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Pure-injective hulls of modules over valuation rings
Francois Couchot
To cite this version:
Francois Couchot. Pure-injective hulls of modules over valuation rings. Journal of Pure and Applied
Algebra, Elsevier, 2006, 207, pp.63–76. �10.1016/j.jpaa.2005.09.002�. �hal-00003336v2�
ccsd-00003336, version 2 - 7 Sep 2006
RINGS
FRANC¸ OIS COUCHOT
Abstract. IfRbis the pure-injective hull of a valuation ringR, it is proved thatRb⊗RM is the pure-injective hull ofM, for every finitely generatedR- moduleM. Moreover Rb⊗RM ∼=⊕1≤k≤nR/Ab kR, where (Ab k)1≤k≤n is the annihilator sequence ofM. The pure-injective hulls of uniserial or polyserial modules are also investigated. Any two pure-composition series of a countably generated polyserial module are isomorphic.
The aim of this paper is to study pure-injective hulls of modules over valuation rings. IfRis a valuation domain andSa maximal immediate extension ofR, then, in [10], Warfield proved that S is a pure-injective hull of R. Moreover, for each finitely generatedR-moduleM, he showed thatS⊗RM is a pure-injective hull of M and a direct sum of genM indecomposable pure-injective modules. We extend this last result to every valuation ringRby replacingSwith the pure-injective hull Rb of R. As in the domain caseRb is a faithfully flat module. Moreover, for each x∈Rb there existr ∈R andy ∈1 +PRb such thatx=ry. This property allows us to prove most of the main results of this paper. We extend results obtained by Fuchs and Salce on pure-injective hulls of uniserial modules over valuation domains ([5, chapter XIII, section 5]). We show that the length of any pure-composition series of a polyserial moduleM is its Malcev rank MrM and its pure-injective hull Mcis a direct sum ofpindecomposable pure-injective modules, where p≤MrM. But it is possible to havep < MrM and we prove that the equality holds for all M if and only if R is maximal (Theorem 4.5). This result is a consequence of the fact thatR is maximal if and only ifR/N and RN are maximal, where N is the nilradical ofR (Theorem 4.4). IfU1, . . . , Un are the factors of a pure-composition series of a polyserial module M then the collection (Rb⊗RUk)1≤k≤n is uniquely determined by M. To prove this, we use the fact that Rb⊗RU is an unshrinkable uniserial T-module for each uniserial R-module U, where T = EndR(R). Whenb R satisfies a countable condition, the collection of uniserial factors of a polyserial moduleM is uniquely determined byM (Proposition 3.7).
In this paper all rings are associative and commutative with unity and all mo- dules are unital. As in [3] we say that anR-moduleEisdivisibleif, for everyr∈R and x∈ E,(0 : r) ⊆(0 : x) implies that x∈rE, and that E is fp-injective(or absolutely pure) if Ext1R(F, E) = 0, for every finitely presented R-module F.
A ring R is called self fp-injective if it is fp-injective as R-module. An exact sequence 0 → F → E → G → 0 is pure if it remains exact when tensoring it with any R-module. In this case we say that F is a pure submodule of E.
Recall that a moduleE is fp-injective if and only if it is a pure submodule of every overmodule. A module is said to beuniserialif its submodules are linearly ordered
1
by inclusion and a ringRis avaluation ringif it is uniserial asR-module. Recall that every finitely presented module over a valuation ring is a finite direct sum of cyclic modules [9, Theorem 1]. Consequently a moduleE over a valuation ringR is fp-injective if and only if it is divisible.
AnR-moduleF ispure-injectiveif for every pure exact sequence 0→N →M →L→0
ofR-modules, the following sequence
0→HomR(L, F)→HomR(M, F)→HomR(N, F)→0
is exact. An R-moduleB is a pure-essential extensionof a submodule A ifA is a pure submodule ofB and, if for each submodule KofB, eitherK∩A6= 0 or (A+K)/K is not a pure submodule ofB/K. We say thatB is apure-injective hull ofAifB is pure-injective and a pure-essential extension ofA. By [10] or [5, chapter XIII] eachR-moduleM has a pure-injective hull and any two pure-injective hulls ofM are isomorphic.
In the sequelRis a valuation ring,P its maximal ideal,Z its subset of zerodi- visors andMcthe pure-injective hull of M, for eachR-moduleM. As in [5, p.69], for every proper idealA, we putA♯={s∈R|(A:s)6=A}.ThenA♯/Ais the set of zerodivisors of R/Awhence A♯ is a prime ideal. In particular{0}♯=Z. When A♯ =P, we say that Ais anarchimedean ideal. Then Ais archimedean if and only if R/Ais self fp-injective.
1. Properties ofRb
The first assertion of the following proposition will play a crucial role to prove the main results of this paper.
Proposition 1.1. The following assertions hold:
(1) For eachx∈Rb there exista∈R, p∈P andy∈Rb such thatx=a+pay.
(2) For each archimedean idealAofR,R/Ab Rbis an essential extension ofR/A.
(3) R/Pb Rb∼=R/P.
Proof. The third assertion is an immediate consequence of the first.
We also deduce the second assertion from the first. SinceRis a pure submodule of R, the natural mapb R/A → R/Ab Rb is monic. Let x∈ Rb\R+AR. We haveb x=a+payfora∈R, p∈P andy∈R. Henceb pa /∈A. SinceA is archimedean, there existsr∈(A:pa)\(A:a). So rx∈R+ARb\AR.b
We proceed by steps to prove the first assertion.
