LOCAL RINGS MASAYOSHI NAGATA Kyoto University, Kyoto, Japan
INTERSCIENCE PUBLISHERS a division of John Wiley & Sons, New York· London
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By MaHayoHhi NagaLa
INTERSCIENCE PUBLISHERS a division of John Wiley & Sons, New York· London
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1. D. Montgomery and L. lIippin-Topological Transfonuation Groups
2. Fritz John-Plan., Waves and Spherical Means Applied to Partial Differential Equations 3. E. Artin-Geonwll'i., Algebra 4. R. D. RichtI1lYI'J'
Dill'erence Methods for Initial-Value Problellls
5. Serge Lanp;- III LI'oduction to Algebraic Geometry 6. Herbert BliHemann-Convex Surfaces 7. Serge Lang-Abelian Varieties 8. S. M. Ulam-A Collection of Mathematical Problems 9. 1. M. Gel'fand-Lectures on Lineal' Algebra 10. Nathan Jacobson-Lie Algebras 11. Serge Lang-Diophantine Geometry 12. Walter Rudin-Foudcr Analysis on Groups 13. Masayoshi Nagata-Local Rings
Additional volumes in preparation
[iJ
COPYRIGHT
© 1962 BY JOHN
WJLJ<;Y
&
80NS, l~.
ALL RIGHTS R1DSffiRVED
lll~ARY OF CONGRESS CATALOG CARD NUMBFJR
62-17459
PRINTED IN THE UNITED STATES OF AMERICA
'l'lu, LlH'OI'Y 01' lo('.nl l'illP;H iH illl(JOI'LHII L ill boLh algebl'ai(~ gooJllcLry Hlld LI)(~ Lheol'Y 01' (~OnlllluLaLivo rillgH, aml ha::; been intensively devplopnd ill Lhe pa:·i!, do(mdo by many authors, But, to the best of the IVI'i(.('I"H kllowblp;n, ollly two books, by Samuel [2] and Akizuki and Nap;aLn III have bonn published on the subject, The former is rather 0\1(.01' dnLo and the latter was written in Japanese, Thp wl'iLm' aimed, therefore, on one hand to give in the present book I\. HYHLmrmtie exposition of an up-to-date theory of local rings, The IVl'iLpl' aimed on the other hand to develop further the theory of local l'inp;H. Among the new methods and new results given in the present I,ook, the following four should be noted: (I) A principle, which is called the principle of idealization, and h.v whieh modules become ideals, is applied manywhere in this book. (2) The primary decomposition of a homogeneous ideal has been LJ'(~ItI;(~d by a different method from that of inhomogeneous ideals, BuL, proving a lemma (our (8,3)), we give a unified treatment which iH an adaption of the one for the inhomogeneous case, The primary doeomposition of a Noetherian graded submodule follows from that of id(~al::; by virtue of the principle of idealization, (3) We give in Chapter II a new theory of the exact tensor 11I'oduct, (4) We give in Chapter IV a new theory of syzygies, which takes L!\() place of homological methods employed in the theory of local rings, Thus we shall never nse homological algebra in the present book, II'urthermore, it should be emphasized here that our theory is simpler Chan the one given by homological algebra even for readers who know Lhe subject, The writer wishes to express his thanks to his colleague Dr, Hideyuki Matsumura for a critical reading of the manuscript, Kyoto, July, 1960
MASAYOSHI NAGATA
[vl
()IJi,V 1,111' i'ollowillg HI'(' II.NNllllIl'd i.o LI' kIIOWII: (I) 1':II'IIICIILN or 1"1('(. I.IIl'OI',V illl"It)(lillg jljOI'II'N 11'111111:1. whil'h
aNHCI'(.H
Lim!' lUI illdlldiv(' Nd haN a IliaxilllaIIiWIlIi>el'. (~)
1\ kllowkdg() of algebra which is common to Van del' Waerden [1] (Vander Waerden [1], Chapters 3-8; ZariskiHHlllild III, Chaptms I-III), and the definitions of Galois extensions alld (laloiH groups including infinite field extensions. (;{) IGlemontary properties of tensor products (see, for instance, Boul'baki [1], §1, §2.) (1) Elementary knowledge of the notions of open sets, neighborhoods, closed sets, To-spaces, and metric spaces.
III Hlld
jljmiHki-NulIluel
()ur use of inclusion symbol: JI S;;; B (or B ;;:;2 A) means that A is a subset of B. il c B (or B :J A) means that A is properly contained in B, namely, that A Q B and A ~ B.
