（i）we put r＝「多1：＝min｛n∈zl多≦n｝． Let n≧obe an integer and assume that In＋1＝QJn． Then（me）n＋1⊆Qsince me⊆1， so that
（n十1）e≧1δ1一ト1＝q一ト4
（・ecall th・t m1δ1望Q）． H・nce−n≧r． Simil・・ly，・ince・r＋1≧箏一牛， w・g・t m（「＋1）e⊆mlδ1＋1⊆Q， so that by Claim we have m（「＋1）e⊆QI「， whence
∬＋1−（Q＋me）「＋1−Q∬＋m（7＋1）e−QJ「．
Thus rQ（1）＝r． Letαi＝＝ max｛αi 11≦i≦d｝． Then， since乏≧αi， we have
Corollary 4．2．4． Tんθ！b♂Joω伽g assertionsんold true．
（1）五et d＝2． Then∫⊆◎⇔ min｛α1，α2｝≧q＋1．
（2）五et d＝3．7「んθη1⊆σ⇔ min｛αi＋αゴ11≦i＜ゴ≦3｝≧q＋2．
（3） 0んoo8θintegersα， q 80 tんαt 2≦α≦（オand（d−1）（α一1）＜q≦d（α一1） and let αi＝α！brαll 1≦i≦〔オ． Then 1⊆A but∫≦茎Q． F（）γ exαmple， let d＝3，α＝2，
αηdl 9＝3． Then
（222コ穿1，コ72，コ93）・m3＝m 9（xl，x3，略）．
4．3 Proof of Theorem 4．1．2−Gorensteinness in G（1）
andπ（1）
The purpose of this section is to prove Theorem 4．1．2．
Pro（ガ（）f Theorem 4．1．2．（1）We put A＝A／Q，∬＝1／Q， and而＝rn／Q． Let fi denote the initial fbrm of婿 with respect to the I−adic filtration of A Let r＝rQ（1）．
Then since！1，ゐ，… ，んfbrm a regular sequence in the Cohen−Macaulay ring G（1）（cf．
Theorem 4．1．1）， we have
G（J）2G（1）／（！1，f2，… ，fd） and r（o）（1）＝r．
H・nce Gのi・aG・・en・t・i・・i・g ifand・nly if・・i・th・・i・g G（7）−e。〉。ア／ア＋1 and the latter condition is equivalent to saying that
（o）・デーア ＋1
fbr all 1≦i≦r（cf㌦［GI］）．
Now assume that Gのis a Gorenstein ring． We then have（0）：T＝ア， that is（0）：iiie＝而fe（recall that I＝Q十me）， whence Q：me＝Q十m「e， so that Q十mq＝Q十m「e by Lemma 4．2．2（3）， Therefore q＝r4， because；m；9＝㎡乏and
〒解＝（0）：1≠（0）．Thus 41（1．
Conversely， suppose that乏lq． Then q＝r4， since r＝「多1 by Theorem 4．1．1（i）．
L・tn−r＋1． Th・n f・・ea・h 1≦i≦r， w・hav・（0）・デー（0）・㎡L［Q・m・e］／Q−
［（？十m（n−i）e］／Qby Lemma 4．2．2（3）， because lδ1十1・＝q十乏＝n乏． Hence （o）・デー徹（・一 ）e一アー ＋1，
54
so that G（1）is a Gorenstein ring．
（2）Because d≧2， R（1）is a Gorenstein ring if and only if G（1）is a Gorenstein ring and a（G（1））＝−2（c￡［1］）， where a（G（1））denotes theα一invariant of G（1）． Notice that a（G（1））＝1多1−d，since G（1）is a cohen−Macaulay ring with rQ（1）＝「多1． Hence R（1）
is a Gorenstein ring if and only if q＝re and r＝d−2！rhis completes the proof of Theorem 4．1．2． 口 Thanks to Theorem 4．1．1 and 4．1．2， we are able to produce quasi−socle ideals I＝
Q：m9， whose Rees algebras are Gorenstein． Let us note some of them．
Corollary 4．3．1． Tんθノ「ollowingαssertions hold true，
（1）五et〔オ＝2αn（iα88賜me thαt∫⊆Q． Then G（1）is notαGorenstein r伽g．
（2）
（3）
Suppose tんαt・d≧3αnd・let・n≧d−1δeαη謝egεr．五et alニd−1，αi＝nプ or all 2≦i≦（オ，αn〔i9＝（d−2）η， Tんen R（1） isα Goren8tein ring．
