R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
U LRICH A LBRECHT
A LBERTO F ACCHINI
Mittag-Leffler modules and semi-hereditary rings
Rendiconti del Seminario Matematico della Università di Padova, tome 95 (1996), p. 175-188
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ULRICH
ALBRECHT (*) -
ALBERTO FACCHINI(**) (***)
1. - Introduction.
In
[2]
it was demonstrated that manyproperties
of torsion-free abelian groups carry over tonon-singular right
modules overright strongly non-singular, right semi-hereditary rings,
where aring
R iscalled
right strongly non-singular
if thefinitely generated non-singu-
lar
right
modules areprecisely
thefinitely generated
submodules of free modules. Acomplete
characterization ofright strongly non-singu-
lar
right semi-hereditary rings
can be found in[9,
Theorem5.18].
Inparticular,
it was shown thatright strongly non-singular, right
semi-hereditary rings
are leftsemi-hereditary too,
so that we shall call suchrings right strongly non-singular semi-hereditary. Examples
of thistype
ofrings
are thesemi-prime semi-hereditary right
and left Goldierings,
for instance Prfferdomains,
as well as infinite dimensionalrings
like Z~’ .
Following [10],
we call aright
R-module A aMittag-Leffler
moduleif the natural map . is a
monomorphism
forall families of left R-modules.
Mittag-Leffler
modules can becharacterized as those modules M for which every finite subset is con- tained in a
pure-projective
pure submodule.Moreover,
theMittag-Lef-
fler torsion-free abelian groups are
precisely
theXl-free
groups[4].
In this note we show that this characterization extends to modules over(*) Indirizzo dell’A.:
Department
of Mathematics, AuburnUniversity,
Auburn, AL 36849, U.S.A.(**) Indirizzo dell’A.:
Dipartimento
di Matematica e Informatica, Univer- sita di Udine, 33100 Udine,Italy.
(***)
Partially supported by
Ministero dell’Universita e della Ricerca Scien- tifica eTecnologica
(Fondi 40% e 60%),Italy.
This author is a member of GNSAGA of CNR.right strongly non-singular semi-hereditary rings.
Our resultsparticu- larly generalize
recent workby
Rothmaler on flatMittag-Leffler
mod-ules over RD-domains
[11].
We show that every RD-Ore-domain is aright strongly non-singular semi-hereditary
Goldiering,
andgive
anexample
that the converse need not to hold.2. -
Non-singularity
andpurity.
It is easy to see that
(1) non-singular right
modules overright strongly non-singular semi-hereditary rings
areflat, (2)
S-closed sub- modules ofnon-singular
modules are pure(recall
that a submodule U ofa module M is said to be S-closed in M if
M/ U
isnon-singular),
and(3) finitely presented
modules over asemi-hereditary ring
haveprojective
dimension - 1. Our first result describes the
right strongly non-singu-
lar
semi-hereditary rings
R for which these three statements can be inverted.THEOREM 1. The
following
conditions areequivalent for
aright strongly non-singular semi-hereditary ring
R:(a)
R has noinfinite
setof orthogonal idempotents.
(b)
R hasfinite right
Goldie dimension.(c)
Afinitely generated right
R-module isfinitely presented if
and
only if it
hasprojective
dimension - 1.(d)
R has no properright
ideals which are essential and pure.(e)
Aright
R-module isflat if
andonly if
it isnon-singu-
Lar.
(, f ~
A submodulesof a non-singular right
R-module is S-closedif
and
onty if it is
pure.PROOF.
(a)==>(6) Suppose
that R has infiniteright
Goldie dimen- sion. Since R is aright non-singular ring,
it contains astrictly
ascend-ing
chain of S-closedright
ideals[9, Proposition
2.4 and Theo-rem
3.14].
For every n theright
R-moduleR/In
isfinitely generated
and
non-singular,
henceprojective,
so thatIn
is a direct summand ofRR .
