！or∫ξHomB（M．，N），m∈Mα棚8，尤∈3．肋θησ乞8αδ一づ80mo叩肋5m．
Pヅ。oヅStraightforward．口
Proposition2．4．12．工θ古B6θαηoe挽θれαηr伽gαηd N．α力η伽佃gεηem亡θ∂
ρη・畑伽θ（B，δ）一m・〃θ・れ㎝ん㎝・．砒η伽〃・ω伽gα・・舳㎝・ん・1 （1） Homδ（ノV，工）皇（ム⑭B」V−1）oαηd Ext芸（ノV，刀）皇乙⑳B／V−1／δ（ム図B
W−1）ω肋W 1＝HomB（〃，B），ω肋。〜3αδ一mo仇τθ．
（2） 伽挽θθ αc亡舵ψeηcθ（2．1），β、：Ext芸（ノV，工2）→Ext芸（ノV，L3）．乞55ぴ一 プec左伽e．
（3） 伽仇θθ αc右3eψeηcθ（2．2），α＊：Ext芸（N2，L）→Ext芸（ノV1，工）台8則ツー
畑伽ぴW・，N。㎝dN・α・・伽的9㎝θm之・∂ρ吻θ舳・（B，δ）一m・〃…∫
ヅαγ乙ん0γ乙e．
Pヅ。oヅ（1） First we sha11show that HomB（N，Z）窒L⑱BW 1as B−modu1es．
Since W is a projective B−modu1e of rank one，there exists an a伍ne open covering｛D（5｛）｝｛of Spec B such that N［54■1］＝到84■1］e｛with e4∈W，
8乞∈B．Then W■1［84−1］＝B［5｛一1］吋with e麦（e乞）：1．For anyづ，ゴ，there existsゐ乞∈B［3ゲ1，3ゴー1］such thatεゴ＝ヵ4εづho1ds in／V［34−1，3ゴー1］、Then e才＝ル書ho1ds in N−1［5｛一1，8ゲ1］．Take anye1ementψ∈HomB（N，ム）．Then ψ（e4）＝Z4∈工［8｛一11．For any乞，ゴ，we have Zゴ＝ψ（eゴ）＝ψ（乃4e｛）＝カ山in ム［・1■1，・ゴー11・・dth…f…l1⑭ε1−l1叫1書＝1ゴ⑳・ラi・加・W−1［ギ1，・ゴー11，
Hence the Z4⑳e㌻correspond to an e1ement of工⑳B W−1．Converse1y，any e1ement ofム⑭B N■1correspond−s to the Z｛⑳e麦and−these e1ements deine ψ∈HomB（N，工）byψ（e乞）＝Z4∈工［84−11．
2．4．THE HOMOLOG∫CAL PROPERTγOFδ一M ODσLES 37
Next we show thatψδ＝δψif and−on1y ifδ（Z｛⑳e才）二0with the above notation．Writeδ（θ乞）＝蝋wi曲。4∈B［84■1］．Then we have
δ（・麦）（・4）＝δ（・才（θ1））一・芸（δ（・1））＝一・芸（蜥）＝一・1
in到8｛■1］andthereforeδ（e才）＝＿c4e麦inN−1［5ゼー1］．We a1sohav岬（δ（e｛））＝
ψ（c｛e｛）＝c山and一δ（ψ（ε｛））＝δ（Z毛）、。Suppose thatψδ＝δψ．Thenδ（Z壱）＝c4Z｛
ho1d−s in工［5乞一1］．Hence we have
δ（1｛肺才）＝δ（1｛）⑭・才十1｛⑧δ（引＝・｛1｛⑭・才十1｛⑭（一・｛・麦）＝0
inム⑱ノV−1［5ゲ1］as desired．Converse1y，the conditionδ（Z乞⑳θ才）＝0for a11台 imp1ies thatδ（Z｛）＝c｛Z｛and−hence gδ（e｛）＝δψ（e｛）for a11づ，which imp1ies that ψδ＝δψ．Hence we conc1ud−e that Homδ（W，ム）皇（ム⑳BN−1）o asλ一modu1es．
It is easy to see that Ext芸（ノV，刀）空Ex砧（B，工㊧B／V−1）via the correspon−
dence of（ルτ，∫，g）∈Extδ（ノV，ム）to（ハ∫⑳B W−1，∫⑭1，g⑳1）∈Extδ（B，工⑭B
／V・1）．Let M ∈Ext芸（B，ム⑳B／V■1）．Then M ＝（ム⑳B／V−1）㊥Bθ and δ（θ ）∈工⑳M−1．Supposethat M ＝（ム⑳BW−1）㊥Bθ is isomorphicto M一 by a（B，δ）一isomorphismθ：M一 →M一 suchthatθ（e ）＝θ 十Z and一θIム酬一1＝id、，
