline{\jmath}_{k}p_{k}=\overline{\jmath}_{k+1}g_{k}p_{k}=\overline{\jmath}_{k+1}p_{k+1}f_{k},$

so they form a cone for the sequence $\\{A_{i}\\}$. There is thus a unique map
$p_{\infty}\colon\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}\to\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}B_{i}$
with $p_{\infty}\overline{\imath}_{k}=\overline{\jmath}_{k}p_{k}$ for all $k$,
as claimed.

  * (a)

Now suppose that all the maps $p_{k}$ are injective. Consider an element
$a\in\ker(p_{\infty})$. By Proposition 10.5 we have
$a=\overline{\imath}_{k}(a^{\prime})$ for some $k$ and some $a^{\prime}\in
A_{k}$. We then have
$\overline{\jmath}_{k}(p_{k}(a^{\prime}))=p_{\infty}(\overline{\imath}_{k}(a^{\prime}))=p_{\infty}(a)=0$.
The same proposition therefore tells us that $g_{km}(p_{k}(a^{\prime}))=0$ for
some $m\geq k$. Now $g_{km}p_{k}=p_{m}f_{km}$ and $p_{m}$ is injective, so
$f_{km}(a^{\prime})=0$. This means that
$a=i_{k}(a^{\prime})=i_{m}(f_{km}(a^{\prime}))=0$. Thus, $p_{\infty}$ is
injective as claimed.

  * (b)

Suppose instead that all the maps $p_{k}$ are surjective. Consider an element
$b\in\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}B_{i}$. By
Proposition 10.5 we have $b=\overline{\jmath}_{k}(b^{\prime})$ for some $k$
and some $b^{\prime}\in B_{k}$. As $p_{k}$ is surjective, we can choose
$a^{\prime}\in A_{k}$ with $p_{k}(a^{\prime})=b^{\prime}$, and then put
$b=\overline{\imath}_{k}(a)$; we find that $p_{\infty}(a)=b$. Thus,
$p_{\infty}$ is also surjective.

  * (c)

If the maps $p_{k}$ are all isomorphisms, then (a) and (b) together imply that
$p_{\infty}$ is an isomorphism.

∎

######  Proposition 10.10.

Suppose we have a commutative diagram as shown, in which all the columns are
exact:

${A_{0}}$${A_{1}}$${A_{2}}$${A_{3}}$${\dotsb}$${B_{0}}$${B_{1}}$${B_{2}}$${B_{3}}$${\dotsb}$${C_{0}}$${C_{1}}$${C_{2}}$${C_{3}}$${\dotsb}$$\scriptstyle{f_{0}}$$\scriptstyle{p_{0}}$$\scriptstyle{f_{1}}$$\scriptstyle{p_{1}}$$\scriptstyle{f_{2}}$$\scriptstyle{p_{2}}$$\scriptstyle{f_{3}}$$\scriptstyle{p_{3}}$$\scriptstyle{g_{0}}$$\scriptstyle{q_{0}}$$\scriptstyle{g_{1}}$$\scriptstyle{q_{1}}$$\scriptstyle{g_{2}}$$\scriptstyle{q_{2}}$$\scriptstyle{g_{3}}$$\scriptstyle{q_{3}}$$\scriptstyle{h_{0}}$$\scriptstyle{h_{1}}$$\scriptstyle{h_{2}}$$\scriptstyle{h_{3}}$

Then the resulting sequence

$\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}\xrightarrow{p_{\infty}}\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}B_{i}\xrightarrow{q_{\infty}}\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}C_{i}$

is also exact.

###### Proof.

First, any element
$a\in\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}$ has
the form $a=\overline{\imath}_{n}(a^{\prime})$ for some $n$ and $a^{\prime}$.
We can then chase $a^{\prime}$ around the diagram

${A_{n}}$${\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}}$${B_{n}}$${\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}}$${C_{n}}$${\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}}$$\scriptstyle{\overline{\imath}_{n}}$$\scriptstyle{p_{n}}$$\scriptstyle{p_{\infty}}$$\scriptstyle{\overline{\jmath}_{n}}$$\scriptstyle{q_{n}}$$\scriptstyle{q_{\infty}}$$\scriptstyle{\overline{k}_{n}}$

to see that $q_{\infty}(p_{\infty}(a))=0$. Conversely, suppose we have an
element $b\in\ker(q_{\infty})$. We then have
$b=\overline{\jmath}_{n}(b^{\prime})$ for some $n$ and some $b^{\prime}\in
B_{n}$. We then have
$\overline{k}_{n}(q_{n}(b^{\prime}))=q_{\infty}(\overline{\jmath}_{n}(b^{\prime}))=q_{\infty}(b)=0$,
so $h_{n,m}(q_{n}(b^{\prime}))=0$ for some $m\geq n$. Now
$h_{n,m}(q_{n}(b^{\prime}))=q_{m}(g_{n,m}(b^{\prime}))$, so
$g_{n,m}(b^{\prime})\in\ker(q_{m})=\operatorname{image}(p_{m})$, so we can
find $a^{\prime}\in A_{m}$ with $p_{m}(a^{\prime})=g_{n,m}(b^{\prime})$. Now
put $a=\overline{\imath}_{m}(a^{\prime})$. We find that

$p_{\infty}(a)=\overline{\jmath}_{m}(p_{m}(a^{\prime}))=\overline{\jmath}_{m}(g_{n,m}(b^{\prime}))=\overline{\jmath}_{n}(b^{\prime})=b,$

so $b\in\operatorname{image}(p_{\infty})$. The claim follows. ∎

######  Proposition 10.11.

Suppose we have a sequence

$A_{0}\xrightarrow{f_{0}}A_{1}\xrightarrow{f_{1}}A_{2}\xrightarrow{f_{2}}A_{3}\xrightarrow{f_{3}}\dotsb$

and a nondecreasing function $u\colon\mathbb{N}\to\mathbb{N}$ such that
$u(i)\to\infty$ as $i\to\infty$. Put

$g_{i}=f_{u(i),u(i+1)}=(A_{u(i)}\xrightarrow{f_{u(i)}}A_{u(i)+1}\xrightarrow{f_{u(i)+1}}\dotsb\xrightarrow{f_{u(i+1)-1}}A_{u(i+1)}),$

so we have a sequence

$A_{u(0)}\xrightarrow{g_{0}}A_{u(1)}\xrightarrow{g_{1}}A_{u(2)}\xrightarrow{g_{2}}A_{u(3)}\xrightarrow{g_{3}}\dotsb$

Then there is a canonical isomorphism
$\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{j}A_{u(j)}=\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}$.

###### Proof.

Let $\overline{\imath}_{n}\colon
A_{n}\to\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}$ and
$\overline{\jmath}_{n}\colon
A_{u(n)}\to\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{j}A_{u(j)}$
be the usual maps. As $\overline{\imath}_{n}=\overline{\imath}_{n+1}f_{n}$ for
all $n$ we find by induction that
$\overline{\imath}_{n}=\overline{\imath}_{m}f_{n,m}$ for all $n\leq m$. By
applying this to the pair $u(k)\leq u(k+1)$, we see that
$\overline{\imath}_{u(k)}=\overline{\imath}_{u(k+1)}g_{k}$, so the maps
$\overline{\imath}_{u(k)}$ form a cone for the sequence
$\\{A_{u(j)}\\}_{j\in\mathbb{N}}$, so there is a unique map
$p\colon\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{j}A_{u(j)}\to\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}$
with $p\overline{\jmath}_{k}=\overline{\imath}_{u(k)}$ for all $k$. In the
opposite direction, suppose we have $n\in\mathbb{N}$. As $u(j)\to\infty$ as
$j\to\infty$, we can choose $k$ such that $n\leq u(k)$, and form the composite

$q_{nk}=(A_{n}\xrightarrow{f_{n,u(k)}}A_{u(k)}\xrightarrow{\overline{\jmath}_{k}}\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{j}A_{u(j)}).$

As
$\overline{\jmath}_{k}=\overline{\jmath}_{k+1}g_{k}=\overline{\jmath}_{k+1}f_{u(k),u(k+1)}$
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 $q_{nk}$ is independent 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 for the
sequence $\\{A_{i}\\}_{i\in\mathbb{N}}$, so there is a unique map
$q\colon\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}\to\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{j}A_{u(j)}$
with $q\overline{\imath}_{n}=q_{n}$ for all $n$.

Now note that any element
$a\in\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}$ has
the form $a=\overline{\imath}_{n}(a^{\prime})$ for some $n$ and $a^{\prime}\in
A_{n}$. If we choose $k$ with $u(k)\geq n$ we have

$pq(a)=pq_{n}(a^{\prime})=p\overline{\jmath}_{k}f_{n,u(k)}(a^{\prime})=\overline{\imath}_{u(k)}f_{n,u(k)}(a^{\prime})=\overline{\imath}_{n}(a^{\prime})=a,$

so $pq$ is the identity. A similar argument shows that $qp$ is the identity. ∎

######  Proposition 10.12.

Suppose we have a commutative diagram

${A_{00}}$${A_{01}}$${A_{02}}$${A_{03}}$${\dotsb}$${A_{10}}$${A_{11}}$${A_{12}}$${A_{13}}$${\dotsb}$${A_{20}}$${A_{21}}$${A_{22}}$${A_{23}}$${\dotsb}$${A_{30}}$${A_{31}}$${A_{32}}$${A_{33}}$${\dotsb}$${\dotsb}$${\dotsb}$${\dotsb}$${\dotsb}$$\scriptstyle{f_{00}}$$\scriptstyle{g_{00}}$$\scriptstyle{f_{01}}$$\scriptstyle{g_{01}}$$\scriptstyle{f_{02}}$$\scriptstyle{g_{02}}$$\scriptstyle{f_{03}}$$\scriptstyle{g_{03}}$$\scriptstyle{f_{10}}$$\scriptstyle{g_{10}}$$\scriptstyle{f_{11}}$$\scriptstyle{g_{11}}$$\scriptstyle{f_{12}}$$\scriptstyle{g_{12}}$$\scriptstyle{f_{13}}$$\scriptstyle{g_{13}}$$\scriptstyle{f_{20}}$$\scriptstyle{g_{20}}$$\scriptstyle{f_{21}}$$\scriptstyle{g_{21}}$$\scriptstyle{f_{22}}$$\scriptstyle{g_{22}}$$\scriptstyle{f_{23}}$$\scriptstyle{g_{23}}$$\scriptstyle{f_{30}}$$\scriptstyle{g_{30}}$$\scriptstyle{f_{31}}$$\scriptstyle{g_{31}}$$\scriptstyle{f_{32}}$$\scriptstyle{g_{32}}$$\scriptstyle{f_{33}}$$\scriptstyle{g_{33}}$

and thus sequences

$\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{j}A_{0j}\xrightarrow{g_{0\infty}}\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{j}A_{1j}\xrightarrow{g_{1\infty}}\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{j}A_{2j}\xrightarrow{g_{2\infty}}\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{j}A_{3j}\xrightarrow{g_{3\infty}}\dotsb$

and

$\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i0}\xrightarrow{f_{\infty
0}}\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i1}\xrightarrow{f_{\infty
1}}\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i2}\xrightarrow{f_{\infty
2}}\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i3}\xrightarrow{f_{\infty
3}}\dotsb$

