@脚
正emma 3．2．4．θ吻po3eオπ∈mq！brαll n∈Zsucんtんα孟η≧c． Thenαg≦cαnd
∫⊆Q．
Proof・We haveαq≦c， since tc∈mg⊆孟αqV（recall m⊆taV， sinceα＝min［H＼｛0｝］）．
Letη∈H and assume t ∈1． We want to show n≧5． Assume the contrary and we see
（8＋c−1）覗＝（c−1）＋（8−n）≧c
because 8＞η， whenceオ（5＋c』1）¶∈mq． Therefbre， since tn∈1＝Q：mq， we get tS＋c−1＝オ（8＋c−1）一πオη∈Q＝（オ8）
wh・nceオc−1∈A＝k［［珂］， whi・h i・imp・ssibl・（・ecall・一・（H））． Th・・オ・∈がy，、。
that∬⊆tsV∩A＝Qas is claimed． 口
The fbllowing result shows that condition（01）in Theorem 3．2．1 is satisfied， if、4＝
k［［H］］is a Gorenstein ring，α≧3， and g＝2．
Proposition 3．2．5． Suppose thαt 4 isαGorenstein ringαnd letα≧3． Then tn∈m2 forα n∈Zsucんthα彦η≧c， Hence（tS）：m2⊆（が）f ）rα〃0＜5∈H，
Proof． We may assume that H is minimally generated by｛αi｝1＜i＜e， Hence e≧2and H≠〈αゴ11≦ブ≦乏，ブ≠i＞for all 1≦i≦乏． We have c≧α≧3， since O〈c∈H．
Notice that c＞α． In fact， assume that c＝α． Then∬∋nfbr all integers n≧α．
Therefbre， becauseαi十αゴーα≧αfbr all 1≦i，ブ≦乏， we have m2＝オαm， so that
eA（、4／孟α．4）≦2， since．4 is a Gorenstein local ring． This is however impossible， because eA（A／孟α、4）＝eX（A）＝α≧3， where eX（A）denotes the multiplicity of．4 with respect to m． Hence c＞α．
Letη≧cbe an integer and assume that tn≠m2， Then n＝αi fbr some 1≦乞≦4．
We have i＞1， since c＞α． Let
．κ＝〈αゴ11≦ゴ≦e，ゴ≠i＞．
Thenα乞≠1（． We have GCD（αゴ11≦ゴ≦乏，ゴ≠の＝1． In fact， let 1≦m＜αbe an integer． Then m￠H but ai十m∈H， sinceα乞≧c． We write
e ・i＋m一Σ・ゴ・ゴ ゴ＝1
with O≦cゴ∈Z． Then ci＝0， because m〜ぎH． Therefbreαi十m∈Kfbr all 1≦m＜α．
Henceαi十1，αi十2∈K， becauseα≧3． Thus GCD（αゴ11≦ゴ≦乏，ブ≠の＝1．
We now apply Lemma 3。2．3 to the numerical semigroup K． Letα＝c−1and let O≦m≦αbe an integer． Then， since O≦m＜c≦ai， we have m∈K＝
〈αゴ［1≦ブ≦乏，ブ≠の，once m∈H（recall thatα＝α1＜α2＜… ＜α∂． SupPose now
thatα一mlぎK． Thenα一m￠」U as O≦α一m≦α， whence m∈Hbecause the
numerical semigroup H is symmetric， so that we have m∈K． Conversely， if m∈K，then m∈H， whenceα一m≠Hso thatα一m￠K． Consequently， becauseα≧αand
α＝min［K＼｛0｝］， by Lemma 3．2．3 we get c（K）＝α十1＝c． Henceα乞∈K， because αi≧c．This is impossible． Thus tn∈In2 for all integers n≧c． The second assertion fbllows from Lemma 3．2．4． □ We are now ready to prove Theorem 3．2．1．
Proof of Theorem 3．2．1．（1）We will show that鳳q−11⊆mgQ：m． We put A＝
｛（α1，α・，…，αの∈Ze lαi≧Of・・all 1≦i≦eandΣ1．、α、−q−1｝． L・t
α＝（α1，α2，… ，αの∈Aand letη∈Hsuch that tn∈∬． Let（ρ＝オΣ1＝1αiαε・tn． Then 9∈m9−1∫⊆Q：m＝（tS）＋（孟5＋c−1），
where the equality Q：m＝（tS）＋（オ8＋c−1）follows from the fact that、4 is a Gorenstein ring． Consequentlyψ∈（tS）or 9∈（が＋c−1）， since g is a monomial inオ． Because オ鵠・が＋c−1＝オm＋（c−1）・が∈mqQ fbr all O＜m∈H（use condition（01）；notice that m＋（c−1）≧c），we have m・が＋c−1⊆mgQ． Hence mg⊆mqQ if（ρ∈（オ5＋c｝1）．