IfJn
is aright
ideal such thatIn fl3 Jn
=RR ,
thenIn fl3 (in
nIn + 1 )
==
In
+ 1, sothat {Jn n In + 1 In
is anindependent
infinite set of directsummands of
RR .
But then R has an infinite set of non-zeroorthogonal idempotents.
(b) ~ (c)
We have to showonly
that if(b) holds,
then every finite-ly generated
module ofprojective
dimension 1 isfinitely presented.
LetM be a
finitely generated
module with Uprojective.
Since R issemi-hereditary,
U is a direct sum offinitely generated
submodules[1].
But
RR
has finite Goldiedimension,
and therefore U c R’~ must have fi- nite Goldie dimension. Hence the direct sum has a finite number ofsummands,
thatis,
U isfinitely generated.
( c ) ~ ( b )
If R has infiniteright
Goldiedimension, RR
contains aninfinite
independent family
of non-zeroprincipal right
idealsri R,
A eThen is a
projective right
ideal ofR,
so thatR/ fli rt R
is a,teA
cyclic right
R-module ofprojective
dimension -1,
which is notfinitely presented
because is notfinitely generated [9, p. 9].
(b) - (d) Suppose
that I is anessential,
pureright
ideal of R.Since R has finite
right
Goldie dimension and isright non-singular,
itsmaximal
right quotient ring Q
issemi-simple
Artinian[9,
Theorem3.17]. Furthermore, IQ
is an essentialright
idealof Q [9, Proposition 2.32].
SinceQ
issemi-simple Artinian,
this isonly possible
ifIQ
=Q.
Hence
(R/I ) QIIQ
= 0. But I is pure inR,
so thatR/I
is flat.Therefore we obtain the exact sequence
0 -~ (R/I ) ®R R ~ (R/I ) ®
®R Q
=0,
whichgives
I = R.(d)==> (e)
It remains to show that a flat module M isnon-singular.
Let x be an element of a flat module M. Since R is
right
semi-hered-itary,
xR is flat[9, p.11].
But xR = so that theright
idealannr ( x )
is pure in R . Therefore either = R or is not es-sential in R. This shows that
Z(M)
= 0.(e) ~ ( f )
Let U be a pure submodule of thenon-singular
moduleM. Since M is
flat,
we know thatM/ U
is a flat R-module.By (e), M/ U
isnon-singular,
i.e. U is S-closed in M.( f ) ~ (a) Suppose
that( f )
holds and R contains an infinitefamily
of non-zero
orthogonal idempotents.
Set. The
right
ideal I is pure in R because it is the union of the di-n
rect summands i
holds,
then I is S-closed inR,
sothat the
non-singular cyclic right
R-moduleR/I
isprojective.
Then I isa direct summand of R. It follows that
RR
is a direct sum ofinfinitely
many non-zero
right ideals,
which is a contradiction.EXAMPLE 2. There exists a
right strongly non-singular
semi-hereditary ring
R that does notsatisfy
theequivalent
conditionsof
Theorem 1.
PROOF. Consider the
strongly non-singular, semi-hereditary ring
R = ZW
(see [2]). Obviously
R does not have finite Goldie dimen- sion.COROLLARY 3. The
following
conditions areequivalent for
aring
R without
infinite families of orthogonal idempotents:
(a)
R isright strongly non-singular
andsemi-hereditary.
(b)
R isleft strongly non-singular
andsemi-hereditary.
Moreover, if
Rsatisfies
theseconditions,
then R is aright
andleft
Goldie
ring.
PROOF. Let R be
right strongly non-singular
andsemi-hereditary.
By
Theorem1,
R has finiteright
Goldie dimension. Since the maximalright quotient ring Q
of R is flat as aright
R-module[9,
Theorem5.18],
we obtain that the left and
right
maximalring
ofquotients
of R coin-cide
[9,
Exercise3.B.23].
Observe that R is aright
p.p.ring
without in- finite families oforthogonal idempotents.
In viewof [5,
Lemma8.4],
such a
ring
has to be left p.p. too. But every left p.p.ring
is left non-sin-gular.
