Using geometrical methods we construct primary monoids whose complete integral closure is not completely integrally closed. Such monoids cannot be realized as multiplicative monoids of integral domains with finitely generated groups of

divisibility.

Complete integral closure, Primary monoids

A. GEROLDINGER

F. HALTER-KOCH

G. LETTL

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Rendiconti di Matematica, Serie VIIVolume 15, Roma (1995), 281-292The complete integral closureof monoids and domains IIA. GEROLDINGER – F. HALTER-KOCH – G. LETTLRiassunto: Utilizzando metodi geometrici, vengono costruiti monoidi primari lacui chiusura integrale completa non è completamente integralmente chiusa. Tali mo-noidi non possono essere ottenuti come monoidi moltiplicativi di domini di integritàcon gruppo di divisibilità finitamente generato.Abstract: Using geometrical methods we construct primary monoids whose com-plete integral closure is not completely integrally closed. Such monoids cannot be re-alized as multiplicative monoids of integral domains with finitely generated groups ofdivisibility.1 – IntroductionIn this note we study the (complete) integral closure of monoids Htogether with their groups of divisibility G(H) = Q(H)/H,. We willshow how under certain assumptions these investigations may be reducedto the case where G(H) = Q(H) is torsion free (Theorem 1 and Corol-lary 1); in particular, this works if G(H) is finitely generated. A monoidwith torsion free quotient group may be considered as a submonoid ofa real vector space. For such a monoid H we characterize its integralclosure EH and the complete integral closure MEH of EH in geometrical terms(Theorem 2).Key Words and Phrases: Complete integral closure – Primary monoidsA.M.S. Classification: 20M14 – 13B20282 A. GEROLDINGER – F. HALTER-KOCH – G. LETTL [2]This allows us to construct primary monoids H whose complete in-tegral closure MH is not completely integrally closed (Theorem 3). In [1;Theorem 4] it was proved thatMEH is completely integrally closed for allprimary monoids H; hence this latter result is sharp.In section 5 we characterize integral domains R with finitely gen-erated groups of divisibility whose multiplicative monoids are primary.In particular it will turn out that the complete integral closure of thesedomains is completely integrally closed.2 – PreliminariesThroughout this paper, a monoid means a commutative and cancella-tive semigroup with unit element. In this section and in the following onewe use multiplicative notation.Let H be a monoid; then H, denotes its group of invertible elementsand Q(H) a quotient group of H with H 9 Q(H); H is called reducedif H, = {1}. H is said to be primary , if H %= H, and if a, b " H andb %" H,, then a|bn for some n " IN+. The integral closure EH 9 Q(H) andthe complete integral closure MH 9 Q(H) are defined byEH = {x " Q(H) | xn " H for some n " IN+}andMH = {x " Q(H) | there exists some c " H s.t. cxn " H for all n " IN+}.H is called integrally closed if H = EH, and it is called completely integrallyclosed if H = MH.Clearly, we haveH 9 EH 9 MH 9 Q(H);furthermore EH and MH are integrally closed but in general MH is not com-pletely integrally closed. If H is primary thenMEH is completely integrallyclosed (cf. [1; Theorem 4]).In analogy to the appropriate notion in ring theory we defineG(H) = Q(H)/H,as the group of divisibility of H.