Supposθ抗α孟〔オ＝5αn（i let ai＝4f ）rαll 1≦i≦5． Let q＝8． Then 1⊆Qαnd G（∫）isαGoren8tein ringωi〃l r（？（1）＝1， buポR（1）is notαGorenstein ring・
Proof．（1）Suppose that q＝r乏with r＝rQ（∫）． Thenα1十α2−1＝1δ1十1＝乏十q＝
4（r十1），which implies， because 4≧αi for i＝1，2， that r＝0． This is impossible．
（2）Since lδ1＝n〔オーn−1， we get乏＝1δ1−q十1＝n． Hence q＝（d−2）e and I＝（？十mn＝（xf−1）十mn， so that R（1）is a Gorenstein ring．
（3）We have Iδ1＝15 and e＝8． Hence 1⊆Q with rQ（1）＝1． Consequently， G（1）
is a Gorenstein ring， but the ring R（1）is not Gorenstein， because q≠（d−2）乏． □
4．4 When I＝Q？
In this section we study the question of when quasi−socle ideals IニQ：mq are integrally closed． For the purpose we need the following．
Proposition 4．4．1． Suppose tんαt d
?Qαnd let n≧2わθan integer．レVe put q＝
（x？−1，x3，…，xZ）． Th・n可＝q＋mn＝＠？−1）＋m ・・nd・ll・th・p・ω・r・＝qM（m≧1）・r・
integrαlly closed．
Pro（ f， Let J＝q＋mn andα＝（x？，x8，… 畷）． Thenα⊆qand mn⊆頁， so that J⊆可．
L，t m≧1b・an i・t・g・・and p・tκ＝（m（n−1Xl），・Yn，…，曜・）． W・will・h・w th・t
π⊆」m・L・tα∈Land assum・th・t m（α1n−1）＋Σ窪、鑑≧・・W・・want t・・h・w th・t xα∈Jm． We may assume thatα1＜m（n−1）． Letα1＝（n−1）￠十ブwith亘，ゴ∈Zsuch th・t・≦ゴ＜（n−1）・Th・n O≦i＜m・Since m（竃と、）＋Σ訟、器≧・， w・g・t
d
ηα1＋（n−1）・Σαi≧蝋η一1），
i＝2 so that
d
（n−1）Σα・≧mn（n・一・1）覗α・一η［（n−1）（m一の一ゴ］，
歪＝2
whence
d Σαi≧n（m一の一π響、・
i＝2
Becau・e暑一ゴ＋詣and O≦ブ＜n−1， w・g・t農一ゴ＋嵩くゴ＋1， wh・nce
Σαi≧n（m一の一ゴ・
i＝2
Therefore we have
Xα＝＝x！n−1）乞・X｛X穿2… X3d∈誘卜1）imn（m−z）⊆Jm
and thus K⊆Jm by Proposition 4．2．1．
Because Jm⊆可n and qm⊆X， we have Jm⊆可η⊆可而⊆K， whence Jm＝可m＝
可五＝K．Letting m＝1， we get J＝可． This completes the proof of Proposition 4．4。1． 口 The following result gives a characterization of quasi−socle ideals∫＝Q：mg to be
integrally closed， in the case whereαi≧2fbr all 1≦乞≦d．
Theorem 4．42． Suppose tんat d≧2αnd ai≧2！brαll 1≦i≦d． Then舌んε／olloω吻
tωO C・nditions are eguivαlent t・eαCん・孟んer，
（1）1＝Q・
（2）Eith・・（・）・i＝e・for all 1≦i≦d，・r（b）tん・…伽孟・1≦ゴ≦伽・んオん蜘＝e if i≠ゴαnd・aゴ＝e−1・
陥，。齢￠，tんecα・e，7万＝1・・f・・α伽≧1，ωん・…即）i・α0・ん・η一M・・α吻・・rm・l domαin．
Proof．（1）⇒（2）Since I＝Q， we get q≦1δl and 1＝Q十me（cf． Lemma 4．2．2）． Notice
that
Q⊆1＝・Q：mq⊆（Q：mq）：m＝Q：mq＋1，
because∫⊆AHence Q：mq＋11Z Q． Consequently 4−1＝1δ1−（《1十1）十1＜ai for
some 1≦i≦d by Theorem 4．1．1， so that， thanks to Theorem 4．1．1 again， we have
4＝αi≧αゴ
fbr all 1≦ゴ≦d， Let△＝｛1≦ブ≦引αゴ＜e｝． We then have the following．
Claim． （1）αゴ＝乏一1，吻∈△・
（2） ＃△≦1．
Proof ofαafm。 Letゴ∈△． Thenαゴく乏＝αi whenceゴ≠iand e≧3． Letα＝
