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Title Torsion-Free Abelian Semigroup Rings, V

Author(s) MATSUDA, Ryuki

Citation Bulletin of the Faculty of Science, Ibaraki University. Series A,Mathematics, 11: 1-37

Issue Date 1979

URL http://hdl.handle.net/10109/2917

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Bull. Fac. Sci., Ibaraki Univ., Math., No. 11, 1979

Torsion-Free Abelian Semigroup Rings, V

Ryuki MATSUDA*

In this paper a ring is a commutative ring with identity. Let X be an indeter-

minate and S a commutative semigroup containing the identity. We consider the

semigroup ring AS[X] of S over a ring A. AS[X] is the ring of elements ΣaαXα,

aα ∈A, α∈S almost all aα are zero. This paper is a continuation of [28] and con-

sists of 10 sections. The respective sections are: 1 The ring AG(X)loc. 2 The

multiplication ideals, 3 The pseudo-prufer and the factorial, 4 The integral, 5

The rank n, 6 The noetherian and the regular, 7 The prime chains, 8 Intersectionsof overrings, 9 The ring AG(X), 10 The Kronecker functions rings. Throughout

the paper S denotes a torsion-free cancellative commutative semigroup properly

containing {0} (additively written), G denotes a torsion-free commutative group

properly containing {0}, A (resp. D) denotes a ring (resp. domain) and E is aunitary A-module. We denote also the nonnegative integers by Z0, the nonnegative

rational numbers by Q0, the total quotient ring of a ring A by q(A), the total quotientring of AS[X] by AS(X), the minimal group containing S by q(S). As to the results

obtained in this paper we introduce at the beginnings of respective sections.

1. The ring. AG(X)loc. In the first section we state properties of the ring

AG(X)loc and the module EG(X)loc. Let E be an A-module. We set E×AAG[X]

=EG[X] and EG[X]×AG[X]AG(X)loc=EG(X)loc. We see that the results of [2]

holds also for modules: for example, if EG(X)loc is finitely generated and locally

principal then it is principal. The other special properties of EG(X)loc are stated

in the continued sections.

Let m' be an ideal of AG(X)loc. Then, m' is a maximal ideal if and only if

m'=mAG(X)loc for a maximal ideal m of A([26](7.6)). We have

(1.1) There is a bijection between the minimal (resp. maximal) prime ideals

of A and the minimal (resp. maximal) prime ideals of AG(X)loc given by p〓

p AG(X)loc.The proof follows from e.g. [26] (7.2).

A ring with only one maximal ideal is called a local ring. An ideal a of A is

called locally principal, if aAm, is a principal ideal of Am for every maximal ideal

Received January 20, 1979. Research partially supported by the Grant-in-Aid for Scientific

Research D-364003 from the Ministry of Education.* Department of Mathematics, Ibaraki University, Mito, Ibaraki 310, Japan .

2 Ryuki MATSUDA

m of A. The following is an analogy of [1] Th. 7.

(1.2) (1) Let f∈AS[X]. If c(f) is locally principal, we have c(f)AS(X)loc

=fAS(X)l oc. (2) Let f, g∈AS[X] such that c(f) is locally principal and c(9)⊂

c(f). Then we have gAs(X)loc⊂fAs(X)loc.

c(f) is the ideal of A generated by the coefficients of f.

PROOF. (1) By (1.1), we may assume that A is local. (2) follows from (1).

From (1.2) (1), we have

(1.3) COROLLARY. Let a be a finitely generated locally principal ideal ofA. Then aAS(X)loc is a principal ideal of AS(X)loc.

(1.4) PROPOSITION. Let f∈AG[X]. The followings are equivalent with

each other.

(1) c(f) is a locally principal ideal of A.

(2) fAG(X)loc=aAG(X)loc.

(3) fAG(X)loc=aAG(X)loc for an ideal a of A.

(4) c(f)AG(X)loc is a principal ideal of AG(X)loc.

(5) c(f)AG(X)loc is a locally principal ideal of AG(X)loc.

PROOF. (1)_??_(2)follows from (1.2) (1). (3)_??_(1) follows from the following

two facts: (1) we have a=aAG(X)loc ∩A, c(f)AG(X)loc ∩A=c(f) by [26] (7.2)

and (ii) for g∈AG[X] such that c(g)=A we have c(fg)=c(f) by [26] (9.8).

(1.5) COROLLARY. Let a be an ideal of A. The followings are equivalentwith each other.

(1) a is finitely generated and locally principal.

(2) aAG(X)loc is a finitely generated locally principal ideal of AG(X)loc.

(3) aAG(X)loc is a principal ideal of AG(X)loc.

(1.6) PROPOSITION. Every finitely generated locally principal ideal ofAG(X)loc is principal.

The proof is similar with that of [2] Th. 2, using (1.4).

Let E be an A-module. Then EG[X] is imbedded into EG(X)loc canonically.

(1.7) (1) Let E1, E2 be submodules of E. Then we have

(E1∩E2)AG(X)loc=E1AG(X)loc∩E2AG(X)loc.

(2) Let E' be a submodule of E. Then we have

EAG(X)loc∩E=E'.

(3) Let E' be a submodule of E. Then we have

EG(X)loc/E'G(X)loc_??_(E/E')G(X)loc.

Torsion-Free Abelian Semigroup Rings, V 3

(4) We make (EG(X)loc)G(Y)loc from AG(X)loc-module EG(X)loc. Then it is canoni-cally isomorphic with EG(X, Y)loc.

In (4), the definition of EG(X, Y)loc is of standard.

(1.8) Let E' be a maximal submodule of E. Then E'G(X)loc is a maximalsubmodule of EG(X)loc.

The proof follows from (1.7) (3).

(1.9) REMARK. Let A be a field, E a 2-dimensional vector space over A.Then there exists a maximal AG(X)loc-submodule _??_ of EG(X)loc such that _??_is never

of the form E'G(X)loc for a maximal submodule E' of E.

PROOF. We set E=Au+Av. EG(X)loc is a 2-dimensional vector space over

AG(X). We set F=uXα+vX2α for 0_??_α∈G and set _??_=FAG(X).

Extending (1.2) (1), we have

(1.10) PROPOSITION. Let f∈ES[X]. If c(f) is locally principal, then we

have

c(f)AS(X)loc=fAS(X)loc.

The proof is the analogy of the ring case.

The following is the extension of (1.4) to modules.

(1.11) THEOREM. Let f∈EG[X]. The followings are equivalent with each

other.

(1) c(f) is locally principal.

(2) fAG(X)loc=c(f)AG(X)loc.

(3) fAG(X)loc=E'AG(X)loc for some submodule E' of E.

(4) c(f)AG(X)loc is a principal AG(X)loc-module.

(5) c(f)AG(X)loc is a locally principal AG(X)loc-module.

PROOF. (3)_??_(1). We may assume that A is local. We have f=Σn1xifi/g

for xi∈E'. By (1.7) (2) we have E'=(x1,…,xn)A. There exists l such that

fAG(X)loc=xiAG(X)loc. We have f=xih/g for h/g∈AG(X)loc. It follows c(f)=

xiA, using (3.14). (5)_??_(1). We may assume A local. We have c(f)AG(X)loc

=xAG(X)loc for x∈c(f). Then it follows c(f)=xA, as in the proof of '(3)_??_(1)' .

(1.12) COROLLARY. Let E be an A-module. The followings are equivalent

with each other.

(1) E is. finitely generated and locally principal.

(2) EG(X)loc is a finitely generated locally principal AG(X)loc-module.

(3) EG(X)loc is a principal AG(X)loc-module.The following is the extension of (1.6) to modules.

4 Ryuki MATSUDA

(1.13) THEOREM. Every finitely generated locally principal AG(X)loc-

submodule of EG(X)loc is principal.

PROOF. The proof goes as that of [2] Th. 2. Let I be such a submodule.

There exist fi∈EG[X] such that I=(f1,…,fn)AG(X)loc. We set f=f1Xα1+…+

fnXαn for suitable 0<α1<…<αn of G. (G becomes an ordered group.) Then

we have c(f)=c(f1)+…+c(fn). We set g=f+f1Yα1+…+fnYαn. It follows

IAG(X, Y)loc=gAG(X,Y)loc by (1.10) and (1.7) (4). We have f=gF/G for F/G

∈AG(X, Y)loc. By (3.14) We have cE(f)=cE(fG). The rest goes as well.

We denote the nilradical of A by√A(or by√0), the Jacobson radical of A by

J-rad (A).

(1.14) PROPOSITION. We have

(1)

(2) J-rad(AG(X)loc)=(J-rad(A))AG(X)loc.

PROOF. (1) comes from [26] (3.3). (2) follows from [26] (7.6).

(1.15) PROPOSITION. Assume that A is not an integral domain. Let a be

the pseudoradical of A. Then the pseudoradical of AG(X)loc is aAG(X)loc.

For an integral domain D we have

(1.16) LEMMA. pseudoradical (D)=∩m pseudoradical (Dm), where m runs all

maximal ideals of D.

PROOF. ⊃. Let a∈ ∩m pseudoradical (Dm). Let p be a height 1 prime ideal

of D. We take a maximal ideal m containing p. There exists s∈D＼m such that

sa∈p. Since a∈D, we have a∈p. Therefore a belongs to pseudoradical (D).

It follows ∩m pseudoradical (Dm)⊂pseudoradical(D).

By the lemma we have a following reduction.

(1.17) pseudoradical (DG(X)loc)=∩m pseudoradical((Dm)G(X)loc), where m runs

all maximal ideals of D.

(1.18) PROPOSITION. Let (D, m) be a local domain which is not a field.

(1) If D is noetherian, we have pseudoradical (DG(X)loc)=(pseudoradical D).

DG(X)loc. (2) If D is not noetherian, the equality does not necessarily hold for

G=Z.

PROOF. (1) The case of dim D=1. Then the pseudoradical (D) is m.

We have dim DG(X)loc=1([28](6.12)). Therefore pseudoradical (DG(X)loc) is

mDG(X)loc. Next, the case of dim D>1. Then pseudoradical (D) is 0([11] p.

58, 3). Suppose pseudoradical (DG(X)loc)_??_0. We take 0_??_f/g∈ pseudoradical

(DG(X)loc). Let a be a nonzero coefficient of f. There exists a nonzero prime

p of D which does not contain a. It follows f/g∈pDG(X)loc; a contradiction.


(2) We take D of [6] Example 1 and the prime ideal P. Since the height of P is 1,

pseudoradical (DP) is PDP. The height of PDZ[X] is 2. It follows

pseudoradical ((DP)Z(X)loc)_??_(pseudoradical DP)((DP)Z(X)loc).

We consider any G. Then there exists (D, m) which is not a field such that

pseudoradical(DG(X)loc)_??_(pseudoradical D)DG(X)loc.

For example, let {αλ; Λ} be a maximal independent subset of G. We set Xα λ=Xλ.

In the proof of (1.18) (2), we have ht (PDZ[Xλ; Λ])_??_2. Since D is integrally closed,

DZ[Xλ; Λ] is also integrally closed ([25](4.4)). It follows ht(PDG[X])_??_2.

Therefore we have

pseudoradical((DP)G(X)loc)_??_(pseudoradical DP)((DP)G(X)loc).

The following property of AG(X)loc is used later.

(1.19) LEMMA. Let H(_??_0) be a subgroup of G. Then we have

AG(X)loc∩AH(X)=AH(X)loc.

PROOF. Take We suppose And

we derive a contradiction. There exists amaximal m of A such that

There exist F, G∈AG[X] such that c(G)=A and fG=hF. AG[X] has a free basis

{Fλ; A} over AH[X] by [5] Lemma 1, (1). We set F=ΣfλFλ, G=ΣgλFλ for fλ,

gλ ∈AH[X]. There exists λ such that gλ ∈mAG[X]. Since fgλ=hfλ, gλ belongs

to (AH(X)loc: ξ)AH(X)loc. Hence gλ belongs to mAH(X)loc, and therefore to mAH[X];

a contradiction.