Step 1. Suppose that R is self fp-injective. In this case, Rb ∼= ER(R) by [5, Lemma XIII.2.7]. We may assume that x /∈R. Then there existsd∈R such that dx∈R anddx6= 0. SinceR is a pure submodule of Rb we havedx=dbfor some b∈ R. By [1, Lemma 2] (0 : x) = (0 :b), whencex=bz for some z∈Rb since Rb is divisible. In the same way, there existsc, u∈Rsuch that cz=cu6= 0. We get that (0 :u) = (0 :z) =b(0 :b) = 0. So uis a unit of R. Since z−u /∈R, there exists s, q∈R and y ∈Rb such that 06=sq =s(z−u)∈R and z−u=qy. We havec∈(0 :z−u) = (0 :q). Soq∈P. Now we puta=buandp=qu−1 and we getx=a+pay.
Step 2. Now we prove that R/rb Rb ∼= ER/rR(R/rR) for each 0 6= r ∈ P. If
∩a6=0aR = 0 then it is an immediate consequence of [3, Theorem 5.6]. Else P is
not faithful, R is self fp-injective andRb∼=ER(R). By Step 1 and the implication 1 ⇒ 2 the second assertion holds. So it remains to show that R/rb Rb is injective overR/rR. LetJ be an ideal ofR such thatRr ⊂J and g:J/Rr→R/rb Rb be a nonzero homomorphism. For eachx∈Rb we denote by ¯xthe image ofxinR/rb R.b Leta∈J \Rr such that ¯y =g(¯a)6= 0. Then (Rr:a)⊆(rRb :y). Let t∈Rsuch thatr=at. Thusty=rz for somez∈R. It follows thatb t(y−az) = 0. So, since at=r6= 0, we have (0 :a)⊂Rt⊆(0 :y−az). The injectivity ofRbimplies that there existsx∈Rb such thaty =a(x+z). We put xa =x+z. Ifb∈J\Ra then a(xa −xb) ∈ rR. Henceb xb ∈ xa+ (rRb :Rb a). Since Rb is pure-injective, by [10, Theorem 4] there existsx∈ ∩a∈Jxa+ (rRb:Rba). It follows thatg(¯a) =a¯xfor each a∈J.
Step 3. Now we prove the first assertion in the general case. If ∩r6=0rR 6= 0, then R is self fp-injective. So the result holds by Step 1. If ∩r6=0rR = 0, we put F =∩r6=0rR. We will show thatb F = 0. Letx∈F∩R. Thenx∈R∩rRb=rRfor eachr∈R, r6= 0. Thereforex= 0 andF∩R= 0. Letx∈R, r, ab ∈R andz∈F such thatrx=a+z. There existsy∈Rbsuch thatz=ry. Sor(x−y) =a, whence there existsb∈Rsuch thatrb=a. It follows thatRis a pure submodule ofR/Fb . Since Rb is a pure-essential extension of R we deduce thatF = 0. Letx∈R. Web may assume thatx /∈R. There exists 06=r∈R such thatx /∈rR. Ifb x∈R+rRb thenx=a+ry, witha∈R andy ∈R. We haveb a /∈rR else x∈rR. Sob r=pa for somep∈P. Ifx /∈R+rRb then, sinceR/Rr is self fp-injective, from Steps 1 and 2 we deduce that x−a−paz ∈rRb for some a∈R, p∈P and z ∈R. It isb
obvious thata /∈rR. Now it is easy to conclude.
As in the domain case we have:
Proposition 1.2. Rb is a faithfully flatR-module.
Proof. Letx∈Rb and r∈R such that rx= 0. By Proposition 1.1 there exist a∈R, p∈P andy∈Rb such thatx=a+pay. Sorpay∈R. It follows that there existsb∈Rsuch thatra(1+pb) = 0. Hencera= 0 andr⊗x=ra⊗(1+py) = 0.
2. Pure-injective hulls of uniserial modules
The following lemma and Proposition 2.2 will be useful to prove the pure- injectivity of some modules in the sequel.
Lemma 2.1. Let U be a module and F a flat module. Then, for each r, s ∈ R, F⊗R(sU :U r)∼= (F⊗RsU :F⊗RU r).
Proof. We putE=F⊗RU. Let φbe the composition of the multiplication by rin U with the natural map U →U/sU. Then (sU :U r) = ker(φ). It follows that F⊗R(sU :U r) is isomorphic to ker(1F ⊗φ) since F is flat. We easily check that 1F ⊗φ is the composition of the multiplication by r in E with the natural map E→E/sE. It follows thatF⊗R(sU :U r)∼= (sE:Er).
Proposition 2.2. Every pure-injectiveR-moduleF satisfies the following property:
if(xi)i∈I is a family of elements ofF and(Ai)i∈I a family of ideals ofRsuch that the family F = (xi+AiF)i∈I has the finite intersection property, then F has a non-empty intersection. The converse holds ifF is flat.
Proof. Leti∈Isuch thatAiis not finitely generated. By [1, Lemma 29] either Ai =P ri or Ai =∩c∈R\AicR. If,∀i∈I such that Ai is not finitely generated, we replacexi+AiF byxi+riF in the first case, and by the family (xi+cF)c∈R\Ai
in the second case, we deduce from F a family G which has the finite intersection property. SinceF is pure-injective, it follows that there existsx∈F which belongs to each element of the family G by [10, Theorem 4]. We may assume that the family (Ai)i∈I has no smallest element. So, if Ai is not finitely generated, there exists j ∈ I such that Aj ⊂Ai. Letc ∈ Ai\P Aj such thatxj+cF ∈ G. Then x−xj ∈cF ⊆AiF andxj−xi ∈AiF. Hencex−xi∈AiF for eachi∈I.