SupPose that（ρ∈（tS）＝Qand write
ゼ
Σ a・ai＋η一ん＋・
i＝1
withん∈H． Then， since n≧8by Lemma 3．2．4， we get ゼ
ん一Σα・α・＋（n−5）
1 ≧Σα・α・
i＝1乏
≧・Σα・一・（q−1），
乞＝1
・・th・t w・hav・オん∈m・−1 by・・nditi・n（0、）． H・・ce．9一オΣ1−・…汁・一オん・孟・∈漁・−1Q and so m（ρ⊆mqQ， Thus mq−II⊆mqQ：m， whence mσ1＝m（mq｝II）⊆mgQ．
Let us show Q∩12＝Q1． Since mql＝mgQ， we have mqln＝mgQπfor all n∈Z．
Let x∈Q∩∫2 and write x＝がy with y∈．4． Then fbr allα∈m9， we have 孟5・αy＝αX∈mql2⊆Q2＝（オ25）．
Henceα〃∈Q＝（ts）so that we have y∈Q：mq＝1． Thusω∈QI whence Q∩12＝QI．
（2）It su伍ces to show∫2⊆Q． Let m，η∈」厚such that孟m，tn∈1． Then m， n≧5≧c by Lemma 3．2．4． We getγη十n−8∈H， since m十n−8＝m十（n−s）≧c， Therefore
tmtn＝tm＋卜8が∈Q， whence∫2⊆Q．
（3）We may assume that I2≠Q∫． Hence I2⊆Q， because Q∩12＝QI． We
have∫⊆mq−1 by condition（02）， since 8≧α（g−1）and 1⊆Q⊆オ8V． Then， since∫2⊆mq−1∬， we get
Q⊆Q＋12⊆Q：m＝Q＋（tS＋c−1）．
32
Therefbre， since乏A（［Q：m｝／Q）＝1（recall that A・is a Gorenstein ring）， we have
Q＋12＝Q・m＝Q＋（tS＋c−1），
whenceが＋卜1∈12 becauseが＋c−11ぎQ． Consequently
12＝（Q∩∫2）＋（オ8＋c−1）＝Q∫＋（が＋c−1）
because Q∩∫2＝Q．τ， whence
∫3＝Q∫2＋∫・オ3＋c−1．
Let us check that I・が＋c−1⊆QI2． Letη∈Hand assume thaげ∈∫． We will show that tnが＋c−1∈QI2． We may assume that n＞8． Letん＝（n十5十c−1）−25＝（n−8）十（c−1）．
Thenん∈Hsinceん＞c． Therefbre
α孟ん・オ28＝α・tntS＋c−1∈mql3⊆Q3＝（オ38）
for allα∈mq and soαオん∈Q． Consequently，オ九∈Q：mq＝∫， whence tnが＋c｝1＝
オ25孟ん∈Q21⊆QI2． Thus I・孟8＋c−1⊆QI2 so that∫3＝Q∫2． Since∫3＝Q12 and Q∩12＝Q∬，we get Q∩Ii＋1＝Qli fbr all i∈Z， whence Gのis a Cohen−Macaulay
ring． 口 Combining Proposition 3．2．5 and Theorem 3．2．1， we readily get Theorem 2．1．1 in the case where the base rings are numerical semigroup rings． Notice that condition（σ2）is automatically satisfied for g＝2．
Corollary 3．2．6（cf． Theorem 2．1．1）． Suppose thαt 4＝k［［珂］isαGorenstein ring αn（オtんαtα≧3．Let O＜8∈Hand put∫＝Q：m2，ωんθγ・e Q＝（tS）． Then tんεfo〃owing α88ertionsんol（t true．
ωm2∫＝m2Q and 13＝QI2．
（2／G（1）＝㊥n＞o ln／ln＋1 is a Cohen−Mαcαulay r吻．
（3？12＝QI，ザ5≧c．
3．3 The case where H＝〈α，α十1＞
In this section let H＝〈α，α十1＞withα≧2． Applying Theorem 3．2．1， we shall explore the numerical semigroup H＝〈α，α十1＞． Let c＝α（α一1）， that is the conductor of H．Similarly as in Section 2， let k be a丘eld and．4＝k［［珂］＝k［［孟α，孟α＋1】］⊆V， where
γ＝k［國is the formal power series ring over k． We denote by m＝（ta，ta＋1）the maximal ideal in．4．
Let O＜5∈H， Q＝㈹， and 1＝Q：mq with q＞Oan integer． We study
the problems of when∬is integral over Q and of when the associated graded ring Gの＝：㊥π＞o ln／ln＋1 is a Cohen−Macaulay ring．Let us begin with the fbllowing．
Lemma 3．3．1． The／blloωingα88er孟乞oπ8ん014蜘ε．
（1）1ンet e，乞≧Obe乞nteger3． Thenαe十乞∈H，ザ乞≦e・The conver8e isαlso true，げ
