in apo1ynomia1expressionofプ，and the monomia1（ 1… 、）±％斗1ユー1apPears in a po1ynomia1expression ofδ（9）、Sinceδ（9）consists of terms
・。b・…・肌b｝斗1バ…μ、。。cれ・1／挑 （3．2）
for 1 ＜づ＜ηand
b1．． b刊 ±．． t・1．． ・帆 ・九十1−1 工1 ． 几 1 二じ几V1 μ几 V九十1 っ
we must have cη十1＝Z−1and−there exists5such that c、＝1and c｛＝0 for a11乞≠8，for七he1ast term camot cance1withδ（以十1）．By comparing the exponents ofエ、in（1）and（2）for乞＝5，we have t＝亡十1＋6，but this is a
COntradiCtiOn．口
Arguing as in the above1emma，we can prove the fo11owing1emma．
1Lemma3．3．8．W肋伽ηo亡α向㎝50∫Tんeorθm3．3．4フ乞∫αm㎝om乞αl oμんθ
∫・㎜工。αμ肌十。吻η。。ω肋4＞0α〃θα・・伽αρ・伽・mlα1θ・岬・・1・η・伽，∈M。
α8㎝θ1θmθ械。ゾRノ伽ηα〉O．
Prooヅ SupPose thatα＝0． Since dμη十1does not a．PPear in anyδ（d均）or anyδ（伽），it fo11ows from the equa1i七y
δ（9几。。切几。。）＝1（・。…・肌）㌔肌。。 一1払。。 （3．3）
十（termS nOt COntaining吻肌十1）
that there exists a monomia1U＝z1b1＿ 、b几g1c1＿蝪十1c・十1吻九十1with
（ろ。，＿，6η，・。ゾ．、，・肌十。）≠（0，＿，0，0＿，0，1）
in a po1ynomia1expression of m，and the monomia1（ 1＿ η）亡％十1〜 1切η十1 apPears in a po1ynomia1expression ofδ（り）．Since a term in a po王ynomia1 expression ofδ（η）containing dg肌十1is
・。b・…・、b｝十1ツ。c・…μれ。。c冗・・吻、。。／挑 （3．4）
for 1 〈づ＜ηor
・。b・…・、bηぺ…・、士眈。・…μれ。几μ、。。c帆・・ 1切、。。，
we must have cn＋1＝Z＿1and there exists5such that c、：1and c｛＝O for a1い≠8．By comparing the exponents of 、in（3）and（4）for乞＝5，
亡＝亡十1＋6，butthisisacontradiction．口
3．3．D工FFERENnAL−M ODσLE8
57Now we give the proof of Theorem3．3．4．
Pヅ。oヅSuppose that Mo＝λm1＋．一一十λm、．There exists a su舐。ient1y Iarge integer q such that no monomia1ofthe form 1α垢十1切、十1with Z≧g appears in a po1ynomia1expression ofany肌｛．Since∂α∈Mo for anyα∈λ，it fo11ows from Lemma3．3．6that Mo contains an e1ement of the form
m＝（ 1％十1q＋（terms of1ower degree in％斗1））勿れ十1 ＋（termS nOt COntaining吻n＋1）．
Then m＝α1m1＋＿十α、肌、for someαづ∈λ．By the choice of g，those termsα〃乞which contribute to prod−uce the term 1㌶斗1吻肌十10f m have 七he coe伍。ien七α｛containing the term所砥十1withα＞0．By Lemma3．3，7 and Lemma3．3．8app1ied toαづand肌respec七ive1y that the coe伍。ient of
帆十14吻叶1inmisnotequa1to 1．Thisisacontradiction．口
We obtain the fo11owing counterexamp1es to Prob1em1．1．2by making use of the counterexamp1es to the fourteenth prob1em of Hi1bert given by Freudenburg［8］and Daig1e−Freudenburg［3］．
Theorem3．3．9．〃B＝ψ，g，8，亡，u，ω16eαρo伽。m乞αZヅ初g㎝ e伽ε
δ∈LNDた（B）6Vδ（ ）＝δ（ひ）：0フδ（5）二 3，δ（亡）＝ひ35，δ（u）＝μ3tαηd δ（り）二π2V2。工θ老M＝ΩB／だろθ流e m0仇Zθ加rωα切0肌ω流ηα之刎mZδ一mOdωθ 5主用伽rθ．Tん㎝MO乞 0老α伽伽佃9θ肌erα亡edんm0〃θ．
Theorem3．3．10．〃B＝ψ，5，乏，u，り1うθαρo伽0mづαけ伽g㎝d加伽θ
