# Algebraic theory of abelian groups

N. P. Strickland

###### Contents

  1. 1 Introduction
  2. 2 Exactness and splittings
  3. 3 Products and coproducts
  4. 4 Torsion groups
  5. 5 Finitely generated abelian groups
  6. 6 Free abelian groups and their subgroups
  7. 7 Tensor and torsion products
  8. 8 Ext groups
  9. 9 Localisation
  10. 10 Colimits of sequences
  11. 11 Limits and derived limits of towers
  12. 12 Completion and derived completion

##  1\. Introduction

This document aims to give a self-contained account of the parts of abelian
group theory that are most relevant for algebraic topology. It is almost
purely expository, although there are some slightly unusual features in the
treatment of tensor products, torsion products and $\operatorname{Ext}$
groups. The book [boka:hlc] is a good reference for Sections 11 and 12.
Earlier sections are more standard and can be found in very many sources.

##  2\. Exactness and splittings

######  Definition 2.1.

Consider a sequence
$A_{0}\xrightarrow{f_{0}}A_{1}\xrightarrow{}\dotsb\xrightarrow{f_{r-1}}A_{r}$
of abelian groups and homomorphisms. We say that the sequence is _exact_ at
$A_{i}$ if $\operatorname{image}(f_{i-1})=\ker(f_{i})\leq A_{i}$ (which
implies that $f_{i}\circ f_{i-1}=0$). We say that the whole sequence is exact
if it is exact at $A_{i}$ for $0<i<r$.

Next, we say that a sequence $A\xrightarrow{f}B\xrightarrow{g}C$ is _short
exact_ if it is exact, and also $f$ is injective and $g$ is surjective.

######  Remark 2.2.

One can easily check the following facts.

  * (a)

A sequence $A\xrightarrow{f}B\xrightarrow{0}C$ is exact iff $f$ is surjective.
In particular, a sequence $A\xrightarrow{f}B\xrightarrow{}0$ is exact iff $f$
is surjective.

  * (b)

A sequence $A\xrightarrow{0}B\xrightarrow{g}C$ is exact iff $g$ is injective.
In particular, a sequence $0\xrightarrow{}B\xrightarrow{g}C$ is exact iff $g$
is injective.

  * (c)

A sequence $A\xrightarrow{0}B\xrightarrow{g}C\xrightarrow{0}D$ is exact iff
$g$ is an isomorphism.

  * (d)

A sequence $0\xrightarrow{}A\xrightarrow{f}B\xrightarrow{g}C\xrightarrow{}0$
is exact iff $A\xrightarrow{f}B\xrightarrow{g}C$ is short exact.

  * (e)

Suppose we have an exact sequence

$A\xrightarrow{f}B\xrightarrow{g}C\xrightarrow{h}D\xrightarrow{k}E.$

Then $g$ induces a map from $\operatorname{cok}(f)=B/f(A)$ to $C$, and $h$ can
be regarded as a map from $C$ to $\ker(k)$, and the resulting sequence

$\operatorname{cok}(f)\xrightarrow{g}C\xrightarrow{h}\ker(k)$

is short exact.

  * (f)

If $A\xrightarrow{f}B\xrightarrow{g}C$ is short exact, then $f$ induces an
isomorphism $A\to f(A)$ and $g$ induces an isomorphism $B/f(A)\to C$. Thus, if
$A$, $B$ and $C$ are finite we have $|B|=|f(A)|.|B/f(A)|=|A||C|$. Similarly,
if $A$ and $C$ are free abelian groups of ranks $n$ and $m$, then $B$ is a
free abelian group of rank $n+m$.

######  Proposition 2.3 (The five lemma).

Suppose we have a commutative diagram as follows, in which the rows are exact,
and $p_{0}$, $p_{1}$, $p_{3}$ and $p_{4}$ are isomorphisms:

${A_{0}}$${A_{1}}$${A_{2}}$${A_{3}}$${A_{4}}$${B_{0}}$${B_{1}}$${B_{2}}$${B_{3}}$${B_{4}.}$$\scriptstyle{f_{0}}$$\scriptstyle{p_{0}}$$\scriptstyle{\simeq}$$\scriptstyle{f_{1}}$$\scriptstyle{p_{1}}$$\scriptstyle{\simeq}$$\scriptstyle{f_{2}}$$\scriptstyle{p_{2}}$$\scriptstyle{f_{3}}$$\scriptstyle{p_{3}}$$\scriptstyle{\simeq}$$\scriptstyle{p_{4}}$$\scriptstyle{\simeq}$$\scriptstyle{g_{0}}$$\scriptstyle{g_{1}}$$\scriptstyle{g_{2}}$$\scriptstyle{g_{3}}$

Then $p_{2}$ is also an isomorphism.

###### Proof.

First suppose that $a_{2}\in A_{2}$ and $p_{2}(a_{2})=0$. It follows that
$p_{3}f_{2}(a_{2})=g_{2}p_{2}(a_{2})=g_{2}(0)=0$, but $p_{3}$ is an
isomorphism, so $f_{2}(a_{2})=0$, so $a_{2}\in\ker(f_{2})$. The top row is
exact, so $\ker(f_{2})=\operatorname{image}(f_{1})$, so we can choose
$a_{1}\in A_{1}$ with $f_{1}(a_{1})=a_{2}$. Put $b_{1}=p_{1}(a_{1})\in B_{1}$.
We then have
$g_{1}(b_{1})=g_{1}p_{1}(a_{1})=p_{2}f_{1}(a_{1})=p_{2}(a_{2})=0$, so
$b_{1}\in\ker(g_{1})$. The bottom row is exact, so
$\ker(g_{1})=\operatorname{image}(g_{0})$, so we can choose $b_{0}\in B_{0}$
with $g_{0}(b_{0})=b_{1}$. As $p_{0}$ is an isomorphism, we can now put
$a_{0}=p_{0}^{-1}(b_{0})\in A_{0}$. We then have
$p_{1}f_{0}(a_{0})=g_{0}p_{0}(a_{0})=g_{0}(b_{0})=b_{1}=p_{1}(a_{1})$. Here
$p_{1}$ is an isomorphism, so it follows that $f_{0}(a_{0})=a_{1}$. We now
have $a_{2}=f_{1}(a_{1})=f_{1}f_{0}(a_{0})$. However, as the top row is exact
we have $f_{1}f_{0}=0$, so $a_{2}=0$. We conclude that $p_{2}$ is injective.

Now suppose instead that we start with an element $b_{2}\in B_{2}$. Put
$b_{3}=g_{2}(b_{2})\in B_{3}$ and $a_{3}=p_{3}^{-1}(b_{3})\in A_{3}$. We then
have $p_{4}f_{3}(a_{3})=g_{3}p_{3}(a_{3})=g_{3}(b_{3})=g_{3}g_{2}(b_{2})=0$
(because $g_{3}g_{2}=0$). As $p_{4}$ is an isomorphism, this means that
$f_{3}(a_{3})=0$, so $a_{3}\in\ker(f_{3})$. As the top row is exact we have
$\ker(f_{3})=\operatorname{image}(f_{2})$, so we can choose $a_{2}\in A_{2}$
with $f_{3}(a_{2})=a_{3}$. Put $b^{\prime}_{2}=b_{2}-p_{2}(a_{2})\in B_{2}$.
We have
$g_{2}(b^{\prime}_{2})=g_{2}(b_{2})-g_{2}p_{2}(a_{2})=b_{3}-p_{3}f_{2}(a_{2})=b_{3}-p_{3}(a_{3})=0$,
so $b^{\prime}_{2}\in\ker(g_{2})=\operatorname{image}(g_{1})$. We can thus
choose $b^{\prime}_{1}\in B_{1}$ with $g_{1}(b^{\prime}_{1})=b^{\prime}_{2}$.
Now put $a^{\prime}_{1}=p_{1}^{-1}(b^{\prime}_{1})\in A_{1}$ and
$a^{\prime}_{2}=f_{1}(a^{\prime}_{1})\in A_{2}$. We find that
$p_{2}(a^{\prime}_{2})=p_{2}f_{1}(a^{\prime}_{1})=g_{1}p_{1}(a^{\prime}_{1})=g_{1}(b^{\prime}_{1})=b^{\prime}_{2}=b_{2}-p_{2}(a_{2})$,
so $p_{2}(a_{2}+a^{\prime}_{2})=b_{2}$. This shows that $p_{2}$ is also
surjective, and so is an isomorphism as claimed. ∎

######  Proposition 2.4.

Suppose we have a commutative diagram as follows, in which the rows are short
exact sequences:

${A}$${B}$${C}$${A^{\prime}}$${B^{\prime}}$${C^{\prime}}$$\scriptstyle{j}$$\scriptstyle{f}$$\scriptstyle{q}$$\scriptstyle{g}$$\scriptstyle{h}$$\scriptstyle{j^{\prime}}$$\scriptstyle{q^{\prime}}$

Then there is a unique homomorphism
$\delta\colon\ker(h)\to\operatorname{cok}(f)$ such that
$\delta(q(b))=a^{\prime}+f(A)$ whenever $g(b)=j^{\prime}(a^{\prime})$.
Moreover, this fits into an exact sequence

$0\to\ker(f)\xrightarrow{j}\ker(g)\xrightarrow{q}\ker(h)\xrightarrow{\delta}\operatorname{cok}(f)\xrightarrow{j}\operatorname{cok}(g)\xrightarrow{q}\operatorname{cok}(h)\to
0.$

###### Proof.

A _snake_ for the above diagram is a list $(c,b,a^{\prime},\overline{a})$ such
that

  * (1)

$c\in\ker(h)\leq C$

  * (2)

$b\in B$ with $qb=c$

  * (3)

$a^{\prime}\in A^{\prime}$ with $j^{\prime}a^{\prime}=gb\in B^{\prime}$

  * (4)

$\overline{a}$ is the image of $a^{\prime}$ in $\operatorname{cok}(f)$.

It is easy to see that the snakes form a subgroup of $\ker(h)\times B\times
A^{\prime}\times\operatorname{cok}(f)$. We claim that for all $c\in\ker(h)$,
there exists a snake starting with $c$. Indeed, as $q$ is surjective, we can
choose $b\in B$ satisfying (2). Then $q^{\prime}g(b)=hq(b)=h(c)=0$, so
$g(b)\in\ker(q^{\prime})=\operatorname{img}(j^{\prime})$, so we can choose
$a^{\prime}\in A^{\prime}$ satisfying (3). Finally, we can define
$\overline{a}$ to be the image of $a^{\prime}$ in $\operatorname{cok}(f)$, so
that (4) is satisfied: this gives a snake as required. Next, we claim that any
two snakes starting with $c$ have the same endpoint. By subtraction we reduce
to the following claim: if $(0,b,a^{\prime},\overline{a})$ is a snake, then
$\overline{a}=0$, or equivalently $a^{\prime}\in\operatorname{img}(f)$.
Indeed, condition (2) says that $b\in\ker(q)=\operatorname{img}(j)$, so we can
find $a\in A$ with $b=ja$. Now $j^{\prime}(fa-a^{\prime})=j^{\prime}fa-gb=gja-
gb=gb-gb=0$, and $j^{\prime}$ is injective, so $fa=a^{\prime}$ as required.
This allows us to construct a map
$\delta\colon\ker(h)\to\operatorname{cok}(f)$ as follows: we define
$\delta(c)$ to be the endpoint of any snake starting with $c$.

We now need to check exactness of the resulting sequence.

  * (1)

As $j\colon A\to B$ is injective, it is clear that the restricted map
$j\colon\ker(f)\to\ker(g)$ is also injective.

  * (2)

As the composite $A\xrightarrow{j}B\xrightarrow{q}C$ is zero, the same is true
of the restricted composite
$\ker(f)\xrightarrow{j}\ker(g)\xrightarrow{q}\ker(h)$. Moreover, suppose we
have $b\in\ker(g)$ with $qb=0$. By the original exactness assumption we can
find $a\in A$ with $ja=b$. Now $j^{\prime}fa=gja=gb=0$ but $j^{\prime}$ is
injective so $fa=0$ so $a\in\ker(f)$. Thus, $b$ is in the image of the map
$j\colon\ker(f)\to\ker(g)$.

  * (3)

Suppose we have $b\in\ker(g)$. Then $(qb,b,0,0)$ is a snake starting with
$qb$, showing that $\delta qb=0$. Conversely, suppose that $c\in\ker(h)$ with
$\delta c=0$, so there exists a snake $(c,b,a^{\prime},0)$. By the last snake
condition, we must have $a^{\prime}\in\operatorname{img}(f)$, say
$a^{\prime}=fa$ for some $a\in A$. Put $b^{\prime}=b-ja\in B$. Snake condition
(2) gives $qb=c$ but also $qj=0$ so $qb^{\prime}=c$. On the other hand, snake
condition (3) gives $gb=j^{\prime}a^{\prime}=j^{\prime}fa=gja$ so
$gb^{\prime}=0$. This means that $c$ is in the image of the map
$q\colon\ker(g)\to\ker(h)$.

  * (4)

Suppose we have $c\in\ker(h)$ with $\delta(c)=\overline{a}$. This means that
there is a snake $(c,b,a^{\prime},\overline{a})$. We claim that the induced
map $j^{\prime}\colon\operatorname{cok}(f)\to\operatorname{cok}(g)$ sends
$\overline{a}$ to $0$, or equivalently that
$j^{\prime}a^{\prime}\in\operatorname{img}(g)$. This is clear because
$j^{\prime}a^{\prime}=gb$ by the snake axioms. Conversely, suppose that
$\overline{a}\in\operatorname{cok}(f)$ and that $\overline{a}$ maps to $0$ in
$\operatorname{cok}(g)$. This means that we can find $a^{\prime}\in A$
representing $\overline{a}$ and that $ja^{\prime}$ lies in the image of $g$,
say $ja^{\prime}=gb$ for some $b\in B$. If we put $c=qb\in C$ we find that
$hc=hqb=q^{\prime}gb=q^{\prime}j^{\prime}a^{\prime}=0$, so $c\in\ker(h)$. By
construction we see that $(c,b,a^{\prime},\overline{a})$ is a snake so
$\overline{a}\in\operatorname{img}(\delta)$.

  * (5)

As the composite
$A^{\prime}\xrightarrow{j^{\prime}}B^{\prime}\xrightarrow{q^{\prime}}C^{\prime}$
is zero, the same is clearly true for the induced maps
$\operatorname{cok}(f)\to\operatorname{cok}(g)\to\operatorname{cok}(h)$.
Conversely, suppose we have an element $\overline{b}\in\operatorname{cok}(g)$
that maps to zero in $\operatorname{cok}(h)$. We can choose $b^{\prime}\in
B^{\prime}$ representing $\overline{b}$, and then $q^{\prime}b^{\prime}$ must
lie in $\operatorname{img}(h)$, say $q^{\prime}b^{\prime}=hc$. As $q$ is
surjective we can choose $b\in B$ with $qb=c$. This gives
$q^{\prime}b^{\prime}=hqb=q^{\prime}gb$, so the element $b^{\prime}-gb$ lies
in $\ker(q^{\prime})$, which is the same as $\operatorname{img}(j^{\prime})$.
We can therefore choose $a^{\prime}\in A^{\prime}$ with
$b^{\prime}=gb+j^{\prime}a^{\prime}$. If we let $\overline{a}$ denote the
image of $a^{\prime}$ in $\operatorname{cok}(f)$, we find that
$\overline{b}=j^{\prime}\overline{a}$ in $\operatorname{cok}(g)$.

  * (6)

Finally, suppose we have $\overline{c}\in\operatorname{cok}(h)$. We can then
choose a representing element $c^{\prime}\in C^{\prime}$. As $q^{\prime}$ is
surjective we can choose $b^{\prime}\in B^{\prime}$ with
$q^{\prime}b^{\prime}=b$, then we can put
$\overline{b}=[b^{\prime}]\in\operatorname{cok}(g)$. We find that
$q^{\prime}\overline{b}=\overline{c}$. This shows that
$q^{\prime}\colon\operatorname{cok}(g)\to\operatorname{cok}(h)$ is surjective.

∎

######  Definition 2.5.

A _split short exact sequence_ is a diagram

${A}$${B}$${C}$$\scriptstyle{i}$$\scriptstyle{p}$$\scriptstyle{r}$$\scriptstyle{s}$

where

$pi=0\hskip 30.00005ptrs=0\hskip 30.00005ptri=1_{A}\hskip
30.00005ptps=1_{C}\hskip 30.00005ptir+sp=1_{B}.$

This can also be displayed as

${A}$${C}$${B}$${A}$${C}$$\scriptstyle{0}$$\scriptstyle{i}$$\scriptstyle{1}$$\scriptstyle{p}$$\scriptstyle{r}$$\scriptstyle{0}$$\scriptstyle{s}$$\scriptstyle{1}$

######  Example 2.6.

Given abelian groups $A$ and $C$, there is a split short exact sequence

${A}$${A\oplus
C}$${C}$$\scriptstyle{i^{\prime}}$$\scriptstyle{p^{\prime}}$$\scriptstyle{r^{\prime}}$$\scriptstyle{s^{\prime}}$

given by

$\displaystyle i^{\prime}(a)$ $\displaystyle=(a,0)$ $\displaystyle
p^{\prime}(a,c)$ $\displaystyle=c$ $\displaystyle s^{\prime}(c)$
$\displaystyle=(0,c)$ $\displaystyle r^{\prime}(a,c)$ $\displaystyle=a.$

The above example is essentially the only example, as we see from the
following result:

######  Proposition 2.7.

Suppose we have a split short exact sequence

${A}$${B}$${C}$$\scriptstyle{i}$$\scriptstyle{p}$$\scriptstyle{r}$$\scriptstyle{s}$

Then there is an isomorphism $f\colon B\to A\oplus C$ given by
$f(b)=(r(b),p(b))$ with inverse $f^{-1}(a,c)=i(a)+s(c)$. Moreover, the diagram

${A}$${B}$${C}$${A}$${A\oplus
C}$${C}$$\scriptstyle{i}$$\scriptstyle{p}$$\scriptstyle{r}$$\scriptstyle{f}$$\scriptstyle{\simeq}$$\scriptstyle{s}$$\scriptstyle{i^{\prime}}$$\scriptstyle{p^{\prime}}$$\scriptstyle{r^{\prime}}$$\scriptstyle{s^{\prime}}$

commutes in the sense that

$fi=i^{\prime}\hskip 40.00006ptfs=s^{\prime}\hskip
40.00006ptpf=p^{\prime}\hskip 40.00006ptrf=f^{\prime}.$

###### Proof.

We can certainly define homomorphisms $B\xrightarrow{f}A\oplus
C\xrightarrow{g}B$ by $f(b)=(r(b),p(b))$ and $g(a,c)=i(a)+s(c)$. We then have
$gf(b)=(ir+sp)(b)=b$ and $fg(a,c)=(ri(a)+rs(c),pi(a)+ps(c))=(a,c)$ so $f$ and
$g$ are mutually inverse isomorphisms. We also have
$fi(a)=(ri(a),pi(a))=(a,0)=i^{\prime}(a)$, and the equations $fs=s^{\prime}$,
$pf=p^{\prime}$ and $rf=f^{\prime}$ can be verified equally easily. ∎

Our terminology is justified by the following observation:

######  Lemma 2.8.

If

${A}$${B}$${C}$$\scriptstyle{i}$$\scriptstyle{p}$$\scriptstyle{r}$$\scriptstyle{s}$

is a split short exact sequence, then

$A\xrightarrow{i}B\xrightarrow{p}C$

is a short exact sequence.

###### Proof.

Suppose that $i(a)=0$. As $ri=1_{A}$ we have $a=r(i(a))=r(0)=0$. This shows
that $\ker(i)=0$, so $i$ is injective. Next, we have $ps=1_{C}$, so for all
$c\in C$ we have $c=p(s(c))\in\operatorname{image}(p)$; so $p$ is surjective.
We also have $pi=0$, so $\operatorname{image}(i)\leq\ker(p)$. Finally, we have
$ir+sp=1_{B}$, so for $b\in B$ we have $b=i(r(b))+s(p(b))$. If $b\in\ker(p)$
this reduces to $b=i(r(b))\in\operatorname{image}(i)$, so
$\ker(p)\leq\operatorname{image}(i)$ as required. ∎

######  Proposition 2.9.

Let $A\xrightarrow{i}B\xrightarrow{p}C$ be a short exact sequence.

  * (a)

For any map $r\colon B\to A$ with $ri=1_{A}$, there is a unique map $s\colon
C\to B$ such that $(i,p,r,s)$ gives a split short exact sequence.

  * (b)

For any map $s\colon C\to B$ with $ps=1_{C}$, there is a unique map $r\colon
B\to A$ such that $(i,p,r,s)$ gives a split short exact sequence.

