LTCC Representation theory

31 Pages·2017·0.236 MB·English

by  Matt Fayers, Markus Linckelmann

LTCC Representation theory
MattFayers,basedonnotesbyMarkusLinckelmann
thankstoDiegoMillanBerdascoforcorrections
Contents
1 Basics 2
1.1 Somelinearalgebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.4 Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.5 Modulehomomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.6 Directsumsofmodules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.7 Decomposablemodules,simplemodulesandmaximalsubmodules . . . . . . . . 10
1.8 Representationsofalgebrasandthestructurehomomorphism . . . . . . . . . . . 13
2 Idempotents 14
3 Semisimplemodulesandsemisimplealgebras 18
4 TheJacobsonradical 22
5 Localalgebrasandindecomposablemodules 25
6 Projectivemodules 27
7 Representationtheoryoffinitegroups 29
1
2 LTCCRepresentationtheory
1 Basics
1.1 Somelinearalgebra
In this course I’ll assume that you’re pretty familiar with basic linear algebra. But we’ll
F
begin by revising a few simple concepts. Throughout these notes we fix a field , and all
F F
vectorspaceswillbeover . dimwillalwaysmeandimensionover .
Somebasicnotation: foranyvectorspaceV,wewriteid (orjustid,ifit’sclearwhatV is)
V
fortheidentityfunctionfromV toV. Also,we’llabusenotationbywriting0insteadof{0}for
thezerovectorspace.
Definition.SupposeV andW arevectorspaces.
1. ThedirectsumV⊕W isthesetofallsymbolsv⊕wforv ∈ V,w ∈ W,withadditionand
scalarmultiplicationdefinedby
(v⊕w)+(v(cid:48)⊕w(cid:48)) = (v+v(cid:48))⊕(w+w(cid:48)), x(v⊕w) = (xv)⊕(xw)
for x ∈ F,v,v(cid:48) ∈ V,w,w(cid:48) ∈ W.
2. The tensor product V⊗W is the set of all formal sums of symbols v⊗w with v ∈ V and
w ∈ W,modulotherelations
(xv)⊗w = v⊗(xw), (v+v(cid:48))⊗w = v⊗w+v(cid:48)⊗w, v⊗(w+w(cid:48)) = v⊗w+v⊗w(cid:48)
for x ∈ F,v,v(cid:48) ∈ V,w,w(cid:48) ∈ W. WemakeV⊗W isavectorspacevia
x(v⊗w) = (xv)⊗w
for x ∈ F,v ∈ V,w ∈ W.
Notethatif BisabasisforV andCisabasisforW,then
{b⊕0| b ∈ B}∪{0⊕c| c ∈ C}
isabasisforV⊕W,and
{b⊗c| b ∈ B, c ∈ C}
isabasisforV⊗W. HenceifV andW arefinite-dimensionalthen
dim(V⊕W) = dim(V)+dim(W), dim(V⊗W) = dim(V)dim(W).
Definition.Suppose V is a vector space and W (cid:54) V. The quotient space V/W is the set of all
cosets
v+W = {v+w| w ∈ W}
withvectorspacestructuregivenby
(u+W)+(v+W) = (u+v)+W, x(v+W) = (xv)+W
foru,v ∈ Mand x ∈ F.
Basics 3
Suppose BisabasisforV andCisabasisforW withC ⊆ B. Then
{b+W | b ∈ B\C}
isabasisforV/W. HenceifV isfinite-dimensional,then
dim(V/W) = dim(V)−dim(W).
Note that we can have u+W = v+W even when u (cid:54)= v, so we need to be careful. A
precise(anduseful)statementisthefollowing.
CosetLemma.SupposeV isavectorspace,u,v ∈ V andW (cid:54) V. Thenu+W = v+W ifandonly
ifu−v ∈ W.
1.2 Rings
Nowlet’srevisesomebasicdefinitionsforrings.
Definition.AringisasetRwithtwospecialelements0and1andtwobinaryoperations+,×
suchthat:
• Risanabeliangroupunder+,withzeroelement0;
• ×isassociative,i.e.r×(s×t) = (r×s)×tforallr,s,t ∈ R;
• 1×r = r×1 = rforallr ∈ R;
• (thedistributivelaw)r×(s+t) = (r×s)+(r×t)and(r+s)×t = (r×t)+(s×t)for
allr,s,t ∈ R.
