Cones in numerical modules

Arch. Math., Vol. 58, 70-80 (1992) 0003-889X/92/5801-0070 $ 3.70/0 ~) 1992 Birkhfiuser Verlag, Basel

Cones in numerical modules

By

ULRICH KRAUSE

0. Introduction. It has been pointed out by Klee [5] that although many of the basic notions in the theory of convexity can be defined for fields other than the field of real numbers R, the behaviour, even in the case of the field of rational numbers ~ , may be different from that in the real case. Replacing fields by rings, new difficulties arise if one tries to establish convexity - like results in modules instead of vector spaces. Whereas, e.g., finitely generated cones over Q are integrally closed, this is no longer true for finitely generated cones over the ring of integers Z, i.e. for finitely generated semigroups. (Cf. Section 2.) There is a need for convexity - like results in Z-modules which stems partly from algebra (cf. [3], [6], [7], [8]) and partly from integer linear programming. (Cf. [1], [4], to mention only a few.) To cover the extreme cases of N and Z, in the present paper we shall deal with cones in numerical modules, i.e. modules over subrings of ]R. A particular case are abelian groups. (A related but quite different approach has been developed by Marcus [8].)

Section 1 treats Stone cones which originate from Dubois' [3] algebraic concept of Stone rings. From a dual description of the infinitesimal hull of a cone it follows that a Stone cone may be characterized by the dual cone of states. This dual characterization of Stone cones yields certain versions of Farkas ' lemma (cf. [9]) and of a theorem of Davis [2], respectively, in possibly infinite dimensional numerical modules.

Whereas Stone cones need not be finitely generated in general, the particularly interesting case of finitely generated Stone cones is treated in Section 2. Employing a version of Carath6odory's theorem for numerical modules it is shown that a finitely generated cone is a Stone cone essentially iff it is integrally closed. This then is used to obtain versions of Farkas ' lemma, involving the concept of a Hilbert basis, and of Davis' theorem for finitely generated cones in numerical modules.

1. Stone cones. Let R be a subring of the ring of real numbers ~ with 1 e R. A numerical

module M is an R-module which is unitary and torsion-free, i.e. ix = x for all x ~ M and r x :I: 0 for all 0 + r ~ R and 0 :I: x ~ M. Let " < " denote the common ordering on R and let R+ = {r ~ R [ r > 0}. An R-cone, or a cone for short, is a subset P of the numerical module M (over R), such that O ~ P , P + P = {x + y [ x , y ~ P } c P and R + P = { r x [ r ~ R + , x ~ P } c P. A cone P induces a partial order on M by x <__y iff y - x ~ P. An archimedean element a of M is an element a ~ P for which ra ~ P with 0 + r ~ R implies r ~ R + and which has the property that for any x E M there exists some r ~ R such that x < ra.

Vol. 58, 1992 Cones in numerical modules 71

In what follows (M, P, a) will be always an R-module M equipped with a cone P c M and an archimedean element a e P. Obviously, M then is directed, i.e. M = P - P. Also, { 0 } ~ P ~ M b e c a u s e o f - a ~ P .

Interesting examples of subrings R of ]R are given by R = R, R = (I~, R = Z. Correspondingly, (M, P) is a vector space over the reals or the rationals, containing a convex cone, or (M, P) is an abelian group together with some subsemigroup. Especially, for some vector space basis (el, ... , e,) of R" the numerical module M may be the lattice

M={~i=~m, e i l m , ~ Z , l < i < n }

which is equipped with a cone

the S z being subsemigroups of the additive group Z. Disregarding the trivial case P = M where an archimedean element cannot exist, in these examples a s P is archimedean precisely if for any x e M there exists some r e R such that x < ra. In the lattice-example with S~ = Z+\{I} for all i one may

choose a = 2 Y" e i as archimedean element. i=1

Given (M, P, a) we will consider the infinitesimal hull 13 of P defined by

t3 = {x e M I for every n e N there exists some s (n) e R + such that s (n) (a + nx) e P}

(N the set of natura l numbers l, 2, 3 . . . . ). Below it will turn out that 13 is a cone too. F o r (M, P, a) the set of (normed) states is given by

P* = { f : M ~ N l f an R-homomorphism, f > 0 on P, f ( a ) = 1}.