where〆∈L⑳B W 1．We haveδθ（e ）＝δ（e 十1 ）＝δ（e ）十δ（1 ）and θδ（e ）＝δ（eノ）．Sinceδθ＝θδ，we haveδ（e ）＝δ（e ）十δ（Z ）．Sinceδ（e ）deteト mines theδ一modu1e structure on ，we conc1ude that Ext芸（B，乙⑳B W■1）室 工⑳B W−1／δ（ム⑳B N−1）aSλ一mOdu1eS．
（2） The exact sequence（2．1）coincides with
0一→（Z。晩W−1）。一÷（Z。⑳。W−1）。一→（ム。⑱。M−1）。
一→工。⑭。W−1／δ（乙。蛎N一ユ）→工。⑱。N■1／δ（工。晩W−1）
→工。⑳。N−1／δ（ム。晩W−1）→0
which fo11ows by the snake1emma app1ied to the fo11owing commutative di−
ag「am・
0一一÷工。晩W 1一→工2⑭BW■1一一÷工3⑳BW−1一一→0
1↓
1⊥ l！
0一→ム。⑭BN■L一→ム。⑳BM−1一一÷ム3⑳BM■1一→0
（3）We have an exact sequence ofδ一modu1es
0→N。一1一→N。一1→W。一1→0
38
CHAPTER2．M ODσLE DER工1岨nONS
and hence an exact sequence ofδ一modu1es
0→＾。N。一1→＾。N。一1一→＾。N。■1一→0．
App1ying the snake1emma to this exact sequence ofδ一modu1es，we obtain
the resu1t． □
2．5 The re1ation betweenδ∈LND（B）andδ一 mOdu1eS
In this section，we consid−er the re1ation betweenδ∈LND（B）andδ一modu1es．
First，the existence of a s1ice can be characterized as fo11ows．
Lemma2．5．1．Foヅδ∈LND（B），伽∫o〃。ω吻。㎝d肋㎝8αヅeψωα1θ仇
（1）δんα・α・1づ・・。
（2）B＝δ（B）
（3）伽岬δ一m。〃θM，ωθんα。θM＝δ（M）．
（4） ハ。ヅαηツδ一mo伽Ze Mフωeんα〃e Ext芸（B，ル7）＝（0）．
ProoヅThe conditions（3）and（4）are equiva．1ent in view of（1）of Proposition
2，4．12．
（2）⇒（1）is c1ear．（1）⇒（2）．Ifδhas a s1ice u∈B，then B＝刈u］．Hence any6∈B is written as6＝∫（u）∈λ［u1．Then6＝δ（わ ）with6 ＝∫∫（u）仇．
（3）⇒（2）is c1ear．（2）⇒（3）．For anyδ一modu1e M，we have M＝Mo⑳λB
（Lemma2．2．3，（2））．Hence any e1ement m∈M「is written as m二Σ｛肌 with m乞∈Mo and6｛∈B．Let肌 ＝Σ二つう ｛m｛withδ（6 づ）＝6｛．Then we have η7＝δ（η7 ）and henceル∫＝δ（M）。 □
Even if there exists aδ一modu1e M satisfyingδ（M）＝M，it d−oes not guaranteeδ（B）二B，which is shown by the fo11owing examp1e．
ExAMPLE2．5．2．Let B＝たレ，μl be a po1ynomia1ring over a丘e1dん。f characteristic zero．Letδ∈LND（B）be de丘ned byδ（ ）＝0andδ（μ）＝1＋ 、 Let M＝B／ B．Then M is aδ一modu1e．SinceδM∈LND（B／ B）has a s1ice ず，we haveδM（M）＝M．On the other ha．nd一，we haveδ（B）＝（1＋ ）B≠B．
However，we have the fo11owing resu1t．
2．5．THE RELAT工ON BETWEENδ∈lLND（B）ANDδ一M．ODσlLES 39
Lemma2．5．3．〃δ∈LND（B），㎝dM一αδ一mo〃θ。〃んθrθθ挑お㎝θ αc右 8eψ㎝Cθ0∫δ一m0〃θ8
0一→W一→M一一→B一→0，
伽ηδ（M）：：M伽伽・流α右δ（B） ＝B．∬θη・θ，ゲM1・α伽θB−m・〃θ㎝d δM乞・切㎝gu1αう1・，挽㎝δ（M）＝Mlmψ・・物オδ（B）＝B．
Prooゾ．By the snake1emmaapp1ied to the aboveexact sequence ofδ一modu1es，
we have an exact sequence of A−modu1es 0→N。→M。→λ→N／δ（W）
