*On leave from Centre National de la Recherche
Scientifiq~e.
**This research was partially supported by the National Science Foundation,
Grant # GP-42325.
STABLE MODULES
**
by
P. Crunion*
D~pa~tment of Statistics
University of North Carolina at ChC'::peZ, Hilt
Institute of Statistics Mimeo Serieo No. 921
Ma.y, 1974··
1. Stable modules.
Definition and main properties
1.1 Subsets of an R-modu1e G verifying four axioms
Let
G be an R-module where
stable set SS
c
ab
E
S &b
E
II:
X
€
A =;> aX
1 :
2
X
€
A & Y € A ~=> X + Y E A
1 :
3
x
€
X &X
4:
S,
R , containing
=>
a
€
S.
A
,
then
-1
Let
S
be a
and verifying the
Finally let
A be a set of
G verifying
subsets of
1
is any commutative ring.
S of nonzero divisors of
prefix condition, i.e.
A ,V a
€
A=;:. {x}
€
S
€
€
A
.
F be any finite subset of
Let
.(:I
x€F
1
R
X
=
cf> =:;>:[
X, Z
€
F wi th
X (1 Z·
= cf>
•
4 is referred to as the "finite intersection property;', or briefly f. Lp.
1.2 Examples
1.2.1
Examples in ordered modules
These exar.~les are recalled from [1] without proofs.
(we suppose
Ag
ordered with
P
=0
V g
€
G ~~ A = 0).
examples
ring.
a
Q or in
a
-Q
€
S,
G is
R is an ordered ring with
Q as
G is an ordered S··module if QP c P. For these
R will be the group of units of
So, for every
is in
If the subjacent group of
as set of positives and if
set of positives, we say that
G is an R-module
aP
equals
R which will be a totally ordered
P
or
-P
according to the fact that
-2-
1.2.1.1 The set A of intervals .with the form [gl,gli] , l ..<Xl,g] , [g, + <Xl] ,
verifies
= 1,2,3,4.
1. , i
~
Where, by definition,
(gl + p) (_, (gil _ p) •
1.2.1.2.
X
€
1.2.2.
A iff
V x,y,z
G , x
€
$
Y
z,
$
and
x,z
€
X , then
y
€
X .
Example in modules over Prufer domains
A
where
If G is totally ordered. A can be the set of all intervals in G:
is the set of all cosets of all submodules of
is a finitely generated ideal of a Pruf~r
h
G ~ aG
We suppose
X
=x
+ hG
12 :
X
=x
+ hG
= ax
Y =y +
aX
(h + h')G :;) hG , h 'G
1 : V
3
1 :
4
x
€
G ,
bJ
+ ahG
= ax
= x + {a}
€
Here
hG,
S
= R\{c
a ~ 0
+ h'G
h'G ) X + Y
and thus
R.
domain
is an injective mapping for every
II:
since
G with the form
=x
+ y + (h + h')G,
(h + hi) G
= hG
+ h IG
A •
This is a consequence of Theorem 6 of [2] of which we state a corollaJ.'
that we need here.
Consider a distributive
seetion ~oj' subgroups of an additive group
lattice~under
G.
addition and inter-
If A is the set of aU eosets
of these subgroups., it has the i.l.p.
So, all we have to verify is t.hat the submodule s h G , where
all finitely generated ideals of
R form a distributive lattice.
h runs over
We solve
this exercise in order that this paper be self contained.
Since every ideal
l
in a Priifer ring has an inverse
(a) , principal ideal, one has:
(1)
i hG
=th IG
~'>
ahG
= ah IG'·".>
hG
= h IG
I,
that is,a
=
-3and
(2 )
.ehG:::>.th 1 G ==;> hG : :> h' G .
We have obviously
(3)
hGnh 'G : :> {h ('h')G ,
and we must prove equality of the two members of (3).
there exist ideals
(4)
.e
and
k, in
S
Since
h:::> h (, h' ,
such that
&=h(1h!
=h
(I
hh 1
= (h
kh'
h'
But
(, 11') (h + h')
= kh'h
+ thh'
= (k
= {h
(I hi) h + {h (I h') h '
+ .e)hh' .
Thus
(5)
k + l
=R
.
Nm?
(6)
l{hG
(,
h'G)
(7)
k{hG
(,
h'G)
= L"lG (,
= khG (-1
£bIG c{h (, h')G
kh'G c{h ('h' )G
By summing (6) and (7) and applying (5) :
(8)
l1G (, h'G c{h
("I
h')G .
From (3) and (8), we have equality of both sides.
We have provei that the considered modules form a
submodules of G.
as we
YUlOW
s~blnttic00f
the lattice of
We have now to check out the law of distributivity, which,
it, holds for the lattice of finitely generated ideals of a Prufe'r
ring (See [3] for a proof without the use of maximal ideals).
-4-
= hG (")(h' + h")G
h':))G = (h (") h r + h (")
hG(")(h'G + hliG)
= (h
(") (h' +
h 7l )G
= (h (") h') G + (h (") h n ) G = hG (") h' G + hG (") h"G .
1.3 Definition and main properties of stable modules
1.3.1
Totally S-stable-matrix
We suppose that the R·-module
G verifies
cf>
:
a
is an injective mapping for every
G a subgroup of RE , E
G
+
aG
a
E
S.
This is true, for example for
being any set, since
have supposed . all
vIe
a E S
were non-zero divisors.
Definition.
An
n x m matrix
S-stable if every non-zero subdeterminant of
The G--rank of any matrix
the largest integer
r
for which one
1.3.2.
such that there is no
B.
E
(10)
,,
f,rn
g
in
Rm .
and range
R
m
R
is the largest
may be considered the module of
So,
we
call support
m
= { j/j
A
A is not zero.
R :
s(x)
G annihilating all
S··stab1e··modul es
"
A span a submodule of
flillctions with domain [l,m)
x
S
Here, clearly, the rank of
r x r subdeterminant of
Primitive eleraents ill
The rows of
is in
A
B is·,by McCoy's definition ,slightly extend.ed in [2],
:r: x r subdeterminants of
r
R is called totally
A with entries in
E
[l,m] ,
Sj ~
O} .
s(x)
of
-5S of minimal supports:
We consider the finite set
(11)
s(x)
1.3.2.1.
S
€
for
x
M iff V Y
€
s(x)
&
y ~ O)~"> s(y) = s(x) •
M.: of J!Tl is S-sta1Jle when for every minimal support s (x) of S
there exists a
u
M with every non-zero aomponant in
€
We call such
a primitive of
u
is the group of units of
S
c
Definition We first consider submodules of Rm •
A submodule
Since
H,(s(y)
€
M.
S
such that
A special case is the one where
= s(
S
Tole caZZed those modules "unimodular modules· i •
R.
is a stable set of non-zero divisors, it is clear that if
ring of fractions of
R with denominators in
unimodular over
But the converse is not true, that is if
R.
s(u)
R is the
S , then a S-stable module is
H is unimodular
in
Rm , the
-1
(thus 1) by definition, but it does not necessarily contain all
is not necessarily S-stable in
R~module
S
~~its
contai
of R.
The following theorem has been proved for unimodular module [1] , that is for
S the group of units of
R •
1.3.2.2. Theorem 1
Ivl
The orthogonal module
of M in the dual of Jfl
is itself S-stable.
Before g:>ing to the proof, let us observe on an example what is the meaning
of the theorem.
Take
R
~m.
direct factor of
is a direct factor)
= 7L m,
~
~
0 , then
and only one primitive with greatest
divise
v.u - u.v
1
Also
1
(For if
1.
=0
and
v.1
u
i
divises
= Vi
.)
u
and
u.v.
1
J
v
S.
cow~on
M
divisor of all
are such primitives and
for every
V
j
~
0
and must
Hence there is a finite number of
primitives and their non-·zero componants generate with
a stable set
M be a
For each minimal support in M we have one (since
non-zero componants equal to
u.
the ring of rational integers, and let
···1
and their divisors
Now the theorem teaches us that for every minimal
s~
-6support in
i
M
componants in
,
there exists a
If
S .
s(u)
u
E
= s(v)
i
M
having this support with all non-zero
L (v
and
1:s;3:S;m
some element in
1.3.2.3.
S , and is consequently in
)
= '71..,
then
j
v
divides
j
S , by the prefix condition.
Numerical example
Primitives in
Primitives in
1,1
(0,6,7,--5)
i
M
( 0 ,13, - 9,3 )
(3,0,2,6)
(52,0,-15,-21)
( ··21 ,12,0, --52)
(12,--5,0,-6)
(15,36,52,0)
(4,7,-6,0)
We see that every integer in the first array divises a product in the second
array and con7er.se'ly;
1.3.2.4.
Proof of Theorem 1.
It is sufficient to prove Theorem 1 in the case where
S
S
R with denominators in
is not a group let
S
alB with
The set of elements
S in R. If
of
R be the ring of fractions of
M
M is S-stable then
is S-stable.
M"i n
Now
exists a non..zero divisor
a
<x,y>
=0 ,
thus
mi!i.imal support in
y
in
-i
M
x
r-r- ,
correspond an
E
M"i
M = RM
m
R
c
Hi
such that
a <x,y>
and
a,~ E
S
is S-stable and the orthogonal
and if
ay
E
x E
rl
and
Mt
y EM, there
H , so
= <x,ay> = 0
!'l = ~?-
•
in
For, if
form a multiplicative group
t l Rm ;
Hi = fJl
it has also minimal support in
ay
is a group.
M
with
s(y)
= s(ay)
If
x
has
~ since to every
Thus there exists a
-7u
W'
€
of
a~~
= s(x)
s(u)
~f
denominatprs
numerator of the
u
, with all non·-zero componants in
zero componants in
• j
{I,
€
is in
.m}
by the prefix condition.
,
j
~
S.
S.
The product
a
S as well as every
Hence
au
€
~I1.l
1 ..
has all non-
This will achieve the proof.
Notations and definitions.
n x m matrix has its rows labeled by the ordered set
An
labeled by the ordered set
[I,m]
labeled by the ordered subset
the ordered subset
with
J
A
J
of
1,2,3,4.5.6.
ccntain~_ng
-1 .
Property 1.
Let A be an n
square ,,:ub;:::atrix of
Det BJ'uK
?roperty 2.
= Det
If AJ
B
and whose columns are labeled by
J
[l.m] .
c
~_~~E$rou£ of
n
m matrix
x
For the following
the
~noup
is a non-singuZar n
= (AJ)...lA
x
of units of
R
m submatrix of A"
M ~ meets J
non-zero
(AJrlDet AJ'UK.
J'
Since
c
J
such that
-1
€
K .
BL ~s
any
If
Det BJ'UK
S , Det B~
€
= tDet
B~
s.
M is spanned by the rows of an echeZon
the identity matrix" then any non-empty support s(x)"
and any row of B has minimaZ support.
y EM.
Let b be a row of B and s(y)
By hypothesis there exists a primitive
c S (y) c S (b)
a zero divisor
is an echelon matrix.
B • there exists a
s(x) meets J , obviously.
s (u)
[l,n]
m matrix> totaZZy S-stabZe (i.e. every non-zero
If an S-stabZe-moduZe
matr1:x B ~ with BJ
€
x
colWlli~
(A J )-lA is a totaZZy S-stabZe echeZon matrix.
(AJ)-lAJ = I ,so
:x:
will be
S
and its
is the submatrix of A whose rows are
An echeZon matrix A is an
[I,m]
suhdeterminant is in B).
But
of
the i~entity matrix for some
Properties
then
I
Ai
[l,n]
s(u)
Necessarily
= s(b)
.
u
= ab
for some
a
€
u
f0~
c
s(b)
E
M with
S and since
some
a
is not
-8If
Property 3.
x
has rrrinimaZ SUpPOy·t in an S-stohZe moduZe
of its aonponants is in
S, every non-zero aomponant of x
By definition there exists a . U E
1;f j
E
[I,m] .
1;f j
E
[I,m] •
Property 4.
Let
Let
J
=I
B
U
€
Now
~
~(u)
-1
x u u
s s
= s(x)
=0
u
and
x
, and
j
s.
is in
E S U
j
= x s u-lU
s ~j
{oJ ,
E
S U {oJ,
B be an eaheZon matrix with a r;iv:.ximum nurriber of rows, aU
M.