P* is a convex set in the real vector space N M and it is compact with respect to pointwise convergence. P* denotes the set of extreme points of P*. The crucial step in establishing a relat ionship between 13 and P* is the following lemma. (Cf. [6] for the case R = Z.)

f r ) functional

k )

p" M -~ ]R is defined, which has the jbllowing properties

(i) p(0) = 0; p(a) = 1; p(x) < p(y) for x < y. (ii) p(x + y) < p(x) + p(y) and p(rx) = rp(x) for r e R+.

(iii) To every x e M there exists an R-homomorphism f : M --* N with f < p on M, 0 < f on P and p (x) = f (x).

P r o o f . Firs t we show that p(x)eTR. If x e M, then - x < ta for some t e R . Let s x < r a for r e R , s e R + . (Since a is archimedean such elements r ,s exist.) Then - s t a ~ s x < r a and O < ( r + s t ) a . Therefore r + s t e R + w { O } because a is

archimedean, and r > _ t. This shows p(x)E 1R. In part icular, choosing t = 0 for S - -

x = 0 gives p (0) > 0 and choosing t = - 1 for x = a gives p (a) > 1. F rom the definition of p it is obvious that p (0) < 0, p (a) N 1 and that p is monotone. This proves par t (i).

72 u . KRAUSE ARCH. MATH.

F o r (ii) let x, y e M with s 1 x < r 1 a and s 2 y < r z a, r i E R, s i ~ R+. Then s z s 1 x < s z r 1 a

and s x s 2 y < s 1 r 2 a which imply s 1 s 2 (x + y) < (s 2 r 1 + s 1 r2) a. Hence

p ( x + y ) < = s2rl + s i r 2 r 1 r2

S1 S 2 S 1 S 2

Taking inf ima this gives p (x + y) < p(x) + p(y). F u r t h e r m o r e ,

p ( t x ) = t i n f s t x < r a , r e R , s e R +

r r ' f o r e v e r y t e R + . S i n c e s t x <= r a m e a n s s ' x <= r ' a w i t h s ' = s t G R + , r ' = r G R a n d s t s'

it fol lows tha t p( tx ) > tp(x). Converse ly , s ' x < r 'a impl ies s t x < ra with s = s' , r = r ' t

r ' r and ~; = ~ . Hence p(x) => p(tx)t which proves p( tx ) = tp(x).

To prove (iii) we app ly an abs t rac t vers ion of the H a h n - B a n a c h theo rem which has been deve loped by Rod~ [10]. F o r this p ick some fixed x o e M and define X = M , F : X ~ [ - o o , oo[ by F ( x ) = p ( x ) and G : X ~ [ - o o , oe[ by G = p on ray(xo) = { r x o t r e R, r > 0}, G(x) -- - oo for x ~ ray(xo). F u r t h e r m o r e , set

F = { g ( r , s ) : X x X --* X l ~ ( r , s ) ( x , y ) = rx + sy for all x, y e M ; r , s e R } .

