Valuation domains with a maximal immediate extension of finite rank

0.2 MB·English

by  Francois Couchot

VALUATION DOMAINS WITH A MAXIMAL IMMEDIATE
EXTENSION OF FINITE RANK.
9
0 FRANC¸OISCOUCHOT
0
2
Abstract. If R is a valuation domain of maximal ideal P with a maximal

n immediateextensionoffiniterankitisproventhatthereexistsafinitesequence
Ja mofapxriimmaelifdoeraelascPhj=,0L0≤⊃j≤L1m⊃−··1·a⊃ndLmRL⊇m0issumcahxtimhaatlRifLLjm/L6=j+01.iTshaelmnowset
2 supposethatthereisanintegern≥1suchthateachtorsion-freeR-moduleof
1 finiterankisadirectsumofmodulesofrankatmost n. ByadaptingLady’s
methods, it is shown that n ≤ 3 if R is almost maximal, and the converse

] holdsifRhasamaximalimmediateextensionofrank≤2.
A
R
. LetRbeavaluationdomainofmaximalidealP,Ramaximalimmediateexten-
h b
t sion of R, R the completion of R in the R-topology, and Q, Q, Q their respective
a e b e
fields of quotients. If L is a prime ideal of R, as in [5], we define the total defect
m
at L, d (L), the completion defect at L, c (L), as the rank of the torsion-free
R R
[
R/L-module (\R/L) and the rank of the torsion-free R/L-module (^R/L), respec-

1 tively. RecallthatalocalringRisHenselianifeachindecomposablemodule-finite
v
R-algebraislocalandavaluationdomainisstronglydiscreteifithasnonon-zero
1
5 idempotent prime ideal. The aim of this paper is to study valuation domains R
6 for which dR(0)< ∞. The first example of a such valuation domain was given by
1 Nagata [11]; it is a Henselian rank-one discrete valuation domain of characteristic
1. p > 0 for which dR(0) = p. By using a generalization of Nagata’s idea, Facchini
0 and Zanardo gave other examples of characteristicp>0, which are Henselian and
9 strongly discrete. More precisely:
0
: Example0.1. [5,Example6]Foreachprimeintegerpandforeachfinitesequence
v
i ofintegersℓ(0)=1, ℓ(1),...,ℓ(m)thereexistsastronglydiscretevaluation domain
X
R with prime ideals P = L ⊃ L ⊃ ··· ⊃ L = 0 such that c (L ) = pℓ(i),
0 1 m R i
r ∀i, 1≤i≤m.
a
So, dR(0)=p(Pii==m1 ℓ(i)) by [5, Corollary 4].
Theorem 0.2. [5, Theorem8] Let α be an ordinal number, ℓ:α+1→N∪{∞} a
mapping with ℓ(0)=1 and p a prime integer. Then there exists a strongly discrete
valuation domain R and an antiisomorphism α+1→Spec(R), λ7→L , such that
λ
c (L )=pℓ(λ), ∀λ≤α.
R λ
So, if ℓ(λ) = 0, ∀λ ≤ α, except for a finite subset, then d (0) < ∞ by [5,
R
Corollary 4].
2000 Mathematics Subject Classification. Primary13F30,13C11.
Key words and phrases. torsion-freemodule,valuationdomain,stronglyflatmodule,content
module.
1
2 FRANC¸OISCOUCHOT
In 1990, Va´mos gave a complete characterization of non-Henselian valuation
domains with a finite total defect, and examples of such rings. His results are
summarized in the following:
Theorem 0.3. [12, Theorem 5] Let R be a non-Henselian valuation domain and
assume that d (0)<∞. Then one of the following holds:
R
(rc) d (0)=2, R has characteristic zero, Qis algebraically closed andits cardi-
R b
nality|Q|ℵ0 =|Q|. Further, Qis real-closed, thevaluation on Qhas exactly
b b
two extensions to Q and R is almost maximal.
b
(y) There isanon-zeroprime ideal Lof Rsuchthat R isamaximalvaluation
L
ring, and R/P and its field of quotients satisfy (rc).
As Va´mos, if R is a domain, we say that fr(R) ≤ n (respectively fro(R) ≤ n)
if every torsion-free module (respectively every submodule of a free module) of
finite rank is a direct sum of modules of rank at most n. By [12, Theorem 3]
fr(R) ≥ d (0) if R is a valuation domain. So, the study of valuation domains
R
R for which d (0) < ∞ is motivated by the problem of the characterization of
R
valuation domains R for which fr(R)<∞.