δ∈LNDん（B）6Vδ（ ）：0，δ（5）＝ 3，δ（亡）＝5，δ（u）＝亡フαηdδ（U）＝・ 2．工θ古 ハ∫＝ΩB／ん6e栃θm0仇iθ加グ切α挽0ηω4流ηα君刎mZδ一m0∂ωθ5かαC切ηθ・r加η M0乞 0亡伽伽妙9θηem考θd0りeザλ．
We can prove the above theorems in the same fashion as Theorem3．3．4
withth・1・mm・・imi1・・t・L・mm・3．3．6一 i…［8，L・mm・21・・d［9，L・mm・
7．51）and−in each case塙（M）gives the new comterexamp1e to the fourteenth prob1em of Hi1bert．
As we have seen above，we use the di丑erentia1mod−u1eΩB／いn ord−er to con弓tmct a counterexamp1e to Prob1em1−1．2．Then R：＝堆（ΩB／冶）gives a
counterexa㎡p1etothefourteenthprob1emofHi1bert．Wecangivethenatura1
δR−modu1e struc七ure to the d．i丘eren七ia1mod−u1eΩR／た，whereδR∈LND（R）is ind−uced byδ．Then we canprove in the same fashion as above thatΩ月／んgives a counterexamp1e to Prob1emユ．1．2and一理（Ω五／た）gives a counbterexamp1e 七〇thefourteenthprob1e皿。fHi1bert．We can conti㎜ethisprocess in丘nite1y many timeS．
Chapter4
A system ofinfinite1y many
generatOrS
4．1 A simp1i丘。ation of Kurod．a，s counterex−
amp1e
In［11］，Kurod−a showed−the fo11owing counterexamp1e to the fourteenth prob−
1em of Hi1bert．
Theorem4．1．1．〃B＝ψ1，＿，叫，μ1ゾ、一，％，％十11ろeαρ0切0m乞αZん一 αZgθうmαηd dφηeδ∈LNDん（B）6Vδ（吻）＝0αη∂δ（挑）＝物2∫oヅα〃1≦
乞≦η，αη∂δ（％十1）＝ 1… 肌．舳〃05θ仇ατ几≧4、肺θηλ：＝Kerδ48ηo舌 伽伽佃9e砒m右θd0ωεヅん。
In ord−er to prove this theorem，he made use of the fo11owing1emma．
Lem岬a4．1．2．㎜肋伽ηo左励。η3α棚α舳m帥。η3伽肋θαろ。〃θ左んθoヅθm，
伽グθe拙左5αρ08舳e加物erα5㏄ん伽左伽ん一舳6αlg伽αλC㎝右α伽8θ1θ一
mθ舳0μ加∫0rm
・。αμれ。ユ4＋（加㎜・・ゾ1・ωθグ佃脇〜、。。）
∫oザθαc㍑≧1．
Kuroda，s construction of the invariant e1ements can be rep1aced−by a more straightforward construction，and according1y the invariant e1emeヰs
aregivenin amoreprecisefom，Weprovethat we cantakeα：1．Name1y，
we prove the fo11owing1
58
4．ユ．
A S工MP∬F工CAnON OF KσRODA 8COσNTEREXAMPLE
59Theorem4．1．3．〃肋伽ηo亡αれ㎝5㎝dα85u岬切㎝8伽伽αう。りe伽。rθm，
伽た一舳ろαlg伽αλC0枇α伽8θ1θ肌卯30μ加∫0ヅm 舳、。。4＋（加㎜・・〃・雌幼冊〜肌十。）
∫oグθαc〃≧1，
Pヅ。oヅIn a subsequent proof，we use the f6110wing resu1t．
I−emma4．1．4．〃B＝ん［ 2ゾ．、， η，μ2ゾ．．，μη］αη〃θ伽eδ∈LNDん（B）6μ δ（均）＝0α〃δ（μゼ）＝叫2！oヅθαcんづ。丁加ηKe工δ乞8αん一αZgθ6m gθηem加d〜
・。，＿，π、㎝れ2的一・ゴ2μ｛（2≦乞，ゴ≦η，1≠ゴ）．
We canprovethis1emmabythesame argument as in［10，Theorem1．21，
Now，for each monomia1m＝ 1α1＿叫α帆μ1b1＿％十1b肌十1，d−e丘ne τ（m）＝［㌢1＋…十［昔1一（6・十・十6η）1
where we write［α］＝max｛η∈Z Iη≦α｝for anyα∈R．Let∫1，η十1＝
1％十1一 2… 〃1and工etノ＝；，ゴ＝軌2的一句2坊for each pair（乞，ゴ）with 1≦づ，ゴ≦ηand乞≠ゴ・It is easy to see that a11ofプ1、η斗1，叫and∫｛，ゴbe1ong七〇 λ＝Kerδ．Letλ be theん一suba1gebra generated−by∫1，肌十1，η（1≦1≦η）and