###### Proof.

We will prove (a) and leave the similar proof of (b) to the reader. Define
$f=1-ir\colon B\to B$. As $ri=1$ we have $fi=i-i(ri)=0$, so $f$ vanishes on
$\operatorname{image}(i)$, which is the same as $\ker(p)$. We therefore have a
well-defined map $s\colon C\to B$ given by $s(c)=f(b)$ for any $b$ with
$p(b)=c$. This means that $sp=f=1-ir$, or in other words $1_{B}=ir+sp$. We
also have $pi=0$ so $psp=p(1-ir)=p$, so $(ps-1)p=0$. As $p$ is surjective this
implies that $ps-1=0$ or $ps=1_{C}$. Finally, we have $ri=1$ so
$rsp=r(1-ir)=r-(ri)r=0$ but $p$ is surjective so $rs=0$. Thus, all the
conditions for a split short exact sequence are verified. If $s^{\prime}\colon
C\to B$ is another map giving a split short exact sequence then we can
subtract the equations $ir+sp=1$ and $ir+s^{\prime}p=1$ to get
$(s-s^{\prime})p=0$ but $p$ is surjective so $s=s^{\prime}$; this shows that
$s$ is unique. ∎

######  Proposition 2.10.

Let $B$ be an abelian group, and let $A$ and $C$ be subgroups such that
$B=A+C$ and $A\cap C=0$. Let $i\colon A\to B$ and $s\colon C\to B$ be the
inclusion maps. The there is a unique pair of homomorphisms
$A\xleftarrow{r}B\xrightarrow{p}C$ giving a split short exact sequence.

###### Proof.

Consider an element $b\in B$. As $B=A+C$ we can find $(a,c)\in A\oplus C$ such
that $b=a+c$. Suppose we have another pair $(a^{\prime},c^{\prime})\in A\oplus
C$ with $b=a^{\prime}+c^{\prime}$. We put $x=a-a^{\prime}$, and by rearranging
the equation $a+c=a^{\prime}+c^{\prime}$ we see that $x=c^{\prime}-c$. The
first of these expressions shows that $x\in A$, and the second that $x\in C$.
As $A\cap C=0$ this means that $x=0$, so $a=a^{\prime}$ and $c=c^{\prime}$.
Thus, the pair $(a,c)$ is unique, so we can define maps
$A\xleftarrow{r}B\xrightarrow{p}C$ by $r(b)=a$ and $p(b)=c$. It is
straightforward to check that these give a split short exact sequence. ∎

######  Proposition 2.11.

Let $B$ be an abelian group, and let $e\colon B\to B$ be a homomorphism with
$e^{2}=e$. Then $\operatorname{image}(e)=\ker(1-e)$ and
$\ker(e)=\operatorname{image}(1-e)$ and
$B=\operatorname{image}(e)\oplus\operatorname{image}(1-e)$.

###### Proof.

First, if $b\in\operatorname{image}(e)$ then $b=e(a)$ for some $a$, so
$(1-e)(b)=e(a)-e^{2}(a)=0$, so $b\in\ker(1-e)$. Conversely, if $b\in\ker(1-e)$
then $b-e(b)=0$ so $b=e(b)\in\operatorname{image}(e)$. This shows that
$\operatorname{image}(e)=\ker(1-e)$ as claimed. Now put $f=1-e$. We then have
$f^{2}=1-2e+e^{2}=1-2e+e=f$, so $f$ is another idempotent endomorphism of $B$.
We can thus apply the same logic to see that
$\operatorname{image}(f)=\ker(1-f)$, or in other words
$\operatorname{image}(1-e)=\ker(e)$.

Now consider an arbitary element $b\in B$. We can write $b$ as
$e(b)+(1-e)(b)$, so $b\in\operatorname{image}(e)+\operatorname{image}(1-e)$;
this shows that $B=\operatorname{image}(e)+\operatorname{image}(1-e)$. Now
suppose that
$b\in\operatorname{image}(e)\cap\operatorname{image}(1-e)=\ker(1-e)\cap\ker(e)$.
This means that $(1-e)(b)=e(b)=0$, so $b=e(b)=0$. This means that
$\operatorname{image}(e)\cap\operatorname{image}(1-e)=0$, so the sum is
direct. ∎

We now give a useful application of Proposition 2.7 to the theory of additive
functors. We recall the definition:

######  Definition 2.12.

A _covariant functor_ from abelian groups to abelian groups is a construction
that gives an abelian group $F(A)$ for each abelian group $A$, and a
homomorphism $f_{*}\colon F(A)\to F(B)$ for each homomorphism $f\colon A\to
B$, in such a way that:

  * (a)

For identity maps we have $(1_{A})_{*}=1_{F(A)}$ for all $A$.

  * (b)

For homomorphisms $A\xrightarrow{f}B\xrightarrow{g}C$ we have
$(gf)_{*}=g_{*}f_{*}\colon F(A)\to F(C)$.

We say that $F$ is a _additive_ if $(f_{0}+f_{1})_{*}=(f_{0})_{*}+(f_{1})_{*}$
for all $f_{0},f_{1}\colon A\to B$.

######  Example 2.13.

Fix an integer $n>0$. We can then define an additive functor $F$ by
$F(A)=A[n]=\\{a\in A\;|\;na=0\\}$, and another additive functor $G$ by
$G(A)=A/nA$. In both cases the homomorphisms $f_{*}$ are just the obvious ones
induced by $f$.

######  Proposition 2.14.

Let $F$ be an additive covariant functor as above. Then for any abelian groups
$A$ and $C$ we have an natural isomorphism $f\colon F(A\oplus C)\to F(A)\oplus
F(C)$ given by $f(b)=(r^{\prime}_{*}(b),p^{\prime}_{*}(b))$ with inverse
$f^{-1}(a,c)=i^{\prime}_{*}(a)+s^{\prime}_{*}(c)$.

###### Proof.

Put $B=A\oplus C$, and recall that
$1_{B}=i^{\prime}r^{\prime}+s^{\prime}p^{\prime}$. As $F$ is an additive
functor we have

$1_{F(B)}=(1_{B})_{*}=(i^{\prime}r^{\prime})_{*}+(s^{\prime}p^{\prime})_{*}=i^{\prime}_{*}r^{\prime}_{*}+s^{\prime}_{*}p^{\prime}_{*}.$

In the same way the equations $r^{\prime}i^{\prime}=1$,
$p^{\prime}s^{\prime}=1$, $p^{\prime}i^{\prime}=0$ and
$r^{\prime}s^{\prime}=0$ give $r^{\prime}_{*}i^{\prime}_{*}=1$,
$p^{\prime}_{*}s^{\prime}_{*}=1$, $p^{\prime}_{*}i^{\prime}_{*}=0$ and
$r^{\prime}_{*}s^{\prime}_{*}=0$, so we have a split short exact sequence

${F(A)}$${F(B)}$${F(C)}$$\scriptstyle{i^{\prime}_{*}}$$\scriptstyle{p^{\prime}_{*}}$$\scriptstyle{r^{\prime}_{*}}$$\scriptstyle{s^{\prime}_{*}}$

Thus, Proposition 2.7 gives us an isomorphism $F(A\oplus C)=F(B)\to F(A)\oplus
F(C)$, and by unwinding the definitions we see that this is given by the
stated formulae. ∎

There is a similar statement for contravariant functors as follows.

######  Definition 2.15.

A _contravariant functor_ from abelian groups to abelian groups is a
construction that gives an abelian group $F(A)$ for each abelian group $A$,
and a homomorphism $f^{*}\colon F(B)\to F(A)$ for each homomorphism $f\colon
A\to B$, in such a way that:

  * (a)

For identity maps we have $(1_{A})_{*}=1_{F(A)}$ for all $A$.

  * (b)

For homomorphisms $A\xrightarrow{f}B\xrightarrow{g}C$ we have
$(gf)^{*}=f^{*}g^{*}\colon F(C)\to F(A)$.

We say that $F$ is _additive_ if $(f_{0}+f_{1})^{*}=f_{0}^{*}+f_{1}^{*}$ for
all $f_{0},f_{1}\colon A\to B$.

######  Proposition 2.16.

Let $F$ be an additive contravariant functor as above. Then for any abelian
groups $A$ and $C$ we have an natural isomorphism $f\colon F(A\oplus C)\to
F(A)\oplus F(C)$ given by $f(b)=((i^{\prime})^{*}(b),(s^{\prime})^{*}(b))$
with inverse $f^{-1}(a,c)=(r^{\prime})^{*}(a)+(p^{\prime})^{*}(c)$.

###### Proof.

Essentially the same as Proposition 2.14. ∎

##  3\. Products and coproducts

If we have a finite list of abelian groups $A_{1},\dotsc,A_{n}$, we can form
the product group $\prod_{i=1}^{n}A_{i}=A_{1}\times\dotsb\times A_{n}$, which
is also denoted by $\bigoplus_{i=1}^{n}A_{i}=A_{1}\oplus\dotsb\oplus A_{n}$.
This should be familiar. These constructions can be generalised to cover
families of abelian groups $A_{i}$ indexed by a set $I$ that may be infinite,
and need not be ordered in any natural way. This is a little more subtle, and
in particular $\bigoplus_{i}A_{i}$ is not the same as $\prod_{i}A_{i}$ in this
context. In this section we will briefly outline the relevant definitions and
properties.

######  Definition 3.1.

Let $I$ be a set, and let $(A_{i})_{i\in I}$ be a family of abelian groups
indexed by $I$. The _product group_ $\prod_{i\in I}A_{i}$ is the set of all
systems $a=(a_{i})_{i\in I}$ consisting of an element $a_{i}\in A_{i}$ for
each $i\in I$. We make this into an abelian group by the obvious rule

$(a_{i})_{i\in I}\pm(b_{i})_{i\in I}=(a_{i}\pm b_{i})_{i\in I}.$

For each $k\in I$ we define $\pi_{k}\colon\prod_{i\in I}A_{i}\to A_{k}$ by
$\pi_{k}((a_{i})_{i\in I})=a_{k}$. This is clearly a homomorphism. We also
define $\iota_{k}\colon A_{k}\to\prod_{i\in I}A_{i}$ by

$\iota_{k}(a)_{i}=\begin{cases}a\in A_{k}&\text{ if }i=k\\\ 0\in A_{i}&\text{
if }i\neq k.\end{cases}$

######  Example 3.2.

If $I=\\{1,2,\dotsc,n\\}$, then $\prod_{i\in I}A_{i}$ is just the set of
$n$-tuples $(a_{1},\dotsc,a_{n})$ with $a_{i}\in A_{i}$, as before.

######  Example 3.3.

Suppose we have a fixed group $U$, and we take $A_{i}=U$ for all $i$. Then
$\prod_{i\in I}A_{i}$ is just the set $\operatorname{Map}(I,U)$ of all
functions from $I$ to $U$, considered as a group under pointwise addition.

######  Remark 3.4.

It is easy to see that a homomorphism $f\colon U\to\prod_{i\in I}A_{i}$ is
essentially the same thing as a family of homomorphisms $f_{i}\colon U\to
A_{i}$, one for each $i\in I$. Indeed, given such a family we define $f\colon
U\to\prod_{i\in I}A_{i}$ by $f(u)=(f_{i}(u))_{i\in I}$, and we can then
recover the original homomorphisms $f_{i}$ as the composites $\pi_{i}\circ f$.
This means that $\prod_{i\in I}A_{i}$ is a product for the groups $A_{i}$ in
the general sense considered in category theory.

######  Definition 3.5.

Given an element $a=(a_{i})_{i\in I}\in\prod_{i\in I}A_{i}$, the _support_ of
$a$ is the set

$\operatorname{supp}(a)=\\{i\in I\;|\;a_{i}\neq 0\\}\subseteq I.$

We put

$\bigoplus_{i\in I}A_{i}=\\{a\in\prod_{i\in
I}A_{i}\;|\;\operatorname{supp}(a)\text{ is a finite set }\\}.$

It is easy to see that $\operatorname{supp}(a\pm
b)\subseteq\operatorname{supp}(a)\cup\operatorname{supp}(b)$, and thus that
$\bigoplus_{i\in I}A_{i}$ is a subgroup of $\prod_{i\in I}A_{i}$. We call it
the _coproduct_ of the family $(A_{i})_{i\in I}$. We also note that
$\operatorname{supp}(\iota_{k}(a))\subseteq\\{k\\}$, so $\iota_{k}$ can be
regarded as a homomorphism $A_{k}\to\bigoplus_{i\in I}A_{i}$.

######  Remark 3.6.

If the index set $I$ is finite then all supports are automatically finite and
so the coproduct is the same as the product. In fact, we only need the set
$I^{\prime}=\\{i\;|\;A_{i}\neq 0\\}$ to be finite for this to hold.

Definition 3.5 is again compatible with the more general definition coming
from category theory, as we see from the following result:

######  Proposition 3.7.

Suppose we have an abelian group $V$, and a system of homomorphisms
$g_{i}\colon A_{i}\to V$ for all $i\in I$. Then there is a unique homomorphism
$g\colon\bigoplus_{i\in I}A_{i}\to V$ such that $g\circ\iota_{k}=g_{k}$ for
all $k\in I$.

###### Proof.

Given a point $a=(a_{i})_{i\in I}\in\bigoplus_{i\in I}A_{i}$, we define

$g(a)=\sum_{i\in\operatorname{supp}(a)}g_{i}(a_{i})\in V.$

The terms in the sum are meaningful because $a_{i}\in A_{i}$ and $g_{i}\colon
A_{i}\to V$, and $\operatorname{supp}(a)$ is finite so there only finitely
many terms so it is not a problem to add them up. If we replace
$\operatorname{supp}(a)$ by some larger finite set $J\subseteq I$ then this
gives us some extra terms but they are all zero so the sum is unchanged. After
taking $J=\operatorname{supp}(a)\cup\operatorname{supp}(b)$ it becomes easy to
see that $g(a+b)=g(a)+g(b)$, so $g$ is a homomorphism. Using
$\operatorname{supp}(\iota_{k}(a))\subseteq\\{k\\}$ we see that
$g\circ\iota_{k}=g_{k}$, as required. Let $g^{\prime}\colon\bigoplus_{i\in
I}A_{i}\to V$ be another homomorphism with $g^{\prime}\circ\iota_{k}=g_{k}$
for all $k$. If we have an element $a$ as before, we observe that
$a=\sum_{i\in\operatorname{supp}(a)}\iota_{i}(a_{i})$, and by applying
$g^{\prime}$ to this we get

$g^{\prime}(a)=\sum_{i\in\operatorname{supp}(a)}g^{\prime}(\iota_{i}(a_{i}))=\sum_{i\in\operatorname{supp}(a)}g_{i}(a_{i})=g(a),$

so $g$ is unique as claimed. ∎

######  Remark 3.8.

It would at worst be a tiny abuse of notation to say that $g(a)=\sum_{i\in
I}g_{i}(a_{i})$. This is a sum with infinitely many terms, which would not
normally be meaningful, but only finitely many of the terms are nonzero, so
the rest can be ignored.

######  Remark 3.9.

Suppose we have an abelian group $A$, and a family of subgroups $(A_{i})_{i\in
I}$. There is then a unique homomorphism $\sigma\colon\bigoplus_{i\in
I}A_{i}\to A$ such that $\sigma\circ\iota_{k}\colon A_{k}\to A$ is just the
inclusion for all $k$. More explicitly, we just have
$\sigma(a)=\sum_{i\in\operatorname{supp}(a)}a_{i}$. If this map $\sigma$ is an
isomorphism, we will say (with another slight abuse of notation) that
$A=\bigoplus_{i\in I}A_{i}$.

##  4\. Torsion groups

######  Definition 4.1.

Let $A$ be an abelian group.

  * (a)

We say that an element $a\in A$ is a _torsion element_ if $na=0$ for some
integer $n>0$.

  * (b)

We write $\operatorname{tors}(A)$ for the set of torsion elements of $A$. This
is easily seen to be a subgroup, because if $na=0$ and $mb=0$ then $nm(a\pm
b)=0$.

  * (c)

We say that $A$ is a _torsion group_ if every element is torsion, or
equivalently $\operatorname{tors}(A)=A$. At the other extreme, we say that $A$
is _torsion-free_ if $\operatorname{tors}(A)=0$.

  * (d)

Now fix a prime $p$. We say that $a$ is a _$p$ -torsion element_ if $p^{k}a=0$
for some $k\geq 0$. We write $\operatorname{tors}_{p}(A)$ for the set of
$p$-torsion elements, which is again a subgroup.

  * (e)

We say that $A$ is a _$p$ -torsion group_ if every element is $p$-torsion, or
equivalently $\operatorname{tors}_{p}(A)=A$. At the other extreme, we say that
$A$ is _$p$ -torsion free_ if $\operatorname{tors}_{p}(A)=0$.

######  Remark 4.2.

We write $n.1_{A}$ for the endomorphism of $A$ given by $a\mapsto na$. Then
$\operatorname{tors}(A)=\bigcup_{n>0}\ker(n.1_{A})$, and $A$ is torsion-free
if and only if the maps $n.1_{A}$ (for $n>0$) are all injective.

######  Example 4.3.

If $A$ is a finite abelian group with $|A|=n$ then Lagrange’s Theorem tells us
that $na=0$ for all $a\in A$, so $A$ is a torsion group. For another
instructive proof of the same fact, consider the element $z=\sum_{x\in A}x$.
As $x$ runs over $A$, the elements $a+x$ also run over $A$, so $z=\sum_{x\in
A}(a+x)=na+z$, so $na=0$. It is a curious fact, which we leave to the reader,
that $z$ itself is actually zero in all cases except when $|A|=2$.

######  Example 4.4.

It is clear that any free abelian group is torsion-free. The groups
${\mathbb{Q}}$ and $\mathbb{R}$ are torsion-free but not free.

######  Example 4.5.

Consider the quotient group $A={\mathbb{Q}/\mathbb{Z}}$. The subset

$A_{n}=\\{\mathbb{Z}=\tfrac{0}{n}+\mathbb{Z},\tfrac{1}{n}+\mathbb{Z},\dotsc,\tfrac{n-1}{n}+\mathbb{Z}\\}$

is a cyclic subgroup of order $n$. Any element $a\in{\mathbb{Q}/\mathbb{Z}}$
can be written as $a=m/n+\mathbb{Z}$ for some $m,n\in\mathbb{Z}$ with $n>0$.
We can also write $m$ as $qn+r$ for some $q,r\in\mathbb{Z}$ with $0\leq r<n$
and observe that $a=m/n+\mathbb{Z}=r/n+q+\mathbb{Z}=r/n+\mathbb{Z}\in A_{n}$.
This proves that $A$ is the union of the subgroups $A_{n}$. As $na=0$ for all
$a\in A_{n}$, we see that $A$ is a torsion group. One can also check that
$A_{n}\leq A_{m}$ if and only if $n$ divides $m$. In particular, for each
prime $p$ we have a chain of subgroups

$A_{p}\leq A_{p^{2}}\leq
A_{p^{3}}\leq\dotsb\leq\bigcup_{k}A_{p^{k}}=\operatorname{tors}_{p}(A).$

######  Example 4.6.

Now consider instead the group $\mathbb{R}/\mathbb{Z}$. Suppose we have a
torsion element $a=t+\mathbb{Z}$. This means that for some integer $n>0$ we
have $nt\in\mathbb{Z}$, which implies that $t$ is rational. It follows that
$\operatorname{tors}(\mathbb{R}/\mathbb{Z})={\mathbb{Q}/\mathbb{Z}}$. A
similar argument shows that $\mathbb{R}/{\mathbb{Q}}$ is torsion-free.

The following result is known as the Chinese Remainder Theorem.

######  Proposition 4.7.

Suppose we have positive integers $n_{1},\dotsc,n_{r}$ any two of which are
coprime, and we put $n=\prod_{i}n_{i}$. Define

$\phi\colon\mathbb{Z}/n\to(\mathbb{Z}/n_{1})\times\dotsb\times(\mathbb{Z}/n_{r})$

by

$\phi(k+n\mathbb{Z})=(k+n_{1}\mathbb{Z},\dotsc,k+n_{r}\mathbb{Z}).$

Then

  * (a)

There exist integers $e_{1},\dotsc,e_{r}$ such that $\sum_{i}e_{i}=1$ and
$e_{i}=1\pmod{n_{i}}$ and $e_{i}=0\pmod{n/n_{i}}$.

  * (b)

The map $\phi$ is an isomorphism.