When R is a valuation domain which is a Q-algebra or not Henselian, then
fr(R) < ∞ if and only if fr(R) = d (0) ≤ 2 by [12, Theorem 10]. Moreover,
R
if fr(R) = 2, either R is of type (rc) and fro(R) = 1 or R is of type (y) and
fro(R) = 2. When R is a rank-one discrete valuation domain, then fr(R) < ∞ if
and only if fr(R)=d (0)≤3 by [13, Theorem 8] and [1, Theorem 2.6].
R
In this paper we complete Va`mos’s results. In Section 1, a description of valua-
tiondomainswithafinitetotaldefectisgivenbyTheorem1.7andProposition1.8.
In Section 2 we give some precisions on the structure of torsion-free R-modules of
finite rank when R satisfies a condition weaker than d (0) < ∞. In Section 3 we
R
extend to every almost maximal valuation domain the methods used by Lady in
[8] to study torsion-free modules over rank-one discrete valuation domains. If R is
an almost maximal valuation domain, we prove that d (0)≤ 3 if fr(R) < ∞ and
R
that fr(R)=d (0) if d (0)≤2.
R R
For definitions and general facts about valuation rings and their modules we
refer to the books by Fuchs and Salce [6] and [7].
1. maximal immediate extension of finite rank
We recall some preliminary results needed to prove Theorem 1.7 which gives a
description of valuation domains with a finite total defect.
Let M be a non-zero module over a valuationdomain R. As in [7, p.338] we set
M♯ = {s ∈ R | sM ⊂ M}. Then M♯ is a prime ideal of R and is called the top
prime ideal associated with M.
Proposition1.1. Let A be a proper ideal of R and let L be a prime ideal such that
A♯ ⊆ L and A is not isomorphic to L. Then R/A is complete in its ideal topology
if and only if R /A is also complete in its ideal topology.
L
Proof. Let (a + A ) be a family of cosets of R such that a ∈ a + A if
i i i∈I L i j j
A ⊂ A and such that A = ∩ A . We may assume that A ⊆ L, ∀i ∈ I. So,
i j i∈I i i
a +L=a +L, ∀i,j ∈I. Letb∈a +L, ∀i∈I. Itfollowsthata −b∈L, ∀i∈I.
i j i i
If R/A is complete in the R/A-topology, ∃c∈R such that c+b−a ∈A , ∀i∈I.
i i
Hence R /A is complete in the R /A-topology too.
L L
A MAXIMAL IMMEDIATE EXTENSION OF FINITE RANK 3
Conversely let (a +A ) be a family of cosets of R such that a ∈ a +A if
i i i∈I i j j
A ⊂ A and such that A = ∩ A . We may assume that A ⊂ A ⊆ L, ∀i ∈ I.
i j i∈I i i
We put A′ = (A ) , ∀i ∈ I. We know that A = ∩ La. Consequently, if
i i L a∈/A
a ∈/ A, there exists i ∈ I such that A ⊆ La, whence A′ ⊆ La. It follows that
i i
A = ∩ A′. Clearly, a ∈ a +A′ if A′ ⊂ A′. Then there exists c ∈ R such
i∈I i i j j i j L
that c ∈ a +A′, ∀i ∈ I. Since A′ ⊂ R, ∀i ∈ I, c ∈ R. From A = ∩ A′ and
i i i j∈I j
A ⊂ A , ∀i ∈ I we deduce that ∀i ∈ I, ∃j ∈ I such that A′ ⊂ A . We get that
i j i
c∈a +A because c−a ∈A′ ⊆A and a −a ∈A . So, R/A is complete in the
i i j j i j i i
R/A-topology. (cid:3)
Proposition 1.2. [7, Exercise II.6.4] Let R be a valuation ring and let L be a
non-zero prime ideal. Then R is (almost) maximal if and only if R/L is maximal
and R is (almost) maximal.
L
Proof. IfRis(almost)maximal,itisobviousthatR/LismaximalandbyProposi-
tion1.1R is (almost)maximal. Converselylet Abe anon-zeroidealandJ =A♯.
L
Suppose that either J ⊂ L or J = L and A is not isomorphic to L. Since R
L
is (almost) maximal it follows that R /A is complete in its ideal topology. From
L
Proposition1.1wededucethatR/Aiscompleteinitsidealtopology. Now,suppose
that L⊂ J. If A⊂ L let t ∈ J \L. Thus A⊂ t−1A. Let s∈ t−1A\A. Therefore
L ⊂ tR ⊆ s−1A. So, R/s−1A is complete in its ideal topology because R/L is
maximal, whence R/A is complete too. Finally if A∼=L the result is obvious. (cid:3)
Proposition 1.3. Let (L ) be a non-empty family of prime ideals of R and let
λ λ∈Λ
L=∪ L . Then L is prime, R =∩ R and R is maximal if and only if
λ∈Λ λ L λ∈Λ Lλ L
R is maximal ∀λ∈Λ.