Obvious ly , G < F on X and F(rE(r, s)(x, y)) <= rF(x ) + s F ( y ) for all r~(r, s)G F with r, s ~ 0. To show the inequal i ty G (~ (r, s) (x, y)) > r G (x) + s G (y) for x, y e X and r, s e R+ u {0} we cons ider var ious cases. If r = s -- 0 then by conven t ion 0 . ( - - oo) -- 0 the inequal i ty holds. S imi lar ly for r > 0 and s = 0. Let r > 0 and s > 0. F o r x r ray (xo) or y r r ay (Xo) the inequal i ty ho lds tr ivially. Hence let x ~ ray (xo) and y e ray (xo), i.e. x = r' Xo, y = s' Xo with r', s' R + u {0}. In this case we get G (rx + sy) = p fir' x o + ss ' xo) = p((rr' + ss ')xo) = (rr' + ss ' )p(xo) = rp(r 'xo) + sp(s 'xo) = rG(x) + sG(y) , which proves the des i red inequal i ty . By Rod~'s t heo rem [10, p. 474], there exists a funct ion ~0 : X --* [ - o0, ~ [ such tha t G < (0 < F on X and q~ (~ (r, s) (x, y)) = r ~0 (x) + s qo (y) for all x, y ~ X and r, s e R + w {0}. F r o m the def ini t ion of G and F it fol lows tha t ~0 < F = p on M and p(xo) = G(x o) < ~o(x o) < F ( x o) = P(Xo). F u r t h e r m o r e , 0 = q~(0) = ~o(x) + q~ ( - x ) implies tha t ~o(x)4= - o e for all x and hence ~0 is real-valued. It fol lows tha t 9 ( r x + s y ) = r q o ( x ) + s ( o ( y ) holds for all x, y e M and r, s e R . If x e P , then - x < 0 and m o n o t o n i c i t y of p gives p ( - x) < p (0) = 0. Hence ~0 ( - x) < p ( - x) < 0, i.e. q~ (x) = - q~ ( - x) > 0. This p roves the lemma. [ ]

Us ing L e m m a 1 we ob ta in the fol lowing geomet r ica l desc r ip t ion of the inf ini tes imal hul l /3 .

T h e o r e m 1 . For any cone P in a numerical module with archimedean element there holds

/3 = {x G M I f ( x ) > 0 for all f G P*} = {x e M I f ( x ) >= 0 for all f G P * } .


P r o o f. It suffices to show the first equality, because the second one follows from the Krein-Milman theorem when applied to the convex compact set P* c F-, u. Suppose first x ~/3, i.e. s(n) (a + nx) ~ P for all n ~ N. It follows s(n) ( f (a) + n f (x ) ) = f ( s (n ) (a + nx)) >= 0 for all f s P*. Hence 1 + n f ( x ) > 0 because of s(n) > 0 and f ( a ) = 1. Since this holds for all n e N we get f ( x ) > 0 for all f ~ P*. Conversely, let f ( x ) > 0 for some x e M and all f e P*. By Lemma 1 there exists an R-homomorphism g : M --, R such that # > 0 on P and p ( - x) = # ( - x). If g (a) = 0 and y e M, then there exists some r e R such that y < r a and g ( y ) < r g ( a ) = O . Hence g = 0 on M and p ( - x ) = 0 . If #(a) > 0 , then

1 p , f = o~a~a~ge and f ( - x ) < O by assumption. Hence p ( - x ) = g ( - x ) < O . By the

definition of p ( - x ) there exist for every n e N numbers r(n)~ R, s (n)e R § such that

r(n) < 1_ and s ( n ) ( - x) < r(n)a. This implies n s ( n ) ( - x) < nr(n)a < s(n)a and there- s (n) - n f o r e s ( n ) ( a + n x ) e P f o r a l l n e N . []

The following definition may be viewed a "linearization" of Dubois ' [3] notion of a Stone ring.

D e f i n i t i o n. A cone P in a numerical module M with archimedean element a is called a Stone cone, if a + nx E P for all n E N implies that x e P.

The following lemma is due to Dubois [3] within the framework of Stone rings. Dubois ' proof however applies also to the case of a Stone cone as defined above.

Lemma 2. I f P is a Stone cone and n o x ~ P for some n o e N and x e M, then x E P.

From Theorem 1 we derive some geometrical description of Stone cones which em- ploys the following assumption. A subring R c R is said to possess property I if for any r ~ R+ there is some s ~ R§ such that rs ~ N.

Obviously, any subfield of R, in particular ~ and R, possesses property I. The ring Z as well as a ring of algebraic integers Z + 2g x/~ for d e N also possesses property I.

Corollary 1. Given a numerical module over a subring of the reals which possesses property I and given an archimedean element, the following statements hold for cone P.

(i) The infinitesimal hull/3 is the smallest Stone cone in the module containing P. (ii) The following properties are equivalent:

(a) P is a Stone cone; (b) P = /3 ;

(c) P = {x e M I f ( x ) >= 0 for all f e P*} (There P* may be replaced also by P*).

