LOCALIZATION, ISOMORPHISMS AND ADJOINT ISOMORPHISM IN THE CATEGORY OF MODULES

Abstract

In this paper the category of modules is the considered category. The aim of this paper is to show at first that the functors E XT n and T or n are adjoint functors and secondly to show the commutativity of the functor S −1 () with the functors tensor product, Hom, E XT n and T or n

Full text

LOCALIZATION, ISOMORPHISMS AND
ADJOINT ISOMORPHISM IN THE
CATEGORY OF MODULES
Bassirou Dembele1, Mohamed Ben Faraj ben Maaouia2and Mamadou Sanghare3
Abstract . In this paper the category of modules is the considered category. The aim
of this paper is to show at first that the functors EXT nand T ornare adjoint func-
tors and secondly to show the commutativity of the functor S−1() with the functors
tensor product, H om,E XTnand Torn
Key words: Ring, saturated multiplicative subset, localization, category of mod-
ules, functor localization S−1().
1 Introduction
In this paper we generalize two results:
1. the result of [8] and [9] about adjonction between E XTnand Tornwithout no
assumption on the module M.
2. the results of [7] about the commutativity of the functor S−1() with homological
functors Hom, tensor product, E XTnand Tornon the objects. We show those
results on the general case.
Let then Aand Bbe two rings, Ma(A−B)bimodule and nan integer, we organize
this paper as following:
we give some preliminary results and definitions in our first section for remainder.
In the second section we show the following results:
1Gaston Berger University , Saint-Louis (UGB), Senegal,
e-mail: bassirou.dembele@aims-senegal.org
·2Gaston Berger University , Saint-Louis (UGB), Senegal,
e-mail: maaouiaalg@hotmail.com
·3Cheikh Anta Diop University of Dakar (UCAD), Senegal
e-mail: mamadou.sanghare@ucad.edu.sn
1

2 Bassirou Dembele, Mohamed Ben Faraj ben Maaouia and Mamadou Sanghare
1. EXT n+1
A(M,−):A−Mod −→ B−Mod and
EXT n
A(K0,−):A−Mod −→ B−Mod are naturally isomorphic;
2. T or A
n+1(M,−):B−Mod −→ A−Mod and
T or A
n(K0,−):B−Mod −→ A−Mod are naturally isomorphic;
3. EXT n
A(M,−):A−Mod −→ B−Mod and
TorB
n(M,−):B−Mod −→ A−Mod are adjoint functors;
4. if Ais a subring of B,Sa saturated multiplicative subset of Aand Bsatisfying
the left Ore conditions then:
EXT n
S−1A(S−1(M),−):S−1A−Mod −→ S−1B−Mod and
TorS−1B
n(S−1(M),−):S−1B−Mod −→ S−1A−Mod are adjoint functors .
And in our last section, we prove the following results:
1. TorS−1A
n+1(S−1(M),S−1()) :B−Mod −→ S−1A−Mod and
TorS−1A
n(S−1(K0),S−1()) :B−Mod −→ S−1A−Mod are naturally isomorphic;
2. S−1(MN−):B−Mod −→ S−1A−Mod and
S−1(M)NS−1() :B−Mod −→ S−1A−M od are naturally isomorphic;
3. S−1T or A
n(M,−):B−Mod −→ S−1A−Mod and
TorS−1A
n(S−1(M),S−1()) :B−Mod −→ S−1A−Mod are naturally isomorphic.
4. EXT n+1
S−1A(S−1(M),S−1()) :A−Mod −→ S−1B−Mod and
EXT n
S−1A(S−1(K0),S−1()) :A−Mod −→ S−1B−Mod are naturally isomor-
phic;
5. If Mis of finite type then S−1H om(M,−):A−Mod −→ S−1B−Mod and
Hom(S−1(M),S−1()) :A−Mod −→ S−1B−Mod are naturally isomorphic;
6. If Mis of fnite type then S−1E XTn
A(M,−):A−Mod −→ S−1B−Mod and
EXT n(S−1(M),S−1()) :A−Mod −→ S−1B−Mod are naturally isomorphic;
2 Definitions and preliminary results
Definition 1 Let Aa ring and Sa subset of A. It is said that:
1) Smultiplicative subset of Aif 1A∈Sand Sis stable by multiplication that is
∀s,s0∈S,ss0∈S.
2) Sis a saturated multiplicative subset of Aif Sis multiplicative subset of Aand if
for all s,s0∈A, if ss0∈Sthen s∈Sand s0∈S.
3) Sis a saturated multiplicative subset of Averifying the left Ore conditions if Sis
a saturated multiplicative subset of Aand then:
1. ∀a∈A,∀s∈S,∃(b,t)∈A×S:ta =bs,
2. ∀a∈A,∀s∈S, if as =0 then there exist t∈Ssuch as t a =0.
Proposition 1 Let A be a ring, S a saturated multiplicative subset of A vÃl’rifying
the left Ore conditions and M a left A module. Then the binary relation defined on
S×M par:

