Rings and modules

39 Pages·2013·0.248 MB·English

by  Ivan Fesenko

Rings and modules
Notation: A ⊂ B means A isasubsetof B, possiblyequalto B.
1. Revision
Allringsarecommutativeringswithunity.
1.1. Let f:A → B bearinghomomorphism.
Theorem on ring homomorphisms. The kernel I of f is an ideal of A, the image
C of f isasubringof B. Thequotientring A/I isisomorphicto C.
Proof. Considerthemap g:A/I → C, a+I (cid:55)→ f(a). Itiswelldefined: a+I = a(cid:48)+I
implies a−a(cid:48) ∈ I implies f(a) = f(a(cid:48)).
The element a+I belongs to the kernel of g iff g(a+I) = f(a) = 0, i.e. a ∈ I,
i.e. a+I = I isthezeroelementof A/I. Thus, ker(g) = 0.
Theimageof g is g(A/I) = {f(a) : a ∈ A} = C.
Thus, g isanisomorphism. Theinversemorphismto g isgivenby f(a) (cid:55)→ a+I.
Correspondence theorem. Let I be an ideal of a ring A. Then there is a bijection
betweenthesetof allideals J of A suchthat I ⊂ J andthesetof allidealsof A/I:
{J : I anidealof A,I ⊂ J} −→ {K : K anidealof A/I}
J −→ J/I
Proof. Denoteby h themorphism h:A → A/I, a (cid:55)→ a+I, itsimageis A/I andits
kernelis I.
For an ideal J of A, I ⊂ J, denote by h| :J → A/I,j (cid:55)→ j +I the restriction
J
of h to J. Its kernel is I. Similarly to the proof of the previous theorem we deduce
that h| (J) isisomorphicto J/I whichisanidealof A/I.
J
For an ideal K of A/I define K(cid:48) = h−1(K) of A. Then K(cid:48) is an ideal of A,
I ⊂ K(cid:48).
Now we have two maps, J (cid:55)→ J/I and K (cid:55)→ h−1(K). They are inverse to
each other, i.e. h−1(J/I) = J and h−1(K)/I = K. Thus, there is a one-to-one
correspondencebetweentheideals.
2 Ringsandmodules
1.2. The intersection of ideals of A is an ideal of A. Given a subset S of A, one
canspeakabouttheminimalidealof A whichcontains S. Thisidealisequalto
{a s + ···+a s : a ∈ A,s ∈ S,m (cid:62) 1}.
1 1 m m i i
Oftenitiscalledtheidealgeneratedby S.
Let I,J beidealsofaring A.
Their sum I +J is the minimal ideal of A which contains both I and J, more
explicitly
I +J = {i+j : i ∈ I,j ∈ J}.
Certainly, I +(J +K) = (I +J)+K. Similarlyonedefinesthesumofseveralideals
(cid:80)
I .
k
Their product IJ is the minimal ideal which contains all ij : i ∈ I,j ∈ J, more
explicitly
IJ = {i j + ···+i j : n (cid:62) 1,i ∈ I,j ∈ J}.
1 1 n n m m
Theproductisassociative:
(IJ)K = I(JK)
anddistributive:
(I +J)K = IK +JK.
Similarlyonedefinestheproductofseveralideals I ...I .
1 n
Notethat (I+J )(I+J ) istheminimalidealwhichcontainsproducts (i +j )(i +
1 2 1 1 2
j ) = (i i +i j +i j )+j j , soitiscontainedin I +J J :
2 1 2 2 1 1 2 1 2 1 2
(I +J )(I +J ) ⊂ I +J J ,
1 2 1 2
buttheinverseinclusiondoesnotholdingeneral.
Foranelement a of A theprincipalidealgeneratedby a is
(a) = aA = {ab : b ∈ A}.
In particular, (0) = {0} is the smallest ideal of A and (1) = A is the largest ideal of
A. Unless A = {0}, thesearetwodistinctidealsof A.
Forseveralelements a , ...,a of A theidealgeneratedbythe a isdenoted
1 n i
(a , ...,a ) = a A+ ···+a A = {a b + ···+a b : b ∈ A}.
1 n 1 n 1 1 n n i
1.3. Aring A isafieldifitcontainsanon-zeroelementandeverynon-zeroelement
of A isinvertiblein A.
