Note
on star-operations
and
semistar-operations
RYUKI MATSUDA*
and IWAO SATO**
0. Introduction
A subsemigroup ⊃≠{0} of a torsion-free abelian (additive) group is called a grading monoid. First we show that [He] holds for a grading monoid S. Especially every ideal of an integrally closed grading monoid S is divisorial if and only if S is a valuation semigroup with the principal maximal ideal. Next we discuss whether [MSu, Proposition 8] holds for n=4, namely, an integrally closed domain D of dimension 4 is or is not a valuation ring if and only if 5〓￨Σ'(D)￨〓9, where Σ'(D) is the set of semistar-operations on D. Next. we study semigroup version of [AA2], [Q1] and [Q2]. Next we study the question of [A]. Finally we show that every cancellation ideal of a grading monoid is principal.
1. Valuation semigroups
A subset I≠ 〓 of a grading monoid S is called an ideal of S if S+I⊂I. For x∈S, set (x)=x+S. An ideal I of S is called a principal ideal if I=(x) for some x∈S. An ideal P of S is called a prime ideal, if s1+s2∈P for s1.
s2∈S⇒s1∈P or s2∈P.
A nonempty subset T of a grading monoid S is called an additive system if it satisfies the following condition: t,t'∈T⇒t+t'∈T. For an additive system T, the quotient semigroup ST is defined as follows:
ST={s-t￨s∈S,t∈T}.
Especially, if T=S, then the quotient semigroup Ss={s1-s2￨s1,s2∈S}
is called the quotient group of S, and is denoted by q(S). Note that q(S) is a group.
Let S be a grading monoid and let G be the quotient group of S. A nonempty subset A of G is said to be a fractional ideal of S if x+A⊂S for some element x of S and S+A⊂A. Each ideal of S is a fractional ideal of S, and is called an integral ideal.
In general, set (I1:I2)I=I1:II2={x∈I￨x+I2⊂I1} for subsets I1,
I2, I of S.
Received February 5, 1996.
1991 Mathematics Subject Classification.
Primary 13A15 Secondary 20M14.
* Department of Mathematical Sciences
, Ibaraki University,
Mito, Ibaraki 310. Japan.
** Oyama National College of Technology
,
Oyama, Tochigi 323, Japan.
6 Ryuki MATSUDA and Iwao SATO
For a fractinal ideal A of S, set A-1=(S:A)q(s). Note that A⊂(A-1)-1 for any fractional ideal A of S. A fractional ideal F of S is called divisorial if
(F-1)-1=F. A grading monoid S is said to be reflexive if every ideal of S is divisorial.
(1.1). If {Aα} is a set of fractinal ideals of a reflexive semigroup S such that ∩ αAα ≠ 〓, then (∩ αAα)-1=∪ αA-1α.
PROOF. We have (∩ αAα)-1⊂Aα, so that A-1α ⊂(∩ αAα)-1. Thus
and UαA-1α is a fractional ideal.
Set A=∪ αA-1α. Then we have A-1α ⊂A. Thus A-1⊂(A-1α)-1=Aα for each α. Therefore A-1⊂ ∩ αAα and
Q.E.D.
(1.2). If M is the maximal ideal of a reflexive semigroup S⊂ ≠q(S), then M-1 properly contains S and is generated by 0 and any element of M-1-S.
PROOF. Since (M-1)-1=M and S-1=S, M-1 properly contains S.
If x∈M-1-S and F=S∪(S+x), then S⊂ ≠F⊂M-1. Thus we have
M=(M-1)-1⊂F-1⊂ ≠S-1=S. Hence M=F-1 and M-1=(F-1)-1=F.
Q.E.D.
(1.3). If A and B are fractional ideals of a grading monoid S and if T is an additive system of S, then (A∩B)+ST=(A+ST)∩(B+ST).
PROOF. It is clear that (A∩B)+ST⊂(A+ST)∩(B+ST). If x∈
(A+ST)∩(B+ST), then x=a-t=b-t', where a∈A, b∈B and t,t'∈T.
Thus we have x=(a+t')-(t+t')=(b+t)-(t+t'). Then a+t'=b+t∈A∩B, and so x∈(A∩B)+ST.
Q.E.D.
(1.4). Let T be an additive system of S. If A is a divisorial ideal of S which is a finite intersection of principal fractional ideals of S, then A+ST is a divisorial ideal of ST.
(1.5). Let S be a reflexive semigroup⊂ ≠q(S), and M its maximal ideal. If (x) is a principal ideal of S, then x+M-1 properly contains (x) and is contained in any fractional ideal which properly contains (x).
PROOF. We choose y∈M-1-S. Then we have x+y∈(x+M-1)-(x+S),
and so x+M-1 properly contains (x). If F is a fractional ideal which properly
contains (x), then S⊂(x+S)-x⊂≠F-x and (F-x)-1⊂≠S. Hence
(F-x)-1⊂M, so that M-1⊂F-x and x+M-1⊂F.
Q.E.D.
We call an ideal irreducible if it is not a finite intersection of ideals strictly containing it.
(1.6). Bach principal ideal of a reflexive semigroup S is irreducible. If each ideal of S is finitely generated, then S is said to be a Noetherian semigroup.
(1.7). If S is a Noetherian semigroup which is reflexive, then S has Krull dimension〓1.
PROOF. (1.6) shows that each principal ideal of S is irreducible, and there-fore, since S is Noetherian, is primary. If S has the maximal ideal M of height 〓2, then there exists a prime ideal P such that M⊃ ≠P. Let x∈M＼P and y∈P. Then we have y〓(x+y) and nx〓(x+y) for every n∈N. Namely, the principal ideai (x+y) is not primary. This is a contradiction.
Q.E.D.
Let G be a torsion-free abelian group and Γ a totally ordered abelian group, where both G and Γ are additive groups. A valuation v:G→ Γ is a function from G into Γ such that v(a+b)=v(a)+v(b) for any a, b∈G. Set V={a∈ G￨v(a)〓0}. Then V is called the valuation semigroup belonging to v. On the other hand, v is called a valuation belonging to V.