Suppose we also put

$h_{i}=g_{i,j+1}f_{ij}=f_{i+1,j}g_{ij}\colon A_{ii}\to A_{i+1,i+1},$

giving a third sequence

$A_{00}\xrightarrow{h_{0}}A_{11}\xrightarrow{h_{1}}A_{22}\xrightarrow{h_{2}}A_{33}\xrightarrow{h_{3}}\dotsb$

Then there are canonical isomorphisms

$\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{j}A_{ij}\simeq\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{ii}\simeq\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{j}\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{ij}.$

######  Example 10.13.

In conjunction with Proposition 10.8, this will give

$A[\frac{1}{n}][\frac{1}{m}]=A[\frac{1}{nm}]=A[\frac{1}{m}][\frac{1}{n}]$

###### Proof.

Put $A_{++}=\bigoplus_{n,m}A_{nm}$, and let $i_{nm}\colon A_{nm}\to A_{++}$ be
the canonical inclusion. Put

$\displaystyle P_{nm}$
$\displaystyle=\operatorname{image}(i_{nm}-i_{n,m+1}f_{nm}\colon A_{nm}\to
A_{++})$ $\displaystyle Q_{nm}$
$\displaystyle=\operatorname{image}(i_{nm}-i_{n+1,m}g_{nm}\colon A_{nm}\to
A_{++}).$

From the definitions we have

$\displaystyle\bigoplus_{n}\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{j}A_{nj}$
$\displaystyle=A_{++}/\sum_{n,m}P_{nm}$
$\displaystyle\bigoplus_{m}\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{im}$
$\displaystyle=A_{++}/\sum_{n,m}Q_{nm}.$

It follows easily that

$\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{j}A_{ij}=A_{++}/\left(\sum_{n,m}P_{nm}+\sum_{n,m}Q_{nm}\right)=\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{j}\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{ij}.$

We write $A_{\infty\infty}$ for this group, and we write
$\overline{\imath}_{nm}$ for the obvious map $A_{nm}\to A_{\infty\infty}$. By
construction, the following diagram commutes:

${A_{nm}}$${A_{n,m+1}}$${A_{n+1,m}}$${A_{\infty\infty}.}$$\scriptstyle{f_{nm}}$$\scriptstyle{g_{nm}}$$\scriptstyle{\overline{\imath}_{nm}}$$\scriptstyle{\overline{\imath}_{n,m+1}}$$\scriptstyle{\overline{\imath}_{n+1,m}}$

It follows that $\overline{\imath}_{kk}=\overline{\imath}_{k+1,k+1}h_{k}$, so
the maps $\overline{\imath}_{kk}$ form a cone for the sequence
$\\{A_{ii}\\}_{i\in\mathbb{N}}$. Thus, if we write $\overline{\jmath}_{k}$ for
the usual map
$A_{kk}\to\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{ii}$,
we find that there is a unique map
$p\colon\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{ii}\to
A_{\infty\infty}$ with $p\overline{\jmath}_{k}=\overline{\imath}_{kk}$ for all
$k$. In the opposite direction, suppose we have $n,m,k\in\mathbb{N}$ with
$n,m\leq k$. By composing $f$’s and $g$’s in various orders we can form a
number of maps $A_{nm}\to A_{kk}$ but they are all the same because the
original diagram is commutative. We write $u_{nmk}$ for this map, and put
$q_{nmk}=\overline{\jmath}_{k}u_{nmk}\colon
A_{nm}\to\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{ii}$.
Now $\overline{\jmath}_{k}=\overline{\jmath}_{k+1}h_{k}$ 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 $k$ (provided that $k\geq\max(n,m)$) so we can denote it by
$q_{nm}$. We can now put the maps $q_{nm}$ together to give a map
$q^{\prime}\colon
A_{++}\to\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{ii}$
with $q^{\prime}i_{nm}=q_{nm}$. 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
$q^{\prime}(P_{nm})=0=q^{\prime}(Q_{nm})$. There is thus an induced map
$A_{\infty\infty}\to\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{ii}$
with $q\overline{\imath}_{nm}=q_{nm}$. We leave it to the reader to check that
$q$ is inverse to $p$. ∎

##  11\. Limits and derived limits of towers

######  Definition 11.1.

A _tower_ is a diagram of the form

$B_{0}\xleftarrow{f_{0}}B_{1}\xleftarrow{f_{1}}B_{2}\xleftarrow{f_{2}}B_{3}\xleftarrow{f_{3}}\dotsb$

Given such a tower an integers $i\geq j$, we write $f_{ij}$ for the composite

$B_{i}\xrightarrow{f_{i-1}}B_{i-1}\xrightarrow{f_{i-2}}\dotsb\xrightarrow{f_{j}}B_{j}.$

Note that $f_{ii}$ is the identity map, and $f_{i+1,i}=f_{i}$, and
$f_{jk}f_{ij}=f_{ik}$ whenever $i\geq j\geq k$.

######  Definition 11.2.

Suppose we have a tower as above. The _limit_ (or _inverse limit_) of the
tower is the group

$\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}B_{i}=\\{a\in\prod_{i}B_{i}\;|\;a_{i}=f_{i}(a_{i+1})\text{
for all }i\\}.$

Equivalently, if we define $D\colon\prod_{i}B_{i}\to\prod_{i}B_{i}$ by
$D(a)_{i}=a_{i}-f_{i}(a_{i+1})$, then
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}B_{i}=\ker(D)$. We
also write
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}{}_{i}^{1}B_{i}$ for
the cokernel of $D$. We define
$p_{n}\colon\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}B_{i}\to
B_{n}$ 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 tower is a group $A$ with a collection of maps $u_{k}\colon
A\to B_{k}$ such that $u_{k}=f_{k}u_{k+1}$ for all $k$. Tautologically, the
maps
$p_{k}\colon\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}B_{i}\to
B_{k}$ form a cone. Moreover, for any cone
$\\{A\xrightarrow{u_{k}}B_{k}\\}_{k\in\mathbb{N}}$ we can define
$u_{\infty}\colon
A\to\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}B_{i}$ by

$u_{\infty}(a)=(u_{0}(a),u_{1}(a),u_{2}(a),\dotsc),$

and this is the unique map with $p_{k}u_{\infty}=u_{k}$ for all $k$.

######  Example 11.4.

Suppose we have a chain of subgroups

$B_{0}\geq B_{1}\geq B_{2}\geq\dotsb$

and we take the maps $f_{i}$ to be the inclusion maps. Then we see from the
definitions that

$\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}B_{i}=\\{(b,b,b,\dotsb)\;|\;b\in\bigcap_{i}B_{i}\\}\simeq\bigcap_{i}B_{i}.$

######  Example 11.5.

Suppose we have a system of groups $C_{i}$ (for $i\in\mathbb{N}$). Put
$B_{k}=\prod_{i=0}^{k}C_{i}$ and let $f_{k}\colon B_{k+1}\to B_{k}$ be the
obvious projection map. If
$b\in\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{k}B_{k}$ then
$b_{k}\in\prod_{i=0}^{k}C_{i}$ so $b_{kk}\in C_{k}$. We can thus define

$d\colon\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{k}B_{k}\to\prod_{k}C_{k}$

by

$d(b)=(b_{00},b_{11},b_{22},\dotsc).$

By the definition of
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{k}B_{k}$, we have
$b_{kj}=b_{jj}$ for $j\leq k$. 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. Then the map $D$ is the identity, so both
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}$ and
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}$ are zero.

######  Example 11.7.

Suppose we have a sequence

$A_{0}\xrightarrow{f_{0}}A_{1}\xrightarrow{f_{1}}A_{2}\xrightarrow{f_{2}}A_{3}\xrightarrow{f_{3}}\dotsb,$

and another group $B$. This gives us a tower

$\operatorname{Hom}(A_{0},B)\xleftarrow{f_{0}^{*}}\operatorname{Hom}(A_{1},B)\xleftarrow{f_{1}^{*}}\operatorname{Hom}(A_{2},B)\xleftarrow{f_{2}^{*}}\operatorname{Hom}(A_{3},B)\xleftarrow{f_{3}^{*}}\dotsb$

The elements of
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}\operatorname{Hom}(A_{i},B)$
are precisely the cones from the sequence $\\{A_{i}\\}_{i\in\mathbb{N}}$ to
$B$, which biject with homomorphisms from
$\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i}$ to $B$. In
other words, we have

$\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}\operatorname{Hom}(A_{i},B)=\operatorname{Hom}(\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{i}A_{i},B).$

######  Example 11.8.

Fix a prime $p$. We then have a tower

$\mathbb{Z}/p\xleftarrow{}\mathbb{Z}/p^{2}\xleftarrow{}\mathbb{Z}/p^{3}\xleftarrow{}\mathbb{Z}/p^{4}\xleftarrow{}\dotsb.$

The inverse limit is called _the ring of $p$-adic integers_, and is denoted by
$\mathbb{Z}_{p}$. We will investigate it in more detail in Section 12. We can
also form a tower

$\mathbb{Z}/0!\xleftarrow{}\mathbb{Z}/1!\xleftarrow{}\mathbb{Z}/2!\xleftarrow{}\mathbb{Z}/3!\xleftarrow{}\dotsb$

The inverse limit is called _the profinite completion of $\mathbb{Z}$_, and is
denoted by $\widehat{\mathbb{Z}}$. Using the Chinese Remainder Theorem
(Proposition 4.7) one can show that
$\widehat{\mathbb{Z}}=\prod_{p}\mathbb{Z}_{p}$. For yet another description,
recall that