CONTENTS 11I11"u ..... lioll ,
' " ' ' ' ' ' ' ' ' ' ' ' ' xi CHAl'Tlm
1
General COllllllutativc Rings I. 2. :\. ·1 . ii. (i. 7. :-l. !l. 10. II. 12. I:;' I~.
Il.i IIJ,(M,
idoal~
and modules .. , ....
1 4
I'rililo ideals and primary ideals. N oeLhorian rings ........... , ..
7
•1acohson radicals, , , , .... , .. , Tho definition of local rings. lUngs of quotients. Prime divisors. Primary decomposition of ideals. The notions of height and altitude. Integral dependence, ,,,,,,.,.,, Valuation rings" ,,,,,,, ,,,,.' ,,,,,,,., Noetherian normal rings, Unique factorization rings" A normalization theorem, CHAPTFJR
12
13 14 19 21
24 28 34
39 42 43
II
Completions Iii. Formal power series ring, An ideal-adic topology. 17. Completions., 18. Exact. tensor products, . , In. The theorem of transition, '
49 50 53
](i.
58 64 CHAPTER
III
Multiplicities 20. Homogeneous rings, 21. A-polynomials, 22. Superficial elements, 23. Multiplicities, ' 24. Syst.em of parameters, 25. Macaulay rings, . , .
67
70 71
75
77 82 CHAPTER
IV
Theory of Syzygies 26. 27. 28. 29.
Definition of syzygies, Change of rings, ' Regular local rings .. ' . , Syzygies of graded modules"
91 94
98 101
, 'II \l1j'bli \'
'I'I,,',",~,
uf C:"'"pldo' 1.01'111 1111111" IIII~I II .. '\1'11111'11111111
:111. :-1111111' PI'OI'I'I·l.i"tI "I' ""1111'1,,1,, I""tli IIIIW' :11, '('h,' ,iI.I·'I,',I,III·" 1.111'01'1'111 "I' '''11111'11'1,1' (''''11,1 I'II'W' :I:.l. 1"ilti"I'III'HH or ,J"riv,'d 111'1'1"",1 I'illl-VI :I:\. I )I'I'i v"d 110"11111.1 I'i II~H or N 1",,,III'riail i 111"1';1'111 d"II,".i IIH :1,1, (~lltI,i IIH or IH'i 1111' id"IIIH. CIIAI"J'lm
VI
Geometric Local Rings
127
:11), ""I~II.J i "inK. :11 .. I'H,'lIdo-gllol""CI'ie ringH .. :17. i\lIldyl,il'.1I1 1I01'l1l1l.lil.y. :IX. :-101"" I'YPI'H or rillg "XLI~IIHi()JIH. :I!l, H"pnrll.hly /2:1'111\1':1.1.(\11 I~XI.""Hio"H. III. MIIJI,iI'Jjl~il.y or 11,101:11.1 l·ill/2:. ·11, 1'lIl'i I.,V or l'I'alll:h 10("j. ,I:.l. 'I'PIiHOI' pl·ollllnI.R .. CIIAI"I'I':H
l:n 135
141 146 153 158
168
VII
1I"IIHt'1 iall IOngK a 11.1 W del'Klt'ass JUngi'\ .J:\. 1II'IIHnli,ml.ioli. ,1,1. ""IIHnl 1,,11111111,. ,Iii. (:ollv"rg"IIL pOWI~l' H()l'ies rin!!;s, . ·10 . ./1I,l:obillll crienl'ion of simple points. ·17. i\ IIalyl.ie tensor product.
179 188
190 194 199
ApPENDIX
i\ I. 1';xIIlIIples of bad Noetherian rings .. i\:.l. II is(.orical note, ..
203
212 223
Tllhl.l Ill' Notation ..
229
Illdex .....