We’llusestandardconventionsofnotation,writingrsinsteadofr×s,andwritingexpres-
sions like rs+t without brackets on the understanding that we do the multiplication first.
We also adopt the following convention: if r ∈ R and X ⊆ R, then we write rX to mean
{rx | x ∈ X}. Similarlywedefine Xr,rXs, XrY etc.
Werefertotheelement1astheidentityelementof R,andto0asthezeroelement. Notethat
(unlike some people) we do not require these two elements to be distinct. The only effect of
this is to permit the trivial ring {0} which has addition and multiplication defined in the only
possible way. Very occasionally we will prove results that only hold for non-trivial rings, but
ingeneralthetrivialringwillbeallowed.
We’llneedthefollowingdefinitionsinordertodefineanalgebra.
Definition.
1. If Risaring,thecentreof Ristheset
Z(R) = {a ∈ R| ab = baforallb ∈ R}.
2. If R and S are rings, a homomorphism from R to S is a function φ : R → S such that
φ(ab) = φ(a)φ(b),φ(a+b) = φ(a)+φ(b)andφ(1 ) = 1 forall a,b ∈ R.
R S
Note that in the definition of a homomorphism, the condition φ(1 ) = 1 rules out, for
R S
example,thetrivialmapwhichsends ato0forall a ∈ R(unlessSisthetrivialring).
4 LTCCRepresentationtheory
1.3 Algebras
Nowwecometoalgebras.
Definition.An F-algebra is a ring A together with a specified ring homomorphism from F to
Z(A).
F
Almost always in these notes we shall be considering algebras over , so we will just say
F
‘algebra’tomean‘ -algebra’.
Remarks.
1. Some people’s definition of an algebra A requires that the homomorphism from F to
A be injective. The only difference between this and our definition is that it rules out
the possibility that A is the trivial ring. Indeed, if A is a ring φ : F → A is a non-
injectivehomomorphism, take0 (cid:54)= x ∈ Fwith φ(x) = 0; then1A = φ(1F) = φ(xx−1) =
φ(x)φ(x−1) = 0φ(x−1) = 0, so that A is the trivial ring. We prefer to allow the trivial
ring;forexample,ifV isanon-zeroF-vectorspace,thenthesetEndF(V)oflinearmaps
fromV toV isanF-algebra;ifwewanttoextendthistothecaseV = 0,weneedtoallow
thetrivialringasanalgebra.
2. In the cases where A is non-trivial (so the corresponding ring homomorphism is injec-
F
tive),itiscustomarytoidentify withitsimageunderthehomomorphism,andthereby
regardFasasubringof Z(A).
3. An algebra A is naturally a vector space over F; if A is trivial this is obvious, while if A
is non-trivial and we regard F as a subring of A as above, then the scalar multiplication
isjustgivenbytheringmultiplicationin A. Thisleadstoanalternativedefinitionofan
algebra: itisanF-vectorspaceequippedwithabinaryoperation×whichwhichsatisfies
theassociative,distributiveandidentitylaws.
Definition.Suppose A and B are algebras. A homomorphism from A to B is an F-linear map
whichisalsoaringhomomorphism.
Note that not every ring homomorphism between two algebras is an algebra homomor-
phism. Forexample,takeF = A = B = C;thencomplexconjugationisaringhomomorphism,
C
butisnot -linear.
Weusestandardterminologyassociatedwithhomomorphisms: anisomorphismisahomo-
morphismwhichisalsoabijection(inwhichcaseitsinverseisalsoanisomorphism),andtwo
algebras are isomorphic if there is at least one isomorphism between them. An automorphism
of an algebra A is an isomorphism from A to A. The kernel of a homomorphism is the set of
elements that map to 0. As is always the case with linear maps, a homomorphism is injective
ifandonlyifitskernelis0.
Definition.Suppose Aisanalgebra.
• A subalgebra of A is an F-subspace which is also a subring (i.e. which is closed under
multiplicationandcontains1).
• Aleftidealof AisanF-subspace I of Asuchthat ai ∈ I forall a ∈ A,i ∈ I.