${\mathbb{Q}/\mathbb{Z}}=\bigcup_{n}({\mathbb{Q}/\mathbb{Z}})[n!]=\operatornamewithlimits{\underset{\longrightarrow}{lim}}_{n}({\mathbb{Q}/\mathbb{Z}})[n!],$

so

$\operatorname{End}({\mathbb{Q}/\mathbb{Z}})=\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{n}\operatorname{Hom}(({\mathbb{Q}/\mathbb{Z}})[n!],{\mathbb{Q}/\mathbb{Z}}).$

Now we have a map
$m\colon\mathbb{Z}\to\operatorname{End}({\mathbb{Q}/\mathbb{Z}})$ defined by
$m(k)=k.1_{{\mathbb{Q}/\mathbb{Z}}}$, and this fits in a commutative diagram

${\mathbb{Z}}$${\operatorname{End}({\mathbb{Q}/\mathbb{Z}})}$${\mathbb{Z}/n!}$${\operatorname{Hom}(({\mathbb{Q}/\mathbb{Z}})[n!],{\mathbb{Q}/\mathbb{Z}}).}$$\scriptstyle{m}$restrict$\scriptstyle{m_{n}}$

Using the fact that $({\mathbb{Q}/\mathbb{Z}})[n!]$ is generated by
$(1/n!)+\mathbb{Z}$ we see that $m_{n}$ is an isomorphism. By passing to
inverse limits, we obtain an isomorphism
$m_{\infty}\colon\widehat{\mathbb{Z}}\to\operatorname{End}({\mathbb{Q}/\mathbb{Z}})$.

######  Proposition 11.9.

Suppose we have a commutative diagram as shown

${A_{0}}$${A_{1}}$${A_{2}}$${A_{3}}$${\dotsb}$${B_{0}}$${B_{1}}$${B_{2}}$${B_{3}}$${\dotsb}$$\scriptstyle{p_{0}}$$\scriptstyle{p_{1}}$$\scriptstyle{f_{0}}$$\scriptstyle{p_{2}}$$\scriptstyle{f_{1}}$$\scriptstyle{p_{3}}$$\scriptstyle{f_{2}}$$\scriptstyle{f_{3}}$$\scriptstyle{g_{0}}$$\scriptstyle{g_{1}}$$\scriptstyle{g_{2}}$$\scriptstyle{g_{3}}$

and we define $p=\prod_{i}p_{i}\colon\prod_{i}A_{i}\to\prod_{i}B_{i}$. Then
the central square below commutes, so there are induced maps $p_{\infty}$ and
$p^{1}_{\infty}$ as shown.

${\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}A_{i}}$${\prod_{i}A_{i}}$${\prod_{i}A_{i}}$${\operatornamewithlimits{\underset{\longleftarrow}{lim}}{}_{i}^{1}A_{i}}$${\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}B_{i}}$${\prod_{i}B_{i}}$${\prod_{i}B_{i}}$${\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}{}^{1}B_{i}.}$$\scriptstyle{p_{\infty}}$$\scriptstyle{p}$$\scriptstyle{D}$$\scriptstyle{p}$$\scriptstyle{p^{1}_{\infty}}$$\scriptstyle{D}$

###### Proof.

Clear from the definitions. ∎

######  Proposition 11.10.

Suppose we have a commutative diagram as shown, in which all the columns are
short exact:

${A_{0}}$${A_{1}}$${A_{2}}$${A_{3}}$${\dotsb}$${B_{0}}$${B_{1}}$${B_{2}}$${B_{3}}$${\dotsb}$${C_{0}}$${C_{1}}$${C_{2}}$${C_{3}}$${\dotsb}$$\scriptstyle{p_{0}}$$\scriptstyle{p_{1}}$$\scriptstyle{f_{0}}$$\scriptstyle{p_{2}}$$\scriptstyle{f_{1}}$$\scriptstyle{p_{3}}$$\scriptstyle{f_{2}}$$\scriptstyle{f_{3}}$$\scriptstyle{q_{0}}$$\scriptstyle{q_{1}}$$\scriptstyle{g_{0}}$$\scriptstyle{q_{2}}$$\scriptstyle{g_{1}}$$\scriptstyle{q_{3}}$$\scriptstyle{g_{2}}$$\scriptstyle{g_{3}}$$\scriptstyle{h_{0}}$$\scriptstyle{h_{1}}$$\scriptstyle{h_{2}}$$\scriptstyle{h_{3}}$

Then there is an associated exact sequence

${\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}A_{i}}$${\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}B_{i}}$${\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}C_{i}}$${\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}_{i}A_{i}}$${\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}_{i}B_{i}}$${\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}_{i}C_{i}}$$\scriptstyle{p_{\infty}}$$\scriptstyle{q_{\infty}}$$\scriptstyle{\delta}$$\scriptstyle{p^{1}_{\infty}}$$\scriptstyle{q^{1}_{\infty}}$

###### Proof.

Apply the Snake Lemma to the diagram

${\prod_{i}A_{i}}$${\prod_{i}B_{i}}$${\prod_{i}C_{i}}$${\prod_{i}A_{i}}$${\prod_{i}B_{i}}$${\prod_{i}C_{i}}$$\scriptstyle{\prod_{i}p_{i}}$$\scriptstyle{D}$$\scriptstyle{\prod_{i}q_{i}}$$\scriptstyle{D}$$\scriptstyle{D}$$\scriptstyle{\prod_{i}p_{i}}$$\scriptstyle{\prod_{i}q_{i}}$

in which the rows are easily seen to be short exact. ∎

In practice the groups
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}_{i}A_{i}$ are
usually either zero, or enormous and untractable. We will thus be very
interested in results that force them to be zero.

######  Proposition 11.11.

Suppose we have a tower in which the maps $f_{i}\colon A_{i+1}\to A_{i}$ are
all surjective. Then
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}_{i}A_{i}=0$, and
the projection maps
$p_{k}\colon\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}A_{i}\to
A_{k}$ are all surjective.

###### Proof.

Consider an element $a\in\prod_{i}A_{i}$. We will choose elements $b_{k}\in
A_{k}$ recursively as follows: we start with $b_{0}=0$, and then take $b_{n}$
to be any element with $f_{n-1}(b_{n})=b_{n-1}-a_{n-1}$. These elements
$b_{k}$ give an element $b\in\prod_{i}A_{i}$ with $D(b)=a$, so $D$ is
surjective and
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}_{i}A_{i}=\operatorname{cok}(D)=0$.

Now suppose we have an element $a\in A_{k}$. Define $c_{i}=f_{k,i}(a)$ for all
$i\leq k$. Then define $c_{i}\in A_{i}$ recursively for $i>k$ by choosing
$c_{i}$ to be any element with $f_{i-1}(c_{i})=c_{i-1}$. This gives an element
$c\in\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}A_{i}$ with
$p_{k}(c)=a$. ∎

######  Definition 11.12.

We say that a tower $A_{0}\xleftarrow{f_{0}}A_{1}\xleftarrow{f_{1}}\dotsb$ is
_nilpotent_ if for all $i$ there exists $j>i$ such that $f_{ji}=0\colon
A_{j}\to A_{i}$.

######  Proposition 11.13.

For a nilpotent tower as above, we have
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}A_{i}=\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}^{1}A_{i}=0$.

###### Proof.

Define $E\colon\prod_{i}A_{i}\to\prod_{i}A_{i}$ by

$E(a)_{i}=\sum_{j=i}^{\infty}f_{j,i}(a_{j}).$

Although the sum is formally infinite, the nilpotence hypothesis means that
there are only finitely many nonzero terms, so the expression is meaningful.
It is then not hard to check that $DE=ED=1$, so the kernel and cokernel of $D$
are zero. ∎

######  Definition 11.14.

Consider a tower $A_{0}\xleftarrow{f_{0}}A_{1}\xleftarrow{f_{1}}\dotsb$, so
for each $i$ we have a descending chain of subgroups

$A_{i}\geq f_{i+1,i}(A_{i+1})\geq f_{i+2,i}(A_{i+2})\geq
f_{i+3,i}(A_{i+3})\geq\dotsb$

We say that the tower is _Mittag-Leffler_ if for each $i$ there exists $j\geq
i$ such that $f_{ki}(A_{k})=f_{ji}(A_{j})$ for all $k\geq j$ (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 is Mittag-
Leffler.

###### Proof.

In the first case, we just choose $j\geq i$ such that the order
$|f_{ji}(A_{j})|$ is as small as possible; it then follows that
$f_{ki}(A_{k})=f_{ji}(A_{j})$ for $k\geq i$. In the second case, use
dimensions instead of orders. ∎

######  Proposition 11.17.

If $A$ is a Mittag-Leffler tower, we have
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}^{1}A_{i}=0$.

###### Proof.

By the Mittag-Leffler condition, there is a subgroup $A^{\prime}_{i}\leq
A_{i}$ such that $f_{ji}(A_{j})=A^{\prime}_{i}$ for all sufficiently large
$j$. Thus, for $j$ very large we have both $A^{\prime}_{i+1}=f_{j,i+1}(A_{j})$
and
$A^{\prime}_{i}=f_{ji}(A_{j})=f_{i}(f_{j,i+1}(A_{j}))=f_{i}(A^{\prime}_{i+1})$.
Thus, the groups $A^{\prime}_{i}$ form a subtower of $A$, with surjective maps
$f^{\prime}_{i}\colon A^{\prime}_{i+1}\to A^{\prime}_{i}$. Now put
$A^{\prime\prime}_{i}=A_{i}/A^{\prime}_{i}$, so there are induced maps
$f^{\prime\prime}_{i}\colon A^{\prime\prime}_{i+1}\to A^{\prime\prime}_{i}$,
giving a third tower. If $j$ is much larger than $i$ we have
$f_{ji}(A_{j})=A^{\prime}_{i}$ and so $f^{\prime\prime}_{ji}=0\colon
A^{\prime\prime}_{j}\to A^{\prime\prime}_{i}$; this shows that the tower
$A^{\prime\prime}$ is nilpotent. Now apply Proposition 11.10 to the short
exact sequence $A^{\prime}\to A\to A^{\prime\prime}$ to give an exact sequence

${\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}A^{\prime}_{i}}$${\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}A_{i}}$${\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}A^{\prime\prime}_{i}}$${\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}_{i}A^{\prime}_{i}}$${\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}_{i}A_{i}}$${\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}_{i}A^{\prime\prime}_{i}}$$\scriptstyle{\delta}$

Here
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}^{1}A^{\prime}_{i}=0$
because the maps in $A^{\prime}$ are surjective, and
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}^{1}A^{\prime\prime}_{i}=0$
because $A^{\prime\prime}$ is nilpotent, so
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}^{1}A_{i}=0$ as
claimed. (We also have
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}A^{\prime\prime}_{i}=0$
and so
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}A^{\prime}_{i}=\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}A_{i}$.)
∎

We also have the following result analogous to Proposition 10.11, which again
indicates that
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}A_{i}$ and
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}_{i}A_{i}$ only
depend on the asymptotic behaviour of the tower.