[3] The complete integral closure etc. 283As usual, we say that an abelian group G is bounded , if there existsan n " IN+ such that gn = 1 for all g " G.Proposition 1. Let H be a monoid, G an abelian group and1 : Q(H) ) G a group epimorphism with kernel K. Then we have:1. If KH,/H, is a torsion group, then 1(EH) = "1(H) and1(H,) = 1(H),.2. If KH,/H, is bounded, then 1(MH) = #1(H).3. If K 9 H, and G is a free abelian group, then H ? K & 1(H).Proof. Clearly, G is a quotient group of 1(H).1. Obviously 1(EH) 9 "1(H) and 1(H,) 9 1(H),. Conversely, lety " "1(H) 9 G be given; then there are x " Q(H), a " H and somen " IN+ such that y = 1(x) and yn = 1(a). Thus xn = as for some s "Ker(1) = K. Since KH,/H, is a torsion group, there exists an m " IN+with sm " H,. This yields xnm " H, x " EH whence y = 1(x) " 1(EH).Next, let 1(a) " 1(H), with a " H. Then there exist a b " H and ans " K such that ab = s. Since there is an m " IN+ for which sm " H,,it follows that a " H,.2. We have 1(MH) 9 #1(H) and in order to verify the opposite inclusionwe take an element y = 1(x) " #1(H) with x = a"1b for some a, b " H.Then there exists an element c " H such that for all n " IN+ cxn = dnsnfor some dn " H and sn " K. Let ( " IN+ be such that s, " H, for alls " K, whence c,x,n " H for all n " IN+. Setting c! = c,a,"1 we inferthat c!xm " H for all m " IN+ whence x " MH.3. Since G is free abelian, the exact sequence1 ) K ) Q(H) ) G ) 1splits; let & : G ) Q(H) be the group monomorphism with 1 ! & = idG.We set H % = &(1(H)); then H % ? 1(H), H % 9 H, Q(H %) = &(G) andQ(H) ? K @ Q(H %). Hence the product K & H % is a direct one, whichobviously is contained in H.Conversely, let a " H be given. Then a = *b with * " K andb " (& ! 1)(Q(H)). Hence b = &(x) with x " G,1(a) = 1(b) = (1 ! &)(x) = x " 1(H)284 A. GEROLDINGER – F. HALTER-KOCH – G. LETTL [4]and thus b = &(x) " &(1(H)) = H %.3 – (Complete) integral closure and groups of divisibilityThe aim of this section is to point out a way how to reduce investi-gations concerning the (complete) integral closure of primary monoids Hto reduced primary monoids H! having torsion free groups of divisibility.The relationship between H and H! will be most intimate if H is reducedand G(H) is a direct sum of a bounded group and a free abelian group.Theorem 1. Let H be a monoid, 1 : Q(H) ) Q(H)/EH, thecanonical epimorphism and H! = 1(H). Then we have1. H! is reduced, G(H!) = G(EH) is torsion free and EH,/H, is thetorsion subgroup of G(H).2. NH! ? EH/EH,, and if G(EH) is free abelian then EH ? EH, & NH!.3. Suppose EH,/H, is bounded. Then OH! ? MH/EH,, and if G(EH) isfree abelian then MH ? EH, & OH!.4. H is primary if and only if H! is primary.Proof.1. H! is reduced by Proposition 1; hence G(H!) = Q(H!) =Q(H)/EH, = G(EH). Clearly G(EH) is torsion free and EH,/H, is thetorsion group of G(H) because G(H)/(EH,/H,) ? G(EH).2. and 3. are consequences of Proposition 1.4. This follows from [2; Lemma 2].Corollary 1. Let H be a reduced monoid and suppose that G(H)is a direct sum of a bounded group T and a free abelian group. ThenEH ? T & NH! and MH ? T & OH!.Proof. Since H, = {1} , T = EH, is the torsion subgroup of G(H) =Q(H) and G(EH) = G(H)/T . Thus the assertion follows immediately fromthe previous theorem.[5] The complete integral closure etc. 2854 – Geometrical methodsIn this section we investigate monoids having torsion free quotientgroups. A torsion free abelian group G is a flat Z-module and hence theinjection Z 7) Q induces an injection G = G > ZZ 7) G > Q. Using baseextension again, the Q-vector space G > Q embeds into the real vectorspace G > IR. Throughout this section we use additive notation for theoperation of monoids. Therefore we may consider a monoid H with Q(H)torsion free as a submonoid of the additive group of the real vector spaceV = Q(H) > IR, and obviously H contains a basis of V . This allows usto study H using geometrical methods in V .Our first aim is to derive geometrical descriptions of EH and MEH andto obtain a geometrical characterization of being primary. From this wesee thatMEH is completely integrally closed, if the quotient group of H isfinitely generated, and it enables us to construct primary monoids H forwhich MH is not completely integrally closed.We recall some geometrical notations. Let V be a real vector space.For two distinct elements x, y " V[x, y] = {y + ((x $ y) | 0 . ( . 