iくα．
（2）me＝（tae＋i l O≦i≦の＝（tn l n∈H， n≧αのノbγ顎αll integers乏≧0．
Pro（ f，（1）If i≦乏， then certainlyα乏十i＝α（6一乞）十（α十1）i∈H． SupPose that
αe＋乞∈Hand乞くα． We write ae＋乞＝αα＋β（α＋1）with O≦α，β∈Z． Then
β＝α［e−（α＋β）］＋乞and so， letting m＝e−（α＋β）， we see m≧0， becauseβ≧O and i＜α． Hencee≧α＋β≧β＝αm＋i≧乞．
Thus乞く乏．
（2）Let e≧Obe an integer． Then since
α（e一の＋（α＋1）i＝αe＋乞 fbr all O≦i≦e， we get
︶ ＃ ︵
mL（ta，ta＋1）e−（孟・e＋乞10≦i≦の．
To see mり（刑n∈H，η≧αの，16t n∈Hsuch thatη≧αゑWe writeη潔αp十乞
with p≧乏and O≦乞くα。 Then p≧乞by assertion（1）， so that tn＝オαP＋z∈m．P by equality（＃）． Hence孟η∈me， because p≧乏． Thus m．e＝（オηln∈H， n≧αの． 口 Proposition 3．3．2． Oonditions（01）and（02）伽Theorem 3．2．1αrεsatisfied／or gげαnd o吻吻くα．
34
Proof． Assume that q＜αand let n≧cbe an integer． Then n≧αq， since q＜α and c＝α（α一1）． Hence tn∈mq by Lemma 3．3．1（2）． Let n∈Hand assume that tn￠mq−1． We then have again by Lemma 3．3．1（2）that n＜α（g−1）． Thus conditions
（01）and（02）in Theorem 32．1 are satisfied． See Lemma 3．2．4 f（）r the onlyげpart．口
The question of when I is integral over Q is now answered in the following way．
Theorem 3．3．3． The！b〃oω吻伽ee conditionsαre equivalent to eαcんotんer．
（1）1⊆Q．
（2） mql＝rnqQ．
（3）q＜α．
Proof．（2）⇒（1）This is clear and well known（［NR］）．
（3）⇒（2）This follows from Proposition 3．3．2． See Theorem 3．2．1．
（1）⇒（3）Assume q≧α． We will check that 8一α￠H． Suppose 8一α∈H and let
n∈Hwith n≧aq． Then
卜α≧αq一α≧α2一α＝c，
whence（η十5一α）−8＝n一α∈H， so that tnが一α＝孟（n＋8一α）−3孟5∈Q． Because
8一α∈Hand m9＝（tn Iη∈H，η≧αq）by Lemma 3．3．1（2）， we getが｝a∈Q：
mq＝1⊆Q⊆tsV by assumption（1）， which is impossible． Thus 5一α￠Hwhence 8＞α．We write 8＝α4十rwith e≧1and O≦r＜α． Then r＞乏一1by Lemma 3．3．1（1）since 8一α＝α（4−1）十r≠H， while r≦6by Lemma 3．3．1（1）since O≦r＜αand 8＝α乏十r∈H． Thus r＝乏so that 8＝（α十1）乏． Hence乏くαbecause
8一α＜c（＝α（α一1））．
Let n∈Hwith n≧αq． Then
αe＋n−s＝n−e≧αq−e≧α2−（a−1）＝c＋1，
whenceα4十n−s∈H， so that tntαe＝tae＋卜5が∈Qfor allη∈Hwith n≧αg． Thus
tαe∈Q：m9＝1since mq＝（tn l n∈H， n≧αg）by Lemma 3．3．1（2）． Consequently tae∈σ⊆オ5V by assumption（1），so thatα乏≧5＝（α＋1）4， which is impossible because e≧1．Thus q＜αas is claimed． 口 Corollary 3．3．4．ノ4ssume thαt q＜α， Then the！blloωingαs8ertions hold true．（1？ 12＝QI， if 8≧αq．
（2／13＝QI2αnd G（1）i8αOohen一ハ4αcαulαy ring， if s≧α（q−1）．
Pro（ f， Since qくα， conditions（01）and（02）in Theorem 3．2．1 are satis丘ed（Proposition 3．3，2）．Hence Q∩12＝QJ by Theorem 3．2．1（1）． Therefore， to see assertion（1）， it suf丑ces to show that I2⊆Q． Let n∈Hwith tn∈1， Then， since tn∈Q⊆tsV by Theorem 3．3．3， we have n≧8≧αq， whence t ∈mg by Lemma 3．3．1（2）． Consequently，
1⊆m9， so that we have 12⊆mql⊆Q as is required． See Theorem 3．2．1（3）f（）r assertion