Conversely, if F is flat then by Lemma 2.1 we have (sF :F r) = (sR : r)F for eachs, r∈R. We use [10, Theorem 4] to conclude.
Proposition 2.3. Let U be a uniserial module andF a flat pure-injective module.
Then F⊗RU is pure-injective.
Proof. Let E = F⊗RU. We use [10, Theorem 4] to prove that E is pure- injective. Let (xi)i∈I be a family of elements of F such that the family F = (xi +Ni)i∈I has the finite intersection property, where Ni = (siE :E ri) and ri, si∈R,∀i∈I.
First we assume thatU =R/AwhereAis a proper ideal ofR. SoE∼=F/AF. If si∈/A thenNi= (siF :F ri)/AF = (Rsi:ri)F/AF. We setAi= (Rsi :ri) in this case. Ifsi∈AthenNi= (AF :F ri)/AF = (A:ri)F/AF. We putAi= (A:ri) in this case. For eachi∈I, letyi∈F such thatxi=yi+AF. It is obvious that the family (yi+AiF)i∈I has the finite intersection property. By Proposition 2.2 this family has a non-empty intersection. ThenF has a non-empty intersection too.
Now we assume thatU is not finitely generated. It is obvious thatF has a non- empty intersection ifxi+Ni=E, ∀i∈I. Now assume there existsi0∈Isuch that xi0+Ni06=E. LetI′={i∈I|Ni⊆Ni0}andF′= (xi+Ni)i∈I′. ThenFandF′ have the same intersection. By Lemma 2.1Ni0 =F⊗R(si0U:U ri0). It follows that (si0U :U ri0)⊂U becauseNi06=E. Hence∃u∈U such thatxi0+Ni0⊆F⊗RRu.
Then, ∀i ∈ I′, xi +Ni ⊆ F ⊗RRu. We have F ⊗RRu ∼= F/(0 : u)F. From the first part of the proofF/(0 :u)F is pure-injective. So we may replaceR with R/(0 : u) and assume that (0 :u) = 0. LetAi = ((siU :U ri) : u), ∀i ∈I′. Thus Ni=AiF, ∀i∈I′. By Proposition 2.2F′ has a non-empty intersection. SoF has
a non-empty intersection too.
LetU be an R-module. As in [5, p.338] we set
U♯={s∈R| ∃u∈U, u6= 0 andsu= 0} andU♯={s∈R|sU ⊂U}.
ThenU♯ andU♯are prime ideals.
Now it is possible to determine the pure-injective hull of each uniserial module.
We get a generalization of [5, Corollary XIII.5.5]
Theorem 2.4. The following assertions hold:
(1) LetU be a uniserialR-module andJ =U♯∪U♯. ThenRcJ⊗RU is the pure- injective hull ofU. MoreoverUb is an essential extension of U if J =U♯. (2) For each proper ideal A of R, R/Ab Rb is the pure-injective hull of R/A.
Moreover R/Ab Rb∼=ER/A(R/A)if Ais archimedean.
Proof. (1) If s ∈ R\J then multiplication by s in U is bijective. So U is anRJ-module. After replacing R with RJ, we may assume that J =P. We put Ue =RcJ⊗RU.
Suppose thatP =U♯. By [10, Proposition 6]Ue =Ub⊕V whereV is a submodule of Ue. Let v ∈V. Then v =x⊗u where u∈U and x∈ R. By Proposition 1.1b x=a+pay, wherea∈R, p∈P and y∈R. Sinceb pU ⊂U,∃u′∈U\(P u∪pU).
Then u=cu′ for some c∈R and v=cau′+pcay⊗u′. We havey⊗u′ =z+w where w∈ V and z ∈Ub. Socau′+pcaz = 0. Since U is pure inUb, there exists z′∈U such thatcau′+pcaz′= 0. Ifv6= 0 the equalityv= (1 +py(⊗cau′ implies cau′ 6= 0. By [1, Lemma 5] we get that u′ ∈ pU, whence a contradiction. Hence V = 0.
Now suppose that P = U♯. If 0 6= z ∈ Ue then z = x⊗u where u ∈ U and x ∈ R. By Proposition 1.1 there existb a ∈ R, p ∈ P and y ∈ Rb such that x=a+pay. So z=au+y⊗pau. Let A= (0 :au). By [1, Lemma 26],A♯=P. So (0 :pau) = (A:p)6=A. Letr∈(A:p)\A. Then 06=rz∈U.
(2) We apply the first assertion by takingU =R/A. In this case,U♯=P. The pure-injective hull of R/A is the same over R and over R/A. Since R/A is self fp-injective whenA is archimedean then we use [5, Lemma XIII.2.7] to prove the
last assertion.
In the previous theorem, ifU is not cyclic and ifU♯⊆U♯thenUb is not necessarily isomorphic toER/(0:U)(U). For instance:
Example 2.5. Assume that P =Z and P is faithful. We choose U =P. Then U♯=U♯=P,Ub =PRb andER(U) =R.b
If U is a non-standard uniserial module over a valuation domain R then Ub is indecomposable by [3, Proposition 5.1] and there exists a standard uniserial module V such thatUb ∼=Vb by [5, Theorem XIII.5.9]. So,Rb⊗RU ∼=Rb⊗RV doesn’t imply U ∼=V. However, it is possible to get the following proposition:
Proposition 2.6. Let U and V be uniserial modules and J =U♯∪U♯. Assume that Rb⊗RU ∼= Rb⊗RV. Then U and V are isomorphic if one of the following conditions is satisfied:
(1) U♯=J andJ 6=J2, (2) U is countably generated.