P r o o f. (i)/3 is a cone by Theorem 1. Moreover, it follows tha t /3 is a Stone cone in the module M with respect to the given archimedean element a. If Q is a Stone cone (for the given a) with P c Q, then P c (~. We show Q c Q. If x e Q, then s(n) (a + nx) ~ Q for s(n)~ R+ and all n E N. By property I there exists r(n)~ R+ such that s(n)r(n) = re(n)~ N. Therefore re(n)(a + nx )~ Q for all n e N.

74 U. KRAUSE ARCH. MATH.

According to Lemma 2, a + nx �9 Q for all n �9 N. Since Q is a Stone cone, x �9 Q. (ii) (a) implies (b) by (i). (b) implies (c) by Theorem 1. Obviously, (c) implies (a). [ ]

Corol lary 1 yields the following general version of the Farkas lemma (cf. also the next section).

Corollary 2. Let E be a subset in a module M over a subring of the reals possessing property I and such that the cone generated by E is a Stone cone. Then an element x ~ M belongs to the cone generated by E if and only if f (e) > 0 for all e �9 E and any R-homomorphism f : M ~ ~ implies that f (x) > O.

P r o o f. The cone P generated by E is given by the sums ~ r e e where r e = 0, except e e E

for finitely many e �9 E with r e �9 R +. The proper ty in Corol lary 2 involving the R-homo- morphisms implies that f ( x ) > 0 for all f e P*. Hence x ~ P by part (ii) of Corol- lary 1. [ ]

Another consequence of Theorem 1 is a character izat ion of pointed Stone cones as certain cones of cont inuous functions, which extends a character izat ion given by Davis [2] for the case of the reals and in finite dimensions. (Cf. also the next section.) A cone P in the module M is pointed, if P n ( - P ) = {0}. ( - P = { - x [ x �9 P}).

Fo r a compact topological space T C(T) denotes the R-module of all real-valued continuous functions on T and C§ (T) denotes the standard cone of all nonnegative functions in C(T). (Thereby R some subring of ~ . )

Theorem 2. The pointed Stone cones, with respect to R-modules over a ring R possessing property I, are up to an isomorphism of cones given precisely by the intersections of a standard cone C+ (T) with an R-submodule of C(T) containing 1.

P r o o f . (1) Consider for some compact space T P = M n C + ( T ) where M is an R-submodule of C(T) with 1 �9 M. P is a cone (over R) and 1 is an archimedean element because for a given f �9 M there is some n �9 N such that n �9 1 - f ~ P and because of r ~ C+ (T) for r < 0. If 1 + n f ~ P for some f e M and all n �9 N, then 1 + n f > 0 pointwise on T for all n �9 N and therefore f �9 C+ (T). Hence P is a Stone cone which is pointed.

(2) Let M an R-module of the kind considered and P c M a pointed Stone cone with archimedean element a. Take T the closure of P* ~ R M with respect to pointwise convergence. Define

qg:M ~ C ( T ) by r for t E T .

~o is an R-homomorphism. The image H = q~ (M) is an R-submodule of C (T), and I r H because of ~o (a) (t) = t (a) = 1 for t e T. By definition, ~o (P) c H c~ C + (T). Conversely, let ~0(x) ~ H n C+ (T) for x e M. It follows 0 < ~0(x) (t) = t(x) for all t �9 T. Since P is a Stone cone part (ii) of Corol lary 1 yields that x e P. Thus tp(P) = H n C+ (T). If ~0(x) = tp (y), then t (x) = t (y) for all t �9 T and hence x - y �9 P n ( - P) = {0}. Thus q~ is injective, and ~o : P ~ H n C + (T) an i somorphism of cones over R. [ ]


In the next sect ion it will be shown how Stone cones m a y arise f rom finitely genera ted

cones. The fo l lowing two qui te different s imple examples i l lustrate the case of no t finitely

genera ted cones.