Let S be a grading monoid and G=q(S) its quotient group. An element x∈G is called integral over S if there exists a positive integer n such that nx∈S. S is called integrally closed if the following condition is satisfied: x is integral over S if and only if x∈S.
(1.8) THEOREM. Let S be an integrally closed semigroup. Then the follow-ing conditions are equivalent.
(1) S is reflexive.
(2) S is a valuation semigroup and the maximal ideal is principal.
PROOF. (1) implies (2): Then S is a valuation semigroup ([MSa, Proposi-tion 17]). If M is the maximal ideal of S, then, since M is divisorial, there exists a principal ideal (y) such that M=S∩(y). If y=a-b with a,b∈S, then we observe that M={d∈S￨d+b∈(a)}. Let v be the valuation associated with S. If v(y)〓0, then S⊂(y). If v(y)>0, then (y)⊂S. It now follows that M is finitely generated. Hence M is principal.
Before proving that (2) implies (1) we first establish the following lemma.
(1.9). Let V be a valuation semigroup with maximal ideal M. Then V is reflexive if and only if M is a principal ideal.
PROOF. That M is principal if V is reflexive is a consequence of (1) implies (2) in (1.8). To prove the converse we assume that M=(x) and that F is an ideal of V. Let F' be the intersection of the principal fractional ideals containing F. If y〓F, then F is properly contained in (y) so that F-y is properly contained in V. Hence F-y⊂(x) and F⊂(x+y)⊂ ≠(y). Thus F'⊂(y+x) and y〓F'. We conclude that F=F'.
We now show that (2)
implies
(1)
in (1.8).
Let S be a valuation
semigroup
with the principal
maximal ideal
M, and let
A be an ideal
of S. Then (1.9)
implies
that A is
divisorial.
8 Ryuki MATSUDA and Iwao SATO
An element x of q(S) is called almost integral over S, if a+nx∈S for some a∈S and any n∈N. S is called completely integrally closed if every almost integral element over S belongs to S.
(1.10). Suppose that S⊂ ≠q(S) is completely integrally closed. Then S is reflexive if and only if S is a discrete valuation semigroup of rank 1.
PROOF. If S is a discrete valuation semigroup of rank 1, S is reflexive by (1.8).
Conversely, assume that S is a completely integrally closed valuation semi-group. Let v be the valuation on q(S)=G belonging to S. Let v(G)=Γ. Suppose that Γ has a convex subgroup H with {0}⊂ ≠H⊂ ≠ Γ. Choose x,y∈S with 0<v(x)∈H and v(y)〓H. Then we have y-nx∈S for each n∈N. Since S is completely integrally closed, we have-x∈S. Hence v(x)=0; a contradiction. Therefore Γ has rank 1, that is, Γ is isomorphic to a subgroup of the real numbers. Since the maximal ideal M of S is principal, Γ is isomorphic to Z.
Q.E.D.
Now we have seen that [He] holds for grading monoids.
2. Known results and problem
Let D be a domain and A a D-submodule of K=q(D). Then A is called a fractional ideal of D if dA⊂D for some 0≠d∈D. Let F(D) be the set of all nonzero fractional ideals of D.
Let S be a grading monoid and F(S) the set of all fractional ideals of S. Let Σ(D) be the set of all star-operations on D. Let Σ(S) be the set of all star-operations on S.
(2.1). (1) For a domain D, D is reflexive if and only if ￨Σ(D)￨=1. (2) For a grading monoid S, S is reflexive if and only if ￨Σ(S)￨=1.
(2.2) ([AA2, PROPOSITION 12]). Let V be a valuation ring with nonzero maximal ideal M. If M is principal, then ￨Σ(V)￨=1. Otherwise ￨Σ(V)￨=2.
(2.3) ([OMSA]). Let V be a valuation semigroup with maximal ideal M. If M is principal, then ￨Σ(V)￨=1. Otherwise ￨Σ(V)￨=2.
Let D be an integral domain with quotient field K. Let F'(D) denote the set of all nonzero D-submodules of K. A mapping*: A→A* on F'(D) is called a semistar-operation on D if the following conditions are satisfied for all a∈K＼{0} and A,B∈F'(D):
(1) (aA)*=aA*;
(2) A⊂A*; if A⊂B, then A*⊂B*; and
(3) (A*)=A*.
Let F'(S) be the set of nonempty subset I of q(S)=G such that S+I⊂I. A mapping *:A→A* on F'(S) is called a semistar-operation on S if the following conditions hold for all a∈G and A,B∈F'(S):
(1) (a+A)*=a+A*;
(2) A⊂A*; if A⊂B, then A*⊂B*; and (3) (A*)*=A*.
Let Σ'(S) be the set of all semistar-operations on S.
(2.4) ([OMSA]). Let S be an integrally closed semigroup of dimension n. Then S is a valuation semigroup if and only if n+1〓￨Σ'(S)￨〓2n+1.
(2.5) (CF. [OMSA, REMARK 57]). The condition "integrally closed" can not be removed in (2.4).
A domain with a unique maximal ideal is called a local domain.
(2.6) ([MSU, COROLLARY 6]). Let D be an integrally closed local domain of dimension n. Then D is a valuation ring if and only if n+1〓￨Σ'(D)￨〓2n+1.
(2.7) (CF. [MSU, EXAMPLE 9]). The condition "integrally closed" can not be removed in (2.6).
(2.8) ([MSU, PROPOSITION 8]). Let D be an integrally closed domain of dimension n〓3. If Σ'(D) satisfies the inequality n+1〓￨Σ'(D)￨〓2n+1,
then D is a valuation ring.
(2.9) PROBLEM. Let D be an integrally closed domain of dimension 4. If Σ'(D) satisfies the inequality 5〓￨Σ'(D)￨〓9, then is D a valuation ring ?