Lλ
Proof. It is obvious that L is prime. Let Q be the field of fractions of R. If
x ∈ Q\R then x = 1 where s ∈ L. Since L = ∪ L , ∃µ ∈ Λ such that
L s λ∈Λ λ
s∈L . We deduce that x∈/ R and R =∩ R .
µ Lµ L λ∈Λ Lλ
If R is maximal, we deduce that R is maximal ∀λ ∈ Λ by Proposition 1.1.
L Lλ
Conversely, by [14, proposition 4] R is linearly compact in the inverse limit
L
topology. Since R isHausdorffinthislineartopologytheneverynonzeroidealis
L
open and also closed. Hence R is linearly compact in the discrete topology. (cid:3)
L
Recall that a valuation domainR is Archimedean if its maximalP is the only
non-zero prime ideal and an ideal A is Archimedean if A♯ =P.
Proposition 1.4. [3, Corollary 9] Let R be an Archimedean valuation domain. If
d (0)<∞, then R is almost maximal.
R
From Propositions 1.1, 1.2, 1.3 and 1.4 we deduce the following:
Proposition 1.5. Let R be a valuation domain such that d (0) < ∞ and R/A
R
is Hausdorff and complete in its ideal topology for each non-zero non-Archimedean
ideal A. Then R is almost maximal.
Proof. LetL, L′beprimeidealssuchthatL′ ⊂L. Since([R )isasummandof(R)
L b L
wehaved (0)≤d (0). Ontheotherhand,bytensoringapure-compositionseries
RL R
of ([R ) with R /L′ we get a pure-composition series of (R\/L′). So, d (L′) ≤
L L L RL
d (0).
R
4 FRANC¸OISCOUCHOT
If R is Archimedean the result follows from Proposition 1.4. Suppose that R
is not Archimedean, let J be a non-zero ideal and let (L ) be the family of
λ λ∈Λ
prime ideals properly containing J and properly contained in P. If Λ = ∅ we
get that R is almost maximal by applying Propositions 1.4 and 1.2. Else, let
L′ = ∪λ∈ΛLλ. By Proposition 1.3, RL′/J is maximal for each non-zero prime J.
If L′ 6= P then R/L′ is maximal by Proposition 1.4 and it follows that R/J is
maximal by Proposition 1.2. If the intersection K of all non-zero primes is zero
then R is almost maximal. If K 6= 0 then R is Archimedean. We conclude by
K
using Propositions 1.4 and 1.2. (cid:3)
Givena ring R,anR-module M andx∈M, the content ideal c(x) ofx in M,
is the intersection of all ideals A for which x∈AM. We say that M is a content
module if x∈c(x)M, ∀x∈M.
Lemma 1.6. Let U be a torsion-free module such that U =PU. Then:
(1) ∀x∈U, x6=0, x∈/ c(x)U;
(2) let 06=x,y ∈U and t∈R such that x=ty. Then c(y)=t−1c(x);
(3) if U is uniserial then, for each x∈U, x6=0, c(x)♯ =U♯.
Proof. (1). If x ∈ c(x)U, there exist a ∈ R and z ∈ U such that x = az and
c(x)=Ra. But, since z ∈PU, we get a contradiction.
(2). Let 06=x,y ∈U such that x=ty. If s∈/ c(y) then x=tsz for some z ∈U
and st∈/ c(x). So, s∈/ t−1c(x). Conversely, if s∈/ t−1c(x) then st∈/ c(x). We have
x=stz for some z ∈U. We get that y =sz. So, s∈/ c(y).
(3). We put A = c(x) and L = A♯. Let s ∈/ L and y ∈ U such that x = ty for
some t ∈ R. Then c(y) = t−1A and t ∈/ A. So, t−1A ⊆ L. Consequently y ∈ sU.
Let s∈L. If s∈A then x∈/ sU. If s∈L\A let t∈s−1A\A. There exists y ∈U
such that x=ty. Since c(y)=t−1A and s∈t−1A we deduce that y ∈/ sU. (cid:3)
Thislemmaandthepreviouspropositionallowustoshowthefollowingtheorem.
Theorem 1.7. Let R be a valuation domain such that d (0) < ∞. Then there
R
exists a finite family of prime ideals P = L ⊃ L ⊃ ··· ⊃ L ⊃ L ⊇ 0 such
0 1 m−1 m
that R /L is almost maximal, ∀k, 0 ≤ k ≤ m−1 and R is maximal if
Lk k+1 Lm
L 6= 0 (or equivalently, for each proper ideal A ≇ L , ∀k, 0 ≤ k ≤ m, R/A is
m k
Hausdorff and complete in its ideal topology). Moreover, d (0)=Qk=mc (L ).