Then these rows form a basis of M.
x be a.ny element in
and let
[l,m]/J
x
XES , then
s
of them in a S-stohZe moduZe
Let
M with
M and if one
s(tB - x)
M with
for
matrix with
u.
?roperty 5.
The matrix
is empty.
s-::';c.~
j
E
M.
J
If
= xj
tB
s(tB - x)
,
=
c J
For, if it were not, there would exist a
J
RepL,.c:n 6 f:'le-:;y row b of B by b - b.u ,
J
v
This is impossible.
B of Property 4 is totally S-stohZe.
By the argument of proof of Property 1, it suffices to show that every square
non-singular
has determinc~t in
S .
Let
DetB
K
B has the largest possible numbers of columns in
where
echelon
~6trix
J\K ~ ~
the i
BJ
and by
and let
th
j
unit vector.
€
4,2
J\K.
If
Ai
3 , every entry of A is in
and
Now there must be a
~
0
j
column by A , obtaining a matrix
{j} U
K\{k}
Then
BJ
k
E
K\J
A~J. =
S
Suppose
and let
1;fj
E J ( \ Ie
T with
AL has determinant in
this is impossible.
This means
k
A
= t.1
we replace, in the identity matrix, the i
Det T E S.
S.
But
Denote by
AL
=
AJ
is singular and since
AJ
th
L the set
(BK)-lBL
B
0
E S,
A is an
L has determinant in S .,hich is impossible since BL has
J
K
Hence A~ = 0 for every j E J/K
But
column that B in B
J.
shows
K
This
more
= (BKJ-IB J
,
-9Property 6.
S~stabZe
If! spanned by the rows of a totaUy
If M is a submoduZe of
matM A", every,· x
in M with minimaZ s,upport is homothetic to a
,row of some echeZon._ matri,x B u.,hose rows span !1.
corresponds such' a B with ~. s (x)
n J
= Jj}", wheI'e.
To every
B'!
X.;;l!
J
0
is. the identity matrix.
M is an'S-stabZe moduZe.
Consider an
n
x
M with minimal support
€
m with rank
x
minimum.
Let
n
k
J n
€
B~
Now we show that
J
Det A
and let
:=
~
j
;;l! 0
T118i:J.
€
s(x)
F' .-
= {j }
U
J/{k.r)
~
Card(K (, s(x»
s(B.)
].
c
'if j
{I, ... ,m} , u
J
€
and
(A<T )-lAk
~
('\
= £.i ,
the
If
B
j
i
;;l!
s(x) ) is
.th
~
unit vector.
we could. bui.ld
0
in the unit ~~trix.
to Bj
£..
Card(J
A is
If
BK -- (AJ)--l,.K
would then b,e.~. ir..vcr'tiblc and so would be
rio
with
s(x)
that
= [l,mlh,(x)
up an invertible matri~ T by changing
K
Now we may suppose
f;1)~~,~ h
1<-
s(x)
o 'if
s(x)
x
< Card (J (, s(x»
= aB . .
].
= B.
Finally since
In conclusion, if
L
M
J
n
belongs to
s(u)
whose rows span
•
M.
L
By Property 6, H
= s(x)
= {j}
s(x)
M is S-stable, there
[I,A]
totally S-stable, span
B.].
is a primitive with
].
may have been chosen in order that
matrix, let us say,
which is iL1possible.
Then
M and
•
B~~
€
S ,
We observe that
for any
j
with
Xj;;l!O.
exists a totally S-stable
Then [AT,_I] , which is itself
is then S-stable) which was
to be proved.
1.3.2.5. Sone consequences
Remark.
of
It is easy to verify that if a S-stable module
Rm (S
being now a general stable semi-group of non-zero divisors with the
prefix property) then
if
(~)L
M is a direct factor
= 1'1
,then
(rl)L
= M.
Also if R is a principal ideal ring and
M is a direct factor of Rm •
basis, it may not have a basis of primitives.
Notice that if
We also have
111 has a
-10-
If a module M is spanned by the rows of an n x m "matrix A
,9orollary 1.
with rank n, a sufficient oondition that it be S-stahZe is that
for every
n
(AJ)-lA
has all its entries in
Now by Property 6 every
x
S
J
Det A
S
M = RM
the group generated by
by the rows of an
every
non-sin~~Zar
S
with M,R defined as in 1.3.2.4 and., denote again by
S in
R.
Then
M is S-stcbl9 iff M is spanned
n xm matrix A with rank
n
and with
Det AJ
€
-
S
for
n x m
M is stable such a basis always exists.
the converse is true.
Finally we have seen in 1.3.2.4 that
"~orollary 3.
S
D:;t A,J , vle:,jnd a
By Properties 4,5, if
iff M = 1&1
in
with minimal support will be found proportional to
primitive with compor-ants in
Let
is invertible
and, moreover, it totally S-stable.
Multiplying this row by
Corollary 2.
S .,
€
m matrix AJ •
x
With the notations of 1.3.2.4,
Then
Det AJ
By Corollary 1
M is S-stable
is S-stable.
To every S-stabZe moduZe
M of Zm, Where
S is the set of odd
integers oorresponds a unimoduZar moduZe.
Let the rmvs of
J
Det A
nations
= 0) .
J
tA
A
"te
Suppose
such that
a basis of :,1
J
Det A
.
is even .
J
s(tA )
J
Let us first show that
J
(Det A
€
(2)
Then among all possible linear combi-
reduces to one element there must be one such
L
(t i) =7l.: Bi;t then tA
l:$i:$n
is a primitive, by Property 2 and since the greatest common divisor of its
that the only componant of
tA
is even with
componants divides that one of any other primitive with the same support,
would not be S-stable.
;::-::.;;..>
Now consider
A modulo
4.
For some invertible
M
-11-
A~B=(AJ)-lA has all its subdeterminants l~-l or zero mod 4.
that matrix
or
-1
B as a matrix over the integers.
mod 4.
Consider now
Every subdeterminant is
But it is known that an echelon matrix with entries
cannot have a subdeterminant greater than
0,1
0,1,-1
without having another equal to
1
±2. * Hence the obtained matrix is totally unimodular and its rows span a uni-
modular Z-module.
1.3.2. A more general definition for
Definit:J.on.
A submodu1e of
m
G
modules
st~b1e
, w~ere
G, defined in 1.3.1, is an R-module,
is Sustab1e if it is tile smallest submodule of
for every
g
in
G and for every
Such a module is denoted by
(M
x
G!'1
containing xg
in as-stable 8ubmodule
G).
M of Rill.
(M. G)
The primitives of
= (x j g )l::::;j::::;m
will be, by
definition the primitives of M.
By 1.3 (9) , if u
in
is a primitive of
M, s(u)
= s(ug)
for every non-zero
g
G.
~epresentatives.
A set of representatives is a finite set of primitives of
(M • G) , one for every minimal support.
Property 7.
ax
= ug
Let
over
R.
minimum.
J
If x
has minimal support in
for a primitive
A
~e
u and some
g
E
(M. GJ
~:!I
S
E:
such that
G .
an echelon matrix, totally S-stab1e and whose rows
We choose it in order that
J
A
=I
for a set
We know that there exists an echelon matrix
= [l,m]/J
a
, whose rows span
Mr
over
S.
J
span
-M
with
B with
J
B
=I
,
x being orthogonal to each row
*This p~operty is not verified if all entries are not in {O,l,-l}. A
totally unimodular matrix in which 4 is added to any entry is a counterexa:m.ple.
(See;for example [4].)
-12-
of
B , it is impossible that
s(x) c
J.
Let
j
argument used for the proof of Property 6, the row
has its support contained in
= s(x)
s(A )
i
x - AiX
u
.
Now
= O.
j
= aA.1
x - AiX
Let
of
Then to eve1'1,j
t3x
= L
(M.G)
ugu'
of
A with
J.
has its support contained in
j
By the
At
=1
€
S , we find
ax
Thus
= ug
with
G .
€
Property 8.
(M. C).
Ai
J.
and it follows from hypothesis that
Multiplying by suitable a
M and g
€
s(x)
r-,
s(x)
€
Su
be an S-stc:bZe module and
x
(M. G)
€
G , ~
€
=0
U a set of representatives
corresponds a
t3
€
S
s(u) ~ s(x) , V u
if
such that
€
U .
U€U
Write
y
=Sx
L
-
u~, choosing
S in
S
and such that
s(y)
s(y)
be minimal.
=~
We show s(y)
Then, by Property 7, there exists a
~ ~.
s(vj y - VYj)
Then
~ ~ 0
and
U€U
v
only if
that is,
U with
€
is properly contained in
y
=0
s(v)
s(y)
s(u)
c
s(x
Suppose
s(y) .
c
Le
But this is
impossible, since
L
v Sx j
13 I
with
Let
g'
Sand
€
uv g
€
U
- \7
j U
U€u
G, V
K be any subset of
U €
restriction of
all
x
in
r1
we suppose now
Property 9.
K
K and
xj
llK at
with
S
x
= xj
•
V
x
j
~>
E
=0
, V j
€
Lug'
UEU
,
U
K= [l,m]\K.
The projection
UK of
h
th e componant s o
f xK are
xK ,were
K .
M, submodule of
j
-
U .
K .
We denote by the same symbol
Gm .
11( the
KerllK' is then the submodule of
We denote by
~K
the operator
KerllK ;
is a group.
For every subset K of
M is S-stC'J; le.
= SiX
[l,m] , and
",m
,
~.onto <G·K,
1S t h e I'1near mapp1ng
labeled in
j
If
[l"m]
"~K(M. G)
is an S-stable module
U is a set of representat-i,ves of M" then
U
n ~KJ1
i.
is
-13--
a set of representatives of 6 (M.G) .
K
We have
Now by Property 8, since
U
the converse inclusion is verified.
~ ,
€
Remark.
If
M is a S-stable module of
S-stable module.
for every K
Rm , it is clear that
[l"m]
c
ax
€
iff M is a dil'ect factor of
IfI. Every x in M
R is a principal ideal ring, it is true thst
M , ex
onto
€
S ~-:> X~E M.
M, P (x)
€
is also a
6J11 is a direct factor of M
a linear cOnWination of primitives with supports contained in
m
a direct factor of R •
If
6~~
Now we have
If R is a principal ideal ring"
Property 10.
The last assertion follows.
For, if
M , (l-p)(x).
MeN = Rm and
N.
€
ex is not a zero divisor, x =p(x.)
€
p
s(x)
if M is
M is a direct factor if
is a projector of
° = (1 - p)(ax) = ex(l-P)(x},
But
is
•
m
R
Since
M
Now the condition is sufficient since it means that the invariant factors of
M are all equal to (1)
Now if
ax
€
6 (1
1
and consequently,
and this means
, since s(x)
6~~
c
H is a direct factor.
s(exx) ,
ex
being a non-zero divisor,
x
€ 6~
is a direct factor.
Before proving the last assertion, we first observe that it may not be true if
M "rere not a direct c\factor •
Example:..
(1,1,5)
(1~·-1,3)
If
M is the module over
are scalar multiples of
71.
spanned by those two elements, its primitives
(2,0,8),(0,2,2)
primitives span a proper submodule of
M.
and
(2,-8,0) , then obviously the
-14We prove the last assertion by recurrence on the cardina.ll..tyy of
the rows of
form a 'basis of
A
6 s (x)M we built up a set
To.
every
(AJ)*A
=B
each minimal support, as follows.
J
(A )*
adjoint
and the rows of
the union of the rows of such
Bls.
B is the determinant of some
1/
of some element in
Now let
'be the subset of
of
and that every
Yj
s
= L
y
U€U
y
Now
u
J
DetA
appears as a componant
Denote by
U with
=0
and so
ctuU ,and
ujx
s(y)
j€s{x)
(13)
j
Y the
is properly contained in
= L
be the
Let
U€U
greatest common divisor of the '..lj ,
13 u
will be
U
Notice that every componant of a row of a
U E
6 (x)I·I.
By recurrence,
= L
u € U for
corresponds an
have minimal supports.