E x a m p l e s . 1) Ice cream cone. Let R -- R, M = R 3 and

P = {X = (X1 ,X2 , X 3 ) ~ 3 I X 1 > 1 Ilxll} = 2

(tl " II the Euclidean norm.) Obviously, P is not finitely generated over R. Any interior point of P may serve as an archimedean element, e.g. a = (1, 0, 0). Since P is closed (for the Euclidean topology), it follows that P is a Stone cone. How does its representation according to Theorem 2 look like? If T = { y ~ R 3 [ IlYll = 1}, it is given by t p : R 3 --*C(T), q~(x)(y)=x 1 - � 8 9 inner product on Px3). Indeed ~ is an •-homomorphism and ~ is injective. Furthermore, because of [I x II = sup { ( x , y)[ll y II = t}, x e e is equivalent to tp (x) > 0 pointwise on T. Hence P is isomorphic to the intersection of the standard cone C+ (T) with the subspace H = r Although P is finite dimensional it does not have a finite representation, i.e. the infinite dimensional C+ (T) cannot be replaced by some standard cone R~+.

2) Multiplicative unit interval. Let R = Z, M = {x ~ P'-I x > 0} equipped with multiplication, and P = {x e R I 0 < x < 1 }. Z operates on M by taking powers, (n, x) ~ x". The 7l-modul M is not finitely generated. P is a cone in the module M. Any element ~ 1 in P may. serve as an

archimedean element, e.g. a = �89 If ax" ~ P for all n e N, then x" < 1_ for all n ~ N and hence x < 1. - - a

Therefore P is a Stone cone. What is the representation of P in the sense of Theorem 2? Consider ~0 : M --* R, ~0 (x) = - log x. tp is a Z-homomorphism and q~ is injective. Furthermore, x e P is equivalent to ~0 (x) > 0. Hence P is isomorphic (as a Z-cone) to the intersection of the standard cone R+ with the subgroup ~o(M) of R, i.e. P is isomorphic to R + equipped with addition. Although P is not finitely generated it does have a finite representation. (T consists of only one point.)

2. Finitely generated cones. Let R be a subr ing of P~ with 1 ~ R and let M be an

R-module . F o r a subse t 0 4: E c M let

c(E) = f ~, rye [ re ~ R, r e >= O, r e 4:0 only for finitely m a n y e~ [eeE )

denote the cone generated by E (over R). A cone P c M is finitely generated, if P = c(E) for some finite set E c M. A subset 0 + E c M is R+-independent, if for any finite

subset 0 4: Eo c E and any r e a R ~ it is impossible that Z re e = 0. (There, e~Eo

R+ = {r ~ R Ir > 0}.) A subset 0 4: E c M is R-independent, if for any finite subset

0 4: Eo c E and any re ~ R\{0} it is imposs ib le that Z re e = 0. O f course, if E is e~Eo

R-independen t , then it is R +- independent ; but not vice versa in general. The cone P is

said to generate M properly, if M = P - P 4: P.

Let P = c(E) a cone which is genera ted by the finite and R + - i n d e p e n d e n t set

0 4: E ~ M and assume M - - P - P. T h e n a = Y~ e is an a r ch imedean element. In e~E

general howeve r P = e(E) need no t be a S tone cone as m a y be seen f rom the fo l lowing example.


E x a m p l e . Let R = Z, M = Z, E = {2, 3} and P = c(E). It is easily seen that

P = {n~Tg[n > O,n ~: 1}.

E is Z +-independent and M = P - P. Hence a = 5 is an archimedean element. (Actually, any natural number > 2 is an archimedean element.) For x = 1, a + nx ~ P for all n e N but 1 d~ P. Hence P is not a Stone cone.

In the above example the cone P is no t integral ly closed in the fo l lowing sense. A cone

P in a numer ica l m o d u l e M is integrally closed (in M ) whenever n x ~ P for n ~ lq and

x ~ M implies that x ~ P.

By L e m m a 2, any Stone cone is integral ly closed. The reverse impl ica t ion is no t t rue

in general as the fol lowing example shows.

E x a m p 1 e. Let R = R, M = ~:~2 and P = {x = (x 1, X2) E •2 I Xt > 0, X 2 ~> 0}. Obviously, P is integrally closed. Any element of P may serve as archimedean element but for none of these P is a Stone cone. E.g. for x = (0, 1) and a ~ P a + nx e P for all n e lq, but x r P.

