Every ideal has a factorization as a product of prime ideals, uniquely determined

EXERC1SES

14. Every ideal has a factorization as a product of prime ideals, uniquely determined

up^to

permutation.

15.

Suppose

^{0 has}

only

^one

prime

^{ideal p. Let tEP and t}

p2.

^Then p ⁼

(t)

^is

principal.

16. Let 0 be any Dedekind

ring.

^{Let p bea}

prime

^{ideal. Let} 0" ^{be thelocal}

ring

^at

p. Then 0"^is Dedekind and has

only

^one

prime

^ideal.

17. As for the

integers,

^wesaythat

alb (a

^divides

b)

^ifthere exists^an ideal c suchthat b⁼ac. Prove:

(a) alb

^{if and}

only

^{if bca.}

(b)

^Let a,b be ideals. Thena+b istheirgreatest ^{commondivisor. In}

particular,

a, bare

relatively prime

^ifand

only

^ifa+b=o.

18.

Every prime

^{ideal p} is maximal.

(Remember,

p :F 0

by convention.)

^In

particular,

if PI' ..., PII ^are distinct

primes,

then the Chinese remainder theorem

applies

^to

h

. '1 r

U h

.

t elrpowers PI' ^...,PII". ^{se t IS to}prove:

19. Let _a, b beideals. Show that there exists an element CEK

(the quotient

^{field of}

0)

^{such thatca is anideal}

relatively prime

^{to b. In}

particular,

everyideal class in

Pic(o)

^contains

representative

^ideals

prime

^toa

given

^ideal.

For a

continuation,

^{see Exercise 7of}

Chapter

^VII.

CHAPTER III

Modules

Although

^this

chapter

^is

logically

self-contained and prepares for future

topics,

in

practice

readers will have had some

acquaintance

^withvector spaces^{over a} field^. We

generalize

this notion heretomodules over

rings.

^{It is a}standard fact

(to

^be

reproved)

^thatavectorspace has^a

basis,

butfor modules this isnot

always

the case. Sometimes

they do;

^mostoften

they

^{do not.} We shall look into cases

where

they

^do.

For

examples

of modules and their relationsto those which havea

basis,

^the reader should look at the comments made at the end of

4.

1. ^BASIC DEFINITIONS

Let A bea

ring.

A left moduleover

A,

^{or a} left A-module M is anabelian group,

usually

^written

additively, together

^withan

operation

^ofAonM

(viewing

A as a

multiplicative

^monoid

by

^RI

2),

^such

that,

^{for all}Q, bEA and _x,yEM

wehave

(a

⁺

b)x

^{= ax + bx and}

a(x

⁺

y)

^{= ax +} aYe

We leave it as anexercise to prove that

a( -x)

⁼

-(ax)

^{and thatOx = O.}

By

definition ofan

operation,

^{wehave 1x = x.}

Inasimilar_way,onedefinesa

right

^{A-module. Weshall deal}

only

^withleft

A-modules,

unless otherwise

specified,

^andhence call these

simply A-modules,

or evenmodulesif the referenceisclear.

117

Let M bean A-module.

By

^{asubmoduleN ofM we mean an} additive sub- group such that AN ^c N. ThenN isamodule

(with

^the

operation

^induced

by

that ofA on

M).

Examples

Wenotethat Ais^amoduleover itself.

Any

commutative_groupisaZ-module.

An additive_group

consisting

^{of0alone isamoduleover} any

ring.

Any

left ideal ofAisamoduleover A.

Let} beatwo-sided ideal of A. Then the factor

ring AI}

^is

actually

^amodule

overA. If^a EA and a + } is a coset of } in

A,

^{then one} defines the

operation

tobea

(x

⁺

})

^{= ax +}}. The reader^can

verify

^atoncethat this definesamodule structureon

AI}.

^More

general,

^{if M isamodule and Na}

^submodule,

^{we shall}

define the factor module below. Thus if L is aleft ideal of

A,

^then

AlLis

^also

a module. Formore

examples

^{in this}

^vein,

^see

^4.

A module over a field is called a vector space. Even

starting

^{with vector}

spaces,^oneis ledtoconsider modulesover

rings. ^Indeed,

^{let V beavector}space

over the field K. The readerno doubt

already

^knows about linear maps

(which

will be recalled below

systematically).

^{Let Rbe the}

ring

^{of all}linear maps of V into itself. Then V is amodule overR.

Similarly,

^{if V = Kn} denotes thevector space of

(vertical) n-tuples

of elements of

K,

and R is the

ring

^{ofn x nmatrices}

with

components

ⁱⁿ

K,

^{then V is a module overR. For more comments}

along

these

lines,

^seethe

examples

^{atthe end of}

^2.

Let S be a non-empty ^{set and M an A-module. Then the set} of maps

Map(S, M)

^{is an}A-module. We have

already

^noted

previously

^{that it is a com-}

mutative group, and for

f

^E

Map(S, M),

^{a E A we define}

af

^to be the map such that

(aj)(s)

⁼

af(s).

The axioms for a module are then

trivially

^verified.

For further

examples,

^see the end of this section.

Fortherestof this

section,

^{wedealwithafixed}

ring ^A,

and hence may omit the

prefix

^A-.