231
INTUOUIICTION
'I'lin liiHCOI',Y or til!: tlW()I',Y or IO('nl l'illgH Ilc/!;illH with 1\l'lIll'c: PHllt'I' 101, Ilcl'(I lie dnlil\(~d It "N(dl(~lll'ing" al:! a Neol;hel'ian ring with only (HIP lIIaximal ideal (a Noetherian ring is a commutative ring with IwiCH whidl Hatisfies the maximum condition for ideals). The name Ntellenring was chosen because such rings are often associated with points on algebraic and analytic varieties. Chevalley [1] renamed them "local rings" since a ring associated with a point on a variety gives local properties of the variety. To illustrate this geometric aspect of local rings let us consider an affine n-dimensional space An over the complex field C. Let Xl, , .. , Xn be a set of coordinates for An and P a point in An ' If Rp is the set of rational functions in Xl, . . . , Xn which are regular at P, then Rp is a Noetherian ring in which mp = {f If E Rp , f(P) = 0 I is the only maximal ideal. Thus Rp is a local ring associated with a point P of An . An irreducible variety V going through P defines a prime ideal of Rp , Ij3 = If If E Rp , f(V) = 0 I and vice versa a prime ideal of Rp defines a variety through P. Further the ring Rp/'jl is again a local ring which we call the local ring of P on V. Thus a ring-theoretic and sheaf-theoretic study of the set of local rings of points on V might be expected to yield properties of V. This can be adapted to algebraic varieties over other defining fields, abstract varieties, and also to analytic varieties if Rp is replaced by the set of analytic (holomorphic) functions of P. Many geometric theorems can be derived from local ring theorems, others may be rephrased in purely ring-theoretic terms, e.g., (1) the irreducibility of the product of two irreducible varieties [algebraic case our (:39,9), analytic case our (47,5)], (2) the normality of the product of two normal varieties [algebraic case our (42,10), analytic case our (47,10)], (:3) the fact that the set of points of a variety whose multiplieities are greater than a given number forms a suhvarif)ty which may be reducible [our (40,:{)1, and (4) the so-c:alled Jacobian criterion for simple points [our (46,:))]. As we have sketched above, the theory of local rings is important for its geometric applications. These local rings whieh OCellI' in geometry are the princ:iple study of this book. Most of nice properties of these local rings are derived from the fact that they are pseudogeometric rings which are homomorphic images of Macaulay rings. The single exception to this aim is §33 where we depart from these [xi]
I N'I'UIIIIPI "1'11 IN
iIlVl'MLi~lI.i,(, 11.11 nl'pll"IIi.l1I1I 01' 1,11f! i.l1I'(lI',V III' 1(1('11,1 l'i1q,r;1'I 1.0 (.1111 1.1I('ol'Y III' No('LI II 'l'ill,lI r1l1l1lt 'I'll(' lil'HI. 1,111'('1\ !d\ll,pl.('I'H oi' tiliH I'!loll 11.1',. dl'I'o!."d to IIII.Hi(', 1'(,HlIII,H for /1:1'111'1'11,1 (~Olllllllll.lIol,ivn l'illgH « !llItpj,(I(' I) II.lld 1,11 1.1\1\ d('vl'lo(1I1I(1111, of two ill'lHII'i,HIII, l,oolH wllidl 11,1'(. IIHl'd 1.111'111111;11(1111" 'I'I\I'H(\ 11.1'(. 1,11(. (\0111pll'tioll of l'inp;H Witll I'('I-IP('(',I, 1,0 II. Hilllpin Lopol0Il:Y ( !lInpl.m' II) Hnd 1.1\11 lIoLioll of multiplicity (UhapLm' III). Tiin rat:!. LhaL I,ho ,:olllplei;ocl l'illf,l;H lIavc a lIit:e I'dationlShip to tho original rillg ii::l ii;self a major 1'{lIi,NOII 1'01' i;ll(\ ui::lduluoss of tho theory of local rings. The notion of 1I111ILipli(,,iLy plays a large role in ideal theory and in algebraic geometry. III Chapter II we have also studied semi-local rings, not because L1\1.y are a p;ellel'ali:.mtioll of lot:ul rings but from theoretical necessity. 