If an integral domain D satisfies the ascending chain condition on integral divisorial ideals of D, then D is called a Mori ring. If a grading monoid S satisfies the ascending chain condition on integral divisorial ideals of S, then S is called a Mori semigroup.
When D or S is not necessarily integrally closed, we have the the following ,
(2.10) ([Q2, THEOREM 3]). Let D⊂ ≠q(D) be a Mori ring. Then the following conditions are equivalent:
(1) D is reflexive.
(2) dim D=1 and M-1 is 2-generated for every maximal ideal M of D. (3) Every 2-generated ideal is divisorial.
(2.11) ([SA] AND [SAM]). Let q(S)⊃ ≠S be a Mori semigroup. Then the followings are equivalent:
(1) S is reflexive.
(2) dim S=1 and M-1 is 2-generated for the maximal ideal M of S. (3) Every 2-generated ideal of S is divisorial.
10 Ryuki MATSUDA and Iwao SATO
(3.1) PROPOSITION. Let D be a 4-dimensional integrally closed domain with exactly two maximal ideals. Then we have ￨Σ'(D)￨〓9.
PROOF. Let M, N be two maximal ideals of D and let ht(M)=4. Since D is integrally closed, D is the intersection of all valuation overrings {Vλ￨λ ∈ Λ} of D. Then *Vλ ∈ Σ'(D) for each λ. Therefore we assume that {Vλ￨λ ∈ Λ} is
a finite set. We assume that Λ={1,…,n} and Vi〓Vj for i≠j. Then D is a Prufer ring with exactly n maximal ideals. By the assumption, we have n=2.
Let M⊃ ≠P3⊃ ≠P2⊃ ≠P1 be a prime ideal chain of D, where ht(Pi)=i for each i. Set DM=V, DPi=Ui and DN=W. Then V, Ui, W are valuation
rings. Set q(D)=K. Then e, *U1, *U2, *U3, *V, *W, d are 7 distinct elements of Σ'(D).
Suppose that ht(N)〓3. Then P3〓N, P3〓N. We set R=U3∩W.
Then R has exactly two maximal ideals and dim R=3. By (2.6), We have ￨Σ'(R)￨〓8. Let *1,…,*8 be distinct elements of Σ'(R). Then δ(*1),…,δ(*8), d, *V are distinct elements of Σ'(D).
Next we assume that ht(N)=4.
Assume that P3〓N. Let Q be a prime ideal of D such that ht(Q)=3 and Q⊂N. Set U3∩DQ=R. Then R has exactly two maximal ideals and dim R=3. Thus ￨Σ'(R)￨〓8 by (2.6). Let *1,…,*8 be distinct elements of Σ'(R). Then it follows that ￨Σ'(D)￨〓10.
Therefore we may assume that P3⊂ ≠N.
By [He, Theorem 5.1], d≠v. Thus e, *U1, *U2, *U3, *V, *W, d, v are 8 distinct elements of Σ'(D).
We consider whether MV, NW are principal. If neither MV nor MW are principal, then by [He, Lemma 5.2], dV≠vV in Σ'(V) and dW≠vW in Σ'(W). Then e, *U1, *U2, *U3, *V, *W, d, v, δ(vV), δ(vW) are distinct.
Therefore we may assume that MV or MW is principal, say MV. There exists an element a∈M such that MV=aV. Then we may take a in M-N. That is, MV=aV, a∈M-N. Then we have M=aD.
Now suppose that NW is also principal. Then N is also principal. Espe-cially, M and N are divisorial ideals. Let {Aλ￨λ ∈ λ} be the set of all ideals A such that P3⊂A〓M. If each Aλ is divisorial, then ∩ λAλ=A is not contained in M similarly to the proof of [He, Lemma 2.3]. If y∈A-M, then y∈(P3,yy2), and so y=x+yy2z(∈P3,z∈D). Thus we obtain a contradiction. Therefore there exists an ideal I such that P3⊂I⊂ ≠Iv, I〓M. Similarly, there exists an ideal J such that P3⊂J⊂ ≠Jv, J〓N. For each A∈F'(D), let A*1=Av∩AV, A*2=Av∩AW. Then it follows that d, v, *1, *2 are distinct each other. Hence it follows that ￨Σ'(D)￨〓10.
Thus we may assume that MV=aV is principal and NW is not principal. We have dW≠vW in Σ'(W). Therefore it follows that Σ'(D)⊃{e,*U1,*U2, *U3,*V,*W,d,v,δ(vW)}.
Q.E.D.
(3.2). Let D be a 4-dimensional integrally closed domain such that ￨Σ'(D)￨ 〓9. Then D is a Prufer ring with at most two maximal ideals.
PROOF. Since D is integrally closed, D is the intersection of all valuation overrings {Vλ￨λ ∈A} of D. Then *Vλ ∈ Σ'(D). It follows that D is a Prufer ring with a finite number of maximal ideals. Let M be a maximal ideal with height 4. Let {M=M1,N=M2,M3,…,Mn} be the set of all maximal ideals and let n〓2. Set DM∩DN=R. Then R is a 4-dimensional integrally closed domain. By (3.1), we have￨Σ'(R)￨〓9. Since ￨Σ'(R)￨〓￨Σ'(D)￨, we have R=D.
Q.E.D.
(3.3). Let D be a valuation ring or a Prufer ring with exactly two maximal ideals which contains a common nonzero prime ideal. Then F'(D)=F(D)∪ {K}, where K=q(D).
PROOF. If D is a valuation ring, the proof is straightforward.
Assume that D is a Prufer ring with exactly two maximal ideals M, N. Set DM=V and DN=W. Let v, w be the valuations of V and W, respectively.
Let I∈F'(D).