→M／δ（M）→B／δ（B）→0．
Henceδ（M ）＝M imp1ies thatδ（B）＝B． 口
The丘xed−point freeness ofδ∈LND（B）can be characterized−as fo11ows．
：Lemma2．5．4．Foグδ∈LND（B）フ海∫o〃。ω伽g c㎝肋㎝3αrθ榊伽α1θ舳 ωδ1・伽d一ρ・舳伽・．
例月・グαη〃・θ一δ一m・伽M，ωθんα・θ〃＝（0）乞アδ（M）＝（0）．
63ソバ・・αWδ一m・〃・M，胱んα・θM＝（0）榊（M）＝（0）．
ωル〃η・一δ珊・〃θM4・g㎝θm右θ Vδ（M）α・αB−m・d仙1θ．
例λWδ一m・〃θMl・g㎝舳舌・れひδ（M）α・αB−m・∂仙1・。
Pグ。oヅ（1）⇒（2）、For6∈B and m∈M，we haveδ（6）m＝δ（6m）＿6δ（m）＝0 by the assumption thatδ（M ）＝（0）．On the other hand，the id−ea1of B generated byδ（B）is a unit idea1beca．use ofthe丘xed一一point freeness．Name1y，
there exist e1ements64and6 40f B for1≦乞≦グsuch thatΣ；＝16 乞δ（6包）＝1．
Then m：（Σ；＝16 ｛δ（6づ））m＝0．Hence M一：（0）．
（2）＝＞（3）and（4）＝÷（5）are c1ear、
（3）⇒（4）．Let M「＝M「／Bδ（M「）、Thenδハ4induces a modu1e derivation δπon M，which is zero．It fo11ows from（3）that M＝（0）．This imp1ies that M is generated as a B−mod−u1e byδ（M）．
（5）⇒（1）．Obvious1y B is aδ一modu1e and therefore（5）imp1ies that the
idea1ofBgeneratedbyδ（B）isauni七id−ea1．口
40
CHAPTER2．一MODσLE DER工帆nONS
We have the fo11owing assertion，
皿emma2．5．5．〃Mろeαη㎝鮒。伽棚 舳m亡ed亡。r5づ㎝一価e（B，δ）一
m・〃θ．8W…伽舌M：BM。㎝dδ（M）＝M．肋η伽伽乞・α君乞・ηδ㎝
Bづ8伽εd一ρ0乞η尤伽e。
。Prooヅ Since M ＝ 一BMo，there exist η71ゾ．．，η7、 ∈ ハ∫o such that ルτ ＝ Σ二；＝1Bm｛．Further，sinceδ（ル∫）＝ハ∫，there exist6幻 ∈B such that m｛＝
δ（Σ二＝16幻m｛）＝Σ二＝1δ（6包ゴ）m4for1≦づ≦r These re1ations are written as
・（ll）一（1），
where
δ（6。。）一1 δ（6。。） δ（6。、）
δ（ろ。。） δ（6。。）一1 δ（6・、）
P＝
δ（う、。） δ（6、。） ＿δ（わ、、）一1
Mu1tipIying the adjoint matrix P＊of P，we have（detP）．m｛＝0for a11づ．
Since M is nonzero and B−torsion−free，we have d−et P＝0，which imp1ies that1∈Bδ（B）．Hence the derivationδon B is丘xed−point free． □
2．6 Projective triangu1abi1ity
In this section，we sha11extend the notion of trianguユabi1ity ofδM0n a free B−mod−u1e M「in七he case where M is a丘nite1y generated一，torsion−free（B，δ）一
㎞odu1e with the restriction that B is an a冊ne norma1C−domain ofdimension tW0、
DEFINITI0N2．6．1．Let B be a d−omain and1et M be a丘nite1y generated一，
torsion−free（B，δ）一mod−u1e，Let r be the rank of M，which is by d−e丘nition equa1to dimQ（B）M⑭B Q（B）・We say thatδM isρ吻θc切吻亡れ㎝g砒励1e if there exists a sequence of projective（B，δ）一submodu1es0⊆M1⊆M；2⊆