R k=1 R k
Proof. Let n=d (0). Then R has a pure-composition series
R b
0=G ⊂R=G ⊂···⊂G ⊂G =R
0 1 n−1 n b
such that, ∀k, 1 ≤ k ≤ n, U = G /G is a uniserial torsion-free module. The
k k k−1
family (L ,...,L ) is defined in the following way: ∀j, 0 ≤ j ≤ m, there exists
0 m
k, 1≤k ≤n such that L =U♯.
j k
Now,letAbe aproperidealsuchthatR/AisHausdorffandnon-completeinits
idealtopology. By [7, Lemma V.6.1]there exists x∈R\R suchthat A=c(x+R)
b
(Clearly c(x + R) = B(x), the breadth ideal of x). Let U be a pure uniserial
submodule of R/R containing x+R and let M be the inverse image of U by the
b
natural map R → R/R. From the pure-composition series of M with factors R
b b
and U, and a pure-composition series of R/M we get a pure-composition series
b
A MAXIMAL IMMEDIATE EXTENSION OF FINITE RANK 5
of R. Since each pure composition series has isomorphic uniserial factors by [7,
b
TheoremXV.1.7],itfollowsthatU ∼=U forsomek, 2≤k ≤n. So,byLemma1.6
k
A♯ =U♯ =U♯.
k
WeapplyProposition1.5anddeducethatR /L isalmostmaximal∀k, 0≤
Lk k+1
k ≤m−1 and R is maximal if L 6=0.
Lm m
Toprovethelastassertionweapply[5,Lemma2](Theconclusionofthislemma
holds if R /L′ is almost maximal, where L and L′ are prime ideals, L′ ⊂L). (cid:3)
L
The following completes the previous theorem.
Proposition1.8. LetRbeavaluationdomainsuchthatd (0)<∞,let(U )
R k 1≤k≤n
be the family of uniserial factors of all pure-composition series of R and let
b
(L ) be the family of prime ideals defined in Theorem 1.7. Then:
j 0≤j≤m
(1) ∀k, 1≤k ≤n, U ∼=R ;
k U♯
k
(2) R has a pure-composition series
b
0=F ⊂R=F ⊂···⊂F ⊂F =R
0 1 m−1 m b
where F /F is a free R -module of finite rank, ∀j, 0≤j ≤m−1.
j+1 j Lj
Proof. (1). Let A be an ideal such that ∃j, 0 ≤ j ≤ m, A♯ = L and A ≇ L . In
j j
the sequel we put L =0 if L 6=0.
m+1 m
First, for each uniserial torsion-free module U, we will show that each family
(x +rU) hasa non-emptyintersectionifx ∈x +tU,∀r,t∈R\A, r ∈tR.
r r∈R\A r t
As in the proofof Proposition1.2 we may assume that L ⊂A. Since R /L
j+1 Lj j+1
is almost maximal and A is an ideal of R the family (x +rU ) has a
Lj r Lj r∈R\A
non-emptyintersection. Ifr ∈L \A,wehaver−1A⊂L . So,ift∈L \r−1Athen
j j j
rt∈/ AandrtU ⊆rU. ItfollowsthatwecandoasintheproofofProposition1.1
Lj
to show that the family (x +rU) has a non-empty intersection.
r r∈R\A
Let 0 = G ⊂ R = G ⊂ ··· ⊂ G ⊂ G = R be a pure-composition series
0 1 n−1 n b
of R whose factors are the U , 1 ≤ k ≤ n. By induction on k and by using
b k
the pure-exact sequence 0 → G → G → U → 0, we get that each family
k−1 k k
(x +rG ) for which x ∈x +tG , ∀r,t∈R\A, r ∈tR, has a non-empty
r k r∈R\A r t k
intersection.
Letk, 2≤k ≤n,be aninteger,let06=x∈G \G andletA=c(x+G ).
k k−1 k−1
Then A♯ =U♯ =L for some j, 1≤j ≤m. We shall prove that A∼=L . For each
k j j
r∈R\A, x=g +ry forsomeg ∈G andy ∈G . Letr,t∈R\Asuchthat
r r r k−1 r k
r ∈tR. Then we get that g ∈g +tG ∩G =g +tG since G is a pure
r t k k−1 t k−1 k−1
submodule. If A≇L the family (g +rG ) has a non-empty intersection.