###### Proof.

For $i\neq j$ we know that $n_{i}$ and $n_{j}$ are coprime, so we can choose
integers $a_{ij}$ and $b_{ij}$ with $a_{ij}n_{i}+b_{ij}n_{j}=1$. We then put
$f_{ij}=b_{ij}n_{j}=1-a_{ij}n_{i}$, so $f_{ij}=1\pmod{n_{i}}$ and
$f_{ij}=0\pmod{n_{j}}$. Now fix $i$, and let $g_{i}$ be the product of the
numbers $f_{ij}$ as $j$ runs over the remaining indices. We find that
$g_{i}=1\pmod{n_{i}}$, but $g_{i}$ is divisible by the product of all the
$n_{j}$, or equivalently by $n/n_{i}$. Thus, the numbers $g_{i}$ almost have
property (a), but we will need a slight adjustment to make the sum equal to
one. However, we are now ready to prove (b). Given any integers
$m_{1},\dotsc,m_{r}$, we have

$\phi(\sum_{i}m_{i}g_{i}+n\mathbb{Z})=(m_{1}+n_{1}\mathbb{Z},\dotsc,m_{r}+n_{r}\mathbb{Z}).$

This proves that $\phi$ is surjective, and the domain and codomain of $\phi$
both have order $n$, so $\phi$ must actually be an isomorphism. By
construction we have $\phi(\sum_{i}g_{i}+n\mathbb{Z})=\phi(1+n\mathbb{Z})$,
and $\phi$ is injective, so $\sum_{i}g_{i}=1+nk$ for some $k$. We define
$e_{i}=g_{i}$ for $i<r$, and $e_{r}=1-\sum_{i<r}g_{i}=g_{r}-nk$; these clearly
satisfy (a). ∎

The following special case is often useful:

######  Corollary 4.8.

Suppose that the prime factorisation of $n$ is $n=p_{1}^{v_{1}}\dotsb
p_{r}^{v_{r}}$, where the primes $p_{i}$ are all distinct. Then there are
integers $e_{1},\dotsc,e_{r}$ such that $\sum_{i}e_{i}=1$ and
$e_{i}=1\pmod{p_{i}^{v_{i}}}$ and $e_{i}=0\pmod{n/p_{i}^{v_{i}}}$. Moreover,
the natural map

$\mathbb{Z}/n\to(\mathbb{Z}/p_{1}^{v_{1}})\times\dotsb\times(\mathbb{Z}/p_{r}^{v_{r}})$

is an isomorphism. ∎

######  Proposition 4.9.

For any abelian group $A$ we have
$\operatorname{tors}(A)=\bigoplus_{p}\operatorname{tors}_{p}(A)$.

###### Proof.

Suppose we have a torsion element $a\in A$, so $na=0$ for some $n>0$. We can
factor this as $\prod_{i=1}^{r}p_{i}^{v_{i}}$ and then choose integers $e_{i}$
as in Corollary 4.8. Now $e_{i}=0\pmod{n/p_{i}^{v_{i}}}$ so
$p_{i}^{v_{i}}e_{i}$ is divisible by $n$, so $p_{i}^{v_{i}}e_{i}a=0$, so
$e_{i}a\in\operatorname{tors}_{p_{i}}(A)$. We also have $\sum_{i}e_{i}=1$, so
$a=\sum_{i}e_{i}a\in\sum_{i}\operatorname{tors}_{p_{i}}(A)$. This shows that
$\operatorname{tors}(A)=\sum_{p}\operatorname{tors}_{p}(A)$.

To show that the sum is direct, suppose we have a finite list of distinct
primes $p_{1},\dotsc,p_{r}$, and elements
$a_{i}\in\operatorname{tors}_{p_{i}}(A)$ with $\sum_{i}a_{i}=0$; we must show
that $a_{i}=0$ for all $i$. As $a_{i}\in\operatorname{tors}_{p_{i}}(A)$ we
have $p_{i}^{v_{i}}a_{i}=0$ for some $v_{i}\geq 0$. We again choose numbers
$e_{i}$ as in Corollary 4.8. As $p_{i}^{v_{i}}a_{i}=0$ and
$e_{i}=1\pmod{p_{i}^{v_{i}}}$ we have $e_{i}a_{i}=a_{i}$. On the other hand,
for $j\neq i$ we have $e_{i}=0\pmod{p_{j}^{v_{j}}}$ and so $e_{i}a_{j}=0$. We
can thus multiply the relation $\sum_{j}a_{j}=0$ by $e_{i}$ to get $a_{i}=0$
as required. ∎

######  Lemma 4.10.

The quotient group $A/\operatorname{tors}(A)$ is always torsion-free.

###### Proof.

Suppose we have a torsion element $a=x+\operatorname{tors}(A)$ in
$A/\operatorname{tors}(A)$. This means that for some $n>0$ we have $na=0$ or
equivalentlt $nx\in\operatorname{tors}(A)$. This in turn means that for some
$m>0$ we have $mnx=0$, which shows that $x$ itself is a torsion element in
$A$. This means that the coset $a=x+\operatorname{tors}(A)$ is zero, as
required. ∎

##  5\. Finitely generated abelian groups

Let $A$ be an abelian group, and let $a_{1},\dotsc,a_{r}$ be elements of $A$.
We then have a homomorphism $f\colon\mathbb{Z}^{r}\to A$ given by

$f(n_{1},\dotsc,n_{r})=n_{1}a_{1}+\dotsb+n_{r}a_{r}.$

In particular, if $e_{i}$ is the $i$’th standard basis vector in
$\mathbb{Z}^{r}$ then $f(e_{i})=a_{i}$. Conversely, if we start with a
homomorphism $f\colon\mathbb{Z}^{r}\to A$ we can put $a_{i}=f(e_{i})\in A$ and
we find that

$f(n_{1},\dotsc,n_{r})=f(\sum_{i}n_{i}e_{i})=\sum_{i}n_{i}a_{i},$

so everything fits together as before. The image of $f$ is the smallest
subgroup of $A$ containing all the elements $a_{i}$, or in other words the
subgroup generated by $\\{a_{1},\dotsc,a_{r}\\}$. This justifies the following
definition:

######  Definition 5.1.

We say that an abelian group $A$ is _finitely generated_ if there exists a
surjective homomorphism $f\colon\mathbb{Z}^{r}\to A$ for some $r$.

######  Example 5.2.

Suppose that $A$ is actually finite, so we can choose a list
$a_{1},\dotsc,a_{r}$ that contains all the elements of $A$. The corresponding
map $\mathbb{Z}^{r}\to A$ is certainly surjective, so $A$ is finitely
generated.

Our main aim in this section is to prove the following classification theorem:

######  Theorem 5.3.

Let $A$ be a finitely generated abelian group. Then $A$ can be decomposed the
direct sum of a finite list of subgroups, each of which is isomorphic either
to $\mathbb{Z}$, or to $\mathbb{Z}/p^{v}$ for some prime $p$ and some $v>0$.
The number of subgroups of each type in the decomposition is uniquely
determined, although the precise list of subgroups is not.

######  Remark 5.4.

Note that Proposition 4.7 gives a decomposition of the stated type for the
cyclic group $\mathbb{Z}/n$.

The groups $\mathbb{Z}^{r}$ themselves are of course finitely generated. It is
convenient to observe that no two of them are isomorphic:

######  Lemma 5.5.

If $\mathbb{Z}^{r}$ is isomorphic to $\mathbb{Z}^{s}$, then $r=s$.

###### Proof.

Any isomorphism $f\colon A\to B$ induces an isomorphism $A/2A\to B/2B$, so in
particular $|A/2A|=|B/2B|$. We have
$\mathbb{Z}^{r}/2\mathbb{Z}^{r}=(\mathbb{Z}/2)^{r}$, which has order $2^{r}$,
and the claim follows easily. ∎

This means that the term ’rank’ in the following definition is well-defined:

######  Definition 5.6.

We say that an abelian group $A$ is _free of rank $r$_ if it is isomorphic to
$\mathbb{Z}^{r}$.

######  Lemma 5.7.

Let $A$ be a subgroup of $\mathbb{Z}$. Then either $A=0\simeq\mathbb{Z}^{0}$
or $A=d\mathbb{Z}\simeq\mathbb{Z}$ for some (unique) $d>0$.

###### Proof.

The case where $A=0$ is trivial, so suppose that $A\neq 0$. As $A=-A$ we see
that $A$ must contain at least one strictly positive integer. Let $d$ be the
smallest strictly positive integer in $A$. It is than clear that
$d\mathbb{Z}\subseteq A$. Conversely, suppose that $n\in A$. As $d>0$ we see
that $n$ must lie between $id$ and $(i+1)d$ for some $i\in\mathbb{Z}$, say
$n=id+j$ with $0\leq j<d$. Now $j=n-id\in A$ and $0\leq j<d$, which
contradicts the defining property of $d$ unless $j=0$. We thus have $n=di$,
showing that $A=d\mathbb{Z}$ as claimed. ∎

######  Proposition 5.8.

Let $A$ be a subgroup of $\mathbb{Z}^{r}$; then $A$ is free of rank at most
$r$.

###### Proof.

For $i\leq r$ we put

$F_{i}=\\{x\in\mathbb{Z}^{r}\;|\;x_{i+1}=\dotsb=x_{r}=0\\}\simeq\mathbb{Z}^{i}.$

We let $\pi_{s}\colon\mathbb{Z}^{r}\to\mathbb{Z}$ be the projection map
$x\mapsto x_{s}$, and put

$J=\\{j\;|\;\pi_{j}(A\cap F_{j})\neq 0\\}=\\{j\;|\;A\cap F_{j}>A\cap
F_{j-1}\\}.$

We can list the elements of this set as $j_{1}<\dotsb<j_{s}$ for some $s\leq
r$ (possibly $s=0$). We see from Lemma 5.7 that $\pi_{j_{p}}(A\cap F_{j_{p}})$
must have the form $d_{p}\mathbb{Z}$ for some $d_{p}>0$ say. We can thus
choose $a_{p}\in A\cap F_{j_{p}}$ such that $\pi_{j_{p}}(a_{p})=d_{p}$ for all
$p$. We claim that the list $a_{1},\dotsc,a_{s}$ is a basis for $A$ over
$\mathbb{Z}$, so that $A$ is a free abelian group as claimed. More precisely,
we claim that $a_{1},\dotsc,a_{p}$ is always a basis for $A\cap F_{j_{p}}$. In
the case $p=0$ we have the empty list and the zero group so the claim is
clear. When $p>0$ we can inductively assume the statement for $p-1$. Consider
an arbitrary element $u\in A\cap F_{j_{p}}$. By the definition of $d_{p}$, we
have $\pi_{j_{p}}(u)=m_{p}d_{p}$ for some $m_{p}\in\mathbb{Z}$. The element
$u^{\prime}=u-m_{p}a_{p}$ then lies in $A\cap F_{j_{p}}$ and satisfies
$\pi_{j_{p}}(u^{\prime})=0$ so in fact $u^{\prime}\in A\cap F_{j_{p-1}}$. By
the induction hypothesis there are unique integers $m_{1},\dotsc,m_{p-1}$ with
$u^{\prime}=m_{1}a_{1}+\dotsb+m_{p-1}a_{p-1}$, and it follows that
$u=u^{\prime}+m_{p}a_{p}=m_{1}a_{1}+\dotsb+m_{p}a_{p}$. This shows that $u$
can be expressed as an integer combination of $a_{1},\dotsc,a_{p}$, and a
similar argument shows that the expression is unique. This completes the
induction step, and after $s$ steps we see that $A$ itself has a basis as
claimed. ∎

######  Remark 5.9.

In the above proof, we can alter our choice of $a_{p}$ by subtracting off
suitable multiples of $a_{p-1}$, $a_{p-2}$ and so on in turn to ensure that
$0\leq\pi_{j_{q}}(a_{p})<d_{q}$ for $1\leq q<p$. One can then check that the
resulting basis satisfying this auxiliary condition is in fact unique.

######  Corollary 5.10.

If $A$ is finitely generated and $B$ is a subgroup of $A$ then $B$ and $A/B$
are also finitely generated.

###### Proof.

Choose a surjective homomorphism $f\colon\mathbb{Z}^{r}\to A$. The composite
$\mathbb{Z}^{r}\xrightarrow{f}A\xrightarrow{\pi}A/B$ is again surjective, so
$A/B$ is finitely generated. Now put $F=\\{x\in\mathbb{Z}^{r}\;|\;f(x)\in
B\\}$, and let $g\colon F\to B$ be the restriction of $f$. For $b\in B\leq A$
we can choose $x\in\mathbb{Z}^{r}$ with $f(x)=b$ (because $f$ is surjective).
Then $x\in F$ be the definition of $f$, and $g(x)=f(x)=b$; this proves that
$g$ is surjective. Moreover, $F$ is a subgroup of $\mathbb{Z}^{r}$, so it is
isomorphic to $\mathbb{Z}^{s}$ for some $s\leq r$ by the proposition. It now
follows that $B$ is finitely generated. ∎

######  Proposition 5.11.

Suppose that $F$ is finitely generated and torsion free; then $F$ is free.

###### Proof.

Choose a surjective homomorphism $f\colon\mathbb{Z}^{r}\to F$ with $r$ as
small as possible. Put $A=\ker(f)$; it will suffice to show that $A=0$. If
not, choose some nonzero element $a=(a_{1},\dotsc,a_{r})\in\ker(f)$. Let $d$
be the greatest common divisor of $a_{1},\dotsc,a_{r}$ (or equivalently, the
number $d>0$ such that $\sum_{i}a_{i}\mathbb{Z}=d\mathbb{Z}$, which exists by
Lemma 5.7). We find that $a/d\in\mathbb{Z}^{r}$ and $d\,f(a/d)=f(a)=0$ in $F$,
so $f(a/d)$ is a torsion element, but $F$ is assumed torsion free, so
$f(a/d)=0$. We may thus replace $a$ by $a/d$ and assume that $d=1$, so
$\sum_{i}a_{i}\mathbb{Z}=\mathbb{Z}$. We can thus choose integers
$b_{1},\dotsc,b_{r}$ with $\sum_{i}a_{i}b_{i}=1$. Now define
$\beta\colon\mathbb{Z}^{r}\to\mathbb{Z}$ by $\beta(x)=\sum_{i}x_{i}b_{i}$ and
put $U=\ker(\beta)$. As $\beta(a)=1$ we find that $x-\beta(x)a\in U$ for all
$x$, and it follows that $\mathbb{Z}^{r}=U\oplus\mathbb{Z}a$. Proposition 5.8
tells us that $U$ is free, and using the splitting
$\mathbb{Z}^{r}=U\oplus\mathbb{Z}a$ we see that $U$ has rank $r-1$. As
$f(a)=0$ we also see that $f(U)=f(U\oplus\mathbb{Z}a)=f(\mathbb{Z}^{r})=F$, so
$f$ restricts to give a surjective homomorphism $U\to F$. This contradicts the
assumed minimality of $r$, so we must have $A=0$ after all, so
$f\colon\mathbb{Z}^{r}\to F$ is an isomorphism. ∎

######  Corollary 5.12.

Let $A$ be a finitely generated abelian group. Then $\operatorname{tors}(A)$
is finite, and there exists a finitely generated free subgroup $F\leq A$ such
that $A=\operatorname{tors}(A)\oplus
F\simeq\operatorname{tors}(A)\oplus\mathbb{Z}^{s}$ for some $s$.

###### Proof.

Firstly, Corollary 5.10 tells us that $\operatorname{tors}(A)$ is finitely
generated, so we can choose a finite list of generators, say
$a_{1},\dotsc,a_{r}$. These must be torsion elements, so we can choose
$n_{i}>0$ with $n_{i}a_{i}=0$. This means that the corresponding surjection
$f\colon\mathbb{Z}^{r}\to\operatorname{tors}(A)$ factors through the finite
quotient group $\prod_{i=1}^{r}(\mathbb{Z}/n_{i})$, so
$\operatorname{tors}(A)$ is finite as claimed. Next, the quotient group
$A/\operatorname{tors}(A)$ is finitely generated (by Corollary 5.10) and
torsion-free (by Lemma 4.10) so it is free of finite rank by Proposition 5.11.
We can thus choose an isomorphism $\overline{g}\colon\mathbb{Z}^{s}\to
A/\operatorname{tors}(A)$. Now choose an element $a_{i}\in A$ representing the
coset $g(e_{i})$ (for $i=1,\dotsc,s$) and define $g\colon\mathbb{Z}^{s}\to A$
by $g(x)=\sum_{i}x_{i}a_{i}$, and put $F=g(\mathbb{Z}^{s})$. If we let $q$
denote the quotient map $A\to A/\operatorname{tors}(A)$ we have
$qg=\overline{g}$, which is an isomorphism. It follows that
$g\colon\mathbb{Z}^{s}\to F$ is an isomorphism, so $F$ is free as claimed.
Next, let $h\colon A\to F$ be the composite

$A\xrightarrow{q}A/\operatorname{tors}(A)\xrightarrow{\overline{g}^{-1}}\mathbb{Z}^{s}\xrightarrow{g}F.$

We find that $qh=q$, so $q(a-h(a))=0$, so $a-h(a)\in\operatorname{tors}(A)$
for all $a$. This implies that $a=(a-h(a))+h(a)\in\operatorname{tors}(A)+F$,
so $A=\operatorname{tors}(A)+F$. Moreover, the intersection
$\operatorname{tors}(A)\cap F$ is both torsion and torsion-free, so it must be
zero, so the sum is direct. ∎

This corollary allows us to focus on $\operatorname{tors}(A)$, which is a
finite group, of order $n$ say. Proposition 4.9 gives a splitting
$\operatorname{tors}(A)=\bigoplus_{p}\operatorname{tors}_{p}(A)$, and it is
clear that $\operatorname{tors}_{p}(A)$ can only be nonzero if $p$ divides
$n$. In that case, $\operatorname{tors}_{p}(A)$ will be a finite abelian group
whose order is a power of $p$.

######  Lemma 5.13.

Let $A$ be an abelian group of order $p^{v}$. Suppose we have an element $c$
of order $p^{w}$, and that every other element has order dividing $p^{w}$, and
that the subgroup $C=\mathbb{Z}c$ has nontrivial intersection with every
nontrivial subgroup. Then $A=C$.

###### Proof.

Consider a nontrivial element $a\in A$. The order of $a+C$ in $A/C$ will then
be $p^{i}$ for some $i\leq w$. We then have $p^{i}a=mc$ for some
$m\in\mathbb{Z}$, and the assumption $(\mathbb{Z}a)\cap C\neq 0$ means that
$mc\neq 0$. We can thus write $mc=up^{j}c$ for some $j<w$ and some $u$ such
that $u\neq 0\pmod{p}$. It follows that the order of $mc$ in $A$ is $p^{w-j}$,
and thus that the order of $a$ in $A$ is $p^{w-j+i}$. By assumption, this is
at most $p^{w}$, so $i\leq j$. We can thus put $b=a-up^{j-i}c$, and observe
that $p^{i}b=0$. Now $b$ is congruent to $a$ mod $C$, so it again has order
$p^{i}$ in $A/C$, but $p^{i}b$ is already zero in $A$, so $(\mathbb{Z}b)\cap
C=0$. As $C$ meets every nontrivial subgroup, we must have $\mathbb{Z}b=0$, so
$a=up^{j-i}c\in C$. This means that $A=C$ as claimed. ∎

######  Corollary 5.14.

Let $A$ be an abelian group of order $p^{v}$, and suppose that the largest
order of any element of $A$ is $p^{w}$. Then $A\simeq
B\oplus(\mathbb{Z}/p^{w})$ for some subgroup $B$ of order $p^{v-w}$.

###### Proof.

Choose an element $c$ of order $p^{w}$, and let $C$ be the subgroup that it
generates, so $C\simeq\mathbb{Z}/p^{w}$. Among the subgroups $B$ with $B\cap
C=0$, choose one of maximal order. Then put $\overline{A}=A/B$, and let
$\overline{C}$ be the image of $C$ in $\overline{A}$, which is isomorphic to
$C$ because $B\cap C=0$. It will suffice to prove that $A=B+C$, or
equivalently that $\overline{A}=\overline{C}$. By the lemma, we need only
check that $\overline{C}$ has nontrivial intersection with every nontrivial
subgroup of $\overline{A}$. Consider a nonzero element
$\overline{a}\in\overline{A}$, and choose a representing element $a\in
A\setminus B$. Then $\mathbb{Z}a+B$ is strictly larger than $B$ and must meet
$C$ nontrivially, so there exists $k\in\mathbb{Z}$ and $b\in B$ with $ka+b\in
C\setminus\\{0\\}$. If $ka+b$ were in $B$ it would give a nontrivial element
of $B\cap C$, contrary to assumption. It follows that $k\overline{a}$ is
nontrivial in $\overline{A}$ and lies in $C$, as required. ∎

######  Corollary 5.15.

Let $A$ be an abelian group of order $p^{v}$. Then $A$ is isomorphic to
$\bigoplus_{i=1}^{r}\mathbb{Z}/p^{w_{i}}$ for some list $w_{1},\dotsc,w_{r}$
of positive integers with $\sum_{i}w_{i}=v$.

###### Proof.

This follows by an evident induction from Corollary 5.14. ∎

######  Definition 5.16.

Let $A$ be a finite abelian group. For any prime $p$ and positive integer $k$,
we put

$F_{p}^{k}(A)=\\{a\in p^{k-1}A\;|\;pa=0\\}.$

This is a finite abelian group of exponent $p$, so it has order $p^{v}$ for
some $v$. We define $f_{p}^{k}(A)$ to be this $v$, and we also put
$g_{p}^{k}(A)=f_{p}^{k}(A)-f_{p}^{k+1}(A)$.

######  Proposition 5.17.

Let $A$ be a finite abelian group.

  * (a)

If $A\simeq A^{\prime}$, then $F_{p}^{k}(A)\simeq F_{p}^{k}(A^{\prime})$ for
all $p$ and $k$, so $f_{p}^{k}(A)=f_{p}^{k}(A^{\prime})$ and
$g_{p}^{k}(A)=g_{p}^{k}(A^{\prime})$.

  * (b)

If $A=B\oplus C$ then $F_{p}^{k}(A)=F_{p}^{k}(B)\oplus F_{p}^{k}(C)$, so
$f_{p}^{k}(A)=f_{p}^{k}(B)+f_{p}^{k}(C)$ and
$g_{p}^{k}(A)=g_{p}^{k}(B)+g_{p}^{k}(C)$.