（・ゴー1）・ゴ＋（・i−・ゴ）・i．Th・n， thank・t・th・m・n・mi・l p・・P・・ty・f id・・1・， w・g・tα∈五
but xα￠Q十me＝1：＝Q， because lα1＝αi−1＝乏一1and xα≠Q． Consequently，
Σ1。1雛く1by P・・P・・iti・n 4・2・1，・・th・t 1＜毒＋箒， becau・e
αゴー1＋αrαゴ＜1．
αゴ αi
L・tn−・i−・ゴ・Th・nαゴ（・i−・ゴ）＜・i a・1＜毒＋舞， wh・nce・ゴn＜・i−・ゴ＋n・・that O≦（・」−1）（n−1）＜1・H・nce・n−1（・ecall th・t・ゴ≧2）and・ゴ＝・i−1−e−1・
Assum，＃△≧2and・h…eブ，κ∈△・・th・tブ≠κ． W・p・t y−・ゴ媛一2・W・
th・・hav・〃e 1−（・1−1）（・匙1）e−2−（鰐ゴ）（・鴛り8−2∈Qe ，・becau・eαゴー・・−e−1・by assertion（1）． Hence y∈（？i＝Q十me， which is impossible because y￠Q（recall that 乏≧3）and y￠me， thanks to the monomial property． Hence tt△≦1．
If△＝の， we then have乏＝αゴfbr all 1≦」≦d． If△≠の， letting△＝｛フ｝， we get αi＝4if i≠ブandαゴ＝乏一1．
（2）⇒（1）S・pP・・e c・nditi・n（b）is sati・丘・d． Th・n∫−Q＋mL（・ITi）＋mL Q by Lemma 4．2．2（3）and Proposition 4．4．1． Suppose condition（a）is satisfied． Then
∬⊆Qby Theorem 4．1．1 and∫＝・Q＋me＝： me by Lemma 4．2．2（3）， whence I＝Q． In
，a，h，、、， all th・p・wers・f∫・・e i・t・g・ally・1・・ed（・f． P・・P・・iti・n 4・4・1 f・・ca・e（b）），
whence the last assertion follows from Theorem 4．1．1． 口
Before closing this chapter， let us state an example．
Example 4．4．3． Suppose that d≧3and letπ≧d−1be an integer． We look at the ideal
Q＝（xg−i，x讐，x讐，… ，x2）
and let q＝n（d−2）． Then e＝n， asρ＝nd−（n＋1）， whence I⊆Qand I＝
Q十mn＝（xl−1）十mη． The ringπ（1）is by Theorem 4．1．2（ii）aGorenstein ring， since q＝（d−2）乏．If n＝・d， then 1＝（x望一1）十md and lm＝Jm for all m≧1by Proposition 4．4．1，so that Rのis a Gorenstein normal ring．
58
CHAPTER 5
QuAsI−socLE IDEALs IN LocAL RINGs
wITH GoRENsTEIN TANGENT CoNEs
5．1 Introduction
This chapter aims at a study of quasi−socle ideals in a local ring with the Gorenstein tangent COne・
In this chapter we shall fbcus our attention on a certain special kind of quasi−socle ideals． We now assume that the tangent cone， that is the associated graded ring
G（搬）一㊥mπ／m ＋1 n≧O
of m， is a Gorenstein ring and that the maximal ideal m contains a systemω1，ω2，…，xd of elements such that the ideal（コ91，コ；2，°°° ，Xd）is a reduction of m（the latter condition is automatically satis且ed， if the且1ed．4／m is in丘nite）． For example， the setting in Chapter 4naturally satisfies our specific situation． So， this chapter is a continuation research of Chapter 4． Letα1，α2，… ，αd， and g be positive integers and we put
Q＝（x？1，x窪2，… ，x匪d） and I＝Q：m9・
Let 2i＝、4／Q，而＝m／Q， and 7＝∫／Q． Letρ＝max｛n∈ZI而n≠（0）｝， that is index
of nilpotency of the ideal頁i， and put 1
e＝ρ十1−q．