As to the condition of AG(X)loc being Hilbert, we have

(1.20) If AG(X)loc is Hilbert, then A is Hilbert.

The proof follows from [26](7.2) and [26](7.6).

(1.21) AG(X)loc is Hilbert if and only if (A/p)G(X)loc is Hilbert for every

minimal prime p of A. AS[X] is Hilbert if and only if (A/p)S[X] is Hilbert for every

minimal prime p of A.

If the equality a∩b=0 for ideals a, b implies a=0 or b=0, the ring is called

irreducible.

(1.22) AG(X)loc is irreducible if and only if AG[X] is irreducible. In this

case, A is irreducible.

(1.23) AG[X] is irreducible if and only if AH[X] is irreducible for every

finitely generated subgroup H_??_0 of G.

PROOF. _??_. Let a, b be nonzero ideals of AH[X]. Take 0_??_f∈a, 0_??_

6 Ryuki MATSUDA

g∈b. We have fAG[X]∩gAG[X]_??_0. It follows fF=gG_??_0 for some F, G

∈AG[X]. Let {Fλ; Λ} be the free basis of AG[X] over AH[X]. We set F=

ΣfλFλ, G=ΣgλFλ for fλ, gλ ∈AH[X]. We have ffλ=ggλ_??_0 for some λ. There-

fore a∩b is not zero.

If the equality E1∩E2=0 for submodules E1, E2 implies E1=0 or E2=0,

the module is called irreducible.

(1.24) PROPOSITION. EG[X] is an irreducible AG[X]-module if and only

if EH[X] is an irreducible AH[X]-module for every finitely generated subgroup

H_??_0 of G.

PROOF. _??_. Let M, N be nonzero submodules of EH[X]. We show M∩N

_??_0. Take 0_??_F0∈M, 0_??_G0∈N. There exist f, g∈AG[X] such that F0f=

G0g_??_0. We make rings by the idealization principle. (A+E)G[X] is isomorphic

with AG[X]+EG[X] and (A+E)H[X] is isomorphic with AH[X]+EH[X]. Let

{fλ+Fλ; Λ} be the free basis of (A+E)G[X] over (A+E)H[X]. We have f=

Σf0λfz, g=Σg0λgλ for f0λ, g0λ∈AH[X]. It follows F0f0λ=G0g0λ_??_0 for some λ.

ES[X] is irreducible if and only if EG[X] is an irreducible AG[X]-module,where we set G=q(S).

2. The multiplication ideals. We concern in this section on some pro-

perties whether or not an ideal is a product of several ideals (or an intersectionof several ideals). Let a be an ideal of A. If for every ideal b contained in a thereexists c such that b=ac, a is called a multiplication ideal. If every ideal of A is a

multiplication ideal, A is called a multiplication ring. If every finitely generated

ideal of A is a multiplication ideal, A is called an arithmetical ring. If every ideal

is an intersection of a finite number of primary ideals. the ring is called lasker. Wesee that the results of [1] on AZ(X)loc hold also for AG(X)loc.

[28] (8.6) asserts that AS[X] is arithmetical if and only if it is semi-hereditary.

(2.1) REMARK. Even if AG(X)loc is arithmetical, it is not necessarily semi-hereditary.

A COUNTER EXAMPLE. Let k be a field. We set A=k[X]/(X2). Then

AG(X)loc is arithmetical, but it is not semi-hereditary.

The following is the analogy of [1] Th. 8, (1).

(2.2) PROPOSITION. AG(X)loc is arithmetical if and only if A is arithmeti-cal. In this case AG(X)loc is actually bezout.

PROOF. Assume A is arithmetical. That AG(X)loc is bezout follows from

(1.2) (1). Conversely, assume AG(X)loc is arithmetical. Let a, b, c be ideals ofA. By [26] (7.5) (2) and the assumption, we have


By [26] (7.2) it follows a∩(b+c)=(a∩b)+(a∩c).

The following is the analogy of [1] Th. 8, (2).

(2.3) PROPOSITION. A is arithmetical if and only if, the set of ideals of A

and that of AG(X)loc canonically corresponds bijectively.

The proof comes also from [26] (7.2) and (1.2) (1).

If for any two submodules E1, E2 of E it follows E1⊂E2 or E1⊃E2, then E is

called chained.

(2.4) Let E be an A-module. The followings are equivalent with each

other.

(1) For any three submodules E1, E2, E3, we have

E1+(E2∩E3)=(E1+E2)∩(E1+E3).

(2) For every maximal ideal m of A, Em is a chained Am-module.

Such modules are the arithmetical. If every finitely generated submodule is

principal, the module is called bezout. Let E be a bezout A-module. Then Em

proves a chained Am-module for every maximal m of A. Therefore E becomes

arithmetical.

The following is the analogy of [1] Th. 9, (1) and is the extension of (2.2) to

modules.

(2.5) THEOREM. E is an arithmetical A-module if and only if EG(X)loc is

an arithmetical AG(X)loc-module. In this case EG(X)loc is actually a bezout

AG(X)loc-module.

PROOF. Assume E arithmetical. That EG(X)loc is then bezout follows from

(2.4) and (1.10). Conversely, assume EG(X)loc is bezout. That E is then arithmeti-

cal follows from (1.7).

The following is the analogy of [1] Th. 9, (2) and is the extension of (2.3) to

modules.

(2.6) THEOREM. Let θ be the canonical mapping from the set of submodules

of E to the set of AG(X)loc-submodules of EG(X)loc.

(1) If E is arithmetical, then θ is a bijection.

(2) If θ is a surjection, then E is arithmetical.

PROOF. (1) By (1.7) (2), θ is injective. By (2.4) and (1.10), θ is surjective.

8 Ryuki MATSUDA

(2) By (2.4) we may assume that A is local. The proof goes analogously as thatof (2.3).

(2.7) If E is arithmetical, then Ep is a chained Ap-module for every prime pof A.

The proof follows from (2.4).

Let E be an arithmetical A-module. Then for every multiplicative systemS of A, Es is an arithmetical As-module. For the proof, we must show that (ES)m'

is a chained (AS)m'-module for every maximal ideal m' of AS. And this follows from

(2.7). Hence we have the following.

(2.8) If the AS[X]-module ES[X] is arithmetical, then (Ap)S[X]-module(Ep)S[X] is arithmetical for every prime p of A.

(2.9) ES[X] is an arithmetical AS[X]-module if and only if (Em)S[X] is anarithmetical (Am)S[X]-module for every maximal ideal m of A.

PROOF. _??_. Let M be a maximal ideal of AS[X]. Set M∩A=p. Let m

be a maximal ideal of A containing p. (Em)S[X] is an arithmetical (Am)S[X]-

module. We see by the localization that ES[X]M is a chained AS[X]M-module.

(2.10) PROPOSITION. If ES[X] is an arithmetical AS[X]-module, then E

is arithmetical.

PROOF. (1) The case where (A, m) is local. Let x, y∈E. We take a maxi-

mal ideal M' containing MAS[X]. ES[X]M' is a chained AS[X]M'-module. Hence

we may suppose gy∈xAS[X] for g∈AS[X]＼M'. It follows y∈xA. Therefore E

is arithmetical. (2) The general case. By (2.9), (Em)S[X] is an arithmetical

(Am)S[X]-module for every maximal ideal m of A. By (1), Em is an arithmeticalAm-module. Therefore E is arithmetical.

From (2.2) and (2.3) we have the following analogy of [1] Th. 8, (4).

(2.11) PROPOSITION. AG(X)loc is a multiplication ring if and only if A isa multiplication ring.

[28] (8.4) asserts that AS[X] is multiplication if and only if AS[X] is heredi-tary. For 'loc', even if AG(X)loc is multiplication, it is not necessarily semi-heredi-

tary. For an example, we may take A in (2.1).

If Am is multiplication for every maximal m of A, A is called almost multi-

plication. By (2.11) we have the following analogy of [1] Th. 8, (3).

(2.12) THEOREM. AG(X)loc is almost multiplication if and only if A is al-most multiplication.

We know the condition of AG[X] being almost multiplication.

(2.13) PROPOSITION. Suppose q(S)_??_S. Then AS[X] is almost multiplica-tion if and only if A is von Neumann regular and S is isomorphic with Z0.


PROOF._??_. We set q(S)=G. By the results of [23], AS[X] is prufer.

By the results of [13], A is von Neumann regular and we may suppose Q⊃G⊃S

=G∩Q0. We take a prime ideal _??_of AS[X] containing {Xα; 0<α ∈S}. Set

_??_∩A=p', Ap'=k and (Xα; 0<α ∈S)kS[X]=p. Then k is a field and AS[X]_??_

is isomorphic with kS[X]p. kS[X]p is a rank 1 discrete valuation ring. We have

pkS[X]p=fkS[X]p for f∈p. Set f=Xα0+a1Xα1+…+anXαn for 0<α0<α1<…

<αn. Let 0<α ∈S. we have Xα=fh/g for h/g∈kS[X]p. It follows α0=min

(S＼0). Therefore we see S_??_Z0.

If every principal ideal is a product of a finite number of prime ideals, the

ring is called π-ring.[27](12.1) states the following proposition, but the proof

seems incomplete.

(2.14) PROPOSITION. If DG[X] is a π-domain, then D is a π-domain.

PROOF. We confer [11](46.7). DG[X] becomes a Krull domain. Hence

D is a Krull domain. Let p be a height 1 prime of D. Dp is the valuation ring of

a discrete valuation v. v is extended canonically to a valuation w of DG(X) as in

[25](3.2). The valuation ring of w is DG[X]pDG[X]. It follows ht(pDG[X])=1.

We see that pDG[X] is invertible. From this, p is proved invertible as in [27]

(12.1).

(2.15) PROPOSITION. If DG(X)loc is a π-domain, then D is a π-domain.

PROOF. DG(X)loc becomes a Krull domain ([11](46.7)). Hence D is a Krull

domain ((8.1)). Let p be a height 1 prime of D. We have DG[X]pDG[X]=

(DG(X)loc)pDG(X)loc. As in the proof of (2.14), we see ht(pDG(X)loc)=1. Hence

we have

p・(DG(X)loc:pDG(X)loc)=DG(X)loc.

We have 1=ΣpiFi, where pi∈p and Fi∈(DG(X)loc:pDG(X)loc). Let o_??_p∈p. We

set Fip=fi/g∈DG(X)loc and fi=ΣaiαXα. It follows p(…,aiα/p,…)=D. Therefore

p is invertible.

AS[X] is a π-ring if and only if A=a1+…+an, (ai)s[X] is a π-domain for all i

([28] (7.10)). For 'loc', if A=a1+…+an, then AG(X)loc is a π-ring if and only if

(ai)G(X)loc is a π-ring for all i. This is because AG(X)loc_??_(a1)G(X)loc+…+

(an)G(X)loc.

As to the intersection representation of an ideal by primary ideals,

(2.16)(1) If AG(X)loc is lasker, then A is lasker. (2) If AG(X)loc is strongly

lasker, then A is strongly lasker.

PROOF. (1) Let a be an ideal of A. We have a primary decomposition in

AG(X)loc: aAG(X)loc=q'1∩ … ∩q'n. Then q'i∩A=qi is a primary ideal of A, and

10 Ryuki MATSUDA

it follows a=q1∩ … ∩qn. (2) Let q be a p-primary ideal of A. By [26](7.2),

qAG(X)loc is pAG(X)loc-primary. Hence pmAG(X)loc⊂qAG(X)loc for some m.

Then it follows pm⊂q.

(2.17) PROPOSITION. Let E be an A-module.

(1) Let E' be a p-primary submodule of E in the sense of [35]. Then E'G(X)loc

is a pAG(X)loc-primary submodule of EG(X)loc in the sense.

(2) Let E' be a p-primary submodule of E in the sense of [3] exc. Then

E'G(X)loc is a pAG(X)loc-primary submodule of EG(X)loc in the sense.