U€U
Sj x
J
A
matrix
x k
one
U ,
If
U •
element ujx - xju
s(x) .
k
s (x) .
running over
U
=R
Then
since, as we have just seen,
m
J
{Det A ) and 6 (X)M is a direct factor of R • Hence,
>:
( 13' ) =
L
S
j€S\x)
j
CardJ=k
by suitable linear combination of the s.x , j € s(x) , we get the res~ut.
J
Let R be a Dadekind domain and let M be a
Proposition 1.
~.
Then evel"y
contained in
x
by
M' .
p.
in
M' , u
a maximal ideal
p
R, M , M'
p
P
Let
at
in M is a linear combination of primi tives wi th supports
s(x) .
We denot.e by
6 M
K
direct factor oj'
p
p
R.
\ve make
be the set of maximal ideals of
minimal support in
M'
can write for every
M' p
Now
p € P
K =
s (x)
and we denote
will denote respectively the localization of
has minimal support in
a primitive in
0:':'
for every
R
p
M'
p
R.
If
u
is a primitive
and thus conversely, if
ct f p
such that
R,M,M!
u(p)
has
ctu(p) € M' , ctu(p)
is
is a principal ideal ring and by Property 10 we
-15(12)
where
a(p)
u
~ 0 implies u(p) is in M'
p
exists a yep) ~ p
~
a(p)
u
such that
y(p)u(p)
and has minimal support.
is a primitive of
0 , and we know that we can find a finite set
L
M'
E of
Then there
for every
PEP
such that
=1
y(p)o(p)
pEE
with suitable o(p)
(14)
E p..
x
where all
y(p)u(p)
Cc~sequently (12)
Rm
and (13) will entail in
= L
pEE
are primitives in
M'
with supports in
sex) •
1.4 The particular case of unimodular modules
Unimodular modules defined ih" [1]
appear as a particular case for S-stable
modules, that is the case where
is the whole group of units of
S
R.
A con-
versation with L. Geissinger during seminar lectures on unimodular modules at th
University of North Carolina lead us to do the following characterization of
unimodular R-modules, for
R a domain.
Let
I1<
be the mapping
x-> (Xj)jEK' 'IXE M.
Theorem 2.
$
~K
c
A
submodule M of Rm ",
[l,m] "' the annihilator A(x)
R
a donnin:; is l.mimodular iff for every
of every
x
€
RK/ni~ is either {o}
or R •
J , there exists a u EM,
We have to prove that for every minimal support
with
s(u)
=J
such that
u
j
is a unit of
R ,V j
E
J.
The hypothesis is
-16that "K
[l.m]
c
exist some
y
if
in
x (1.0,
j
~u
J
= {j} u
= (xj,O,
J
a •
;t
then
, j
E:
K
R is
J , with
E:
IT YI
K
is in
u
=1
, hence
II~
E:
(1.0,
>
6 M
J
, that is. there
Indeed. if this occurs
IIIr' •
=J
s(x)
, a v is found in t.}1 with v k = 1
Vj~
IIrc
and then contains
{oJ
,0)
but such a
M
• U E
IIr!1 • a
is not
V
K
,0)
E:
M whose projection in
the annihilator of
Now, let
aIIrc
1
.
, minimal support.
...
,0)
llKM , (1,0,
E:
Now if
k
Since
s(v)
Since
...
=
,0)
any other element in
is
= s(u) = J
, vju - v
=C
is a unit as i t was to be shown.
~
Conversely, if M is unimodular, it is spanned by the rows of a totally
unimodular matrix.
Then
II}1
Now II~ is. spanned,:by the rOvlS of an echelon matrix
Card K
matrix;
II~ , X E:
ax
E:
ax
= ay
+ az
~
Then obviously,
1 .
J
R .
We write
thus
az
v
= O.
=y
Since
IIj~ e RJ\K
+ Z , Y
E:
A submoduZe
cf>
;t
K
Then
y
c
[l,m] "
Let
s{x)
0 N
j
= RK
E:
Z E:
R is a domain,
z
•
the unit
Let
RJ\K x folK •
=0
if
a;t
0 , x
=
R is a domain).
If .
factor of
support.
We make
Let us denote
ITE!1.
=y
+
Z
•
Y
E:
ITrcM ,
K
= {j } u J ,
(1,0, ... ,0)
,
P.. j
= RJ
x folK
K
A
M of lrfl , R a domain, is unimodular, 1:ff for every
llJ!1 is a direct
= J be a minimal
,0)
(x ,0,
with
A
ITJM ,
Another characterization is the following (again
Theorem 3.
J
R , " J c [I,m] .
is a unimodular module of
z E: N
when
by lj .
j
E:
Since
J •
-17Since
xjl j
is in
and
llKM
xj
~
0
is not a zero divisor, lj
=y
ll~
E
The proof now follows with the same argument as for the previous theorem.
2. The Theorem of Compatibility
2.1 Propertl 11.
Let
M an S·-s·table module with a n-bas-is.
•
.,
D
•
• •
cannon'l.oatll-'/..f;wmotapn-z.sm
0
f .Rn onto
h' d
an .,t-et
the
Denote by
t(a) _- .
,,-la.
'" 1M
. , .....v a ~!~
•
Denote?
)
by
X
U a set of representatives of 11 and by
E
rfl
[l"m] " an
verifies
I
a.x
l:>;j:>;m J j
t ~ a )b • , b
= L
l:>;i:>;n
J.
n • "if
a E
iff
G
E
J.
LUX
=
j j
(1)
If
K any subset of
l:>;j:>;m
(1)
is verified, we have, by Property 8, if
(2)
f3
L
l:>;j:>;m
ajx j
=
But, applying
II
a
L
b.
-1
to (2)
,
u l:>;j:>;m
J.
UEUM~
L
uEun~r1
L
l:>;i:>;n
St(a) =
L
Hence
=
a t~u)
U J.
ue:Un~r!'1
~KM
'
uau
lUX.
j J
L
ue:UMKM
=
(3)
=
Sa
a E
L
ue:un~K!-1
=<
a t(u)
~
b •
a
L t~u)b.
u l:>;i:>;n
J.
J.
L
UEUMKM
a t(u) >
'ij
-18-·
and since
13
G + 13G
is injective,
L
l:5i:5n
2.2 First statement and proof
Let
\
( r jJj
€
1.1) relative to a stable set S of R.
Finally let
be an
S-stable module
A
be a family of subsets belonging to
[I,m]
n
Let also
{b }
i
m matrix with rank
x
M with representative set
U.
n
€
A, i
(defined in
€
[l,n]
whose rows span a
We have
Theorem 4. 'The lineal" system
I
(4)
l~j:5m
has a soLution
a~x
1.
(x j )l:5j:5m
= b. , i = 1, ... ,n •
j
1.
=x
L
= (r j )l:5j:5m
r
€
tiu)b
l:5i:5n
i
L
€
l:5j:5m
iff for every u = t(u)A in U
u r
j j
We prove that if (5) entails existence of an
'if
c
€
6KM' l~j:5m
L c x~K)
j J
= I
l:5i:5n
t~c)b
1.
i
x(K)
€
r
such that
(see Property 11 for notations) for
every Kc[l ,m] with Card K<r, then (5 )also entails existence of such an x
any K'c[l,m];.rith Card K'
exists an
K'
=K
in
K'
x
€
r
U
as claimed.
= Ul
Let
This will prove that for
s
be any element in
K
K'
= [I,m]
U U
2
where
U
2
fOJ"
there
and write
U be the subset of all representatives of
Let
u {s}
= r.
(K' )
M with support
is the set of all representatives of
M
with
-19supports in
K.
By Property 11, we have to show that there exists a
such that for every
x(K')€ r
u € U ;
1
Yill be constructed as follows. We show that r
may be reduced to one
s
i
point {x(K )} = r' and that the hypothesis (5) is still verified for every
s
s
i
U€U(K ) with that new set of f is. We will then consider another s in K'
X(K
)
j
r j 's
and change aga.in the
set of
f
u's in
j
(8)
~
, j
•
U
l
Then , u' s
are in
f
j
is
that we finally obtain is the
We first observe that whatever
for which (5) is then verified.
f
it will be, the new
old
The set of
S
will verify (5) for every
s , since
u € U
2
together with the
We have thus only to consider the
Let us define some notations,
= a.a
-
S.
i
a constant, for every
Notice that the
u € U .
l
eu 's
We have
I
I
lS;jS;m
l~iS;n
vu
ujr j + afs '
€
U1 .
j~s
which means that
ar s
meets
I
IsiS;n
(u' )
I
t.1.
u?j , for every
lS;uS;m
j~s
If we could prove that
'I
E= ( 1 (
U€U
I
I
lS;i::5:n
I
l~jS;m
uir.) ~
j
J
j~s
by the finite intersection property, we would have
</>
,
u
and
a
-20·-
(10)
and,
a x
a.
(K' )
E a.r
n E
s
s
being a non zero divisor verifying 1.2 (9),
r'
s
= {X(K')}
s
would meet
our requirement.
Now, by recurrence, we know that there exists a
x~K)
x(K) ,
verifying
x~K) for u 4 K are irrelevant.
Notice that the componants
(K)
x
s
Let us consid.:=r
J
in
verifying
SG
v )K) =
s s
(12)
for some
v
Now for any
picked up in
U .
1
u EU ,
VU-UVE
Li14,
s
s
UK
and, by (11), from Property (11)
L
(14)
jEK
(v
x(K)
U v )
s j
U
s j
j
=
(v
t~U)
S 1
1'Thich means that
v
v
S
(16)
is in
U
x(K)
s s
S
=v
L
S
jEK
u.x(K) - v
J j
L t~v)b . .
S l~i~n
1
1
and we may conclude that
S
x(K)
U
S
S
E
L
jEK
u.r .
J j
L
l~i~n
t~U)b.
~
l'
V
U E
U
l'
E r
j
,
j €
K,
-21-
(15) by Su ,we obtain (9).
Multiplying both members of
(For clarity, we point out that
(K)
(Ki)
X
of (10) is not necessarily the same as
S
X(K)
which allowed the proof of (9) is not required to be in
s
'
(K)
since it is even not required to be in G but in SG. However, ax
s
, since
xs
rs
is in
G.)
2.3 A more general statement
In 2.2, we have considered
basis.
a
problem concerning an S-stable module
An S-stable module has not necessarily a basis since a submodule of
for
R a donuin. has not necessarily a basis.
Let
R(M)
be the se+ of almost zero
v
the submodule of
Property
onto
M with a
t - (t a ) aEM' t a E R ' V a E ,.
1'1
-
consisting of all
t
11, there is still a R-isoIDorphism
N
•
l t aa = 0
aEM
of the quotient module
is
As for
such that
~
RID
R(M) IN
M:
Now, if
b
= (b a) aEM
H
E G ,and
l
aEM
t b
a a
=0
,
for every
tEN , then
l t(C)b
aEM a
a
(18)
is also defined for every
t(c)
in
R(M)/N
The arguments of proof of Propert,
11 and of Theorem 4 hold for proving
;1
Theorem 5.
Let M be an S-stab Ze moduZe of
here above.
Then if r j E A ,V j E (l"m] , i f {ba} EA. V a EM,
exists an x
=
(xi l~j~m
E:
(r j) 15,js.m
I.Jith
and N and
b
defined as
t~J.el'e
-22-
L
lSjSm
iff for every
U
= L
t(u)a
ad1
Let
A.
Let
r.
J
€
fit
V a € M
in a finite set U of representatives of M,
L
Theorem 6.
a
a
t(u)b
a€M a
a
(20)
=b
ajx j
L
€
u.r
lSjSm
.
J j
be an S-stdble module spanned by the rows of a n x m matrix
A ~ for all j , and {g}
E
A , Vg
€
G.
Let U be a set of
representatives of M and T1 any set of t's for whiah
U = {u I tA = U , t E 'T 1 } , let b € ef ; then there exists a finite set
T2
c
~ = ~ suah that
(21)
Ax
=b
, X
€
r
= (r)
[
]
j j€ I,m
has a solution iff
(22)
tA
=
L
idl,n]
t . b.