Proof. Letφ:Rb⊗RU →Rb⊗RV be the isomorphism. Let 06=u∈U. Then φ(u) =x⊗v for somex∈Rb andv ∈V. By proposition 1.1 we may assume that x= 1 +py for somep∈P and y∈R. First we shall prove that (0 :b u) = (0 :v).
It is obvious that (0 : v)⊆(0 :u). Letr ∈(0 :u). Then x⊗rv = 0. From the flatness of Rb we deduce that there exists ∈R and z ∈ Rb such that x=sz and srv = 0. If s∈P then we get that 1 = qefor someq ∈P ande∈R. Sinceb R is pure in R, it follows that 1b ∈P. This is absurb. Hencesis a unit andr∈(0 :v).
Letv, v′ be nonzero elements ofV andx, x′∈1 +PRbsuch thatx⊗v=x′⊗v′. There existst∈Rsuch thatv=tv′. Now we shall prove thattis a unit ofR. We get that (x′−tx)⊗v′ = 0. Ift ∈ P, as above we deduce thatv′ = 0, whence a contradiction.
Let u∈ U and v ∈ V as in the first part of the proof. By [1, Lemma 26] we have U♯ = (0 : u)♯ = (0 : v)♯ = V♯. Let p∈ P. We shall prove that u ∈ pU if
and only if v ∈ pV. If v =pw for some w∈ V then φ(u) =px⊗w= pφ(z) for somez ∈ Rb⊗RU. Since U is a pure submodule, then u=pu′ for some u′ ∈U. Conversely, if u = pu′ for some u′ ∈ U and φ(u′) = x′ ⊗v′ where v′ ∈ V and x′ ∈1 +PR, we get thatb x′⊗pv′ =x⊗v. From above, we deduce thatv ∈pV. So,U♯=V♯.
Now we can prove that U and V are isomorphic when the first condition is satisfied. In this case U and V are modules over RJ. Since J 6= J2 , JRJ is a principal ideal ofRJ. Since JU ⊂U and JV ⊂V, U and V are cyclic overRJ. Letu∈U andv∈V as in the first part of the proof, and suppose thatU =RJu.
Ifv=rwfor somer∈RJ andw∈V then we get, as above, thatu=ru′ for some u′∈U. Soris a unit and U and V are isomorphic.
Let{ui}i∈I be a spanning set ofU. For eachi∈I, letvi∈V andxi ∈1 +PRb such that φ(ui) = xi ⊗vi. Suppose that (0 : U) ⊂ (0 : u), ∀u ∈ U. From the first part of proof we deduce that (0 : V) ⊂ (0 : v), ∀v ∈ V. We have
∩i∈I(0 :ui) = (0 :U). Thus∩i∈I(0 :vi) = (0 :V). So, for eachv ∈V there exists i∈I such that (0 :vi)⊂(0 :v). Hencev ∈Rvi. Now, suppose∃u∈U such that (0 : u) = (0 : U). By [5, Lemma X.1.4] J =U♯. We may assume that J = J2 and I is infinite. Then JU =U and JV = V. Let v ∈ V. There exists p ∈ J such thatv ∈pV. But there existsi∈I such thatui ∈/ pU. So, vi ∈/ pV. Hence v ∈RJvi. Now suppose thatI =N. Let (an)n∈N be a sequence of elements ofP such that un =anun+1, ∀n∈N. We put ϕ(u0) =v0. Suppose that ϕ(un) =snvn
wheresn is a unit. By the second part of the proof there exists a unittn such that anvn+1 =tnϕ(un). Hence we setϕ(un+1) =t−1n vn+1. So, by induction on n, we
get an isomorphismϕ:U →V.
LetT = EndR(R). Thenb T is a local ring by [3, Proposition 5.1] and [5, Theorem XIII.3.10]. For each R-module M, Rb⊗RM is a left T-module. As in [4] we say that a left uniserialT-moduleF isshrinkableif there exists twoT-submodulesG andH of F such that 0⊂H ⊂G⊂F andF ∼=G/H. OtherwiseF is said to be unshrinkable.
Proposition 2.7. Let U be a uniserialR-module. Then:
(1) Rb⊗RU is a left unshrinkable uniserialT-module.
(2) EndT(Rb⊗RU) is a local ring.
Proof. (1) Letx∈1 +PR. First we prove thatb Rx is a pure submodule ofR.b Leta, b∈Randy∈Rb such thatby=ax. By Proposition 1.1y=c+pczfor some c∈R, p∈P and z∈R. Suppose thatb a /∈Rbc. Thenbc=ra for somer∈P. If x= 1 +qx′ for someq∈P andx′∈R, we get thatb a(1−r) =a(rpz−qx′) =aty′ for somet ∈P andy′ ∈R. Sinceb R is a pure submodule ofRb there exists s∈R such that a(1−r−ts) = 0. We deduce that a= 0, whence a contradiction. So a∈Rbc. By using similar arguments we easily show thatRxis faithful.
Letz, z′∈Rb⊗RU. We havez=x⊗uandz′=x′⊗u′wherex, x′ ∈1 +PRband u, u′∈U. Assume thatu′=rufor somer∈R. The homomorphismφ:Rx→Rrx′ such thatφ(x) =rx′ is well defined and can be extended toR. We get thatb φz=z′. HenceRb⊗RU is uniserial overT.