Lemma. Anon-zeroringisafieldiff ithasexactlytwodifferentideals, (0) and (1).
3
Proof. If I is a non-zero ideal of a field F, then I contains a non-zero element a.
Therefore it contains aa−1 = 1 and therefore it contains 1b = b for every b in F; so
I = F.
Conversely, if a non-zero ring has only two distinct ideals then it is a field: for
every nonzero element aA must be equal to (1), hence a multiple of a is 1 and a is
invertible.
An ideal I of a ring A is called maximal if I (cid:54)= A and every ideal J such that
I ⊂ J ⊂ A eithercoincideswith A orwith I. By1.1thisequivalentto: thequotient
ring A/I has no proper ideals. By the previous lemma this is equivalent to A/I is a
field. Soweproved
Lemma. I isamaximalidealof A iff A/I isafield.
1.4. Aring A isanintegraldomainif A (cid:54)= 0 andforevery a,b ∈ A ab = 0 implies
a = 0 or b = 0.
Example: everyfieldisanintegraldomain: ab = 0 and a (cid:54)= 0 implies b = a−1ab =
0. Z is an integral domain. More generally, every non-zero subring of an integral
domainisanintegraldomain.
If A isanintegraldomain,onecanformthefieldoffractions F of A as
{a/b : a ∈ A,b ∈ A\{0}}.
Bydefinition a/b = c/d iff ad = bc.
This is an equivalence relation: if a/b = c/d and c/d = e/f then ad = bc and
cf = ed so adf = bcf = bed, d(af −be) = 0. As d isnotzero, af = be.
Definetworingoperations a/b+c/d = (ad+bc)/(bd) and (a/b)(c/d) = (ac)/(bd).
Thezeroof F is 0/1 = 0/a foranynon-zero a. Everynonzeroelement a/b of F is
invertible: if a/b (cid:54)= 0 then (a/b)−1 = b/a. Thus F isafield. Theringhomomorphism
A → F, a (cid:55)→ a/1 is injective: a/1 = 0/1 implies a = 0. Thus A can be identified
withthesubring A/1 of F. Then a/b canbeidentifiedwith ab−1 givingthemeaning
offractiontothesymbol a/b.
Thus, every integral domain is a non-zero subring of a field, and the latter is an
integral domain. So the class of integral domains coincides with the class of non-zero
subringsoffields.
1.5. An ideal I of a ring A is called prime if I (cid:54)= A and for every a,b ∈ A the
inclusion ab ∈ I impliesthateither a ∈ I or b ∈ I.
Example: everyfieldhasaprimeideal: (0).
Lemma. I isaprimeidealof A iff A/I isanintegraldomain.
Proof. Let I be a prime ideal of A. Let (a+I)(b+I) = 0+I, then ab ∈ I. So at
leastoneof a,b isin I whichmeansthateither a+I = 0+I or b+I = 0+I. Thus,
A/I isanintegraldomain.
4 Ringsandmodules
Conversely,let A/I beanintegraldomain. If ab ∈ I then (a+I)(b+I) = I = 0+I,
hence either a+I = I and so a ∈ I, or b+I = I and so b ∈ I. Thus, I is a prime
idealof A.
Example: foraprimenumber p theideal pZ isaprimeidealof Z. Thezeroideal
(0) isaprimeidealof Z.
Corollary. Everymaximalidealisprime.
Proof. Everyfieldisanintegraldomain.
Remark. In general, not every prime ideal is maximal. For instance, (0) is a prime
idealof Z whichisnotmaximal.
1.6. For rings A define their product A × ···×A as the set theoretical product
i 1 n
endowedwiththecomponentwiseadditionandmultiplication.
Chinese Remainder Theorem. Let I , ...,I be ideals of A such that I +I = A
1 n i j
forevery i (cid:54)= j. Then
(cid:89)
A/(I1...In) (cid:39) A/Ik, a+I1...In (cid:55)→ (a+Ik)1(cid:54)k(cid:54)n.
1(cid:54)k(cid:54)n
Proof. Firstlet n = 2. Then
I I ⊂ I ∩I = (I ∩I )A = (I ∩I )(I +I ) ⊂ (I ∩I )I +(I ∩I )I ⊂ I I .