We then have the lbllowing， which are the answers to Problem A in our specific setting．
Theorem 5．1．1． The／b〃oω吻伽θe condition8αrθequivalent toεαcんothθr．
59
（1）1⊆Q．
（2）mql＝mqQ．
（3）e≧αi for a〃1≦i≦d．
陥eπ棚528オんεCα3e， tんe fo〃Oωing assertions hold true．
（i）rQ（1）＝「多1：＝min｛n∈Zl婁≦n｝・
（ii）The grαded rings G（1）αnd F（1）αre Oohen−Mαcαulαy．
Theorem 5．1．2． Suppose彦hαt e≧αi／bTα〃1≦i≦d． Thθn we have the folloωing．
（i）G（1）isαGorenstein ring if and only if乏lq．
（ii）7ヒ（1）isαGorenstein ring ifαnd onl3ノザ（1＝（（オー2）乏．
The present research is， more or Iess， a continuation of Chapter 2，3，4， and［Wan］．
Our Theorems 5．1．1 and 5．1．2 might have some significance towards the second step，
providing new insight with Problem A．
Our setting naturally contains not only the case where A is a regular local ring with xl，x2，…，xd a regular system of parameters but also the case where A is an abstract hypersurface with the infinite residue class丘eld and the case where A＝RM is the localization of the homogeneous Gorenstein ring R＝k［R1］over an infinite field k＝ Ro at the irrelevant maximal ideal M＝R＋． In Section 3 we will explore a few examples in order to see how Theorems 5．1．1 and 5．1．2 work for the analysis of concrete examples．
The proofs of Theorems 5．1．1 and 5．1．2 themselves shall be given in Section 2．
5．2 Proof of Theorems 5．1．1 and 5．1．2
The purpose of this section is to prqve Theorems 5．1．1 and 5．1．2． First of all， let us restate our setting， which we shall maintain throughout this chapter．
Let A be a Noetherian local ring with the maximal ideal m and d ・ dim A＞0． We assume that the associated graded ring
G（m）一㊥mn／mn＋1
n≧0
60
of m is Gorenstein and that the maximal ideal m contains a systemω1，x2，…，xd of elements which generates a reduction of m（the latter condition is satisfied if the filed A／m is infinite）． Hence A is a Gorenstein ring and the initial fbrms｛Xi｝1＜i＜d of｛ci｝1≦i≦d with respect to m constitute a regular sequence in G（m）and we have a
canonical isomorphism
G（m／（Xl，ω2，…，Xd））E）1 G（m）／（X1，X2，…，Xの
of graded、4−algebras（［VV］）． Letα1，α2，…，ad， and q be positive integers and we put
Q・＝（x呈i，x睾2，… ，x2d） and I＝Q：m9・
Let A＝ノ1／Q，而＝m／Q， and I＝1／Q． Then
G（m）yG（m）／（X1α1，X2α2，…，Xdad），
whence G侮）is a Gorenstein ring． Letρ＝max｛n∈Zl：m：：n≠（0）｝， that is index of nilpotency of the ideal而， and we have
d ρ一・（G（司）一・（G（m））＋Σ・i，
i＝1
where a（＊）denotes theα一invariant of the corresponding graded ring（［GW，（3．1．4）］）．
Let
e＝ρ十1−q．