Assume that v(I), w(I) are bounded to the below. Then there exists a∈D
such that We have aI⊂ ≠V. Similarly there exist 0≠b∈D such
that bI⊂ ≠W. Then abI⊂ ≠V∩W=D. Therefore it follows that I∈F(D). Next, assume that both v(I) and w(I) are not bounded to the below. Let 0≠x∈K. Then there exists i∈I such that v(i)<v(x). We have x∈iV⊂ ≠IV. Similarly we have x∈IW, and so x∈I. Therefore K=I.
Now, suppose that one of v(I), w(I) is bounded to the below, and the other is not bounded to the below, say, v(I) is not bounded to the below. Then there
is 0≠a∈D such that Let 0≠ π ∈P1, where P1 is a nonzero
prime ideal with P1⊂M∩N. Let i∈I with
Since
we have Furthermore, since s〓P1, we have
Therefore
This is a contradiction to the choice of
Q.E.D.
(3.4). Let D be a 4-dimensional integrally closed domain and ￨Σ'(D)￨〓9. Then ￨Σ(D)￨〓2. Furthermore, if D is not local, then ￨Σ(D)￨=2 .
PROOF. In the case that D has exactly one maximal ideal, D is a valuation ring. Then, by (2.2), we have ￨Σ(D)￨〓2.
By (3.2), we may assume that D has exactly two maximal ideals M , N. Then, by the proof of (3.1), one of MV, NV, say MV is principal, and NV is not principal. Furthermore we have Σ'(D)={e,*U1,*U2,*U3,*V,*W ,d,v, δ(vW)}.
Let * be a star-operation on D. Then a semistar-operation *∈ Σ'(D) is naturally induced by *. Namely we set as follows:
12 Ryuki MATSUDA and Iwao SATO
Thus it follows that *=d or *=v. We have *=d or *=v. Therefore ￨Σ(D)￨ =2, and so Σ(D)={d,v}.
Q.E.D.
(3.5) PROPOSITION. Let D be a 4-dimensional integrally closed domain which is not local. Then ￨Σ'(D)￨〓9 if and only if the following conditions hold:
(1) D is a Prufer ring with exactly two maximal ideals M, N. (2) There exist prime ideals P1, P2 and P3 of D such that
(3) P1DP1, P2DP2, P3DP3 are principal.
(4) One of MDM and NDN is principal and the other is not principal. (5) ￨Σ(D)￨=2.
PROOF. Sufficiency: Then Σ'(U1), Σ'(U2), Σ'(U3) consists of two, three, four elements respectively, where Ui=DPi for each i. We have Σ'(V)=
δ(Σ'(U3))∪{dV}, Σ'(W)=δ(Σ'(U3))∪{dW,vW}, where V=DM, W=DN.
Therefore Σ'(D) contains the following nine elements: e, *U1, *U2, *U3, *V, *W, d, v, δ(vW). Now, let *∈ Σ'(D). If D*=R⊃ ≠D, then * is a descent to D of a semistar-operation on R. Next, if D*=D, then * ￨F(D) is a semistar-operation on D. We have Σ(D)={d,v}. Using (3.3), * coincides with d or v.
Q.E.D.
4. Note on [AA2]
in this section we study a semigroup version of [AA2].
Let S be a grading monoid with q(S)=G. We let f(S) be the set of finitely generated members of F(S).
We may define a partial order〓on Σ(S) by *1〓*2 if and only if I*1⊂I*2 for every I∈F(S).
For *1,*2∈ Σ(S), it is easily seen that the following conditions are equiva-lent: (1) *1〓*2, (2) (I*2)*1=I*2 for every I∈F(S), (3) (I*1)*2=I*2 for every I∈F(S), and (4) F*2(S)⊂F*1(S). Here we denote F*(S)={I*￨I∈F(S)}
for any *∈ Σ(S). Under this partial order, there is a smallest element, the d-operation and a greatest element, the v-operation. A star-operation * on S is said to have finite character if for each I∈F(S), I*=∪{J*￨f(S)∋J,J⊂I}.
Given a star-operation * on S, we can define a new star-operation *f on S as follows: I*f=∪{J*￨J∈f(S),J⊂I}. Then *f is a finite character star-operation on S. We denote the set of all finite-character star-operations on S by
Σf(S).
We note that [A] holds for S([M,ｧ4]).
Now Σ(S) is a partially ordered set with the greatest element and the small-est element, and is closed under arbitrary meets and arbitrary joins. Namely we have the following,
(4.1) PROPOSITION. Σ(S) is a complete lattice with respect to the partial order〓.
(4.2) LEMMA. A nonempty subset F'⊂F(S) has F'=F*(S) for some star-operation * on S if and only if (1) S∈F', (2) I∈F' implies that x+I∈ F' for each x∈q(S), and (3) 〓 ≠{Iα}⊂F' with ∩ αIα ≠ 〓 implies that ∩ αIα ∈F'.
In this case, * may be defined by I*=∩{J∈F'￨J⊃I}.
(4.3) PROPOSITION. Σf(S) is a complete lattice with respect to the partial order〓.
Suppose that the intersection of any two finitely generated ideals of a semi-group S is finitely generated. Then S is called a coherent semigroup (this is a natural semigroup version of the definition of a coherent ring (c.f. [C, Theorem 2.2])).
Let Y={Pλ￨λ ∈ Λ} be a family of prime ideals of S such that ∩ λSPλ=S. If, for each I∈F(S), we set I*=∩ λ(I+SPλ), * is a star-operation on S. Let I(S) be the set of all star-operations of this form.
(4.4). Let S be a coherent semigroup and *∈ Σ(S) which satisfies (I∩J)*= I*∩J* for all I,J∈f(S). Then (I∩J)*f=I*f∩J*f for all I,J∈F(S). Hence *f∈I(S).
Let * be a star-operation on S. A fractional ideal I of S is called a *-ideal, if I*=I. An ideal I is called a prime *-ideal, if I is a prime ideal and I is a *-ideal.
(4.5) THEOREM. Let S be a coherent semigroup such that each prime ideal of S is a direct union of finitely generated prime ideals. If *∈I(S), then *f∈I(S). Moreover, {P￨P is a prime *f-ideal}={P￨P is a direct union of finitely generated Prime *-ideals of S}.