• Arightidealof AisanF-subspace I of Asuchthatia ∈ I forall a ∈ A,i ∈ I.
Basics 5
• Anidealof Aisaleftidealwhichisalsoarightideal.
Wewrite I A, I A, I Atomeanthat I isaleftideal,arightidealoranidealof A,
L R
(cid:80) (cid:80) (cid:80)
respectively.
Some key examples: if A is a non-trivial algebra, then F is a subalgebra of A. If X is any
subset of A, let AX = {ax | a ∈ A, x ∈ X}. Then the subspace of A spanned by AX is a left
ideal of A. Similarly the subspace spanned by XA is a right ideal, and the subspace spanned
by AXA = {axb| a,b ∈ A, x ∈ X}isanideal.
Lemma1.1.Suppose A isanalgebraand I A. Thenthequotientvectorspace A/I isanF-algebra,
withmultiplicationdefinedby(a+I)(b+I(cid:80)) = ab+I andwithidentityelement1+I.
Proof. Exercise.
The next result is called the First Isomorphism Theorem for algebras. If you’ve seen the
First Isomorphism Theorem for groups, rings, or modules, then you should have no trouble
provingit.
Theorem 1.2.Suppose A and B are algebras and φ : A → B is a homomorphism. Then ker(φ) is an
idealof A,im(φ)isasubalgebraofB,andthereisanalgebraisomorphism A/ker(φ) → im(φ)given
by a+ker(φ) (cid:55)→ φ(a).
Proof. Exercise.
Herewecollectafewmorebitsofterminology.
Definition.If Aisanalgebraand a ∈ A,then aisinvertibleifthereisanelement a−1 ∈ Asuch
that aa−1 = 1 = a−1a.
Remarks.
1. Theinverseofaninvertibleelementisunique.
2. Thesetofinvertibleelementsof Aisagroupundermultiplication.
3. Note that we require an inverse to be two-sided, i.e. aa−1 = 1 and a−1a = 1. However, if
Aisfinite-dimensionalthenthisisunnecessary,aswenowshow.
Lemma1.3.Suppose Aisafinite-dimensionalalgebra,and a,b ∈ Asuchthat ab = 1. Thenba = 1.
Proof. Considerthemap
φ : A −→ A
c (cid:55)−→ ca.
φ isalinearmap,andweclaimthat φ isinjective: if φ(c) = 0,then ca = 0,so0 = 0b = cab =
c1 = c.
Nowalinearmapfromafinite-dimensionalvectorspacetoitselfisinjectiveifandonlyif
itissurjective(bytheRank–NullityTheorem),soφissurjectiveandhencethereisc ∈ Asuch
thatca = 1. Andnow
c = c1 = cab = 1b = b
sothatba = ca = 1.
6 LTCCRepresentationtheory
Definition.Suppose A is an algebra. The opposite algebra Aop is the same vector space with
the opposite multiplication: a·b = ba, where a·b is the multiplication in Aop and ba is the
multiplicationin A.
Alotofthealgebraswe’llseeareisomorphictotheiropposites. Buttherearesomewhich
R
aren’t. Forexample,considerthe -algebra
(cid:26)(cid:18) (cid:19)(cid:12) (cid:27)
A = a b (cid:12)(cid:12) a ∈ R, b,c ∈ C
0 c (cid:12)
withtheusualmatrixmultiplication. Toseewhy A (cid:29) Aop,notethat Ahasaleftidealwhichis
R
one-dimensional (over ), but doesn’t have a one-dimensional right ideal (you should check
thesestatements!). Hence Aop doesn’thaveaone-dimensionalleftideal,so Aand Aop can’tbe
isomorphic.
Nowweseetwowaystocombinealgebrastomakelargerones.
Definition.Suppose Aand Barealgebras.
1. Thedirectsum A⊕Bisanalgebrawithmultiplication(a⊕b)(a(cid:48)⊕b(cid:48)) = (aa(cid:48))⊕(bb(cid:48)).
2. Thetensorproduct A⊗Bisanalgebrawithmultiplicationdefinedby(a⊗b)(a(cid:48)⊗b(cid:48)) =
(aa(cid:48))⊗(bb(cid:48)),extendingbilinearly.
1.4 Modules
Nowwecometothedefinitionofamodule.