(3) Let E' be a p-primary submodule in the sense of [35]. Then E'S[X] is a

pAS[X]-primary submodule of ES[X] in the sense.

(4) Let E' be a p-primary submodule in the sense of [3] exc.. Then E'S[X]

is a pAS[X]-primary submodule of ES[X] in the sense.

PROOF. (1) We use (3.14). We have

Assume (f/g)(F/g')∈E'G(X)loc, f/g∈AG(X)loc, F/g'∈EG(X)loc, f/g∈pAG(X)loc. We

have g"fF=F'∈E'G[X] for g"∈AG[X] such that c(g")=A. Considering in

Ep, we have CEq(F/1)∈E'p. It follows sF∈E'G[X] for s∈A＼p, and hence F/g'∈

E'G(X)loc. (2) We have

0_??_EG(X)loc/E'G(X)loc_??_(E/E')G(X)loc

by (1.7)(2) and (1.7)(3). In the rest we use (3.14). (3) and (4) follows also from

(3.14).

Finally in this section, as to the behaviour of prime divisors for scalar exten-

sions.

(2.18) LEMMA. Let a, b be ideals of A such that b⊂a, and let p be a prime

ideal containing a. Then p is a prime divisor of a if and only if p/b is a prime

divisor of a/b in A/b.

We use 'prime divisor' in the sense of [29]. The proof is straightforward

from the definition.

Let K be a field containing a field k, and let q be a p-primary ideal in k[X1,…,

Xn]. [36] p.225, Cor. 3 asserts that the prime divisors of pK[X1,…,Xn] coincide

with them of qK[X1,…,Xn].

(2.19) PROPOSITION. Let q be a p-primary ideal of kG[X]. We consider

the statement, 'every prime divisor of qKG[X] is a prime divisor of pKG[X]'.

Then if the statement is true for free G, then it is true for any G.

PROOF. Let {Xα;G} be a set of indeterminates over K. We define a k-


homomorphism θ:kZ[Xα; G]→kG[X] by Xα →Xα. Set θ-1(p)=p', θ-1(q)=q',

θ-1(0)=_??_.Then q' is p'-primary. Similarly we have a natural isomorphism

φ:Kz[Xα; G]/_??_e_??_KG[X]. We have φ-1(pe)=P'e/_??_e, φ-1(qe)=q'e/_??_e. Let

n be a prime divisor of qe and set φ-1(n)=_??_/_??_e. By (2.18), we see that n is a

prime divisor of pe.

'e' denotes the extension ideal.

The situation is the same for the statement, 'every prime divisor of pe is a

prime divisor of qe'. As to 'S', the situations are similar. Let q be a p-primary

ideal in kS[X]. We consider the statement, 'every prime divisor of qe (resp. pe)

is a prime divisor of pe (resp. qe). If the statement is true for G=Σ+Z0, then it

is true for any S.

(2.20) Let a be an ideal of kG[X], d a cardinal number. Assume for every

prime divisor p of a, transc. degk q(kG[X]/p) is d. Consider the statement, 'for

every prime divisor _??_ of ae, transc. degK q(KG[X]/_??_) is d'. Then if the state-

ment is true for every free G, it is true for any G.

The proof is similar with that of (2.19), using (2.18) and [28] (1.7).

3. The pseudo-prufer and the factorial. It is obvious that a principal ideal

domain is a factorial domain and a factorial domain is a pseudo-bezout (i.e.

GCD-) domain. If the monoid of divisors of finite type over a domain D makes

a group, D is called pseudo-prufer. A pseudo-bezout domain is pseudo-prufer.

If we have c(fg)=c(f)c(g) for every f, g∈A[X], A is called Gaussian. A domain

D is Gaussian if and only if D is prufer. For every a, b of a prufer domain D,

there exists n such that (a, b)n=(an, bn). We say that a ring A has property (n),

if (a, b)n=(an, bn) for every a, b of A. We concern these in this section. For

example, we show that DG(X)loc is pseudo-prufer if and only if D is pseudo-prufer.

The equivalence of (1) and (2) in the following proposition is [27](2.6).

(3.1) PROPOSITION. Let D be a v-domain. The followings are equivalent

with each other.

(1) D is a pseudo-prufer domain.

(2) Dv is a quotient ring of DG[X].

(3) Dv is a quotient ring of DG(X)loc.

PROOF. (3)_??_(2). We set N={g∈DG[X]; 1/g∈Dv}. Then we see Dv=

DG[X]N.

(3.2) THEOREM. DG(X)loc is pseudo-prufer if and only if D is pseudo-

prufer.

PROOF. _??_. By [26](2.10), DG[X] is pseudo-prufer. Set _??_=(ζ1,…,

ζn)DG(X)loc for ζ1,…, ζn∈DG(X). We show that div _??_ has an inverse element of

12 Ryuki MATSUDA

finite type. We have

div(ζ1,…,ζn)DG[X]+div(η1,…,ηm)DG[X]=div DG[X]

for ηj∈DG(X). It follows

div(ζ1,…,ζn)DG(X)loc+div(η1,…,ηm)DG(X)loc=div DG(X)loc._??_

. Let a1,…,an∈K＼{0}, K=q(D). We have

div(a1,…,an)(F1,…,Fm)DG(X)loc=div DG(X)loc

for Fi∈KG[X]. Set Fi=ΣaijXαij. Then it follows

div(a1,…,an)+div(…,aij,…)=div D.

(3.3) PROPOSITION. DG(X)loc is prufer if and only if D is prufer.

(3.4) PROPOSITION. AG(X)loc is (n) if and only if (Am)G(X)loc is (n) for

every maximal m of A.

PROOF. _??_. Let f1,f2∈AG[X]. Suppose

(f1,f2)nAG(X)loc_??_(fn1,fn2)AG(X)loc.

We have fk1fn-k2∈(fn 1,fn2)AG(X)loc for some k. There exists a maximal ideal M

of AG(X)loc containing ((fn1,fn2)AG(X)loc:fk1fn-k2)AG(X)loc. M is of the form

mAG(X)loc for a maximal m of A. By the assumption we have gfk1fn-k2∈(fn1,

fn2)AG[X] for g∈AG[X]＼mAG[X]. It follows g∈((fn1,fn2)AG(X)loc: fk1fn-k2)AG(X)loc

and g∈M; a contradiction.

Let n be a positive integer. If xn∈an implies x∈a for every x∈A and every

ideal a, then we say that A has the property (n)'.

(3.5) A has (n)' if and only if Am has (n)' for every maximal m of A.

PROOF. _??_. Let b be an ideal of Am and let a∈A. Suppose (a/1)n∈bn.

set a=f-1(b), where f is the natural map of A into Am. we have (sa)n∈an for

s∈A＼m. It follows a/1∈b. _??_. Let a be an ideal of A and let x∈A. Suppose

xn∈an and x∈a. There exists a maximal m of A containing (a:x)A. By as-

sumption x/1 belongs to aAm. We have sx∈a for s∈A＼m; a contradiction.

(3.6) COROLLARY. AG(X)loc is (n)' if and only if (Am)G(X)loc is (n)' for


(3.7) PROPOSITION. Let (A, m) be a local ring. Then, AG(X)loc is (n) if

and only if AH(X)loc is (n) for every finitely generated subgroup H_??_0 of G.

PROOF._??_. Let ζ,η ∈AH[X] and let 0<k∈Z. We have ζkηn-k=ζnF1/G

+ηnF2/G for Fi/G∈AG(X)loc. We set AG[X]=ΣΛ+AH[X]Hλ, Fi=Σfλ1Hλ, G=


ΣgλHλ. We have gλ ∈mAH[X] for some λ. Since ζkηn-kgλ=ζnfλ 1+ηnfλ2, it

follows ζkηn-k∈(ζn,ηn)AH(X)loc.

The similar proof shows

(3.8) PROPOSITION. Let (A, m) be a local ring. Then AG(X)loc is (n)' if

and only if AH(X)loc is (n)' for every finitely generated subgroup H_??_0 of G.

The similar reduction applies for the Gaussian property as well. If for every

f∈A[Y], F∈E[Y] we have cA(f)cE(F)=cE(fF), E is called a Gaussian A-module.

(3.9) PROPOSITION. EG[X] is a Gaussian AG[X]-module if and only if

EH[X] is a Gaussian AH[X]-module for every finitely generated subgroup H_??_0

of G.

PROOF._??_.Let EH[X][Y]∋ η=ΣFiYi, AH[X][Y]∋ζ=ΣfjYj. We have

for every i, j, fiFj=ΣlhlΣkfkFl-k for hl∈AG[X]. From the equality and from

the fact that (A+E)G[X] is a free (A+E)H[X]-module, we obtain the assertion.

(3.10) Let A be a local ring. Then AG(X)loc is Gaussian if and only if AH(X)loc

is Gaussian for every finitely generated subgroup H_??_0 of G.

PROOF._??_. Let ζ, η∈AH(X)loc[Y]. Set ζ=ΣφiYi, η=Σ ψjYj. We have

φiψj=Σlφ'lΣkφkψl-k for φ'l∈AG(X)loc. We set φi=fi/g, ψj=gj/g, φ'l=f'l/g'.

We have figjg'=Σlf'lΣkfkgl-k. By [5] Lem. 1, (1) we have AG[X]=ΣΛ+

AH[X]Xα λ. Set g'=ΣgλXα λ. We have gλ0∈mAH[X] for some λ0. Comparing

the λ0-components of both sides, we see

figj∈(…,Σkfkgl-k,…)AH(X)loc.

(3.11) PROPOSITION. Let A be a local ring. Then EG(X)loc is a Gaussian

AG(X)loc-module if and only if EH(X)loc is a Gaussian AH(X)loc-module for every

finitely generated subgroup H_??_0 of G.

PROOF._??_. Let AH(X)loc[Y]∋ζ=Σ φiYi, EH(X)loc[Y]∋ η=Σ ψjYj and set φi

=fi/g, ψj=gj/g. We have φiψj=Σ φ'lΣφkψl-k for φ'l=f'l/g'∈AG(X)loc. It fol-

lows g'f igj=Σf'lΣfkgl-k. (A+E)G[X] is a free (A+E)H[X]-module. We com-

pare the suitable components of the both sides.

(3.12) PROPOSITION. (1) EG[X] is a Gaussian AG[X]-module if and only

if (Em)G[X] is a Gaussian (Am)G[X]-module for every maximal m of A.(2)

AG(X)loc is Gaussian if and only if (Am)G(X)loc is Gaussian for every maximal m

of A.

PROOF. (1) _??_. Set EG[X][Y]∋ η=ΣFiYi, AG[X][Y]∋ζ=ΣfjYj. We

suppose c(ζ)c(η)_??_c(ζη). We have fiFj∈c(ζ η) for some i, j. There exists a maxi-

mal ideal m of A containing (c(ζη):fiFj)A. Set (Em)G[X][Y]∋ η'=Σ(Fi/1)Yi,

14 Ryuki MATSUDA

(Am)G[X][Y]∋ ζ'=Σ(fj/1)Yj. From c(ζ')c(η')=c(ζ'η'), we have sfiFj∈c(ζ η) for

s∈A＼m; a contradiction. (2) follows from [11] p. 351, 7 (a).

E is a Gaussian A-module if and only if Em is a Gaussian Am-module for


Let m be a maximal ideal of A. Then (AG(X)loc)mAG(X)loc-module

(EG(X)loc)mAG(X)loc is canonically isomorphic with AG[X]mAG[X]-module EG[X]mAG[X].

From the above remarus we have the following reduction.

(3.13) PROPOSITION. EG(X)loc is a Gaussian AG(X)loc-module if and only

if (Em)G(X)loc is a Gaussian (Am)G(X)loc-module for every maximal m of A.

The following Gaussian property we have already used before.

(3.14) Let f∈EG[X], g∈AG[X]. If c(g)=A, then we have cE(fg)=cE(f).