Title Suppressed Due to Excessive Length 3
(s,m)R(s0,m0)⇐⇒ ∃ x,y∈S:(xm =ym0
xs =ys0
is an equivalence relation.
Proof [see [1],theorem 2.7]
we denote by S−1Mthe set of classes and (s,m)by m
s.
Proposition 2 Let A be a ring and S a saturated multiplicative subset of A vÃl’rify-
ing the left Ore conditions. Then:
(1) S−1A can be equipped to a ring structure as following:
If a
t,b
s∈S−1A then:
•a
t+b
s=xa +yb
y s where x,y∈S/xt =ys
•a
t×b
s=zb
wt where (z,w)∈A×S/wa=zs
(2) S−1M can be equipped to a S−1A−module structure as following::
If a
t∈S−1A,m
s,m0
s0∈S−1M then:
•m
s+m0
s0=xm+ym 0
y s0where x,y∈S/xs =ys0
•a
t.m
s=zm
wt where(z,w)∈S×A/wa=zs
Proof [see [1],theorem 2.8]
Proposition 3 Let A be a ring, S a saturated multiplicative subset of A vÃl’rifying
the left Ore conditions and f :M−→ M0a morphism of left A-modules. Then,
S−1f:S−1M−→ S−1M0
m
s7−→ f(m)
s
is a morphisme of left S−1A-modules.
Proof
Let m
sand m0
s0be two elements of S−1M, we have:
S−1f(m
s+m0
s0)=S−1f(xm +ym0
ys0)=f(xm +ym0)
ys0=x f (m)+yf(m0)
ys0where xs =ys0
=⇒S−1f(m
s+m0
s0)=x f (m)
ys0+yf(m0)
ys0=f(m)
s+f(m0)
s0=S−1f(m
s)+S−1f(m0
s0)
S−1f(a
t.m
s)=S−1f(zm
wt)=z f (m)
wtwhere wa=zs
On the other part a
t.f(m)
s=z f (m)
wt
Hence S−1f(a
t.m
s)=a
t.f(m)
s

4 Bassirou Dembele, Mohamed Ben Faraj ben Maaouia and Mamadou Sanghare
Proposition 4 Let A be a ring and S a saturated multiplicative subset of A verifying
the left Ore conditions. Then the following relation:
S−1() :A−Mod −→ S−1A−M od such that
1. if M is an object of A −M od then :
S−1(M)=S−1M
2. if f :M−→ M is a morphism of A −Mod then
S−1(f):S−1M−→ S−1N.
Then S−1() is an exact covariant functor.
Proof see [1], theorem 4.2 
Definition 2 Let Mbe a left A-module and consider a projective resolution of M:
M•:. . . −→ Pn+1
dn+1
−→ Pn−→ . . . −→ P2
d2
−→ P1
d1
−→ P0

−→ C−→ 0.
Then Ker(dn)is called the nth kernel ofM•and we denote it by Kn.
Definition 3 Cand Dare two categories, Fand Gtwo functors with same variance
from Cto D. A natural transformation from Fto Gis a map Φ:F−→ Gsuch as:
•If Fand Gare covariant, then
Φ:Ob(C)−→ Mor(D)
M7−→ ΦM
is a map such that ΦM:F(M)−→ G(M)and for any f∈Mor (C)so that
f:M−→ N, then we have the following commutative diagram :
F(M)
ΦM

F(f)//F(N)
ΦN

G(M)G(f)//G(N)
•If Fand Gare contravariant then the following diagram is commutative:
F(N)
ΦN

F(f)//F(M)
ΦM

G(N)G(f)//G(M)
If ΦMis an isomorphism for all Mthen Φis called functorial isomorphism.