  * (c)

If $A$ has order not divisible by $p$, then $F_{p}^{k}(A)=0$ and so
$f_{p}^{k}(A)=g_{p}^{k}(A)=0$.

  * (d)

Suppose that $A$ has a decomposition as a sum of subgroups
$\mathbb{Z}/p_{i}^{v_{i}}$ (with $v_{i}>0$). Then $g_{p}^{k}(A)$ is the number
of times that $\mathbb{Z}/p^{k}$ occurs in the decomposition.

###### Proof.

Parts (a) to (c) are straightforward and are left to the reader. Given these,
part (d) reduces to the claim that $g_{p}^{k}(\mathbb{Z}/p^{j})$ is one when
$j=k$, and zero otherwise. One can see from the definitions that
$f_{p}^{k}(\mathbb{Z}/p^{j})$ is one when $k\leq j$, and zero when $k>j$; the
claim follows easily from this. ∎

###### Proof of Theorem 5.3.

Let $A$ be a finitely generated abelian group. Corollary 5.12 and subsequent
remarks show that
$A\simeq\mathbb{Z}^{s}\oplus\bigoplus_{i=1}^{m}\operatorname{tors}_{p_{i}}(A)$
for some finite list of primes $p_{i}$. After applying Corollary 5.15 to each
of the groups $\operatorname{tors}_{p_{i}}(A)$, we get the claimed splitting
of $A$ as a sum of copies of $\mathbb{Z}$ and $\mathbb{Z}/p_{j}^{w_{j}}$. The
number $s$ is the rank of the quotient group $A/\operatorname{tors}(A)$, which
is well-defined by Lemma 5.5. We can also apply the last part of Proposition
5.17 to $\operatorname{tors}(A)$ to see that the number of summands of each
type is independent of the choice of splitting. ∎

##  6\. Free abelian groups and their subgroups

In various places we have already used the free abelian group $\mathbb{Z}[I]$
generated by a set $I$. We start with a more careful formulation of this
construction. One approach is to define $\mathbb{Z}[I]=\bigoplus_{i\in
I}\mathbb{Z}$ as in Definition 3.5. That is essentially what we will do, but
we will spell out some details.

######  Definition 6.1.

Let $I$ be any set. We write $\operatorname{Map}(I,\mathbb{Z})$ for the set of
all maps $u\colon I\to\mathbb{Z}$. These form an abelian group under pointwise
addition. For any map $u\colon I\to\mathbb{Z}$, the _support_ is the set

$\operatorname{supp}(u)=\\{i\in I\;|\;u(i)\neq 0\\}\subseteq I.$

We put

$\operatorname{Map}_{0}(I,\mathbb{Z})=\\{u\colon
I\to\mathbb{Z}\;|\;\operatorname{supp}(u)\text{ is finite }\\}.$

It is easy to see that $\operatorname{supp}(u\pm
v)\subseteq\operatorname{supp}(u)\cup\operatorname{supp}(v)$, and thus that
$\operatorname{Map}_{0}(I,\mathbb{Z})$ is a subgroup of
$\operatorname{Map}(I,\mathbb{Z})$.

Next, for any $i\in I$ we define $\delta_{i}\colon I\to\mathbb{Z}$ by

$\delta_{i}(j)=\begin{cases}1&\text{ if }j=i\\\ 0&\text{ if }j\neq
i.\end{cases}$

Note that $\operatorname{supp}(\delta_{i})=\\{i\\}$ so
$\delta_{i}\in\operatorname{Map}_{0}(I,\mathbb{Z})$.

######  Remark 6.2.

If $I$ is a finite set with $n$ elements then we see that
$\operatorname{Map}_{0}(I,\mathbb{Z})=\operatorname{Map}(I,\mathbb{Z})\simeq\mathbb{Z}^{n}$.
The situation is a little more subtle when $I$ is infinite.

The following lemma shows that $\operatorname{Map}_{0}(I,\mathbb{Z})$ is
generated freely, in a certain sense, by the elements $\delta_{i}$.

######  Lemma 6.3.

Let $A$ be an abelian group. Then for any function $f\colon I\to A$ there is a
unique homomorphism $\overline{f}\colon\operatorname{Map}_{0}(I,\mathbb{Z})\to
A$ such that $\overline{f}(\delta_{i})=f(i)$ for all $i\in I$.

###### Proof.

We put

$\overline{f}(u)=\sum_{i\in\operatorname{supp}(u)}u(i)\,f(i).$

The terms are meaningful, because each $u(i)$ is in $\mathbb{Z}$ and each
$f(i)$ is in $A$ so we can multiply to get an element of $A$. The sum is
meaningful because $u\in\operatorname{Map}_{0}(I,\mathbb{Z})$, so
$\operatorname{supp}(u)$ is finite, so there are only finitely many terms to
add. More explicitly, if $\operatorname{supp}(u)=\\{i_{1},\dotsc,i_{r}\\}$ and
$u(i_{t})=n_{t}\in\mathbb{Z}$ for all $t$ then

$\overline{f}(u)=n_{1}f(i_{1})+\dotsb+n_{r}f(i_{r})\in A.$

Note that it would be harmless to replace $\operatorname{supp}(u)$ by any
finite set $J$ with $\operatorname{supp}(u)\subseteq J\subseteq I$; this would
introduce some extra terms, but they would all be zero. After taking
$J=\operatorname{supp}(u)\cup\operatorname{supp}(v)$ we can check that
$\overline{f}(u+v)=\overline{f}(u)+\overline{f}(v)$, so $\overline{f}$ is a
homomorphism. From the definitions it is clear that
$\overline{f}(\delta_{i})=f(i)$. Now let
$\alpha\colon\operatorname{Map}_{0}(I,\mathbb{Z})\to A$ be another
homomorphism with $\alpha(\delta_{i})=f(i)$. Put $\beta=\alpha-\overline{f}$,
so $\beta(\delta_{i})=0$ for all $i$. It is not hard to see that a general
element $u$ as above can be expressed in the form

$u=n_{1}\delta_{i_{1}}+\dotsb+n_{r}\delta_{i_{r}}.$

It follows that

$\beta(u)=n_{1}\beta(\delta_{i_{1}})+\dotsb+n_{r}\beta(\delta_{i_{r}})=n_{1}.0+\dotsb+n_{r}.0=0.$

This shows that $\beta=0$, so $\alpha=\overline{f}$ as required. ∎

The notation used so far is convenient for giving the definition, and the
proof of the above freeness property, but not for the applications. We thus
introduce the following alternative:

######  Definition 6.4.

We write $\mathbb{Z}[I]$ for $\operatorname{Map}_{0}(I,\mathbb{Z})$, and $[i]$
for $\delta_{i}$. We say that an abelian group $A$ is _free_ if it is
isomorphic to $\mathbb{Z}[I]$ for some $I$.

The freeness property now takes the following form:

######  Lemma 6.5.

Let $A$ be an abelian group. Then for any function $f\colon I\to A$ there is a
unique homomorphism $\overline{f}\colon\mathbb{Z}[I]\to A$ such that
$\overline{f}([i])=f(i)$ for all $i\in I$.

One key result is as follows:

######  Theorem 6.6.

If $A$ is a free abelian group, then every subgroup of $A$ is also free.

This is a generalisation of Proposition 5.8, which covered the case where $A$
is finitely generated. Below we will state and prove some more refined
statements. To prove Theorem 6.6 itself, we can use Remark 6.14 and the case
$k=\top$ of Lemma 6.16 (in the notation of Definition 6.13).

To extend the proof of Proposition 5.8 to cover infinitely generated groups,
we need two ingredients. Firstly, we need to modify the inductive argument so
that it works for a suitable class of infinite ordered sets. Next, we need to
show that any set can be ordered in the required way. The precise structure
that we need is as follows:

######  Definition 6.7.

A _well-ordering_ on a set $I$ is a relation on $I$ (denoted by $i\leq j$)
such that

  * (a)

For all $i\in I$ we have $i\leq i$.

  * (b)

For all $i,j\in I$ we have either $i\leq j$ or $j\leq i$, and if both hold
then $i=j$.

  * (c)

For all $i,j,k\in I$, if $i\leq j$ and $j\leq k$ then $i\leq k$.

  * (d)

For any nonempty subset $J\subseteq I$ there is an element $j_{0}\in J$ such
that $j_{0}\leq j$ for all $j\in J$. (In other words, $j_{0}$ is smallest in
$J$.)

######  Remark 6.8.

We have stated the axioms in a form that is conceptually natural but
inefficient. Axiom (a) follows from (d) by taking $J=\\{i\\}$, the first half
of (b) follows from (d) by taking $J=\\{i,j\\}$, and with a little more
argument one can deduce (c) by taking $J=\\{i,j,k\\}$ and appealing to the
second half of (b). Thus, we really only need (d) together with the second
half of (b).

######  Example 6.9.

The obvious ordering of $\mathbb{N}$ is a well-ordering, as is the obvious
ordering on $\mathbb{N}\cup\\{\infty\\}$. The obvious ordering on $\mathbb{Z}$
is not a well-ordering, because the whole set does not have a smallest
element. We can choose a bijection $f\colon\mathbb{N}\to\mathbb{Z}$ (for
example, by setting $f(2n)=n$ and $f(2n+1)=-n-1$) and use this to transfer the
standard ordering of $\mathbb{N}$ to a nonstandard ordering of $\mathbb{Z}$
that is a well-ordering. Alternatively, we can specify a well-ordering on
$\mathbb{Z}$ by the rules

$0<1<2<3<4<\dotsb<-1<-2<-3<\dotsb.$

By constructions such as these, one can give explicit well-orderings of most
naturally occurring countable sets.

######  Remark 6.10.

Let $I$ be a well-ordered set. If $I$ is nonempty then it must have a smallest
element, which we denote by $\bot_{I}$ or just $\bot$. Now suppose that $i\in
I$ and $i$ is not maximal, so the set $I_{>i}=\\{j\in I\;|\;j>i\\}$ is
nonempty. Then $I_{>i}$ must have a smallest element. We denote this by
$s(i)$, and call it the _successor_ of $i$. We say that an element $j\in I$
_is a successor_ if $j=s(i)$ for some $i$. For example, in
$\mathbb{N}\cup\\{\infty\\}$ the elements $0$ and $\infty$ are not successors,
but all other elements are successors.

######  Theorem 6.11.

Every set admits a well-ordering.

The proof will be given after some preliminaries.

There is no known well-ordering of $\mathbb{R}$, and indeed it is probably not
possible to specify a well-ordering concretely, although the author does not
know of any precise theorems to that effect. Similarly, there is no known
well-ordering of the set of subsets of $\mathbb{N}$, or of most other
naturally occurring uncountable sets. The problem is that one needs to make an
infinite number of arbitrary choices, which cannot be done explicitly.
However, we shall assume the Axiom of Choice, a standard principle of Set
Theory, which says that such choices are nonetheless possible. More precisely,
we shall assume that every set has a choice function, in the following sense:

######  Definition 6.12.

Let $I$ be a set, and let $P^{\prime}(I)$ denote the set of nonempty subsets
of $I$. A _choice function_ for $I$ is a function $c\colon P^{\prime}(I)\to I$
such that $c(J)\in J$ for all $J\in P^{\prime}(I)$. (In other words, $c(J)$ is
a “chosen” element of $J$.)

If $I$ is well-ordered, we can define a choice function by taking $c(J)$ to be
the smallest element of $J$. Conversely, if we are given a choice function
then we can use it to construct a well-ordering, as we now explain. In the
literature it is more common to do this by proving Zorn’s Lemma as an
intermediate step, but here we have chosen to bypass that.

###### Proof of Theorem 6.11.

Let $I$ be a set, and let $c$ be a choice function for $I$. Let $P(I)$ be the
set of all subsets of $I$, and put $P^{*}(I)=P(I)\setminus\\{I\\}$. Define
$d\colon P^{*}(I)\to I$ by $d(J)=c(I\setminus J)$, and then define $e\colon
P(I)\to P(I)$ by

$e(J)=\begin{cases}J\cup\\{d(J)\\}&\text{ if }J\neq I\\\ I&\text{ if
}J=I.\end{cases}$

We call this the _expander_ function. Clearly we have $J\subseteq e(J)$ for
all $J$, with equality iff $J=I$. Now say that a subset
${\mathcal{A}}\subseteq P(I)$ is _saturated_ if

  * (a)

Whenever $J\in{\mathcal{A}}$, we also have $e(J)\in{\mathcal{A}}$.

  * (b)

For any family of sets in ${\mathcal{A}}$, the union of that family is also in
${\mathcal{A}}$.

We say that a set $J$ is _compulsory_ if it lies in every saturated family.
For example, by applying (b) to the empty family we see that the set
$J_{0}=\emptyset$ is compulsory. It follows using (a) that the sets
$J_{n}=e^{n}(\emptyset)$ are compulsory for all $n$. Axiom (b) then tells us
that the set $J_{\omega}=\bigcup_{n=0}^{\infty}J_{n}$ is compulsory, as is the
set $J_{\omega+1}=e(J_{\omega})$. If we had developed the theory of infinite
ordinals, we could use transfinite recursion to define compulsory sets
$J_{\alpha}$ for all ordinals $\alpha$. As we have not discussed that theory,
we will instead use an approach that avoids it. We let ${\mathcal{C}}$ denote
the family of all compulsory sets. This is clearly itself a saturated family.
We say that a set $J\in{\mathcal{C}}$ is _comparable_ if for all other
$K\in{\mathcal{C}}$ we have either $J\subseteq K$ or $K\subseteq J$. Let
${\mathcal{D}}\subseteq{\mathcal{C}}$ be the set of all comparable sets; we
will show that this is saturated, and thus equal to ${\mathcal{C}}$. Consider
a family of comparable sets $J_{\alpha}$, with union $J$ say, and another set
$K\in{\mathcal{C}}$. For each $\alpha$ we have $J_{\alpha}\subseteq K$ or
$K\subseteq J_{\alpha}$, because $J_{\alpha}$ is comparable. If
$J_{\alpha}\subseteq K$ for all $K$ then clearly $J\subseteq K$. Otherwise we
must have $K\subseteq J_{\alpha}$ for some $\alpha$ but $J_{\alpha}\subseteq
J$ so $K\subseteq J$. This shows that $J$ is comparable, so ${\mathcal{D}}$ is
closed under unions. Now consider a comparable set $J$; we claim that $e(J)$
is also comparable. To see this, put

${\mathcal{E}}_{J}=\\{K\;|\;e(J)\subseteq K\text{ or }K\subseteq J\\}.$

By a similar argument to the previous paragraph, this is closed under unions.
Suppose that $K\in{\mathcal{E}}_{J}$.

  * (a)

If $e(J)\subseteq K$ then clearly $e(J)\subset e(K)$, so
$e(K)\in{\mathcal{E}}_{J}$.

  * (b)

If $K=J$ then $e(J)=e(K)$, so $e(K)\in{\mathcal{E}}_{J}$.

  * (c)

Suppose instead that $K\subset J$. As $K\in{\mathcal{C}}$ we also have
$e(K)\in{\mathcal{C}}$, and $J$ is comparable so either $e(K)\subseteq J$ or
$J\subset e(K)$. In the latter case we have $K\subset J\subset e(K)$, so
$|e(K)\setminus K|\geq 2$, but $|e(K)\setminus K|\leq 1$ by construction, so
this is impossible. We therefore have $e(K)\subseteq J$, so again
$e(K)\in{\mathcal{J}}$.

We now see that ${\mathcal{E}}_{J}$ is a saturated subset of ${\mathcal{C}}$,
so it must be all of ${\mathcal{C}}$. It follows easily from this that $e(J)$
is comparable. This means that ${\mathcal{D}}$ is a saturated subset of
${\mathcal{C}}$, so it must be all of ${\mathcal{C}}$, so all compulsory sets
are comparable, or in other words ${\mathcal{C}}$ is totally ordered by
inclusion.

We now claim that ${\mathcal{C}}$ is in fact well-ordered by inclusion. To see
this, consider a nonempty family of compulsory sets $K_{\alpha}$. Let $J$ be
the union of all compulsory sets that are contained in
$\bigcap_{\alpha}K_{\alpha}$. As ${\mathcal{C}}$ is closed under unions, we
see that $J$ is actually the largest compulsory set that is contained in
$\bigcap_{\alpha}K_{\alpha}$. In particular, the larger set $e(J)$ cannot be
contained in $\bigcap_{\alpha}J_{\alpha}$, so for some $\alpha$ we have
$e(J)\not\subseteq K_{\alpha}$. Now $K_{\alpha}\in{\mathcal{C}}$ and we have
seen that ${\mathcal{C}}={\mathcal{E}}_{J}$ and using this we see that
$K_{\alpha}\subseteq J$. From this it follows easily that $K_{\alpha}$ is the
smallest set in the family, as required.

Next, put ${\mathcal{C}}^{*}={\mathcal{C}}\setminus\\{I\\}$, which is again
well-ordered by inclusion. For $i\in I$ we let $p(i)$ be the union of all
compulsory sets that do not contain $i$. As ${\mathcal{C}}$ is closed under
unions this defines a map $p\colon I\to{\mathcal{C}}^{*}$. We can also
restrict $d$ to get a map $d\colon{\mathcal{C}}^{*}\to I$ in the opposite
direction. Note that $e(p(i))=p(i)\cup\\{d(p(i))\\}$ is a compulsory set not
contained in $p(i)$, so we must have $i\in e(p(i))$, but $i\not\in p(i)$ by
construction, so we must have $d(p(i))=i$. In the opposite direction, suppose
we start with a compulsory set $J\in{\mathcal{C}}^{*}$, and put $i=d(J)$, so
$e(J)=J\amalg\\{i\\}$. Now $p(i)\in{\mathcal{C}}$ and we have seen that
${\mathcal{C}}={\mathcal{E}}_{J}$ so either $e(J)\subseteq p(i)$ or
$p(i)\subseteq J$. The first of these would imply that $i\in p(i)$,
contradicting the definition of $p(i)$, so we must instead have $p(i)\subseteq
J$. On the other hand, $J$ is one of the sets in the union that defines
$p(i)$, so $J\subseteq p(i)$, so $J=p(i)=p(d(J))$. This proves that the maps
$I\xrightarrow{p}{\mathcal{C}}^{*}\xrightarrow{d}I$ are mutually inverse
bijections. We can thus introduce a well-ordering of $I$ by declaring that
$i\leq j$ iff $p(i)\subseteq p(j)$. ∎

We can now start to prove as promised that subgroups of free abelian groups
are free. It will be enough to prove that every subgroup of $\mathbb{Z}[I]$ is
free, and by Theorem 6.11 we may assume that $I$ is well-ordered. We will need
some auxiliary definitions.

######  Definition 6.13.

Let $I$ be a well-ordered set, and let $A$ be a subgroup of $\mathbb{Z}[I]$.
We put $I_{\top}=I\amalg\\{\top\\}$, ordered so that $i\leq\top$ for all $i\in
I$ (which is again a well-ordering). Put $I_{<j}=\\{i\in I\;|\;i<j\\}$ and
$A_{<j}=A\cap\mathbb{Z}[I_{<j}]$, and similarly for $I_{\leq j}$ and $A_{\leq
j}$. Let $\pi_{j}\colon\mathbb{Z}[I]\to\mathbb{Z}$ be the $j$’th projection,
which is characterised by the fact that $\pi_{j}([j])=1$ and $\pi_{j}([i])=0$
for all $i\neq j$. Put

$J=\\{j\in I\;|\;\pi_{j}(A_{\leq j})\neq 0\\}=\\{j\in I\;|\;A_{<j}<A_{\leq
j}\\},$

and $J_{<k}=\\{j\in J\;|\;j<k\\}$, and similarly for $J_{\leq k}$. For $j\in
J$, we let $d_{j}$ be the positive integer such that $\pi_{j}(A_{\leq
j})=d_{j}\mathbb{Z}$. We put

$B_{j}=\\{x\in A_{\leq j}\;|\;x_{j}=d_{j}\\}\subseteq A,$

which is nonempty by the definition of $d_{j}$. For any $k\in I_{\top}$, we
define an _adapted basis for $A_{<k}$_ to be a map $a\colon J_{<k}\to A$ such
that for all $j\in J_{<k}$ we have $a(j)\in B_{j}$. We also put

$T_{k}=\\{u\in\mathbb{Z}[I_{<k}]\;|\;0\leq u_{j}<d_{j}\text{ for all }j\in
J_{<k}\\}.$

Note that this contains zero but is not a subgroup (unless $d_{j}=1$ for all
$j\in J$).

######  Remark 6.14.

As each $B(j)$ is nonempty, we see that there exist adapted bases. Implicitly
we are using the Axiom of Choice here: there exists a choice function $c$ for
$A$, and then we can take $a(j)=c(B(j))$ for all $j$. However, we will prove
as Proposition 6.20 that there is a _unique_ adapted basis satisfying a
certain normalisation condition, which enables us to avoid this use of choice.