1
1
=0
,V t
€
T
2
and
Since the rows of
A form a set of generators of
11, we may replace (20) in
the statement of Theorem 5 by
(24)
(a € A means a runs over the set of rows of
exists a finite set
T
2
c
A
R spanning over
A).
We first
R the module
sho~
L
l
that there
of
tis
verifyir
-23-
=0
t a
a
•
If this is proved, we will have
(L
(26)
=0
t b
a a
aE: A
~ Vt
€
since the denominators in
will have
L
adI
t b
a a
b
c
=0
L
=0V
L ).<",=,:> . (
t b
l
aE:A a a
t
€
T )
2
R are non zero divisors (1.324).
for every
=
ct
b
a a
Furthermore, we
N defined in Th80rem 5, ~ith
t € N,
,whenever
ct
a
a
=c
and the sufficient condition for Theorem 6 will follow.
By corollary 2 in #1, we know that
matrix
B with
J
B
.
=I
M = RM
Since every support
.
L t a = 0 iff a€A
l t a a.J = 0 , j € J
a€A a
n
the columns of A in R with indexes in
iff
t € L~
Then
~ of ~ onto ~
an isomorphism.
Then
Ll = L~.
k
p
is spanned by the rows of an echelon
Since
meets
Let
L
2
J .
We have just seen that
J ,
be the R-module spanned by
t E: L
l
J
B = I , there exists a R-homomorphism
the dimension of
= v -1o~
sex) , x € M\{O}
M whose restriction v
is a projector of
-·n
R
onto
RL
2
at
RL
which is a
direct factor:
(28)
But then
::n
R .
is the projector of
2
::n
R
onto
is
-24the cannonical basis of I(l
met in
Finally if
{g(li)}i€[l,n] , we make
Remark.
is the j
We notice that
th
componant of
T2
T2
a
is the product of the denominators
= {aq(li)}i€[l,n]
, and (25) is proved.
may be actually constructed:
q(li)
3. Consequences of the The0rem of Compatibility.
'Theorem 5 is the theorem of compatibility in its most convenient form for deducing known and new results.
3.1
Consequences through known results.
It generalizes
"Le
theoreme de compatibilite " of [1] from which were deduced
Ii
the usual theorems of consistancy of systems of linear inequalities,
Le theor(
Ii
de decomposition pour les groups reticules
N. Bourbaki [5], and results in
Graph Theory.
3.2 The generalized Lemma of Farkas
Let
ring
R be a linear ordered domain.
R
o~
Then if
all fractions with denominators in
ordered ring of fractions with denominators in
of this order to
vals in
units in
S
is any stable set in
S
is a subring of the linear
R\{o}
R makes it a linear ordered ring.
R verifies the axioms
S • We first prove the
R, the
Then the restriction
A of all
The set
1 ,1 ,1 ,1 4 of 1.1, since
1 2 3
S
inter~
is a set of
-25Let U be a set of representatives of the S-stab le module
3.2. 1 Lemma
-u
We denote
l
x=
by
u U
V.
Then for any
x
in M" we may wri te
au,a
u
U€V
u
The denominators of aU
au
possibly involved in expressions Uke
from a finite subset of S" the set of non zero aomponants of the
x ~ O.
We proof (1) for
e:~
{E: }l<j< '
j - -m
<oJ
M.
€
(1)
are take
u's.
The lemma will follow since for any set
{l,-l} ,
(2)
M onto an S-stable module.
defines a mapping of
We first show that there exists a u ~ 0 "
V " with
U €
is m1nimal,it is true, since then, there exists a
thus,
x
j
x~
-1
s
= Us-1xsuj
for all
is
u
y
~
x u
s
=0
0 , y
~
0
still valid for any
with
S €
k
s(v) c s(x) .
x
~
for
with
J
V
c
with
Then for any
s(x).
s(u)
j
€
S
It
s{XJ
= s(x)
(u) ,
We suppose now the property verified
Card s(x)
for which such an
x
=k
i
s(y)
,
and we prove that it
where
does exist.
ki
is the smalles
There exists a
v
in
Let
v.
S €
€
elements in
k
= {j I v j
>
O} .
is not empty since we would have taken
Now let
(4)
0
> 0 .
B
j > O.
with at most
J
J
s(u) , u
> 0 , consequently, u
integer larger than
V
with
u
s(u)
be such that
-v
in
V if it occured that
J
=~
-26It is clear that
y
and since
=x
~
s(y) c sex) , s(y)
exists with
s(u)
c
s(y)
c
- v
-1
s
x v
s
~
0 •
s(x)., Card s(y)
sex) .
Thus a
:s; k
u
~
0 , U
€
V
Knowing the existence of such a u , we prove
(1) by recurrence on Card sex) , x
O. We consider again s € s(u) with
~
(6)
Now
Y
=x
a v
v
d.
- u
-1
s
x u
s
~
0 ,
and, by recurrence,
(8)
y
= L
VEV
~O,.a
v
v
€
Then, (7) and (2) achieve the proof of (1).
R
d v
v
~
0 •
Now, to prove the last assertion,
observe by (7) that the denominators concerned are taken from the finite set of"
componants of
u's
in
V.
3.2.2 Notations and definitions
Let
Rn
A be a finite subset of
with rank r .
The elements of
columns of a matrix that we denote by the same letter
span a module
S.
M(A)
over
S whenever every factorization in
A and
Now the rows of
A
S that we may consider a S-stable module for suitable
For example we may build up
dependant rows in
A.
A form the
J
{AI}J€E
S as follows if we want a finitely generated
R is finite.
Let
AI
be a set of
the set of all non-singular
r
x
r
r
in-
submatri·
-27of ~.'
Denote
opposites while
by
E the set of all determinants of all
multiplicative group generated by
=E
make
S
1.1.
By multiplying AI
n R , and clearly
in the field
E
K of fractions of
R, we
S meets the requirements of its definition in
on the left by all adjoints of
obtain a set of rows of matr.ices which form a set
S-stable module
and their
E. E also contains -1. If E is now the
runs over
J
Ails
Ai's,
J
E , we
€
of representatives of the
U
M(A).
Definitions.
We denote by
Va> 0 , a
When
C
c
C any
€
n
R
R and
~
over
C+ C
A*
,the Polar C
A*
C
= {y I V
X
C,
c
of
€
aC
C ,
c
C being a subset of a R-module.
C
is by definition the cone
L
C
xiYi SO) Y
€
n
R }
lSiSn
F is a set of non zero elements in R, we denote by F-IR the set of all
If
fractions with denominators in
E
set verifying
R, that is any
= ER
F.
Given a subset
C of
n
R
and a subset
A
of the field
(10)
K of fractions of
Ix = L
{x
c€C
a c , a
c
c
€
R, we denote by
E , a
c
~
C(E)
the cone
O} .
We are now ready to state the generalizations of the Lemma of Farkas.
again the smallest field containing
K is
R .
Theorem 7.
n
Let A be a finite subset of R
that makes
M(A)
and let
S be any suitable set
an S-stable module, then there exists a finite subset
A*
A
n
S such that the po lar A of the cone A = RnA (K) has the form
"*
A 1
A* *
A*
A = ~ n B(P- R), B a finite se~. Moreover the polar (A)
of A
A
Moreover
A
.
=~
A
n AtF
-1
R) .
P of
is
A
-28Before proving the theorem, let us observe that if
A is the cone hull of
A, since
F-IR
=R
R is an ordered field,
and the theorem's meaning in that
particular case is that the polar of A is a finitely generated cone
polar is
A itself.
This is the well known lemma of Farkas.
We first define
each time we
u
~
F
0 •
first prove:
F.
When forming a set
have the choice between a
w~.ll
B whose
u
~
U of representatives of
0
and its ..,opposite, we take the
be the set of non zero componants of elements in
A= Rn n A(p-1R )
,.
n
R n A(K)
is the cone
Ax
independant columns of A, where
J
non-singular submatrix of A •
=b
r
, with
s(x)
c J ,
AI
J
A
E
~
,
x
~
0 ,
a set of r
AIJ
Let
is the rank of A .
Multiplying
We
U •
of aU eZements in Rn
whiah beZong to the cone hun of A in If.
,.
n
If b E R n A(K) , it is well known that there exists an x
such that, in matrix notations,
M(A) ,
be an
r xr
on the left by the adjoint of
~ we obtain a matrix B whose rows have minimal supports and may be replaced
J.
by elements in
C , with
J
C
U with corresponding supports.
an
r x r
~
~
~
t
(11)
L
j EJ
For every row
c
Now every row c
diagonal matrix.
t~C)A, , with t~c)
in
E
R , for all
CjX
j
t
= .<'<
L
1 --l-n
The matrix finally obtained is
c , for all
(c)
t,
1
i
in
.
b. •
~
C , the second member of (11) is in
one summand in the first member does not vanish.
which proves this first assertion.
C has the forn
Hence
xj
E
R and at most
F-~ , V
j E J
We now come back to the ordered ring
,
R
defined in the introduction of 3.2 and what we have just proved allows us in
particular to consider
A
cone hull of A over
R.
the cone of all elements in
n
R belonging to the
This will allow us to apply Theorem 6 in which
S
-29has to be replaced by the group
in
R.
= [a,m
rj
ment and where
The present
if
b
n
R
€
c R , for all
in the present state-
S
A is the set of all intervals
j.
U is still a set of representatives for
M in theorem 6.
will be the module
(12)
generated by the
S
b
A
€
RM(A)
Then, we may characterize
iff:!Ix <:: a , x
€
FfIl ,
Ax
=b
which
A by saying
.
The consequ.ence of Theorem 6 is
iff V t
€
l
-T
u T u Ti:
t b sa,
2
lSiSn i i
2
where
U €
Ti
c
T
l
is the subset of t(u)
B
= -T2
corresponding to non-positive primitive
U •
Let
u
T2 u T'l '
As in the proof of Theorem 6, we denote by
Ll
the R-module spanned by
Now t
€
T ' that is, the cone hull over R of -T u T .
2
2
2
au, a u S a
c = tA S a and by the lemma, c =
u
u
r
"* means
A
U€U
This means
tA
=
r
U€U
t
=
a t(u)A , or
u
r
t
r
-
a t(u) mod L ,
U
l
U€U
or
Consequently t
, with
belongs
U€U
to
n
~
R n B(F
-1:
sfnc~
R)
the denominators of the
a 's
u
belong to
F and a usa.
u
,,*
"
Then ACRnnE(F~lR) and the first assertion follaws.
By (13) ,V'e know that
for all
is a cone
t
in
"*
A
A ; hence
C such that
is the set of
(A"* ) *
-C n C
=A
•
= {a}
cone A has full rank iff its polar
b's
for which
r
lSiSn
t.b. s a ,
~ ~
A pointed cone (or proper cone)
It is immediately checked out that a
,,*
A
is a pointed cone.
We have the fol-
lowing stronger statement, in this case, which follows the one of Theorem 7.
-30If A is a pointed cone,
Theorem 8.
oomponants of u "s
F may be ohosen the set of aU non-zero
~ 0 , for aU
s
representatives for the S-stabte modute M(A) .
Since
over
in U verifying u .• u
J
j,3,
U a set of
A'
of generators
A is a finite set, we may find in it a subset
K; that is a set whose cone hull over
element in
A'
is not in the cone hull of others in
by successively deleting from
of others in
K contains
K
A over
A and such that any
A' .
A'
A elements which are positive linear combinations
Now, since
B has full rank, there is an isomorphisrr.
of
n
R
onto
in
A'
whose image under this isomorphism is
of
b'B
M(B) , the R-module spanned by the rows of B.
is minimal.
is simply obtaine
For, if not,
biB
b'B
~
o.
Consider a
The support
b'
s(b'B)
would be, by the lemma, a positive
K
of
~ 0
whose reciprocal images in
Rn
should be proportional to generators of A' ,
linear combination over
elements in
M(B)
with minimal supports
since any positive linear combination of more than one element in
image in
ab ' E A' , a > 0 , a
b'
Hence
b'B
E
€
K.
A' c A ,b'B
all denominators of ai'
0
lB
F.