Title Suppressed Due to Excessive Length 5
3 ADJOINT ISOMORPHISM BETWEEN E X T AND T or in
Comp(A−Mod)
Definition 4 Cand Dare two categories, F:C −→ D and G:D −→ C are two
functors. The couple (F,G)is said to be adjoint if for any A∈Ob(C)and for any
B∈Ob(D), there is an isomorphism:
rA,B:HomC(A,G(B)) −→ HomD(F(A),B)
so that:
a) For any f∈HomC(A0,A), the following diagram is commutative:
HomC(A,G(B))
rA,B

H om (f,G(B)) //HomC(A0,G(B))
rA0
,B

HomD(F(A),B)H o m(F(f),B)//H omD(F(A0),B)
b) For any g∈HomD(B,B0), the following diagram is commutative:
HomC(A,G(B))
rA,B

H om (A,G(g)) //H omC(A,G(B0))
rA,B0

HomD(F(A),B)H o m(F(A),g)//H omD(F(A),B0)
Lemma 1 Let M be a left A-module and M•a projective resolution of M of nt h
kernel Ker(dn)=Kn. Then the functors
T or A
n+1(M,−)T or A
n(K0,−)
Proof Since . . . −→ Pn+1
dn+1
−→ Pn−→ . . . −→ P2
d2
−→ P1
d1
−→ P0

−→ M−→ 0 is
a projective resolution of Mthen . . . −→ Pn+1
dn+1
−→ Pn−→ . . . −→ P2
d2
−→ P1
d1
−→
K0−→ 0 is a projective resolution of K0. So on one hand:
T or A
n+1(M,N)T or A
n(K0,N),∀N∈Ob(A−Mod )
On other hand by doing the same thing for morphisms we get the result. 
Lemma 2 Let M be a left A-module and M•be a projective resolution of M of
nt h kernel Kn. Then the functors EXT n+1
A(M,−)and EXT n
A(K0,−)are naturally
isomorphic.
Proof The proof is similar to the proof of lemma 1.

6 Bassirou Dembele, Mohamed Ben Faraj ben Maaouia and Mamadou Sanghare
Lemma 3 :
Let M be a (A−B)-bimodule. Then the functors
HOmA(M,−):A−Mod −→ B−Mod and
MN−:B−Mod −→ A−Mod are adjoint functors.
Proof See [3] 
Theorem 1 Let M be a (A−B)-bimodule. Then the functors
EXT n
A(M,−):A−Mod −→ B−Mod and
TorB
n(M,−):B−Mod −→ A−Mod are adjoint functors.
Proof For n=0, we have on one part EXT0
A(M,−)HOmA(M,−)and on t other
part TorB
0(M,−)MN−. And by lemma 3 HOmA(M,−)and MN−are ad-
joint functors. Hence E XT0
A(M,−)and TorB
0(M,−)are adjoint functors.
Suppose by induction that the result is verified for all k<nand prove that it is
verified for k=n. That’s EXT n
A(M,−)and TorB
n(M,−)are adjoint functors.
By lemma 2 E XTn
A(M,−)EXT n−1
A(K0,−)and by lemma 1 Tor B
n(M,−)
TorB
n−1(K0,−). By the hypothesis E XTn−1
A(K0,−)and TorB
n−1(K0,−)are adjoint
functors then E XTn
A(M,−)and TorB
n(M,−)are adjoint functors. 
Theorem 2 Let B be a ring, A a sub-ring of B and S a saturated multiplicative
subset of A and B satisfying the left and right Ore conditions and M a (A−B)-
bimodule. Then the functors EXT n(S−1(M),−):S−1A−M od −→ S−1B−M od
and
TorS−1B
n(S−1(M),−):S−1B−Mod −→ S−1A−Mod are adjoint functors.
Proof Since Mis a (A−B)bimodule then S−1(M)is a (S−1A−S−1B)bimodule.
Then by theorem 1 the functors E XTn
S−1A(S−1(M),−):S−1A−Mod −→ S−1B−
Mod and
TorS−1B
n(S−1(M),−):S−1B−Mod −→ S−1A−Mod are adjoint functors. 
4 Isomorphisms and localization in A−Mod
Lemma 4 Let M be a left A-module and M•a projective resolution of M. Then
TorS−1A
n+1(S−1(C),S−1()) TorS−1A
n(S−1(K0),S−1())
Proof As the proof of lemma 1. 
Theorem 3 Let B be a ring, A a sub-ring of B, S a suturated multiplicative subset
of A and B verifying the left Ore conditions and M a A −B bimodule.
Let be the functors S−1(MN−):B−Mod −→ S−1A−Mod and S−1(M)NS−1() :
B−Mod −→ S−1A−Mod such as:
1. for all left B-module N we have:

Title Suppressed Due to Excessive Length 7
a. S−1(MN−) (N)=S−1(MNN)
b. S−1(M)NS−1() (N)=S−1(M)NS−1(N)
2. for all morphism f :N1−→ N2we have:
a.
S−1(MOf):S−1(MON1)−→ S−1(MON2)
m⊗n1
s7−→ m⊗f(n1)
s
b.
S−1(M)OS−1(f):S−1(M)OS−1(N1)−→ S−1(M)OS−1(N2)
m
s⊗n1
s07−→ m
s⊗f(n1)
s0
Then S−1(MN−)and S −1(M)NS−1() are isomorphic.
Proof ΦN:S−1(MNN)−→ S−1(M)NS−1(N)definied as following
ΦN:S−1(M⊗N)−→ S−1M⊗S−1N
m⊗n
s7−→ m
s⊗n
s
is an isomorphism.
Now it remind to prove for all morphism f:N1−→ N2that the following diagram
is commutative:
S−1(MNN1)
ΦN1

S−1(MNf)//S−1(MNN2)
ΦN2

S−1(M)NS−1(N1)S−1(M)NS−1(f)//S−1(M)NS−1(N2)
So let m⊗n1
s∈S−1(MN(N1)).We have on one hand:
ΦN2◦S−1(MOf)( m⊗n1
s)=ΦN2(m⊗f(n1)
s)=m
s⊗f(n1)
s
And other hand we have:
(S−1(M)OS−1(f)) ◦ΦN1(m⊗n1
s)=(S−1(M)OS−1(f)) ( m
s⊗n1
s)
=m
s⊗f(n1)
s
Theorem 4 Let B be a ring, A a sub-ring of B, S a suturated multiplicative subset of
A and B verifying the left Ore conditions and M a (A−B)bimodule. Then the fuctors

8 Bassirou Dembele, Mohamed Ben Faraj ben Maaouia and Mamadou Sanghare
S−1T or A
n(M,−):B−Mod −→ S−1A−Mod and TorS−1A
n(S−1(M),S−1()) :
B−Mod −→ S−1A−Mod are naturally isomorphic.
Proof Let us prove it by induction on n.
On one part :
MO−T or A
0(M,−)
and then
S−1(XO−)S−1T or A
0(M,−)
and on other part:
S−1(M)OS−1() T orS−1A
0(S−1(M),S−1())
By theorem 3 S−1(MN−)S−1(M)NS−1() and so
S−1T or A
0(M,−)TorS−1A
0(S−1(M),S−1()) and the result is true for k=0.
Asume that the result is true for all k<nand show that it is true for n.
By lemma 1 we have:
T or A
n(M,−)T or A
n−1(K0,−)
then
S−1T or A
n(M,−)S−1TorS−1A
n−1(K0,−)
We have also by 4
TorS−1A
n(S−1(M),S−1()) TorS−1A
n−1(S−1(K0),S−1())
By hypothesis we have:
S−1TorS−1A
n−1(K0,−)TorS−1A
n−1(S−1(K0),S−1())
Thus −1T or A
n(M,−)TorS−1A
n(S−1(M),S−1()).
Lemma 5 Let M be a left A-modules and M•be a projective resolution of M. Then
the functors E XTn+1
S−1A(S−1(C),S−1()) and EXT n
S−1A(S−1(K0),S−1()) are naturally
isomorphic.
Proof As the proof of lemma 1 
Theorem 5 Let B be a ring, A a sub-ring of B, S a suturated multiplicative subset
of A and B verifying the left Ore conditions and M a finite type (A−B)bimodule.
Let be the functors S−1Hom A(M,−):A−Mod −→ S−1B−M od and H omA(S−1(M),S−1()) :
A−Mod −→ S−1B−Mod such that:
1. for all left A-module N we have:
a. S−1H om A(M,−)(N)=S−1HomA(M,N)
b. Hom A(S−1(M),S−1())(N)=HomA(S−1(M),S−1(N))