######  Remark 6.15.

Let $a$ be an adapted basis for $A_{<k}$. We then have a unique homomorphism

$f_{k}\colon\mathbb{Z}[J_{<k}]\to A_{<k}$

such that $f_{k}([j])=a(j)$ for all $j\in J_{<k}$. If $m<k$ then $a|_{J_{<m}}$
is easily seen to be an adapted basis for $A_{<m}$. It therefore gives rise to
a homomorphism $f_{m}\colon\mathbb{Z}[J_{<m}]\to A_{<m}$, which is just the
restriction of the map $f_{k}\colon\mathbb{Z}[J_{<k}]\to A_{<k}$.

We can also define functions $g_{m}\colon\mathbb{Z}[J_{<m}]\times
T_{m}\to\mathbb{Z}[I_{<m}]$ by $g_{m}(x,y)=f_{m}(x)+y$.

The terminology is justified by the following result:

######  Lemma 6.16.

Let $a$ be an adapted basis for $A_{<k}$. Then the corresponding map
$f_{k}\colon\mathbb{Z}[J_{<k}]\to A_{<k}$ is an isomorphism, and the map
$g_{k}\colon\mathbb{Z}[J_{<k}]\times T_{k}\to\mathbb{Z}[I_{<k}]$ is a
bijection. In particular, the group $A_{<k}$ is free.

We will deduce it from the following auxiliary result:

######  Lemma 6.17.

Let $a$ be an adapted basis for $A_{<k}$, and suppose that for all $m<k$ the
maps $f_{m}$ and $g_{m}$ are bijective. Then $f_{k}$ and $g_{k}$ are also
bijective.

###### Proof.

There are three cases to consider:

  * (a)

$k$ is not a successor.

  * (b)

$k=s(m)$ for some $m\not\in J$.

  * (c)

$k=s(m)$ for some $m\in J$.

Suppose that case (a) holds. Then for $m<k$ we have $s(m)<k$ and so
$f_{s(m)}\colon\mathbb{Z}[J_{<s(m)}]\to A_{<s(m)}$ is an isomorphism. Now
$J_{<k}$ is easily seen to be the union of these sets $J_{<s(m)}$, and
$A_{<k}$ is the union of the groups $A_{<s(m)}$. It therefore follows that
$f_{k}$ is also an isomorphism $\mathbb{Z}[J_{<k}]\to A_{<k}$, as required.
Essentially the same argument proves that $g_{k}$ is a bijection.

Next, in case (b) we see from the definition of $J$ that $A_{<k}=A_{\leq
m}=A_{<m}$ and similarly $J_{<k}=J_{<m}$ so $f_{k}=f_{m}$ and this is an
isomorphism as required. We also have $T_{k}=T_{m}\times\mathbb{Z}.[m]$ and
$\mathbb{Z}[I_{<k}]=\mathbb{Z}[I_{<m}]\times\mathbb{Z}.[m]$ so the bijectivity
of $g_{k}$ follows from that of $g_{m}$.

Finally, in case (c), we know that $f_{m}$ and $g_{m}$ are bijective by
assumption. Suppose that $u\in\mathbb{Z}[I_{<k}]$. We then have
$u_{m}=r\,d_{m}+s$ for some $s$ with $0\leq s<d_{m}$. Put
$u^{\prime}=u-r\,a_{m}-s\,[m]$, so $u^{\prime}_{m}=0$, so
$u^{\prime}\in\mathbb{Z}[I_{<m}]$. As $g_{m}$ is a bijection we see that there
is a unique pair $(x^{\prime},y^{\prime})\in\mathbb{Z}[J_{<m}]\times T_{m}$
with $u^{\prime}=f_{m}(x^{\prime})+y^{\prime}$. If we put
$x=x^{\prime}+r\,[m]\in\mathbb{Z}[J_{<k}]$ and $y=y^{\prime}+s\,[m]\in T_{k}$
we find that $(x,y)$ is the unique pair with $g_{k}(x,y)=u$. It follows that
$g_{k}$ is a bijection. In the case where $u\in A_{<k}$ we must have $u_{m}\in
d_{m}\mathbb{Z}$ so $s=0$ and $u^{\prime}=u-r\,a_{m}\in A_{<m}$ so
$y^{\prime}=0$; using this we see that $f_{k}$ is also an isomorphism. ∎

###### Proof of Lemma 6.16.

We actually claim that more generally, the restricted maps
$f_{n}\colon\mathbb{Z}[J_{<n}]\to A_{<n}$ are isomorphisms for all $n\leq k$.
If not, as $I_{\leq k}$ is well-ordered, there must be a smallest $n$ for
which $f_{n}$ is not an isomorphism. This means that $f_{m}$ is an isomorphism
for $m<n$, so we can apply Lemma 6.17 to $a|_{J_{<n}}$ to see that $f_{n}$ is
an isomorphism, which is a contradiction. The claim follows. ∎

######  Remark 6.18.

The method that we used to deduce Lemma 6.16 from 6.17 is called _transfinite
induction_ ; it is evidently an extension of the usual kind of induction over
the natural numbers. We will use transfinite induction again below without
spelling it out so explicitly.

As we remarked previously, Theorem 6.6 follows from Theorem 6.11, Lemma 6.16
and Remark 6.14. Because we need Theorem 6.11, the proof is unavoidably
nonconstructive. Nonetheless, we can remove one set of arbitrary choices by
pinning down a specific adapted basis, as we now explain.

######  Definition 6.19.

Let $a$ be an adapted basis for $A_{<k}$. We say that $a$ is _normalised_ if
for all $i,j\in J_{<k}$ with $i<j$ we have $0\leq a(j)_{i}<d_{i}$.

######  Proposition 6.20.

There is a unique normalised adapted basis for $A$.

This follows by transfinite induction from the following lemma:

######  Lemma 6.21.

Suppose that for all $m<\top$ there is a unique normalised basis for $A_{<m}$.
Then there is a unique normalised basis for $A$.

###### Proof.

We must again separate three cases:

  * (a)

$\top$ is not a successor.

  * (b)

$\top=s(m)$ for some $m\not\in J$.

  * (c)

$\top=s(m)$ for some $m\in J$.

We first consider case (a). For each $m<\top$ we see that $s(m)<\top$, so by
the inductive assumption we have a unique normalised adapted basis
$a_{m}\colon J_{<s(m)}\to A_{<s(m)}$. Now for $n<m$ we see that
$a_{m}|_{J_{<s(n)}}$ is an adapted basis for $A_{<s(n)}$ so it must be the
same as $a_{n}$. It follows that there is a unique map $a\colon J=J_{<\top}\to
A$ such that $a|_{J_{<s(m)}}=a_{m}$ for all $m$. (Explicitly, it is given by
$a(m)=a_{m}(m)$ for all $m<\top$.) It is straightforward to check that this is
a normalised adapted basis for $A$, and that it is the unique one.

Now consider instead case (b). Here we have $A_{<\top}=A_{<m}$ and
$J_{<\top}=J_{<m}$ so everything is trivial.

Finally, consider case (c). Let $a\colon J_{<m}\to A_{<m}$ be the unique
normalised adapted basis for $A_{<m}$. By the definition of $d_{m}$, we can
choose $b\in A$ with $b_{m}=d_{m}$. Put
$b^{\prime}=b-d_{m}\,[m]\in\mathbb{Z}[I_{<m}]$. As
$g_{m}\colon\mathbb{Z}[J_{<m}]\times T_{m}\to\mathbb{Z}[I_{<m}]$ is a
bijection, there is a unique pair $(x,y)$ with $f_{k}(x)+y=b^{\prime}$. We put
$a(m)=b-f_{k}(x)=y+d_{m}\,[m]$. The description $a(m)=b-f_{k}(x)$ shows that
$a(m)\in A$, and the description $a(m)=y+d_{m}\,[m]$ shows that $a(m)$
satisfies the conditions for a normalised adapted basis. Now suppose we have
another normalised adapted basis for $A$, say $a^{\prime}$. Then $a|_{J_{<m}}$
and $a^{\prime}|_{J_{<m}}$ are both normalised adapted bases for $A_{<m}$, so
they are the same by the induction hypothesis, so $a(j)=a^{\prime}(j)$ for all
$j<m$. We also have $a(m)_{m}=d_{m}=a^{\prime}(m)_{m}$, so the element
$u=a^{\prime}(m)-a(m)$ lies in $A_{<m}$, so $u=f_{m}(t)$ for some
$t\in\mathbb{Z}[J_{<m}]$. If $t$ is nonzero, then there are only finitely many
indices $j$ with $t_{j}\neq 0$, so we can let $k$ be the largest one. We then
find that $u_{k}=a^{\prime}(m)_{k}-a(m)_{k}=t_{k}d_{k}$, so
$a^{\prime}(m)_{k}=a(m)_{k}\pmod{d_{k}}$. On the other hand, the normalisation
condition means that $0\leq a^{\prime}(m)_{k},a(m)_{k}<d_{k}$, and this can
only be consistent if $a^{\prime}(m)_{k}=a(m)_{k}$, so $t_{k}=0$,
contradicting the choice of $k$. Thus $t$ must actually be zero, showing that
$a=a^{\prime}$ as required. ∎

##  7\. Tensor and torsion products

######  Definition 7.1.

Let $A$ be an abelian group. We make the free abelian group $\mathbb{Z}[A]$
into a commutative ring by the rule

$(\sum_{i}n_{i}[a_{i}]).(\sum_{j}m_{j}[b_{j}])=\sum_{i,j}n_{i}m_{j}[a_{i}+b_{j}]$

(so in particular $[a][b]=[a+b]$). We define a ring homomorphism
$\epsilon\colon\mathbb{Z}[A]\to\mathbb{Z}$ by
$\epsilon(\sum_{i}n_{i}[a_{i}])=\sum_{i}n_{i}$, and we define $I_{A}$ to be
the kernel of $\epsilon$. We write $I^{2}_{A}$ for the ideal generated by all
products $xy$ with $x,y\in I_{A}$. We also define a group homomorphism
$q\colon\mathbb{Z}[A]\to A$ by $q(\sum_{i}n_{i}[a_{i}])=\sum_{i}n_{i}a_{i}$.

######  Proposition 7.2.

  * (a)

The abelian groups $I_{A}$ and $I_{A}^{2}$ are both free.

  * (b)

More specifically, the elements $\langle a\rangle=[a]-[0]$ for $a\in
A\setminus 0$ form a basis for $I_{A}$.

  * (c)

Put

$\langle a,b\rangle=\langle a\rangle\langle b\rangle=[a+b]-[a]-[b]+[0]=\langle
a+b\rangle-\langle a\rangle-\langle b\rangle.$

Then the set of all elements of this form generates $I^{2}_{A}$ as an abelian
group.

  * (d)

There is a natural short exact sequence
$I^{2}_{A}\xrightarrow{j}I_{A}\xrightarrow{q}A$ (where $j$ is just the
inclusion).

###### Proof.

  * (a)

Both $I_{A}$ and $I^{2}_{A}$ are subgroups of $\mathbb{Z}[A]$, so they are
free by Theorem 6.6.

  * (b)

In the case of $I_{A}$ it is easy to be more concrete. Suppose we have an
element $x=\sum_{i}n_{i}[a_{i}]\in I_{A}$. Then $\sum_{i}n_{i}=0$, so $x$ can
also be written as $\sum_{i}n_{i}([a_{i}]-[0])$, and it is clearly harmless to
omit any terms where $a_{i}=0$, so we see that $x$ is in the subgroup
generated by the elements $[a]-[0]$ with $a\neq 0$. It is easy to see that all
such elements lie in $I_{A}$ and that they are independent over $\mathbb{Z}$,
so they form a basis for $I_{A}$ as claimed.

  * (c)

Let $M$ be the subgroup of $I_{A}$ generated by all elements of the form
$\langle a,b\rangle$. As the elements $[a]-[0]$ generate $I_{A}$ as an ideal,
it is clear from the description $\langle a,b\rangle=\langle a\rangle\langle
b\rangle$ that $M$ generates $I^{2}_{A}$ as an ideal, so it will be enough to
check that $M$ itself is already an ideal. This follows easily from the
identity $[x]\langle a,b\rangle=\langle a,b+x\rangle-\langle a,x\rangle$,
which can be verified directly from the definitions.

  * (d)

It is clear from the definitions that $q(\langle a,b\rangle)=a+b-a-b+0=0$, so
$qj=0$ by part (c), so we have an induced map $\overline{q}\colon
I_{A}/I^{2}_{A}\to A$. In the opposite direction, we can define $s\colon A\to
I_{A}/I_{A}^{2}$ by $s(a)=[a]-[0]+I_{A}^{2}$. We have

$s(a+b)-s(a)-s(b)=[a+b]-[a]-[b]+[0]+I_{A}^{2}=\langle
a,b\rangle+I_{A}^{2}=I_{A}^{2},$

which means that $s$ is a homomorphism. It is visible that
$\overline{q}s=1_{A}$, so $s$ is injective and $\overline{q}$ is surjective.
It is also clear from (b) that $s(A)$ generates $I_{A}/I_{A}^{2}$ but $s$ is a
homomorphism so $s(A)$ is already a subgroup of $I_{A}/I_{A}^{2}$, so $s$ is
surjective. This means that $s$ is an isomorphism, with inverse
$\overline{q}$. As $\overline{q}$ is an isomorphism we see that
$I^{2}_{A}\xrightarrow{j}I_{A}\xrightarrow{q}A$ is exact.

∎

######  Definition 7.3.

Let $A$ and $B$ be abelian groups. We regard $A$ and $B$ as subgroups of
$A\times B$ in the obvious way. In $\mathbb{Z}[A\times B]$ we let $J$ be the
ideal generated by all elements $\langle a\rangle$ with $a\in A$, and we let
$K$ be the ideal generated by all elements $\langle b\rangle$ with $b\in B$.
We then put $A\otimes B=JK/(J^{2}K+JK^{2})$, and write $a\otimes b$ for the
coset $\langle a,b\rangle+J^{2}K+JK^{2}\in A\otimes B$. We also write
$\operatorname{Tor}(A,B)=(J^{2}K\cap JK^{2})/(J^{2}K^{2})$.

######  Remark 7.4.

Note that $A\otimes B$ is generated by elements of the form $a\otimes b$.
Moreover, because

$\displaystyle\langle a+a^{\prime},b\rangle-\langle a,b\rangle-\langle
a^{\prime},b\rangle$ $\displaystyle=\langle a\rangle\langle
a^{\prime}\rangle\langle b\rangle\in J^{2}K$ $\displaystyle\langle
a,b+b^{\prime}\rangle-\langle a,b\rangle-\langle a,b^{\prime}\rangle$
$\displaystyle=\langle a\rangle\langle b\rangle\langle b^{\prime}\rangle\in
JK^{2}$

we see that these satisfy

$\displaystyle a\otimes(b+b^{\prime})$ $\displaystyle=(a\otimes b)+(a\otimes
b^{\prime})$ $\displaystyle(a+a^{\prime})\otimes b$ $\displaystyle=(a\otimes
b)+(a^{\prime}\otimes b).$

It follows easily that $a\otimes 0=0$ for all $a\in A$, and $0\otimes b=0$ for
all $b\in B$, and $(na)\otimes(mb)=nm(a\otimes b)$ for all $n,m\in\mathbb{Z}$.
In fact, $A\otimes B$ can be defined more loosely as the abelian group
generated by symbols $a\otimes b$ subject only to the relations
$a\otimes(b+b^{\prime})=(a\otimes b)+(a\otimes b^{\prime})$ and
$(a+a^{\prime})\otimes b=(a\otimes b)+(a^{\prime}\otimes b)$.

######  Remark 7.5.

Suppose we have homomorphisms $f\colon A\to A^{\prime}$ and $g\colon B\to
B^{\prime}$. These give a homomorphism $f\times g\colon A\times B\to
A^{\prime}\times B^{\prime}$, which induces a ring map $(f\times
g)_{\bullet}\colon\mathbb{Z}[A\times B]\to\mathbb{Z}[A^{\prime}\times
B^{\prime}]$. This sends the ideals $J$ and $K$ to the coresponding ideals in
$\mathbb{Z}[A^{\prime}\times B^{\prime}]$ and so induces a homomorphism
$A\otimes B\to A^{\prime}\otimes B^{\prime}$, which we denote by $f\otimes g$.
By construction we have $(f\otimes g)(a\otimes b)=f(a)\otimes g(b)$. It is not
hard to see that this construction is functorial, in the sense that
$1_{A}\otimes 1_{B}=1_{A\otimes B}$ and that $(f^{\prime}\otimes
g^{\prime})(f\otimes g)=(f^{\prime}f)\otimes(g^{\prime}g)$ for all
$f^{\prime}\colon A^{\prime}\to A^{\prime\prime}$ and $g^{\prime}\colon
B^{\prime}\to B^{\prime\prime}$. It is also bilinear in the following sense:
if $f_{0},f_{1}\colon A\to A^{\prime}$ and $g_{0},g_{1}\colon B\to B^{\prime}$
then

$(f_{0}+f_{1})\otimes(g_{0}+g_{1})=(f_{0}\otimes g_{0})+(f_{0}\otimes
g_{1})+(f_{1}\otimes g_{0})+(f_{1}\otimes g_{1})$

as homomorphisms from $A\otimes B$ to $A^{\prime}\otimes B^{\prime}$.

######  Remark 7.6.

As in Proposition 7.2, one can check that

  * (a)

$J$ is freely generated as an abelian group by the elements $\langle
a\rangle[b]=[a+b]-[b]$ with $a\in A\setminus\\{0\\}$ and $b\in B$.

  * (b)

$K$ is freely generated as an abelian group by the elements $[a]\langle
b\rangle=[a+b]-[a]$ with $a\in A$ and $b\in B\setminus 0$.

  * (c)

$JK$ is freely generated as an abelian group by the elements $\langle
a,b\rangle$ with $a\in A\setminus 0$ and $b\in B\setminus 0$.

  * (d)

$J^{2}K$ is generated as an abelian group by the elements $\langle
a\rangle\langle a^{\prime}\rangle\langle b\rangle$ with $a,a^{\prime}\in A$
and $b\in B$.

  * (e)

$JK^{2}$ is generated as an abelian group by the elements $\langle
a\rangle\langle b\rangle\langle b^{\prime}\rangle$ with $a\in A$ and
$b,b^{\prime}\in B$.

We can generalise the identities in Remark 7.4 as follows:

######  Definition 7.7.

Let $A$, $B$ and $V$ be abelian groups. We say that a function $f\colon
A\times B\to V$ is _bilinear_ if for all $a,a^{\prime}\in A$ and all
$b,b^{\prime}\in B$ we have

$\displaystyle f(a,b+b^{\prime})$ $\displaystyle=f(a,b)+f(a,b^{\prime})$
$\displaystyle f(a+a^{\prime},b)$ $\displaystyle=f(a,b)+f(a^{\prime},b).$

More generally, we say that a map $g\colon A_{1}\times\dotsb\times A_{n}\to V$
is _multilinear_ (or more specifically _$n$ -linear_) if for each $k$ and each
$a_{1},\dotsc,a_{k-1},a_{k+1},\dotsc,a_{n}$, the map

$x\mapsto g(a_{1},\dotsc,a_{k-1},x,a_{k+1},\dotsc,a_{n})$

is a homomorphism from $A_{k}$ to $V$.

######  Example 7.8.

  * (a)

Matrix multiplication defines a bilinear map $\mu\colon
M_{n}(\mathbb{Z})\times M_{n}(\mathbb{Z})\to M_{n}(\mathbb{Z})$ by
$\mu(M,N)=MN$.

  * (b)

The dot product defines a bilinear map
$\mathbb{R}^{3}\times\mathbb{R}^{3}\to\mathbb{R}$, the cross product defines a
bilinear map $\mathbb{R}^{3}\times\mathbb{R}^{3}\to\mathbb{R}^{3}$, and the
triple product $(u,v,w)\mapsto u.(v\times w)$ defines a trilinear map
$\mathbb{R}^{3}\times\mathbb{R}^{3}\times\mathbb{R}^{3}\to\mathbb{R}$.

  * (c)

By construction, we have a bilinear map $\omega\colon A\times B\to A\otimes B$
defined by $\omega(a,b)=a\otimes b$.

Example (c) above is in a sense the universal example, as explained by the
following result:

######  Proposition 7.9.

Let $f\colon A\times B\to V$ be a bilinear map. Then there is a unique
homomorphism $\overline{f}\colon A\otimes B\to V$ such that
$\overline{f}(a\otimes b)=f(a,b)$ for all $a\in A$ and $b\in B$ (or
equivalently $\overline{f}\circ\omega=f$).

###### Proof.

As $A\otimes B$ is generated by the elements $a\otimes b$, it is clear that
$\overline{f}$ will be unique if it exists.