B
= T1 U T2
U
-T2 ' that for
l'EA'
being in
S
~
= R\{O}
al,l'B, ai' ~ 0 ,
F.
u
0 ,
Now consider any lEAA = (AA*)*
M(B) , which is S-stable for
= l
We
with minimal support corresponds an
has as componants the componants of some
has its componants in
o~
~
But we know, since here
have, by the Lemma, applied to
(14)
has an
M(B) whose support is the union of at least two distinct supports.
thus also observe that to any b'B
every
A'
This means, by isomorphism,
U €
We
•
U
-31-
which shows that if A
t
= Rn
A
n A(K) , then
A*
€
Rn which belongs to A
non-negative
u's
in
M(A)
tive linear combination of
for elements in
A*
Hence
A
= Rn
tA
~
0
A
= Rn
1
A
n A(F- R) .
Now consider any
is by the lemma a linear combination of
with coefficients in' F~lR. This means
some elements in
t
is a posi-
Ti u T2 u -T 2 ' the coefficients
-1
Ti
being in F R,and, for elements in
A
1
n B(F- R) •
Before gjving an application of
Th~orem
7,we study the inclusions
T
A(R) c M(A ) n A(K)c Rn n A(K),
A
T
where
M(A)
A
A
is the R-module spanned by the columns of A •
We may have equality, for example 'Then the columns of
basis of
n
R •
A form the canonica
The first inclusion, may be a proper one, if
1 -1
;3
A =
112
(O,l)T
belongs to
T
A
M(A ) and to A(ID )
A
but not to
A(Z).
We will give the example of an infinite class of
A's
for which we have
equality in place of the first inclusion and a proper inclusion in the second
place.
The submonoid of k··valencies in
Jf
As it comes out clearly from [6] the natural framework for the study of
cones over the ring
of
~n
II 19 the commutative monoid since a cone in a submodule
may be considered a submonotd of a commutative monoid.
-32Let
d
EO:
IN n
X be a finite set.
We identify X with
may thus be considered a mapping
the set of k~subsets of
a mapping
f: P
k
-+
:m
X,
d : X -+ :N.
= (di)iEX
d
{1,2, ••• ,n}.
An el€men~
Now if we denote by Pk
is called a k-valency if there exist~
such that
(16)
=
d.
~
f( u) , 'if i EX.
The set of k-valencies form clearly a submonord of
previous notations, consider the set
A
c:m n
INn. Coming' back to the
consisting of all elements with
E
X the canonical basis of ~n.
t.~ with Card J = k and {t.},
~ lE
iEJ
~
T
Then the considered cone is A( 7l ). We show that ACo..) = M(A ) n A~D by provi
the form
~
Indeed Theorem 9 will show that
Theorem 9.-
~n
denotes a submodule of
containing
T
M(A )
~
.
..
A(ll) = MI n A'(b:) where
and
Al
HI
is a finite set con-
~
taining A.
AI (~)
Actually
will appear as the polar of a cone, polar who
T
A(l):> M(A ) n A(Q)
~
A
contains
A,
This will give the inclusion
from which
follows the announced equality.
THEOREt1 9.
d.t-
~
1
...
The mapping
~
d.t- •
J
2:
d: X
d.t-
(k - r)
IN is a vaZenay iff when ordering
-+
one has
n
:;; r
L
j >r
d.
, r
= 1,2,
..• k - 1 •
1. j
and
(18)
M'
d. - 0 mod k
~
is the module of
x E
verifying
r
l:;;i:;;n
x. - 0 mod k which actually
1.
-33T
M(A )
contains
elements
"
and
A' (i1) is the cone in
r
ie:I
n
polar to all opposites of
71. n and to all elements with the form
of the canonical basis of
(k - r)
"ll.
L~ -
\
l..
r
ie:I
CardI=r
This last fact follows from (17) where we relabel the componants of
d
in order
that
(20)
~
d
n
by
(21)
r
(k - r)
d.~ $; (k - r)
ie:I
CardI=r
L
i$;r
We prove the theorem by r..eCNrr.Bnce on
X e: ~n,
L
x =
ii ' Card J = k
d
r
i>r i
d.~ $; r
d.
r
d.
~
iEI
We show that there exists an
•
~
$; r
such that
ie:J
i = 1, ... ,n ,
and
d'
~
~
~>
(23) (k - r)
L
i$;r
d' i .. r
L
i>r
d' . = (k - r)
~
If
d
.
If
~ = ~+l > 0
...
i = 1,
,n
d j +l =~
x.]. = 0
k
= ~+l
and by
except for
i $; j
i
L
i$;r
x
...
as follows.
,k , Xi
d. - r
~
r
i>r
=0
, i >k
d.:L $; 0
(17) for r = k - 1 gives
=0
As it shall be.
that
We built up
~+l , we make x.~ = 1 , i = 1,
, n - 1 , and
d'.1+1 , i = 1,
If
d'
verifies the hypothesis.
,
denote by
j
i
Then
.
d.~ = 0
,
the smallest integer such
the largest integer with
(maybe this set of
.
di = <1~
.
We make
is empty) and for
-34-
= t,
i
t - 1 , ••• , t - k + j + 1 , where
we have to show that (17) holds for
I
iSr
d. = qr
~
I
and
i>r
di
= Pr
xi
= 1.
We write again (22) and
For easier notations, we ,~ite
d' .
By hypothesis
•
(24)
Then
Vr<k,rp
r - (k - r)~ -
rp
r
- kq
-r
- rp
- O(k) •
r
So we may write
(26)
= khr
rPr - (k - r)~
, r = 1, ... ,k - 1 .
It follows that
h
- h
r
r-l
+ d
r
But
(28)
drk
= dkk
< ~t S ~ + Pr ' for
r S k - 1 < t
j <
then
d < k-l(~ + P )
r
r
for
j
P + ~
r
which means by (27) (note here that
h
Now if we had made
had (23).
r
xi
~
r - j
=1
, j
, i
< r S k - 1 < t
docs not depend on
< r S k - 1 < t
= 1,
... ,k
and
r )
.
xi
=0
, i > k , we would have
Comparing the present situation to this one we have
(30) (k - r)
L
iSr
d'. - r
~
+ r(r - j)
L
d!.
i>r
= khr
~
= (k
- r)
L
iSr
d. - r
~
L
i>r
+ k(r - j) SO, by (29).
d
i
+ (k - r) (r - j)
-35(30) completes the proof.
Remark:
Th. 9
is a corollary of Theorem 1.1 Chapter 6 of H. Ryser [7].
this result is made part of linear algebra.
Observe that this argument could
be general: an element
d
is in a finitely generated cone
written
dt
€
d' + x ,with
C and
x
a generator of
to check out is that the inclusion
A(ll) c 7l. n
Consider in A the set
,x(k+l)} .
{x(l),
xU)
(31)
=
r
But
n
C.
C iff it may be
What is now left
A(~) is a proper inclusion.
£.i •
l~i~+l
i;1!lj
Then
k-l
r
x( j)
is in
...
Zn n A(Q)
and is certainly not in
"
A( 1l ) .
j~k+l
A relation between the three aonsidered aones and finitely generated monotds.
We have seen in Theorem 7 that
n
R n A(K)
= Rn
the product of all elements in
F, it is clear that
n A(F-1R)
Rn n A(K) ; Rn ri f-1A(R) ~ Rn n A(F-1S) ~ Rn n A(K) .
generated cone over
R in the R-module
f-1R
n
.
Now
Now if
f-1A(R)
f
denotes
is a finite
On the other hand,
Thus the three cones that we consider are all intersections of at most three
finitely generated cones in the R-module
When
f
--1 n
R .
R = 1l , we may apply Theorem VI of [6], page 188 which teaches us in
particular that if
C' ,c"
are two finitely generated cones over
7l.. in a com-
mutative group, or in other vTords"two stable sets under addition in an additive
group, then their intersection is finitely generated over
consequence is the following
~.
An immediate
-36If A is a finite set in 7Z n -' M a 7Z -submodule of "l1. n over
Property 1:
7Z
"
,A(~)
the cone spanned by
•
rt\n
1-n
;'l<
over
A
~-'
M n A(~)
then
is finitely
generated over "l1..
As well, all cones considered may be imbedded in the group generated in
ItIn
'"
f-IA
by the union of
and a basis of
7l n •
In particular, the three cones that we consider are finitely generated
for
R
= 71..
•
The problem of finding out the generators of such cones may be eased by an
argument which generalized the one used for the cone of valencies and which
follows the herebelow exploitation of some results in [6].
''Ie call ordered mono'l.d of non-negative elements a commutative monoId
C
in which the cancellation low holds and whose ordered set of elements verifies
x,y,z
C , &x s y
E
implies
x + z
S
y +
Z
and
X
€
C
0 s x •
implies
We first observe that a commutative monoId is an ordered set of non-negative
elements iff
x,y
E
C &x + y
=0
implies
x
=y = 0
Z
=Y
•
If (34) is verified, we define
x
S
Y
i ff
~
z
€
C ,
X
+
The relation defined is reflexive and transitive.
y + z!·
=X
,
then
x +
Z
i
+
z'~
= x ,
Now if x +
Z'
=Y
and
-37and, by the cancellation law, and (34)
z' + zYV
=0
, z'
y
y
x
= zn = 0
.
This means
(38)
x
~
and
~
implies
x
=y
.
(32) is verified for the ordered relation introduced and also (33) since
x
=x
+
o.
,.c~er.sely'.
if an ordered monotd verifies (32) and (33) this order
is such that (34) is true.
= z.
Let x + y
particular if
x
=0
, y
=0
Since
x + Y
=0
~
y
, then
0 , by (32), x + y
~
x , thus
x
x
~
0 , and since
~
z
~
x.
In
0 , by antisymmetry,
.
Finding out a minimal set of genepatops of a mono'id C in which holds the canceUation
l(JJ;)~
when
C is a submono'ld of a finitely genepated mono'ld
Zarogest submono'ld which is a gPoup is contained in
be the submonotd of
whos(
C.
C verifies (34).
We reduce the problem to that one where
C'
Indeed let
G
C for which
t:J x
€
G , :!I Xl
€
C :
x + x' = 0 •
G is a group and the congruence of monotd
Q= {(x,y)l(x,y)
(39)
defines a
in
(40)
C'
q~t
CI /Q
has the form
€
= C'
x + G ,
x~y,z €
CI
in ,,'hieh
X E
C' .
and
z
,:!I
C' xC'
C/Q
G, x+ z =y}
€
=C
is a submonotd.
Every elemen'
The cancellation law holds in
x + y
=x
+
Z
C
I
si~e(
-38means
x + y
= x'
x' +
and since
(41)
XU
+ z' , for some
=X
,
x ll
by the cancellation law in
(42)
Now
+ z' , then
C' .
y n z
x + Y
=0
G, this means
x,y E G.
means
x + y
Finally
x + y
E
B of
E
~
~
=z
, y
G for some
x,y E C' & x + Y = 0
In particular, if
its opposite.
finitely generated, since
7.1, page 179.
of C\{O}
,
•
x
Y
EX,
E
C , then also
~?
x,y
=0
y.
By definition of
x
and
, that is
y , thus
C'
is an ordered
C.
group, it has a basis.
any representative in each element of
negative
= z'
Knowing a set of generato:
C') we will get a set of generators of C (or C' ) by taking
C (or
THEOREM 10.
Y , z + z
This means
has an opposite in
G being an abelian
E
y + x"
mono!d of non-negative elements as well as
Now
x,y
G
E
=x
x + y + x"
x,x'
We are now
Let
elements~
C'
B and the union of the basis of G and
is finitely generated,
C' and G are also
C' , [6] Propositio1
G is a subtractive submono!d of
in a situation of applying the following.
C be a submono'id of an abelian ordered mono'id C'
of non-
C' finitely generated. Then the intersection of all subsets B
verifying B + C = C\{O}
~
that is
v X E C\{O} , ayE B , a z E C with
is the unique minimal
set~
x
=Y +
Z
for set inclusion, of generators of C.
-39For a better understanding of the meaning of this statement we suggest that
the reader go back to the example dealt with in Theorem 9 and the remark after
its proof.
for
C.