⊆ ハ4、 ＝ ルτsuch that for each 1 ≦ づ ≦ r，ルZ｛十1／ル4 is a projective
（B，δ）一mod−u1e of rank one．
2．6．PROJEσT工VE TR工ANGσLAB几工TY
41Note that the triangu1abi1ity imp1ies the projective triangu1abi1ity for a modu1e d−erivation on a free modu1e．Furthermore，note that if払／払＿1is a free B−modu1e of rank one for anyづ，then the projective trianguIabi1ity imp1iさs the triangu1abi1ity on a free modu1e−
We consider the case where B is a norma1a伍ne C−domain of dimen−
sion two，δ∈LND（B）is丘xed−point free，and−M is a丘nite1y generated，
torsion−free（B，δ）一modu1e．Then we show thatδM is projective1y triangu1a−
b1e（Theorem2．6．7）．
DEF1N1T10N2．6．2．Let M be a丘nite1y generated B−modu1e．We ca11〃＊＝
HomB（M ，B）the dua1」of M and M．＊＊＝Hom」B（M＊，B）the bidua1of M「．
Deine a B−modu1e homomorphismα：M→M＊＊byα（m）＝而，where
而is de丘ned by而（∫）＝∫（m）forプ∈M＊．We say that M is re且exive if α：M→M＊＊is an isomorphism．IfB is anoetherian ring，then M＊and−M＊＊
are a1so丘nite1y genera七ed．Hence if B is a noetherian ring，δ∈LND（B），and M is a丘nite1y generated一（B，δ）一modu1e，then M＊and−M＊＊are a1so丘nite1y generated一（B，δ）一modu1es．
：Lemma2．6．3．工θ舌Bうθαη舳θgml♂omα伽㎝〃θ舌M うεα伽伽妙ge胱m左e∂
r批痂eB−m0〃θ．r加ηM乞8む0ザ5乞㎝一価θ．
PηooヅWe have an exact sequence F→M→0for some free modu1e F．
The sequence0→M＊→F＊is exact．Since F＊is isomorphic to．F，it fo11ows that M＊is torsion−free．In a simi1ar fashion，M＊＊is torsion−free．Hence M
istorsion−freeifMisre且exive1口
In the subsequent arguments，we need the fo11owing resu1t，which is weI1 known but we give a short proof for the1ack of suitab1e references．
lLemma2．6．4．〃B6eαreg此rαが胱C−domα伽。ゾ励m舳乞㎝ωo。肋㎝
㎝V伽伽ψ9eηεm亡eれψε ωeB−m0〃θM48αρ吻eC痂eB−m0〃θ．
ProoヅWe have on1y to show that M戸is a free Bグmodu1e for everyρ∈
Spec B．If htρ：1，then M拒is a free B月一modu1e since Bp is a discrete va1uation ring and−M拒is Bグtorsion−free by Lemma2．6．3．In the case htp＝2，
we have an equa1ity
proj．dim Mp＋depth M拒：depth B拒
by a theorem ofAus1and−er−Buchsbaum（see［2，p．17，Theorem1．3．31），where we note that proj，d−im M草＜○o since B is regu1ar（see［2，p．66，Theorem
42
CHAPTER2．MODσLE DER∫帆nONS
2．2．7］）．Meanwhi1e，since耳is regu1ar，we have d−epth B拒＝dim8炉＝2．On the other hand，we know that d−epth仏＝depthM言＊≧min（2，d−epthB沖）＝2