Title Suppressed Due to Excessive Length 9
2. for all morphism f :N1−→ N2we have:
a.
S−1HomA(X,f):S−1H om A(M,N1)−→ S−1HomA(M,N2)
g
s7−→ f◦g
s
b.
HomA(S−1(M),S−1(f)) :HomA(S−1(M),S−1(N1)) −→ HomA(S−1(M),S−1(N2))
g7−→ S−1(f)◦g
Then S−1H om A(M,−)and HomA(S−1(M),S−1()) are naturally isomorphic.
Proof ΨN:S−1HomA(M,N)−→ HomA(S−1(M),S−1(N)) such as:
ΨN(g
σ)( m
s)=1
s.g(m)
σ
Now let f:N1−→ N2be a morphism, let us prove that the following diagram is
commutative:
S−1Hom(M,N1)
ΨN1

S−1H om A(M,f)//S−1H om A(M,N2)
ΨN2

HomS−1A(S−1M,S−1N1)H om S−1A(S−1M,S−1f)//HomS−1A(S−1M,S−1N2)
Then let g
σ∈S−1Hom(M,N1). At first we have
ΨN2◦S−1HomA(M,f) ( g
σ)( m
s)=ΨN2(f◦g
σ)( m
s)=1
s.f◦g
σ(m)
And secondly:
HomS−1A(S−1M,S−1f)◦ΨN1(g
σ)( m
s)=S−1(f)◦ΨN1(g
σ)( m
s)=1
s.f◦g
σ(m)
Theorem 6 Let B be a ring, A a sub-ring of B, S a suturated multiplicative subset
of A and B verifying the left Ore conditions and M a (A−B)bimodule of finite type.
Then the functors S−1EX Tn
A(M,−):A−Mod −→ S−1B−Mod and
EXT n
S−1A(S−1(M),S−1()) :A−Mod −→ S−1B−Mod are naturally isomorphic.
Proof Let us prove it by induction on n.
On one part we have:
HomA(M,−)E XT0
A(M,−)

10 Bassirou Dembele, Mohamed Ben Faraj ben Maaouia and Mamadou Sanghare
and then
S−1HomA(M,−)S−1E XT0
A(M,−)
and other part we have:
HomS−1A(S−1(M),S−1()) EXT 0
S−1A(S−1(M),S−1())
By theorem 5 S−1HomA(M,−)HomS−1A(S−1(M),S−1()) and then S−1EXT 0
A(M,−)
EXT 0
S−1A(S−1(M),S−1()). Hence the relation is true for k=0.
Assume that it is true for all k<nand prove that it is true for n.
By lemma 2 we have:
EXT n
A(M,−)EXT n−1
A(K0,−)
and then
S−1EXT n
A(M,−)S−1EXT n−1
S−1A(K0,−)
And by lemma 5 we have:
EXT n
A(S−1(M),S−1()) EXT n−1
S−1A(S−1(K0),S−1())
By the hypothesis we have:
S−1EXT n−1
A(K0,−)EXT n−1
S−1A(S−1(K0),S−1())
Thus S−1EXT n
A(M,−)EXT n
S−1A(S−1(M),S−1()).

Title Suppressed Due to Excessive Length 11
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