1 2 1 2 1 2 1 2 1 2 1 2 1 1 2 2 1 2
So I I = I ∩I . Thekernelofthehomomorphism
1 2 1 2
(cid:89)
A → A/I , a (cid:55)→ (a+I ,a+I )
k 1 2
1(cid:54)k(cid:54)2
is I ∩I = I I . Itissurjective: since I +I = A, thereareelements x ∈ I ,y ∈ I
1 2 1 2 1 2 1 2
such that x + y = 1 and hence bx + ay = a + (b − a)x ∈ a + I and similarly
1
bx+ay ∈ b+I .
2
Now proceed by induction on n. Denote J = I ,J = I ...I , so J J =
1 1 2 2 n 1 2
I ...I . Since I +I = A forall k > 1, wededuceusing1.2that
1 n 1 k
J +J = I +I ...I ⊃ (I +I )...(I +I ) = A,
1 2 1 2 n 1 2 1 n
so J +J = A. Nowinthesamewayasinthepreviousparagraphonegets A/(J J ) (cid:39)
1 2 1 2
(cid:81) (cid:81)
A/J . Bytheinductionhypothesis A/J (cid:39) A/I . Thus,
1(cid:54)k(cid:54)2 k 2 2(cid:54)k(cid:54)n k
(cid:89)
A/(I ...I ) (cid:39) A/I .
1 n k
1(cid:54)k(cid:54)n
5
Example. Let p bedistinctprimesand r positiveintegers. Then
i i
(cid:89)
Z/(pr1...prnZ) (cid:39) Z/priZ.
1 n i
6 Ringsandmodules
2. Modules over rings
2.1. Let A be a ring. An abelian group M is called an A-module if there is a
multiplication A×M → M suchthat a(x+y) = ax+ay,(a+b)x = ax+bx,a(bx) =
(ab)x,1x = x.
Examples. Everyabeliangroupisa Z-module,sotheclassofabeliangroupscoincide
withtheclassof Z-modules.
Everyvectorspaceoverafield F isan F-module.
2.2. A map f:M → N is called a homomorphism of A-modules if f(x + y) =
f(x)+f(y) forevery x,y ∈ M and f(ax) = af(x) forevery a ∈ A, x ∈ M. Aho-
momorphism f of A-modulesiscalledanisomorphismof A-modules,oralternatively
an A-isomorphism,if f isbijective.
2.3. Asubgroup N ofan A-module M iscalledan A-submoduleof M if an ∈ N
forevery a ∈ A,n ∈ N.
Example: Submodulesofthe A-module A areidealsof A.
For an A-module M and its A-submodule N define the quotient module M/N
as the quotient set of cosets m + N with the natural addition and multiplication by
elementsof A.
Similarly to 1.1 one proves: If M,N are A-modules and f:M → N is an
A-module homomorphism, then the kernel of f is a submodule of M and the image
of f isasubmoduleof N, and M/ker(f) is A-isomorphicto im(f).
Similarlyto1.1submodulesofthequotientmodule M/N arein1–1correspondence
withsubmodulesof M containing N.
Inparticular,if f:M → N isan A-modulehomomorphism,and K isasubmodule
of ker(f), then f inducesan A-modulehomomorphism g:M/K → N, m+K (cid:55)→
f(m).
2.4. For A-modules M,N the intersection M ∩N is an A-module. So if M,N
arecontainedinalargermodule L, onecanspeakabouttheminimal A-modulewhich
containsafixedsetofelementsrelatedto M and N.
Then the M +N = {m+n : m ∈ M,n ∈ N} is the minimal A-module which
containsallallelementsof M and N.
Define the direct sum of modules as the set theoretical product with the natural
additionandmultiplicationbyelementsof A.
Lemma. Let N,K be A-submodulesofan A-module M. Amap f:N⊕K → N+K,
f((n,k)) = n+k isasurjective A-modulehomomorpismwhosekernelis A-isomorphic
tothesubmodule N∩K. Therefore,if N∩K = {0}, N⊕K isisomorphicto N+K.
7
Proof. Clearly f is surjective. Its kernel is {(n,k) : n+k = 0}. Then n = −k ∈
N ∩K. Amap {(n,k) : n+k = 0} → N ∩K, (n,−n) (cid:55)→ n isabijection.