We know that
x
y
$
iff
z e C , x + z = y
~
B of
Then the intersection of all subsets
C verifying (43) con-
tains all minimal elements for
$
in
of minimal elements in
$
verifies (43) and that
C for
C.
is an order relation
B generates
C.
Observing that any possible set of generators of
C would verify (43), the
proof will be done.
s(x) c C of all
y € C\{O}
IN
We first prove that the set
is finite for all
m + C'
x.
y
$
This will prove the first assertion.
be a surjective morphism.
~
B
We will then prove that the set
-1
x
x ,
Let
is finite, since, if not, there
would exist t' ,t"ecf>-lx,t '>t" ,'as. it is well kno.wn, and ~(t '-t")=O. Then the
y + z = 0 , y , z
tradict the fact that
C'
~
implies
y = z = O.
Then the
running over
T n ~-lC
is finite.
S(x)
is finite as well.
Now
-1
x
contains the set
which is thus finite.
It immediately follows by recurrence on
~
hull over
by
cf>(T n cf>-lC)
~
Card S(x) that
spanned by the minimal elements in
S(x) .
x
is in the cone
Then
C is generated
B.
If a cone
COROLLARY.
T in
lln
has rank
r -' then its polar
C has a unique
minimal set of generators.
T contains a cone
elements in
theorem.
T.
T' spanned by a linearly independent set
The polar
C'
of
T'
and
L of
n
C verify the hypothesis of the
Notice that an easy direct argument begins with defining a linear
isomorphism of
T'
onto
:N n •
-40-
If"
An appliaation of Theorem? to bounded polyhed:t>ons in
the field of
(x
fraations of a linear ly· ordered domain R).
Property 2:
Let A be an n x m matrix with entries in R.
p
= { t I tA
~ c , t
~}
€
be a non-empty bounded polyhed:t>on not reduaed to one point.
matrix formed .by the
is a bijeation of
in
l'01JS
of A and -a.
A*
B (K) ,
~:
t ~> {at,a}}
~
the polar of
is injective, since if
(t,l) € {(8t',S)}8€K+ ' (t,l)
{at,a)}
= (t',l)
We now have to check out that
point in
atA - ac
~
if not, we would have
tA
~
B(X)
for
t
A*
B (K)\ {OJ
~
0
and
Let
t'
E
P.
A*
B (K) .
1
= (-t,-S)
Since
t'
~,
P
Then
since
is in
(-St' ,-8)
A*
B (K)
was any point in
(t' - tll)A
On the other hand,
P
bijection as claimed.
=0
8
Suppose
~*
E
(t - 8t')A ~ 0 , and by boundedness,
(-88- t,-8)
tA - c
€
Clearly
~ 0
with
8
0 , since,
~
would not be bounded.
< 0 .
-1
B t
€ ~
Hence
If
and
By hypothesis
But also
= 8-l t .
(t,8)
maps every
~
implies
We have
B (K) •
t'
K ' then
+
•
is surjective.
$
(t,S)
0
is a ray of
Thus
= t'
t
B > 0 , the proof will be done, since then
we show that
not empty.
and
B (K) , since
Let
0
~€
~*
onto some ray of
P
= {(8t',8)}o
K
a€ +
(t,S)
is
P
A*
E
B (K) .
This entails
tA
= Bc
, t'A
=c
.
P, we see that we could have another point t"
with
t i_ttl
'muld mean that
cannot reduce to one point; hence
S
>
P
0
is unbounded.
and
9 is a
.
aEJ
IfI
$
in
Let B be the
Then the mapping
onto the set of aU rays in
P
Let
-41THEOREM 11:
Let
P
= { t I tA
Let B = [ A]
be a bounded poZyhedron.
c
S suah that every
t
in
€
~}
as previousZy and Zet S be any suitaJ:.
-a
subset of R that makes M(B)
F
~ c , t
an S-stable module.
P n ~
Then there exist a finite
is a aonvex Zinear aombination with
C!O-
effiaients in p-1R of vertiaes of P.
If
P
is empty or redt:ced to one point,the statement is obvious.
"'* (K) •
B
P verifies the hypothesis of Property 2 and we apply Theorem 7 to
The restriction of
"'*
$
to the vertices of
P
If not,
is a bijection onto the set of
rays of B (K)
which are not positive linear combinations of other rays and
which generate
"'* (K). In each of those rays we find a point in Rm such
B
"'* (K) n Rm by a linear combination
B
that we obtain, by Theorem 7 every point in
of that finite set
of F-IR.
in
T of points with coefficients in the set of
In particular, to every point
t
in
~
0
elenents
P n Rm corresponds a
(t,l)
"'* (K) n RID which is such a linear combination and may be seen a convex
B
linear combination of images under
~ with coefficients in F-IR of vertices
of P •
Corollary:
If R = 11
and if M(B)
is a unimodular moduZe, only vertices of
P may be integer points.
Hence
F- I 71.= 11. •
are trivial.
M(B)
is a stable module with
S
= {l,-l}
and since
F c S ,
Then all convex linear combinations with coefficients in
F-I 71.
-424.
Lineazo equations over a PrUfel' domain
We will here apply Theorem 5 with for
A is the set of all cosets of all submodules of the R-module
Thus
PrUfer domain, with the form
R.
A the example dealt with in 1.2.2.
We assume as before
obtain Theorem
: G ~ aG
a
h
R a
is a finitely generated ideal of
is injective for
a
~
0 .
Our aim is to
4 of [2] as a corollary of the present #2. Theorem 5. The axiom
of choice, or more
the axiom of maximal ideals, will not be avoided
pre~isely
in the general case.
whenever
~
hG, where
G,
However, we will give a proof which does not require it
R is a principal ideal ring or the domain of integers of an algebraic
number field.
We first show how the first assertion of Theorem 4 of [2] is a particular
case of the one #2. Theorem 5.
denote by
B the
n
B has the form
fj
=G ,
where
b
a
H.~
=0
j
m matrix whose rows span
x
[A,-I] where
= 1,
= h.G
... ,m - nand
, h.~
~
Let M in #2. Th. 5 be finitely generated and
I
is the
rj
=r i
n
x
n
+ Hi j
M.
identity matrix.
=m -
a non-zero f.g. ideal, for all
n + i , i
i.
Now make
= 1,
••. ,n
Finally we make
va€M.
The first assertion of Th. 5 becomes, writing
(1)
Suppose, moreover that
:a: x
€
mn
G- ,
l
l~j~m-n
a. jX
1
j
:: r. (H.) i
~ ~
A = (a
= 1,
ij
) ,
••• ,n .
Now in Th. 4 of [2] this assertion is proved to be equivalent to the following
Let
{oJ
;t
H
= n H. •
be the submodule of
(2)
We know that
~
n
R
of those elements
y. H.
~
1
c
H
= hG , where
Y = (y i )
H , i _. 1, ... ,n .
l~i~n
11 = n h . •
1
verif1ing
v
Now let
(~
y
-43v Y
Y , yA - O(h)
€
implies
We prove the following specializa.tion of #2. Theorem 5 for
R a PrUfer
domain.
THEOREM 12:
be two supplementary modules of ~(M
Let M and M'
Let be given
r.
We denote by
}~ the submodule of those elements
L
a
Os,jSn
oX.
J
=
+ H. , H.
1" •
J
J
= 0 , V a
€
M.
= h.G , h.
J
J
Denote by
(h:h.).<.~~
J 'L--J=1l
by
M' =~) •
a f. g. ideal of S, for all j •
J
x
€
am
verifying
h: h • the annihilator of h • mod h
J
J J
We denote
ED
J
Y.
r
(4)
J.
n M ;e {tP}
G
iff
L
V Y € M n Y,
lS,j::>m
yjr
j
-
0 (H) .
For the sake of better understanding, we OJserve
S-stable module for
S
= R\{O}
, any finite set U
c
that, since M is an
M with one element in U
for each minimal support of M is a set of representatives of
show that (5) implies (20) of Th. 5.
M.
We will
We first check directly:
The condition is necessary.
x
€
r
n
.. "
).
MCli ,
then
x
j
=
and
(6)
,.
.
•.
'.
: ... ~
i:
.,
."
We first· prove Th. 12. for-n° a finit~ly ·pr:i.ncipth ideal ring, that is a
ring where every t.g.i. is principal.
This will be used as a lemma but it aJ.so
-44gives a statement of special interest in itself, since whenever
M n Y has a
finite basis, (5) needs to be verified for the element of this basis only.
H
This is being possible only by the fact that
~
{Ole
~~at
we have to prove is
We have
I
(8)
l$j$m
uh
j j
I
::>
l$j$m
ah ,
uh =h
j
or
a,a -1U j
with
Since
y
€
= Rm
M E9 M'
(h : hj)l~j~m
€
for all j •
R ,
Ba
t..
-1
l~j$m
which implies
= (y )
j l$j$m
and (5) applies.
~
(10)
,y
ujr j
€
-- ( Ba -1U j ) l$j$m
€
M
•
By (9),
We then have, by (9) again
H=
(7).
..
Now, if we prove that Theorem 12 is valid for every localization
a maximal ideal of
R ,p
every element not in
p
R
(R
p
is the smallest ring containing
has an inverse) with
R
R
p
of
in which
Gp , Mp ,h ,p defined as in [3],
j
then Lemma 1 of [3], slightly modified, will obviously prove Th. 12.
We know that
of
R
M
E9 M'
P
P
R
p
is principal.
is a
Talu~tion
Also if
M E9 MI
ring, then every finitely generated ideal
= Rm
, it is easily verified that
P
We now prove Theorem 12,withou+, the help of assumed maximal ideals, for
the ring of integer( of an algebraic number field.
Every proper ideal is here
R
-45a product of prime ideals. [8].
If
h
is a product of prime ideals, here is a proof that leads to an
algorithm for actually finding a solution to
(11)
(12)
having convenient notetions, we write
for some
x(k)
t
~
s.
is a finitely 3enerated prime ideal.
~l
i t _l
it
.
Pl'" Pt-l = k and Pt = q = p1
For
By recurrence we assume that we have a solution
to (11) with
If we prove that there exists a solution
to (11) with
x(q)
we will then be through: (14) will give us the basis for recurrence and, moreover, since, by properties of PrUfer rings,
(15)
we have
k + q
a
€
k ,
e€
q with
= (1)
, k n q
= kq
,
a + B = 1 , and we will obtain
and, by recurrence,
(16)
Hence we prove (14).
q
= pi
and we consider the localization of
R at
p.
-46As in Lemma 1 of [3] we find an
Now w
= (yep»)
This implies
+ pi
x(p)
is a f.g.i. which divides
w is the whole ring or a power of
does not belong to any power of p.
NY(P) -_-
(18)
~
Making
ax(p)
solution of (11) with the property
= x(q)
Then
1
w
pi
pi
p.
But
=R
is a f.g. ideal.
yep)
€
w , and yep)
and there exists
a
€
R :
(i)
p.
, (14) will be verified.
An algorithm for the case R
=G
The preceeding proof gives a hint for an algorithm for actually solving
(11) and (12).
Suppose that we are provide.d a solution
xjq)
q
for every highest power
= Pl
= rj(q
x(q)
with
+ hj ) ,
of a prime (ideal) factor of h .
Then (19)
satisfies (14) and the argument of the proof allows the build-up of a solution
to (11) and (12).
l
Now
q + h
j
=P
with the condition
p
a prime ideal.
j
,
0 ~ lj ~ l , and our prob~em reduces to solving (11)
.-47Q denote the quotient ring RIp!.
Let
We will show how a solution to
(11) and (20) modulo p! provides a solution.
!
union of classes of the ideal
tion to (20) mod
M
is the ideal
Det (AJ )
of
with determinant not in
(21)
Ax
p
tj
of R so a solu-
pt is a solution to (20). We consider a basis of the module
and make it the rows of a matrix A.
determinants
Let us first point out that the
M over the local ring Rp of R.
spanned by
M
P
p j mod p
t
=0
,
X
m
R
\
L
=;>
O::;j::;o
ajx j
so that there exists an
R
p
The rows of A
p •
E:
The ideal spanned by the principal
is the whole
A
We may take this basis in
span
=0
,
Mc M
p
"if a
E:
We make a left unimodular transformation of A over
over
R
P
J
A
and thus
M •
Rp
in order that
J
A
becomes the identity matrix and we multiply each row of the resulting matrix by
suitable" elements
entries in
So
·not·~-:in
p
in order that the obtained matrix
R but verifies (21) like
A does.