In the fo l lowing we shall show that in a numer ica l m o d u l e over a r ing which has

p roper ty I a finitely genera ted cone is a S tone cone if and only if it is integral ly closed and

it generates p roper ly the module . (The cone in the example above is no t finitely genera ted

over R. ) F o r this we need the fol lowing vers ion of Cara th60dory ' s t heo rem (cf. [9]) for

numer ica l modules .

L e m m a 3. Le t P be a cone in the R-module M and let x ~ P \ { 0 } and E c P\{0} , E

finite, such that

r x = ~, r~e with r, r e E R + . e ~ E

Then E contains an R-independent subset U such that

s x = ~, s ,u with s, s u e R +. uEU

P r o o f. I f E is a l ready R- independent , then take U = E. Otherwise , there holds an

equa t ion Z t e e = 0 with t e E R and t e > 0 for at least one e ~ E. Hence there exists some e ~ E

t e t c c ~ E such that 0 < m a x - = - . It fol lows that

e E E r e r c

t c r x = E t c r e e - E r c t e e = E se e e ~ E e ~ E e ~ E '

where s e = tc re - rcte ~ R +, E ' c E \ { c } , E ' ~ 0 since x 4= 0. If E ' is R- independen t , then

take U = E' , s -=- t c r. Otherwise , the above p rocedure can be appl ied to E'. Since E is

finite, the process stops wi th a representa t ion of the k ind required. [ ]

L e m m a 4. In a numerical module M over a subring o f the reals which has property I

a f ini tely generated cone P is a Stone cone i f and only i f P is integrally closed and P generates M properly.


P r o o f. Let P = c(E), E a finite subset of M with 0 r E. Suppose first, P is a Stone

cone. P generates M properly, i.e. M = P - P 4: P. By Lemma 2, P is integrally closed. Conversely, assume P = c (E) integrally closed and M = P - P oe P. First we show that

a = Z e is an archimedean element. If x = Y~ x e e e P - P = M, then with r the maxi- e c E e~E

mum of the numbers xe r a - x = ~, ( r - x ~ ) e e P . If - r a = ~, - r e a P for some eEE eEg

r ~ R + , then - r e o = ~ - r e + Z r e a P for any e o e E . For n ~ l q such that eEE eEE, e * e o

n r > l it follows that - e o = - n r e o + ( n r - 1 ) e o ~ P . Hence - E c P and M = P - P c P which contradicts M * P. Thus ra ~ P is possible only for r = 0 or r ~ R + and a is an archimedean element. It remains to show, that P is a Stone cone for

a = ~ e. Suppose a + nx ~ P for some x ~ M and for all n ~ N. If a + nx - 0 for some eEE

n, then x = a + (n + 1) x ~ P. Hence we may assume a + nx 4:0 for all n and by Lemma 3 there exist for every n E ~q s (n) ~ R + and an R-independent set U (n) c E such that

s ( n ) ( a + n x ) = ~ SuU, where s u ~ R + for u ~ U ( n ) . u~U(n)

Since E is finite, the collection {U (n)}n~N is finite too. Hence there exists some k ~ N and some infinite subset N of N such that for all n ~ N

s ( n ) ( a + n x ) ~ C where C={~v~(u k ) r u u l r u ~ R + u{O}}"

For m, n e N with m < n it follows that ns (n ) s (m)x - ms (m)s (n )x e C - C and hence r x ~ C - C for r = s(m)s(n) (n - m) ~ R+. Similarly, ra ~ C - C. Therefore

r x = ~. x~u, r a = ~ auu with x ~ , a u e R . ueU(k) ueU(k)

From r s ( n ) ( a + n x ) e C for n e N we have r s ( n ) ( a + n x ) = ~. r~(n)u with

r u (n)e R+ u {0}. Putt ing together we get u~ v(k)

E (s(n)au + s(n)nxu)u = 7~ ru(n)u. u~U(t) ue U (k)