######  Proposition 11.18.

Suppose we have a tower

$A_{0}\xleftarrow{f_{0}}A_{1}\xleftarrow{f_{1}}A_{2}\xleftarrow{f_{2}}A_{3}\xleftarrow{f_{3}}\dotsb$

and a nondecreasing function $u\colon\mathbb{N}\to\mathbb{N}$ such that
$u(i)\to\infty$ as $i\to\infty$. Put

$g_{i}=f_{u(i+1),u(i)}=(A_{u(i+1)}\xrightarrow{f_{u(i+1)-1}}A_{u(i+1)-1}\dotsb\xrightarrow{f_{u(i)}}A_{u(i)}),$

so we have a tower

$A_{u(0)}\xleftarrow{g_{0}}A_{u(1)}\xleftarrow{g_{1}}A_{u(2)}\xleftarrow{g_{2}}A_{u(3)}\xleftarrow{g_{3}}\dotsb$

Then there are canonical isomorphisms
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{j}A_{u(j)}=\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}A_{i}$
and
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}_{j}A_{u(j)}=\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}_{i}A_{i}$.

###### Proof.

Define $v\colon\mathbb{N}\to\mathbb{N}$ by $v(i)=\min\\{j\;|\;u(j)\geq i\\}$.
We will construct a diagram as follows:

${\prod_{j}A_{u(j)}}$${\prod_{i}A_{i}}$${\prod_{j}A_{u(j)}}$${\prod_{j}A_{u(j)}}$${\prod_{i}A_{i}}$${\prod_{j}A_{u(j)}}$$\scriptstyle{\phi}$$\scriptstyle{D^{\prime}}$$\scriptstyle{\psi}$$\scriptstyle{D}$$\scriptstyle{D^{\prime}}$$\scriptstyle{\lambda}$$\scriptstyle{\mu}$

The maps are:

$\displaystyle D^{\prime}(b)_{j}$
$\displaystyle=b_{j}-f_{u(j+1),u(j)}(b_{j+1})$ $\displaystyle D(a)_{i}$
$\displaystyle=a_{i}-f_{i+1,i}(a_{i+1})$ $\displaystyle\phi(b)_{i}$
$\displaystyle=f_{uv(i),i}(b_{v(i)})$ $\displaystyle\psi(a)_{j}$
$\displaystyle=a_{u(j)}$ $\displaystyle\lambda(b)_{i}$
$\displaystyle=\sum_{u(j)=i}b_{j}$ $\displaystyle\mu(a)_{j}$
$\displaystyle=\sum_{u(j)\leq i<u(j+1)}f_{i,u(j)}(a_{i}).$

Thus $D$ and $D^{\prime}$ are the usual maps whose kernels and cokernels are
the $\operatornamewithlimits{\underset{\longleftarrow}{lim}}$ and
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}$ groups under
consideration. We claim that the diagram commutes. To see this, consider a
point $b\in\prod_{j}A_{u(j)}$. We then have

$\displaystyle D(\phi(b))_{i}$
$\displaystyle=\phi(b)_{i}-f_{i+1,i}(\phi(b)_{i+1})$
$\displaystyle=f_{uv(i),i}(b_{v(i)})-f_{uv(i+1),i}(b_{v(i+1)})$
$\displaystyle\lambda(D^{\prime}(b))_{i}$
$\displaystyle=\sum_{u(j)=i}D^{\prime}(b)_{j}=\sum_{u(j)=i}(b_{j}-f_{u(j+1),u(j)}b_{j+1}).$

If $u^{-1}\\{i\\}=\emptyset$ we find that $v(i+1)=v(i)$ and so
$D(\phi(b))_{i}=0=\lambda(D^{\prime}(b))_{i}$. If $u^{-1}\\{i\\}$ is nonempty
then it will be an interval, say $u^{-1}\\{i\\}=\\{j_{0},\dotsc,j_{1}-1\\}$.
In our expression for $\lambda(D^{\prime}(b))_{i}$, the map $f_{u(j+1),u(j)}$
is just the identity except when $j=j_{1}-1$. The expression therefore cancels
down to $b_{j_{0}}-f_{u(j_{1}),i}(b_{j_{1}})$. On the other hand, we also find
that $v(i)=j_{0}$ and $v(i+1)=j_{1}$, so

$D(\phi(b))_{i}=f_{ii}(b_{j_{0}})-f_{u(j_{1}),i}(b_{j_{1}})=\lambda(D^{\prime}(b))_{i}.$

Thus, the left square commutes. For the right square, consider an element
$a\in\prod_{i}A_{i}$. We have

$\displaystyle\mu(D(a))_{j}$ $\displaystyle=\sum_{u(j)\leq
i<u(j+1)}f_{i,u(j)}(D(a)_{i})$ $\displaystyle=\sum_{u(j)\leq
i<u(j+1)}(f_{i,u(j)}(a_{i})-f_{i+1,u(j)}(a_{i+1}))$
$\displaystyle=a_{u(j)}-f_{u(j+1),u(j)}(a_{u(j+1)})$
$\displaystyle=\psi(a)_{j}-f_{u(j+1),u(j)}(\psi(a)_{j+1})=D^{\prime}(\psi(a))_{j}$

as required. We therefore have induced maps

$\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{j}A_{u(j)}\xrightarrow{\phi^{\prime}}\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}A_{i}\xrightarrow{\psi^{\prime}}\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{j}A_{u(j)}$

and

$\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}_{j}A_{u(j)}\xrightarrow{\lambda^{\prime}}\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}_{i}A_{i}\xrightarrow{\mu^{\prime}}\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}_{j}A_{u(j)}.$

It is straightforward to check that $\phi^{\prime}\psi^{\prime}$ and
$\psi^{\prime}\phi^{\prime}$ are the respective identity maps. Now define
$\sigma\colon\prod_{i}A_{i}\to\prod_{i}A_{i}$ by

$\sigma(a)_{i}=\sum_{i\leq h<uv(i)}f_{h,i}(a_{h}).$

We claim that $\lambda\mu+D\sigma=1$ (which implies that
$\lambda^{\prime}\mu^{\prime}=1$). The proof that
$\lambda(\mu(a))_{i}+D(\sigma(a))_{i}=a_{i}$ splits into two cases, depending
on whether $i\in\operatorname{image}(u)$ or not. If
$i\not\in\operatorname{image}(u)$ we find that the inequality $i\leq u(v(i))$
cannot be an equality, so $v(i+1)=v(i)$. Using this we deduce that
$D(\sigma(a))_{i}$ is the difference between two sums that mostly have the
same terms, and thus that $D(\sigma(a))_{i}=a_{i}$. On the other hand, we have
$\lambda(\mu(a))_{i}=\sum_{u(j)=i}\mu(a)_{j}$ which is zero as the sum has no
terms. Now consider instead the case where $i\in\operatorname{image}(u)$. Let
$j_{1}$ be the largest integer such that $u(j_{1})=i$, and put
$i^{\prime}=u(j_{1}+1)>i$. From the definitions we have

$\lambda(\mu(a))_{i}=\sum_{u(j)=i}\sum_{u(j)\leq h<u(j+1)}f_{h,u(j)}(a_{h}).$

However, the inner summation is empty unless $j=j_{1}$, so the formula reduces
to $\lambda(\mu(a))_{i}=\sum_{i\leq h<i^{\prime}}f_{h,i}(a_{h})$. We also have
$uv(i)=i$, so $\sigma(a)_{i}=0$, and $uv(i+1)=u(j_{1}+1)=i^{\prime}$, so
$\sigma(a)_{i+1}=\sum_{i<h<i^{\prime}}h_{h,i+1}(a_{h})$. From this it is easy
to see that $\lambda(\mu(a))_{i}+D(\sigma(a))_{i}=a_{i}$ as required.

We now consider instead the map $\mu\lambda$. Define
$\tau\colon\prod_{j}A_{u(j)}\to\prod_{j}A_{u(j)}$ by

$\tau(b)_{j}=\sum_{k<j,\;u(k)=u(j)}b_{k}.$

We claim that $\mu\lambda=1+D\tau$ (which implies that
$\mu^{\prime}\lambda^{\prime}=1$). The proof that
$\mu(\lambda(b))_{j}=b_{j}+D(\tau(b))_{j}$ again splits into two cases. First
suppose that $u(j+1)=u(j)$. It is then immediate from the definitions that
$\mu(\lambda(b))_{j}=0$. On the other hand, the sums defining $\tau(b)_{j}$
and $\tau(b)_{j+1}$ differ only by a single term, so $D(\tau(b))_{j}=-b_{j}$,
as required. Now suppose instead that $u(j+1)>u(j)$. In this case we have
$\tau(b)_{j+1}=0$, so

$b_{j}+D(\tau(b))_{j}=b_{j}+\tau(b)_{j}=b_{j}+\sum_{k<j,\;u(k)=u(j)}b_{k}=\sum_{u(k)=u(j)}b_{k}.$

On the other hand, we have

$\mu(\lambda(b))_{j}=\sum_{u(j)\leq i<u(j+1)}\sum_{u(k)=i}f_{i,u(j)}(b_{k}).$

The inner sum has no terms unless $i=u(j)$, and in that context $f_{i,u(j)}$
is the identity, so the above reduces to

$\mu(\lambda(b))_{j}=\sum_{u(k)=u(j)}b_{k}=b_{j}+D(\tau(b))_{j}$

as required. ∎

##  12\. Completion and derived completion

######  Definition 12.1.

Let $p$ be a prime. For any abelian group $A$ we have a tower of surjections

$0=A/p^{0}\xleftarrow{}A/p\xleftarrow{}A/p^{2}\xleftarrow{}A/p^{3}\xleftarrow{}\dotsb$

We write $A_{p}$ for the inverse limit of this tower, and call this the _$p$
-completion_ of $A$. In particular, we have a group $\mathbb{Z}_{p}$, whose
elements are called _$p$ -adic integers_. For any $A$ we have a homomorphism
$\eta\colon A\to A_{p}$ given by

$\eta(a)=(a+p^{0}A,a+p^{1}A,a+p^{2}A,a+p^{3}A,\dotsc).$

We say that $A$ is _$p$ -complete_ if $\eta$ is an isomorphism.