Let A be an entire

ring

^{and let M be an} A-module. We define the torsion submodule M_tor to be the subset of elements x E M such that there exists

a E

A,

^a =f=. 0 suchthatax⁼o. It is

immediately

verified that M_torisasubmodule.

Its structurein an

important

^case will be determined in 7.

Let a be a left

ideal,

^{and M a} module. Wedefine aM to be the set ofall elements

atXt ^{+ ... +} anxn

with_aiE aand_XiEM. Itis

obviously

^asubmodule ofM. If_a,b areleft

ideals,

then wehave

associativity, namely

a{bM)

⁼

{ab)M.

We also have ^some obvious

distributivities,

^like

(a

⁺

b)M

^{= aM + bM. If}

N,

^N'aresubmodules of

M,

^then

a(N

⁺

N')

^{= aN + aN'.}

Let M be an

A-module,

^{and N a submodule. We} shall define a module structure on the factor _group

M/N (for

^{the additive} group

structure).

^Let

x + N be a coset of N in

M,

^{and let aEA. We define}

a(x

⁺

N)

^{to be the}

coset ax + N. It is trivial to

verify

that this is well defined

(i.e.

^{ify is in the}

samecoset asx, then _ayis inthe same coset as

ax),

and that this is anopera- tion of A on

M/

^N

satisfying

^the

required condition, making M/

^{N into a}

module,

^{called the}factor moduleofM

by

^N.

By

^a

module-homomorphism

one means amap

f:

^{M -.M'}

ofonemodule into another

(over

^thesame

ring A),

^{which isanadditive}group-

homomorphism,

and such that

f(ax)

⁼

af(x)

forall a EA and xEM. It is then clear that the collection of A-modules is a

category,

^whose

morphisms

^{are the}

module-homomorphisms usually

^also

called

homomorphisms

^for

simplicity,

^ifno confusion is

possible.

^Ifwewish

to referto the

ring A,

^{we also} say that

f

^isan

A-homomorphism,

^{or also that}

it isanA-linear map.

If M is a

module,

^{then the}

identity

map is a

homomorphism.

For any module

M',

^the map

(:

^{M-. M' such that}

(x)

^{= 0 for all xEM is a homo-}

morphism,

^calledzero.

In the next

section,

^we shall discuss the

homomorphisms

^ofa module into

itself,

^{and as aresultwe shall}

give

^further

examples

of modules which arise in

practice.

^{Herewecontinueto}tabulate thetranslationof basic

properties

of groups to modules.

Let M be a module and N asubmodule. Wehave the canonical additive

group-homomorphism

f:M

^-.

M/N

andoneverifies

trivially

^{that it isa}

module-homomorphism.

Equally trivially,

^one verifies that

f

is universal in the

category

^{of homo-}

morphisms

^{of Mwhose}kernel containsN.

If f:

^{M -. M' is a}

module-homomorphism,

^{then its kernel and}

^image

^are

submodules

of

^MandM'

respectively (trivial verification).

Let!:

^{M M' bea}

homomorphism. By

the cokernel

of!we

^{meanthefactor}

module

M'/Im!

^{= M'}

/!(M).

One may also ^meanthe canonical

homomorphism

M'

M'/!(M)

^{rather than the module} itself. The context should make clear which is meant. Thus the cokernel is ^a factor module ofM'^.

Canonical

homomorphisms

discussed in

Chapter ^{I, 3} apply

^{to modules}

mutatis mutandis. For the convenience of the

reader,

^{we summarise these}

homomorphisms:

Let

N,

^{N' be two submodules}

of

^{a module M. Then N + N' is also asub-}

module,

^andwehavean

isomorphism

Nj(N

ⁿ

N') (N

⁺

N')jN'.

If

^{M ::J M' ::J Milare}

^modules,

^then

(MjM")j(M'jM") MjM'.

Iff:

^{M-. M' isa}

module-homomorphism,

^{and N' isasubmodule}

of M',

^then

f

-

l(N')

^isasubmodule

of

^{Mandwehaveacanonical}

injective homomorphism J:Mjf-l(N')

^-.

M'jN'.

Iff

^is

surjective,

^then

J

^isa

module-isomorphism.

The

proofs

^{are obtained}

by verifying

^{that all}

homomorphisms

^which ap-

peared

^when

dealing

with abelian _groups ^{are now}

A-homomorphisms

^of

modules. We leave the verification ^tothe reader.

Aswith_groups,weobserve that^a

module-homomorphism

^whichis

bijective

is a

module-isomorphism.

^Here

again,

^the

proof

^{is the same as} for groups,

adding only

the observation that the inverse _map,whichweknow is agroup-

isomorphism, actually

^{is a}

module-isomorphism. Again,

^{weleavethe verifica-}

tion to the reader.

Aswith abelian_groups,^wedefineasequenceof

module-homomorphisms

M'

1.

M Mil

to be exact if

Imf

⁼ Ker g. We have an exact sequence associated with a

submoduleN ofamodule

M, namely

o^-. N^-. M^-. M

j

^N-.

0,

the _map ofN into M

being

^the

^inclusion,

^{and the}

subsequent

map

being

^the

canonical_map. The notion ofexactnessis dueto

Eilenberg-Steenrod.

Ifa

homomorphism

^{u :N -.M} is such that

O-.NM

is exact, then^wealso _say thatu is a

monomorphism

^{or an}

embedding. Dually,

if

u

N-.M-.O