（see［2，p．26，Exercise1，4．19］）．Hence we have proj．d−im M担：0，and−
therefore l拒is a free Bグmodu1e一 □
Eemma2．6．5．〃Bうθα3加工θmmα身．σ4フδ∈LND（B）㎝d Mα伽伽佃
g㎝θm古・dフ君・グ・1㎝一加θ（B，δ）一m・〃θ．3W・・θ伽亡δ帥〃一ρ・伽亡伽・。珊・η
M48αr枇 1ηeB−m0〃eαη＾㎝㏄α川畑伽θB−m0〃θ．
Pグ。oヅLetα：M「→M ＊＊be a B−mod−u1e homomorphism as in De丘ni七ion 2．6．2．We c1aim thatαis aδ一isomorphism．First，we sha11show thatαis aδ一homomorphism，i．e、，αδ（m）＝δα（m）for any m∈M「．Ind−eed，for any m∈M and∫∈M ＊，we have
α（δ（m））（∫）＝δ（m）（ア）＝∫（δ（m）），
δ（α（m））（∫）＝δ（α（m）（ゾ））一α（η7）（δ（∫））＝δ（プ（m））一（δ∫）（m）
＝δ（∫（m））一δ（∫（m））十∫（δ（m））：∫（δ（m））．
Next we sha11show thatαis an isomorphism．We consider an exact sequence ofδ一modu1es
0一→K一一＞Mr㌧M＊＊一一〉0一→0，
where K＝Kerαand0＝Cokerα．Since bothκand O are丘nite1y genera七ed−
B−modu1es，we have Supp K＝γ（A㎜K）and Supp0：γ（Am0）．We
have on1y to show that K：0＝（O）．For this，it su伍。es to show that K早＝0沖二（0）for any prime idea1p of B．Suppose丘rst that htρ＝1．
Then叫is a free B拒一modu1e since B拒is a d−iscrete va1uation ring and M悼
is a丘nite1y generated．，torsion−free B拒一modu1e．HenceαパM沖→M芹＊is an isomorphism，and therefore K疸＝0戸：（0）．Hence if Sppp K≠⑦，
then any p∈Supp K has height2and−therefore we have a minima1prime
d−ecomposition扁：肌1∩＿∩m，with maxima1idea1s帆｛、Since
Am K isムδ一idea1（Lemma2．2．10，（1）），the m乞are a1soδ一idea1s（Lemma 2．1．8，（1）），which contradicts the丘xed−point freeness ofδ（Lemma2，1．11）．
Hence SupP K＝・②and K＝（0）．In a simi1ar fashion，we have0＝（0）． □ In Theorem2．6．7，we need the fo11owing resu1t，which is obvious because the singu！ar1ocus is a丘nite G、一stab1e set for a two−d−imensiona1norma1va−
riety．
2．7． 工NVERT工BLEδ一M ODσlLES 43 Lemma2．6．6．〃B6εαηormα1α痂ηe C一∂omα伽。ゾd乞m㎝8乞㎝地。㎝d
Zθτδ∈LND（B）．3u〃05e枕α亡δ45ガ e∂一ρo初む介eε一丁加ηBづ5αザegωαrヅ初g．
Theorem2．6．7．〃B，δうθα8加工θmmα2．6．6，㎝♂1θ右M6eα伽伽佃
9舳m￠θdフτ・ザ・1㎝一価・（B，δ）一m・〃・．砒ηδM帥・・ゴ・舳物納㎝〃αう1θ．
ProoヅLet｛m1⑭1，．．．，m、⑭1｝be a basis of the Q（・B）一vector spaceルτ⑳B Q（B）．Let3＝λ一（0）．Then we have
〃⑳B Q（B）：〃図B8−1B⑳8−1B Q（B）
一8−1M・亭。（λ）3■1肋・一・・Q（B）一3 1M・⑳。（λ）Q（B）
＝必⑳〃（λ）⑳Q（λ）Q（B）＝M・⑳λQ（B）・
Hence we can take a11the m｛in Mo．Let W＝Bm1＋＿十Bm、＿1．Then