######  Example 12.2.

Let $A$ be a finite abelian group, so $A$ splits as $B\oplus C$ 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 $A_{p}=B$.
In particular, 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
${\mathbb{Q}}_{p}=0$ and $({\mathbb{Q}}/\mathbb{Z})_{p}=0$.

######  Remark 12.4.

The symbol ${\mathbb{Q}}_{p}$ is often used for
${\mathbb{Q}}\otimes\mathbb{Z}_{p}$, which is not zero; it is known as the
field of $p$-adic rationals. However, this is different from the group that we
have called ${\mathbb{Q}}_{p}$, which is trivial.

######  Proposition 12.5.

For all $A$ and $k\geq 0$ the projection $\pi_{k}\colon A\to A/p^{k}A$ 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

$A/p^{k}\xleftarrow{}A/p^{2k}\xleftarrow{}A/p^{3k}\xleftarrow{}\dotsb.$

First, the projection $\pi_{1}\colon\prod_{i}A/p^{i}\to A/p$ restricts to give
a homomorphism $\phi\colon
A_{p}=\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}A/p^{i}\to
A/p$. Proposition 11.11 tells us that this is surjective. As $A/p$ has
exponent $p$, the subgroup $pA_{p}$ is contained in the kernel. We need to
prove that the kernel is precisely $pA_{p}$. Suppose that $a\in\ker(\phi)$.
Choose $a_{i}\in A$ representing the component of $a$ in $A/p^{i}A$. As
$A/p^{0}A=0$ and $\phi(a)=0$ we can take $a_{0}=a_{1}=0$. By the definition 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 $b_{i}\in A$ (and we may take
$b_{0}=0$). Now put $c_{n}=\sum_{i=1}^{n}p^{i-1}b_{i}\in A$. It is visible
that $c_{n+1}=c_{n}+p^{n}b_{n+1}=c_{n}\pmod{p^{n}A_{n}}$, so the cosets
$c_{n}+p^{n}A$ define an element $c\in A_{p}$. We also see by induction that
$pc_{k}=a_{k+1}=a_{k}\pmod{p^{k}A_{k}}$, so $pc=a$. Thus $a\in pA_{p}$ as
claimed. ∎

######  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 natural maps
$A_{p}\to(A_{p})_{p}$ are the same, and they are both isomorphisms.

We next examine the structure of $\mathbb{Z}_{p}$ in more detail. First, as
the groups $\mathbb{Z}/p^{k}$ have canonical ring structures, and the maps
$\mathbb{Z}/p^{k+1}\to\mathbb{Z}/p^{k}$ are ring maps, we see that
$\mathbb{Z}_{p}$ is a subring of $\prod_{k}\mathbb{Z}/p^{k}$. We will write
$\pi_{k}$ for the projection $\mathbb{Z}_{p}\to\mathbb{Z}/p^{k}$, which is a
surjective ring homomorphism. Note also that for $n\in\mathbb{Z}$ we have
$\eta(n)=0$ iff $\pi_{k}\eta(n)=0$ for all $k$ iff $n$ is divisible by $p^{k}$
for all $k$ iff $n=0$. Thus, $\eta$ gives an injective ring map
$\mathbb{Z}\to\mathbb{Z}_{p}$. We will usually suppress notation for this and
regard $\mathbb{Z}$ as a subring of $\mathbb{Z}_{p}$.

######  Definition 12.7.

For $a\in\mathbb{Z}_{p}$ we define $v(a)=\min\\{k\;|\;\pi_{k}(a)\neq 0\\}$ (or
$v(a)=\infty$ if $\pi_{k}(a)=0$ for all $k$, which means that $a=0$). We also
define $d(a,b)=p^{-v(a-b)}$, with the convention $p^{-\infty}=0$.

######  Proposition 12.8.

The function $d$ defines a metric on $\mathbb{Z}_{p}$ (called the _$p$ -adic
metric_), with respect to which it is complete and compact. Moreover, the
subspace $\mathbb{Z}$ is dense.

###### Proof.

It is clear that $d(a,b)=d(b,a)$, and that this is nonnegative and vanishes if
and only if $a=b$. This just leaves the triangle inequality $d(a,c)\leq
d(a,b)+d(b,c)$. This is clear if $a=b$ or $b=c$, so suppose that $a\neq b\neq
c$. Put $m=\min(v(a-b),v(b-c))$. For $k<m$ we have
$\pi_{k}(a)=\pi_{k}(b)=\pi_{k}(c)$. It follows that $v(c-a)\geq m$, so

$d(a,c)\leq p^{-m}=\max(d(a,b),d(b,c))\leq d(a,b)+d(b,c)$

as required.

Now consider a Cauchy sequence $(a_{0},a_{1},a_{2},\dotsc)$ in
$\mathbb{Z}_{p}$. Given $k\in\mathbb{N}$ we can choose $m$ such that
$d(a_{i},a_{j})<p^{-k}$ for all $i,j\geq m$. From the definition of $d$, this
means that the element $\pi_{k}(a_{i})\in\mathbb{Z}/p^{k}$ is independent of
$i$ for $i\geq m$. Let $b_{k}$ denote this element. If $i$ is large enough we
will have $b_{k}=\pi_{k}(a_{i})$ and also $b_{k+1}=\pi_{k+1}(a_{i})$; using
this we see that the projection $\mathbb{Z}/p^{k+1}\to\mathbb{Z}/p^{k}$ sends
$b_{k+1}$ to $b_{k}$. Thus, sequence $(b_{0},b_{1},b_{2},\dotsc)$ is an
element $b\in\mathbb{Z}_{p}$, and by construction $a_{i}\to b$ as
$i\to\infty$. This shows that $\mathbb{Z}_{p}$ is compact.

Next, suppose we are given $k\in\mathbb{N}$, and we put
$T_{k}=\\{0,1,\dotsc,p^{k}-1\\}$ and $a\in\mathbb{Z}_{p}$. The map
$T_{k}\to\mathbb{Z}/p^{k}$ is a bijection, so for any $a\in\mathbb{Z}_{p}$
there is a unique $m\in T_{k}$ with $\pi_{k}(m)=\pi_{k}(a)$, so
$d(m,a)<p^{-k}$. Using this we see that $\mathbb{Z}$ is dense in
$\mathbb{Z}_{p}$. Next, recall that an _$\epsilon$ -net_ in a metric space $X$
is a finite set $F\subseteq X$ such that every point is within $\epsilon$ of a
point in $F$, that $X$ is _totally bounded_ if it has an $\epsilon$-net for
every $\epsilon>0$, and that a complete metric space is compact if and only if
it is totally bounded. The set $T_{k}$ is a $2^{-k}$-net, so $\mathbb{Z}_{p}$
is totally bounded, so it is compact. ∎

######  Proposition 12.9.

Put $D=\\{0,1,2,\dotsc,p-1\\}$ (the set of $p$-adic digits). Then there is a
bijection

$\sigma\colon\prod_{i=0}^{\infty}D\to\mathbb{Z}_{p}$

given by $\sigma(u)=\sum_{i}u_{i}p^{i}$ (a convergent sum with respect to the
$p$-adic metric). In particular, $\mathbb{Z}_{p}$ is uncountable. We call
$\sigma^{-1}(a)$ the _base $p$ expansion_ of $a$.

###### Proof.

It is elementary that the corresponding map
$\prod_{i=0}^{k-1}D\to\mathbb{Z}/p^{k}$ is a bijection, and the claim follows
by passing to inverse limits. ∎

######  Proposition 12.10.

The ring $\mathbb{Z}_{p}$ is torsion free and is an integral domain. It is
also a local ring, with $p.\mathbb{Z}_{p}$ being the unique maximal ideal. The
group of units is

$\mathbb{Z}_{p}^{\times}=\\{a\in\mathbb{Z}_{p}\;|\;\pi_{0}(a)\neq
0\\}=\mathbb{Z}_{p}\setminus p\mathbb{Z}_{p},$

and every nonzero element is a unit times $p^{k}$ for some $k$.

###### Proof.

First, consider an element $a\in\mathbb{Z}_{p}$ with $\pi_{0}(a)=1$. We see
from Proposition 12.5 that $a=1-px$ for some $x\in\mathbb{Z}_{p}$. It follows
easily that the series $\sum_{i}(px)^{i}$ is Cauchy, so it converges to some
$b\in\mathbb{Z}_{p}$, and we find that $ab=1$. More generally, suppose merely
that $\pi_{0}(a)\neq 0$ in $\mathbb{Z}/p$. As $\mathbb{Z}/p$ is a field, we
can find $b\in\mathbb{Z}$ such that $\pi_{0}(b)$ is inverse to $\pi_{0}(a)$.
We then find that $ab$ is invertible by the previous case, and it follows that
$a$ is invertible. Conversely, as $\pi_{0}$ is a ring map it certainly sends
units to nonzero elements. We therefore see that

$\mathbb{Z}_{p}^{\times}=\\{a\in\mathbb{Z}_{p}\;|\;\pi_{0}(a)\neq
0\\}=\mathbb{Z}_{p}\setminus p\mathbb{Z}_{p}$

as claimed. Note also that $\mathbb{Z}_{p}/p\mathbb{Z}_{p}$ is the field
$\mathbb{Z}/p$, so $p\mathbb{Z}_{p}$ is a maximal ideal. If $\mathfrak{m}$ is
any maximal ideal we must have
$\mathfrak{m}\cap\mathbb{Z}_{p}^{\times}=\emptyset$, so $\mathfrak{m}\leq
p\mathbb{Z}_{p}$, so $\mathfrak{m}=p\mathbb{Z}_{p}$ by maximality. Thus,
$\mathbb{Z}_{p}$ is a local ring.