(4.6). Let S be a grading monoid. Then
If S is Noetherian, then ￨Σ(S)￨〓2￨S￨.
PROOF. In the case of S=q(S), it is clear.
Now, let S⊂q(S). Then we have ￨F(S)￨=∞. Thus the cardinality of the product set F(S)×F(S) is equal to that of F(S). Therefore we have 2￨F(S)×F(S)￨=2￨F(S)￨. Since a star-operation * on S is a mapping from F(S) to F(S), ￨Σ(S)￨〓2￨F(S)×F(S)￨. Hence ￨Σ(S)￨〓2￨F(S)￨.
Q.E.D.
Next, we consider a semigroup version of [AA2, Theorem 14]. If R is an affine domain with dim R〓2, then [AA2, Theorem 14] shows that ￨Σ(R)￨=2￨R￨.
Now, let k be a torsion-free abelian group ⊃≠{0}. Let X, Y be
inde-terminates.
We consider
the polynomial
semigroup
R of X, Y over k, i.e.,
R=k[X,Y]=k+Z0X+Z0Y.
14 Ryuki MATSUDA and Iwao SATO
(4.7). R=k[X,Y] is of dimension 2. Furthermore, there exists k such that ￨Σ(R)￨<2￨R￨.
PROOF. M=(X,Y) is the maximal ideal of R. Next, (X) and (Y) are prime ideals of R with height 1, and there do not exist prime ideals except them. Thus we have dim R=2. Furthermore, M=(X)∪(Y) and M〓(X), M〓(Y). Any element of R is associated with an element of the form mX+nY (m,n∈ Z0). Let A be the set of all subsets of Z0+Z0. Then we have ￨F(R)￨〓￨A￨. Let T be the set of all mappings from A to A. Then we have ￨Σ(R)￨〓￨T￨. of course, ￨T￨ is not dependent of k. But, there exists k such that ￨k￨>￨T￨. For such k, we have ￨Σ(R)￨<2￨R￨.
Q.E.D.
Now the semigroup version of [AA2] has finished.
5. On [Q1]
(5.1) (SEMIGROUP VERSION OF [Q1, THEOREM 1]). The followings are equivalent:
(1) S is a Mori semigroup.
(2) For any integral ideal I, there exists a finitely generated ideal J such that J⊂I and I-1=J-1.
(5.2) (SEMIGROUP VERSION OF [Q1, COROLLARY 1]). For any fractional ideal A of a Mori semigroup S, there exists a finitely generated fractional ideal I such that I⊂A and A-1=I-1.
(5.3) (SEMIGROUP VERSION OF [Q1, COROLLARY 2]). Let V〓q(V) be a valuation semigroup which is Mori. Then V is discrete of rank 1.
PROOF. The maximal ideal M of V is principal: M=(x). If there exists y∈V such that v(nx)<v(y) for all n∈N, we have
a contradiction. Q.E.D.
Let A be an ideal or an extension semigroup of a grading monoid S. Then the followings are equivalent:
(1) For any ideal I, J of S, (I∩J)+A=(I+A)∩(J+A).
(2) For any ideal I of S and for any s∈S, (I:s)S+A=(I+A:s)A.
If an ideal A of S satisfies the above conditions, then A is called a flat ideal of S. If an oversemigroup A of S satisfies the above conditions, then A is called a flat oversemigroup of S. If a flat A satisfies A+M〓A, then A is called faithfully flat over S, where M is the maximal ideal of S.
(5.4) (SEMIGROUP VERSION OF [Q1, THEOREM 2]). Let S be a Mori semi-group and T an extension semigroup which is flat over S. Then there exist
nat-ural mappings φ:F(S)→F(T), j:D(S)→D(T) and j:C(S)→C(T).
If T is faithfully flat over S, then φ and j are injective. If T is an oversemigroup of S, then φ, j and j are surjective.
PROOF. Assume that Av=Bv in F(S). We may assume that A= (al,...,an) and B=(b1,...,bm) are finitely generated by (5.2). Since A-1=B-1. we have ∩n1(-ai)=∩m1(-bj). Since T is flat, ∩(-ai+T)=∩(-bj+T). Hence
(A+T)-1=(B+T)-1. Hence (A+T)v=(B+T)v. Two mappings j and j
are well-defined.
Assume that T is faithfully flat over S. Assume that A+T=B+T. If there exists a∈A with a〓B, then we have (B:a)S⊂M. Then
T=(A+T:a)T=(B+T:a)T=(B:a)S+T⊂M+T〓T;
a contradiction. Therefore A=B, and hence φ is injective.
Assume that (A+T)v=(B+T)v. We will derive Av=Bv. A=(a1,...,an) and B=(b1,...,bm) may be assumed finitely generated. Since (A+T)-1= (B+T)-1, it follows that ∩(-ai+T)=∩(-bj+T) and ∩(-ai)+T=∩(-bj)+T. And hence ∩(-ai)=∩(-bj). It follows that A-1=B-1 and Av=Bv.
Assume that T is an oversemigroup of S. Let A'∈F(T). We have d+A'⊂T for some d. Set(-d)∩A'=A. Then A+T=(-d+T)∩A'. Hence φ is surjective. Q.E.D.
(5.5). Let A be a v-ideal of a Mori semigroup S. Then A is the intersection of a finite number of principal fractional ideals of S.
PROOF. We have A=(x1,...,xn)-1 by (5.2). Then A=∩n1(-xi). Q.E.D.
(5.6). Let S be a Mori semigroup and T an extension semigroup which is flat over S. If A is a v-ideal of S, then A+T is a v-ideal of T.
PROOF. By (5.5), we have A=∩n1(xi). Then A+T=∩n1(xi)+T=∩n1(xi+T).
Q.E.D.
(5.7). Let S be a Mori semigroup and T an oversemigroup which is flat over S. Then T is a Mori semigroup.