c1ear1y N is aδ一modu1e and−hence L：＝（M／N）／（M「／N）t。、is aδ一mod−u1e by（2）of Lemma2．2．10，where（M「／W）七。、is the torsion part of M／N，In add−ition，ムis a丘nite1y generated，torsion−free B−mod−u1e．Henceムis a projective B−modu1e by Lemma2，6．5．Then we have an exact sequence of
（B，δ）一m・du1・・
0一→〃→M→五一合0，
where Mソis the keme1of the natura1homomorhpism from M to刀．Since rank M＝r and rankム＝1，we have ra．nk M∫＝r＿1．If we use induction on rank M，we conc1ude thatδM is projec七ive1y triangu1abIe． 口
2．7 Invertib1eδ一modu1es
In this section，we consid−er an伽りer肋Ze（B，δ）一mod−u1e M，i．e．，aδ一mod−u1e wh三。h is a projective B−mod−u1e of rank one．If M＝Be is a free B−mod−u1e
of rank one，thenδ〃（ε）二0and−henceルτ：BルZo（Proposition2．2．14）。The genera1case is not so simp1e．We give an examp1e of an invertib1e（B，δ）一
modu1e M such that M≠BMo．
ExAMPLE2．7．1．Let B＝C［ ，μ，z1／（z2−2〃一2）、De丘neδ∈LND（B）
byδ（ ）＝0，δ（μ）＝z andδ（z）＝ ．Then it is easi1y veri丘ed thatλ＝
Kerδ＝C同，and the quotient morphism元：Spec B→Specλhas a unique singu1ar丘ber over the point ＝0，which is a disjoin七sum of two a伍ne1ines
Spec（B／（ ，z＋〉り）B），Spec（3／（ ，z一〉り）B）．LetM＝ B＋（z一〉O）B，
which is aδ一id−ea1of B．It is easy to show that M is a non−free，projective
44 CHAPTER2．MIODσLE DER工帆nONS
B−modu1e ofrankone．We can a1so show easi1ythat Mo＝M∩λ＝汕and
M
bRMo．
Now consider an invo1utionσon B（1e丘ned−by
σ（・，μ，・）＝（一 ，一μ，一・）．
Let B＝Bσ＝C［ 2，ひ2，榊，肌，脾］，where we denote the residue c1asses of π，μ，z by ，μ，z respective1y for simp1icity．Wri七e X＝ 2，γ：μ2，Z＝ μ，
σ： z andγ＝脾．Thenδd−escends to a1oca11y ni1potent derivationδon B…hth・tδ（X）＝0，δ（γ）：2γ，δ（Z）＝ひ，δ（σ）＝X・・dδ（γ）＝2＋3Z．
Furthermore，it is easy to see thatλ＝Kerδ＝qX］and the quotient morphismπ：Spec B→Spec A has a unique；ingu1ar mu1tip1e丘ber dver the point X＝0，which is a non−reduced 奄窒窒?р浮モ奄bPe scheme．Let邪be the id−ea1 of3generated by X，Z，σ．Set M＝準，which is an invertibIeδ一modu1e with
Mo二M ∩λ＝XA Hence〃≠BMo．
The above two examp1es imp1y that an invertib1e（B，δ）一modu1e M is re1ated to irreducib1e components of the singu1ar丘bers of the quotient mor−
phism7r：Spec B→Specλ．The fo11owing resu1t shows，in the case of
d−imension two with some ad−d−itiona1cond−itions，that this is true．