Next, using base $p$ expansions we see easily that multiplication by $p$ is
injective, and that every nonzero element $a\in\mathbb{Z}_{p}$ can be written
as $a=p^{k}b$ for some $k\geq 0$ and $b$ with $\pi_{0}(b)\neq 0$, so
$b\in\mathbb{Z}_{p}^{\times}$. As multiplication by $p^{k}$ is injective and
$b$ is invertible we see that multiplication by $a$ is injective. This means
that $\mathbb{Z}_{p}$ is an integral domain. By considering
$a\in\mathbb{Z}\subset\mathbb{Z}_{p}$ we also see that $\mathbb{Z}_{p}$ is
torsion free. ∎

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 $\mathbb{Z}_{p}$. We
say that $f$ is _asymptotically zero_ 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 $\mathbb{Z}[I]_{p}$ is naturally isomorphic to $AZ(I)$.

###### Proof.

First, we put

$(\mathbb{Z}/p^{k})[I]=\\{f\colon I\to\mathbb{Z}/p^{k}\;|\;\\{i\;|\;f(i)\neq
0\\}\text{ is finite }\\}.$

We write $\pi_{k}$ for the projection $\mathbb{Z}\to\mathbb{Z}/p^{k}$, or the
projection $\mathbb{Z}_{p}\to\mathbb{Z}/p^{k}$. We then write
$\pi^{\prime}_{k}(f)=\pi_{k}\circ f$; this defines a map
$AZ(I)\to(\mathbb{Z}/p^{k})[I]$, which we can restrict to $\mathbb{Z}[I]\leq
AZ(I)$. Note that $\pi^{\prime}_{k}(f)=0$ iff $f(i)\in p^{k}\mathbb{Z}_{p}$
for all $i$, in which case we can define $g=f/p^{k}\colon I\to\mathbb{Z}_{p}$
and we find that $g$ is again asymptotically zero, so $f\in p^{k}AZ(I)$. This
shows that $\pi^{\prime}_{k}$ induces an isomorphism
$AZ(I)/p^{k}AZ(I)\to(\mathbb{Z}/p^{k})[I]$. By a similar argument, it also
induces an isomorphism
$\mathbb{Z}[I]/p^{k}\mathbb{Z}[I]\to(\mathbb{Z}/p^{k})[I]$. Thus, we have
$\mathbb{Z}[I]_{p}=\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{k}(\mathbb{Z}/p^{k})[I]$.
The maps $\pi^{\prime}_{k}\colon AZ(I)\to(\mathbb{Z}/p^{k})[I]$ therefore
assemble to give a homomorphism $\pi^{\prime}\colon
AZ(I)\to\mathbb{Z}[I]_{p}$. In the opposite direction, suppose we have
$g\in\mathbb{Z}[I]_{p}$. For each $k\geq 0$ we therefore have
$\pi_{k}(g)\in\mathbb{Z}[I]/p^{k}\mathbb{Z}[I]=(\mathbb{Z}/p^{k})[I]$, so
there is a unique map

$g_{k}\colon I\to\\{0,1,\dotsc,p^{k}-1\\}$

with $\pi_{k}(g)(i)=g_{k}(i)\pmod{p^{k}}$ for all $k$. Note that the set
$\\{i\;|\;g_{k}(i)\neq 0\\}$ is finite for all $k$. If we fix $i$, we find
that the sequence $\\{g_{k}(i)\\}_{k\geq 0}$ is Cauchy, converging to some
element $g_{\infty}(i)\in\mathbb{Z}_{p}$ say, and we have
$g_{\infty}(i)=g_{k}(i)\pmod{p^{k}}$ for all $k$. This means that
$g_{\infty}\in AZ(I)$ and $\pi^{\prime}(g_{\infty})=g$, so $\pi^{\prime}$ is
surjective. We also have $\pi^{\prime}(h)=0$ iff $h(i)$ is divisible by
$p^{k}$ for all $i$ and $k$, which implies that $h=0$. Thus, the map
$\pi^{\prime}$ is an isomorphism. ∎

######  Proposition 12.13.

For any abelian groups $A$ and $B$ there is a natural map $\mu\colon
A_{p}\otimes B_{p}\to(A\otimes B)_{p}$, which induces an isomorphism
$(A_{p}\otimes B_{p})_{p}\to(A\otimes B)_{p}$.

###### Proof.

Note that $p^{k}.1_{A\otimes B}=(p^{k}.1_{A})\otimes
1_{B}=1_{A}\otimes(p^{k}.1_{B})$. Using the right exactness of tensor products
we see that

$(A\otimes B)/p^{k}=(A/p^{k})\otimes
B=A\otimes(B/p^{k})=(A/p^{k})\otimes(B/p^{k}).$

Combining this with Proposition 12.5 gives

$(A_{p}\otimes
B_{p})/p^{k}=(A_{p}/p^{k})\otimes(B_{p}/p^{k})=(A/p^{k})\otimes(B/p^{k})=(A\otimes
B)/p^{k}.$

Passing to inverse limits gives an isomorphism $(A_{p}\otimes
B_{p})_{p}=(A\otimes B)_{p}$. We can compose this with the map $\eta\colon
A_{p}\otimes B_{p}\to(A_{p}\otimes B_{p})_{p}$ to get a map $\mu\colon
A_{p}\otimes B_{p}\to(A\otimes B)_{p}$, which is what we most often need for
applications. ∎

######  Proposition 12.14.

There is a natural map $\mathbb{Z}_{p}\otimes A\to A_{p}$, which is an
isomorphism when $A$ is finitely generated.

###### Proof.

The map is just the composite

$\mathbb{Z}_{p}\otimes A\xrightarrow{1\otimes\eta}\mathbb{Z}_{p}\otimes
A_{p}\xrightarrow{\mu}(\mathbb{Z}\otimes A)_{p}=A_{p}$

(or it can be defined more directly by the method used for $\mu$). By the
classification of finitely generated abelian groups, it will suffice to prove
that we have an isomorphism when $A=\mathbb{Z}$ or $A=\mathbb{Z}/p^{k}$ or
$A=\mathbb{Z}/q^{k}$ for some prime $q\neq p$. The case $A=\mathbb{Z}$ is
clear. When $A=\mathbb{Z}/p^{k}$ we have $A_{p}=A$ as in Example 12.2. We also
$\mathbb{Z}_{p}\otimes A=\mathbb{Z}_{p}/p^{k}\mathbb{Z}_{p}$ by the right
exactness of tensoring, and this is the same as $\mathbb{Z}/p^{k}=A$ by
Proposition 12.5, so $\mathbb{Z}_{p}\otimes A=A_{p}$ as claimed. Finally, if
$q$ is different from $p$ then $q^{k}$ is invertible in $\mathbb{Z}_{p}$ so
$\mathbb{Z}_{p}\otimes\mathbb{Z}/q^{k}=\mathbb{Z}_{p}/q^{k}\mathbb{Z}_{p}=0$,
and similarly $(\mathbb{Z}/q^{k})_{p}=0$ as in Example 12.2 again. ∎

######  Corollary 12.15.

If $A\to B\to C$ is an exact sequence of finitely generated abelian groups,
then the resulting sequence $A_{p}\to B_{p}\to C_{p}$ is also exact.

###### Proof.

As $\mathbb{Z}_{p}$ is torsion free, Proposition 7.16 tells us that the
sequence $\mathbb{Z}_{p}\otimes A\to\mathbb{Z}_{p}\otimes
B\to\mathbb{Z}_{p}\otimes C$ is exact. ∎

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 $\eta\colon A\to A_{p}$ 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
$\mathbb{Z}_{p}$.

  * (d)

Suppose that $f\colon A\to B$ is a homomorphism, and that $B$ is $p$-complete.
Then there is a unique homomorphism $f^{\prime}\colon A_{p}\to B$ such that
$f^{\prime}\circ\eta=f\colon A\to B$.

###### 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 Proposition 9.14.

  * (c)

Proposition 12.13 gives a map

$\mathbb{Z}_{p}\otimes A=\mathbb{Z}_{p}\otimes A_{p}\to(\mathbb{Z}\otimes
A)_{p}=A_{p}=A.$

For $r\in\mathbb{Z}_{p}$ and $a\in A$ we can thus define $ra=\mu(r\otimes a)$.
Equivalently, this is characterised by the fact that
$\pi_{k}(ra)=\pi_{k}(r)\pi_{k}(a)$ in $A/p^{k}A$ (where we have used the
obvious structure of $A/p^{k}A$ as a module over $\mathbb{Z}/p^{k}$). From
this description it is clear that our multiplication rule is associative,
unital and distributive, so it makes $A$ into a module over $\mathbb{Z}_{p}$.

  * (d)

The map $f\colon A\to B$ induces an isomorphism $f_{p}\colon A_{p}\to B_{p}$,
and we have an isomorphism $\eta\colon B\to B_{p}$. We can and must take
$f^{\prime}=\eta^{-1}\circ f_{p}$.

∎

While our definition of completion is quite natural and straightforward, its
exactness properties for infinitely generated groups are very delicate, and
they do not relate well to topological constructions. We will therefore
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 $v(x)=\sum_{i=0}^{\infty}a_{i}x^{i}$ with $a_{i}\in A$ for all
$i$. This is a module over $\mathbb{Z}[\\![x]\\!]$ by the obvious rule

$\left(\sum_{i}n_{i}x^{i}\right)\left(\sum_{j}a_{j}x^{j}\right)=\sum_{k}\left(\sum_{i=0}^{k}n_{i}a_{k-i}\right)x^{k}.$

We define

$\displaystyle L_{0}A$ $\displaystyle=A[\\![x]\\!]/((x-p).A[\\![x]\\!])$
$\displaystyle L_{1}A$ $\displaystyle=\\{v(x)\in
A[\\![x]\\!]\;|\;(x-p)v(x)=0\\}.$

We can identify $A$ with the set of constant series in $A[\\![x]\\!]$, and
then restrict the quotient map $A[\\![x]\\!]\to L_{0}A$ to $A$ to get a
natural map $\eta\colon A\to L_{0}A$. We say that $A$ is _Ext- $p$-complete_
if $\eta\colon A\to L_{0}A$ is an isomorphism and $L_{1}A=0$. We also call
$L_{0}A$ the _derived completion_ of $A$.

######  Remark 12.18.

Readers familiar with the general theory of derived functors should consult
Corollary 12.27 to see why the term is appropriate here.