By［Wat］（see［0， Theorem 1．6】also）we then have the following．
Proposition 5．2．1．（0）：㎡＝而ρ＋1−i forα〃i∈Z． In pαrticulαr 7＝（0）：而P＝iiie
ωhence・1＝Q＋me．
The key for our proof of Theore】〔耳5．1．1 is the fbllowing．
Lemma 5．2．2． SUPP・se thαt・e≧αi！∂rα〃1≦i≦d． Then
Q∩rnne＋m⊆rnmQI −1
for all m≧0αnd・n≧1．
Pro（ f， We have
d
Q∩m e＋m一Σ蜘πゑ＋m−a 歪＝1
since xl，x2，… ，xd is a super regular sequence with respect to m・・Because ne＋m一αi＝（n−1）e＋m＋（e一α乞）≧（η一1）e＋m for each 1≦i≦d， we get
mne＋m−ai⊆m（n−1）e＋m＝m」m・（me）n−1．
Therefore， since me⊆∫by Proposition 52．1， we have
d
Q∩m e＋m一Σ蜘 乏＋ α
￠：1
⊆Σ蜘m（me） 1 iニユ
⊆m」mQln−1
as is claimed， 口 Let us now prove Theorem 5．1．1。
Proof of Theorem 5，1．1．（2）⇒（1）This is well−known． See［NR］．
（3）⇒（2）By Proposition 5．2．1 we get mqI＝mqQ＋mq＋e， whence mq＋e⊆Q， so that mq＋e＝Q∩mq＋e⊆mqQ by Lemma 5．2．2， because e≧αi for all 1≦i≦d． Thus
mql＝mqQ．
（1）⇒（3）Let 1≦i≦d be an integer． Then媛∈me⊆1⊆Q． We look at the
integral eqUatiOn（ω乏 t）n十c1（姥）n−1十… 十Cn＝O
with・n》Oand・、∈Q・． Th・n・〆∈Σ乳、 Qゴ・1 ゴ）乏and・・， thank・t・th・m・n・mi・l property of the sequenceω1，ω2，… ，ωd （recall that the sequence x1，x2ヂ・・，xd is AL regular）， we have
犀∈Qゴ・1π一ゴ）e
for some 1≦ゴ≦n． Let五＝｛（α1，α2，…，αd）∈Zd 1αた≧Ofbr al11≦k≦d｝and
Aゴ＝｛α∈LlΣ1．1α・一ゴ｝． Th・n・inced
Qゴー（fi ・・akαk 1 ce∈Aゴ），
k＝1
thanks to the monomial property of xl，x2，…，ωd again， we get
d
敢［rレ・α・｝・1 ゴ）eA k＝1
fbr someα∈Aゴ． Hence
d
（ne）・1＝＝Σ（・k…）…＋［（π一ゴ）司・i＋β k；1
withβ∈ゐ， where｛ei｝1＜i＜d denotes the standard basis of Zd． Consequentlyαk＝β，＝
Oif k≠iand soαi＝ゴ． Hence O＝α￠ゴーゴ4＋β￠， so that we have e≧αi as is claimed．
Let us now consider assertions（i）and（ii）． Let n≧1be an integer． Then In＝
Qln−1十mne since∬＝Q十me（Proposition 5．2．1）， so that （〜∩ln＝Qln『i＋［Q∩mne］⊆Qln−1
because Q∩Inne⊆Qln−i by Lemma 5．2．2． Therefore Q∩In＝QIη一1 fbr a11 n≧1，
whence G（1）is a Cohen−Macaulay ring（［VV， Corollary 2．7］）．
We will show that rQ（1）＝「ll． Notice that
rQ（∫）＝min｛n≧OlIn＋1⊆Q｝，
because ln＋1＝Qln if and only if ln＋1⊆（〜． Suppose that Jn＋1⊆Q．
而＠＋1）e＝（0）（recall that 7＝飯乏）， whence（n十1）4≧ρ十1． Therefbre η＋1≧ρ壱1−q吉乏一1＋1，
because 6＝ρ十1−q， so that we have n≧多．
If・n≧多， th・n（n＋1）e≧（多＋1）e−q＋e一ρ＋1and・・ア＋1−
whence In＋1⊆Q． Thus r（〜（1）＝「多1．
To see that F（1）is a Cohen−Macaulay ring， it su伍ces to show that Q∩mlη＝mQ∫η一1
We then have
而（n＋1）e ・．（o），
for all n≧1． By Lemma 5．2．2 we have
Q∩mln−Q∩［rnQI −1＋m e＋1］
−mα −1＋［Q∩m e＋1］
⊆ m（〜ln−1，
whence（［ll∩mln＝mQln−i．