PROOF. We may suppose that (A, m) is local. Next, we have c(fg)+mc(f)

=c(f). Then we use Nakayama's lemma.

As a generalization of (3.14) and [26] (9.8),

(3.15) PROPOSITION. Let F∈EG[X], f∈AG[X]. Then there exists m such

that

c(f)m+1c(F)=c(f)mc(fF).

PROOF. We make a ring A+E and apply [26](9.8).

A Krull domain D such that Cl(D) is torsion is called almost factorial. [25]

(5.4) states that DG[X] is almost factorial if and only if D is almost factorial.

(3.16) REMARK. Even if DZ(X)loc is a principal ideal domain, D need not

be almost factorial.

A COUNTER EXAMPLE. We take a Dedekind domain D such that Cl(D)_??_Z.

That DZ(X)loc is a principal ideal domain follows as in the below (3.20).

From [27](13.1) and [27](13.3) we have

(3.17) PROPOSITION. DG(X)loc is a principal ideal domain if and only if

D is a Dedekind domain.

If Dm is factorial for every maximal m of D, D is called locally factorial.

(3.18) THEOREM. DG(X)loc is locally factorial if and only if D is locally

factorial.

PROOF._??_. Let m be a maximal ideal of D. Then DG[X]mDG[X] is factorial.

By [27](8.16)(2) we see that Dm is factorial. _??_ follows from [26](7.6) and [27]

(8.16)(1).

(3.19) DS[X] is locally factorial if and only if (Dp)S[X] is locally factorial

for every prime p of D.


(3.20) If D is pseudo-bezout, then DG(X)loc is pseudo-bezout. But, even if

DZ(X)loc is pseudo-bezout D need not be pseudo-bezout.

PROOF. The first assertion follows by [6](1.4). For the second assertion,

we take the ring D of algebraic integers which is not factorial. Then D is not

pseudo-bezout. By (3.17), DZ(X)loc is a principal ideal domain.

Let D be an integrally closed domain. If every prime ideal _??_ of DS[X] such

that _??_ ∩D=0 is principal, then D becomes pseudo-bezout ([28](3.5)). For 'loc',

even if every prime ideal _??_ of DG(X)loc such that _??_ ∩D=0 is principal, D need not

be pseudo-bezout. For an example, we take again the ring D of algebraic integers

which is not factorial. (Then by [28](6.12), dim DG(X)loc is 1.)

We add one more proposition in this section.

(3.21) PROPOSITION. If DG(X)loc is regularly integrally closed, then D is

regularly integrally closed.

PROOF. We confer [27](1.8) and [27](1.9). Suppose div a is equal to div b

in D. Then we see div aDG(X)loc=div bDG(X)loc. Suppose we have

div a+div(x1,…,xn)=div b+div(x1,…,xn)

for xi∈q(D). By the above remark we have

div a(x1,…,xn)DG(X)loc=div b(x1,…,xn)DG(X)loc.

It follows div aDG(X)loc=div bDG(X)loc. This derives to div a=div b.

4. The integral. This section concerns with the integral elements. For

instance, we determine the complete integral closure of A in AG(X)loc.

(4.1) PROPOSITION, DG(X)loc is integrally closed if and only if D is inte-

grally closed.

The proof comes from [26](7.4).

If AG[X] is integrally closed, then √A=0 ([27](1.3)). For 'loc', even if

AG(X)loc is integrally closed √A need not be zero. A=Z/(4) is a counter example.

(4.2) Set q(A)=B. Then AG(X)loc is integrally closed if and only if A and

BG(X)loc are integrally closed.

_??_ follows from [26] (7.3).

[28](4.9) states that if A is von Neumann regular, then AG[X] is integrally

closed. The assertion implies (1) of the following

(4.3) PROPOSITION. (1) If A is von Neumann regular, then AG(X)loc is

integrally closed. (2) If A is von Neumann regular, then AG(X)loc is completely

integrally closed.

PROOF. (2) Let ζ=f/g∈AG(X) be almost integral over AG(X)loc. We

16 Ryuki MATSUDA

show ζ∈AG(X)loc by the induction on the number of terms of g. Let b be the

highest coefficient of g. There exists e∈A such that bA=eA, e2=e, We set

eA=B, (1-e)A=C. We have AG[X]=BG[X]+CG[X], AG(X)loc=BG(X)loc+

CG(X)loc and AG(X)=BG(X)+CG(X). (1-e)f/(1-e)g is an element of CG(X)

which is almost integral over CG(X)loc, and the number of terms of (1-e)g is less

than that of g. It follows (1-e)f/(1-e)g∈CG(X)loc. Hence f/g=((ef)/(eg),

(1-e)f/((1-e)g))∈BG(X)loc+CG(X)loc.

Let A be von Neumann regular. Then AS[X] is completely integrally closed

if and only if: for s∈S and α∈q(S), s+nα ∈S for all n_??_0 implies α∈S ([28] (4.16)).

We remark that such S need not be equal to q(S). Z0 or Z0+Z is an example.

If xn∈A, x∈q(A), 0<n∈Z implies x∈A, then A is called root closed. The

following proposition comes from the fact : AG(X)loc∩q(A)=A.

(4.4) Set q(A)=B. If AG(X)loc is root closed, then A and BG(X)loc are root

closed.

PROOF. Let ζ∈AG(X), ζn∈BG(X)loc. We have ζn=F/G for F, G∈BG[X]

such that CB(G)=B. Let s be the common denominator of the coefficients of F

and G. It follows (ζGs)n=Fs(Gs)n-1∈AG[X]. Hence ζGs∈AG(X)loc, and

hence ζ∈BG(X)loc.

A noetherian ring A is called D-ring if all of its overrings are integrally closed.

(4.5) PROPOSITION. If AG(X)loc is a D-ring, then A is a D-ring.

PROOF. By [26](7.2), A is noetherian. Let B be an overring of A. Then

BG(X)loc is an overring of AG(X)loc. Hence BG(X)loc is integrally closed. By

(4.2), B is integrally closed.

(4.6) PROPOSITION. DG(X)loc is a D-domain if and only if D is a Dedekind

domain.

PROOF._??_. By (4.2), D is integrally closed. By [8] Th. 5, dim D is less

than or equal to 1. _??_. By (4.1), DG(X)loc is integrally closed. The assertion

follows from again [8] Th. 5.

If the integral closure of a domain D is a finite D-module, D is called N-1.

(4.7) PROPOSITION. DG(X)loc is N-1 if and only if D is N-1.

PROOF. Let E be the integral closure of D. Then EG[X] is integrally closed

([25](4.4)). Set T={g∈DG[X]; c(g)=D}. EG[X]T is also integrally closed.

Therefore EG[X]T is the integral closure of DG(X)loc. _??_. We have E=Σn1eiD.

It follows EG[X]T=Σn1eiDG(X)loc. _??_. We have EG[X]T=Σn1FiDG(X)loc. We

set Fi=ΣxiαXα. Then we have E=Σi,αxiαD.

The similar proof shows that if DG(X)loc is Japanese ([18]), then D is Japanese.


If every domain D which is A-algebra of finite type is Japanese, A is called

universally Japanese.

(4.8) PROPOSITION. If DG(X)loc is universally Japanese, then D is univer-sally Japanese.

PROOF. Let a domain E be a D-algebra of finite type, and let e be the identity

of E.E is of the form (De) [x1,…,xn]. Set T'={ge; g∈DG[X], c(g)=D}. By

the assumption, EG[X]T' is Japanese. Let L be a finite extension field of q(E),

and let F be the integral closure of E in L. Then FG[X]T' is the integral closure of

EG[X]T' in LG(X). It follows FG [X]T'=Σm1fiEG[X]T' for fi∈FG[X]. Set fi=

ΣaijXaj. Then we have F=ΣaijE. Hence E is Japanese, and hence D is uni-

versally Japanese.

(4.9) (1) AG(X)loc is universally Japanese if and only if (A/p)G(X)loc isuniversally Japanese for every minimal prime p of A. (2) AG[X] is universallyJapanese if and only if (A/p)G[X] is universally Japanese for every minimal prime p

of A.

PROOF. (1) _??_. Let a domain D be AG(X)loc-algebra of finite type. The

kernel _??_ of the natural homomorphism AG(X)loc→D is a prime ideal. We take a

minimal prime _??_0 contained in _??_. We have _??_0=pAG(X)loc for a minimal prime

p of A ((1.1)). D is a (A/p)G(X)loc-algebra of finite type. Hence D is Japanese,

and hence AG(X)loc is universally Japanese.

We note a localization reduction.

(4.10) DZ[X] is root closed if and only if (Dm)Z[X] is root closed for every

maximal m of D.

(4.11) THEOREM. The integral closure of A in AG(X)loc is A+nAG(X)loc,

where n=√A.

PROOF. (1) That the elements of A+nAG(X)loc are integral over A is ob-

vious. (2) The case where A is a domain D. Set K=q(D). The assertion

follows from the fact: K is algebraically closed in KG(X) ([25](6.1)). (3) The

case where A is a finite direct sum of domains. The assertion follows from (2).

(4) The case where √A=0 and A is noetherian. Let p1,…,pn be the set of mini-

mal primes of A. Set B=A/p1+…+A/pn. Then B contains A. Let f/g∈

AG(X)loc is integral over A. By (3), f/g=b belongs to B. Since c(g)=A, b be-

longs to A. (5) The case where √A=0. Let f/g∈AG(X)loc be integral over A.

Set f=ΣaiXαi, g=ΣbjXβj. We have Σbjxj=1 for xj∈A. We have

(f/g)k+(f/g)k-1yk-1+…+(f/g)y1+y0=0

for yi∈A. Let B be the subring of A generated by {a0,…,an;b0…,bm;x0,…,

18 Ryuki MATSUDA

xm;y0,…,yk-1}. By (4), f/g belongs to B, and hence it belongs to A. (6) The

general case. Let f/g∈AG(X)loc be integral over A. We have

fn+fn-1gan-1+…+gnan=0

for ai∈A. We set A/n=A. From (5), considering on A we have f=gb+h for

b∈A, h∈AG[X] such that c(h)⊂n. It follows f/g=b+h/g∈A+nAG(X)loc.

(4.12) THEOREM. The complete integral closure of A in AG(X)loc is A+

nAG(X)loc, where n=√A.

PROOF. (1) It is obvious that A+nAG(X)loc is contained in the complete

integral closure. (2) When A is a field K. The assertion follows by the fact:

K is algebraically closed in KG(X). (3) When A is a domain D. The assertion

follows from (2). (4) Let ПΛDλ denote the complete direct product of domains.

Then there exists a canonical homomorphism θ: (ПDλ)G[X]→П((Dλ)G[X]). θ

is the map Σ(aα)Xα_??_(…,Σaα λXα,…). (5) There exists a canonical homomor-

phism τ: (ПDλ)G(X)loc→П((Dλ)G(X)loc). τ is defined by τ(f/g)=θ(f)/θ(g). (6)

When A is a complete direct product ПDλ. Let ξ∈AG(X)loc be almost integral

over A. We have {ξ, ξ2,…}⊂Aξ1+…+Aξm for ξi∈AG(X)loc. We set ξn=an1ξ1

+…+anmξm for anj∈A, ξ=f/g, ξi=fi/g, f=Σ(aα)Xα, g=Σ(bα)Xα and fi=

Σ(aiα)Xα. Taking the λ-components of both sides of the equality: (τ(ξ))n=

Στ(anj)τ(ξj), we have

((ΣaαλXα)/(ΣbαλXα))n=Σ(anjλ(ΣajαλXα)/(ΣbλXα)).

Hence(ΣaαλXα)/(ΣbαλXα) is almost integral over Dλ. We have (ΣaαλXα)/(ΣbαλXα)

=cλ ∈Dλ by (3). It follows

ξ=f/g=(Σ(aα)Xα)/(Σ(bα)Xα)=(Σ(c)(bα)Xα)/(Σ(bα)Xα)

=((c)Σ(bα)Xα)/(Σ(bα)Xα)=(c)∈ПDλ .