2.5. The submodule M generated by elements x is the minimal submodule which
i
containsallofthem,itconsistsoffinite A-linearcombinationsof x ; elements x ∈ M
i i
arecalledgeneratorsof M.
Theminimalnumberofgenerators(ifitexists)of M iscalledtherankof M.
M issaidtobeoffinitetypeifithasafinitenumberofgenerators.
(cid:80)
An A-module M is called free if M has generators x such that a x = 0
i i i
implies a = 0 forall i. Thesetof x iscalledthenabasisof M.
i i
2.6. Lemma.
(1)Themodule An = ⊕1(cid:54)i(cid:54)nA isfreeof rank n.
(2)Let M bean A-moduleof finitetypeandlet x , ...,x begeneratorsof M.
1 n
Defineahomomorphism
(cid:88)
f:An → M, (a , ...,a ) (cid:55)→ a x .
1 n i i
Itissurjective. If N isthekernelof f, then M isisomorphictothequotientmodule
(A)n/N. Thus, every A-module of finite type is isomorphic to a quotient of a free
module.
(3)Everyfreemoduleof finiterank n isisomorphicto An.
Proof. (1),(2)followfromthedefinitions. If M isfreeandthenumberofgenerators
isfiniteequalto n, thenthehomomorphism (A)n → M issurjectiveandinjective.
Elementsof N serveasrelationsforgeneratorsof M.
Asacorollarywededucethatthedirectsumoffreemodulesisfree: An ⊕Am (cid:39)
An+m.
Examples. 1. Fromlinearalgebraitisknownthateverymoduleoffiniterankovera
fieldhasabasisandisfree.
2. Let A = Z and M = Z/nZ for n > 1. Then M has rank 1 and is not a free
A-module,sinceif M (cid:39) (Z)1 then M wouldhavebeeninfinite.
3. Polylinear constructions
3.1. Thesetof A-modulehomomorphismsfroman A-module M to N isan A-module:
(af)(m) = a·f(m), (f +g)(m) = f(m)+g(m). Itisdenoted Hom (M,N).
A
8 Ringsandmodules
Example. Let A = F be a field, and let M be an F-vector space of dimension d
1
and N be an F-vector space of dimension d . Fix a basis {m } in M and a basis
2 i
{n } in N. Let C be a matrix of order d ×d with entries in F. Define a map
j 1 2
(cid:80) (cid:80)
f:M → N, f( a m ) = b n where (b ,...,b ) = (a ,...,a )C. The map f
i i j j 1 d2 1 d1
is an F-linear map. Conversely, every F-linear map M → N is determined by its
values on {m : 1 (cid:54) i (cid:54) d }. Write f(m ) = (cid:80)c n and define C = (c ). This
i 1 i ij j ij
givesaninversemaptothepreviousmap. Thus, Hom (M,N) is F-isomorphictothe
F
F-moduleofmatricesoforder d ×d withentriesin F.
1 2
Examples–Exercises. Hom (0,N) = Hom (M,0) = 0. Hom (A,N) (cid:39) N,
A A A
Hom (M,N ⊕N ) (cid:39) Hom (M,N )⊕Hom (M,N ).
A 1 2 A 1 A 2
3.2. Amap f:M ×N → R iscalled A-bilinearifforall m,m ,m ,n,n ,n ,a
1 2 1 2
f(m,n +n ) = f(m,n )+f(m,n ), f(m,an) = af(m,n)
1 2 1 2
f(m +m ,n) = f(m ,n)+f(m ,n), f(am,n) = af(m,n).
1 2 1 2
So for every m the map N → R,n (cid:55)→ f(m,n) is a homomorphism of A-modules
andforevery n themap M → R,m (cid:55)→ f(m,n) isahomomorphismof A-modules.
Note that an A-bilinear map f does not induce a homomorphism of A-modules
M ⊕N → R, since f(a(m,n)) = f(am,an) = a2f(m,n) isnotequalto af(m,n) in
general.
Denote the set of all A-bilinear maps f:M ×N → R by Bil (M,N;R). The
A
latter is an A-module with respect to the sum of maps and multiplication of a map by
anelementof A.