Proposition2．7．2．8ψρ05θ挽α亡B乞8αZocα吻∫αc左0れαZ，η0ヅmαZαがηe C−
domα初一工θ左X：Spec B．8他〃05e加グ抗θr抗α左B＊＝C＊αη∂Pic（X）づ5α
励8cηe老e gro刎p，んeηce eりerμθZθme械。∫pic（X）乞5G、一切りαれα枇．Tんe肌θηθザμ 伽〃e竹泌Ze5んeα！∠二∈Pic（X）45ザeρザθ5θ枇edうμα0αグ切θr砺切50r D＝・Σ二づηづγ；
舳Cん妨α左Xづ3α0α一3肋1θかη幽C伽θ3伽α伽左40∫C0流mθη84㎝0ηθ∫0ヅθ㏄ん
4．
Pヅ。oヅIt su伍。es to show the case where D≧0．For eachλ∈C，since λD〜D，there exists〈such thatλD＝D＋（〈）．Then for eachλ，μ∈C，
we have
λ十μD＝μ（λD）＝μ（D＋（∫λ））＝μD＋（μム）
＝D＋（∫μ）十（μム）＝D＋（∫パμ∫λ）一
Hence（∫λ十μ）＝（∫μ．μ∫入）on X and there exists c（λ，μ）∈F（X，0x＊）＝C＊
such that∫人十μ＝c（λ，μ）∫μ．μ∫λ．Note that the morphism c：A2→C＊：A圭 is constant．Hence there exists c∈C＊such that∫人十μ：c∫μ．μ∫λ一Then c∫λ十μ＝（c∫μ）．μ（c∫λ）and therefore we may assume that
∫λ十μ：∫パμム・ （2．3）
2．7．工NVERnBLEδ一MIODσlLES
45 If we write D＝Σ〃山，then each篶。orresponds to an idea1∫篶。f3．SinceλD：D＋（〈），we have欧入（∫篶）例＝∫λ⑭乞∫篶η｛．This i㎞p1ies that
∫入＝F（λ）／6for some F（λ），6∈B．Write F（左）＝ろ。＋61士十…十6mtm∈B［t1 with6｛∈B．Then the equa1ity（2．3）imp1ies the re1ation
F（亡十u） F（伽） ψ仙（F（左））
b 6 ψ。（6）
i耳ρ（B［亡，u］），whereψ刎：B［t］→B［亡，u］is de丘ned by
灼（シ）一シ（㍍）が・
This gives rise to
ψ伽（6）（6。十6。（t＋u）十…十ろ、、（亡十u）m）
＝（う。十わ。u＋…十わmum）（9 （6。）十9 （6。）十…十ψω（6m）亡m）．
The comparison of the constant terms imp1ies6o：6．Now compare the top
乏一termS：
ψ、（う）6m：（ろ十61u＋… 十らmum）ψu（6m）。
Hence we have
牝（6） 6 61 らm m ＝一十一u＋… 十一u ． 牝（6m） 6m うm うm On the other hand，we have
甜）一イ∴）一言去・（÷）・
Hence we deduce
Hlて÷）
for0≦乞≦m．Then F（亡）＝6＋61f＋… 十6m亡m＝6m帖（6／6m）and therefore
F（亡）イ缶）
6 ⊥
ろmThusλD＿D＝（F（λ）／6）＝（軌（6／6m））＿（6／6m）．This imp1ies thatλ（D＿
（6／6m））＝D一（6／6m）． □
Chapter3
The14th prob1em of Hi1bert
3．1 Su舐icient conditions f；or inite generation
In this section we give some su舐。ient conditions for the丘nite generation of MO．
In the beginning of§2．3，we investigate how the torsion of a丘nite1y gener−
ated一（B，δ）一modu1e M a丘ec七s the丘nite generation of Mo in the case where B is a po1ynomia1ring刈μ］over a noetherian domain A Arguing as in Lemma 2．3．1，we can prove the fo11owing．
lLemma3．1．1．〃B＝0kμ］，6eαρo伽。m乞α1ヅ初g oりeヅαた一αlg伽α0 αηd dφηεδ∈LNDo（B）吻δ（ ）＝0フδ（μ）二∫αηdδ（z）＝g，ω加ヅθプ乞5 αη㎝・θη・θ1θm㎝亡・！0［ 1αηdg乞・川㎝鮒・θ1θm・械・μ［μ1．Zθ右〃6θα