1}denotes the line segment joining x and y; we set [x, y) = [x, y] \ {y} and(x, y) = [x, y) \ {x}.Let M 9 V be a subset; M is called (algebraically) open if for alla " M and for all a %= x " V there exists some b " (a, x) for which[a, b] 9 M . M is called convex if [a, b] 9 M for all a, b " M . We denotebyC(M)=6 %x'M(xx)))(x " IR-0 and (x =0 for all but finitely many x"M7.the convex cone with apex 0 " V which is generated by M ; C(M) 9 V isa convex set. Finallylin (M) = M 0 1x " V | [a, x) 9 M for some a " M with a %= x2is the set of points which are either in M or linearly accessible from M .286 A. GEROLDINGER – F. HALTER-KOCH – G. LETTL [6]If M is a convex subset with non empty interior in a topological vectorspace then M is algebraically open if and only if M is topologically open,and lin (M) is just the topological closure M of M ( [5; §11 A, Lemma]).Theorem 2. Let H be a nontrivial monoid such that Q(H) istorsion free and suppose that Q(H) 9 Q(H) > Q 9 Q(H) > IR = V .Then we have1. EH = C(H) 4 Q(H).2. C(EH) = C(H).3.MEH = lin (C(H)) 4 Q(H).4. H is primary and reduced if and only if C(H) \ {0} is open andC(H) %= V .Proof.1. Let a " EH 9 Q(H); then we have na " H 9 C(H) for somen " IN+, which yields a " C(H) 4 Q(H).Conversely, let 0 %= a " C(H) 4 Q(H) be given. Hence there existhi " H and (i " IR>0 with a =3ki=1 (ihi. Since a and hi are containedin the rational vector space Q(H)>Q, ((1, . . . , (k) may be interpreted asa solution of a system of linear equations over Q. Since this system has apositive solution, it has a positive rational one, and thus we may assumethat all (i " Q>0. Let n " IN+ be such that n(i " IN+ for 1 . i . k;then na " H and hence a " EH.2. We haveC(H) 9 C(EH) 9 C(C(H)) = C(H).3. Let a " MEH 9 Q(H) be given; by definition there exists some c " EHsuch that c+aIN+ 9 EH 9 C(H). Therefore 1nc+a " C(H) for all n " IN+.Thus[c + a, a) =Pn-14c + a, 1nc + a5 9 C(H)which implies a " lin (C(H)).To verify the opposite inclusion, we take an element a " lin (C(H))4Q(H). Hence there is an element c% " C(H) such that[c%, a) =1a + ((c% $ a) | 0 < ( . 12 9 C(H),[7] The complete integral closure etc. 287which implies that (n + 1)(a + 1n+1(c% $ a)) = c% + na " C(H) for alln " IN+. If c% =3k/=1 (/c/ with c/ " H and (/ > 0, choose m " IN+with 0 < (/ < m for all 1 . ) . k. Thenm(k%/=1c/) + na " C(H) 4 Q(H) = EHfor all n " IN+, and thus a " MEH.4. Suppose that C(H) \ {0} is open and let a, b " H \ {0} be given.Then there exists some c " [a, a $ b) = 1a + (($b) | 0 . ( < 12 suchthat [a, c] 9 C(H), and therefore a $ 1nb " C(H) for some n " IN+. Thenna $ b " C(H) 4 Q(H) = EH and hence there exists some m " IN+ suchthat m(na$ b) " H which yields mna$ b " H. Since a, b " H \{0} werearbitrary, we conclude that either H is reduced and primary or H = H,,which yields the contradiction C(H) = V .Conversely, suppose that H is a reduced, primary monoid; then ob-viously C(H) %= V . Let 0 %= a " C(H) and x " V be given. Since H con-tains a basis of V , there are h1, . . . , hn " H such that a =3ni=1 5ihi, x $a =3ki=1 µihi $3ni=k+1 µihi with all 5i, µi 2 0. We may assume that51 > 0.If k = n everything is clear; otherwise we have[a, x] =651h1 +& n%i=25ihi + (k%i=1µihi' $ (n%i=k+1µihi))) 0 . ( . 17=6 51n $ kn%i=k+1&h1 $ (µi(n $ k)51hi'+ c,)))0 . ( . 