11'01' i IIKl,all(:(., a I'illp; wbidl iH H finito integral extension of a local iIlL('p;ral dOlltaill Ilr Illom gnrwmlly n ring which is a finite module over a lonal rillg iK 1101, ill general n lomd rillg but rather a semi-local ring. ( h,ollldl'i,:ally, 1.l1iK iH [L(\eo\llli;ed for by tho fact, that if V'is a finite (:ovnl'illg vltl'idy of a vat'ioi;y V Own 1.0 eHeh point P of V there corI'l'.HPOlldK a lilliL(I 1I\lIl1bn!" of pointH of V', but in general more than one. 1\H lIoL(.d ill I.hn pldtwe, Chapter IV gives a theory of syzygies whidl iH Ililhol't'H notable contribution to the theory of invariants. (I;x(\(\pl; a fow elementary results from Chapter I, this chapter can be I'(lad without reference to the preceding ones. (~hapter V is devoted to the theory and application of complete h:nl rings. This is of particular importance for applications and leads 1.0 Olll" theory of pseudo-geometric rings. Next in Chapter VI we take up the subject of pseudo-geometric l'illP;H which, as we have noted, is our main object of study, Chapter VII is devoted to the general theory of convergent power KCl'ies rings, 1\ few words on our general approach are in order. If our only con(',nl'lI had been the construction of a theory of local rings which appear ill algebraic geometry, then other methods might have been used. For illKLanee, some of the necessary ring-theoretic results in Chapter IV (\/1,11 Iw derived without the use of syzygies and the existence of subfinld;; Hometimes makes for easier treatment. But these known methodH (:Il""lOi, be applied even to localities over the ring of rational integers, whidl i::lhould certainly be included for their geometric interest. Thus (HlHpLnrs IV and V (except §33) are important for the theory of local l'illgH. it might be thought that we are indulging in unnecessary gen-
l'illJ.\1'I 1.0
~"II(\I'lti
IN'I'IIIIIIIII"I'IIIN l'ndil.y, hilL, HH
1'11,1'
xiii
HH I.liiH WI'iI.I'I' kIIOIVH, Hlly I'PHI.I'il"j,pd LI'l:aI.IIH',ld, wliit:1i
Hl.ill 1I111,lIn"';('H Co illl',liul., IO('.lI.1il.i('H OV(:I' I·illl-!:H 01' illl.q!;!:J'H doeH not yield
Hli hHl.allLial Hi 1\I pi ifi(',al.ioIlH. HOlllo l'o!-;ulLH have proofs which use ele1\ In II Lal'Y idonH !III L wliieh are nevertheless quite difficult. For instance, 1.1i('. validity of Lho dlain <:ondition for prime ideals in a locality over a Ikdnkilld domain can be proved directly, (see Nagata [13]), but the 1>1'001' iH not easy and our generalized treatment in §34 is simpler. N umOl'ous exercises are included in the book but the reader should not bo discouraged if he cannot do all of them. Except for a very few enHY problems the results are not used in the text except perhaps for other exercises. Thus the reader need not solve all the problems in order to understand the text. Nevertheless it is advisable to attempt all the problems in order to get a better understanding of the general theory of local rings,
CHAPTER I
General COIDlllntative Rings 1. Rings, ideals, and modules A ring will always mean a commutative ring with a unit. When R is a ring and Xl, . . . , X" are either indeterminates or elemelits of a ring containing R, then the ring R[XI , ... , Xn] is well defined. This ring may be denoted simply by R[x] if there will be no confusion. When R is a ring, an R-module .M itl Duch that 1m = m for any m E M (1 being the identity of R). When M is an R-module, a submodule of J];[ will mean R-submodules, unless otherwise stated explicitly. A homomorphism (or an isomorphism) from an R-module M into another R-module N means an R-homomorphism (or an Risomorphism). A homomorphism cf> from Minto N is said to be surjective if cf> ( M ) = N. Adapting the definition of zero divisors and nilpotent elements, we define: An element a of a ring R is called a zero divisor with respeet to a given R-module 111 if there is a non-zero element m of M such that am = 0; a is said to be nilpotent with mopect to M if a'M = 0 for some natural number r; and a is said to be an annihilator of ill[ if aM = o. The intersection of submodules is again a submodule. We say that the intersection NI n ... n N, is irredundant if nir'j Ni 7'" ni Ni for any j = 1, 2 , ... , r. In the following, let R be a ring, M an R-module and let N, N).. be submodules of M. The sum of Embmodules N).. is the snbmodule generated by the N, , and is denoted by L N)... The sum of NI , ... , N, may be denoted by NI N,. Note that NI N, is set-theoroti(:ally nr ni ENd· (nl A subset {u,} of N which generates N is called a basis for N. A
+ ... + + ... +
+ ... +
I
llJ
·)
GENERAL COMMUTATIVE RINGS