In order to show that R is leftstrongly non-singular,
it therefore remains to showthat Q
is flat as a left R-moduleby [9,
Theorem5.18]
since the
multiplication map Q ®R Q - Q
is anisomorphism. By [9,
The-orem
3.10],
a sufficient condition for this is that everyright
ideal of R isessentially finitely generated, i.e.,
R has finiteright
Goldie dimension.Thus,
R is leftstrongly non-singular.
It remains to show that R has the a.c.c. for
right
annihilators. But this followsimmediately
from Theorem 1 and[5,
Lemma1.14].
In view of Theorem 1 and the
left/right symmetry proved
in Corol-lary
3 we shall call therings
characterized in Theorem 1strongly
non-singular semi-hereditary
Goldierings.
Note that theleft/right
symme-try
may fail if R has an infinite set oforthogonal idempotents
(see [9]).
,EXAMPLE 4. A
strongly non-singular semi-hereditary
Goldiering
need not be
PROOF. Let R be the
ring
of lowertriangular
2 x 2-matrices over afield
F,
so that R isright
and lefthereditary
and Artinian[3].
It is easy to see that R is essential as aright
and as a left submodule ofQ
==
Mat2 (F). By [9, Proposition 2.11], Q
is the maximalright
and the maxi- mal leftring
ofquotients
of R . Since R isright Artinian,
we have thatevery
right
ideal of R isessentially finitely generated. [9,
Theorem3.10] yields
thatRQ
is flat and that themultiplication
mapQ ©R Q - Q
is an
isomorphism. Thus,
R isright
and leftstrongly non-singular,
butis not
semi-prime.
3. -
Mittag-Leffler
modules.We now turn to the discussion of
Mittag-Leffler
modules overstrongly non-singular semi-hereditary rings.
In order toadapt
thenotion of an
Xl-free
module to modules overstrongly non-singular
semi-hereditary
Goldierings,
a reformulation of the definition used in abelian groups becomes necessary. Otherwise it mayhappen
that R it-self may be not
Xl-free
unless R isright hereditary.
We say that a non-singular right
module M over aright strongly non-singular
Goldiering
R is
Xl-projective
if the S-closure of everycountably generated
submod-ule of M is
projective.
From the next result it followsimmediately
thatevery
projective
module over astrongly non-singular semi-hereditary
Goldie
ring
isixl-projective.
THEOREM 5. The
following
three conditions areequivalent for
aright strongly non-singular right
Goldiering
R:(a)
R issemi-hereditary.
(b)
Aright
R-module M ispure-projective if
andonly if M/Z(M)
is
projective
andZ(M)
is a direct summandof
a rrcoduleof
theform
Ni
where eachNi
is afinitely generated singular
moduleof projec-
tive dimension 1.
(c)
Thefollowing
conditions areequivalent for
aright
R-mod-ule M:
(i)
M is anon-singular Mittag-Leffler
module.(ii)
M isXl-projective.
(iii) Every finite
subsetof
M is contained in a S-closedprojec-
tive submodule
of
M.PROOF.
(a)
=>(b)
Let M be apure-projective
module. We know that M is a direct summand of a direct sum offinitely presented
mod-ules,
say . for some R-module N where eachVi
isfinitely presented.
Since R isstrongly non-singular, Vi /Z(Vi)
isprojective,
sayThen I
’i yields
thatM/Z(M)
isprojective. Moreover, Z(M) ® Z(N)
=== 0153 Z( Vi )
where eachZ( Vi )
isfinitely presented
as a direct summand ofa
finitely presented
module. We writeZ(Vi ) = R ni / Ui
for some n2 cv andfinitely generated submodule Ui
of R ni . Since R is anon-singular semi-hereditary ring, Ui
isprojective,
andZ( Vi )
hasprojective
dimen-sion 1.