By Lemma 6.5, there is a unique homomorphism $f_{0}\colon\mathbb{Z}[A\times
B]\to V$ such that $f_{0}([a,b])=f(a,b)$ for all $a$ and $b$. Note that for
$a\in A$ we have $f_{0}([a])=f_{0}([a,0])=f(a,0)=0$, and similarly
$f_{0}([b])=0$ for $b\in B$, so

$f_{0}(\langle
a,b\rangle)=f_{0}([a,b])-f_{0}([a,0])-f_{0}([0,b])+f_{0}([0,0])=f(a,b).$

Note also that

$f_{0}(\langle a\rangle\langle a^{\prime}\rangle\langle
b\rangle)=f_{0}(\langle a+a^{\prime},b\rangle)-f_{0}(\langle
a,b\rangle)-f_{0}(\langle
a^{\prime},b\rangle)=f(a+a^{\prime},b)-f(a,b)-f(a^{\prime},b)=0,$

so (using Remark 7.6(d)) we see that $f_{0}(J^{2}K)=0$. Similarly, we have
$f_{0}(JK^{2})=0$, so $f_{0}$ induces a homomorphism

$\overline{f}\colon A\otimes B=\frac{JK}{J^{2}K+JK^{2}}\to V$

with

$\overline{f}(a\otimes b)=\overline{f}(\langle
a,b\rangle+J^{2}K+JK^{2})=f_{0}(\langle a,b\rangle)=f(a,b)$

as required. ∎

######  Remark 7.10.

For example, we have a bilinear map $\mu\colon M_{n}(\mathbb{Z})\times
M_{n}(\mathbb{Z})\to M_{n}(\mathbb{Z})$ given by $\mu(M,N)=MN$, so there is a
unique homomorphism $\overline{\mu}\colon M_{n}(\mathbb{Z})\otimes
M_{n}(\mathbb{Z})\to M_{n}(\mathbb{Z})$ such that $\overline{\mu}(M\otimes
N)=MN$. Rather than spelling this out explicitly, we will usually just say
that $\overline{\mu}\colon M_{n}(\mathbb{Z})\otimes M_{n}(\mathbb{Z})\to
M_{n}(\mathbb{Z})$ is defined by $\overline{\mu}(M\otimes N)=MN$.

It will be convenient to reformulate Proposition 7.9 in a different way. We
write $\operatorname{Hom}(A,B)$ for the set of homomorphisms from $A$ to $B$,
considered as a group under pointwise addition. Similarly, we write
$\operatorname{Bilin}(A,B;V)$ for the group of bilinear maps from $A\times B$
to $V$.

######  Proposition 7.11.

For any abelian groups $A$, $B$ and $V$, there are natural isomorphisms

$\operatorname{Hom}(A\otimes
B,V)\simeq\operatorname{Bilin}(A,B;V)\simeq\operatorname{Hom}(A,\operatorname{Hom}(B,V))\simeq\operatorname{Hom}(B,\operatorname{Hom}(A,V)).$

More specifically, if we have elements

$f_{0}\in\operatorname{Hom}(A\otimes B,V)\hskip
20.00003ptf_{1}\in\operatorname{Bilin}(A,B;V)\hskip
20.00003ptf_{2}\in\operatorname{Hom}(A,\operatorname{Hom}(B,V))\hskip
20.00003ptf_{3}\in\operatorname{Hom}(B,\operatorname{Hom}(A,V))$

then they are related by the above isomorphisms if and only if for all $a\in
A$ and $b\in B$ we have

$f_{0}(a\otimes b)=f_{1}(a,b)=f_{2}(a)(b)=f_{3}(b)(a).$

###### Proof.

This is mostly trivial. For any map $f_{1}\colon A\times B\to V$, we can
define a map $f_{2}\colon A\to\operatorname{Map}(B,V)$ by
$f_{2}(a)(b)=f_{1}(a,b)$. If $f_{1}$ satisfies the right-linearity condition
$f_{1}(a,b+b^{\prime})$ we see that
$f_{2}(a)(b+b^{\prime})=f_{2}(a)(b)+f_{2}(a)(b^{\prime})$, so $f_{2}(a)$ is a
homomorphism from $B$ to $V$, or in other words $f_{2}$ is a map from $A$ to
$\operatorname{Hom}(B,V)$. If $f_{1}$ also satisfies the left linearity
condition $f_{1}(a+a^{\prime},b)=f_{1}(a,b)+f_{1}(a^{\prime},b)$ then we see
that $f_{2}(a+a^{\prime})$ is the sum of the homomorphisms $f_{2}(a)$ and
$f_{2}(a^{\prime})$, so $f_{2}$ itself is a homomorphism, or in other words
$f_{2}\in\operatorname{Hom}(A,\operatorname{Hom}(B,V))$. All of this is
reversible, so we have an isomorphism
$\operatorname{Bilin}(A,B;V)\simeq\operatorname{Hom}(A,\operatorname{Hom}(B,V))$.
We can define $f_{3}(b)(a)=f_{1}(a,b)$ to obtain a similar isomorphism
$\operatorname{Bilin}(A,B;V)\simeq\operatorname{Hom}(B,\operatorname{Hom}(A,V))$.
Finally, Proposition 7.9 gives an isomorphism
$\operatorname{Bilin}(A,B;V)\simeq\operatorname{Hom}(A\otimes B,V)$. ∎

######  Proposition 7.12.

There are natural isomorphisms as follows:

$\displaystyle\eta_{A}\colon\mathbb{Z}\otimes A$ $\displaystyle\to A$
$\displaystyle\eta_{A}(n\otimes a)$ $\displaystyle=na$
$\displaystyle\tau_{AB}\colon A\otimes B$ $\displaystyle\to B\otimes A$
$\displaystyle\tau_{AB}(a\otimes b)$ $\displaystyle=b\otimes a$
$\displaystyle\alpha_{ABC}\colon A\otimes(B\otimes C)$
$\displaystyle\to(A\otimes B)\otimes C$
$\displaystyle\alpha_{ABC}(a\otimes(b\otimes c))$ $\displaystyle=(a\otimes
b)\otimes c.$

In other words, the operation $(-)\otimes(-)$ is commutative, associative and
unital up to natural isomorphism.

###### Proof.

First, we certainly have a bilinear map
$\eta^{\prime}_{A}\colon\mathbb{Z}\times A\to A$ given by
$\eta^{\prime}_{A}(n,a)=na$, and by Proposition 7.9 this gives a homomorphism
$\eta_{A}\colon\mathbb{Z}\otimes A\to A$ as indicated. We can also define a
homomorphism $\zeta_{A}\colon A\to\mathbb{Z}\otimes A$ by
$\zeta_{A}(a)=1\otimes a$, and it is clear that $\eta_{A}\zeta_{A}=1_{A}$. In
the opposite direction, we must show that the map

$\xi=1_{\mathbb{Z}\otimes A}-\zeta_{A}\eta_{A}\colon\mathbb{Z}\otimes
A\to\mathbb{Z}\otimes A$

is zero. By Proposition 7.11, it will suffice to show that the corresponding
homomorphism
$\xi^{\prime}\colon\mathbb{Z}\to\operatorname{Hom}(A,\mathbb{Z}\otimes A)$ is
zero. This is given by

$\xi^{\prime}(n)(a)=(n\otimes a)-(1\otimes na)$

so visibly $\xi^{\prime}(1)=0$ but $\xi^{\prime}$ is a homomorphism so
$\xi^{\prime}(n)=n.\xi^{\prime}(1)=0$ for all $n$ as required.

Next, Lemma 6.5 tells us that there is a unique homomorphism
$\tau^{\prime}_{AB}\colon\mathbb{Z}[A\times B]\to\mathbb{Z}[B\times A]$ with
$\tau^{\prime}_{AB}[a,b]=[b,a]$. It is visible that this is an isomorphism
(with inverse $\tau^{\prime}_{BA}$) and that
$\tau^{\prime}_{AB}(R_{AB})=R_{BA}$ so there is an induced isomorphism
$\tau_{AB}\colon A\otimes B\to B\otimes A$, with inverse $\tau_{BA}$.

Now fix $a\in A$, and define $\alpha^{\prime\prime}(a)\colon B\times
C\to(A\otimes B)\otimes C$ by $\alpha^{\prime\prime}(a)(b,c)=(a\otimes
b)\otimes c$. This is bilinear, and moreover
$\alpha^{\prime\prime}(a+a^{\prime})=\alpha^{\prime\prime}(a)+\alpha^{\prime\prime}(a^{\prime})$,
so we have a homomorphism

$\alpha^{\prime\prime}\colon A\to\operatorname{Bilin}(B,C;(A\otimes B)\otimes
C).$

We also have an isomorphism

$\operatorname{Bilin}(B,C;(A\otimes B)\otimes
C)\simeq\operatorname{Hom}(B\otimes C,(A\otimes B)\otimes C)$

and using that we obtain a homomorphism

$\alpha^{\prime}\colon A\to\operatorname{Hom}(B\otimes C;(A\otimes B)\otimes
C).$

characterised by $\alpha^{\prime}(a)(b\otimes c)=(a\otimes b)\otimes c$. As in
Proposition 7.11 this corresponds to a homomorphism $\alpha\colon
A\otimes(B\otimes C)\to(A\otimes B)\otimes C$, given by

$\alpha(a\otimes(b\otimes c))=\alpha^{\prime}(a)(b\otimes c)=(a\otimes
b)\otimes c.$

In the same way, we can construct $\beta\colon(A\otimes B)\otimes C\to
A\otimes(B\otimes C)$ with $\beta((a\otimes b)\otimes c)=a\otimes(b\otimes
c)$. This means that $\beta\alpha(x)=x$ whenever $x$ has the form
$a\otimes(b\otimes c)$, but elements of that form generate $A\otimes(B\otimes
C)$, so $\beta\alpha=1$. A similar argument shows that $\alpha\beta=1$, so
$\alpha$ is an isomorphism as required. ∎

######  Proposition 7.13.

For any families of abelian groups $(A_{i})_{i\in I}$ and $(B_{j})_{j\in J}$
there is a natural isomorphism

$\left(\bigoplus_{i\in I}A_{i}\right)\otimes\left(\bigoplus_{j\in
J}B_{j}\right)\simeq\bigoplus_{(i,j)\in I\times J}(A_{i}\otimes B_{j}).$

In particular, for any $A$, $B$ and $C$ we have $A\otimes(B\oplus
C)\simeq(A\otimes B)\oplus(A\otimes C)$.

###### Proof.

For brevity, let $L$ and $R$ be the left and right hand sides of the claimed
isomorphism. For each $i$ and $j$ we can define a bilinear map
$g^{\prime}_{ij}\colon A_{i}\times B_{j}\to L$ by
$g^{\prime}_{ij}(a,b)=\iota_{i}(a)\otimes\iota_{j}(b)$. This gives a
homomorphism $g_{ij}\colon A_{i}\otimes B_{j}\to L$, and Proposition 3.7 tells
us that there is a unique $g\colon R\to L$ with $g\circ\iota_{ij}=g_{ij}$ for
all $i$ and $j$. In the opposite direction, we can define a bilinear map
$f^{\prime}\colon(\bigoplus A_{i})\times(\bigoplus B_{j})\to R$ by

$f^{\prime}(a,b)=\sum_{i\in\operatorname{supp}(a)}\sum_{j\in\operatorname{supp}(b)}\iota_{ij}(a_{i}\otimes
b_{j}).$

This corresponds in the usual way to a homomorphism $f\colon L\to R$. We leave
it to the reader to check that $fg=1_{R}$ and $gf=1_{L}$. ∎

######  Corollary 7.14.

Given sets $I$ and $J$ and an abelian group $B$, we have natural isomorphisms
$\mathbb{Z}[I]\otimes B\simeq\bigoplus_{i\in I}B$ and
$\mathbb{Z}[I]\otimes\mathbb{Z}[J]\simeq\mathbb{Z}[I\times J]$. In particular,
this gives $\mathbb{Z}^{r}\otimes B\simeq B^{r}$ and
$\mathbb{Z}^{r}\otimes\mathbb{Z}^{s}\simeq\mathbb{Z}^{rs}$.

###### Proof.

By applying the Proposition to the family $\\{\mathbb{Z}\\}_{i\in I}$ and the
family consisting of the single group $B$, we obtain $\mathbb{Z}[I]\otimes
B\simeq\bigoplus_{i\in I}B$. If instead we use the family
$\\{\mathbb{Z}\\}_{j\in J}$ on the right hand side, we obtain
$\mathbb{Z}[I]\otimes\mathbb{Z}[J]\simeq\mathbb{Z}[I\times J]$. ∎

Another straightforward example is as follows:

######  Proposition 7.15.

For any integer $n$ and any abelian group $A$ there is a natural isomorphism
$(\mathbb{Z}/n)\otimes A\simeq A/nA$. In particular, we have
$(\mathbb{Z}/n)\otimes(\mathbb{Z}/m)=\mathbb{Z}/(n,m)$, where $(n,m)$ denotes
the greatest common divisor of $n$ and $m$.

###### Proof.

We can define a bilinear map $f^{\prime}\colon(\mathbb{Z}/n)\otimes A\to A/nA$
by $f^{\prime}(k+n\mathbb{Z},a)=ka+nA$, and this induces a homomorphism
$f\colon(\mathbb{Z}/n)\otimes A\to A/nA$. In the opposite direction, we can
define $g\colon A/nA\to(\mathbb{Z}/n)\otimes A$ by
$g(a+nA)=(1+n\mathbb{Z})\otimes a$. We leave it to the reader to check that
these are well-defined and that $fg$ and $gf$ are identity maps. In
particular, this gives
$(\mathbb{Z}/n)\otimes(\mathbb{Z}/m)=\mathbb{Z}/(n\mathbb{Z}+m\mathbb{Z})$,
but it is standard that $n\mathbb{Z}+m\mathbb{Z}=(n,m)\mathbb{Z}$. ∎

We saw in Section 5 that every finitely generated abelian group is a direct
sum of groups of the form $\mathbb{Z}$ or $\mathbb{Z}/p^{v}$. We can thus use
Proposition 7.13, Corollary 7.14 and Proposition 7.15 to understand the tensor
product of any two finitely generated abelian groups.

We next consider the interaction between tensor products and exactness.

######  Proposition 7.16.

Let $A$, $B$, $C$ and $U$ be abelian groups.

  * (a)

If we have an exact sequence

$A\xrightarrow{j}B\xrightarrow{q}C\to 0,$

then the resulting sequence

$U\otimes A\xrightarrow{1\otimes j}U\otimes B\xrightarrow{1\otimes q}U\otimes
C\to 0$

is also exact. (In other words, tensoring is _right exact_.)

  * (b)

If we have an injective map $j\colon A\to B$ and $U$ is torsion-free, then
$1\otimes j\colon U\otimes A\to U\otimes B$ is also injective.

  * (c)

If we have a short exact sequence

$0\to A\xrightarrow{j}B\xrightarrow{q}C\to 0,$

and $U$ is torsion-free, then the resulting sequence

$0\to U\otimes A\xrightarrow{1\otimes j}U\otimes B\xrightarrow{1\otimes
q}U\otimes C\to 0$

is also short exact.

For the first part, we will use the following criterion:

######  Lemma 7.17.

A sequence $A\xrightarrow{j}B\xrightarrow{q}C\to 0$ is exact iff for every
abelian group $V$, the resulting sequence
$0\to\operatorname{Hom}(C,V)\xrightarrow{q^{*}}\operatorname{Hom}(B,V)\xrightarrow{j^{*}}\operatorname{Hom}(A,V)$
is exact.

######  Remark 7.18.

The evident analogous statement with short exact sequences is not valid. We
will investigate this in more detail later.

###### Proof.

Let ${\mathcal{S}}$ denote the first sequence, and write
$\operatorname{Hom}({\mathcal{S}},V)$ for the second one.

Suppose that ${\mathcal{S}}$ is exact, so $q$ is surjective and
$\ker(q)=\operatorname{image}(j)$. Suppose that $f\in\ker(q^{*})$, so $f\colon
C\to V$ and $fq=0\colon B\to V$. This means that $f(q(b))=0$ for all $b\in B$,
but $q$ is surjective, so $f(c)=0$ for all $c\in C$, so $f=0$. This proves
that $\ker(q^{*})=0$, so $q^{*}$ is injective. Now suppose that
$g\in\ker(j^{*})$, so $g\colon B\to V$ and $gj=0$, or equivalently
$g(\operatorname{image}(j))=0$, or equivalently $g(\ker(q))=0$. We thus have a
well-defined map $f\colon C\to V$ given by $f(c)=g(b)$ for any $b$ with
$q(b)=c$. Now $f\in\operatorname{Hom}(C,V)$ and $q^{*}(f)=fq=g$, so
$g\in\operatorname{image}(q^{*})$. This proves that
$\ker(j^{*})=\operatorname{image}(q^{*})$, so
$\operatorname{Hom}({\mathcal{S}},V)$ is exact.

Conversely, suppose that $\operatorname{Hom}({\mathcal{S}},V)$ is exact for
all $V$. Take $V=\operatorname{cok}(q)=C/q(B)$ and let $f\colon C\to V$ be the
evident projection (which is surjective). By construction we have $q^{*}(f)=0$
but $q^{*}$ is assumed to be injective so $f$ is zero as well as being
surjective. This implies that $C/q(B)=0$ so $q$ is surjective.

Now instead take $V=C$. As $\operatorname{Hom}({\mathcal{S}},C)$ is exact, we
certainly have
$j^{*}q^{*}=0\colon\operatorname{Hom}(C,C)\to\operatorname{Hom}(A,C)$. In
particular, we see that $j^{*}q^{*}(1)=0$ in $\operatorname{Hom}(A,C)$, or in
other words that $qj=0\colon A\to C$. This implies that
$\operatorname{image}(j)\leq\ker(q)$.

Finally, take $V=\operatorname{cok}(j)=B/j(A)$, and let $g\colon B\to V$ be
the projection. Then $j^{*}(g)=gj=0$, so
$g\in\ker(j^{*})=\operatorname{image}(q^{*})$, so there exists $f\colon C\to
B/j(A)$ with $fq=g\colon B\to B/j(A)$. Now if $q(b)=0$ then
$b+\operatorname{image}(j)=g(b)=f(q(b))=0$, so $b\in\operatorname{image}(j)$.
This proves that $\ker(q)=\operatorname{image}(j)$, so ${\mathcal{S}}$ is
exact as claimed. ∎

###### Proof of Proposition 7.16.

  * (a)

By Lemma 7.17, the sequence

$0\to\operatorname{Hom}(C,V)\xrightarrow{q^{*}}\operatorname{Hom}(B,V)\xrightarrow{j^{*}}\operatorname{Hom}(A,V)$

is exact for all $V$. As $V$ is arbitrary we can replace it by
$\operatorname{Hom}(U,V)$, where $U$ and $V$ are both arbitrary. This gives an
exact sequence

$0\to\operatorname{Hom}(C,\operatorname{Hom}(U,V))\xrightarrow{q^{*}}\operatorname{Hom}(B,\operatorname{Hom}(U,V))\xrightarrow{j^{*}}\operatorname{Hom}(A,\operatorname{Hom}(U,V)),$

and we can use Proposition 7.11 to rewrite it as

$0\to\operatorname{Hom}(U\otimes C,V)\xrightarrow{(1\otimes
q)^{*}}\operatorname{Hom}(U\otimes B,V)\xrightarrow{(1\otimes
j)^{*}}\operatorname{Hom}(U\otimes A,V).$

Finally we apply Lemma 7.17 in the opposite direction to see that the sequence
$U\otimes A\to U\otimes B\to U\otimes C\to 0$ is exact.

  * (b)

Now suppose instead that we have an injective map $j\colon A\to B$, and that
$U$ is torsion-free. We must show that $(1\otimes j)\colon U\otimes A\to
U\otimes B$ is injective. If $U$ is actually free then we may assume that
$U=\mathbb{Z}[I]$ for some set $I$. In this case Corollary 7.14 tells us that
$1\otimes j$ is just a direct sum of copies of $j$ and the claim is clear. In
particular, this holds whenever $U$ is finitely generated and torsion-free, as
we see from Proposition 5.11. The real issue is to deduce the infinitely
generated case from the finitely generated case. Suppose we have an element
$x\in U\otimes A$ with $(1\otimes j)(x)=0$. We can write $x$ in the form
$x=\sum_{i=1}^{n}u_{i}\otimes a_{i}$ say. We then have
$\sum_{i=1}^{n}u_{i}\otimes j(a_{i})=0$. Going back to Definition 7.3, we
deduce that $\sum_{i}[u_{i},j(a_{i})]$ can be expressed in $\mathbb{Z}[U\times
B]$ as a finite $\mathbb{Z}$-linear combination of terms of the form
$[u+u^{\prime},b]-[u,b]-[u^{\prime},b]$ or
$[u,b+b^{\prime}]-[u,b]-[u,b^{\prime}]$. Choose such an expression, and let
$S$ be the (finite) set of elements of $U$ that are involved in that
expression, together with the elements $u_{1},\dotsc,u_{n}$ occuring in our
expression for $x$. Let $U_{0}$ be the subgroup of $U$ generated by $S$, which
is finitely generated and torsion-free. We now have an element
$x_{0}=\sum_{i}u_{i}\otimes a_{i}$ in $U_{0}\otimes A$ and we find that
$(1\otimes j)(x_{0})=0$ in $U_{0}\otimes B$. By the finitely generated case we
see that $x_{0}=0$, but $x$ is the image of $x_{0}$ under the evident
homomorphism $A\otimes U_{0}\to A\otimes U$, so $x=0$ as required.

  * (c)

This is just a straightforward combination of (a) and (b).

∎

Note that in part (b) of the Proposition, it is definitely necessary to assume
that $U$ is torsion-free. Indeed, we can take $j\colon A\to B$ to be
$n.1_{\mathbb{Z}}\colon\mathbb{Z}\to\mathbb{Z}$, and then $1\otimes j$ is just
$n.1_{U}$, and these maps are injective for all $n>0$ if and only if $U$ is
torsion free.