Suppose that Rb⊗RU is shrinkable over T. By [4, Lemma 1.17] there exists z∈Rb⊗RU such thatT zis shrinkable. We havez=x⊗uwherex∈1 +PRb and
u∈U. SoT z=Rb⊗RRu. There existz′ ∈T zand a non-injectiveT-epimorphism α : T z′ → T z. Let K = Ker α. We may assume that α(z′) = z. We have z′=x′⊗ruwherex′∈1 +PRbandr∈R. Letybe a nonzero element ofK. Thus y=tz′=ay′⊗rufor somet∈T,y′∈1 +PRb anda∈R. But there exists, s′∈T such that x′ = sy′ and y′ = s′x′. So 0 6= ax′ ⊗ru ∈ K. Since y 6= 0 we have aru 6= 0. On the other hand x⊗aru =α(ax′⊗ru) = 0. It follows that aru = 0 whence a contradiction. SoRb⊗RU is unshrinkable.
(2) is an immediate of (1) and [4, Proposition 9.24].
Proposition 2.8. Let cbe a cardinal. Consider ac-generatedR-moduleM andU a pure uniserial R-submodule ofM. Then U isc-generated.
Proof. We easily check that Rb⊗R U is a pure submodule of Rb ⊗RM. By Proposition 2.3Rb⊗RU is pure-injective. HenceRb⊗RU is a summand ofRb⊗RM. On the other hand Rb⊗RM is a c-generated T-module. Then Rb⊗RU is also c- generated overT. We may assume thatRb⊗RU is generated by (1⊗ui)i∈I, where I is a set whose cardinal is c and ui ∈ U, ∀i ∈ I. Let V be the submodule of U generated by (ui)i∈I. Then the inclusion map V →U induces an isomorphism Rb⊗RV →Rb⊗RU. SinceRb is faithfully flat it follows thatV =U. From Theorem 2.4 we deduce the following corollary on the structure of inde- composable injective modules.
Corollary 2.9. LetE be an indecomposable injective module,J =E♯andA(E) = {(0 :RJ x)|06=x∈E}. Then:
(1) ∀A, B∈ A(E), A⊆B there exists a monomorphism ϕA,B:RcJ/BRcJ →RcJ/ARcJ
such that ϕA,C =ϕA,B◦ϕB,C,∀A, B, C∈ A(E), A⊆B⊆C.
(2) E∼= lim−→{(RcJ/ARcJ, ϕA,B)|A, B∈ A(E), A⊆B}.
(3) E∼=RcJ/(0 :RJ e)RcJ if (0 :RJ e) = (0 :RJ E)for somee∈E.
(4) Suppose thatEcontains a uniserialRJ-moduleU such thatA(E) =A(U)1. Then E∼=RcJ⊗RU. Moreover, ∀A, B∈ A(E), A⊆B, there exists r∈R such that one can choose ϕA,B = 1Rc
J ⊗¯r where r¯ : RJ/B → RJ/A is defined by ¯r(a+B) =ar+A, ∀a∈R.
Proof. (1) IfA∈ A(E) thenA♯=J by [1, Lemma 26]. SoAis an archimedean ideal ofRJ. By Theorem 2.4 there exists an isomorphism
φA:RcJ/ARcJ→(0 :EA).
LetuA,B: (0 :EB)→(0 :EA) be the inclusion map ,∀A, B∈ A(E), A⊆B. We setϕA,B=φ−1A ◦uA,B◦φB. It is easy to check the first assertion.
(2) and (3) These assertions are now obvious.
(4) First we prove that U is fp-injective. Let x ∈ E and s ∈ R such that 0 6= sx∈ U. We put u=sx. From A(E) = A(U), it follows that ∃v ∈U such that (0 :RJ v) = (0 :RJ x) and consequently u = tv for some t ∈ R. We set A= (0 :RJ x). We get that (0 :RJ u) = (A:RJ t) = (A:RJ s). By [1, Lemma 26]
1We know that this condition holds ifRsatisfies an additional hypothesis: see [1, Corollary 22], [8, Theorem 5.5] or Remark 3.6
A♯ =E♯ =J. It follows that RJs =RJt. So U is a pure submodule ofE. We conclude by Theorem 2.4 and [5, Lemma XIII.2.7] thatE ∼=RcJ⊗RU.
Let u, v∈ U such that (0 :RJ u) = A and (0 :RJ v) = B. There existsr ∈ R such that v =ruand B = (A:r) (ifA =B we takev=uandr= 1). So ¯r is a
monomorphism.
3. Pure-injective hulls of polyserial modules
We say that a moduleM ispolyserialif it has a pure-composition series 0 =M0⊂M1⊂ · · · ⊂Mn=M,
(i.e. Mk is a pure submodule of M, for each k, 0 ≤ k ≤ n) where Mk/Mk−1 is uniserial for eachk, 1≤k≤n. By [5, Lemma I.7.8], ifM is finitely generated,M has a pure-composition series, where Mk/Mk−1 ∼=R/Ak andAk is a proper ideal, for eachk, 1 ≤ k ≤n. We denote by gen M the minimal number of generators of M. By [5, Lemma V.5.3]n= gen M. The following sequence (A1,· · ·, An) is called theannihilator sequence of M and is uniquely determined by M, up to the order (see [5, Theorem V.5.5]).
Now we can extend the result obtained by Warfield[10] in the domain case for finitely generated modules.
Theorem 3.1. LetM be a finitely generatedR-module. ThenR⊗b RM ∼=Mc. More- over, Mc∼=R/Ab 1Rb⊕ · · · ⊕R/Ab nRb where(A1,· · · , An)is the annihilator sequence of M.