The converse holds
by
Theorem 1.(b) ~ (a)
Let I be afinitely generated right
ideal of R. SinceR/I
is
finitely presented,
it is the direct sum of aprojective
module and amodule of
projective
dimension at most 1by (b).
Hence I has to beprojective.
(a) ==> (c): (I) - (it)
Let U be acountably generated
submodule ofa
non-singular Mittag-Leffler
module M.By [10]
there is apure-projec- tive, countably generated
pure submodule V of M that contains U.By
Theorem 1 and the
already proved implication ( a ) ~ ( b )
of thistheorem,
V is an S-closed
projective
submodule of M. Inparticular,
V contains thes-closure
U *
of U.By [2, Proposition 2.2]
the moduleV/ U *
hasprojec-
tive dimension at most 1. Since V is
projective,
thisyields
thatU *
hasto be
projective
too.is obvious.
By [10]
it isenough
to show that every finite subset of M is contained in apure-projective
pure submodule of M. But S-closed submodules are pureby
Theorem 1.(c) ~ (a)
Let I be afinitely generated right
ideal of R. Consideran exact sequence 0 -~ U -~ R n ~ I -~ 0 of
right
R-modules where ncv . Since R has finite
right Goldie-dimension,
U contains afinitely
gen- erated essential submodule V.Furthermore,
R n is anon-singular
Mit-tag-Leffler
module.By (c),
the S-closure W of V in Rn isprojective [5, Proposition 8.24] yields
that W isfinitely generated.
Since U is s closed in Rn and V is essential inU,
it follows that U = W. Thus I isfinitely presented,
and inparticular,
aMittag-Leffler
module.By (c), finitely generated non-singular Mittag-Leffler
modules areprojec-
tive.
Since every ideal of a Noetherian
integral
domain is aMittag-Lef-
fler
module,
thering
is anexample
of a domain over which thereexist torsion-free
Mittag-Leffler
modules which are notN1-projective.
In
[11,
Section6.3]
Rothmaler studies the structure of flat Mit-tag-Leffler
modules over aright hereditary RD-Ore-domain, i.e.,
aright hereditary right
and left Ore-domain for whichpurity
andrelative
divisibility
coincide. An RD-Ore-domain isright
and leftsemi-hereditary,
hence it is astrongly non-singular semi-hereditary
Goldie
ring.
FromExample
4 we thus haveEXAMPLE 6.
Every
RD-Ore-domain is astrongly non-singular semi-hereditary
Goldiering,
but the converse is not true in gener- aL.We can use Theorem 5 to determine the
projective
dimension ofMittag-Leffler
modules:COROLLARY 7. Let R be a
ring.
(a) R is right semi-hereditary if
andonly if for
everyMittag-Lef fler right
R-module M and everyinteger
n ~0, if
M can begenerated
elements then
proj, dim. M % n
+ 1.(b) If
R is astrongly non-singular semi-hereditary
Goldiering
and M is a
non-singular Mittag-Leffler
modulegenerated by ~
ele-ments,
thenproj. dim. M ~
n.PROOF. If every
countably generated Mittag-Leffler right
R-mod-ule M has
projective
dimension £1,
then 1 for everyfinitely generated right
ideal I ofR,
so that R isright
semi-heredi-tary.
Conversely,
suppose that R isright semi-hereditary
and argueby
induction
on n ~ 0. If n = 0, a Mittag-Leffler right
R-modulegenerated by £ No
elements ispure-projective,
and therefore it hasprojective
di-mension K 1 because every
finitely presented right
R-module over aright semi-hereditary ring
hasprojective
dimension £ 1. And if n = 0 and M is anon-singular Mittag-Leffler
module over astrongly
non-sin-gular
Goldiering generated by £ No elements,
then M isprojective by
Theorem 5.