∫ ，ゴ（1≦乞，ゴ≦η，4≠ゴ）．Sinceλis factoria11y c1osed in8，i．e．，α＝ろ162∈λ with61，62∈B imp1ies61，62∈λ，it su伍。es to show that there exists∫∈
…hth・t∫。，、。。L∫i・・fth・f・・皿
・。勾、。。4＋バ1（t・・m・・f1・w・・d・g…i・μ、十。）
二・・4■1（何れ十・乏十（t・・m・・f1・w・・d・g…i・ツ、。。））．
We have
∫。，、。。㌧・。㌦肌。。L佃4■1g、。。乏一1・。・、μ。
・（1）＾・（・2・・沽・…
十（一1）〉・ぺμ。4
and we construct∫∈λ which，when subtracted−from∫1，、斗14，cance1s the terms in∫1，、十140fdegree＜4＿1inπ1and produces on1y the．terms ofdegree
≧4＿1in 1．Name1y，as the e1ementプ1，、十14＿プ，we construct an e1ement in〃。f七he form
4 4 4−1 乏一2
19れ十1 ＋9！＿1μれ十1 ＋94＿2V肌十1 ＋・ 十90，
60
σHAPTER4．工NF工NITELY−MANY GENERATOR8
where g4∈吋 1，．．．，叫，μ1ゾ、．，ひ帆］and 1ゼ■1divides every g壱．
By the d−escend−ing induction on r，we suppose that we obtain an e1ement in〃。f the form
G、：・。㌦、。。老十9に。μ、。。ゼー1＋9ゼー。ツ、。。ゼー2＋…十9、ツ、十。「十…十9．
with g｛∈ψ1ゾ．．， 、，眈，．．．，g、］and g4−1，．．。，g、十1divisib1e by 14■1．We show that G，is mod−i丘ed by an e1ement of〃so that a new g，is divisib1e by
14−1without changing the terms g4＿1ゾ．．，g、十ユ．Furthermore，we suppose the fo11owing conditions are sa七is丘ed一．
（1）F・・0≦1≦4−1，ifw・w・it・g1一Σゴ州・％・仇，ゴwit川，ゴ∈
ん［ 2ゾ。．， η，g2ゾ．．，gη］，thenづ十ゴ十2％，ゴ＝24．
（2）We haveん｛，o＝…＝11｛，｛一1＝0for0≦づ≦4−1，i．e．，for each
1〜1q｛・ゴ〜，ゴapPearing in g｛，we haveゴ≧乞．
（3）For each monomia1肌＝エ1ゴ 2α2＿ 几α冊μ1わユ．．．μ、6冊μ肌十1｛with61＝酌，ゴ in％斗1㌦1〜1〜ん｛，ゴ，we have
（i）2τ（m）≧トト3a・dα。ゾ．．，α。a…11oddint・g・・sifゴ≡4−1
（m・d2），
（ii）2τ（m）≧4一ゴandα2ゾー．，αれare a11even integers ifゴミ4（mod−2）、
In order to improve the term g，in such a way thatん、，o＝．．一＝ん、，4＿2＝O，we suppose by a d−oubIe induction thatん、，o＝＿：ん、，ρ＿1＝0andん、，ρ≠0with
r≦ρ≦4＿2．Withthishypothesistakenintoaccount，wedenotethepo1yno−
mia1G，by G、，、．Thebeginningpo1ynomia1for ind−uctionis G乏＿2，目＝！1，肌十ユゼ，
f・・whi・hgF（一1）4−4（1）・・乞（α・・、μ・）4−4，伽，F（一1）4．｛（1）（… 、）4−4，
ん｛，ゴ：0（4≠ゴ）a．nd軌，｛＝4一乞for0≦乞，ゴ≦4＿1．One can check easi1y that th・、・b・・・…diti・・・・・…ti・台・df・・G。．。，。．。：∫。，れ。。4．
We exp1ain the process ofimproving g、．Since g、十1is divisib1e byぺ一1
and
δ（ 。ρμ。q叫、，幽。。「）＝・・ρバ，仏。・「δ（ん、，ρ）
十（terms of degree＞ρin 1）十（terms of degree＜r in脇十1），
we have
0＝δ（G、，ρ）＝工・pバ，仏。・「δ（ん、，ρ）
十（terms of degree＞ρin 1）十（terms of degree≠r i耳％斗1）
4．ユ．A8工MP∬F工CAT工ON OF KσRODA SσOσNTEREXAMPLE
61and−henceδ（ん、，ρ）：O．Lemma4．1．4imp1ies thatん。，ρis a sum ofpo1ynomia1s of the form
…あ…・肌d一■九，ゴ〜
包，ゴ∈｛2，＿，肌｝
with c∈たan−d non−negative integers凶，ち，ゴ．Note that a11of d2，＿，dη are odd integers（resp．even integers）if p≡4＿1（mod2）（resp．ifρ≡4