Similarlyonecandefine A-n-linearmaps.
Example. Let A = F be a field, and let M be an F-vector space of dimension
d and N be an F-vector space of dimension d . Fix a basis {m } in M and a
1 2 i
basis {n } in N. Let C be a matrix of order d ×d with entries in F. Define
j 1 2
a map f:M × N → F, f(m,n) = mCn◦ where m is written as a row and n◦
as a column. The map f is an F-bilinear map. Conversely, every F-bilinear map
M × N → F is determined by its values on {(m ,n ) : 1 (cid:54) i (cid:54) d ,1 (cid:54) j (cid:54)
i j 1
(cid:80) (cid:80) (cid:80)
d }: f( a m , b n ) = a b f(m ,n ). Now form a matrix C whose entries
2 i i j j i j i j
are f(m ,n ). Thus, there is a one-to-one correspondence between bilinear maps
i j
M ×N → F andmatricesoforder d ×d withentriesin F.
1 2
3.3. To study A-bilinear maps from M ×N to R it is useful to introduce another
A-module T andabilinearmap g:M×N → T suchthatbilinearmaps f:M×N →
R are in one-to-one correspondence with homomorphisms of A-modules T → R
via g. In other words, we define an isomorphism of A-modules Bil (M,N;R) (cid:39)
A
Hom (T,R); T doesnotdependon R butonlyon A-modules M and N.
A
Todefine T firstdenoteby L thefree A-modulewithabasisconsistingofelements
l indexed by elements of M ×N. So an arbitrary element of L is a finite sum
m,n
9
(cid:80)
a l with a ∈ A, m ∈ M and n ∈ N. Let K be the A-submodule of L
i mi,ni i i i
generatedbyelements
l −l −l , l −l −l ,
m1+m2,n m1,n m2,n m,n1+n2 m,n1 m,n2
l −al , l −al
am,n m,n m,an m,n
(forall a ∈ A, m ∈ M and n ∈ N).
Denote T = L/K. The image of l in T, i.e. the coset l +K is usually
m,n m,n
denotedby m⊗n.
Since L isgeneratedby l , themodule T isgeneratedby m⊗n, i.e.
m,n
(cid:8)(cid:88) (cid:9)
T = a m ⊗n : a ∈ A,m ∈ M,n ∈ N .
i i i i i i
Thesesatisfyrelations:
(m +m )⊗n = m ⊗n+m ⊗n, (am)⊗n = a(m⊗n),
1 2 1 2
m⊗(n +n ) = m⊗n +m⊗n , m⊗(an) = a(m⊗n)
1 2 1 2
(forall a ∈ A, m ∈ M and n ∈ N).
The module T is denote M ⊗ N and is called the tensor product of M and N
A
over A.
If M, N arefinitelygenerated A-modules,withgenerators m ,n , then M⊗ N
i j A
isafinitelygenerated A-modulewithgenerators m ⊗n .
i j
Wehave n⊗0 = 0(n⊗1) = 0.
Nowdefineamap g:M×N → M⊗ N by (m,n) (cid:55)→ m⊗n. Itisan A-bilinear
A
map.
3.4. Theorem. For an A-bilinear map f:M ×N → R define f(cid:48):M ⊗ N → R
A
as f(cid:48)((cid:80)a m ⊗n ) = (cid:80)a f(m ,n ). It is a well defined map and it is a homomor-
i i i i i i
phism of A-modules. The correspondence f (cid:55)→ f(cid:48) is an isomorphism of A-modules
Bil (M,N;R) and Hom (M ⊗ N,R).
A A A
Proof. Extend f to a homomorphism L → R defined on elements of the basis of L
by l (cid:55)→ f(m,n).
m,n
Since f is bilinear, all generators of K are mapped to zero, so we get f(cid:48) =
α(f):M ⊗ N → R, f(cid:48)((cid:80)a m ⊗n ) = (cid:80)a f(m ,n ). Themap α isahomomor-
A i i i i i i
phismof A-modules.
Conversely, if f(cid:48):M ⊗ N → R is a homomorphism of A-modules, then define
A
f = β(f(cid:48)):M ×N → R as f(m,n) = f(cid:48)◦g(m,n). Then f isan A-bilinearmap.