Suppose n
> 0. Let M be aMittag-Leffler right
R-modulegenerated by
aon )
c M. For every finite subset X of M fix a pure,countably generated, pure-projective
submoduleVx
of Mcontaining
X.Define a submodule
W,,
of Mgenerated by £ Xn - 1 elements by
transfi-nite induction where (On X (oo denotes the
lexicograph-
ic
product
ofwn and w0,
in thefollowing
way. Setx is a limit
ordinal,
setWa
=U
not a limit ordi-nal,
then a =( v, r
+1)
for some v (On and some r wo. If v is a limitordinal,
setWa
=Ww,
r) . If v is not a limitordinal,
then a =(p
+1,
r ++
1).
In this case letXCIJ.
+ 1,~ be a set ofgenerators
ofW~~ + 1, r~
of cardinal-ity ~
and set Note thatWa
has a set ofgenerators
ofcar£inality £
It is clear that
Wo
cWI
c ... cWa
c ... , a E x to 0, is anascending
chain of submodules of M. We claim that is pure in M for every ordinal v on . In order to prove the
claim,
let A be a k x m matrixover
R,
Z a 1 matrix over M and Y =( yl ,
... ,ym)
a 1 x m matrixover
W(v, 0)
such that ZA = Y. We must show that there exists a 1 x k matrix Z’ overW(v, 0)
such that Z’ A = Y. Since yi , 9 ... , ym EWe v,
0) and( v, 0)
E cv n x too is a limitordinal,
there exists~8 ( v, 0)
such that yi , ... , ym eWp .
Let~8 ~ ~8
be the least ordinal such that yl , 9 ... , ym EE
W~ .
Then~8
is not a limitordinal, and #
must be of the form(o
+1,
9r +
1).
Let X be a finite subset of the setX~~,
+ 1, r + 1> ofgenerators
ofWem
+1, r +1) such that y1, ..., ymbelong
to the submodule XR of M gen-erated
by
X. The pure submodule I of M is contained inWcw + ~, r + 2>
and contains yl , 9 ... , ym . Therefore there exists a 1 ma-trix Z’ over
VX u
such that Z’ A = Y. This concludes theproof
of theclaim,
becauseSince the modules are
generated by ~
1 elements and pure submodules ofMittag-Leffler
modules areMittag-Leffler
mod-ules,
it follows that the inductivehypothesis
can beapplied,
so thatproj.
dim.W(,,
o) ~ n(and proj.
dim.Ww, strongly
non-singular semi-hereditary
Goldiering
and M isnon-singular)
for everyv o n .
By
Auslander’sTheorem,
theprojective
dimension of M cannotexceed n + 1
(or n
if R is astrongly non-singular semi-hereditary
Goldie
ring
and M isnon-singular).
If we restrict our discussion to
semi-prime rings,
theequivalences
in Part
(c)
of Theorem 5 can be furtherimproved.
Observe that the se-mi-prime strongly non-singular semi-hereditary rings
without infinite sets oforthogonal idempotents
areprecisely
thesemi-prime right
andleft
semi-hereditary
Goldierings. Moreover,
if R is asemi-prime right
Goldie
ring,
then aright
ideal of R is essential if andonly
if it contains aregular element [5,
Lemma 1.11 andCor. 1.20],
so thatZ(M) =
- ~ x
E = 0 for someregular
element c ER }
for anyright
R-mod-ule M. In
particular
if N is a submodule of anon-singular right
moduleM over a
semi-prime semi-hereditary
Goldiering,
then N is pure in M if andonly
if Mc ON = Nc for everyregular
element c E R.COROLLARY 8. Let R be a
semi-prime, right
and semi-hered-itary
Goldiering.
Thefollowing
conditions areequivalent for
an R-modute M:
(a)
M is aMittag-Leffler
module.(b) Z(M)
is aMittag-Leffler module,
andM/Z(M)
isjective.
PROOF. Since the class of
Mittag-Leffler
modules is closed with re-spect
to pure submodules and pureextensions,
Theorems 1 and 5 re-duce the
problem
toshowing
thatM/Z(M)
isMittag-Leffler
wheneverM is
Mittag-Leffler.
Forthis,
let U be afinitely generated
submoduleof
M,
and choose apure-projective
pure submodule V of M which con-tains U.