(7) when √A=0. Let {pλ;Λ} be the set of minimal primes of A. set A/pλ=

Dλ and ПDλ=B. We have

A⊂AG(X)loc⊂BG(X)loc⊃B⊃A.

Let ξ=f/g∈AG(X)loc be almost integral over A. By (6), we see ξ=b∈B. From

f=gb, we have b∈A. (8) The general case. Let ξ=f/g∈AG(X)loc be almost

integral over A. We have {ξ, ξ2,…}⊂Aξ1+…+Aξm for ξi∈AG(X)loc. Set

A/n=B. There exists a canonical homomorphism ρ:AG(X)loc→BG(X)loc. By

(7), we see ρ(ξ)=a∈B for a∈A. It follows ξ∈A+nAG(X)loc.

Of course (4.11) becomes a corollary of (4.12).

It is obvious that, if A is the integral closure of A then the integral closure of

A in AG(X) coincides with the integral closure of A in AG(X).


(4.13) PROPOSITION. Let {Hλ;Λ} be the set of finitely generated sub-

groups_??_0 of G, and let Iλ be the integral closure (resp. complete integral closure)

of A in AHλ(X). Then the integral closure (resp. complete integral closure) of

A in AG(X) is ∪ Iλ.

(4.14) PROPOSITION. Let {Hλ;Λ} be the set of finitely generated sub-

groups of G, and let Iλ be the (resp. complete) integral closure of AHλ(X)loc.

Then the (resp. complete) integral closure of AG(X)loc is ∪ Iλ.

(4.15) Let {Sλ;Λ} be the set of finitely generated subsemigroups _??_0 of S,

and let Iλ be the integral closure of ASλ[X]. Then the integral closure of AS[X]

is ∪ Iλ.

(4.16) PROPOSITION. AG(X)loc is (resp. completely) integrally closed, if

and only if AH(X)loc is (resp. completely) integrally closed for every finitely

generated subgroup H_??_0 of G.

The proof follows from (1.19).

(4.17) AG(X)loc is root closed if and only if AH(X)loc is root closed for every

finitely generated subgroup H of G.

5. The rank n. Let n be a positive integer. If every ideal of A is generated

by n elements, A is called to have rank n. This section concerns with the rank.

We determine the condition that AS[X] has rank n.

(5.1) EXAMPLE. Let S be a subsemigroup of Z0 containing 2, and let k be

a field. Then kS[X] has rank 2.

PROOF. We show that every ideal a=(f, g, h) generated by three elements

is generated by two elements. We take the generators f, g, h the sum of which

degrees is minimal among all three generators of a. We may assume deg(f)≡

deg(g) (2), and may assume deg(f)≦deg(g). Seeing the sum of degrees of

generators f, g-fXdeg(g)-deg(f), h, we get the assertion.

Let S={0, 3, 4, 5,…}. Then kS[X] does not have the rank 2. Because

a={f;o(f)>0} is not generated by two elements. ('o' is the order.)

(5.2) If AS[X] has the rank 2, then S is generated by two elements over Z0.

PROOF. Let m be a maximal ideal of A and set A/m=k. By [28] (2.3), we

see that S is isomorphic with Z or a subsemigroup of Z0. The later case. We

set min (S＼0)=α. If S_??_Z0α, we set min (S＼Z0α)=β. Let f, g be generators of

the ideal a={f;o(f)>0} of kS[X]. We may assume f=Xα+ΣasXs. Then

we may assume o(g)=β. Suppose S_??_Z0α+Z0β. Then for γ=min(S＼(Z0α+

Z0β)), Xγ does not belong to (f, g); a contradiction. Therefore S=Z0α+Z0β.


(5.3) If AS[X] has rank n, then S is generated by n elements over Z0.

20 Ryuki MATSUDA

(5.4) Let k be a field. kS[X] has rank 2 if and only if S is isomorphic with

Z or a subsemigroup of Z0 containing 2.

PROOF. _??_. Suppose the assertion is false. By [28] (2.3), we may assume

S is a subsemigroup of Z0 such that q(S)=Z. By (5.2), S is of the form Z0α+Z0β

for α<β. There exists n such that nα<β<(n+1)α. Every open interval] nα,

(n+1)α[,](n+1)α, (n+2)α[,…contains elements of S. Now suppose each these

interval contains exactly one element of S. Then it follows

S={0,α,2α,3α,…;β,β+α,β+2α,…}.

Since s-s'=1 for s, s'∈S, we have β=mα ±1 for m>0. In the case '+', we see

β=nα+1. Considering 2β, we get a contradiction. In the case '-', we see

β=(n+1)α-1, and get the similar contradiction. Therefore some interval]pα,

(p+1)α[contains more than one element of S. Let p be the minimal such p.

We consider the ideal (Xβ+(p-n-1)α,Xpα,Xs') of kS[X], where S∋s'_??_β+

(p-n)α and S'∈]pα,(p+1)α[. Let f, g be the generators of the ideal. They are

of the forms: f=Xβ+(p-n-1)α+…, g=Xpα+…. It follows then that Xs' is

not of the form fF+gG for F, G∈kS[X]; a contradiction.

The similar proof with (5.1) shows

(5.5) Let n be a positive integer. If a subsemigroup S of Z0 contains n,

then kS[X] has rank n.

(5.6) PROPOSITION. Let 0<n∈Z. DS[X] has rank n if and only if D is

a field and S is isomorphic with Z or a subsemigroup of Z0 containing some

positive integer m_??_n.

PROOF. _??_. By [28] (2.3), D is a field and S is isomorphic with Z or a sub-

semigroup of Z0. We may assume Z0⊃S, q(S)=Z. Set

l=min{n';Ds[X] has rank n'};

n_??_l_??_1, and set m=min(S＼0). We may assume m>1. The greatest common

divisor of S＼0 is 1. Therefpre there exists h such that {h,h+1,h+2,…}⊂S(e.g.

[34] (3.2.3)), We take such h with h>m. And set a=(Xh,Xh+1,…,Xh+m-1).

Suppose a is generated by (m-1) elements f1,…,fm-1. We may assume

f1=Xh+…

f2=Xh+1+…

fm-1=Xh+m-2+….

Then there do not exist Fi∈DS[X] such that

Xh+m-1=f1F1+…+fm-1Fm-1.


Hence DS[X] is not rank (m-1), and hence m-1<l(_??_n).

Let AS[X] be of rank n. By [28] (2.3), A is noetherian of dimension 0.

Henoe we have the irredundant primary decomposition (0)=q1∩ … ∩qn, and hence

the direct product decomposition: A_??_A/q1+…+A/qn. Then (A/qi)S[X] has rank

n.

(5.7) Let (A,√0) be a noetherian local ring. A[X] has rank 2 if and only

if √0 is principal with √02=0.

PROOF. _??_. We have √0=uA for u∈A. The case √0_??_0. We suppose

that an ideal a of A[X] generated by three elements is not generated by two ele-

ments. We take f, g,h such that (f,g,h)=a and the sum of their degrees is

minimal. Since we may assume that the highest coefficients of f, g, h are 1 or u,

we can make new generators the sum of which degrees is smaller. _??_. A becomes

of rank 2. We suppose first that m=√0 is not principal. Let m=(u,v) and let

a=(uX,vX2,X3). There exist f, g such that a=(f,g).

f=na1X+(ua2+va3)X2+…

g=ub1X+(ub2+vb3)X2+…(ai,bi∈A).

Since uX∈(f,g), we may assume a1 unit. Hence we may set

f=uX+(ua1+va2)X2+…

g= vbX2+… (ai,bi∈A).

Since vX2∈(f,g), b is a unit. Hence we may set

f=uX+uaX2+…

g= vX2+….

Then X3∈(f,g) implies a contradiction. Hence m=(u) is principal. Next sup-

pose min {m>0;um=0}=n_??_3. a=(u2X,uX2,X3) is generated by two f, g.

f=u2a1X+ua2X2+…

g=u2b1X+ub2X2+….

Since u2X∈(f,g), we may assume a1 unit. Hence we may assume

f=u2X+uaX2+…

g= ubX2+….

Since uX2∈(f,g), b is a unit. Hence we may assume

f=u2X +aX3+…

g= uX2+….

22 Ryuki MATSUDA

Then X3∈(f,g) derives a contradiction.

(5.8) A[X] has rank 2 if and only if A is a finite direct sum of local rings

(Ai,mi) such that mi is principal with m2i=0.

5.9) Let (A,√0) be noetherian local. AZ[X] has rank 2 if and only if

√0 is principal with √02=0.

PROOF. The proof is similar with (5.7). _??_. It follows then that A[X] is

rank 2. Hence AZ[X] has rank 2. _??_. We set √0=m, 1-X=Y. We suppose

m is not principal. Let m=(u,v) and a=(uY,vY2,Y3). There exist f, g such

that a=(f,g). We may assume

f=ua1Y+(ua2+va3)Y2+…

g=ub1Y+(ub2+Vb3)Y2+….

Since u Y∈(f,g), we may assume

f=uY+(ua1+va2)Y2+…

g= vbY2+….

Since vY2∈(f,g), we may assume

f=uY+uaY2+…

g= vY2+….

Then Y3∈(f,g) derives a contradiction. Hence m=(u) is principal. Next, sup-

pose u2_??_o. a=(u2Y, uY2, Y3) is generated by two f, g. We may assume

f=u2a1Y+ua2Y2+…

g=u2b1Y+ub2Y2+….

Since u2Y∈(f,g), we may assume

f=u2Y+uaY2+…

g= ubY2+….

Since uY2∈( .f,g), we may assume

f=u2Y +aY3+…

g= uY2+….

Then Y3∈(f,g) derives a contradiction.

5.10) AG[X] is rank 2 if and only if G_??_Z and A is a finite direct sum of local

rings (Ai,mi) such that mi is principal with m2i=0.

(5.11) PROPOSITION. AS[X] is rank 2 if and only if A is a finite direct

Torsion-Frec Abelian Semigroup Rings, V 23

sum of local rings (Ai,mi) such that mi is principal with m2i=0, and one of the

followings holds; (i) S is isomorphic with Z, (ii) S is isomorphic with Z0, (iii)S is neither isomorphic with Z nor Z0, and isomorphic with a subsemigroup of Z0

containing 2, and each Ai is a field.

PROOF. _??_. We may assume q(S)=Z. We suppose S_??_Z, S_??_Z0. By

(5.6), we may assume Z0⊃S∋2. Let 2n+1 be the least odd number contained in

S. Suppose mi=uAi_??_0 for some i. Let f, g be generators of the ideal a=(uX2n,

X2n+1,X2n+2) of (Ai)S[X].

f=ua1X2n+a2X2n+1+…

g=ub1X2n+b2X2n+1+… (aj,bj∈Ai).

Since uX2n∈(f,g), we may assume

f=uX2n+aX2n+1+…

g= bX2n+1+….

Since X2n+1∈(f,g), we may assume

f=uX2n +aX2n+2+…

g= X2n+1+bX2n+2+….

Then X2n+2∈(f,g) derives a contradiction.

Let (A,√0) be a local ring. If AS[X] has rank n, then √0 is generated by n-1

elements. For the proof, we see at first that AG[X] is rank n for G=q(S). Next

we may assume G=Z. Suppose √0=(u0,…,un-1) is not generated by n-1

elements. The ideal a=(u0,u1Y,…,un-1Yn-1,Yn) of AZ[X] is generated by n

elements f1,…,fn, where Y=1-X. We determine the forms of the fi from ujYj

∈(fi,…,fn). Then Yn∈(f1,…,fn) derives a contradiction.

(5.12) EXAMPLE. Let (A,√0) be a local ring. A[X] is rank 3 if and only

if (1) √0=(u) with u3=0, or (2) √0=(u,v) with u2=v2=uv=0.

PROOF. _??_. The case where √0=(u) is principal. If u3_??_0, the ideal (u3,

u2X,uX2,X3) is not generated by 3 elements. The case where √0=(u,v) is not

principal. If u2_??_0, the ideal (u2,uX,vX2,X3) is not generated by 3 elements.