伽的9㎝εm古・∂（B，δ）一m・〃θ．舳〃…亡んα影θθ1舳θηけんα ・古・ヅ・乞㎝
伽M ．肋㎝Mo乞5α伽伽佃9㎝舳亡e∂んmo〃θ．
Prooゾ．Let cヅbe the highest degree term ofg．We have
l・／l（・）一
¥・茅。・・1・・（、土1）、（サ1・（一）1一・
Note thatδ（ψ1／∫（z））＝0withδextended natura11y to B［∫一1コ．The coe田一
・i・・t・｝十1i・ルーリ／∫（・）i・・q・・1t・
／誉（一1）小一1）、一（卜z＋2）ト、11・・
46
3．ユ．SσFF工C工ENTσOND∫丁工ONS FOR F工NITE GENERAT工ON 47
Thus we haveλ⊃ん［ ，ヅ十1＋．、、］and−hence B＝刈z］十λ［z］μ十＿十λ［z］ヅ．
We c1aim that B必＝（㊥二〇ハん）㊥…①（㊥二〇カ肌必）．Indeed since μ（μ）＝1andμ（z）：η十1，we haveμ（ぴzゴ）＝乞十（η十1）ゴ．For0≦乞≦〆≦η andゴ，ア≧0such that（づ，ゴ）≠（〆，ゴ ），we have〃（ψzゴ）≠レ（〆ノ）It fo11ows easi1y thatΣ乏。／mo｛十＿十ΣZoメゾmη乞＝0with m者ゴ∈Mo imp1ies that m幻＝0for a11乞，ゴ，where we use the assumption that the e1ement∫has no
t…i・ninM・ndth・f・・tth・tifδ「（〃）∈λ，th・nδ「（ψ）＝・戸f・…m・
5∈Q and一方∈N．Hence we have BMo＝（㊥二〇ノMo）θ．．．㊥（㊥二〇ノガM；o）
and Mo：BMo／（μ，z）BMo is a丘nite1y generatedλ一mod−u1e． □
In the rest of this section，we consid−er the case where M is a torsion−free B−modu1e．
lLemma3．1．2．〃δ∈LNDん（B）α〃1θ左Mろeα伽伽1 舳m亡θ〃。ヅ5乞㎝一 価θ（B，δ）一m・〃θ．切ρ・・θ伽1δ1川㎝・θ・・㎝〃1・㎝棚θgm㍑・mα伽．
肋・η伽・θθ洲・α伽・（B，δ）一m・〃・F＝B∫。㊥…θB∫ηω肋九∈昂α棚
F C0η。左α乞η一3ノレ∫ α3αδ＿5伽うη70d砒Zθ、
Pヅ。oヅSinceδ≠0，there exists a nonzero eIementαi叫λ∩δ（B）so that the derivation on B［α■1］induced−byδhas a s1ice．Hence we have Mo［α一1］
is a丘nite1y genera七ed．4［α一1］一modu1e by Lemma2．2．3．Since M；o⑳λρ（λ）
is a freeρ（λ）一modu1e，there exists c∈λsuch that Mlo［c−1］is a丘nite1y generated−free刈。■1］一modu1e and the derivation on B［c 1］induced byδhas a s1ice．Then we have M mc−1］＝B［c−1］⑭刈、＿。］Mo［c−1］．Let｛e1ゾ．．，e、｝be a free basis of M；o［c−11．Suppose that m1，．．．，m，generate M「as a3−modu1e
and−that m｛＝c η｛Σ仁16切eゴwith non−negative integers〜and6打∈B Let W＝max乞（η壱）．Then we have M⊂（D二1B（θづ／cw）． □
In the above1emma，if the rank of F is one，then we can regard〃as a
δ一id−ea1．Indeed，if we write F＝Bθ，then M「is isomorphic to∫：＝｛6∈B1 6e∈M｝as aδ一mod−u1e．In particu1ar，M is a free B−modu1e of rank one if and on1y if∫is principa1．
Next we consider the case whereλis a noetherian domain．In this case we have the positive answer to Prob1em1．1．2．
Theorem3．1．3．〃δ∈LNDん（3）α棚Ze6Mろεα伽伽1〃舳ηα6θ〃。グ3台。η一 価θ（B，δ）一m・ωθ．8W・・θ伽立山・αη・肋θれ㎝d・mα伽．Tん・ηM。乞・α 伽伽佃9㎝em亡εdλ一m0〃e．