PROOF. Let. B1⊂B2⊂ … be integral v-ideals of T. Set Bi∩S=Ai. Of course, (Ai)v=(Avi)v. By the first part of the proof of (5.4), we have (Ai+T)v=
(Avi+T)v. Thus Avi⊂(Ai+T)v⊂Bvi=Bi. Therefore Avi⊂Bi∩S=Ai. Hence
Avi=Ai. Then Ai is a v-ideal of S with Ai+T=Bi . We have An=An+1=… for some n. It follows that Bn=Bn+1=….
Q.E.D.
(5.8). Let S be a Krull semigroup and T an oversemigroup which is flat
16 Ryuki MATSUDA and Iwao SATO
PROOF. Since j is surjective, D(T) is a group. Hence T is completely integrally closed. By (5.7), T is a Mori semigroup. Therefore T is a Krull semigroup.
Q.E.D.
If a fractinal ideal A of S satisfies Av=A and A+A-1=A, then A is called strongly divisorial.
We denote the complete integral closure of S by S*. An element A of F(S) is called an idempotent if A+A=A.
(5.9). The followings are equivalent:
(a) F(S) contains a maximum idempotent E under the inclusion. (b) F(S) contains a completely integrally closed oversemigroup E of S. (c) S*∈F(S).
(d) There exists a minimum member I among strongly divisorial fractional deals of S under the inclusion.
PROOF. (a)⇒(b): E is an oversemigroup of S. And E is completely inte-grally closed.
(b)⇒(c): Clearly E⊂S*. And S*=E∈F(S). (c)⇒(a): S* is a maximum idempotent.
(a)⇒(d): (S:S*)S is a minimum member among strongly divisorial frac-tional ideals of S under the inclusion.
(d)⇒(a): I-1 is a maximum idempotent. Q.E.D.
(5.10). S is completely integrally closed if and only if S is a minimum member under the inclusion among the strongly divisorial fractinal ideals of S.
PROOF. Necessity: In the proof of (a)⇒(d) of (5.9), we have I=(S:S)S= S. If A is strongly divisorial, then S=I⊂A.
Sufficiency: (a)∼(d) of (5.9) hold for S. In the later half of the proof of (d)⇒(a) of (5.9), we have I=S and S*=I-1. Hence S*=S. Therefore S is completely inetgrally closed.
Q.E.D.
The intersection of all prime ideals of a grading monoid S is called the pseudo-radical of S. Let T be an oversemigroup of S. Then, if the pseudo-radical of S is nonempty, then that of T is also nonempty.
(5.11). Let the pseudo-radical of S be nonempty. Then (1) If V is a valuation oversemigroup of S, then V*∈F(V). (2) If S is the integral closure of S, then (S)*∈F(S).
PROOF. (1) Let a be an element of the pseudo-radical of S. There exists a prime ideal Q such that V*=VQ. Then we have a∈Q. Therefore a+V*⊂V. Hence we have V*∈F(V).
(2) Let {Vλ￨λ ∈ Λ} be the set of all valuation oversemigroups of S. Let a be an element of the pseudo-radical of S. By (1), we have a+(S)*⊂a+∩ λV*λ ⊂ ∩ λVλ=S.
Q.E.D.
For n∈N and a subset I of S, set nI=I+…+I={a1+…+an￨ai∈I}.
(5.12). Let S be a Mori semigroup, M a maximal ideal of S and dim S=1. Let x∈S. Then,
(1) There exists an n∈N such that nM⊂(x). (2) M-1〓S.
PROOF. (1) There exists a finitely generated ideal I=(c1,…,ck)⊂M such that M-1=I-1. Since dim S=1, there exists an n∈N such that nI⊂(x). Then we have nM⊂nMv=nIv⊂(nI)v⊂(x).
(2) Let x∈M. Since nM⊂(x), we have (nMv)v⊂(x). Therefore Mv=M. Hence we have M-1〓S.
Q.E.D.
(5.13). Let S be a Mori semigroup, M a maximal ideal of S and dim S>1. Then M-1=S or M is strongly divisorial.
PROOF. Let M-1〓S. Then M+M-1=M or M+M-1=S.
Now, let M+M-1=S. Then M is principal. Set M=(x). There exists a prime ideal P such that M〓P. Let p∈P. There exists r1∈P such that
p=x+r1. There exists r2∈P such that r1=x+r2…. Thus we have
p=ix+ri(i=1,2,…), ri∈P. Then (p-x)〓(p-2x)〓(p-3x)〓 …. This
is a contradiction to the fact that S is Mori. Therefore we have M+M-1=M. Hence M is strongly divisorial.
Q.E.D.
6. On [Q2]
Let S be a Mori semigroup. If every ideal generated by two elements is divisorial, then S is called an M-semigroup.
(6.1) ([SA] AND [SAM]). Let S≠q(S) be a Mori semigroup and M the maximal ideal of S. Then the following conditions are equivalent:
(1) dim S=1 and M-1 is 2-generated. (2) S is reflexive.
(3) Each 2-generated ideal of S is a v-ideal.
(6.2) (A PART OF [MSA, PROPOSITION 17]). Let S be integrally closed. Then the followings are equivalent:
(1) S is a valuation semigroup.
(2) Each 2-generated ideal of S is divisorial.
7. On the question of [A]
Let D be an integral domain. We obtain necessary and sufficient conditions for (A∩B)v=Av∩Bv for all nonzero ideals A, B of D. Among other theorems D.D. Anderson [A] proved the following,
18 Ryuki MATSUDA and Iwao SATO
(7.1) ([A, THEOREM 4]). Let * be a star-operation On D. Then the follow-ing conditions on * are equivalent:
(1) There is a collection S={Pλ￨λ} of prime ideals of D with D=∩ λDPλ so that A*=∩ λ AD Pλ for each A∈F(D).
(2) (a) (A∩B)*=A*∩B* for all (integral) A, B∈F(D).