basis eonsisting of a finite number of elements is called a finite basis. X i;;: called a finite module if N has a finite basis. A basis for N is said to be a minimal basis for N if any proper subset of the basis is not a ba;;:is for N. Note that the submodule of a module generated by the empty set consists of zero. If a is an ideal of R, we define the product aN to be the sub module of S generated by {an I a E a, n E N}, this last set itself may not be a module. Xote that if x is an element of R, then xN = {xn In E N} is an R-module. 'lYe call the following well-known theorem the isomorphism theorem: 1·_Y1 + N 2 )/N1 is naturally isomorphic to N 2 /(N 1 N 2 ). Xext we observe some facts about ideals. In the following, German letters denote ideals of R, and A, B, HI>, C, D, E, denote subsets of R. [a:B]c denotes the set of elements c of C such that cB C a; this set is obyiouslyan ideal if C is an ideal. [a:B]R is denoted simply by a:B. The following equalities can be verified easily:
n
. 1.1 )
nC
[a:B]c
=
(a:B)
[[a:B]b:CJE
=
En [a:BClb:E
[(na}.) :BJD = n[(a,:B)]D
[a: (L B,R)]D
=
n[a:B']D
·We generalize tht above to the case of modules, using the following priueiple, which we call the principle of idealization: When an Rmodule lVf is given, let R* be the direct sum R EEl M as modules. In R*, we introduce multiplication defined by (r + m)(r' + m') = rr' + rm' + r'm, and R* becomes a ring containing Rand M, in which 111 is an ideal and M2 = O. Furthermore, submodules of Mare nothing but ideals of R* contained in 111; the structure of 1J.1. as an R-module is substantially the same as that of 111 as an R*-module beea.use R* / 111 = Rand 1112 = O. Xow, for submodules N, N' of 111, we defined N :N' to be {x I x E R, xX' eN}; if A is a subset of R, then N:A, which may better be denoted by [N:A]M, is defined to be {m I m E 111, am eN}. Then, in R* = R EEl 111 as above, our N:N' is nothing but [N:N'JR and our X:A. is nothing but [N :A]M' Therefore (1.1) can be generalized to the ..ase of modules in the following forms:
3
CHAPTER I
(l.2)
(N:A):B
=
N:AB,
(N:A) :N'
(nN,.):A
=
n(N,,:A),
(nN,,):N
(L: A,,)
=
n(N:A,,),
N:
N:
=
(L: N,,) a+ 0 =
N:AN'
=
n(N,,:N) =
n(N:N,,)
a + e = R. Then We go back to ideals: Assume that R = R2 = (a + 0) (a + c) C a + oe C R, hence we have a + oe = R. On the other hand, since (a + 0) (a no) C ao Can 0 in general, we have, by our assumption that a + 0 = R, that ao = an o. Furthermore, the isomorphism theorem shows that R/ao = (a + 0) / (a n 0) = (a/(a no» + (o/(a no» = ((a + 0)/0) + ((a + o)/a) = R/o + R/a; this sum is a direct sum because, in R/ao, (a/(a no» n (o/(a n h» = O. Thus we have: (l.3) If a + 0 = R, a + e = R, then a + oe = R (hence a + 02 = R), ao = an 0, R/ao = R/a EEl R/o. From this, we deduce: 0.4) If, for given ideals al , . . . , an; it holds that (ni",j ai) + aj = R for every j = ], ... , n, then, for any power a: of ai, we have (a) (n i ",] a:) + a; = R for any j, (b) II a: = na', and (c) R/(na:) = R/a~ EEl ... EEl R/a~ . . " Proof: If n = 2, then a repeated application of a + 02 = R in (l.3) gives (a); and (b) and (c) follow from (a) and (l.3); the general case can be proved easily by induction on n. N ext we give an application of the isomorphism theorem: (l.5) Let M be a module over a ring R. Let N be a submodule of M and let m be an element of M. Then (N + mR)/N is isomorphic to R/(N:mR). Proof: (N + mR)/N = mR/(mR n N) by the isomorphism theorem. Let cf> be the homomorphism from R/(N:mR) onto mR/ (mR N) such that cf>(x modulo N:mR) = (mx modulo mR N). cf>(x modulo N:mR) = 0 implies that mx E mR N, whence mx E N and x E N:mR, and therefore (x modulo N:mR) = O. Thus cf> is an isomorphism, and the assertion is proved. Similarly we have: (l.6) With M, N, and R as above, if a E R, then (N + aM)/N is isomorphic to M/(N:aR). We add here some definitions and remarks concerning tensor products. Let R be a ring and let R' be a ring which is an R-module. Then for an R-module M, M ® R R', which is usually denoted simply
n
n
n
GENERAL COMMUTATIVE RINGS