（mod−2））．Infact，sincethe contributions ofthe九，ゴto the exponent d2，＿，dη are even，the remark fo11ows from the conditions（i）and一（ii）of（3）．Now we
・h・・・…y・…fth・・b…p61y・・mi・1…d1・tH一π、，ゴ、｛。，．．．，几｝カ，！｛・ゴ Then，for each monom三a1m in妬十1「 1ρμ1q・町2d2＿ 几dη∬，we have in view
・f（i）・nd（ii）of（3），
・伽・㌧・）一・τ（・）・ ^lll−31㌶：1一＾。 ；d2），
where mu1tip1ying山畑 2あ＿ 肌dn by any物2坊（づ≠1），μ、十10r 1does not change the va1ue ofτ．Note thaけ≦ρ≦4−2and一七hat ifρ…4−1（mod2），
thenρ≦4−3and hence4一ρ一3≧0．Thus we haveτ（μ！q… 2d2… 、∂れ）≧0 and there exists an e1ement F∈〃。f the form
F一・・。ρ一「∫。，、。。「允，。q2…∫、，。q一・。あ一2q2…・、d一一2％
＝・・。ρﾊ。伽・・μ、。。「・。あ…・、、d・
十（terms of degree＞ρin 1）十（terms of degree＜r in％十1），
where g2＋＿十％＝g、，ρ．We can prove that G、，p＿FH satisies the same cond−itions as G、，ρdoes except for the conditionん、，ρ≠0but the number of nonzero terms inん、，ρgets sma11er，We prove this be1ow．By repeating this proces§丘nite1y many times，we obtain a new G，satisfying the condition ん、，ρ＝O．Further，continuing this process丘nite1y many times，we obtain a modi丘ed G，satisfying the conditionん、，o＝…＝ん、，4＿2＝0，i．e、，g，is divisib1e by 14−1．Hence by induction on r，we comp1ete a proof．
Now we show that G、，ρ＿FH satis丘es the same conditions as G。，ρdoes but the number of nonzero monomia1terms inん、，ρbecomes1ess．We have on1y to show that each monomia1in F satis丘es七he cond−itions（！）一（3）since none of％十1， 1andμ1appears in∬and mu1tip1ication of any monomia1in H to a monomia1does not chage the va1ue ofτ．Each nonzero monomia1mF
62 CHAPTER4．工NF工N工TELγMANy．GENERATOR8
in F is of the form
π
π・ρ一「（舳。・）「・（・・…舳）「2・・あ一2q2・が一2q¶（凶・）α｛（π・2ωβ｛
4＝2
withヅ1＋ザ2斗andα｛十β4＝％for乞＝2，．、．，η．We choose one岬and
1etω，z1，and zn＋1be the exponents of 1，μ1，a．ndμη斗1inη7F respective1y・
Then we have
ω＝ρ一r＋ヅ1＋2β2＋… 十2βn＝ρ一ザ2＋2β2＋… 十2βη，
Z1＝ヅ2＋α2＋…十αηand Zれ十1＝r1．
First we prove mF satis丘es七he cond−i七ions（1）and一（2）．Indeed一，we have zη斗1＋ω十2z1＝ρ十ヅ1＋r2＋2（α2＋β2）十… 十2（αη十βη）
＝ρ十÷十2g。十…十2q几＝叶・十2g、，ρ＝24
and一
ω一z九十！＝ρ一（r1＋r2）十2（β2＋… 十βη）
二ρ一ヅ十2（β2＋…十βη）≧P−r≧0．
In ord−er to prove that mF sa．tis丘es the cond−ition（3），we consider four
CaSeS
（a）ρ…4＿1（mod−2）and r2＝2u＋1 （b）ρ≡4＿1（mod2）and r2：2伽 （c）・ρ≡4（mod2）and−r2＝2刎十1 （d）ρ≡4（mod2）and−r2＝2刎，
where u is an integer．We on1y consider the case（a）．The remaining cases can be treated in a simi1ar fashion．Then we have
ω≡4−1−2u−1＋2β2＋…十2βη≡4 （mod2）．
The exponent of each吻（乞≠1）in mF is equa1to2u＋1＋4−2g乞十2α｛．Since each凶is an odd integer by the condition（i）of（3），it is an even integer．In
4．2．