If uv_??_0, the ideal (uv, uX, vX2, X3) is not generated by 3 elements.

(5.13) PROPOSITION. Let (A,√0) be a noetherian local ring and k be the

least integer such that √0k+1=0. Let un(i-1)+1+√0i+1,…,un(i)+√0i+1 be

the least number of generators of √0i/√0i+1 where we set n(0)=0. Then A[X]

has rank n if and only if n(k)<n.

24 Ryuki MATSUDA

PROOF. _??_. We set ui=vn(k)+1-i. We suppose n(k)_??_n and set a=(v1X,

v2X2,…,vnXn,Xn+1). Let f1,…,fn be generators of a. We have

fi=ai1X+ai2X2+…,aih∈(v1,…,vh) for h_??_n.

Since v1X∈(f1,…,fn), we may assume a11=v1, a21=…=an1=0,{a2h,…,anh}⊂

(v2,…,vh) for h_??_n. Since v2X2∈(f1,…,fn), we may assume a22=v2, a32=…=

an2=0, a12∈(v1),{a3h,…,anh}⊂(v3,…,vh) for h_??_n. Since v3X3∈(f1,…,fn), we

may assume a33=v3, a13∈(v1), a23∈(v2),{a4h,…,anh}⊂(v4,…,vh) for h_??_n. We

continue the operation.

Then Xn+1∈(f1,…,fn) derives a contradiction. _??_. Let a be an ideal generated

by n+1 elements f1,…,fn+1. Every nonunit of A is written as the form Σuiεi

with units εi. We may assume deg (f1)_??_…_??_deg (fn+1). For i<j, we replace

fj by fj-εfiXdeg(fj)-deg(fi) with some unit ε. Thus we see that a is generated by

n elements.

(5.14) A[X] has rank n if and only if A is a finite direct sum of local rings

(Ai, √Ai) such that each Ai[X] has rank n.

(5.15) PROPOSITION. Let (A, √0) be a noetherian local ring. AG[X]

has rank n if and only if G_??_Z and A[X] has rank n.

PROOF. _??_. The similar proof with (5.13) applies as well, using Y=1-X.

_??_. We show every ideal a generated by n+1 elements f1,…,fn+1 is generated by

n elements. We may assume oX(fi)=0 and degY(fi)_??_degY(fi+1). Express the

Y-highest coefficient ai of fi by ε or Σuiεi with units ε, εi. For i<j, we replace fj

by fj-εfiYdegY(fj)-degY(fi). We may assume that the first ui1 appearing in a1

does not appear in a2, a3,…,an+1. Next we may assume that the first ui2 appear-

ing in a2 does not appear in a3, a4,…,an+1…

(5.16) AG[X] has rank n if and only if A is a finite direct sum of local (Ai,

√Ai) such that (Ai)G[X] has rank n.

(5.17) THEOREM. Let (A,√0) be a noetherian local ring, k the least integer

such that √0k+1=0 and S a subsemigroup of Z0 with q(S)=Z. Let un(i-1)+1+

√0i+1,…,un(i)+√0i+1 be the least number of generators of √0i/√0i+1 (we set

n(0)=0) and let min (S＼0)=l. Then AS[X] has rank n if and only if (1+n(k))l_??_n.


PROOF. _??_. We set n(k)=m, vm+1-i=ui. Suppose (1+m)l>n. There

exists h such that {h, h+1, h+2,…}⊂S. Let a be the ideal of As[X] generated

by the following*:

a is generated by n elements f1,…,fn. We set

Since v1Xh∈(f1,…,fn), we may assume a1h=v1, a2h=…=anh=0. Since v1Xh+1

∈(f1,…,fn), we may assume a2h+1=v1, a1h+1=0, a3h+1=…=anh+1=0…. Since

v1Xh+l-1∈(f1,…,fn), we may assume alh+l-1=v1, a1h+l-1=…=al-1h+l-1=0,

al+1h+l-1=…=anh+l-1=0. Since v2Xh+l∈(f1,…,fn), we may assume al+1h+l=v2,

a2h+l=…=alh+l=0, al+2h+l=…=anh+l=0. We continue this operation.

We take the monomial ζ of degree h+n in *. Then ζ∈(f1,…,fn) derives a con-

tradiction. _??_. We show that every ideal a generated by n+1 elements f1,…,

fn+1 is generated by n elements. Classify deg(f1),…, deg(fn+1) by mod l. There

exists a class containing more than m+1 elements, say {f1,f2,…,fm+2}. We

may assume deg(f1)_??_…_??_deg(fm+2). For i<j, we replace fj by fj-

εfiXdeg(fj)-deg(fi) (ε unit).….

Finally in this section we remark on 'loc'.

(5.18) PROPOSITION. AG(X)loc has finite rank if and only if A is noetherian

and (A/p)G(X)loc has finite rank for every minimal prime p of A.

26 Ryuki MATSUDA

PROOF. _??_. Then it follows AG(X)loc is noetherian ((6.3)). The rest comes

from [12] Th. 2.

6. The noetherian and the regular. This section consists of several sim-

ple remarks on the noetherian property and the regular rings.

If Ap is noetherian for every prime p of A, A is called locally noetherian.

(6.1) (1) AG(X)loc is locally noetherian if and only if (Am)G(X)loc is

noetherian for every maximal m of A. (2) AS[X] is locally noetherian if and

only if (Am)S[X] is locally noetherian for every maximal m of A.

PROOF. (2) _??_. Let _??_ be a prime ideal of AS[X]. Let m be a maximal

ideal of A containing A ∩ ??_. Then we see

AS[X]_??__??_(Am)S[X]_??_(Am)S[X].

It follows AS[X]_??_ is noetherian-

If Ap is regular local for every prime p of A, then A is called locally regular.

(6.2) (1) AG(X)loc is locally regular if and only if (Am)G(X)loc is regular local

for every maximal m of A. (2) AS[X] is locally regular if and only if (Am)S[X]

is locally regular for every maximal m of A.

Let k be a field. Then even if kq(S)[X] is locally regular, kS[X] need not be

locally regular. For example, set S={0,3,4,5,…}. We have dim kS[X]=1

(e.g.[28](6.5)). For p=(X3,X4,…)kS[X], kS[X]p is not regular.

(6.3) PROPOSITION. (1) AG(X)loc is noetherian if and only if A is

noetherian and (A/p)G(X)loc is noetherian for every minimal prime p of A. (2)

AS[X] is noetherian if and only if A is noetherian and (A/p)S[X] is noetherian

for every minimal p of A.

PROOF. (1) _??_. It is sufficient to show that every prime ideal _??_ of AG(X)loc

is finitely generated. We take a minimal _??_0 of AG(X)loc contained in _??_. _??_0 is

of the form pAG(X)loc for a minimal p of A ((1.1)). Set p=(a1,…,an) and

_??_/pAG(X)loc=(F1+pAG(X)loc,…,Fm+pAG(X)loc).

Then we have _??_=(a1,…,an,F1,…,Fm)AG(X)loc.

An ideal a of A is called regular ideal, if a contains a regular element. If the

maximal condition holds for regular ideals, the ring is called r-noetherian.

(6.4) PROPOSITION. If As[X] is r-noetherian, then A is noetherian and

q(S) is finitely generated.

PROOF. We set q(S)=G. Let a1⊂a2⊂a3⊂ … be an ascending chain of

regular ideals of AG[X]. Then a1 ∩AS[X]⊂a2∩AS[X]⊂ … is an ascending chain

of regular ideals of AS[X]. It follows an∩AS[X]=an+1∩AS[X]=… for some n.


Then we see an=an+1=…. Therefore AG[X] is r-noetherian. Our assertion

follows from [27](11.11).

If the maximal condition holds for divisorial ideals of a domain D, D is called

Mori. If every ideal of a Mori ring D generated by two elements is divisorial, D

is called M-ring.

(6.5) If DG(X)loc is a Mori-ring, then D is a Mori-ring.

(6.6) EXAMPLE. Let k be a field and S={0,2,3,4,…}. Then kS[X] is a

M-ring.

PROOF. We set kS[X]=D, m=(X2,X3)D and a=(f,g)D for nonzero f, g

∈D. (1) First m is divisorial. (2) Next mm2 is divisorial. (3) When GCDk[X](f,

g)=1. If a_??_D, then a=m. (4) (X3,X4)D is divisorial. (5) When f=aX2+

X3 for k∋a_??_0 and g=X4. Then a is divisorial. (6) When GCDk[X](f, g)=X2.

(7) The general case.

We know that if DS[X] is a M-ring, then D is a field and S is isomorphic with

a subsemigroup of Z ([28](2.10)).

If every subring containing A of q(A) is noetherian, A is called a C-ring.

(6.7) PROPOSITION. DG(X)loc is a C-domain if and only if D is noetherian

and dim D_??_1.

The assertion follows from [7] Th. 1.

[27](1.21) asserts that AG[X] is a D-ring if and only if AG[X] is a C-ring. For'loc'

, even if DG(X)loc is a C-domain, it need not be a D-domain. For example,

we set S={0,2,3,…} and D=kS[X] for a field k.

(6.8) If AG(X)loc is a C-ring, then A is a C-ring.

One more noetherian property in this section on AG(X).

(6.9) PROPOSITION. If AG(X) is noetherian semi-local, then q(A) is

noetherian semi-local.

PROOF. We may assume q(A)=A. It is proved in (9.2) that A is semi-local.

Let a be a proper ideal of A. Let b∈aAG(X)∩A. We have bg∈aAG[X] for a

regular g of AG[X]. It follows c(g)=A also from (9.2). Hence b∈a, and hence

aAG(X)∩A=a. Therefore A is noetherian.

7. The prime chains. This section concerns with the lengths of prime chains.

Let Q be a prime ideal of AG[X] and set Q∩A=P. We prove the following

analogy of [6]:

ht(Q)=ht(Q/PAG[X])+ht(PAG[X])_??_torsion-free rank (G)+ht(PAG[X]).

(7.1) PROPOSITION. If AG(X)loc is noetherian biequidimensional ([18]),

then so is A.

28 Ryuki MATSUDA

PROOF. We have dim A=dimAG(X)loc ([28](6.12)). We show that, if

p0_??_p1_??_…_??_pm is a maximal chain of prime ideals of A then p0AG(X)loc⊂

p1AG(X)loc⊂… ⊂pmAG(X)loc is a maximal prime chain of AG(X)loc. For the proof

we show that, if ht(p/p') is 1 in A/p' then ht(pAG(X)loc/p'AG(X)loc) is 1. We may

assume p'=0, and may assume A=D is a domain. We see

ht(pDG(X)loc)=dim(DG(X)loc)pDG(X)loc=dim(Dp)G(X)loc=dimDp

=1 .

(7.2) LEMMA. Let G=ΣΛ+Qα λ,Q' be a prime ideal of AG[X] and set

Q'∩A=P. Then we have

ht(Q')=ht(Q'/PAG[X])+ht(PAG[X]).

PROOF. The case the cardinality |Λ|<∞ follows by e.g. [28](6.11). The

case |Λ|=∞. we may assume ht(Q'/PAG[X])=n<∞, ht(PAG[X])=m<∞.

Suppose ht(Q')>n+m. We take a large finite subset {λ(1),…,λ(k)}⊂Λ and set

H=Qα λ(1)+…+Qα λ(k) and set Q"=Q'∩AH[X]. Then we see ht (Q"/PAH[X])

=n , ht(PAH[X])=m and ht(Q")>n+m; which contradicts with the first half.

(7.3) THEOREM. Let Q be a prime ideal of AG[X] and set A∩Q=P. Then

we have the analogy of [6] Th. 1:

ht(Q)=ht(Q/PAG[X])+ht(PAG[X]).