B has a diagonal of units in
of those units of Q.
Let
Q.
Then we may define
0
= Det
rj , by
(22)
Now suppose we find a solution to
Bx'
Then, denoting
=0
(Q)
xj
t
= rj
(p j) .
B by (b ij ), we have the relations in R
(24)
We thus find a y, s(y)
c J ,
with
L
B has all its
l::;j~m
b
ij
y
j
= oy.
J.
,
J
B
'be thci product
-48and for
x
= ox'
- y , we have in
(25)
Bx
R
=0
x.
hence (11) is satisfied by
l
oXj :: rj(p j)
We observe that
l.
x j - rj(p j) , j
(26)
,
l
l
pcp j
and
Y
j
= 1,
•.• ,m .
€
Then
Since we made sure that (23) solves the problem, we are left with solving (23),
which is equivalant to
L
(27)
l::>j::>m
(Ct ) :: p
j
with
lj
b. jCtjZ j :: ~
solution
Z
b. r' (Q)
~j j
l
modp , since we know that
and is thus a principal ideal ring.
Any
L
l::>j::>m
= (Zj)l::>j::>m
solves the problem.
€
Q has a finite chain of ideals
It may occur that
t j = 0 , then Ct j
=1
•
Qm to (27) gives a solution to (23) and hence
There is no problem to diagonolize
(bijCt j ) mod Q , with
a left unimodular transformation.
Remark:
is an
If M is the module spanned by the rows of an echelon matrix A, that
n x m matrix
o of (22) is one and if the problem has a solution, then if
ideal for some
j
¢
J
A having the identity matrix as a submatrix A , then
h
j
is the zero
J , the algorithm will provide a solution as well.
-495.
Some properties of suhmodules of Rn .) R a linearly ordered domain.
Denote by
R the linearly ordered field of fractions of a linearly ordered
R .
domain
5.1 Definition:
R separates its fraations whenever
(1)
as -1
"If
€
R , 3: Y
€
R :
y ~ as -1 .
5.2 Examp les
Notice that this does not mean that for any two elements in
R , then there
exists an element in
R larger than the first one smaller than the other, since
this is not true for
71., and
Let us show that if
polynomials in
in
K[X]
X over
> 1
a >
a,s
>
a .
y=a
If
S
as -1 S
y
=a
If
€
as -1
>
if
a
Y
we write the fraction in
= as -1 .
If
8
4
K , .re
as -1 .
verifies (1).
its term with lowest degree is positive, then
K[X] •
a
K , we make
However, if a polynomial is positive in
fractions.
>
and
(2)
Then
K[X] , the ring of
K separates its fractions if a polynomial is positive
we make
order that
8
K is a linearly ordered field,
whenever its leading coefficient is positive.
If
have
7l verifies (1).
Let us show indeed that
We know that
1
is
is any polynomial in
>
a ,
-1
X
K[X]
K[X]
whenever the coefficient of
does not separate its
is larger than any polynomial in
by the axioms of an ordered
K[X] , we have
dOID~in.
Then
-501 > aX ,
since the constant term of
aX
is
O.
Now
-1
> 0 ,then
X is
X
is
> 0
and
-1
(4)
X
-1
. 1 > X ~ . aX
=a
.
5.3 The ideaZ of insepar'abiZity
Property 1:
Let
Tp
Then Tp
€
V a
S € Tp , then -S
€
R , oS < p}
€
If p
is an ideal-.
If
S' ,S
= {Si
~
-So
since
Tp ,13' s 13 , and let us show that
(S' + S)o
Now if
a
is any element in
S
R,
> 0 •
Tp is a prime ideaZ.
1 ,
Tp
p
for
2130
S(-o)
implies
p
S + S'
= So'
(as)o
~
€
~
p
.
Let
Tp'
< P •
= So'
< p.
p ~ 1 ,Tp
If
is a
prime ideal since,
ao'
which means
Note:
If
given of
ao'Se"
~
ao'p
~
2
P
~
P
as ¢ T p •
p < 1
K[X]
Definition:
general
~ p > 0 , SOil ~ P ~
1.
,Tp may not be a prime ideal. In the example previously
with
X < 1
We call
, TX is not a prime ideal since then TX = (X2 )
Tp the ideal of inseparability if p
Tp is the ideal of p-inseparability of
2.
He say that
R
R
.
p-separates its fractions if
= 1
.
In
-51-
(6)
THEOREM 13: A ring p-separates its fraations iff its ideal. of p-inseparabiZity
Tp Nduaeo to
Tp is an interval..
{a}.
set of p(>sitives of R, Q = {a + T Ia
p ~
NOI.J l.et
If R+ is the
1.
is a set of positives for RIT
+
p
R/T p the structure of a l.inearl.y ordered domain whichp-separates
E
p
which gives
R}
its fractions.
any
{a} , then, :!I BET ,6> o • We see that
p
If Tp
;t
oE
since
R
o~
13
-1
p
would mean
130
~
13
-1
p
is larger than
p
On the other hand if R does not separate its fractions, there exists at
least one
a;t
a
with
130
'ifoER
since, if not, to every
for any
a > 0 , aoe
13
a
E
ap , ao
~
would correspond a
aB
~
-1
~
130
Tp
yo
aB
-1
p •
is an interval.
0 <
p ~
If
as -1
relations for
R+.
-a +Tp
E
-Q
a + 13 E Tp •
c
a
~
a.
a,
130
~ p
and
•
~
ao
U
~
is
which is
o> a •
Then for every
Consequently
< p.
Tp + a
where
Q = R/Tp
We must have
R/Tp
130
y = ao
Tp
is a domain, since
Q , -Q
Then V 0 > a , ao
Now the order relation for
R with
corresponds a
Now we have to show that
n Q ,
E
y • with
Q the set of residue classes
Q + Q c Q , QQ
or
~
1 , R/Tp
The relations
Let
6
a ~
and for every
< p
denote by
Let
0
p •
This means that to any fraction
larger than
< p ,
Tp
E
and
is a prime ideal
runs over
follows from the
Tp
13
a
13
R+.
correspondi~
-Q n
is the only element in
E
(a + a) 0 < p
-a + Tp , for some
and
a E Tp •
13
~
a,
-52a ~ 8 iff V a
€
a , V 8
B € R/T p
Thus if (7) occurs for some
if ,
€
a ~ B in
R.
e~ e
this means every
,
that is,
B = O. Hence R/T p P:separates its fractions.
5.4 Let
Y be a finite set.
= R+Y
P
of positives.
Y
R
of
Let J
Lerruna:
TI
the linear mapping
J
The support of
MeRY is the set of
y
be the support of a submoduZe M of R
its fractions;} then
M is a directed group iff
(8)
av
to say that V x,y
addition of
0 , y
with
v
i
€
Y for
€
M ,
vi > 0 , V i
€
If R separates
•
J .
There are two equivalent ways of defining a directed group
~
'
TI.M
~ {a} •
].
which
z
Tp
Y
R will be ordered canonically by its set
We denote by
J
R , J c Y •
onto
is in
z
~
~
M.
0
x , z
a
M,
€
z
€
M with
It says that any
x,y
~
€
-x.
M.
If
z
x
~
x , z
~
y.
M is directed, to every
Card J
The first is
The other uses the
x
M may be written
in
Thus by summing
M.
of such
=z
- y , where
x corresponds a
z
one obtains the
z
of (8) .
We now prove that (8) is a sufficient condition.
thus
d
for some
is the difference of two elements in
i , and there exists an
-1
v a
s s
S
= min
a >0
i
such that
-1
v.a.
l.
J..
P
If
a
€
-P , a
Now we may S'Qppose
= O-(-a),
a
i
> 0
-53Since
R separates its fractions, there exists a
y €
R
(10)
Thus
(11)
P n M
= ya s v
~
v
- (ya v - a)
€
3 W
-1
s
a v
s
~
a •
Finally,
= ya s v
a
CounterexampZe for
Let
module
s
P n M- P n M .
R not separating its fractions.
K[X] be the previous example.
M spanned by
a
= (-1
X < 1.) Consider the
(That is,
+ X , 1 + X)
and
b
= (1,-1).
a + b
= 6x,x)
verifies
(8). We show that nevertheless M is not directed. There should
exist a
z
= aa
+ l3b
~
b
and
~
-b.
That is
-a + aX + l3
(12)
a + aX - l3
~
~
1
1
or
1 + a - aX S l3 S a + aX - 1 .
But
1 - aX
is
>
o.
Thus (13) entails
(14)
a <
B< a ,
which is impossible.
Let us denote by
~J
the operator
iff it is spanned by a subset of
could be finite.
P
KerITY\J.
Notice that
We investigate a case
M is directe'
~here
this subset
-54THEOREM 14: Let
fmctions.
R be a 'lineal'ly ol'del'ed Noethe1'ian domain which sepa1'ates its
A submodule
M of
II
(fol'
spanned by a finite subset of P = R;.
Card Y
< 00)
is dil'eeted iff it is
MOl'eovel', if M is dir>ected, then
everry I7'l(J,X'i,mal sequence
(16)
{a} ~ 6
Jo
Me 6 . Me ••• e 6
Me ••• e M
J1
Js
of distinct d1:l'ected moduZes has length l'ank
a p1'incipal ideal 1'ing3
This implies
M.
MOl'e speaificaZZy if R is
M is a dil'ect factor of 6 J.
6J
M , s < dim M - 1 .
M has a basis in P.
The condition ·is
sufficient~
A
Let;.. Be P ·.:be a finite set spanning
and
M.
Then
M
A
= B(R)-B(R)ePnM-pnM
as dircrcted.'
M:
Pl'oof of the main assertions.
where
is any maximal subset of
J
directed.
If we also prove that
by a finite set in
we
'II see.
w~
Y for which
M/6JM
I nd ee d th e vec t or space spanne d by
Every maximal sequence (16) ends by some
Y for which
6~I
and that
= rank
6 M,
J
M over
Ml
R •4n
M \6 M and a basis of
l J
rank M = 1 + rank
(16)
that rank
1
M, is
is spanned
P , the proof of the first two assertions will be done, as
a basis formed with any element in
set of
has rank
6}'1, distinct from
distinc~
from
6 M.
J
-R~ has then
Then
6JA
6 M e M where
J
J
M, is directed.
is a maximal subSince we claim
M - 1 , the proof of the second assertion will be done by
-55recurrence on the rank.
set in
P.
Let
M = r\ + l::.JM
Hence
S €
Y\J
But by recurrence as well,
is spanned by a finiti
is spanned by a finite set in P •
IT M ~ {a} •
s
such that
l::.iM
By hypothesis, the ideal
IT M is
s
finitely generated and
l
2
ITM= (d
d
s
s' s'
(17)
where
M'
D
= {dl ,d2 , ...
,d.t } eM.
,
, dt s )
We may choose
dJ
s
0
'>
,V
.
j
We denote bjT
1 the module spanned by D and we first show that.
Since every element in
M is the difference of two elements in
enough to show that any
show that either
if
Ji
c
J
€
Mn P
or
is the support of
Consider any
-1
wjd j ~ 0
Ws > 0
w
d
€
Mi
where
w
€
M n P , it is
is in the second member of (18).
l::.JM.
We know, since
l::.JM, there exists a
\l::.~~ with d s
>
O.
Let
v
Wid;l
runs over the componants of
d
l::.~~
€
We first
is directed, thai
l::.JM , v
j
> 0 , Vj € ;
be the smallest fractio:
for which
see that
Now,
bj
>
0 , Vj
E
Ji
a directed submodule of
J'
b~
= O.
Since there does not exist a support
M properly contained in the support of
as a proper subset, by the maximality of
lemma, we must have
(20)
and
J
for
l::.JM
0
M and havi!)
directed by the
-56In particular, if w
chosen as
.e
~
and for
wj > 0 ,
> 0 , contradicting (20).
d
s
= w
=0
since then
, d.ewj - w.edj
= d.ewj = b j
[).JM , we cannot have
j
4J
ws
By (17) we are thus able to take
could be
s
would be'·
d E M
l
such that
Finally, by (19) and (20)
s
(21)
Making
=s
j
in (21) shows that
d.e
= w.e
' and consequently, since
d.e > 0 ,
(22)
(22) entails
which proves (18).