For various purposes it is important to understand the kernel of $1\otimes j$
in more detail. We will first discuss the case $U=\mathbb{Z}/n$, which is
quite straightforward.

######  Definition 7.19.

We write $A[n]$ for $\\{a\in A\;|\;na=0\\}$, which can also be identified with
$\operatorname{Hom}(\mathbb{Z}/n,A)$. (A homomorphism $u\colon\mathbb{Z}/n\to
A$ corresponds to the element $u(1+n\mathbb{Z})\in A[n]$.)

######  Proposition 7.20.

Fix an integer $n>0$. For any short exact sequence
$A\xrightarrow{j}B\xrightarrow{q}C$, there is a unique homomorphism
$\delta\colon C[n]\to A/nA=(\mathbb{Z}/n)\otimes A$ such that
$\delta(q(b))=a+nA$ whenever $nb=j(a)$. Moreover, this fits into an exact
sequence

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

###### Proof.

This is just the Snake Lemma (Proposition 2.4) applied to the diagram

${A}$${B}$${C}$${A}$${B}$${C}$$\scriptstyle{j}$$\scriptstyle{n.1_{A}}$$\scriptstyle{q}$$\scriptstyle{n.1_{B}}$$\scriptstyle{n.1_{C}}$$\scriptstyle{j}$$\scriptstyle{q}$

∎

It turns out that there are similar six-term exact sequences in much greater
generality, involving the groups $\operatorname{Tor}(A,B)$ introduced in
Definition 7.3. We start by recording an obvious fact:

######  Lemma 7.21.

There is an isomorphism $\tau_{AB}\colon\mathbb{Z}[A\times
B]\to\mathbb{Z}[B\times A]$ given by $\tau_{AB}([a,b])=[b,a]$, and this
induces an isomorphism
$\tau_{AB}\colon\operatorname{Tor}(A,B)\to\operatorname{Tor}(B,A)$ with
inverse $\tau_{BA}$. ∎

######  Lemma 7.22.

Let $A$ and $B$ be abelian groups, and let $J$ and $K$ be ideals in
$\mathbb{Z}[A\times B]$ as in Definition 7.3. Then there are natural exact
sequences

$0\to\operatorname{Tor}(A,B)\to A\otimes I_{B}^{2}\xrightarrow{1\otimes
j_{B}}A\otimes I_{B}\xrightarrow{1\otimes q_{B}}A\otimes B\to 0$

and

$0\to\operatorname{Tor}(A,B)\to I^{2}_{A}\otimes B\xrightarrow{j_{A}\otimes
1}I_{A}\otimes B\xrightarrow{q_{A}\otimes 1}A\otimes B\to 0.$

###### Proof.

Let $J$ and $K$ be as in Definition 7.3. As $JK^{2}\leq JK$ and
$J^{2}K^{2}\leq J^{2}K$ we have a map
$f\colon(JK^{2})/(J^{2}K^{2})\to(JK)/(J^{2}K)$ given by
$f(x+J^{2}K^{2})=x+J^{2}K$. The cokernel is $JK/(J^{2}K+JK^{2})$, which is
$A\otimes B$ by definition. The kernel is $(JK^{2}\cap J^{2}K)/(J^{2}K^{2})$,
which is $\operatorname{Tor}(A,B)$ by definition. In other words, we have an
exact sequence

$0\to\operatorname{Tor}(A,B)\to\frac{JK^{2}}{J^{2}K^{2}}\xrightarrow{f}\frac{JK}{J^{2}K}\to
A\otimes B\to 0.$

Next, we have $\mathbb{Z}[A\times B]=\mathbb{Z}[A]\otimes\mathbb{Z}[B]$ by
Corollary 7.14. For $p,q\geq 0$ we have ideals $I_{A}^{p}\leq\mathbb{Z}[A]$
and $I_{B}^{q}\leq\mathbb{Z}[B]$. These are free abelian groups by Theorem
6.6, so using Proposition 7.16(b) we see that the evident homomorphisms

${I_{A}^{p}\otimes
I_{B}^{q}}$${I_{A}^{p}\otimes\mathbb{Z}[B]}$${\mathbb{Z}[A]\otimes
I_{B}^{q}}$${\mathbb{Z}[A]\otimes\mathbb{Z}[B]}$

are all injective. This means that $I_{A}^{p}\otimes I_{B}^{q}$ can be
identified with its image in $\mathbb{Z}[A\times B]$, which is just
$J^{p}K^{q}$. Our exact sequence now takes the form

$0\to\operatorname{Tor}(A,B)\to\frac{I_{A}\otimes I_{B}^{2}}{I_{A}^{2}\otimes
I_{B}^{2}}\xrightarrow{f}\frac{I_{A}\otimes I_{B}}{I_{A}^{2}\otimes I_{B}}\to
A\otimes B\to 0.$

Next, as tensoring is right exact (Proposition 7.16(a)) we can identify
$(I_{A}\otimes I_{B}^{2})/(I_{A}^{2}\otimes I_{B}^{2})$ with
$(I_{A}/I^{2}_{A})\otimes I_{B}^{2}$, which is $A\otimes I_{B}^{2}$ by
Proposition 7.2. Similarly, we can identify $(I_{A}\otimes
I_{B}^{2})/(I_{A}^{2}\otimes I_{B}^{2})$ with $A\otimes I_{B}$, so our exact
sequence becomes

$0\to\operatorname{Tor}(A,B)\to A\otimes I_{B}^{2}\xrightarrow{}A\otimes
I_{B}\xrightarrow{}A\otimes B\to 0$

as claimed. The other sequence is obtained symmetrically, or by appealing to
Lemma 7.21. ∎

######  Corollary 7.23.

If $A$ or $B$ is torsion-free then $\operatorname{Tor}(A,B)=0$. In particular,
this holds if $A$ or $B$ is free.

###### Proof.

If $A$ is torsion-free then the homomorphism

$A\otimes I_{B}^{2}\to A\otimes I_{B}$

is injective by Proposition 7.16(b), but $\operatorname{Tor}(A,B)$ is the
kernel so $\operatorname{Tor}(A,B)=0$. The other case follows symmetrically. ∎

We next discuss the functorial properties of Tor groups. Suppose we have
homomorphisms $f\colon A\to A^{\prime}$ and $g\colon B\to B^{\prime}$. As
discussed in Remark 7.5, these give a homomorphism $f\times g\colon A\times
B\to A^{\prime}\times B^{\prime}$, which induces a ring map $(f\times
g)_{\bullet}\colon\mathbb{Z}[A\times B]\to\mathbb{Z}[A^{\prime}\times
B^{\prime}]$, sending $J$ and $K$ to the coresponding ideals in
$\mathbb{Z}[A^{\prime}\times B^{\prime}]$. It therefore induces a homomorphism
$\operatorname{Tor}(A,B)\to\operatorname{Tor}(A^{\prime},B^{\prime})$, which
we denote by $\operatorname{Tor}(f,g)$. This makes $\operatorname{Tor}(A,B)$ a
functor of the pair $(A,B)$.

Now suppose we have two homomorphisms $f_{0},f_{1}\colon A\to A^{\prime}$. We
then have ring maps $(f_{0})_{\bullet}$, $(f_{1})_{\bullet}$ and
$(f_{0}+f_{1})_{\bullet}$ from $\mathbb{Z}[A]$ to $\mathbb{Z}[A^{\prime}]$,
and it is not true that
$(f_{0}+f_{1})_{\bullet}=(f_{0})_{\bullet}+(f_{1})_{\bullet}$. Because of
this, it is not obvious that
$\operatorname{Tor}(f_{0}+f_{1},g)=\operatorname{Tor}(f_{0},g)+\operatorname{Tor}(f_{1},g)$.
However, this does turn out to be true, as we now prove.

######  Proposition 7.24.

  * (a)

Given homomorphisms $f_{0},f_{1}\colon A\to A^{\prime}$ and $g\colon B\to
B^{\prime}$ we have
$\operatorname{Tor}(f_{0}+f_{1},g)=\operatorname{Tor}(f_{0},g)+\operatorname{Tor}(f_{1},g)$.

  * (b)

Given homomorphisms $f\colon A\to A^{\prime}$ and $g_{0},g_{1}\colon B\to
B^{\prime}$ we have
$\operatorname{Tor}(f,g_{0}+g_{1})=\operatorname{Tor}(f,g_{0})+\operatorname{Tor}(f,g_{1})$.

  * (c)

Given families of groups $\\{A_{i}\\}_{i\in I}$ and $\\{B_{j}\\}_{j\in J}$
there is a natural isomorphism

$\bigoplus_{i,j}\operatorname{Tor}(A_{i},B_{j})\to\operatorname{Tor}\left(\bigoplus_{i}A_{i},\bigoplus_{j}B_{j}\right).$

###### Proof.

Lemma 7.22 allows us to describe $\operatorname{Tor}(A,B)$ as the kernel of
the map $1\otimes j\colon A\otimes I_{B}^{2}\to A\otimes I_{B}$, and claim (a)
follows easily from this. Claim (b) is proved similarly.

For (c), let us put $A^{*}=\bigoplus_{i}A_{i}$ and $B^{*}=\bigoplus_{j}B_{j}$
and $T=\bigoplus_{i,j}\operatorname{Tor}(A_{i},A_{j})$. We have inclusions
$\iota_{i}\colon A_{i}\to A^{*}$ and $\iota_{j}\colon B_{j}\to B^{*}$, which
give homomorphisms
$\operatorname{Tor}(\iota_{i},\iota_{j})\colon\operatorname{Tor}(A_{i},B_{j})\to\operatorname{Tor}(A^{*},B^{*})$.
We also have inclusions $\iota_{ij}\colon\operatorname{Tor}(A_{i},B_{j})\to
T$. By the universal property of coproducts (Proposition 3.7) there is a
unique homomorphism $\phi\colon T\to\operatorname{Tor}(A^{*},B^{*})$ with
$\phi\circ\iota_{ij}=\operatorname{Tor}(\iota_{i},\iota_{j})$ for all $i$ and
$j$. It is this map that we claim is an isomorphism.

For fixed $j$, we can identify $\operatorname{Tor}(A^{*},B_{j})$ with the
kernel of the evident map $A^{*}\otimes I^{2}_{B_{j}}\to A^{*}\otimes
I_{B_{j}}$. From this it follows easily that
$\operatorname{Tor}(A^{*},B_{j})=\bigoplus_{i}\operatorname{Tor}(A_{i},B_{j})$.
A similar argument shows that
$\operatorname{Tor}(A^{*},B^{*})=\bigoplus_{j}\operatorname{Tor}(A^{*},B_{j})$,
and by putting these together we obtain $\operatorname{Tor}(A^{*},B^{*})\simeq
T$ as claimed. We leave it to the reader to check that the isomorphism arising
from this argument is the same as $\phi$. ∎

######  Proposition 7.25.

For any abelian group $U$ and any short exact sequence $A\to B\to C$ there is
a natural exact sequence

$0\to\operatorname{Tor}(U,A)\to\operatorname{Tor}(U,B)\to\operatorname{Tor}(U,C)\to
U\otimes A\to U\otimes B\to U\otimes C\to 0.$

###### Proof.

Apply the Snake Lemma (Proposition 2.4) to the diagram

${I^{2}_{U}\otimes A}$${I^{2}_{U}\otimes B}$${I^{2}_{U}\otimes
C}$${I_{U}\otimes A}$${I_{U}\otimes B}$${I_{U}\otimes C,}$

noting that the rows are short exact because $I^{2}_{U}$ and $I_{U}$ are free.
∎

######  Remark 7.26.

Let $A$ and $B$ be abelian groups. We can always choose a free abelian group
$F$ and a surjective homomorphism $p\colon F\to B$. Indeed, one possibility is
to use the natural map $q\colon I_{B}\to B$ from Proposition 7.2, but we can
also make a less natural choice that is typically much smaller. We then let
$F^{\prime}$ denote the kernel of $p$, and note that this is again free by
Theorem 6.6. Now the Proposition gives an exact sequence

$0\to\operatorname{Tor}(A,F^{\prime})\to\operatorname{Tor}(A,F)\to\operatorname{Tor}(A,B)\to
A\otimes F^{\prime}\xrightarrow{1\otimes i}A\otimes F\to A\otimes B\to 0.$

As $F^{\prime}$ and $F$ are free we have
$\operatorname{Tor}(A,F^{\prime})=\operatorname{Tor}(A,F)=0$, so we conclude
that $\operatorname{Tor}(A,B)$ is isomorphic to the kernel of $1\otimes i$.

A more traditional approach to $\operatorname{Tor}$ groups is to _define_
$\operatorname{Tor}(A,B)$ as the kernel of $1\otimes i$. In this approach one
has to work to prove that the resulting group is well-defined up to canonical
isomorphism, and that $\operatorname{Tor}(A,B)\simeq\operatorname{Tor}(B,A)$
and so on. However, one makes more contact with the general techniques of
homological algebra, which are important for other reasons.

######  Remark 7.27.

Suppose we have subgroups $A^{\prime}\leq A$ and $B^{\prime}\leq B$. Using
Proposition 7.25 we see that the maps

${\operatorname{Tor}(A^{\prime},B^{\prime})}$${\operatorname{Tor}(A^{\prime},B)}$${\operatorname{Tor}(A,B^{\prime})}$${\operatorname{Tor}(A,B)}$

are all injective, so we can regard
$\operatorname{Tor}(A^{\prime},B^{\prime})$ as a subgroup of
$\operatorname{Tor}(A,B)$.

######  Proposition 7.28.

$\operatorname{Tor}(A,B)$ is the union of the subgroups
$\operatorname{Tor}(A^{\prime},B^{\prime})$ for finite subgroups
$A^{\prime}\leq A$ and $B^{\prime}\leq B$.

###### Proof.

Any element of $x\in\operatorname{Tor}(A,B)$ has the form $x=y+z+J^{2}K^{2}$
for some $y\in J^{2}K$ and $z\in JK^{2}$. We can write $y$ as
$\sum_{i=1}^{r}n_{i}\langle a_{i}\rangle\langle a^{\prime}_{i}\rangle\langle
b_{i}\rangle$ for some $n_{i}\in\mathbb{Z}$ and $a_{i},a^{\prime}_{i}\in A$
and $b_{i}\in B$. Similarly, we can write $z$ as $\sum_{j=1}^{s}m_{j}\langle
c_{j}\rangle\langle d_{j}\rangle\langle d^{\prime}_{j}\rangle$ for some
$m_{j}\in\mathbb{Z}$ and $c_{j}\in A$ and $d_{j},d^{\prime}_{j}\in B$. Let
$A_{0}$ be the subgroup of $A$ generated by the elements
$a_{i},a^{\prime}_{i}$ and $c_{j}$, and let $B_{0}$ be the subgroup of $B$
generated by the elements $b_{i},d_{j}$ and $d^{\prime}_{j}$. It is then clear
that $x\in\operatorname{Tor}(A_{0},B_{0})\leq\operatorname{Tor}(A,B)$.
Moreover, $A_{0}$ and $B_{0}$ are finitely generated, so the subgroups
$A^{\prime}=\operatorname{tors}(A_{0})$ and
$B^{\prime}=\operatorname{tors}(B_{0})$ are finite. It follows from Theorem
5.3 that $A_{0}=A^{\prime}\oplus P$ and $B_{0}=B^{\prime}\oplus Q$, where $P$
and $Q$ are free. This means that
$\operatorname{Tor}(P,B^{\prime})=\operatorname{Tor}(P,Q)=\operatorname{Tor}(A^{\prime},Q)=0$
by Corollary 7.23, so
$\operatorname{Tor}(A_{0},B_{0})=\operatorname{Tor}(A^{\prime},B^{\prime})$.
The claim follows. ∎

######  Corollary 7.29.

$\operatorname{Tor}(A,B)$ is always a torsion group.

###### Proof.

In view of the proposition, it will suffice to prove this when $A$ is finite.
In that case there exists $n$ such that $n.1_{A}=0$, and so
$\operatorname{Tor}(n.1_{A},1_{B})=0\colon\operatorname{Tor}(A,B)\to\operatorname{Tor}(A,B)$.
However, using Proposition 7.24 we see that
$\operatorname{Tor}(n.1_{A},1_{B})=n.\operatorname{Tor}(1_{A},1_{B})=n.1_{\operatorname{Tor}(A,B)}$.
This means that $nx=0$ for all $x\in\operatorname{Tor}(A,B)$, so
$\operatorname{Tor}(A,B)$ is a torsion group. ∎

We next show how to construct elements of $\operatorname{Tor}(A,B)$ more
explicitly.

######  Proposition 7.30.

  * (a)

For $n>0$ and $a\in A[n]$ and $b\in B[n]$ there is an element
$e_{n}(a,b)\in\operatorname{Tor}(A,B)$ given by

$e_{n}(a,b)=n\langle a\rangle\langle b\rangle+J^{2}K^{2}\in\frac{J^{2}K\cap
JK^{2}}{J^{2}K^{2}}=\operatorname{Tor}(A,B).$

  * (b)

The map $e_{n}\colon A[n]\times B[n]\to\operatorname{Tor}(A,B)$ is bilinear,
so it induces a homomorphism $A[n]\otimes B[n]\to\operatorname{Tor}(A,B)$.

  * (c)

Suppose we have elements $a\in A[n]$ and $b\in B[m]$. Let $d$ be any common
divisor of $n$ and $m$. Then $(n/d)a\in A[m]$ and $(m/d)b\in A[n]$ and we have

$e_{n}(a,(m/d)b)=e_{nm/d}(a,b)=e_{m}((n/d)a,b).$

  * (d)

Suppose we have a short exact sequence

$0\to A\xrightarrow{j}B\xrightarrow{q}C\to 0$

giving rise to an exact sequence

$0\to\operatorname{Tor}(U,A)\to\operatorname{Tor}(U,B)\to\operatorname{Tor}(U,C)\xrightarrow{\delta}U\otimes
A\to U\otimes B\to U\otimes C\to 0$

as in Proposition 7.25. Suppose that $u\in U[n]$ and $c\in C[n]$. Then
$\delta(e_{n}(u,c))$ can be described as follows: we choose $b\in B$ with
$q(b)=c$, then there is a unique $a\in A$ with $j(a)=nb$, and
$\delta(e_{n}(u,c))=u\otimes a$.

###### Proof.

  * (a)

First, as $na=0$ in $A\simeq I_{A}/I_{A}^{2}$ we see that $n\langle
a\rangle\in I_{A}^{2}$ and so $n\langle a\rangle\langle b\rangle\in
I_{A}^{2}\otimes I_{B}=J^{2}K$. Similarly we have $n\langle b\rangle\in
I_{B}^{2}$ so $n\langle a\rangle\langle b\rangle=\langle a\rangle.(n\langle
b\rangle)\in I_{A}\otimes I_{B}^{2}=JK^{2}$. Thus, we have $n\langle
a\rangle\langle b\rangle\in J^{2}K\cap JK^{2}$ and the definition of $e_{n}$
is meaningful.

  * (b)

Next, recall that $\langle a+a^{\prime}\rangle-\langle a\rangle-\langle
a^{\prime}\rangle=\langle a\rangle\langle a^{\prime}\rangle\in I_{A}^{2}$, so

$n\langle a+a^{\prime}\rangle\langle b\rangle-n\langle a\rangle\langle
b\rangle-n\langle a^{\prime}\rangle\langle b\rangle=(\langle a\rangle\langle
a^{\prime}\rangle).(n\langle b\rangle)\in I_{A}^{2}\otimes
I_{B}^{2}=J^{2}K^{2},$

so $e_{n}(a+a^{\prime},b)=e_{n}(a,b)+e_{n}(a^{\prime},b)$. By a symmetrical
argument we see that $e_{n}(a,b)$ is also linear in $b$, as required.

  * (c)

Suppose we have elements $a\in A[n]$ and $b\in B[m]$. Let $d$ be any common
divisor of $n$ and $m$, so $n=pd$ and $m=qd$ for some $p,q>0$. Now
$m.(n/d)a=pqda=q.na=0$, so $(n/d)a=pa\in A[m]$, and similarly $(m/d)b=qb\in
B[m]$. Next note that

$pd\langle a\rangle\langle qb\rangle-pqd\langle a\rangle\langle
b\rangle=(n\langle a\rangle)(\langle qb\rangle-q\langle b\rangle)\in
I_{A}^{2}\otimes I_{B}^{2}=J^{2}K^{2},$

so $e_{pd}(a,qb)=e_{pqd}(a,b)$, or $e_{n}(a,(m/d)b)=e_{nm/d}(a,b)$. By a
symmetrical argument, we also have $e_{nm/d}(a,b)=e_{m}((n/d)a,b)$.