Proof. It is easy to verify that M is a pure submodule of Rb⊗RM. We have that Rb⊗RM1 is a pure submodule ofRb⊗RM too. By Proposition 2.3 Rb⊗RM1
is pure-injective. It follows that Rb ⊗R M ∼= (Rb⊗RM1)⊕(Rb⊗RM/M1). By induction on n we get that Rb⊗RM ∼= R/Ab 1Rb⊕ · · · ⊕R/Ab nR. Sob Rb⊗RM is pure-injective. By [10, Proposition 6] Mc is a direct summand of Rb⊗RM. So Rb⊗RM ∼=Mc⊕V, whereV is a submodule ofRb⊗RM. From Proposition 1.1 we deduce that, for each x∈Rb⊗RM, there exist m∈M, p ∈P and y ∈Rb⊗RM such thatx=m+py. Assume thatx∈V. There existsz ∈Mcand v ∈V such thatx=m+pz+pv. It follows thatx=pv, whenceV =P V. On the other hand, R/Ab Rb is indecomposable by [3, Proposition 5.1] and EndR(R/Ab R) is local by [11,b Theorem 9] or [5, Theorem XIII.3.10], for every proper idealA. By Krull-Schmidt TheoremV ∼=R/Ab k1Rb⊕ · · ·⊕R/Ab kpRbwhere{k1,· · · , kp}is a subset of{1,· · ·, n}.
If V 6= 0, by Proposition 1.1 we get V 6=P V. This contradiction completes the
proof.
TheMalcev rankof a moduleN is defined as the cardinal number MrN = sup{genM |M ⊆N, genM <∞}.
The following proposition is identical to the first part of [5, Proposition XII.1.6].
Here we give a different proof.
Proposition 3.2. The length of any pure-composition series of a polyserial module M equalsMrM.
Proof. Let 0 = M0 ⊂ M1 ⊂ · · · ⊂ Mn =M be a pure-composition series of M with uniserial factors. As in [5, Corollary XII.1.5] we prove that Mr M ≤ n.
Equality holds for n = 1. From the pure-composition series of M, we deduce a pure-composition series ofM/M1 of lengthn−1. By induction hypothesisM/M1
contains a finitely generated submoduleY with genY =n−1.
Assume thatY is generated by{y2, . . . , yn}. Letx2, . . . , xn ∈M such thatyk= xk+M1 andF be the submodule ofM generated by x2, . . . , xn. IfF∩M1=M1
thenM1⊆FandM1is a pure submodule ofF. In this caseM1is finitely generated by Proposition 2.8. It follows that the following sequence is exact:
0→ M1
P M1
→ F P F → Y
P Y →0.
So we have gen Y = gen F −gen M1 ≤ n−2. We get a contradiction since gen Y = n−1. Hence F ∩M1 6= M1. Let x1 ∈ M1\F ∩M1. Let X be the submodule ofM generated byx1, . . . , xn. ClearlyRx1=M1∩X. We will show that P x1=Rx1∩P X. Letx∈Rx1∩P X. Thenx=pPk=n
k=1akxk=rx1 wherep∈P andr, a1, . . . , anare elements ofR. It follows thatpPk=n
k=2akxk= (r−pa1)x1. So (r−pa1)x1∈M1∩F ⊂Rx1. We deduce thatr−pa1 ∈P whence r∈P. Hence x∈P x1. Consequently the following sequence is exact:
0→ Rx1
P x1
→ X P X → Y
P Y →0.
Then genX =n.
Now we study the pure-injective hulls of polyserial modules.
Theorem 3.3. Let M be a polyserial module with the following pure-composition series:
0 =M0⊂M1⊂ · · · ⊂Mn=M For each integer k,1≤k≤nwe putUk =Mk/Mk−1. Then:
(1) There exists a subsetI of {k∈N|1≤k≤n}such that Mc∼=⊕k∈IUck. (2) Rb⊗RM is pure-injective and isomorphic to⊕k=nk=1Rb⊗RUk.
(3) The collection (Rb⊗RUk)1≤k≤n is uniquely determined by M.
Proof. (1) LetNbe a pure submodule ofM. The inclusion mapN→Nb can be extended tow:M →N .b Letf :M →Nb⊕M/N\defined byf(x) = (w(x), x+N), for each x ∈ M. It is easy to verify that f is a pure monomorphism. It follows that Mcis a summand of Nb ⊕\M/N. So, by induction on n, we easily get that Mcis a summand of ⊕k=nk=1Uck. Since, ∀k ∈ N, 1 ≤ k ≤ n, Uck is indecomposable by [3, Proposition 5.1] and EndR(Uck) is local by [11, Theorem 9] or [5, Theorem XIII.3.10], we apply Krull-Schmidt Theorem to conclude.
(2) We do as in the proof of Theorem 3.1.
(3) SinceRb⊗RM andRb⊗RUk areT-modules, we conclude by Proposition 2.7
and Krull-Schmidt theorem.
Corollary 3.4. If M is polyserial and countably generated then any two pure- composition series of M are isomorphic.
Proof. By Theorem 3.3 the collection (R⊗b RUk)1≤k≤nis uniquely determined by M. It remains to show that, ifU andV are uniserial modules such thatRb⊗RU ∼=
Rb⊗RV then U ∼= V. It is an immediate consequence of Proposition 2.8 and
Proposition 2.6.
Recall that anR-moduleM isfinitely(respectivelycountably cogenerated) ifM is a submodule of a product of finitely (respectively countably) many injective hulls of simple modules.
The following proposition completes [1, Corollary 35].
Proposition 3.5. The following conditions are equivalent:
(1) Every finitely generated R-module is countably cogenerated and every ideal of Ris countably generated.
(2) For each prime idealJ which is the union of the set of primes properly con- tained inJ there is a countable subset whose union isJ,and for each prime idealJ which is the intersection of the set of primes containing properlyJ there is a countable subset whose intersection isJ.
(3) Each uniserial module is countably generated.