口Assume that乏≧αi for all 1≦i≦dand Iet｝奄，s be the initial fbrms of尋，s with respect to 1． Then yi，Y2，…，Yd is a homogeneous system of parameters of G（1）， since Qis a reduction of I（Theorem 5．1．1）． It theref（）re constitutes a regular sequence in Gの，because G（1）is a Cohen−Macaulay ring by Theorem 5．1，1（ii）， so that we have a
canonical isomorphism
G（1）i≧G（1）／（Yl，Y2，… ，Yd）
of graded A−algebras（［VV］）． Hence a（G（1））＝a（G（1））十d． Let r be the index of nilpo．
tency of∫， that is r＝a（G（∫））． Then since r＝rQ（1）（recall that xlal，x2a2，…，xdad is asuper regular sequence with respect to∫）and a（Gの）＝a（G（1））−d（［GW，（3．1．6）］），
by Theorem 5．1．1（i）we have the following．
Lemma 5．2．3．3吻po8ε伽紹≧αi！br all 1≦i≦d． Thθn a（G（1））＝「多1−d．
Corollary 5・2・4・・48sume thαt e≧ αi／brαll 1 ≦ i≦ （i． Then 7？．（1）i8αCohen−
1レ1αcαulayγ゜ing ifαnd only if「多1 ＜（オ・ When this is the cα8θ， d≧2・
1）roof． Since G（1）is a Cohen−Macaulay ring by Theorem 5．1．1（ii）， R（1）is a Cohen−
Macaulay ring if and only ifα（G（1））＜0（［TID． By Lemma 5．2．3 the latter condition is equivalent to saying that「多1＜d（c￡［GSh1， Remark（3．10）D． When this is case，
d≧2because O＜「多1． 口
We are now in a position to prove Theorem 5．1．2．
Proo∫o／Theorem 5．1．2．（i）Notice that G（1）is a Gorenstein ring if and only if so is the graded ring
G（∫）＝G（∫）／（Yl，】隆，… ，｝急），
where Yi s stand f（）r the initial fbrms of礁 s with respect to I． Let r be the index of nilpotency of I． Then r＝rQ（一τ）＝「多1，and G（1）is a Gorenstein ring if and only if the equality
（o）・デーア＋1−
holds true fbr a11 i∈Z（［0， Theorern 1．6］）． Hence if G（1）is a Gorenstein ring， we have
（0）：7＝ア＝而「e．On the other hand， sinceア＝iiie and q＝ρ十1一乏， by Proposition 5．2．1we get
（o）・7＝（o）・菰L補．
Therefbre q＝r4， since㎡乏＝而lq≠（0）． Thus e l g and r＝多．
Conversely， suppose that引q． Hence r＝多by Theorem 5．1。1（i）． Let i∈Z． Then
，inceアー賊w， g・tア＋1 −hi（・＋1− ）e， whilb
（0）・ア＝（0）・斎L而 ＋1− e
by P，。P。，iti。n 5．2．1． W， th・n hav・（0）・デーア＋1− f・・all・i∈Z，・ince
（r十1一の乏＝・9十乏一i乏＝ρ十1−i乏・
Thus G（ア）is a Gorenstein ring， whence so is G（1）．
（ii）The Rees algebraπ（1）of 1 is a Gorenstein ring if and only if G（1）is a Gorenstein ring andα（G（1））＝−2， provided d≧2（［1， Corollary（3・7）］）・SupPose that 7己（1）is a Gorenstein ring． Then d≧2by Corollary 5．2．4． Sinceα（G（∫））＝rQ（1）一（1＝−2， by assertion（i）and Theorem 5．1．1（i）we have多＝rQ（J）＝・d−2， whence q＝（d−2）乏．
Conversely， suppose that g＝（d−2）6． Then d≧3since q≧1． By assertion（i）