j r k−1 r∈R\A
Let g ∈ g +rG ∀r ∈ R\A. Then (x−g) ∈ rG , ∀r ∈ R\A. Since G is a
r k−1 k k
pure-essentialextensionofafreemodule,G isacontentmoduleby[4,Proposition
k
23]. Itfollowsthat(x−g)∈AG . Sox+G ∈AU . But,sincek ≥2,U =PU
k k−1 k k k
because R/PR ∼= R/PR. So, x+G ∈/ AU . From this contradiction we get
b b k−1 k
thatA=sL forsome06=s∈R. IfsL 6=L thenx+G =sy+G forsome
j j j k−1 k−1
y ∈ G because s ∈/ A. If follows that c(y+G ) = L . We put y′ = y+G .
k k−1 j k−1
Then, for each z ∈U \Ry′ there exists t∈R\L such that y′ =tz. We get that
k j
U =R y′.
k Lj
(2). Let M =R/R. Then L =M♯. From above we get that M/L M 6=0. By
b 1 1
[4, Proposition 21] M contains a pure free R -submodule N such that N/L N ∼=
L1 1
6 FRANC¸OISCOUCHOT
M/L M. It follows that (M/N)♯ = L . We set F the inverse image of N by the
1 2 2
natural map R→M. We complete the proof by induction on j. (cid:3)
b
2. Torsion-free modules of finite rank.
Inthissectionwegivesomeprecisionsonthestructureoftorsion-freeR-modules
of finite rank when R satisfies a condition weaker than d (0) <∞. The following
R
lemmas are needed.
Lemma 2.1. Let R be a valuation ring (possibly with zerodivisors), let U be a
uniserial module and let L be a prime ideal such that L⊂U♯. Then U is a cyclic
L
R -module.
L
Proof. Let s∈U♯\L and let x∈U \sU. Let y ∈U \Rx. There exists t∈P such
that x=ty. Then t∈/ Rs, whence t∈/ L. It follows that U =R x. (cid:3)
L L
Lemma 2.2. Let R be a valuation ring (possibly with zerodivisors), let U and V
be uniserial modules such that V♯ ⊂ U♯. Assume that U is a faithful R -module,
L L
where L=V♯. Then Ext1(U,V)=0.
R
Proof. Let M be an extension of V by U. By lemma 2.1 U is a free cyclic R -
L L
module. Since V is a module over R , it follows that V is a summand of M . We
L L
deduce that V is a summand of M too. (cid:3)
Lemma 2.3. Let R be a valuation domain for which there exists a prime ideal
L 6= P such that R/L is almost maximal. Then Ext1(U,V) = 0 for each pair of
R
ideals U and V such that L⊂U♯∩V♯.
Proof. Let M be an extension of V by U. It is easy to check that U/LU and
V/LV are non-zero and non divisible R/L-modules. Since R/L is almost maximal
M/LM ∼=U/LU⊕V/LV by [6, PropositionVI.5.4]. If L6=0, it follows that there
existtwosubmodulesH andH ofM,containingLM,suchthatH /LM ∼=U/LU
1 2 1
and H /LM ∼=V/LV. For i=1,2 let x ∈H \LM and let A be the submodule
2 i i i
of H such that A /Rx is the torsion submodule of H /Rx . Then A +LM/LM
i i i i i i
is a non-zero pure submodule of H /LM which is of rank one over R/L. It follows
i
that H =A +LM. By Lemma 2.1 M ∼=V ⊕U . We deduce that LM ∼=LM
i i L L L L
is a direct sum of uniserial modules. Since A ∩LM is a non-zero pure submodule
i
of LM there exists a submodule C of LM such that LM =(A ∩LM)⊕C by [7,
i i i
Theorem XII.2.2]. It is easy to check that H =A ⊕C . From M =H +H and
i i i 1 2
LM =H ∩H we deduce that the following sequence is pure exact:
1 2
0→LM →H ⊕H →M →0,
1 2
where the homomorphism from LM is given by x 7→ (x,−x), x ∈ LM, and the
one onto M by (x,y)7→x+y, x∈H , y ∈H . Since H ⊕H is a direct sum of
1 2 1 2
uniserialmodules,so is M by [7,TheoremXII.2.2]. Consequently M ∼=V ⊕U. (cid:3)
Proposition 2.4. Let R be a valuation domain. Let G be a torsion-free R-
module of finite rank. Then G has a pure-composition series with uniserial factors
(U ) such that U♯ ⊇U♯ , ∀k, 1≤k ≤n−1.
k 1≤k≤n k k+1
Proof. G has a pure-composition series
0=H ⊂H ⊂···⊂H ⊂H =G
0 1 n−1 n
A MAXIMAL IMMEDIATE EXTENSION OF FINITE RANK 7
such that, ∀k, 1 ≤ k ≤ n, V = H /H is a uniserial torsion-free module.