######  Remark 12.19.

It will follow from Proposition 12.28 that the condition $L_{1}A=0$ is
actually automatic when $\eta\colon A\to L_{0}A$ is an isomorphism. However,
it is easier to develop the theory if we have both conditions in the initial
definition.

######  Remark 12.20.

There is an evident product map

$\mu\colon A[\\![x]\\!]\otimes B[\\![x]\\!]\to(A\otimes B)[\\![x]\\!]$

given by

$\mu\left(\left(\sum_{i}a_{i}x^{i}\right)\otimes\left(\sum_{j}b_{j}x^{j}\right)\right)=\sum_{k}\left(\sum_{k=i+j}a_{i}\otimes
b_{j}\right)x^{k}.$

This induces a map $\mu\colon(L_{0}A)\otimes(L_{0}B)\to L_{0}(A\otimes B)$,
which fits in a commutative diagram

${A\otimes B}$${(L_{0}A)\otimes(L_{0}B)}$${L_{0}(A\otimes
B).}$$\scriptstyle{\eta\otimes\eta}$$\scriptstyle{\eta}$$\scriptstyle{\mu}$

######  Remark 12.21.

If we have a short exact sequence $A\xrightarrow{}B\xrightarrow{}C$, we can
apply the Snake Lemma to the diagram

${A[\\![x]\\!]}$${B[\\![x]\\!]}$${C[\\![x]\\!]}$${A[\\![x]\\!]}$${B[\\![x]\\!]}$${C[\\![x]\\!]}$$\scriptstyle{x-p}$$\scriptstyle{x-p}$$\scriptstyle{x-p}$

to obtain an exact sequence

$L_{1}A\rightarrowtail L_{1}B\to L_{1}C\to L_{0}A\to L_{0}B\twoheadrightarrow
L_{0}C.$

######  Proposition 12.22.

  * (a)

If we have a short exact sequence $A\to B\to C$ in which two of the three
terms are 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 $p^{k}.1_{A}=0$ 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}$${B}$${C}$${L_{1}A}$${L_{1}B}$${L_{1}C}$${L_{0}A}$${L_{0}B}$${L_{0}C}$

  * (b)

Clear.

  * (c)

Consider a homomorphism $f\colon A\to B$ between Ext-$p$-complete groups, and
the resulting short exact sequences $\operatorname{img}(f)\to
B\to\operatorname{cok}(f)$ and $\ker(f)\to A\to\operatorname{img}(f)$. These
give diagrams

${\operatorname{img}(f)}$${B}$${\operatorname{cok}(f)}$${L_{1}\operatorname{img}(f)}$${0}$${L_{1}\operatorname{cok}(f)}$${L_{0}\operatorname{img}(f)}$${L_{0}B}$${L_{0}\operatorname{cok}(f)}$$\scriptstyle{\simeq}$

and

${\ker(f)}$${A}$${\operatorname{img}(f)}$${L_{1}\ker(f)}$${0}$${L_{1}\operatorname{img}(f)}$${L_{0}\ker(f)}$${L_{0}A}$${L_{0}\operatorname{img}(f).}$$\scriptstyle{\simeq}$

From the first diagram we see that $L_{1}\operatorname{img}(f)=0$ and that the
map $\operatorname{img}(f)\to L_{0}\operatorname{img}(f)$ is injective, and
from the second we see that the map $\operatorname{img}(f)\to
L_{0}\operatorname{img}(f)$ is surjective; thus $\operatorname{img}(f)$ is
Ext-$p$-complete. Given this, it is a special case of (a) that $\ker(f)$ and
$\operatorname{cok}(f)$ are also Ext-$p$-complete as claimed.

  * (d)

This is clear, because
$(\prod_{i}A_{i})[\\![x]\\!]=\prod_{i}A_{i}[\\![x]\\!]$.

  * (e)

If $k=1$ the definitions give $L_{0}A=A[\\![x]\\!]/(x.A[\\![x]\\!])=A$ and
$L_{1}A=\\{v(x)\in A[\\![x]\\!]\;|\;xv(x)=0\\}=0$ as required. The general
case follows by induction using (a) and the short exact sequence $pA\to A\to
A/pA$.

  * (f)

As the tower $\\{A/p^{k}\\}$ consists of surjections, the
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}$ term is zero. As
$A$ is $p$-complete, we therefore have a short exact sequence

$A\rightarrowtail\prod_{k}A/p^{k}\twoheadrightarrow\prod_{k}A/p^{k}.$

The second and third terms are Ext-$p$-complete by parts (e) and (d), so $A$
is Ext-$p$-complete by part (a).

∎

######  Proposition 12.23.

Let $A$ be any abelian group. Then
$L_{1}A=\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{k}A[p^{k}]$
and there is a natural short exact sequence

${\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{k}^{1}A[p^{k}]}$${L_{0}A}$${A_{p}.}$$\scriptstyle{\xi}$$\scriptstyle{\zeta}$

The limit symbols here refer to the tower

$0=A[p^{0}]\xleftarrow{p}A[p]\xleftarrow{p}A[p^{2}]\xleftarrow{p}A[p^{3}]\xleftarrow{p}\dotsb$

###### Proof.

First, an element of $L_{1}A$ is a series $v(x)=\sum_{i}a_{i}x^{i}$ with
$(x-p)v(x)=0$, which means that $pa_{0}=0$ and $pa_{i+1}=a_{i}$ for all $i\geq
0$. It follows inductively that $p^{i+1}a_{i}=0$ for all $i$, so the sequence
$(0,a_{0},a_{1},\dotsc)$ is an element of
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}A[p^{i}]$. All
steps here can be reversed so
$L_{1}A=\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}A[p^{i}]$.
Next, define $\zeta^{\prime}_{i}\colon A[\\![x]\\!]\to A/p^{i}A$ by

$\zeta^{\prime}_{i}(\sum_{j}a_{j}x^{j})=\sum_{j<i}a_{j}p^{j}+p^{i}A.$

It is clear that
$\zeta^{\prime}_{i}(v(x))=\zeta^{\prime}_{i+1}(v(x))\pmod{p^{i}A}$, so the
maps $\zeta^{\prime}_{i}$ fit together to give a homomorphism
$\zeta^{\prime}\colon A[\\![x]\\!]\to A_{p}$, which can be described
heuristically as $\zeta^{\prime}(v(x))=v(p)$. It is also easy to check that
$\zeta^{\prime}((x-p)w(x))=0$ for all $w(x)$, so there is an induced map
$\zeta\colon L_{0}A\to A_{p}$. Given an arbitrary element $b\in A_{p}$ we can
choose $b_{i}\in A$ representing the coset $\pi_{i}(b)\in A/p^{i}A$. These
will then satisfy $b_{i+1}=b_{i}\pmod{p^{i}A}$, so we can choose $a_{i}\in A$
with $b_{i+1}=p^{i}a_{i}+b_{i}$. The series $v(x)=\sum_{i}a_{i}x^{i}$ then has
$\zeta(v(x))=b$, so we see that $\zeta$ is surjective.

Next, for any $c\in\prod_{i}A[p^{i}]$ put
$\xi^{\prime\prime}(c)=\sum_{i}c_{i}x^{i}\in A[\\![x]\\!]$, and let
$\xi^{\prime}(c)$ denote the image of $\xi^{\prime\prime}(c)$ in $L_{0}A$. It
is clear by construction that $\zeta\xi^{\prime}=0$. Suppose that $c$ lies in
the image of the map $D\colon\prod_{i}A[p^{i}]\to\prod_{i}A[p^{i}]$, so there
is a sequence $(d_{i})_{i\geq 0}$ with $p^{i}d_{i}=0$ and
$c_{i}=d_{i}-pd_{i+1}$ for all $i$. Note that $d_{0}=p^{0}d_{0}=0$ and put
$w(x)=\sum_{i}d_{i+1}x^{i}$; we find that $\xi^{\prime\prime}(c)=(x-p)w(x)$
and so $\xi^{\prime}(c)=0$. We thus have an induced map
$\xi\colon\operatorname{cok}(D)=\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}_{i}A[p^{i}]\to
L_{0}A$ with $\zeta\xi=0$. Consider an arbitrary element
$v(x)=\sum_{i}a_{i}x^{i}\in A[\\![x]\\!]$ with $\zeta^{\prime}(v(x))=0$. This
means that for all $i\geq 0$ we can choose $b_{i}\in A$ with
$\sum_{j<i}a_{j}p^{j}=p^{i}b_{i}$. It follows that $b_{0}=0$ and
$p^{i}b_{i}+p^{i}a_{i}=p^{i+1}b_{i+1}$, so the element
$c_{i}=b_{i}+a_{i}-pb_{i+1}$ has $p^{i}c_{i}=0$, so $c\in\prod_{i}A[p^{i}]$.
If we put $w(x)=\sum_{i}b_{i+1}x^{i}$ we find that

$v(x)=\xi^{\prime\prime}(c)-(x-p)w(x),$

so in $L_{0}A$ we have $v=\xi(c)$. This proves that the map
$\xi\colon\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}_{i}A[p^{i}]\to\ker(\zeta)$
is surjective.

Finally, suppose we have $c\in\prod_{i}A[p^{i}]$ with $\xi^{\prime}(c)=0$.
This means that there exists $u(x)=\sum_{i}b_{i}x^{i}\in A[\\![x]\\!]$ with
$\xi^{\prime\prime}(c)=(x-p)u(x)$, so $c_{0}=-pb_{0}$ and
$c_{i+1}=b_{i}-pb_{i+1}$ for all $i\geq 0$. Form this it follows easily that
$p^{i+1}b_{i}=0$ for all $i\geq 0$, so we have an element
$b^{\prime}=(0,b_{0},b_{1},\dotsc)\in\prod_{i}A[p^{i}]$. We now see that
$c=D(b^{\prime})$, so $c$ represents the zero element of
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}_{i}A[p^{i}]$,
thus, the map
$\xi\colon\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}_{i}A[p^{i}]\to\ker(\zeta)$
is injective. ∎

######  Remark 12.24.

It is clear from the definitions that the diagram

${A}$${L_{0}A}$${A_{p}}$$\scriptstyle{\eta}$$\scriptstyle{\eta}$$\scriptstyle{\zeta}$

commutes.

######  Definition 12.25.

We say that an abelian group $A$ has _bounded $p$-torsion_ if there exists
$k\geq 0$ such that $p^{k}.\operatorname{tors}_{p}(A)=0$.