Because U(k) is an R-independent set this implies s(n)(au + nx~) = r~(n) > 0 for all

u e U(k) and all n e N. It follows that a~ + nxu > 0 and xu >__ - a~ for all u and all n. n

Hence x~ ->_ 0 for all u and r x = Y. x~ u ~ C c P. By assumption the ring R has prop- ' ueU(k)

erty I and hence nx e P for some n ~ P. Since P is integrally closed we obtain x e P. Thus P is a Stone cone. []

The property of integral closedness of a cone finitely generated by E can be expressed in terms of E by employing the concept of a Hilbert basis. Since in the following different subrings of P, will occur we denote the cone c (E) generated by a subset E of the R-module M more precisely by cR (E). Let R c S c R, R and S subrings of ~, (containing 1, as


always) and let M be an R-module which is contained in some S-module. A finite set E c M is called an S-Hilbert basis in M, if

c s (E) n M ~ c R (E).

The not ion of a Hilbert basis as it is introduced in [1] (cf. also [4]) is the part icular case where R = Z, S = R and M = :E n. The following theorem describes in terms of the Hitbert basis integral closedness as well as the holding of some version of Fa rkas lemma.

Theorem 3. Let R be a subring of R with property 1 and let M be generated as an R-module by the finite set E. Then the following statements are equivalent:

(i) E is an S-Hilbert basis in M for the ring of quotients

(ii) P = cR (E) is integrally closed (in M). (iii) For any x e M, x e CR(E ) if and only if f (e) > O for all e e E and any R-homomor-

phism f : M ~ R implies that f (x) >= O.

P r o o f. (i) =~ (ii): Suppose nx e P for some x e M and some n e N, i.e. nx = ~., ree e~E

with re e R, r e => 0. Since M is contained in some S-module we may divide by n to obtain

re x = Z - - e e c s ( E ) c ~ M c CR(E) = P

e~E n

because E is assumed to be an S-Hilbert basis.

(ii) ~ (iii): M = P - P because M is assumed to be generated by E. If P = M, then (iii) holds trivially. If P t M then Lemma 4 applies and P is a Stone cone. By Corol lary 2 of Section 1 statement (iii) follows.

(iii) ~*- (i): Suppose x e cs(E) c~ M, i.e. x Z re = - - e with r e e R, r e > 0 and n e e N. F o r e c E He

n = I~n~ , me ,= 17 n e it follows that n x = ~ meree. Hence f ( x ) > O for any e~E e~g ,e :~e ' e~E

R-homomorphism f : M - ~ R for which f ( e ) > 0 for all e ~ E. By (iii) therefore

x e CR (E). []

Consider a subring R of IR which has proper ty I and which is divisible in the sense that for any r e R and any n ~ Z~{0} there exists some s e R such that sn = r. Then any finite subset E of an R-module M is an S-Hilbert basis in the R-module lin (E) generated by E in the sense of s tatement (i) if Theorem 3. By Theorem 3 therefore P = CR (E) is integrally d o s e d and, by Lemma 4, a Stone cone. Since any subfield of R is divisible, it follows, for example, that every finitely generated cone in a real vector space is a Stone cone. Therefore, and since R - h o m o m o r p h i s m s on a subspace of R n are induced by the inner product , the common Fa rkas lemma in R n (cf. [9]) is a special case of Theorem 3. The case of a non-divisible R is i l lustrated by the following simple example.


E x a m p l e . Let R = Z , S = R , M = Z " . Theorem3 may be also used to check i f a given generating set E of Z" is a Hilbert basis. For example E = {(1, 2), (l, l)} generates Z 2 and cz(E) = {(m, n) ~ ;EEl0 _< m < n). Since cz(E ) is integrally dosed in ~E 2, E is a Hilbert basis (cf. also [4]). On the other hand, coming back to a previous example, E = {2, 3} generates the Z-modul Z and cz(E) = {n e Z ln __> 0, n # 1} is not integrally closed in ;l. By Theorem 3 therefore E cannot be a (l~-Hilbert basis and afort ior i no Hilbert basis. Indeed, 1 -~- x. 2 + g 31. e co(E ) c~ 2t, but 1 ~ cz(E ).