Definition.Let R be a ring. A left R-module is an abelian group M together with a function
× : R×M → Msuchthat
• r×(m+n) = (r×m)+(r×n),
• (r+ s)×m = (r×m)+(s×m),
R
• (r× s)×m = r×(s×m),
R
• 1×m = m
forallr,s ∈ R,m,n ∈ M.
Thereisacorrespondingdefinitionofaright R-module. Inthesenoteswe’llalmostalways
considerleftmodules,sowe’lljustsay‘module’tomean‘leftmodule’.
Of course, if A = F then you already know what an A-module is – it’s the same thing as
a vector space over F. In general, if A is an algebra and B is a subalgebra of A then any A-
moduleisa B-moduleinanaturalway. Inparticularif A isnon-trivialthenFisasubalgebra
of A so an A-module is automatically an F-module, i.e. a vector space. So we can talk about
basesformodules,finite-dimensionalmodulesetc.
Twokeyexamplesofmodulesarethezeromodule0,withtheobviousvectorspacestructure
andactionofA,andtheregularmodule: thisisAitselfasavectorspace,withtheA-actiongiven
by the multiplication in A. We often write the regular A-module as A to make it clear that
A
we’rethinkingofitasamoduleratherthananalgebra.
Nowwelookatsubmodules.
Basics 7
Definition.Suppose A is an algebra and M is an A-module. A submodule of M is a subspace
which is also a module under the same operations. We write N (cid:54) M to mean that N is a
submoduleof M.
Note that a submodule of the the regular module A is just the same thing as a left ideal
A
of A.
Definition.Suppose M is an A-module and N (cid:54) M. Then the quotient space M/N is an
A-modulevia
a(m+N) = (am)+N
for a ∈ Aandm ∈ M.
1.5 Modulehomomorphisms
Aswithrings,wemaketheusualdefinitionsregardinghomomorphisms. Ahomomorphism
betweentwo A-modules Mand Nisafunctionφ : M → Nsuchthatφ(m+n) = φ(m)+φ(n)
and φ(am) = aφ(m) for m,n ∈ M and a ∈ A. We use the terms isomorphism, automorphism,
kernelandimageintheusualway.
If Mand N are A-modules,wewriteHom (M,N)forthesetofallhomomorphismsfrom
A
MtoN. Anexampleofsuchahomomorphism(whichisoftentheonlyhomomorphism)isthe
zerohomomorphism,whichmapsm (cid:55)→ 0foreverym ∈ M. ThesetHom (M,N)isnaturallyan
A
F
-vectorspacevia
(φ+ψ)(m) = φ(m)+ψ(m), (xφ)(m) = x(φ(m))
forφ,ψ ∈ Hom (M,N),m ∈ Mand x ∈ F.
A
Anendomorphismofan A-module Mjustmeansahomomorphismfrom Mto M. Wewrite
End (M) for the vector space Hom (M,M). For every x ∈ F, the map xid is an endomor-
A A M
phismof M.
Aswellasbeingavectorspaceasdescribedabove,End (M)hasanotherbinaryoperation
A
◦definedbycomposition:
(φ◦ψ)(m) = φ(ψ(m)).
Proposition1.4.Suppose Aisanalgebraand Man A-module. ThesetEnd (M)isanalgebraunder
A
theoperations+,◦,with x ∈ Fmappingto xid .
M
Proof. Exercise.
End (M)isreferredtoastheendomorphismalgebraof M.
A
Thecasewhere M = Aisofparticularinterest.
A
Proposition1.5.Suppose Aisanalgebra. ThenEnd (A) ∼= Aop.
A
Proof. Given b ∈ A, we have an endomorphism φ of A defined by φ (a) (cid:55)→ ab; this gives a
b b
function
Φ : A −→ End (A)
A
b (cid:55)−→ φ
b
andsince φ = φ ◦φ , Φ isahomomorphismfrom Aop toEnd (A). Φ hasaninversegiven
bc c b A
byφ (cid:55)→ φ(1),soisanisomorphism.
8 LTCCRepresentationtheory
Aswithalgebras,wehavetheFirstIsomorphismTheoremformodules.
Theorem1.6.Suppose Aisanalgebra, Mand Nare A-modulesandφ : M → Nisahomomorphism.