  * (d)

For $u\in U[n]$ and $c\in C[n]$ we note that $n\langle u\rangle\in I_{U}^{2}$
so we can put $e^{\prime}_{n}(u,c)=(n\langle u\rangle)\otimes c\in
I_{U}^{2}\otimes C$. Note that it is not legitimate to rewrite this as
$\langle u\rangle\otimes nc$ because $\langle u\rangle\not\in I_{U}^{2}$.
However this rewriting becomes legitimate if we work in $I_{U}\otimes C$, and
the result is zero because $nc=0$. In other words, $e^{\prime}_{n}(u,c)$ lies
in the kernel of the map $I_{U}^{2}\otimes A\to I_{U}\otimes A$, which was
identified with $\operatorname{Tor}(U,A)$ in Lemma 7.22. It is not hard to
check that $e^{\prime}_{n}(u,c)$ corresponds to $e_{n}(u,c)$ under this
identification. Now consider the diagram

${I^{2}_{U}\otimes A}$${I^{2}_{U}\otimes B}$${I^{2}_{U}\otimes
C}$${I_{U}\otimes A}$${I_{U}\otimes B}$${I_{U}\otimes
C.}$$\scriptstyle{1\otimes j}$$\scriptstyle{i\otimes 1}$$\scriptstyle{1\otimes
q}$$\scriptstyle{i\otimes 1}$$\scriptstyle{i\otimes 1}$$\scriptstyle{1\otimes
j}$$\scriptstyle{1\otimes q}$

By inspecting the proof of Proposition 2.4 we find the following prescription
for the connecting map $\delta\colon\operatorname{Tor}(U,C)\to A\otimes C$.
Given $z\in I^{2}_{U}\otimes C$ with $(i\otimes 1)(z)=0$ we choose $y\in
I^{2}_{U}\otimes B$ with $(1\otimes q)(y)=z$, then we check that $(i\otimes
1)(y)\in\ker(1\otimes q)=\operatorname{image}(1\otimes j)$ so there exists
$x\in I_{U}\otimes A$ with $(1\otimes j)(x)=(i\otimes 1)(y)$, and the image of
$x$ in $U\otimes A=(I_{U}/I_{U}^{2})\otimes A$ is by definition $\delta(z)$.
Now suppose again that we are given $u\in U[n]$ and $c\in C[n]$ and take
$z=e^{\prime}_{n}(u,c)$. As $q$ is surjective we can choose $b\in B$ with
$q(b)=a$. We then have $q(nb)=na=0$ so $nb\in\ker(q)=\operatorname{image}(j)$,
so there exists $a\in A$ with $j(a)=nb$. We can thus put $y=(n\langle
u\rangle)\otimes b\in I^{2}_{U}\otimes B$ and $x=\langle u\rangle\otimes a\in
I_{U}\otimes A$, and we find that $(1\otimes q)(y)=z$ and $(i\otimes
1)(y)=\langle u\rangle\otimes nb=(1\otimes j)(\langle u\rangle\otimes a)$.
This means that $\delta(e_{n}(u,a))$ is the image of $\langle u\rangle\otimes
a$ in $U\otimes A$, which is just $u\otimes a$ as claimed.

∎

######  Corollary 7.31.

For any abelian group $U$ and $n>0$, the map $u\mapsto e_{n}(u,1+n\mathbb{Z})$
gives an isomorphism $U[n]\mapsto\operatorname{Tor}(U,\mathbb{Z}/n)$.

###### Proof.

Consider the short exact sequence
$\mathbb{Z}\xrightarrow{n}\mathbb{Z}\to\mathbb{Z}/n$. As
$\operatorname{Tor}(U,\mathbb{Z})=0$ and $U\otimes\mathbb{Z}=\mathbb{Z}$ the
six-term exact sequence reduces to

$0\to\operatorname{Tor}(U,\mathbb{Z})\xrightarrow{\delta}U\xrightarrow{n}U\xrightarrow{}U/n\xrightarrow{}0,$

which means that $\delta$ gives an isomorphism
$\delta\colon\operatorname{Tor}(U,\mathbb{Z})\to U[n]$. Part (d) of the
proposition tells us that $\delta(e_{n}(u,1+n\mathbb{Z}))=u$ for all $u\in
U[n]$, and the claim follows from this. ∎

######  Corollary 7.32.

For any abelian groups $A$ and $B$, the group $\operatorname{Tor}(A,B)$ is
generated by all the elements $e_{n}(a,b)$ for $n>0$ and $a\in A[n]$ and $b\in
B[n]$.

###### Proof.

In view of Proposition 7.28, it is enough to prove this when $A$ and $B$ are
finite. Now $A$ and $B$ can be written as direct sums of finite cyclic groups,
and using Proposition 7.24(c) we reduce to the case where $A$ and $B$ are
themselves cyclic. That case is immediate from Corollary 7.31. ∎

######  Corollary 7.33.

For any abelian group $B$, there is a natural isomorphism
$\operatorname{Tor}({\mathbb{Q}/\mathbb{Z}},B)\simeq\operatorname{tors}(B)$.

###### Proof.

Let $A_{m}$ be the subgroup of ${\mathbb{Q}/\mathbb{Z}}$ generated by
$1/m+\mathbb{Z}$, and define $\epsilon_{m}\colon
B[m]\to\operatorname{Tor}(A_{m},B)\leq\operatorname{Tor}({\mathbb{Q}/\mathbb{Z}},B)$
by $\epsilon_{m}(b)=e_{n}(1/m+\mathbb{Z},b)$. Note that $A_{m}$ is cyclic of
order $m$, so the previous result tells us that $\epsilon_{m}$ is an
isomorphism. Now suppose that $m$ divides $n$, so $B[m]\leq B[n]$. We can take
$d=m$ in Proposition 7.30(c) to see that $e_{n}(a,b)=e_{m}((n/m)a,b)$ whenever
$na=0$ and $mb=0$. Taking $a=1/n+\mathbb{Z}$ we deduce that
$\epsilon_{n}(b)=\epsilon_{m}(b)$ whenever $mb=0$, so
$\epsilon_{n}|_{B[m]}=\epsilon_{m}$. It follows that there is a unique
homomorphism

$\epsilon\colon\operatorname{tors}(B)=\bigcup_{n>0}B[n]\to\operatorname{Tor}({\mathbb{Q}/\mathbb{Z}},B).$

Every finite subgroup of ${\mathbb{Q}/\mathbb{Z}}$ has the form $A_{n}$ for
some $n$, and it follows from this using Proposition 7.28 that $\epsilon$ is
an isomorphism.

There is an explicit construction of the inverse which is quite instructive.
An element $x\in{\mathbb{Q}/\mathbb{Z}}$ is a coset of $\mathbb{Z}$ in
${\mathbb{Q}}$, and any such coset intersects the interval $[0,1)$ in a single
point, which we call $\lambda(x)$. The function
$\lambda\colon{\mathbb{Q}/\mathbb{Z}}\to{\mathbb{Q}}$ is not a homomorphism,
but $\lambda(x+x^{\prime})$ and $\lambda(x)+\lambda(x^{\prime})$ are both
representatives of the same coset $x+x^{\prime}$, so we at least have

$\lambda(x+x^{\prime})-\lambda(x)-\lambda(x^{\prime})\in\mathbb{Z}.$

We can extend $\lambda$ to give a group homomorphism
$\mathbb{Z}[{\mathbb{Q}/\mathbb{Z}}]\to{\mathbb{Q}}$ by
$\lambda(\sum_{i}n_{i}[x_{i}])=\sum_{i}n_{i}\lambda(x_{i})$. We then find that

$\lambda(\langle x\rangle\langle
x^{\prime}\rangle)=\lambda([x+x^{\prime}]-[x]-[x^{\prime}]+[0])=\lambda(x+x^{\prime})-\lambda(x)-\lambda(x^{\prime})\in\mathbb{Z},$

and thus that $\lambda(I_{{\mathbb{Q}/\mathbb{Z}}}^{2})\leq\mathbb{Z}$. We
therefore have a homomorphism $\lambda\otimes 1\colon
I_{{\mathbb{Q}/\mathbb{Z}}}^{2}\otimes B\to\mathbb{Z}\otimes B=B$. Lemma 7.22
allows us to regard $\operatorname{Tor}({\mathbb{Q}/\mathbb{Z}},B)$ as a
subgroup of $I^{2}_{{\mathbb{Q}/\mathbb{Z}}}\otimes B$, so we can restrict to
this subgroup to get a homomorphism
$\mu\colon\operatorname{Tor}({\mathbb{Q}/\mathbb{Z}},B)\to B$. We leave it to
the reader to check that the image of $\mu$ is $\operatorname{tors}(B)$, and
that
$\mu\colon\operatorname{Tor}({\mathbb{Q}/\mathbb{Z}},B)\to\operatorname{tors}(B)$
is inverse to $\epsilon$. ∎

######  Proposition 7.34.

Define maps

${\bigoplus_{n,m,d}(A[nd]\otimes B[md])}$${\bigoplus_{p}(A[p]\otimes
B[p])}$${\operatorname{Tor}(A,B)}$$\scriptstyle{\lambda}$$\scriptstyle{\rho}$$\scriptstyle{\epsilon}$

by

$\displaystyle\lambda(i_{n,m,d}(a\otimes b))$ $\displaystyle=i_{nd}(a\otimes
mb)$ $\displaystyle\rho(i_{n,m,d}(a\otimes b))$
$\displaystyle=i_{md}(na\otimes b)$ $\displaystyle\epsilon(i_{p}(a\otimes b))$
$\displaystyle=e_{p}(a,b).$

Then the sequence

${\bigoplus_{n,m,d}(A[nd]\otimes B[md])}$${\bigoplus_{p}(A[p]\otimes
B[p])}$${\operatorname{Tor}(A,B)}$${0}$$\scriptstyle{\lambda-\rho}$$\scriptstyle{\epsilon}$

is exact, so $\operatorname{Tor}(A,B)$ is the cokernel of $\lambda-\rho$.

###### Proof.

Let $T(A,B)$ be the cokernel of $\lambda-\rho$, and let $e^{\prime}_{p}(a,b)$
be the image of $i_{p}(a\otimes b)$ in $T(A,B)$, so by construction we have
$e^{\prime}_{nd}(a,mb)=e^{\prime}_{md}(na,b)$ whenever $nda=0$ and $mdb=0$.
Part (c) of Proposition 7.30 shows that $\epsilon\lambda=\epsilon\rho$, so
$\operatorname{img}(\lambda-\rho)\leq\ker(\epsilon)$, so there is a unique
homomorphism $\overline{\epsilon}\colon T(A,B)\to\operatorname{Tor}(A,B)$ such
that $\overline{\epsilon}(e^{\prime}_{p}(a,b))=e_{p}(a,b)$ for all $p\geq 1$
and $a\in A[p]$ and $b\in B[p]$. This is surjective by Corollary 7.32. We must
show that this is also injective.

Consider the special case where $A$ and $B$ are finitely generated. It is easy
to see that there are natural splittings

$\displaystyle T(A\oplus A^{\prime},B\oplus B^{\prime})$ $\displaystyle\simeq
T(A,B)\oplus T(A,B^{\prime})\oplus T(A^{\prime},B)\oplus
T(A^{\prime},B^{\prime})$ $\displaystyle\operatorname{Tor}(A\oplus
A^{\prime},B\oplus B^{\prime})$
$\displaystyle\simeq\operatorname{Tor}(A,B)\oplus\operatorname{Tor}(A,B^{\prime})\oplus\operatorname{Tor}(A^{\prime},B)\oplus\operatorname{Tor}(A^{\prime},B^{\prime})$

so we can reduce to the case where $A$ and $B$ are cyclic. If $A=\mathbb{Z}$
or $B=\mathbb{Z}$ it is clear that $T(A,B)=0=\operatorname{Tor}(A,B)$, so we
may assume that $A=\mathbb{Z}/r$ and $B=\mathbb{Z}/s$ say. Let $t$ be the
least common multiple of $r$ and $s$, so we have an element
$u=e^{\prime}_{t}(1+r\mathbb{Z},1+s\mathbb{Z})\in T(A,B)$. Note that $T(A,B)$
is generated by elements of the form $v=e_{p}(a+r\mathbb{Z},b+s\mathbb{Z})$
with $pa\in r\mathbb{Z}$ and $pb\in s\mathbb{Z}$, which implies that $pab=tc$
for some $c\in\mathbb{Z}$. We thus have an element

$x=i_{c,1,t}(1+r\mathbb{Z},1+s\mathbb{Z})+i_{1,b,ap}(1+r\mathbb{Z},1+s\mathbb{Z})-i_{a,1,p}(1+r\mathbb{Z},b+s\mathbb{Z})\in\bigoplus_{n,m,d}(A[nd]\otimes
B[md]),$

and we find that

$(\lambda-\rho)(x)=i_{p}((a+r\mathbb{Z})\otimes(b+s\mathbb{Z}))-i_{t}(c+r\mathbb{Z},1+s\mathbb{Z})=i_{p}((a+r\mathbb{Z})\otimes(b+s\mathbb{Z}))-c\,i_{t}(1+r\mathbb{Z},1+s\mathbb{Z})$

so $v=cu$. This proves that $u$ generates $T(A,B)$. It is also clear that
$ru=su=0$, so if we put $d=(r,s)$ we have $du=0$. This means that $T(A,B)$ is
cyclic of order dividing $d$ but the map
$T(A,B)\to\operatorname{Tor}(A,B)\simeq\mathbb{Z}/d$ is surjective so it must
be an isomorphism.

We now revert to the general case, where $A$ and $B$ may be infinitely
generated. Consider an element $w\in\ker(\overline{\epsilon})\leq T(A,B)$. We
can write $w$ as $\sum_{i=1}^{N}e^{\prime}_{p_{i}}(a_{i},b_{i})$ say, and then
let $A^{\prime}$ be the subgroup of $A$ generated by $a_{1},\dotsc,a_{N}$, and
let $B^{\prime}$ be the subgroup of $B$ generated by $b_{1},\dotsc,b_{N}$.
There is then an evident element $w^{\prime}\in T(A^{\prime},B^{\prime})$ that
maps to $w$ in $T(A,B)$. Now consider the diagram

${T(A^{\prime},B^{\prime})}$${\operatorname{Tor}(A^{\prime},B^{\prime})}$${T(A,B)}$${\operatorname{Tor}(A,B).}$$\scriptstyle{\overline{\epsilon}}$$\scriptstyle{\simeq}$$\scriptstyle{\overline{\epsilon}}$

The top map is an isomorphism by the special case considered above, and the
right hand map is injective by Remark 7.27. By chasing $w^{\prime}$ around the
diagram we see that $w=0$. Thus, the map $\overline{\epsilon}\colon
T(A,B)\to\operatorname{Tor}(A,B)$ is injective as required. ∎

##  8\. Ext groups

######  Definition 8.1.

Let $A$ and $B$ be abelian groups, and let $j$ be the inclusion $I^{2}_{A}\to
I_{A}$. We define $\operatorname{Ext}(A,B)$ to be the kernel of the map
$j^{*}\colon\operatorname{Hom}(I_{A},B)\to\operatorname{Hom}(I^{2}_{A},B)$.
Now suppose we have homomorphisms $f\colon A^{\prime}\to A$ and $g\colon B\to
B^{\prime}$. We then have compatible homomorphisms $f_{\bullet}\colon
I_{A^{\prime}}\to I_{A}$ and $f_{\bullet}\colon I^{2}_{A^{\prime}}\to
I^{2}_{A}$, and we can use these in an evident way to construct a commutative
square of maps

${\operatorname{Ext}(A,B)}$${\operatorname{Ext}(A,B^{\prime})}$${\operatorname{Ext}(A^{\prime},B)}$${\operatorname{Ext}(A^{\prime},B^{\prime})}$$\scriptstyle{g_{*}}$$\scriptstyle{f^{*}}$$\scriptstyle{f^{*}}$$\scriptstyle{g_{*}}$

######  Remark 8.2.

Let $q$ be the usual map $I_{A}\to A$, with kernel $I_{A}^{2}$. Consider the
sequence

$0\to\operatorname{Hom}(A,B)\xrightarrow{q^{*}}\operatorname{Hom}(I_{A},B)\xrightarrow{j^{*}}\operatorname{Hom}(I^{2}_{A},B)\xrightarrow{}\operatorname{Ext}(A,B)\to
0.$

By combining Lemma 7.17 with Definition 8.1, we see that this is exact.

######  Remark 8.3.

If we have two homomorphisms $g_{0},g_{1}\colon B\to B^{\prime}$, it is clear
that

$(g_{0}+g_{1})_{*}=g_{0*}+g_{1*}\colon\operatorname{Ext}(A,B)\to\operatorname{Ext}(A,B^{\prime}).$

If we have two homomorphisms $f_{0},f_{1}\colon A^{\prime}\to A$, it is not
obvious from the definitions that $(f_{0}+f_{1})^{*}=f_{0}^{*}+f_{1}^{*}$, but
later we will give a different description of the Ext groups that makes this
fact clear.

######  Lemma 8.4.

Let $F$ be a free abelian group. Then for every surjective homomorphism
$q\colon B\to C$, the resulting map
$q_{*}\colon\operatorname{Hom}(F,B)\to\operatorname{Hom}(F,C)$ is also
surjective. Equivalently, for every pair of homomorphisms $f$ and $q$ as shown
with $q$ surjective, there exists $g$ with $qg=f$.

${F}$${B}$${C.}$$\scriptstyle{g}$$\scriptstyle{f}$$\scriptstyle{q}$

Conversely, every group $F$ with this property is free.

###### Proof.

Firstly, there exists a map $g$ as shown if and only if the element
$f\in\operatorname{Hom}(F,C)$ lies in the image of
$q_{*}\colon\operatorname{Hom}(F,B)\to\operatorname{Hom}(F,C)$; so the two
versions of the statement are indeed equivalent. To prove them, we may assume
that $F=\mathbb{Z}[I]$ for some $I$. We then have elements $f([i])\in C$ and
$q\colon B\to C$ is surjective so we can choose $b_{i}\in B$ with
$q(b_{i})=f([i])$. Now there is a unique homomorphism $g\colon\mathbb{Z}[I]\to
B$ with $g([i])=b_{i}$ for all $i$, and $qg=f$ as required.

Conversely, let $F$ be any abelian group that has the property under
consideration. Take $C=F$ and $B=I_{F}$, let $q\colon B\to C$ be the usual
surjection $I_{F}\to F$, and let $f\colon F\to C$ be the identity. Then there
must exist $g\colon F\to I_{F}$ with $qg=1_{F}$. This means that $g$ is
injective, so $F$ is isomorphic to a subgroup of the free group $I_{F}$, so
$F$ is free by Theorem 6.6. ∎

######  Remark 8.5.

In the more general context of modules over an arbitrary ring, the property in
the lemma is called _projectivity_. Thus, we have shown that an abelian group
is projective if and only if it is free. The same argument shows that an
$R$-module is projective if and only if it is a direct summand in a free
$R$-module. The analogue of Theorem 6.6 is not valid for modules over a
general ring, so projective modules need not be free.

######  Proposition 8.6.

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

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

###### Proof.

We first claim that the sequence

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

is exact. Indeed, if $j_{*}(\alpha)=0$ then $j(\alpha(u))=0$ for all $u$, but
$j$ is injective so $\alpha(u)=0$ for all $u$, so $\alpha=0$; this proves that
$j_{*}$ is injective. Next, suppose that $q_{*}(\beta)=0$, so $q(\beta(u))=0$
for all $u\in U$, so $\beta(u)\in\ker(q)=\operatorname{image}(j)$, so there
exists $\alpha(u)\in A$ with $\beta(u)=j(\alpha(u))$. This element $\alpha(u)$
is in fact unique, because $j$ is injective. We also have

$j(\alpha(u+u^{\prime})-\alpha(u)-\alpha(u^{\prime}))=j(\alpha(u+u^{\prime}))-j(\alpha(u))-j(\alpha(u^{\prime}))=\beta(u+u^{\prime})-\beta(u)-\beta(u^{\prime})=0,$

but $j$ is injective so $\alpha(u+u^{\prime})-\alpha(u)-\alpha(u^{\prime})=0$,
so the map $\alpha\colon U\to A$ is a homomorphism. Clearly
$j_{*}(\alpha)=\beta$, so we see that
$\ker(q_{*})=\operatorname{image}(j_{*})$ as required.

Now consider the sequence

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

As $U$ was arbitrary we can replace it by $I_{U}$ to see that $j_{*}$ is
injective and $\operatorname{image}(j_{*})=\ker(q_{*})$. As $I_{U}$ is free we
also see from Lemma 8.4 that $q_{*}$ is surjective, so the sequence is short
exact. The same argument applies with $I_{U}$ replaced by $I_{U}^{2}$.

Now consider the diagram

${\operatorname{Hom}(I_{U},A)}$${\operatorname{Hom}(I_{U},B)}$${\operatorname{Hom}(I_{U},C)}$${\operatorname{Hom}(I^{2}_{U},A)}$${\operatorname{Hom}(I^{2}_{U},B)}$${\operatorname{Hom}(I^{2}_{U},C).}$$\scriptstyle{j_{*}}$$\scriptstyle{q_{*}}$$\scriptstyle{j_{*}}$$\scriptstyle{q_{*}}$

We have just seen that the rows are short exact, so we can apply the Snake
Lemma (Proposition 2.4) to get a six-term exact sequence involving the kernels
and cokernels of the vertical maps. Remark 8.2 identifies these kernels and
cokernels as $\operatorname{Hom}$ and $\operatorname{Ext}$ groups, as
required. ∎

######  Corollary 8.7.