k k k−1
Suppose there exists k, 1 ≤ k ≤ (n−1) such that V♯ ⊂ V♯ . By Lemma 2.2,
k k+1
H /H ∼= V ⊕V . So, if H′ is the inverse image of V by the surjection
k+1 k−1 k k+1 k k+1
H → H /H , if U = H′/H and U = H /H′ then U♯ ⊃ U♯ .
k+1 k+1 k−1 k k k−1 k+1 k+1 k k k+1
So, in a finite number of similar steps, we get a pure composition series with the
required property. (cid:3)
Proposition 2.5. Let R be a valuation domain. Assume that there exists a finite
family of prime ideals P =L ⊃ L ⊃···⊃ L ⊃ L =0 such that R /L
0 1 m−1 m Lk k+1
is almost maximal ∀k, 0 ≤ k ≤ m−1. Let G be a torsion-free R-module of finite
rank. Then G has a pure-composition series
0=G ⊆G ⊆···⊆G ⊆G ⊆G =G
0 1 m−1 m m+1
where G /G is a finite direct sum of ideals of R , ∀j, 0≤j ≤m.
j+1 j Lj
Proof. By Proposition 2.4 G has a pure-composition series
0=H ⊂H ⊂···⊂H ⊂H =G
0 1 n−1 n
such that, ∀k, 1 ≤ k ≤ n, U = H /H is a uniserial torsion-free module and
k k k−1
U♯ ⊇U♯ , ∀k, 1≤k ≤n−1. Now, for each j, 1≤j ≤m, let k be the greatest
k k+1 j
index such that L ⊂ U♯ . We put G = H . Then G /G is an R -module
which is a direct sujm ofkijdeals by Lemjma 2.3k.j j+1 j Lj (cid:3)
3. Valuation domains R with fr(R)<∞.
Firstweextendtoeveryalmostmaximalvaluationdomainthemethodsusedby
Ladyin[8]tostudytorsion-freemodulesoverrank-onediscretevaluationdomains.
So,except inTheorem3.6, we assume that R is analmostmaximalvaluationring.
We put K = Q/R. For each R-module M, d(M) is the divisible submodule of
M which is the union of all divisible submodules and M is said to be reduced if
d(M)=0. We denote by M the pure-injective hull of M (see [7, chapter XIII]).
c
IfU isauniserialmodulethenU ∼=R⊗ U becauseRisalmostmaximal. LetGbe
b b R
a torsion-free module of finite rank r. By Proposition 2.5 G contains a submodule
B which is a direct sum of ideals and such that G/B is a Q-vector space. We put
corank G=rank G/B. Now, it is easy to prove the following.
Proposition3.1. Let G be a torsion-free R-module of rank r and corankc. Then:
(1) G contains a pure direct sum B of ideals, of rank r−c, such that G/B is
a Q-vector space of dimension c.
(2) G contains a pure direct sum B′ of ideals, of rank r such that G/B′ is
isomorphic to a quotient of Kc.
An element of Q⊗ Hom (G,H) is called a quasi-homomorphism from G
R R
to H, where G and H are R-modules. Let C be the category having weakly
ab
polyserialR-modules(i.e moduleswithcompositionserieswhosefactorsareunis-
erial) as objects and quasi-homomorphisms as morphisms and let C be full sub-
category of C having torsion-free R-modules of finite rank as objects. Then C
ab ab
is abelian by [7, Lemma XII.1.1]. If G and H are torsion-free of finite rank, then
the quasi-homomorphisms from G to H can be identified with the Q-linear maps
φ:Q⊗ G→Q⊗ H suchthatrφ(G)⊆H forsome06=r ∈R. WesaythatGand
R R
8 FRANC¸OISCOUCHOT
H arequasi-isomorphiciftheyareisomorphicobjectsofC. Atorsion-freemodule
of finite rank is said to be strongly indecomposable if it is an indecomposable
object of C.
Lemma 3.2. Let φ : Q⊗ G → Q⊗ H be a Q-linear map. Then φ is a quasi-
R R
homomorphism if and only if (R⊗ φ)(d(R⊗ G))⊂d(R⊗ H).
b R b R b R
Proof. Assume that φ is a quasi-homomorphism. There exists 0 6= r ∈ R such
that rφ(G) ⊆ H. It successively follows that r(R⊗ φ)(R⊗ G) ⊆ R⊗ H and
b R b R b R
(R⊗ φ)(d(R⊗ G))⊆d(R⊗ H).
b R b R b R
Conversely, let B be a finite direct sum of ideals which satisfies that G/B is a
Q-vector space. There exists a free submodule F of Q⊗ G such that B ⊆ F.