By
Theorem5,
V = P ? for someprojective
submodule P of M. Since[ U
+Z(M)]/Z(M)
C[P
EDZ(M)]/Z(M),
thecorollary
willfollow once we have shown that P 0
Z(M)
is S-closed in M.Suppose
that x E M satisfies xc E
[P fl3 Z(M)]
for someregular
element c E R. Wecan find
YEP
and aregular
d E R such that xcd -pd
= 0. But P is pure in V and V is pure inM,
so that P is pure in M. Thus xcd = P nfl Mcd =
Pcd,
and x E P fl3Z(M).
The rest of this Section is devoted to
completely
recover Lemmas6.10, 6.11, 6.12,
Theorem 6.13 andCorollary
6.14 of[11]
for the more gen- eral class ofrings
discussed in this paper.PROPOSITION 9. Let R be a
strongly non-singular semi-hereditary
GoLdie
ring
and M an R-module with theproperty
that everycountably generated
submoduleof M is projective.
Then M is anon-singular
Mit-tag-Leffler
R-module and everyfinite
subsetof
M is contained in afinitely generated projective
pure submoduleof
M.PROOF. Let M be a module
satisfying
thehypotheses
of the state-ment. It is obvious that M is
non-singular.
We claim that if X is a
finitely generated
submodule ofM,
then theS-closure C of X in M is
finitely generated.
In order to prove the claim it is sufficient to show that eachcountably generated
submodule N of Ccontaining X
isfinitely generated. Any
such N isprojective,
hence N =, where the
Ni
areisomorphic
tofinitely generated right
idealsSo it is
enough
to show that for some n cv . Choosen A - 11
n to such
thatXc
andNi.
We have X c N’c N c C.i=O
Since C modulo the submodule
generated by
X issingular, NIN’
==; also is
singular.
But theNi’s
areisomorphic
toright
idealsof R,
and therefore . i is
non-singular.
Therefore N’ =N,
and Nis
finitely generated.
This proves our claim.Since every
finitely generated
submodule of M isprojective,
it isnow clear that the S-closure of every
finitely generated
submodule of M is afinitely generated projective
pure submodule of M. Inparticular
Mis a
Mittag-Leffler
module(Theorem 5).
LEMMA 10. Let R be a
right strongly non-singular right
Goldiering,
C anon-singular right
R-module and P afinitely generated
sub-module
of
C.If C/P
issingular,
then C hasfinite
Goldie dimen-sion.
PROOF. Since P is a
finitely generated non-singular
module over aright strongly non-singular ring,
P is a submodule of afinitely generat-
ed free module. Inparticular,
P has finite Goldie dimension. Since C isnon-singular
andC/P
issingular,
P is an essential submodule of C. This shows dim C = dim P 00.THEOREM 11. The
following four
conditions areequivalent for
aright strongly non-singular right
Goldiering
R:(a)
R is aright hereditary ring.
(b)
R is aright noetherian, right hereditary ring.
(c)
R isa right semi-hereditary ring
and all submodulesof
non-singular Mittag-Lef,fLer right
R-modules areMittag-Leffler
mod-ules.
(d)
Thefollowing
conditions areequivalent for
aright
R-module M:(i)
M is anon-singular Mittag-Leffler
module.(ii) Every countably generated
submoduleof
M isprojective.
(iii) Every finite
subsetof M
is contained in afinitely generat-
ed
projective
pure submoduleof
M.(iv)
M isnon-singular
and everyfinite
subsetof
M is con-tained in a
finitely presented
pure submoduleof
M.(v)
M isnon-singular
and every submoduleof
Mof finite
Goldie dimension is a
finitely generated projective
module.PROOF.
(a) ~ (d) Suppose
that R isright hereditary.
(i)
~(ii)
isproved
in[11,
Cor.6.3].
(ii)
~(iii)
isproved
inProposition
9.If every element of M is contained in a
projective module,
M must benon-singular. Moreover,
everyfinitely generated projective
submodule isfinitely presented.