Now α◦β(f(cid:48)) = α(f(cid:48)◦g) andso α◦β(f(cid:48))((cid:80)a m ⊗n ) = (cid:80)a f(cid:48)◦g(m ,n ) =
i i i i i i
f(cid:48)((cid:80)a m ⊗n ). We also have β ◦α(f)(m,n) = α(f)◦g(m,n) = α(f)(m⊗n) =
i i i
f(m,n).
Thus, α and β areisomorphisms.
10 Ringsandmodules
Thus,usingthetensorproductonecanreducethestudyofbilinearmapstothestudy
oflinearmaps.
Example. Let A = F be a field. Let M,N be two F-vector spaces of dimensions
d and d . In accordance with the previous theorem the vector space of linear maps
1 2
M ⊗N → F is isomorphic to the vector space of bilinear maps M × N → F. In
F
accordance with Example in 3.2 the dimension of the space Bil(M,N;F) is d d ; if
1 2
m , ...,m is a basis of M and n , ...,n is a basis of N, then every bilinear
1 d1 1 d2
map f:M ×N → F isdeterminedbyitsvalueson {(m ,n )}.
i j
Therefore, the dimension of the vector space Hom (M ⊗ N,F) is d d . It
F F 1 2
is known from linear algebra that the dimension of a vector space V equals to the
dimensionof Hom (V,F). Sothedimensionof M ⊗N is d d ; the F-vectorspace
F 1 2
M ⊗ N hasabasis m ⊗n , 1 (cid:54) i (cid:54) d ,1 (cid:54) j (cid:54) d .
F i j 1 2
Note that in the particular case of M = N the space N ⊗ N has dimension
F
equaltothesquareofthedimensionof N. Inphysics, N over F = C representsthe
statevectorofaparticle,and N ⊗C N representsthestatevectorsoftwoindependent
particlesofthesamekind.
3.5. Firstpropertiesofthetensorproduct:
Lemma. (i) M ⊗ A (cid:39) M,
A
(ii) M ⊗ N (cid:39) N ⊗ M,
A A
(iii) (M ⊗ N)⊗ R (cid:39) M ⊗ (N ⊗ R),
A A A A
(iv) M ⊗ (N ⊕R) (cid:39) (M ⊗ N)⊕(M ⊗ R),
A A A
(v) Hom (M⊗ N,K) (cid:39) Hom (M,Hom (N,K)) (cid:39) Hom (N,Hom (M,K)).
A A A A A A
Proof. To prove (i) we first define an A-homomorphism f:L → M,l (cid:55)→ am
m,a
where L is a free A-module with a basis l , m ∈ M,a ∈ A. Then K (which
m,a
is the submodule of L defined as in 3.3) is in the kernel of f. So f induces an
A-homomorphism g:M ⊗ A = L/K → M, m ⊗ a (cid:55)→ am. Define h:M →
A
M ⊗ A,m (cid:55)→ m⊗1. Then g and h areinversetoeachother.
A
To prove (ii) use an A-homomorphism f:M ⊗N → N ⊗M,m⊗n (cid:55)→ n⊗m
whichcorrespondstoamap l (cid:55)→ n⊗m andan A-homomorphism g:N ⊗M →
m,n
M ⊗N,n⊗m (cid:55)→ m⊗n. f and g areinversetoeachother.
Toprove(iii)use m⊗(n⊗r) (cid:55)→ (m⊗n)⊗r,(m⊗n)⊗r (cid:55)→ m⊗(n⊗r).
For (iv) use m⊗(n,r) (cid:55)→ (m⊗n,m⊗r), (m ⊗n,m ⊗r) (cid:55)→ m ⊗(n,0)+
1 2 1
m ⊗(0,r).
2
For(v)use h ∈ Hom (M⊗ N,K) (cid:55)→ h(cid:48) ∈ Hom (M,Hom (N,K)), h(cid:48)(m)(n) =
A A A A
h(m⊗n) and h(cid:48) ∈ Hom (M,Hom (N,K)) (cid:55)→ h ∈ Hom (M⊗ N,K), h(m⊗n) =
A A A A
h(cid:48)(m)(n).
3.6. Examples.
(1) An⊗ Am = Anm.
A