PROOF. Let {αλ;Λ} be a maximal independent subset of G. We may assume

ht(Q/PAG[X])=n<∞, ht(PAG[X])=m<∞. Suppose ht(Q)>n+m and set

G'=ΣΛ+Qα λ. We take a prime Q' of AG'[X] lying over Q such that ht(Q')>n+m.

By (7.2) we have a contradiction:

ht(Q')=ht(Q'/PAG'[X])+ht(PAG'[X])_??_ht(Q/PAG[X])+ht(PAG[X]).

(7.4) COROLLARY. Let Q be a prime ideal in AG[X] and set Q∩A=P.

Then we have

ht(Q)_??_torsion-free rank (G)+ht(PAG[X]).

A prime chain p0_??_p1_??_…_??_pn of AS[X] is called special, if (pi∩A)AG[X]

coincides with some pj fbr every i.

(7.5) PROPOSITION. If dim AG[X] is finite, then dim AG[X] is realized as

the length of a special chain.

The proof comes from e.g. [28](6.11).

(7.6) PROPOSITION. Let Q be a prime ideal of AG[X] such that ht(Q)<∞.

Then ht(Q) is realized as the length of a special chain descending from Q.


The proof comes from (7.3).

Let k be a field. Then kG[X] satisfies the first chain condition. The case

torsion-free rank (G)<∞ is proved in [5] Th. C. We assume torsion-free rank

(G)=∞ and suppose kG[X] has a maximal chain P0_??_P1_??_…_??_Pn=0 of finite

length. Let {αλ;Λ} be a maximal independent subset of G and set G'=ΣΛQα λ.

Since kG'[x] is integral over kG[X], there exists a maximal chain P'0_??_P'1_??_…_??_

P'n=0 such that P'i∩kG[X]=Pi. We take a large finite subset {λ(1),…,λ(m)}⊂

Λ and set Qα λ(1)+…+Qα λ(m)=H and P'i∩kH[X]=Q'i. It follows Q'0_??_Q'1_??_…_??_

Q'n=0 and Q'ikG'[X]=P'i. Therefore P'0=Q'0kG'[X] is not maximal in kG'[X];

a contradiction.


(7.7) PROPOSITION. Let P be a prime ideal of kG[X](k a field). Then

kG[X]P satisfies the first chain condition.

PROOF. At first we may assume that G is of the form ΣΛ+Qα λ. Next we

may assume that |Λ|=∞ and ht(P)=∞. Suppose there exists a saturated chain

P=P0_??_P1_??_…_??_Pn=0. We make a chain P=_??_0_??__??_1_??_…_??__??_n+1=0 of

length n+1. We take a large finite subset {λ(1),…,λ(h)}⊂Λ and set H=Qαλ(1)

+…+Qαλ(h) and set PikH[X]=Qi and _??_j∩kH[X]=_??_j. Then we have Qi_??_Qi+1

and _??_j_??__??_j+1 for every i,j. We see that QikG[X] coincides with Pi and Q0_??_Q1

_??_…_??_Qn=0 is a saturated chain. Since kH[X] satisfies the first chain condition,

we have ht(Q0)=n. On the other hand we have a chain Q0=_??_0_??__??_1_??_…_??__??_n+1

=0 of length n+1; a contradiction.

(7.8) If AG[X] satisfies the saturated chain condition for every free G, then

AG[X] does for every G. AS[X] satisfies the saturated chain condition if and only

if (A/p)S[X] does for every minimal prime p of A. And DS[X] satisfies the satu-

rated chain condition if and only if (Dm)S[X] does for every maximal ideal m of D.

Also we have a similar reduction for 'loc'.

(7.9) AG(X)loc satisfies the saturated chain condition if and only if (A/p)G(X)loc

does for every minimal prime p of A. And DG(X)loc satisfies the saturated chain

condition if and only if (Dm)G(X)loc does for every maximal m of D.

(7.10) (1) If the chain condition for primes holds in a noetherian DG(X)loc

in the sense of [36], then it does for D. (2) The condition holds in a noetherian

DG[X] if and only if D is noetherian and the condition holds in noetherian (Dp)G[X]

for every prime p of D.

A similar way of proof with (7.1) implies (1).

A ring A satisfies the chain condition (of [29]) if and only if A/p does for

every minimal prime p.

(7.11) (1) AS[X] satisfies the chain condition if and only if (A/p)S[X] doesfor every minimal prime p of A. AG(X)loc satisfies the chain condition if and

30 Ryuki MATSUDA

only if (A/p)G(X)loc does for every minimal prime p of A. (2) DS[X] satisfies the

chain condition if and only if (Dm)S[X] does for every maximal m of D.

Since a domain D satisfies the chain condition if and only if Dm does fbr every

maximal m of D, it follows

(7.12) DG(X)loc satisfies the chain condition if and only if (Dm)G(X)loc does for

every maximal m of D.

The following is the analogy of [6] Cor. 4, (b).

(7.13) THEOREM. Let A be a noetherian Hilbert ring or a semi-hereditary

Hilbert ring such that all maximal ideals have a same height. Then all maximal

ideals of AG[X] have a same height.

PROOF. Since the proofs are the same for two cases of A, we assume A

noetherian Hilbert. Let M, N be maximal ideals of AG[X] and suppose ht (M)

>ht (N). We may assume that G is of the form ΣΛ+Qα λ. The case |Λ|=∞.

We take a chain N=N0_??_N1_??_…_??_Nn with n=ht(N). We take a large finite sub-

set {λ(1),…,λ(l)}⊂Λ and set H=Qα λ(1)+…+Qα λ(l) and Ni∩AH[X]=N'i. It

follows N'0_??_N'1_??_…_??_N'n and Ni=N'iAG[X]. And there arises a contradiction

that N is not maximal. Therefore we assume |Λ|<∞; Λ={1,2,…,t}. We set

G'=Zα1+…+Zαt anew and M ∩AG'[X]=M'. M' is a maximal ideal of AG'[X].

By [27](6.8), M'∩A is a G-ideal of A. Hence M∩A=m is a maximal ideal of A.

Similarly N ∩A=n is a maximal ideal of A. If ht(mAG[X])>ht(m), we have

ht(m)=ht(mA[X1,…,Xt])_??_ht(mAG'[X])_??_ht(mAG[X])>ht(m);

a contradiction. Therefore we see ht(mAG[X])=ht(m); similarly ht(nAG[X])

=ht(n). Using (7.3) we have

ht(M)=ht(M/mAG[X])+ht(mAG[X])=t+ht(mAG[X])

=t+ht(m)=t+ht(n)=t+ht(nAG[X])

=ht(N/nAG[X])+ht(nAG[X])=ht(N).

(7.14) PROPOSITION. Let A be a Hilbert ring such that for every maximal

M of A, ht(MAG[X]) is equal to ht(M) and for all maximal M, MAG[X] has

a same height. Then for all maximal M' of AG[X], M' has a same height.

The proof is similar with (7.13).

Even if all maximal ideals of AZ[X](resp. A[X]) have a same height, A need

not be Hilbert. For an example, let R be a countable ring and set RZ[X1,X2,

X3,…]=A. A is not Hi bert ([27](6.7)).

The following is the analogy of [6] Cor. 4, (a) and the proof is similar.

(7.15) PROPOSITION. If all maximal ideals of AG[X] have a same finite

height, then A is Hilbert.


PROOF. We see that torsion-free rank (G) is finite. We may assume G=

Zα1+…+Zαt. Next we may assume t=1; G=Z. Let M' be a maximal ideal of

AZ[X] and set M' ∩A=M. Suppose M is not maximal and take a prime P_??_M.

We take a maximal ideal N' of Az[X] containing P. Using (7.4), we have

ht(M')_??_1+ht(MAZ[X])<1+ht(PAZ[X])_??_ht(N').

This is a contradiction. Hence M is maximal and by [27] (6.2), A is Hilbert.

8. Intersections of overrings. This section concerns with rings which

become some kinds of intersections of overrings; domains with finite character,

Krull type,… We determine a necessary and sufficient condition of As[X] being

Dedekind ring (with zerodivisors).

(8.1) If DG(X)loc has the finite character ([11])(resp. the finite rational

character ([4]), the finite real character ([4])), then D has the respective character.

If DG(X)loc is a Krull (resp. almost Krull) domain, then so is D.

If Dp is a Krull domain for every prime p of D, then D is called almost Krull.

(8.2) PROPOSITION. If DG(X)loc is Krull type ([11]) (resp. generalized

Krull), then so is D respectively.

PROOF. Assume DG(X)loc is Krull type. Let υ be one of the valuations of

DG(X) defining the Krull type of DG(X)loc. Let q(D)=K, υ|K=w and let _??_ be the

center of υ in DG(X)loc, p the center of w in D, Rυ the va uation ring of υ and Rw

the valuation ring of w. We have Rυ=(DG(X)loc)_??_. Let x∈Rw. We have xf'/g'

=f/g for f/g,f'/g'∈DG(x)loc with f'/g'_??__??_. Since υ(g')=0, we see υ(f')=0 .

Therefore there exists a coefficient a of f' g such that w(a)=0. It follows x∈Dp,

and Rw=Dp.

(8.3) PROPOSITION. If DG(X)loc is a K domain, then D is a K domain.

PROOF. The proof of the essentiality is similar with (8.2). Let υ be a valua-

tion of DG(X) defining K domain of DG(X)loc and _??_ be the center of υ in DG(X)loc.

We set q(D)=K, υ|K=w and let p be the center of w in D. _??_ is of the form ∩Λ(fλ/

hλ)DG(X)loc. We may assume p_??_0 and take 0_??_a∈p. Then we may assume

_??_=∩Λ(a/hλ)DG(X)loc. We have p=∩0_??_bλα(a/bλα)D for hλ=Σbλ αXα. Hence p is

divisorial.

As to the Krull ring in the sense of [22], it is stated in [27] (10.25) that if AG[X]

is a Krull ring then q(A) is A or A is a Krull ring. Though the proof is not stated,it does not seem trivial.

(8.4) If As[X] is a Krull ring, then q(A) is A or A is a Krull ring.

PROOF. We assume q(A)_??_A. Let{υ λ;Λ} be the family of valuations of

As(X) defining the Krullness of As[X], and set υλ|K=wλ and q(A)=K . We set

32 Ryoki MATSUDA

Σ={σ ∈Λ;wσ is a valuation of K}. Let x=a/b∈K＼A. We have υλ(x)<0 for

some λ; and hence 0_??_υλ(a)<υ λ(b). υλ(K)＼∞ is isomorphic with a subsemigroup

of Z which contains proper positive integers and proper negative integers, and

hence a subgroup of Z. Therefore λ belongs to Σ and hence Σ is not empty. It

follows A=∩Σ{x∈K;υ σ(x)_??_0}. We set pσ={x∈K;wσ(x)>0} for σ ∈Σ. We

take x=a/b with wσ(x)<0. Then as above b∈pσ. Therefore pσ is a regular ideal

of the valuation ring of wσ.


(8.5) PROPOSITION. If AG(X)loc is a Krull ring, then q(A) is A or A is a

Krull ring.

A ring A is called a Dedekind ring (with zerodivisors) if it is prufer and satisfies

the maximal condition on regular ideals.

(8.6) THBOREM. Aε[X] is a Dedekind ring if and only if A is a finite

direct sum of fields and S is isomorphic with Z or Z0.

PROOF. _??_. By the results of [13], A is von Neumann regular and S is

isomorphic with a subgroup of Q or a prufer subsemigroup of Q0. Then seeing the

form of elements of q(S), we see that Aq(S)[X] is r-noetherian. By [27] (11.11),

A is noetherian and q(S) is finitely generated.

(8.7) DS[X] is almost Krull ([11]) if and only if (Dm)s[X] is almost Krull for

every maximal m of D. And DG(X)loc is almost Krull if and only if (Dm)G(X)loc

is Krull for every maximal m of D.

(8.8) PROPOSITION. Let DG(X)loc be local Krull. Then, (1) D is a local

Krull ring, (2) There exists a natural homomorphism Cl(D)→Cl(DG(X)loc), (3)

This homomorphism is injective.