(b) Each proper integral *-ideal of D is contained in a prime *-ideal. (3) (a) (A:DB)*=(A*:DB*) for all (integral) A∈F(D) and B∈f(D). (b) Each proper integral *-ideal of D is contained in a prime *-ideal. Moreover, if either (2) or (3) is satisfied, then we can take S={Pλ￨λ} to be any collection of prime *-ideals of D with the property that each proper intetgral *-ideal of D is contained in some Pλ.
It is stated in[A, p. 2550] that we had been unable to character the integrally closed domains D for which (A∩B)v=Av∩Bv for all A, B∈F(D).
Let I be an integral ideal of D and let A∈F(D). Then we denote the subset {x∈K￨sx∈A for some s∈D-I} of K by AI, where K=q(D).
Recall that D is a v-domain (respectively, completely integrally closed) if, for all A∈f(D) (respectively, A∈F(D)), there exists a B∈F(D) with (AB)v=D. In this case, Bv=A-1.
The necessity of the following is proved in [AACDMZ]. The sufficiency is proved in [MO].
(7.2). Let D be an integrally closed domain. Then (A1∩...∩An)v=(A1)v∩ … ∩(An)v for all (integral) A1,…,An∈f(D) if and only if D is a v-domain.
So we may naturally conjecture that the followings are equivalent for an integrally closed domain D: (1) (A1∩...∩An)v=(A1)v∩...∩(An)v for all (integral) A1,…,An∈F(D), (2) D is completely integrally closed. But (1) does not necessarily imply (2). For example, a valuation domain D of dimension >1 satisfies (1). But D does not satisfy (2). However, (2) does imply (1) as the following proposition shows.
(7.3) (A PART OF [M, (1.5)]). Let * be a star-operation on D. Assume
that (AA-1)*=D for each A∈F(D). Then we have (A1∩...∩An)*=
(A1)*∩ …∩(An)* for every Ai∈F(D).
(7.4) LEMMA. Let * be a star-operation on D. Let {Iλ￨λ} be the set of proper integral *-ideals of D. Then we have ∩ λAIλ ⊂A* for each A∈F(D).
PROOF. Let 0≠x∈ ∩ λAIλ. Then we have (A:Dx)〓Iλ for each λ.
Hence (A*:Dx)〓Iλ. We have (A*:Dx)=x-1 A*∩D. It follows that
Namely
(A*:Dx) is a *-ideal of D. It follows that (A*:Dx)=D. 1∈(A*:Dx) implies x∈A*. We have proved ∩ λAIλ ⊂A*.
Q.E.D.
(7.5) THEOREM. Let * be a star-operation on D. Then the following con-ditions are equivalent.
(2) (A∩B)*=A*∩B* for all (integral) A, B∈F(D). (3) (A∩B)*=A*∩B* for all (integral) A∈F(D) and B∈f(D).
(4) (A∩(x))*=A*∩(x) for all (integral) A∈F(D) and x∈K-{0}. (5) (A∩D)*=A*∩D for all A∈F(D).
(6) (A:DB)*=(A*:DB*) for all (integral) A∈F(D) and B∈f(D). (7) (A:Dx)*=(A*:Dx) for all (integral) A∈F(D) and x∈K-{0}. (8) (A:DD)*=(A*:DD) for all A∈F(D).
(9) Let S={Iλ￨λ} be the set of proper integral *-ideals of D. Then A*=∩ λAIλ for each A∈F(D).
(10) There is a collection S={Iλ￨λ} of proper integral *-ideals of D with the property that each proper integral *-ideal of D is contained in some Iλ so that A*=∩ λAIλ for each A∈F(D).
(11) There is a collection S={Iλ￨λ} of proper integral *-ideals of D so that A*=∩ λAIλ for each A∈F(D).
(12) Let S={Iλ￨λ} be the set of proper integral *-ideals of D. Then A*={x∈K￨(A:Dx)〓Iλ for each λ} for each A∈F(D).
(13) There is a collection S={Iλ￨λ} of proper integral *-ideal of D with the property that each proper integral *-ideal of D is contained in some Iλ so that A*={x∈K￨(A:Dx)〓Iλ for each λ} for each A∈F(D).
PROOF. The following implications are clear: (1)⇒(2)⇒(3)⇒(4)⇒(5), (6)⇒(7)⇒(8), (9)⇒(10)⇒(11), (12)⇒(13).
The following implications are straightforward: (2)⇒(1), (8)⇒(5), (9) ⇒(12), (11)⇒(13).
(5)⇒(9): By (7.4), we have ∩ λAIλ ⊂A*. Conversely, let 0≠x∈A*. Then
(x)=(x)∩A*=(x)(D∩x-1A*)=(x)(D∩x-1A*)*. Hence D=(D∩x-1A)*.
Therefore D∩x-1A〓Iλ for each λ. It follows x∈AIλ for each λ. (1)⇒(6): set B=ΣbiD. Then we have
(A*:DB*)=∩i(b-1iA*∩D)=(∩ib-1iA∩D)*=(A:DB)*.
(13)⇒(2): Let x∈A*∩B*. There exist sλ, tλ ∈D-Iλ such that sλx∈A, tλx∈B for each λ. Then tλsλx∈A∩B. Then sλx∈(A∩B)* for each λ. It follows x∈((A∩B)*)*. Hence x∈(A∩B)*.
Q.E.D.
(5) is a condition on only one fractional ideal A of D for every A. The same is true on (8).
(7.6) COROLLARY. Let D be an integral domain. Then the following con-ditions are equivalent.
(1) (A1∩ … ∩An)v=(A1)v∩ … ∩(An)v for all (integral) A1,…,An∈F(D). (2) (A∩B)v=Av∩Bv for all (integral) A, B∈F(D).
(3) (A∩B)v=Av∩Bv for all (integral) A∈F(D) and B∈f(D). (4) (A∩(x))v=Av∩(x) for all (integral) A∈F(D) and x∈K-{0}. (5) (A∩D)v=Av∩D for all A∈F(D).