Since
find
e>
R is a ring which separates its fractions 'We may
0
such that
This comes from
W
EM,
span
W
j
M
l
d~ ~ 0 ,
j
> 0 , for eve!y
j
4J
as it is entailed from (19), (20) using
h
in the support of M. The d + Sv will now
anti clearly
(24)
What is left to be shown is that for any two
there exists a fraction
h
E(d +Sv)
= dk +Sv
E in
modulo
h
d
k
+ Sv, d
+ Sv with
dh,d
R such that
-
the R-vector space spanned by
[).~
in
-Y
R
•
k
E
D
-57We will see that
h h
= dj/d
j
€
V j 4 J ,j
,
consider (19) and (20) with w
of
>
° componants.
For, we
M n P ,w having the pargest possible number
We have
{
(26)
€
in the support of M.
h
h
dj = dlwj/w , V j
l
k
j
d
4J
k
= dl,w/wl, , V j 4 J .
or
which proves (25).
Let us now investigate the more special case of
Here,
D
reduces to one element.
D.~
in
Since
changed into the
of rank M
$
is spanned by one element which is not
M
l
R is a domain,
R a principal ideal ring.
Ml n D.jM
of a direct sum.
= {a}
and the
By recurrence,
M
+
of (18) may be
is then the direct sum
submodules, each one generated by an element in
P\{O}.
This
.,
means that
M has a basis in
A question arises.
Is Theorem 1 the best statement, or could we have
D. M a direct factor of
J
s
principal ideal ring.
Let
b , a
6.
J s+l
M,
when
= (l,X
maximal for
c
+ 1), b
= (X
R is a Noetherian ring but not a
Here is a counterexample.
M be the module spanned in
directed.
D.~
P , which was the last assertion of the theorem.
= (O,X-l)
- l)a - (X + l)b
II [X]
is the submodule of
is ordered with
= (X
D.jM being distinct from
M spanned by
by the two elements
(ll [x])2
1,0)
€
P.
X > 1.
Then for
J
For, if x
= aa
and
M is
= {I}
M and directed, by the lemma.
c.
a
+ Sb
€
,J
is
Actually
D.JM , we
-58must have
and
a(l) = 0
6 belongs to
depend on
a,
and
o.
6(-1) =
it is a multiple of
belongs to the ideal
(28)
M
We see that necessarily
b
X-I.
(X - 1)
The second componant of aa + 6b
is a multiple of
6 M
J
cannot be a direct factor of M.
= {oJ
a
(X + 1) . Since the first componant of aa + 8b does not
being zero, every element in
~ n 6~
Thus
c.
We show that
(c)
Suppose
= (c)
ED~ •
belongs then to
MI.
Then, by the requirement
, we have
The conclusion is
(30)
This is impossible since
a ( (c)
ED
6{2}M •
,
"
~:
.
R is now a principal ideal ring.
CoroUary 1.
Let M be a directed submodule of
factor of M" directed as weU.
that LeN
= M.
II
and let L be a direct
Then" there exists a directed module
N such
L has a basis contained in P whiah may be aomp Zeted into
a basis of M contained in P.
By hypothesis,
let
Let
v E M be the element verifying (8) and
w E L be the one obtained by substituting
B be a basis of
(31)
M = L e L' .
L' :
v
= L
xEB
a x + t , tEL
x
L to
M in the lemma.
Let
-59and
B may be chosen in order that
Now
e>
all
> 0
is chosen in order that the set
0
componants in the support of
L •
13w + B = B'
l
Then
XEB
has all
componants in the support of
> 0
by B'
is directed and
CoroUaPy 2.
a (Bw + x)
x
Hence the module
l
~
XEB
a x
x
N spanned
LeN=M.
Let A be an m
the colwrrns of A span
M.
has all elements with
FfI,
x
n matrix,
m': n , with entries in R+.
If
then A may be completed into a square invertible
matrix with entries in R+.
5.5 In #3, we have considered finitely generated cones in ~ and their intern
R , K the field of fractions of
sections with
~tudying the intersections of their faces with
Definition.
subset
A face of dimension
F with dimension
not mee-t;
i
i
R.
We will use Th.14 for
Rn •
of a convex set C of
It
of C such that the convex hull of C\F does
F.
Let
C
be a finitely generated pointed cone of dimension
(That is
C
is not reduced to zero and
morphism
<p
of
containing
in
ik
under
~
-1
onto
in
k
. .0
l\.
-C n C
= {OJ).
k
in
~.
We may then define a
~ whose restriction at the smallest subspace
C is an isomorphism onto
has rank
</>
is a I77ClX"imal
ik.
Consequently, the polar of
as the polar of a pointed cone should, and its image
,
which we may define straightforward by
E
<pC
-60-
*
CE
has a finite set
~
and
of E
= {xix
€ E , V y € C , <x,y> ~ O} ,
U of generators the rank of which is
~E~
may be defined in order that
~ ; rank
in
E~
=n
We have
, where E~ is the orthogonal
= {oJ
- k .
V be the union of U and a basis of E~.
Let
k.
We denote by
L the
K-vector space of the (a)
u u€ V' a u € K , such that
l
(34)
u€V
a u = 0 .
u
Since rank V = n , that is, the maximum rank for a subset of ~ , the morphism
ljJ : ~
V
K
-+
defined by
ljJx
since the polar of C* in
E
uk
V'
u€
KV .
L~ , the orthogonal of L in
onto
~-
= «x,u»
of a cone
~C
ll~ n K:
means
E is
with rank k ,
= {oJ
C.
*
~CE
n
VX € K
is an isomorphism of
We observe that
*
Now, since
~CE
,
= <x,u> ,
V U € V\U
for all
u € U .
Property 2.
JL ~
II
VU€U.
F
.
is the polar cone in
This
a.nd, by a well known theorem on linear inequalities
But we may find a
Then for
This means that
~
y € E
x € ~
such that
such that
<Y,u>
V
z = x - y , ljJz € ll~~ n K+ , and
II L~
U
<z,u> > 0 ,
is directed.
is a faoe of C ifl -ljJF = ll.f~ n K: ~ for some
is direoted.
C* •
E
is a pointed cone, and so is
(which may be deduced from #2 Th. 4) there exists an
<x,u> > 0
r
J
c
U
where
-61The condition is necessary.
If
F
is a face of
C , -WF
union of the supports of the elements in
an element
v
in
with support
-'l'F
-WF c ~JL.l n K~
We have
V
.1
w € 6 L n K+
J
and w
such that
CJ.W~
v -
4
J .
-WF.
Then
0
i
-W-1w € C
Then
V -
For example
CtW
Ct
= min
is not in
VjW
-WF.
This implies
-1
is directed.
Suppose
-WF.
a > 0
the definition of a
By
j
has rank i + 1 , since
-WF , since, if it were,
Finally if such a
v/2 would be in the middle of the segment [(v are not in
J
V
F , F u {w-lw}
linear combination of elements in
~ L.l
sew) c s(v) = J , there would exist
Since
is the dimension of
be the
and there exists
-WF = ~J L.l n K+
wj>O
face, if
J
Let
J c U
This means that
and we must show that
-WF.
V
n K+
.1
~uL
is a subset of
CJ.W)
,
Ctw]
W
would be a
w would exist,
whose extremities
F could not be a face.
The condition is sufficient.
~JL.l is directed, J
-w-l(~JL.l
n
K~)
=F
is contained in
is a face.
F
maximal set with dimension rank F
is not contain-ed in
Now every element in
J
U and we must show that
is certainly contained in
since, if
Cane it is a
X€C\F, the support of
-W x
and consequently
-W(C\F)
has a support which is not
cont~ined
in
J .
Then a convex linear combination of such elements has a support which cannot
be contained in J .
CoraoUary 5.
This completes the proof.
Let C be a finiteZy generaated pointed cone in If" wherae K
is the fieZd of fraactions of a Ping R vePifying the hypothesis of Theoraem 2.
-62Denote by
the moduZe spanned over
M.
~
with dimension
M.
Then
i
If
by
where
n c. ,
~
is the faae
c.
~
of the sequenae
(whiah has rank i ) is spanned by a finite subset of c. n
~
~
is a prinaipaZ ideaZ ring, M.
R
R
is a direat faator of Mi +1 ' i < k.
~
a direat faator of
If.
If
Mk
is
If .
Let us denote by
N the module
ljJR
n
and by
M the module
At!.
Now
by Property 2
-ljJC.
~
where
V
€
a
€
A L~
J
= AJ
V
L~
n K+
= 0,
i
••• ,k
i
is directed.
is directed as well, since if
i
A L~,
J.~
V
j
>
R such that
o
for all
aljJ -1 v
€
be a finite set generating
j
Rn
in the support of
and then
av
M n A L~
, since ljJ(Mi
J
M.
€
Then
) c M t'): A L~
.
-1
(M
i
M.~
since to
linearly independant elements in
linearly independant elements in
Let
i
,'hich spans
i
ljJ
J.1
n
C.1 n R )
.
there exists an
C.~
B.
(-B.)
~
~
c
V (Theorem 2)
R+
is a subset of
has the same rank
over
K correspond
C.nR
n
~
as
i
C.~
i
This entails
,k .
M n A L~
J
i
The sequence
onto a sequence of directed modules
n AJ.L
~
1
is thus mapped by
ljJ
-63(39)
A MC
J
•••
C
o
A MC A
J
J
i
M c ••• c A M = M
J
1+1
k
which is easily seen to be maximal since rank AJ M
=i
, Vi.
Hence the first
i
assertion regarding principal ideal
rin8~
follows from Theorem 2.
wi th
VU
f
which means
If
cone
V\U , <ax,u>
=0
, thus
Vu
IPx
x
IP -1M.
Hence
f
M , or
f
f
a
f
R ,
V\U , <x,u>
-1
IP M
n
R
X f
= Mk
=0
, then
,
is a direct factor of
R does not separate its fractions, corollary 3 fails.
C spanned by
(1,1 - X)
and
(l,l+X)
X < 1.
2
in K ,where
linearly ordered field
Q and where
If
it is easily seen that
a(X - 1 , X+ 1) + S(l , -1)
Now
(a,S)
is
R
Consider the
=Q[X]
for any
is any element in
~
O.
C,
If the corollary
2
were true, there would be in C n R a basis of a direct factor with dimension
2
2 of R , that is, of R2 itself. To this basis would then correspond a
basis of
~
0
elements of the R-module spanned by (X - 1 , X + 1)
and
(1, -1).
This would entail that any element is the difference of two
elements.
This fails to be true as proved in 5.6.
~ 0
-64References
[1]
P. Camion, Modules Unimodulaires.
Journal of Combinatorial T"neory, Vol.
4. 1968.
[2J
P. Cmnion, L. S. Levy and H. B. Mann.
Ring.
[3 J
Linear Equations over a Commutative
Journal of Algebra, VOl. 18 p. 432 - 446 (1971).
P. Camion, L. S. Levy and H. B. Mann.
PrUfer Rings.
Journal of Number
Theory, Vol. 5 No.2, April 1973.
[4J
P. Cmnion, Characterization of Totally Unimodular Matrices. Proc. Amer.
Math. Soc. 16, No.5 (Oct. 1965).
[5]
N. Bourbaki, Algebra, Chap. VI (Groupes et corps ordonn~s) Herman, Paris.
[6]
S. Ei1enberg and M. P. SchUtzenberger, Rational sets in Commutative Monotds.
Journal of Algebra 13, 113 - 191 (1963).
[7]
H. J. Ryser, Combinatorial Mathematics. The Caruso Mathematical Monographs,
John Wiley and Sons, Inc. (Ch. 6. Theorem 1.1).
[8]
H. Mann, Introduction to Algebraic Number Theory. The Ohio State University,
Columbus.
[9]
P. Camion and J. F. Maurras, Po1yedres
unit~.
(To be published).
a sommets
entiers daUB 1e cube