A further proper ty of finitely generated cones which allows to get some finite representat ion from Theorem 2 is given in the following lemma.

Lemma 5. Let P be a cone in the R-module M such that P = c R (E) for some finite set E and M = P - P. Then for any archimedean element a (e.g. a = ~, e) the set P* is finite.

e E E

P r o o f. Let E ( f ) = {e �9 E I f (e) > 0} for f �9 P*. Since E is finite, there exists a finite family F c P* such that { E ( f ) l f �9 P*} -- { E ( f ) I f �9 f } . Suppose f �9 P* and choose

f ~ e / x

g �9 F with E ( f ) = E(#). Then min } J t e ) [ e � 9 E ( g ) / = r with r, s � 9 g + . It follows that ( g ( e ) ~ s

rg (e) <_ s f ( e ) for all e ~ E and hence rg ~ s f on P or tg <= f on P with 0 < t ~ 1. Suppose t = 1. To x �9 P there exists n ~ N with x < n a. Hence 0 < f (x) - g (x) < n f (a) - n g (a) = 0

which implies f = g. I f t < 1 t h e n f = tg + (l - t) with g, ~ P * . I f f �9 P~*,

i.e. if f is extremal, then f = g. Thus f = g e F. [ ]

Theorem 4. A cone P in a numerical module over a ring possessing property I which is pointed, finitely generated and integrally closed in P - P is isomorphic to the intersection of a standard cone R"+ with an R-submodule o f hR'.

P r o o f . I f P - P = P then P = P c ~ ( - P ) = { 0 } and P = g + c ~ { 0 } . I f P - P # P then by Lemma 4 P is a Stone cone. Hence by Theorem 2 P is isomorphic to the intersection of C+ (T) with an R-submodule of C (T) where T is the pointwise closure of Pe* in N M, M = P -- P. By Lemma 5 P* is finite and hence C(T) = N" where n is the number of elements in P* . [ ]

In case the ring R is divisible, by Theorem 3 the assumption of integral closedness may be omit ted in Theorem 4. F o r the special case R = N Theorem 4 specializes to a theorem of Davis [2]. (Cf. also [8] for a different approach.) In general however the assumption of integral closedness cannot be cancelled in Theorem 4 as is shown by the cone P = Cz({2, 3}) in 71 which is not integrally closed and not i somorphic to the intersection of a s tandard cone with a group. F o r R = 7 / a cone P as in Theorem 4 is a special kind of a monoid of finite character and the representat ion Theorem 4 is related to the divisor theory for such monoids (cf. [7]).

References

[1] W. COOK, L. Lov~,sz and A. SCHRIJVER, A polynomial time test for total dual integrality in fixed dimension. Math. Programming Stud. 22, 64-69 (1984).

[2] C. DAVIS, Remarks on a previous paper. Michigan Math. J. 2, 23-25 (1953).


[3] D.W. DtrBoxs, A note on David Harrison's theory of preprimes. Pacific J. Math. 21, 15-19 (1967).

[4] J. JEGIER and A. RYCERZ, An integer version of Weyl's theorem. Opuscula Math. 4, 63-72 (1988).

[5] V. KLF~, Extreme points of convex sets without completeness of the scalar field. Matematika 11, 59-63 (1964).

[6] U. KRAtJSE, On Stone groups. Arch. Math. 42, 260-266 (1984). [7] U. KRAUSE, On monoids of finite real character. Proc. Amer. Math. Soc. 105, 546-554 (1989). [8] D. A. MARCUS, Normal semimodules: A theory of generalized convex cones. Canad. J. Math.

36, 156-177 (1984). [9] R.T. ROCICA~LLAR, Convex Analysis. Princeton 1970.

[10] G. ROD~, Eine abstrakte Version des Satzes yon Hahn-Banach. Arch. Math. 31, 474-481 (1978).

Eingegangen am 21.6. 1990

Anschrift des Autors:

Ulrich Krause Fachbereich Mathematik und Informatik Universit/it Bremen Postfach 330440 DW-2800 Bremen 33

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Cones in numerical modules