(1)⇔(2) holds by [1, Corollary 35]
(3)⇒(2) Let J be a prime ideal. ThenJ andRJ are uniserialR-modules. So they are countably generated. If RJ is generated by {t−1n |n∈N}, where tn ∈/ J
∀n∈N, then J=∩n∈NRtn. Now it is easy to get the second condition.
(1) ⇒ (3) Let U be a uniserial module and J = U♯∪U♯. Then U is an RJ- module. ButR/J countably cogenerated is equivalent toRJ countably generated.
HenceU is countably generated overRif and only ifU is countably generated over RJ. So we may assume thatJ=P.
First assume thatU♯=P. IfP U ⊂U thenU =Ruwhere u∈U \P U. Now suppose that P U = U. Let r, s ∈ P such that rU 6= 0. If rU = rsU then by [1, Lemma 5] we have U = sU, hence a contradiction. Let {pn | n ∈ N} be a spanning set of P such that pn+1 ∈/ Rpn. Then U =∪n∈NpnU. We may assume that pnU 6= 0, ∀n∈N. So pnU ⊂pn+1U for eachn∈N. Let un ∈pn+1U\pnU for eachn∈N. ThenU is generated by{un|n∈N}.
Now suppose that U♯=P. Assume that (0 :u) = (0 :U) for someu∈U. Let v ∈U such that u=av for some a∈R. By [1, Lemma 2] (0 :u) = ((0 :v) :a).
We get that (0 :v) = ((0 :v) :a) = (0 :U). Since (0 :v)♯=P by [1, Lemma 26]
a is a unit, and consequentlyU is cyclic. Now we assume that (0 :U) ⊂(0 : u) for each u∈U. We have (0 :U) = ∩u∈U(0 :u). By [1, Lemma 30] there exists a countable family (un)n∈N of elements ofU such that (0 :U) = ∩n∈N(0 : un) and un+1∈/Run,∀n∈N. Ifu∈U, since (0 :u)6= (0 :U), then there existsn∈Nsuch that (0 :un)⊂(0 :u). Henceu∈Run andU is generated by{un|n∈N}.
Remark 3.6. In the same way, one can prove that the two first conditions of [1, Proposition 32] (respectively [1, Corollary 34]) are equivalent to the following:
each indecomposable injective module E such that E♯ = P contains a uniserial pure submodule which is countably generated (respectively each indecomposable injective module contains a uniserial pure submodule which is countably generated).
Proposition 3.7. Suppose that R satisfies the equivalent conditions of Proposi- tion 3.5. Then any two pure-composition series of a polyserial R-module are iso- morphic.
Proof. It is an immediate consequence of Proposition 3.5 and Corollary 3.4.
4. Two criteria for maximality of R
By Theorem 3.1, ifM is finitely generated, thenMcis a direct sum of genM in- decomposable pure-injective modules and genM = MrM by [5, Corollary XII.1.7].
But Theorem 4.5 proves that, ifM is polyserial, thenMcis not necessarily a direct sum of MrM indecomposable pure-injective modules.
As in [7], ifx∈Rb\R, we say that B(x) ={r∈R|x /∈R+rR}b is thebreath ideal of x. Then Proposition 4.2 is a generalization of [7, Proposition 1.4]. The following lemma is useful to prove this proposition.
Lemma 4.1. Let J be a proper ideal such thatJ =∩c /∈JcR. Then JRb=∩c /∈JcR.b Proof. By Theorem 2.4R/Jb Rbis the pure-injective hull ofR/J. In the proof of Step 3 of Proposition 1.1 it is already shown that∩a6=0aRb = 0 if∩a6=0aR= 0. So
we apply this result toR/J to get the lemma.
Recall that theideal topologyofRis the linear topology which has as a basis of neighborhoods of 0 the nonzero principal ideals.
Proposition 4.2. Let A be a proper ideal. Then R/A is Hausdorff and non- complete in its ideal topology if and only if A= B(x)for somexinRb\R.
Proof. To show thatR/B(x) is Hausdorff, we do as in [7, Proposition 1.4], we prove that a /∈B(x) implies thatpa /∈B(x) for some p∈P. We havex=r+ay where r ∈ R and y ∈ R. By Proposition 1.1,b Rb =R+PR. Sob y = s+pz, for somes∈R,p∈P andz∈R. Therefore we getb x=r+as+paz∈R+paR. Forb each a /∈ B(x), x∈ ra+aRb for some ra ∈ R. If the family (ra+aR)a /∈B(x) has a non-empty intersection then, by using Lemma 4.1, we get that x∈R+ B(x)R,b whence a contradiction. SoR/B(x) is non-complete.
Conversely, assume thatR/Ais Hausdorff and non-complete. Then there exists a family (ra+aR)a /∈Awhich has the finite intersection and an empty total intersection.
SinceRb is pure-injective, the total intersection of the family (ra+aR)b a /∈Acontains an elementxwhich doesn’t belong toR. Clearly B(x)⊆A. Ifx=r+bRbfor some r∈R andb∈Athen r∈ra+aR for eacha /∈A, sinceR is a pure submodule of R. We get a contradiction. Sob A= B(x).
The following lemma is a generalization of [7, Lemma 1.3]. It will be useful to prove Theorem 4.4.
Lemma 4.3. Let x∈Rb such thatx=r+ay for some r, a∈R andy ∈R. Thenb B(y) = (B(x) :a).
Proof. Lett /∈B(y). Theny=s+tz for somes∈Randz∈R. It follows thatb x=r+as+aty. Sot /∈(B(x) :a).