R
So, ∃0 6= r ∈ R such that rφ(B) ⊆ rφ(F) ⊆ H. Since R⊗ G = (R⊗ B)⊕
b R b R
(d(R⊗ G)), it follows that r(R⊗ φ)(R⊗ G) ⊆ (R⊗ H). We deduce that
b R b R b R b R
rφ(G)⊆(R⊗ H)∩(Q⊗ H)=H. (cid:3)
b R R
Proposition 3.3. Let G be a torsion-free R-module of rank r and corank c.
(1) IfGhas nosummandisomorphic toanideal, thenEnd(G)canbeembedded
in the ring of c×c matrices over Q. In particular if c = 1, End(G) is a
b
commutative integral domain.
(2) If G is reduced, then End(G) can be embedded in the ring of (r−c)×(r−c)
matrices over R. In particular if c = r − 1, End(G) is a commutative
b
integral domain.
Proof. See the proof of [8, Theorem 3.1]. (cid:3)
Inthesequelweassumethatn=c (0)<∞. So,therearen−1unitsπ ,...,π
R 2 n
in R\R such that 1,π ,...,π is a basis of Q over Q. By [7, Theorem XV.6.3]
b 2 n b
thereexistsanindecomposabletorsion-freeR-moduleE withranknandcorank1.
WecandefineE inthefollowingway: if(e ) isthecanonicalbasisofRn−1,if
k 2≤k≤n b
e =Pk=nπ e and V is the Q-vector subspace of Qn−1 generated by (e ) ,
1 k=2 k k b k 1≤k≤n
then E = V ∩Rn−1. Then a basis element for d(R ⊗ E) can be written u +
b b 1
π u +···+π u , whereu ,...,u ∈E. Since E isindecomposableitfollowsthat
2 2 n n 1 n
u ,...,u is a basis for Q⊗E ∼=V.
1 n
Theorem 3.4. Let G be a torsion-free R-module of rank r and corank c. Then
the following assertions hold:
(1) The reduced quotient of G is isomorphic to a pure submodule of B where B
b
is a direct sum of (r−c) ideals.
(2) G is the direct sum of ideals of R with a quasi-homomorphic image of Ec.
Proof. (1) can be shown as the implication (1) ⇒ (2) of [8, Theorem 4.1] and (2)
as the implication (1)⇒(3) of [8, Theorem 4.1]. (cid:3)
Corollary 3.5. Let G be a torsion-free R-module of rank r and corank c. Then:
(1) If G has no summand isomorphic to an ideal, then r≤nc.
(2) If G is reduced, then nc≤(n−1)r.
Proof. This corollary is a consequence of Theorem 3.4 and can be shown as [8,
Corollary 4.2]. (cid:3)
A MAXIMAL IMMEDIATE EXTENSION OF FINITE RANK 9
Theorem3.6. LetRbeavaluation domain suchthat d (0)=2. Then fr(R)=2.
R
Moreover fro(R)=1 if c (0)=2 and fro(R)=2 if c (0)=1.
R R
Proof. First suppose that c (0) = 2. So, R is almost maximal and fro(R) = 1.
R
Let G be an indecomposable torsion-free module with rank r and corank c which
is not isomorphic to Q and to an ideal. Then G is reduced and has no summand
isomorphic to an ideal of R. From Corollary 3.5 we deduce that r = 2c. By
Theorem 3.4 G is isomorphic to a pure submodule of B where B is a direct sum
b
of c ideals. Since rank B = 2c it follows that G ∼= B. So, c = 1 and G ∼= A for a
b b b
non-zero ideal A.
If c (0) = 1 let L be the non-zero prime ideal such that c (L) = d (0) = 2.
R R R/L
Then fr(R/L) = 2. Since R is maximal it follows that fr(R) = fro(R) = 2 by
L
[12, Lemma 9 and Lemma 4]. (cid:3)
Lemma 3.7. Every proper subobject of E in C is a direct sum of ideals.
Proof. Let G be a proper object of E in C and let H be the pure submodule of E
such that H/G is the torsion submodule of E/G. Since E is indecomposable, E
has no summand isomorphic to a direct sum of ideals. So, corank E/H = 1 and
corankH =0. AscorankH ≥corankGwegetthatGisadirectsumofideals. (cid:3)
Proposition 3.8. E is an indecomposable projective object of C.