Then ker(φ) (cid:54) M, im(φ) (cid:54) N and there is an isomorphism from M/ker(φ) to im(φ) given by
m+ker(φ) (cid:55)→ φ(m).
Proof. Exercise.
We also prove the Second and Third Isomorphism Theorems for modules. Unfortunately,
thereisnogeneralagreementaboutwhichiswhich.
Theorem1.7.Suppose A isanalgebraand M isan A-module, and N,P (cid:54) M. Then N/(N∩P) ∼=
(N+P)/P.
Proof. Wedefineahomomorphism
φ : N −→ M/P
n (cid:55)−→ n+P.
Thekernelofφisthesetofalln ∈ N suchthatn+P = 0+P,i.e.n ∈ P;sothekernelis N∩P.
We claim that the image is (N +P)/P. An element of im(φ) has the form n+P for n ∈ N;
since N ⊆ N+P, this coset lies in (N+P)/P. Conversely, an element of (N+P)/P can be
writtenas(n+p)+Pforn ∈ N and p ∈ P,but(n+p)+P = n+P ∈ im(φ).
NowtheFirstIsomorphismTheoremgivestheresult.
Theorem1.8.Suppose Aisanalgebraand Misan A-module,andP (cid:54) N (cid:54) M. ThenN/P (cid:54) M/P,
∼
and(M/P)/(N/P) = M/N.
Proof. Defineahomomorphism
φ : M/P −→ M/N
m+P (cid:55)−→ m+N.
φ is well-defined: if l + P = m+ P, then l −m ∈ P, so l −m ∈ N, so l + N = m+ N. φ
is obviously surjective. The kernel of φ is the set of all m+P such that m+ N = 0+ N, i.e.
m ∈ N. Soker(φ) = {n+P| n ∈ N} = N/P.
NowtheFirstIsomorphismTheoremgivestheresult.
1.6 Directsumsofmodules
Definition.Suppose Mand N are A-modules. Thedirectsum M⊕N isalsoan A-modulevia
a(m⊕n) = (am)⊕(an)
for a ∈ A,m ∈ M,n ∈ N.
Thereisanaturalwaytothinkof Mand N assubmodulesof M⊕N: therearehomomor-
phisms
ι : M −→M⊕N ι : N −→M⊕N
M N
m (cid:55)−→m⊕0 n (cid:55)−→ 0⊕n
Basics 9
calledthecanonicalinjections. BytheFirstIsomorphismTheorem, M isisomorphictothesub-
module
im(ι ) = {m⊕0| m ∈ M}
M
of M⊕N, and we often identify M with this submodule. Similarly, we often identify N with
the submodule {0⊕n | n ∈ N}. Then M and N are submodules of M⊕N with M∩N = 0
and M+N = M⊕N.
Conversely, if L is an A-module and M,N are submodules of L with M ∩ N = 0 and
M+ N = L, then there is an isomorphism from M⊕ N to L given by m⊕n (cid:55)→ m+n, and
we often identify L with M⊕N via this isomorphism, and say that L is the direct sum of M
and N.
Of course, we can extend the idea of a direct sum to more than two modules. Given a
∼
module M and a family of submodules M ,...,M , when do we have M = M ⊕···⊕ M ?
1 r 1 r
It’s not enough to have M = M +···+ M and M ∩ M = 0 for all i (cid:54)= j (take A = F, and
1 r i j
considerthreeone-dimensionalsubspacesof M = F2).
Proposition1.9.Suppose A isanalgebraand M isan A-module,and M ,...,M aresubmodulesof
1 r
M. Thenthereisahomomorphism
φ : M ⊕···⊕M −→ M
1 r
m ⊕···⊕m (cid:55)−→ m +···+m .
1 r 1 r
φisanisomorphismprovided M +···+M = Mand M ∩(∑ M ) = 0foreveryi.