(iv) ~ (v) Suppose
that(iv)
holds and let N be a submodule of M of finite Goldie dimension. Let X ={xl ,
x2 , ... ,xn}
be a finite subsetof N such that is an essential submodule of N. Then
n
’ is a
singular
submodule of so that if C denotesthe S-closure of in
M,
then the subset X is con-tained in a
finitely presented
pure submodule D of M. Since M is non-singular,
D also isnon-singular,
henceflat,
henceprojective.
Thus X iscontained in the
finitely generated projective
pure submodule D of M.Therefore C is a submodule of D. Hence N is contained in the
projective
module
D,
and N isprojective
because R isright hereditary. By [1]
N isisomorphic
to a direct sum offinitely generated right
ideals. But N hasfinite Goldie
dimension,
and therefore N itself is afinitely generated projective
module.(v)
~(i) Suppose
that(v)
holds. In order to prove that M is aMittag-Leffler
module it is sufficient to show that every finite subset X of M is contained in apure-projective
pure submodule of M[4, Th. 6].
Let P be the submodule of M
generated by
a finite subset X of M and C be the S-closure of P inM,
so that C is pure in M.By
Lemma 10 themodule C has finite Goldie dimension.
By Hypothesis (v)
C is afinitely generated projective
module.( d ) ~ ( c )
Assume that R has theproperty
that the five conditionsare
equivalent
for everyright
R-module M. Let us show that R isright hereditary.
If I is aright
ideal ofR,
then I is a submodule of the non-singular Mittag-Leffler
moduleRR ,
which is of finite Goldie dimension.By (d)
I is afinitely generated projective
module.Since M is a
non-singular Mittag-Leffler
module if andonly
if everycountably generated
submodule of M isprojective,
every submodule ofa
non-singular Mittag-Leffler
module is anon-singular Mittag-Leffler
module.
( c ) ~ ( b )
In order to show that R isright noetherian,
it is suffi-cient to show that if I is a
countably generated right
ideal ofR,
then I isfinitely generated.
SinceRR
is anon-singular Mittag-Leffler module,
every
right
ideal of R is anon-singular Mittag-Leffler
module. Henceevery
countably generated right
ideal I of R is anon-singular
pure-pro-jective module,
thatis,
it isprojective.
ThenR/I
is afinitely generated
module of
projective
dimension -1,
and therefore it isfinitely present-
ed
(Theorem 1).
Hence I isfinitely generated.
(b) - (a)
is obvious.4. - Prffer
rings
andindecomposable Mittag-Leffler
modules.Recall that a commutative
integral
domain issemi-hereditary
if andonly
if it is aPrufer ring,
thatis,
all its localizations at maximal idealsare valuation domains. If R is an
integral domain,
for every R-module M the submoduleZ(M)
isexactly
the torsion submodulet(M)
ofM,
sothat a module is
non-singular
if andonly
if it is torsion-free. Hence The-orem 5
gives
acomplete description
of torsion-freeMittag-Leffler
mod-ules over Pr3fer domains: a torsion-free module over a Prüfer domain is
a
Mittag-Leffler
module if andonly
if it isX1-projective.
Moregeneral- ly,
a module M over a Pr3fer domain is aMittag-Leffler
module if andonly
ifM/t(M)
isX1-projective
andt(M)
is a torsionMittag-Leffler
module. The structure of torsion
Mittag-Leffler
modules over a Priiferdomain R
depends heavily
on theproperties
of R. Forinstance,
in[4, Prop. 7]
it is shown that a torsion abelian group G is aMittag-Leffler
Z-module if and
only
ifn
nG = 0. In the nextProposition
we describetorsion
Mittag-Leffler
modules over almost maximal valuation domains andarbitrary Mittag-Leffler
modules over maximal valuationrings.
Recall that an R-module is
cyclically presented
if it isisomorphic
toRIAR
for some a E R[8].