Proof. Let φ : H → E be a quasi-epimorphism where H is a torsion-free module
of finite rank. Suppose that H = F ⊕G where F is a direct sum of ideals. By
Lemma 3.7, φ(G) is quasi-isomorphic to E. So, we may assume that H has no
summand isomorphic to an ideal. By Theorem 3.4 there is a quasi-epimorphism
ψ : Ec → H where c = corank H. It is sufficient to see that φ ◦ ψ is a split
epimorphism in C. But by Proposition 3.3(1), Q⊗End(E) is a subfield of Q, so
b
everyquasi-homomorphismE →E iseitheraquasi-isomorphismortrivialandthe
splitting follows immediately. (cid:3)
In the sequel, Q⊗ Hom (R⊕E,M) is denoted by M¨ for each R-module M
R R
and the ring Q⊗ End (R⊕E) by Λ.
R R
Theorem 3.9. The foncteur Q⊗ Hom (R⊕E, ) is an exact fully faitful functor
R R
from C into mod−Λ, the category of finitely generated right Λ-modules.
Proof. ByTheorem3.4andProposition3.8,R⊕E isaprogeneratorofC. Foreach
finite rank torsion-free R-module H, the natural map Q⊗ Hom (R⊕E,H) →
R R
Hom (R¨⊕E¨,H¨)isanisomorphismbecauseΛ=R¨⊕E¨. ThusQ⊗ Hom (F,H)→
Λ R R
Hom (F¨,H¨)isanisomorphismifF isasummandofafinitedirectsumofmodules
Λ
isomorphic to R ⊕ E. Let G be a finite rank torsion-free R-module. We may
assume that G has no summand isomorphic to an ideal of R. By Proposition 3.8
and Lemma 3.7, there is an exact sequence 0→Rnc−r →Ec →G→0 in C. Since
both functors are left exact, we get that Q⊗ Hom (G,H)∼=Hom (G¨,H¨). (cid:3)
R R Λ
Lemma 3.10. If M is a right Λ-module and M ⊆G¨ for some finite rank torsion-
free R-module G, then M ∼=H¨ for some torsion-free R-module H.
Proof. See the proof of [8, Lemma 5.2]. (cid:3)
10 FRANC¸OISCOUCHOT
Proposition 3.11. The ring Λ is a hereditary Artinian Q-algebra such that
(rad Λ)2 =0. There are two simple right Λ-modules, R¨ which is projective and K¨
which is injective.
Proof. See the proof of [8, Proposition 5.3]. (cid:3)
Proposition 3.12. Q¨ is an injective hull for R¨ and E¨ is a projective cover for K¨.
Proof. See the proof of [8, Proposition 5.4]. (cid:3)
Theorem 3.13. The image of C under the functor Q⊗ Hom (R⊕E, ) is the
R R
full subcategory of mod− Λ consisting of modules with no summand isomorphic to
K¨.
Proof. See the proof of [8, Theorem 5.5]. (cid:3)
LetM beafinitelygenerated(i.e.,finitelength)rightΛ-module. Wedefinerank
M to be the number of factors in a composition series for M isomorphic to R¨ and
corank M to be the number of composition factors isomorphic to K¨.
Proposition3.14. The foncteur Q⊗ Hom (R⊕E, ) preserves rank and corank.
R R
Proof. See the proof of [8, Proposition 5.6]. (cid:3)
We now consider the functors D=Hom ( ,Q) and Tr=Ext ( ,Λ) which take
R Λ
right Λ-modules to left Λ-modules and conversely. It is well known that D is an
exact contravariant lenght preserving functor taking projectives to injectives and
conversely, and that D2 is the identity for finitely generated Λ-modules. Since Λ
is hereditary, Tr is right exact and Tr2 M ∼=M if M has no projective summand,
Tr M = 0 if M is projective. We consider the Coxeter functors C+ = DTr and
C− = TrD. Thus C+ : mod−Λ → mod−Λ is left exact and C− : mod−Λ →
mod−Λ is right exact. If M has no projective (respectively injective) summand, it
iseasytocheckthatM isindecomposableifandonlyifC+ M (respectivelyC− M)
is indecomposable.
Proposition 3.15. Let M be a right Λ-module with rank r and corank c.
(1) If M has no projective summand, then corank C+ M = (n−1)c−r and
rank C+ M =nc−r.
(2) If M has no injective summand, then rank C− M = (n−1)r −nc and
corank C− M =r−c.
Proof. See the proof of [8, Proposition 5.7]. (cid:3)
Theorem 3.16. The following assertions hold:
(1) If n = 3, then, up to quasi-homomorphism, the strongly indecomposable
torsion-free R-modules are R, Q, E, R and an R-module with rank 2 and
b
corank 1 (corresponding to C+ Q¨ =C− R¨).
(2) If n ≥ 4, there are strongly indecomposable torsion-free R-modules with
arbitrarily large rank.
Proof. We show (1) and (2) as Lady in the proof of [8, Theorem 5.11] by using
Proposition 3.15 and [10, Theorem 2]. (cid:3)