17with c, " C(H).Since H is reduced, h1 /" H,, and since H is primary, there is an m " IN+such that mh1 $ hi " H for k + 1 . i . n. Thus h1 $ (µi(n"k).1 hi " C(H)if ( . .1mµi(n"k) for all k + 1 . i . n with µi > 0, which implies that[a, b] 9 C(H) for some b " (a, x).Remark. Let H be a monoid such that Q(H) is finitely generatedand torsion free. Then Q(H) > IR has finite dimension, the interior of288 A. GEROLDINGER – F. HALTER-KOCH – G. LETTL [8]C(H) is non empty and henceMEH = C(H) 4 Q(H).This immediately shows thatMEH is completely integrally closed. Sinceevery monoid with finitely generated quotient group is a G-monoid, thisalso follows from [1; Theorem 4].This result cannot be enbettered even for primary monoids: the nexttheorem exhibits primary monoids H 9 ZZs with 2 . s . / for whichall inclusionsH 3 EH 3 MH 3 MEHare strict; in particular MH %= MH.Theorem 3. For every d " IN+ 0 {/} there exist reduced primarymonoids H with dim IR&Q(H)> IR' = d+1 such that MH is not completelyintegrally closed.Proof.1. Let d " IN+ 0 {/} be given and put I = {i " IN | 0 . i < d + 1}.For 0 %= i " I choose %i " IR>0 with3di=1 %i < / and defineH =6(xi)i'I " IN I+))) supi'I|xi| < / andd%i=1%ixi . x207.Thus H consists of all bounded sequences x " IN I+ for which x0 is suf-ficiently large. If (xi)i'I , (yi)i'I " H then3di=1 %i(xi + yi) . x20 + y20 .(x0 + y0)2, hence H 9 ZZI is a monoid which obviously is reduced, andwe claim that(1) Q(H) =6(xi)i'I " ZZI))) supi'I|xi| < /7.Indeed, a sequence (xi)i'I " ZZI is bounded if and only if it is the di!erenceof two bounded sequences in IN I+ with su$ciently large first component.2. In the second step we will determine EH and MEH. PutH1 =6(xi)i'I " IN I+))) supi'I|xi| < /70 102[9] The complete integral closure etc. 289andH2 =6(xi)i'I " INI))) supi'I|xi| < /7.Let x = (xi)i'I " H1 and m " IN+ with m 23di=1 %ixi. Then we obtainmx " H and hence 1nx " C(H) for all n " IN+. Thus we have(2)Pn-11nH1 9 C(H) 9 IR I>0 0 {0},where the second inclusion is obvious. Now Theorem 2.1 and (1) yieldEH = C(H) 4 Q(H) = H1.Now let y = (yi)i'I " H2. Then x = (max{yi, 1})i'I " H1 9 C(H)and for any n " IN+ x(n) = (max{yi, 1n})i'I " 1nH1 9 C(H), which yields[x, y) =Fn-1 [x, x(n)] 9 C(H). Thus we have shown thatH2 9 lin&C(H)' 9 IR I-0.Using Theorem 2.3 and (1) yieldsMEH = lin&C(H)' 4 Q(H) = H2.3. Now we will show that H is a primary monoid. If d is finite wededuce from (2) that C(H) = IRd+1>0 0 {0}, thus by Theorem 2.4 H isprimary.For d = / we will give a direct proof which is easier than show-ing that C(H) is (algebraically) open. Let x = (xi)i'I , y = (yi)i'I "H \ {0}. For su$ciently large n " IN we have nxi > yi for all i " I and3di=1 %i(nxi $ yi) . (nx0 $ y0)2, which shows nx $ y " H.4. In the final step we show thatMH =6(xi)i'I " H2)))x0 > 070 102.This yields MH %= MEH and thus MH is not completely integrally closed. SinceMH 9 MEH = H2 we must prove that for 0 %= x " H2 we have: x " MH if andonly if x0 %= 0.290 A. GEROLDINGER – F. HALTER-KOCH – G. LETTL [10]First consider x = (xi)i'I " H2 with x0 = 0 and xj %= 0 for somej " I. For any c = (ci)i'I " H we have for su$ciently large n " IN+ :3di=1 %i(ci + nxi) 2 %jnxj > c20, thus c + nx /" H and x /" MH.Now let x = (xi)i'I " H2 with x0 2 1. Put m = max i'I |xi|, choosec0 " IN+ with c0 2 m3di=1 %i and let c = (c0, 1, 1, . . . ) " H. Thenone easily checks that c + nx " H holds for all n " IN which provesx " MH.Remark.1. Considering the finite dimensional case d " IN+, a quick glance atthe above proof exhibitsC(H) = IRd+1>0 , lin&C(H)' = IRd+1-0 , EH = INd+1+ ,MH = IN+ & INd and MEH = INd+1 .2. For every submonoid H 9 ZZ with Q(H) = ZZ we have EH "{$IN, IN,ZZ}. Thus EH is completely integrally closed and the assumptiond + 1 2 2 in the previous theorem cannot be improved.5 – Integral domainsFor an integral domain R let R• = R \ {0} denote the multiplicativemonoid of R, R, = R•, its group of units, G(R) = G(R•) the group ofdivisibility and R the integral closure of R in its quotient field.We shall make use of the following simple observation: let A, B betwo domains with A 9 B whose quotient fields coincide; then they giverise to the exact sequence of abelian groups(3) 1 ) B,/A, ) G(A) ) G(B) ) 1Theorem 4. Let R be an integral domain. Then the followingconditions are equivalent:1. R• is primary and G(R) is finitely generated.2. R is a one-dimensional local noetherian domain, its integral clo-sure R is a finitely generated R-module and (R,: R,) < /.