(6) (A:DB)v=(Av:DBv) for all (integral) A∈F(D) and B∈f(D). (7) (A:Dx)v=(Av:Dx) for all (integral) A∈F(D) and x∈K-{0}. (8) (A:DD)v=(Av:DD) for all A∈F(D).
20 Ryuki MATSUDA and Iwao SATO
(9) Let S={Iλ￨λ} be the set of proper integral v-ideals of D. Then Av=∩ λAIλ for each A∈F(D).
(10) There is a collection S={Iλ￨λ} of proper integral v-ideals of D with the property that each proper integral v-ideal of D is contained in some Iλ so that Av=∩ λAIλ for each A∈F(D).
(11) There is a collection S={Iλ￨λ} of proper integral v-ideals of D so that Av=∩ λAIλ for each A∈F(D).
(12) Let S={Iλ￨λ} be the set of proper integral v-ideals of D. Then Av={x∈K￨(A:Dx)〓Iλ for each λ} for each A∈F(D).
(13) There is a collection S={Iλ￨λ} of proper integral v-ideals of D with the property that each proper integral v-ideal of D is contained in some Iλ so that Av={x∈K￨(A:Dx)〓Iλ for each λ} for each A∈F(D).
(7.7). (7.5) and (7.6) hold for S.
(7.8) COROLLARY. Let q(S)〓S be a grading monoid and assume that the maximal ideal M of S is divisorial. If (A∩B)v=Av∩Bv for all A, B∈F(S), then S is reflexive.
PROOF. By (7.7), (7.6) holds for S. Thus, let {1λ￨λ} be the set of proper integral v-ideals of S. Then Av=∩ λAIλ for each A∈F(S). Since AM=A, we have Av=A for every A∈F(S).
Q.E.D.
Similarly, we have
(7.9). Let q(D)〓D be a local domain and assume that the maximal ideal M of D is divisorial. If (A∩B)v=Av∩Bv for all A, B∈F(D), then D is reflexive.
(7.10). Ler q(D)〓D be a domain and assume that each maximal ideal M of D is divisorial. If (A∩B)v=Av∩Bv for all A, B∈F(D), then D is
reflexive.
8. Cancellation is principal
[AA1] posed the problem whether a cancellation ideal of a local domain is principal.
(8.1) LEMMA. Let a∈S. If there exists an ideal of S which does not contain a, then there exists the maximum ideal which does not contain a.
PROOF. Let {Jλ￨λ} be the set of ideals of S which do not contain a. Set J=∪ λJλ. Then J is the maximum ideal which does not contain a.
(8.2) THEOREM. Every cancellation ideal of a grading monoid S is princi-pal.
PROOF. Let I be a cancellation ideal of S. If I=S, then I=(0) is principal.
Let I〓S. And, let M be the maximal ideal of S. Then we have I= I+S⊃I+M. If I=I+M, then, since I is a cancellation ideal, S=M; a contradiction. Therefore we have I⊃I+M. Let x be an element of I such that x〓I+M. Then we have (x)=S+x⊂I. If(x)=I, then I is principal. Next, let (x)〓I. Let y be an element of I such that y〓(x). Set x+y=a.
If a∈(2x), then we have x+y=2x+s for some s∈S, i.e., y=x+s∈(x); a contradiction. Hence we have a〓(2x). By (8.1), there exists the maximum ideal J which does not contain a; 2x∈J. Of course, a 〓J.
Let b∈I.
Case 1 b∈(x):b=s+x for some s∈S. We have b+a=(s+y)+2x∈I+J, i.e., b+a∈I+J.
case 2 b〓(x): If a∈(b+y), then a=b+y+s for some s∈S. Since x〓I+M, we have s〓M. Namely, s is a unit of S. Then b=x-s∈(x);
a contradiction. Therefore we have a〓(b+y). Hence b+y∈J. Then
b+a=x+(b+y)∈I+J, i.e., b+a∈I+J.
By the above, we have I+a⊂I+J. Since I is a cancellation ideal, a∈J; a contradiction. Therefore I=(x) is a principal ideal.
Q.E.D.
[AA1] holds for S ([SuM, Proposition 5]). For example,
(8.3). All ideals of S are cancellation ideals if and only if S=q(S) or S is a discrete valuation semigroup of rank 1.
(8.4). (1) Assume that S has a unique maximal ideal M and satisfies the ascending chain condition on principal ideals. If M is a cancellation ideal, then
S is a discrete valuation semigroup of rank 1.
(2) Assume that a valuation semigroup S has a unique maximal ideal M which is principal. Then S need not be a discrete valuation semigroup of rant 1.
PROOF. (1) By (8.2), M=(x) is principal. Since S has a.c.c.p., we have ∩∞n=1(nx)=0. Furthermore, {(0),(x), (2x), (3x), …} is the set of all ideals of S. By (8.3), S is a discrete valuation semigroup of rank 1.
(2) We take the lexicographic order in G=Z+Z: (1,0)<(0,1). The identity mapping v of G is a valuation on G. Let S be the valuation semigroup belonging to v. Then the maximal ideal M=(1,0)+S of S is principal. But, S is not a discrete valuation semigroup of rank 1.
Q.E.D.
References
[A] D.D. Anderson, Star-operations induced by overrings, Comm. Alg. 16 (1988), 2535-2553. [AA1] D.D. Anderson and D.F. Anderson, Some remarks on cancellation ideals, Math. Japon 29 (1984), 879-886.
[AA2] D.D. Anderson and D.F. Anderson, Examples of star operations on integral domains, Comm. Alg. 18 (1990), 1621-1643.
[C] S. Chase, Direct products of modules, Trans. Amer. Math. Soc. 97 (1960), 457-473. [G] R. Gilmer, Multiplicative Ideal Theory, Marcel Dekker, 1972.
[He] W. Heinzer, Integral domains in which each non-zero ideal is divisorial, Mathematika. 15 (1968), 164-170.
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