PROOF. DG(X)loc is a faithfully flat D-module ([26](7.1)), From e.g. [10]

(6.10), we see that D is a Krull domain. And since every invertible ideal of D is

principal, we get (3).

If Am is a Dedekind ring for every regular maximal ideal m, A is called almost

Dedekind.

(8.9) AG(X)loc is an almost Dedekind ring if and only if (Am)G(X)loc is Dede-

kind for every maximal ideal m of A that contains a nonempty finite subset F with

(0:F)A=0.

PROOF. _??_. Assume m be such a maximal ideal of A. For F={a0, a1,…,

an} and 0_??_α∈G, we set f=a0+a1Xa+…+anXnα. Since f is regular in AG(X)loc,

(AG(X)loc)mAG(X)loc is Dedekind.

(8.10) DG(X)loc has the QQR-property ([11]) if and only if (Dm)G(X)loc has

the QQR-property for every maximal m of D. And if Ds[X] has the QQR-proper-


ty, then D is a field and S is isomorphic with a subgroup of Q or a subsemigroup

of Q0.

PROOF. The first half comes from [11] p. 339, 2 and [26] (7.6). If Ds[X]

has the QQR-property, then DG[X] has the property (cf. [11] p. 340, 4) for q(S)

=G. By [27] (14.8), D is a field and G is isomorphic into Q. If S is not a group,

it is isomorphic with a subsemigroup of Q0 (e.g. [34] (3.5.1)).

We see from (3.3) that if DG(X)loc has the QR-property, then D is prufer.

9. The ring AG(X). This section is notes on AG(X) concerning some proper-

ties but special ones stated in other sections. We determine a necessary and

sufficient condition of AG(X) being semi-local.

(9.1) PROPOSITION. AG(X) is a local ring if and only if, (q (A), m) is a

local ring such that for every finite subset F of m we have (0:F)q(A)_??_0.

PROOF._??_. We set q(A)=B. Let M be a maximal ideal of BG(X) and set

m=M∩B. Every element s of B＼m is regular in B. And hence m is a unique

maximal ideal of B. We take any F={p1,…,pn}. We set f=p1Xα1+…+pnXαn

for α1<…<αn. Since f∈M,f is a zerodivisor of BG(X). By [26] (9.5), we have

(0:f)B_??_0. _??_. First we have mBG(X)_??_BG(X). Let ζ=f/g∈BG(X)＼mBG(X).

We set f=a1Xα1+…+anXαn, ai_??_0 for every i and for α1<…<αn. Since some

ai_??_m, f is regular in BG[X]. Hence ζ is a unit of BG(X).

A ring with only a finite number of maximal ideals is called semi-local.

(9.2) THEOREM. AG(X) is a semi-local ring if and only if, q(A) is a semi-

local ring such that for every finitely generated proper ideal a of q(A) we have

(0:a)q(A)_??_0.

PROOF. We may assume q(A)=A. If I is an ideal of AG(X) such that I∩A

=m is a maximal ideal of A, then it follows I=mAG(X). For the proof , we sup-

pose I_??_mAG(X). We take f of AG[X] with f_??_mAG(X). Set f=a1Xα1+…+

anXan for ai_??_m for every i. We have aan+p=1 for a∈A and p∈m. af+pXan

is a unit of AG(X) contained in I. Hence I=AG(X); a contradiction. _??_. Let

M1,…, Mk be the set of maximal ideals of AG(X). We set Mi∩A=pi. Let m be

a maximal ideal of A. m is contained in ∪ipi. Hence m is pi for some i. There-

fore A is semi-local. Let a=(a1,…,an) be a finitely generated proper ideal of A.

We take a maximal m of A containing a. By the above, we see m=pj for some j.

By the fact at the beginning, we have pjAG(X)=Mj. We set f=a1Xa1+…+

anXan for α1<…<αn. Since f∈Mj, f is a zerodivisor of AG[X]. It follows

(0:a)_??_0. _??_. Let m1,…, mk be the set of maximal ideals of A, and set miAG(X)

=Mi. Mi is a maximal ideal of AG(X). Let M be any maximal ideal of AG(X)

and let ξ∈M∩AG[X]. We set ξ=a1Xa1+…+anXan and set (a1,…,an)A=a.

34 Ryuki MATSUDA

Since ξ is not a unit, we have a_??_A. a is contained in some mi. Hence ζ is con-

tained in miAG(X)(=Mi), and hence M is contained in ∪iMi. It follows M⊂Mj

for some j. Therefore AG(X) is semi-local.

In the above, the numbers of maximal ideals of A and AG(X) coincide. And

(9.1) becomes a corollary of (9.2).

(9.3) REMARK. There exists a local ring (A, m) with q(A)=A such that for

a nonempty finite subset F of m, (0:F) is zero.

In fact, we make a similar ring with [21] Ch. 2, 2-2, Exer. 7. Let k be a field

and let x, y be indeterminates. We set R=k[x,y](x ,y). Let {pλ; Λ} be the com-

plete representatives of non-associated prime elements of R. We set E=Σ+R/

pλR and make a semi-direct sum R+E=A. Then q(A) coincides with A, and A

is local. For F={x,y}, we see (0:F)=0.

(9.4) PROPOSITION. (1) If q is a primary ideal of A such that qAG(X)_??_

AG(X), then qAG(X)∩A coincides with q. (2) Let A ｂe a local ring such that

q(A)=A, and a an ideal of A such that aAG(X)_??_AG(X). Then aAG(X)∩A need

not be a. (3) Assume A=q(A) and let p be a prime ideal such that pAG(X)_??_

AG(X). Then AG(X)/pAG(X) need not be isomorphic with (A/p)G(X). (4) Let

a1,…,an be ideals of A. Then we have

(a1∩ … ∩an)AG(X)=a1AG(X)∩ … ∩anAG(X).

(5) Let m be a maximal ideal of A such that mAG(X)_??_ΛG(X). Then mAG(X)

is a maximal ideal of AG(X).

PROOF. (1) Let a∈qAG(X)∩A. qAG[X] is a primary ideal of AG[X]

(e.g. [26] (9.7)). It follows a∈qAG[X] and hence a∈q. (2) We consider A of

(9.3). Let p be a prime element of R which is not associated with x, y, and set

(px,py)A=a. Suppose that aAG(X)=AG(X). Then pxf+pyh is a regular ele-

ment of AG[X] for f,h∈AG[X]. For (…,0,1,0,…)=e, we have (pxf+pyh)e=

0; a contradiction. Hence aAG(X)_??_AG(X). We set g=xXα1+yXα2 for α1_??_α2.

Since pg∈ αAG[X], we have p∈aAG(X)∩A. On the other hand, we have p_??_a.

Hence aAG(x)∩A_??_a. (3) We take the first A of the examples in the introduction

of [28] 4. We have dim A_??_1. Let m be the maximal ideal and p a minimal prime

of A, mAG(X) and pAG(X) are prime ideals of AG(X), and we have AG(X)_??_

mAG(X)_??_pAG(X). Hence AG(X)/pAG(X) is not a field, and while (A/p)G(X) is

a field, (5) follows similarly as (9.2).

(9.5) REMARK. There exists a ring A with q(A)=A such that for every

maximal ideal M of AG(X), M∩A is not a maximal ideal of A.

We take the ring of [21] Ch. 2, 2-2, Exer. 7 again. Let k be a field and x, y

indeterminates and set R=k[x,y]. Let {pλ; Λ} be the complete representatives


of non-associated prime elements of R, and make E=Σ+R/(pλ) and set A=R+E.

Then we see q(A)=A. Let M be a maximal ideal of AG(X) and set M∩A=p.

Suppose p is maximal in A. Then p contains E, and R∩p=m is maximal in R.

We have Pλ1, Pλ2∈m for some λ1_??_λ2. Set f=Pλ1Xa1+Pλ2Xa2 for α1_??_α2.

f is regular in AG[X]. Hence M=AG(X); a contradiction.

(9.6) PROPOSITION. (1) Let n be the nilradical of A. Then the nilradi-

cal of AG(X) is nAG(X). (2) The pseudoradical of AG(X) is also nAG(X).

The proof follows from [28] (5.1).

(9.7) PROPOSITION. AG(X) is semi-hereditary if and only if A is von

Neumann regular.

The proof follows from [9] Th. 2 and [28] (8.5).

10. The Kronecker functions rings. We treated the G-powered Kronecker

functions rings of domains in [26] 8 and [27] 2. We note in this section that the

theory of Kronecker functions rings with zerodivisors by Hinkle and Huckaba

([19]) also holds for G-powered ones.

(10.1) Let f∈As[X]. In the followings, (1) and (2) are equivalent with each

other and (3) implies (2).

(1) f is regular in As[X].

(2) (0:c(f))=0.

(3) c(f) is a regular ideal of A.

If A satisfies the condition 'if f is regular in A[X] then c(f) is regular in A', A

is called to have property (A).

(10.2) Assume A satisfies property (A). If g is regular in As[X], then c(g)

is regular in A.

PROOF. We set g=a1Xα1+…+anXαn, ai_??_0 forα1<…<αn. Then a1X

+…+anXn is regular in A[X]. Hence c(g) is regular in A.

If every regular ideal of A is generated by regular elements, A is called to have

property (P).

(10.3) PROPOSITION. Let A be an integrally closed ring with properties

(P) and (A). Then for f,g∈As[X] with g regular, we have c(fg)'=(c(f)c(g))'.

Let {Vλ; Λ} be the set of valuation overrings of A on q(A). For a fractional

ideal a of A we set a'=∩ΛaVλ.

PROOF OF (10.3). We have c(f)c(g)k+1=c(fg)c(g)k for some k by [26]

(9.8). By (10.2), c(g) is regular. By [19](3.5)(c), we have c(fg)'=(c(f)c(g))'.

(10.4) PROPOSITION. Assume A is integrally closed with properties (P)

and (A). Then the following set makes a ring:

36 Ryuki MATSUDA

AK={f/g;f,g∈As[x], g regular, c(f)'⊂c(g)'}.

The proof is similar with the usual polynomial case, using (10.2) and (10.3).

(10.5) PROPOSITION. We have AK∩q(A)=A. And every. finitely gener-

ated regular ideal of AK is principal.

(10.6) THBOREM. Assume A be integrally closed with properties (P) and

(A). Then the followings are equivalent with each other.

(1) A is a prufer ring.

(2) AS(X)loc=AK.

(3) AS(X)loc is a prufer ring.

(4) AK is a regular quotient ring of AS[X].

PROOF. (2)_??_(3) follows from (10.5). (3)_??_(1) follows by using [14] Th. 13,

11.

(10.7) Let V be a valuation ring (not necessarily be a domain) and υ the

corresponding valuation. Then υ is extended to a valuation of VS(X) if and only

if every finite subset of υ-1(∞) has the nonzero annihilator.

Using (10.2), we have as a corollary that if V has property (A), υ is extended to

a valuation of Vs(X).

(10.8) PROPOSITION. Let V be a valuation ring with properties (P) and

(A), v the corresponding valuation, v* the extension of v to Vs(X) and V* the

valuation ring of v*. Then we have V*=VK=Vs(X)loc.

VK=VS(X)loc follows from (10.6).

Finally we add a remark on the property (P).

(10.9) PROPOSITION. The ring As[X] has property (P).

PROOF. Let f/g∈AS(X). We set f=ΣaiXαi, g=ΣbjXβj, and set q(S)=G.

Let H be the subgroup of G generated by {α1,…,α.;β1,…,βm}. We have H_??_

Z+…+Z. We may suppose g_??_A. We see that the k-component βkl of βl is not

zero for some l, k. In the case of βkl>0, we take n such that maxi.j(αki-βkj)<nβkl.

In the case of βkl<0, we take n such that mini,j(αki-βkj)>nβkl. It follows f+Xnβlg

is regular in As[X]. Hence f/g+Xnβl is regular in As(X).

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