Conversely, if t /∈ (B(x) : a) then we get the following equalitiesx=r+ay = s+taz for some s ∈ R and z ∈ R. Sinceb R is a pure submodule of Rb it follows that a(y−tz −b) = 0 for some b ∈ R. From the flatness of Rb we deduce that (y−tz−b)∈(0 :a)R. Butb ta /∈B(x) implies that ta6= 0, whence (0 :a)⊂Rt.
Hencet /∈B(y).
Theorem 4.4. Let N be the nilradical of R. Then R is maximal if and only if R/N andRN are maximal.
Proof. Suppose that Ris maximal. It is obvious that R/N is maximal. By [6, Lemma 2]RN is maximal too.
Conversely assume thatR/N andRN are maximal. LetK be the kernel of the natural mapR→RN. Letr∈K. Thus there existss∈R\N such thatsr= 0.
It follows that K ⊆N ⊂(0 :r). Then K2 = 0. So K is a uniserialR/K-module which is linearly compact ifR/K is maximal. Consequently R is maximal if and only ifR/K is maximal. In the sequel we may assume thatK= 0. SoN =Z and it is anRN-module. It is enough to show thatN is a linearly compact module. Let (Ai)i∈I be a family of ideals contained inN and (xi)i∈I a family of elements ofN such that the familyF = (xi+Ai)i∈I has the finite intersection property. We put A=∩i∈IAi. We may assume thatA⊂Ai,∀i∈I.
First suppose that N ⊂A♯. Assume that the total intersection ofF is empty.
Then R/A is non-complete in its ideal topology. By Proposition 4.2 there exists x ∈ Rb\R such that B(x) = A. Let b ∈ A♯\N. There exists a ∈ (A : b)\A.
Since B(x) =A we havex=r+ay for some r∈ R and y ∈R. By Lemma 4.3b B(y) = (A :a). Since b∈B(y) we have N ⊂B(y). By Proposition 4.2R/B(y) is non-complete in its ideal topology. This contradicts thatR/N is maximal. So the total intersection ofF is non-empty in this case.
Now we assume that N = A♯. Then A is an ideal of RN. By [1, Lemma 29]
eitherA=N afor somea∈N orA=∩a /∈AaRN.
First we assume thatA=N a. We may suppose thatAi⊆aRN,∀i∈I. SinceF has the finite intersection property,xi+aRN =xj+aRN,∀i, j∈I. Lety∈xi+aRN
for each i∈I. Then (xi−y+Ai)i∈I is a family of cosets ofaRN which has the finite intersection property. But aRN/aN is a uniserial module over R/N. Then aRN/aN is linearly compact sinceR/N is maximal. Thus ∩i∈I(xi−y+Ai)6=∅.
Hence the total intersection ofF is non-empty.
Now suppose thatA=∩a /∈AaRN. By Proposition 3.5 and [1, Lemma 30] there exists a countable family (an)n∈Nof elements of N\Asuch that A=∩n∈NanRN
andan∈/an+1RN, ∀n∈N. By induction onnwe get a subfamily (Ain)n∈Nof the family (Ai)i∈I such thatAin⊂anRN in the following way: we choose i0∈I such that Ai0 ⊂ a0RN and, ∀n ∈ N, we pick in+1 such that Ain+1 ⊂ Ain ∩an+1RN. Then the family (xin+anRN)n∈N has the finite intersection property. SinceRN is maximal there exists x∈ xin+anRN, ∀n ∈N. But the equality A =∩a /∈AaRN
implies that, ∀n∈N, there exists an integerm > n such thatamRN ⊆Ain. Since x−xim ∈amRN and xim −xin ∈Ain we get thatx∈xin+Ain,∀n∈N. Hence F has a non-empty total intersection. The proof is now complete.
Theorem 4.5. ThenRis maximal if and only if, for each polyserialR-moduleM, Mcis direct sum of MrM indecomposable pure-injective modules.
Proof. If R is maximal, then each polyserial module M is a direct sum of MrM pure-injective uniserial modules by [5, Proposition XII.2.4] (even ifR is not a domain, this proposition holds, with the same proof).
IfRis not maximal thenR/N orRN is not maximal by Theorem 4.4.
Assume thatR′=R/N is not maximal. Then E=Rc′/R′ is a nonzero torsion- freeR′-module. Let x∈cR′\R′, ¯xbe its image in E and U the submodule of E such thatU/R′x¯ is the torsion submodule ofE/R′x. Then¯ U is a pure submodule
ofE, a rank one torsion-free module and a uniserial module. LetM be the inverse image ofU by the natural mapRc′→E. ThenM is a pure submodule ofRc′ and a non-uniserial polyserial module with the two following (standard) uniserial factors:
R′andU. We have MrM = 2. LetW be a submodule ofRc′ such thatM∩W = 0 and M →Rc′/W is a pure monomorphism. ThusR′∩W = 0 andR′ →Rc′/W is a pure monomorphism too. SinceR′ is pure-essential incR′ it follows thatW = 0.
We conclude that M is pure-essential in Rc′, so that Mc=cR′ ⊂ cR′⊕Ub. (Let us observe thatM andU are not finitely generated by Theorem 3.1.)
Suppose that R′ =RN is not maximal. After replacingR′ withR′/rR′, where r is a non-unit ofR′, we may assume that R′ is coherent and self fp-injective by [1, Theorem 11]. Then E = cR′/R′ is a nonzero fp-injective R′-module. By [2, Lemma 6]Econtains a pure uniserial submoduleU. We defineM as above. Then MrM = 2 andM is an essential submodule ofcR′. SoMc=cR′.
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Laboratoire de Math´ematiques Nicolas Oresme, CNRS UMR 6139, D´epartement de math´ematiques et m´ecanique, 14032 Caen cedex, France
E-mail address: [email protected]