1 r i j(cid:54)=i j
Proof. It is easy to check that φ is a homomorphism. Clearly im(φ) = M +···+M , so φ is
1 r
surjective if and only if M +···+ M = M. Now we claim that φ is injective if and only if
1 r
M ∩(∑ M ) = 0foreveryi. Ifφfailstobeinjective,takeanon-zeroelementm ⊕···⊕m
i j(cid:54)=i j 1 r
initskernel. Sincethisisnon-zero,theremustbesomeisuchthatm (cid:54)= 0. Butnow
i
∑ ∑
m = − m ∈ M ∩( M ),
i j i j
j(cid:54)=i j(cid:54)=i
so M ∩(∑ M ) (cid:54)= 0. Conversely,supposethatforsomeiwehave M ∩(∑ M ) (cid:54)= 0. This
i j(cid:54)=i j i j(cid:54)=i j
meanswecanfindm ∈ M foreach jsuchthat0 (cid:54)= m = −∑ m . Butthenm ⊕···⊕m is
j j i j(cid:54)=i j 1 r
anon-zeroelementofker(φ),soφisnotinjective.
Often we will want to consider a direct sum M⊕···⊕ M of copies of the same module,
andwemaywrite M⊕n forthedirectsumof n copiesof M. Givensuchadirectsum,welet ι
i
betheithcanonicalinjectionfrom Mto M⊕n,i.e.themapwhichsends
m (cid:55)−→ 0⊕···⊕0⊕m⊕0⊕···⊕0
with m in the ith position. We also let π : M⊕n → M be the ith canonical projection, which
i
maps
m ⊕···⊕m (cid:55)−→ m .
1 n i
Thenι ,π arehomomorphisms,andwehave
i i
π ◦ι = id ,
i i M
π ◦ι = 0 ifi (cid:54)= j,
i j
n
∑
ιi◦πi = idM⊕n.
i=1
10 LTCCRepresentationtheory
Theendomorphismringof M⊕n iseasytodescribe;if Bisanyalgebra,letMat (B)denotethe
n
algebra of n×n matrices over B, with the usual addition and multiplication. (n.b. You have
tobeslightlycarefulwithmultiplication,if Bisnotcommutative: theproductoftwomatrices
m,nisgivenby(mn) = ∑ m n ,not∑ n m .
ij k ik kj k kj ik
Proposition 1.10.Suppose A is an algebra and M is an A-module, and n is a non-negative integer.
Then
End (M⊕n) ∼= Mat (End (M)).
A n A
Proof. Supposeα ∈ End (M⊕n). DefineamatrixΦ(α)bysettingΦ(α) = π ◦φ◦ι foreach
A ij i j
i,j. ThenΦ(α) ∈ End (M),sowehaveamapΦ : End (M⊕n) → Mat (End (M)). It’seasy
ij A A n A
tocheckthatΦislinear,andinfactit’sanalgebrahomomorphism,sinceforα,β ∈ End (M⊕n)
A
wehave
Φ(α◦β) = π ◦α◦β◦ι
ij i j
(cid:32) (cid:33)
n
∑
= π ◦α◦ ι ◦π ◦β◦ι
i k k j
k=1
n
∑
= (π ◦α◦ι )◦(π ◦β◦ι )
i k k j
k=1
n
= ∑ Φ(α) ◦Φ(β)
ik kj
k=1
= (Φ(α)Φ(β)) .
ij
To show that Φ is bijective, we construct its inverse: given a matrix φ ∈ Mat (End (M)),
n A
defineα ∈ End (M⊕n)by
A
∑
α = ι ◦φ ◦π .
i ij j
i,j
ThenαistheuniqueelementofEnd (M⊕n)satisfyingΦ(α) = φ.
A
1.7 Decomposablemodules,simplemodulesandmaximalsubmodules
Definition.SupposeAisanalgebraandMisanA-module. Misreducibleifithasapropernon-
zerosubmodule,andirreducible(orsimple)ifitisnon-zeroandnotreducible. Misdecomposable
if it can be written as the direct sum of two non-zero submodules, and indecomposable if it is
non-zeroandnotdecomposable.
Note that we do not consider the zero module to be either reducible or irreducible, or
decomposable or indecomposable (in the same way that 1 is neither a prime number nor a
compositenumber).
Example.A decomposablemodule must bereducible, but the converseis not truein general.
Take Atobethealgebra
(cid:26)(cid:18) (cid:19)(cid:12) (cid:27)
T (F) = a b (cid:12)(cid:12) a,b,c ∈ F
2 0 c (cid:12)
andlet MbethenaturalmoduleF2.