[11] The complete integral closure etc. 291Proof.1. =- 2. R is one-dimensional and local because R• is primary ([4;Theorem 4.1]). Since G(R) is finitely generated, R is a finitely generatedR-module and R,/R, is finite by [3; Theorem 3.9].It remains to verify that R is noetherian. G(R) has finite torsionfree rank and therefore, by [3; Theorem 2.1], R is the intersection ofjust finitely many valuation overrings, say R =Qni=1 Vi with Vi $ Vj fori %= j. Then, for 1 . i . n, we have Vi = RPi for Pi 8 R prime ( [6;Theorem 12.2]). Since R is one-dimensional, all Vi have rank one; settingA = R and B = Vi in (3) we infer that G(Vi) is finitely generated andhence isomorphic to ZZ; again from [6; Theorem 12.2] it follows that Ris a principal ideal domain. So finally R is noetherian by the theorem ofEakin-Nagata.2. =- 1. R• is primary by [4; Theorem 4.1]. R is a Dedekinddomain with finitely many prime ideals, whence it is a principal idealdomain. ThusR•= F & R,where F is a free abelian monoid with finite basis, and hence G(R) =Q(F ) is a free abelian group of finite rank. Thus the exact sequence (3)(with A = R and B = R ) splits andG(R) = G(R) @ R,/R,is finitely generated.Remarks. Let R be an integral domain satisfying the equivalentconditions of the previous theorem.1. Since R is noetherian we infer R = RR and being a Dedekinddomain RR is completely integrally closed.2. The number of prime ideals of R equals the torsion free rank ofG(R).3. R• is a finitely primary monoid (cf. [2; Theorem 2]).In a final example we discuss a class of domains satisfying the con-ditions of Theorem 4.Example. Let o be an order in a Dedekind domain R (i. e. o 9 Ris a subring, the quotient fields of o and R coincide, and R is a finitely292 A. GEROLDINGER – F. HALTER-KOCH – G. LETTL [12]generated o-module). Then for every (0) %= p " spec(o) op is a one-dimensional local noetherian domain, op is a finitely generated op-moduleand (op, : o,p ) . (R : F ) where F is the conductor of o 9 R (cf. [7; Kap.I, Satz 12.11]). Hence if R has the finite norm property, then op satisfiescondition 2 of Theorem 4.REFERENCES[1] A. Geroldinger: The complete integral closure of monoids and domains,PU.M.A. 4 (1993), 147 - 165.[2] A. Geroldinger: On the structure and arithmetic of finitely primary monoids,submitted[3] B. Glastadt – J. L. Mott: Finitely generated groups of divisibility , Contemp.Math. 8 (1982), 231 - 247.[4] F. Halter-Koch: Divisor theories with primary elements and weakly Krull do-mains, Boll. U.M.I. (1995), to appear.[5] R.B. Holmes: Geometric functional analysis and its applications, Springer GTM24, 1975.[6] H. Matsumura: Commutative ring theory , Cambridge University Press, 1986.[7] J. Neukirch: Algebraische Zahlentheorie, Springer, 1992.Lavoro pervenuto alla redazione il 23 settembrea 1994modificato il 27 dicembre 1994ed accettato per la pubblicazione il 15 febbraio 1995.Bozze licenziate il 30 marzo 1995INDIRIZZO DEGLI AUTORI:Alfred Geroldinger – Franz Halter-Koch – Günter Lettl – Institut für Mathematik – Karl-Franzens-Universität – Heinrichstraße 36 – A-8010 Graz – Austria.

The complete integral closure of monoids and domains II

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The complete integral closure of monoids and domains II

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