PROPOSITION 12. Let M be a
torsion
module over an almost maxi- mal valuation domain R or anarbitrary
module over a maximal valu- ationring
R. Thefollowing
conditions areequivalent:
(a)
M is aMittag-Lef fler
R-module.(b) Every finite
subsetof
M is contained in a direct summandof
M that is a direct sum
of cyclically presented
modules.(c) Every
elementof M is
contained in a direct summandof M
that is a direct sum
of cyclically presented
modules.PROOF.
(a)
~(b)
Let X be a finite subset of M. Then X is contained in apure-projective
pure submodule P of M[4,
Th.6].
Thepure-projec-
tive module P is a direct sum of
cyclically presented
modules[8,
Th. 11.3.4 and
Prop. 11.4.3].
Hence Pdecomposes
asP = P’ E9 P" ,
where X c P’ and P’ is a finite direct sum of
cyclically presented
mod-ules.
By [8,
Th.XI.4.2]
P’ ispure-injective.
Since P’ is pure inM,
P’must be a direct summand of M.
(b) - (c)
is obvious.(c)
=>(a)
Let X be a finite subset of M.By [4,
Th.6]
it is sufficient to prove that X is contained in apure-projective
pure submodule of M.By [8, Prop. XIII.2.4]
the module M isseparable,
thatis,
every finite set of elements of M can be embedded in a direct summand which is adirect sum of uniserial modules. Hence it is
enough
to prove that every uniserial direct summand U of M iscyclically presented.
Let x be a non-zero element of a uniserial direct summand U of M.
By (c)
there exists adirect summand P of M such that P is a finite direct sum of
cyclically presented
modules and x E P. Let Wand Q
be directcomplements
of Uand
P,
so thatM = U ® W = P ® Q.
Since P has theexchange property
([12]
and[8,
Cor.VII.2.7]),
there are submodules U’ ~ U and W’ ~ Wsuch that M W’ . Since an essential
submodule of the uniserial module U. But
(P
nU)
f1 U’ =0,
so that U’ = 0 andM = P fl3 W’ .
Then U is a direct summand of Ufl3 fl3( W/W ’ )
=( U fl3 W )/W ’
=M/W ’
= P. Inparticular
U ispure-projec- tive,
thatis,
U is a direct sum ofcyclically presented
modules. Hence the uniserial module U must becyclically presented.
Therefore over an almost maximal valuation domain R the indecom-
posable
torsionMittag-Leffler
modules areonly
thecyclically present-
ed modules a ~
0,
and over a maximal valuationring
R the in-decomposable Mittag-Leffler
modules areonly
thecyclically presented
modules a E R. The last result of this paper addresses the ques- tion whether there exist
arbitrarily large indecomposable non-singular Mittag-Leffler
modules.EXAMPLE 13. Let R be a
strongly non-singular semi-hereditary
Goldie
ring
whose additive group iscotorsion-free.
Then there exists aproper class
of pairwise non-isomorphic, indecomposable, non-singu-
lar
Mittag-Leffler
R-modules.PROOF. Let K be an infinite cardinal. Since R has a cotorsion-free additive group, there exists an
X1-projective
left R-module M of cardi-nality
at least K such thatEndz (M) =
R OPby [6]. By
Theorem5,
M is anon-singular Mittag-Leffler
module whoseR-endomorphism ring
isCenter (R).
Since R does not contain any infinitefamily
oforthogonal idempotents,
the same holds forCenter (R).
We write 1 = el + ...+ en
... ,
en }
is afamily
oforthogonal, primitive idempotents
ofCenter (R).
ThenMi
=ei (M)
is anindecomposable Mittag-Leffler
mod-ule. Since
I
, at least one of theMi’s
has cardinal-ity
at least K.’
The
ring
ofalgebraic integers
is anexample
for aring
as inExample
13.REFERENCES
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projective
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ucts, Rend. Sem. Mat. Univ. Padova, 85 (1991), pp. 119-132.[8] L. FUCHS - L.
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