and Theorem 5．1．1（i）G（1）is a Gorenstein ring with rQ（1）二多＝d−2， whence α（G（1））＝（d−2）−d＝−2，so thaポR（1）is a Gorenstein ring． □ Example 5．2．5． Suppose thatρ≧5is an odd integer， sayρ＝27十1with T≧2． Let
q＝ρ一1．Then乏＝ρ十1−q＝2． Hence， choosingαi≦2fbr all 1≦i≦d， we have
I＝Q＋rn2⊆σwith rQ（1）＝T by Theorem 5．1．1． Since引q， by Theorem 5．12（i）the ring G（1）is Gorenstein， despite Q望mq（compare the condition with those in Wang，s Theorem E）． The ring R（1）is by Theorem 5．1．2（ii）aGorenstein ring， if d＝T十2．5．3 Examples and apPlications
In this section we shall discuss some applications of Theorems 5．1．1 and 5．1．2． Let us b・gi・with・th…as・wh・・e A i・anum・・i・al・emig・・up・i・g・f H−〈α，b〉， th・t i・a continuous research of quasi−socle ideals in one−dimen8ionα1 Cohen−Macaulay rings．
5．3．1 The case where．A＝ん［［tα，tb］］
Let 1＜α＜わbe integers with GCD（α，b）＝1． We look at the ring A＝k［［オα，孟b］］⊆k［［オ］］，
wh・・eん［［オ］］d・n・t・・th・f・・m・l p・wers se・ies ri・g・v・・a且・ld鳶・W・p・厩＝オαand y＝オb．Then、4 is a one−dimensional Gorenstein local ring alld凱＝（x，y）． Because
腔ん［［X，y］］／（XL yα）wh・・eκ［［X， y］］d・n・t・・th・f・・m・l p・wers se・ies ri・g・v・・th・
fieldん， we get
G（m）望k［X，y】／（yα）・
Let n， q≧1be integers， and put Q＝（xn）and I＝Q：m9． Then because
α（G（m））＝α一2，we haveρ＝α＋n−2and e＝（α＋n）一（g＋1）． Consequently J⊆（？i if and only if q＜α（Theorem 5．1．1）， whence the condition that I⊆Qis independent of the choice of the integer n≧1． When this is the case， by Theorems 5．1．1 and 5．1．2 we have the following．Proposition 5．3．1． Tんεノblloωingαssertionsんold true．
ω・・（∫）＝「（。＋。）呈（，＋1）1・
（2？The gra（ied rings G（1）and F（1）αre Oohθn一ルfacαulay rings．
（3ソTんering G（∫）isα（］orenstein ring if and only if（α十n）一（q十1）divi（ies q．
Hence， if q＝α一1， we then have， for each integer n≧1such thatηlq， that G（1）is a Gorenstein ring．
5．3．2 The case where．4ニRM
Our setting naturally contains the case where A＝RM is the localization of the ho−
mogeneous Gorenstein ring R＝k［R1］over an in￡nite field k＝Ro at the irrelevant maximal ideal M＝R＋． Let us note one example，
Example 5．3．2． Let 5＝k［X，｝ ， Z］ be the polynomial ring over an infinite field k and let R＝S／fS， where O≠！∈S is a form with degree n≧2． Then R is a homogeneous Gorenstein ring with dim R＝2． Let x1，x2 be a linear system of parameters in．R and letハ4＝R＋， We look at the local ring A＝RM． Letα1＝2，α2 ・・ n， and q＝n． Let
Q＝（xl，錫）A and I＝Q：mq， where m＝MA． Then
ρ＝a（R）十（α1十α2）＝2n−1．
Hence乏＝q＝n， so that I⊆Q，1＝Q＋mn＝（xそ）＋mn， and G（1）is a Gorenstein
ring with rQ（1）ニ1（Theorems 5．1．1 and 5．1．2）． We have Q⊆ mq， if n≧3．