######  Corollary 12.26.

If $A$ has bounded $p$-torsion (in particular, if $A$ is a free abelian group)
then $L_{1}A=0$ and $L_{0}A=A_{p}$.

###### Proof.

The tower $\\{A[p^{i}]\\}$ is nilpotent, so
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}A[p^{i}]=\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{i}^{1}A[p^{i}]=0$
by Proposition 11.13. ∎

######  Corollary 12.27.

Suppose we have a short exact sequence $P\xrightarrow{f}Q\to A$ where $P$ and
$Q$ are free abelian groups. Then $L_{0}A$ and $L_{1}A$ are the cokernel and
kernel of the induced map $P_{p}\to Q_{p}$.

###### Proof.

This is immediate from Corollary 12.26 and Remark 12.21. ∎

######  Proposition 12.28.

For any abelian group $A$, the groups $L_{0}A$, $L_{1}A$, $A_{p}$ and
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{k}^{1}A[p^{k}]$ are
all Ext-$p$-complete.

###### Proof.

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 $\prod_{k}A[p^{k}]$ and
$\prod_{k}A/p^{k}$ are Ext-$p$-complete, and then by part (c) that the groups
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{k}A[p^{k}]=L_{1}A$,
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}_{k}A[p^{k}]$ and
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{k}A/p^{k}=A_{p}$ are
Ext-$p$-complete. We can thus apply part (a) to the short exact sequence

${\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{k}^{1}A[p^{k}]}$${L_{0}A}$${A_{p}}$$\scriptstyle{\xi}$$\scriptstyle{\zeta}$

to deduce that $L_{0}A$ is Ext-$p$-complete. ∎

######  Proposition 12.29.

For any abelian group $A$, we have $L_{0}A=0$ iff $A_{p}=0$ iff $A/p=0$.

###### Proof.

Proposition 12.23 shows that $A_{p}$ is a quotient of $L_{0}A$, and
Proposition 12.5 shows that $A/p$ is a quotient of $A_{p}$. Conversely, if
$A/p=0$ then $p.1_{A}$ is surjective, so $A/p^{k}=0$ for all $k$ and the maps
in the tower $\\{A[p^{k}]\\}$ are all surjective. It follows that
$A_{p}=\operatornamewithlimits{\underset{\longleftarrow}{lim}}_{k}A/p^{k}=0$
and (using Proposition 11.17) that
$\operatornamewithlimits{\underset{\longleftarrow}{lim}}^{1}_{k}A[p^{k}]=0$,
so the short exact sequence in Proposition 12.23 shows that $L_{0}A=0$. ∎

We next explain a more traditional construction of the functors $L_{0}$ and
$L_{1}$. This involves a group known as $\mathbb{Z}/p^{\infty}$.

######  Definition 12.30.

Define $f_{k}\colon\mathbb{Z}/p^{k}\to\mathbb{Z}/p^{k+1}$ by
$f_{k}(a+p^{k}\mathbb{Z})=pa+p^{k+1}\mathbb{Z}$, so we have a sequence

${0=\mathbb{Z}/p^{0}}$${\mathbb{Z}/p}$${\mathbb{Z}/p^{2}}$${\mathbb{Z}/p^{3}}$${\mathbb{Z}/p^{4}}$${\dotsb}$$\scriptstyle{f_{0}}$$\scriptstyle{f_{1}}$$\scriptstyle{f_{2}}$$\scriptstyle{f_{3}}$$\scriptstyle{f_{4}}$

We define $\mathbb{Z}/p^{\infty}$ to be the colimit of this sequence.

######  Proposition 12.31.

There are canonical isomorphisms

$\mathbb{Z}/p^{\infty}=\mathbb{Z}[1/p]/\mathbb{Z}=\operatorname{tors}_{p}({\mathbb{Q}/\mathbb{Z}})=({\mathbb{Q}/\mathbb{Z}})_{(p)}={\mathbb{Q}}/\mathbb{Z}_{(p)}.$

###### Proof.

First, consider the diagram

${\mathbb{Z}}$${\mathbb{Z}}$${\mathbb{Z}}$${\mathbb{Z}}$${\mathbb{Z}}$${\dotsb}$${\mathbb{Z}}$${\mathbb{Z}}$${\mathbb{Z}}$${\mathbb{Z}}$${\mathbb{Z}}$${\dotsb}$${\mathbb{Z}/p^{0}}$${\mathbb{Z}/p^{1}}$${\mathbb{Z}/p^{2}}$${\mathbb{Z}/p^{3}}$${\mathbb{Z}/p^{4}}$${\dotsb}$$\scriptstyle{1}$$\scriptstyle{p}$$\scriptstyle{p^{2}}$$\scriptstyle{p^{3}}$$\scriptstyle{p^{4}}$$\scriptstyle{p}$$\scriptstyle{p}$$\scriptstyle{p}$$\scriptstyle{p}$$\scriptstyle{p}$$\scriptstyle{f_{0}}$$\scriptstyle{f_{1}}$$\scriptstyle{f_{2}}$$\scriptstyle{f_{3}}$$\scriptstyle{f_{4}}$

Using Propositions 10.10 and 10.8 we obtain a short exact sequence
$\mathbb{Z}\to\mathbb{Z}[1/p]\to\mathbb{Z}/p^{\infty}$, so
$\mathbb{Z}/p^{\infty}=\mathbb{Z}[1/p]/\mathbb{Z}$. Next, for
$a\in{\mathbb{Q}}$ we note that $a+\mathbb{Z}$ is a $p$-torsion element in
${\mathbb{Q}/\mathbb{Z}}$ iff $p^{k}a\in\mathbb{Z}$ for some $k$, iff
$a\in\mathbb{Z}[1/p]$. It follows that
$\operatorname{tors}_{p}({\mathbb{Q}/\mathbb{Z}})=\mathbb{Z}[1/p]/\mathbb{Z}$.
We also know from Proposition 9.17 that
$\operatorname{tors}_{p}({\mathbb{Q}/\mathbb{Z}})=({\mathbb{Q}/\mathbb{Z}})_{(p)}$.
It is clear that ${\mathbb{Q}}$ is $p$-local, so
${\mathbb{Q}}_{(p)}={\mathbb{Q}}$. We can thus apply Proposition 9.10 to the
sequence
$\mathbb{Z}\rightarrowtail{\mathbb{Q}}\twoheadrightarrow{\mathbb{Q}/\mathbb{Z}}$
to see that $({\mathbb{Q}/\mathbb{Z}})_{(p)}={\mathbb{Q}}/\mathbb{Z}_{(p)}$. ∎

######  Proposition 12.32.

There are natural isomorphisms
$L_{0}A=\operatorname{Ext}(\mathbb{Z}/p^{\infty},A)$ and
$L_{1}A=\operatorname{Hom}(\mathbb{Z}/p^{\infty},A)$.

###### Proof.

In this proof we will identify $\mathbb{Z}/p^{\infty}$ with
$\mathbb{Z}[{\textstyle\frac{1}{p}}]/\mathbb{Z}$. Put
$F=\bigoplus_{i=0}^{\infty}\mathbb{Z}$, and let $e_{i}$ be the $i$’th basis
vector in $F$. Define maps

$F\xrightarrow{\phi}F\xrightarrow{\psi}\mathbb{Z}/p^{\infty}$

by

$\phi(e_{i})=e_{i-1}-pe_{i}\hskip 60.00009pt\psi(e_{i})=p^{-i-1}+\mathbb{Z}$

(where $e_{-1}$ is interpreted as $0$), or equivalently

$\displaystyle\phi(n_{0},n_{1},n_{2},\dotsc)$
$\displaystyle=(pn_{0}-n_{1},pn_{1}-n_{2},pn_{2}-n_{3},\dotsc)$
$\displaystyle\phi(m_{0},m_{1},m_{2},\dotsc)$
$\displaystyle=\sum_{i}m_{i}p^{-i-1}+\mathbb{Z}.$

By considering the first nonzero entry in $n=(n_{0},n_{1},\dotsc)$, we see
that $\phi$ is injective. Any element of $\mathbb{Z}/p^{\infty}$ can be
written as $k/p^{i+1}+\mathbb{Z}$ for some $i\geq 0$ and $k\in\mathbb{Z}$, and
this is the same as $\psi(ke_{i})$, so $\psi$ is surjective. It is clear from
the definitions that $\psi\phi=0$, so
$\operatorname{img}(\phi)\leq\ker(\psi)$. Conversely, suppose we have $n\in F$
with $\psi(n)=0$, so the number $q=\sum_{i}n_{i}p^{-1-i}$ actually lies in
$\mathbb{Z}$. Put $m_{i}=\sum_{j<i}n_{j}p^{i-j-1}-p^{i}q$, and note that
$m_{i}=0$ for $i\gg 0$, so $m\in F$. We find that $\phi(m)=n$, so the sequence
$F\xrightarrow{\phi}F\xrightarrow{\psi}\mathbb{Z}/p^{\infty}$ is short exact.
As $F$ is free, this gives us an exact sequence

${\operatorname{Hom}(\mathbb{Z}/p^{\infty},A)}$${\operatorname{Hom}(F,A)}$${\operatorname{Hom}(F,A)}$${\operatorname{Ext}(\mathbb{Z}/p^{\infty},A).}$$\scriptstyle{\psi^{*}}$$\scriptstyle{\phi^{*}}$

Next, for any series $v(x)=\sum_{i}a_{i}x^{i}\in A[\\![x]\\!]$ we have a
homomorphism $\alpha(v(x))\colon F\to A$ given by $\alpha(v(x))(e_{i})=a_{i}$
for all $i\geq 0$. This construction gives an isomorphism
$A[\\![x]\\!]\to\operatorname{Hom}(F,A)$. If we make the conventions
$a_{-1}=0$ and $e_{-1}=0$ we also have

$\alpha((x-p)v(x))(e_{j})=\alpha\left(\sum_{i}(a_{i-1}-pa_{i})x^{i}\right)(e_{j})=a_{j-1}-pa_{j}=\alpha(v(x))(e_{j-1}-pe_{j})=\alpha(v(x))(\phi(e_{j}))$

Thus, multiplication by $x-p$ on $A[\\![x]\\!]$ corresponds to $\phi^{*}$ on
$\operatorname{Hom}(F,A)$, and the claim follows from this. ∎

## References