by M ® R ' , is naturally an R'-module, though it is still an R-module. ,Ye treat M ® R' as an R' -module, unless the contrary is explicitly stated. Let cf> be a homomorphism from an R-module M into an Rmodule N. Then there is a uniquely determined homomorphism cf>* from M ® R' into N ® R' such that cf>*(m ® r') = cf>(m) ® r'. This cf>* is denoted by cf> ® R'. It is obvious that: (1. 7) If cf> is surjective, then so is cf> ® R'. We say that ® R ' , or more exactly ® R R ' , is exact if for any identity map cf> from a finite R-module iVI into a finite R-module N such that JI eN, the tensor product cf> ® R' is an isomorphism. (This definition can be adapted to the case where R' is just an R-module. ) In closing this section, we define the length of a module. If an Rmodule M has a composition series M = Mo:J M1:J ... :J lJ;In = 0, then the length n is independent of the choice of the composition series by the well known Jordan-Hoelder-Schreier theorem. The length n is called the length of the R-module M and is denoted by length R M or simply by length M. E:S:ERCISE: Let R be a ring. A sequence of R-modules Ni accompanied by homomorphiBms d i , which is denoted by No ~ N, ~ Nt ~ ... ... ~ N,_! ~ N r , is called exact if the kernel of di-! is the image of di for i = 2, .,. , r. Prove that, in such a case, if ®R' is' exact, then No ® R'~N! ®R'~ '" ~ N,®R'isexact.
"-e
2. Prime ideals and primary ideals
observe first that the following conditions for an ideal \J of a ring R (lJ ~ R) are equivalent to each other: (1) \J is a prime ideal of R. ':2) If ab E \J(a, b E R), then either a E \J or b E \J. (:3) If a and b are ideals of R such that ab C \J, then either a C \J or b C \J. (4) If 1) C a and \J C b, then ab %\J. Though the ring R itself satisfies the above conditions, we exclude it from th6 set of prime ideals. Let S be a multiplicatively closed subset of a ring R. An ideal \J of R is called a maximal ideal with respect to S if \J does not meet S and if eyery ideal of R properly containing \J meets S. :2.1 i THEOREM: Let S be a multiplicatively closed subset of a ring R. If an ideal a of R does not meet S, then there is a maximal ideal \J with "t--,-[yd tl} S sllch that \J contains a. Such a \J is necessarily prime. P1<J.')i: Let F be the set of ideals b containing a such that S b= ;o.::np,,:y_ Then F is an inductive set (the order being given by the in-
n
CHAPTER I
5
elusion relation). Hence, Zorn's lemma implies the existence of p. If p C 0, P C c, then 0 and c meet S. Hence OC meets S and therefore OC $ p. This implies that p is prime and the theorem is proved. We say that an ideal m is a maximal ideal if it is a maximal ideal with respect to {I}. By this definition, we have the following corollary to (2.1): (2.2) If u is an ideal of R and if u ~ R, then there is a maximal ideal m of R which contains u. Hence, if a is a non-unit of R, then there exists a maximal ideal of R which contains a. A maximal ideal is a prime ideal. Furthermore, since a field is characterized as a ring which has no ideals except the zero ideal and the ring itself, we see that an ideal m of a ring R is maximal if and only if Rjm is a field. For an ideal a of a ring R, the intersection of all prime ideals containing a is called the radical of a. The radical of the zero ideal is called the radical of R. An ideal a is called semi-prime if the radical of a coincides with a itself. (2.3) THEOREM: The radical of an ideal a of a ring R is the set of all elements of R which are nilpotent modulo u. Proof: Assume that xn E a (x E R) for a natural number n. Then every prime ideal p containing a contains x n , hence x E p, which implies that x is in the radical of a. Conversely, assume that xn EE a for any natural number n. Then the set S of powers of x is a multiplicatively closed set which does not meet a. Therefore (2.1) implies that there is a prime ideal p of R containing a but not containing x. Hence we see that x is not in the radical of a. Thus the proof is complete.
(2.4) COROLLARY: An ideal a of a ring R is semi-prime if and only if Rja has no nilpotent element except o. Let a be an ideal of a ring R. A prime ideal p of R is called a minimal prime divisor of u if it is minimal among prime ideals containing u. (2.5) THEOREM: If an ideal a of a ring R is contained in a prime ideal p, p contains a minimal prime divisor of a. Proof: In the set of prime ideals containing a and contained in p, we introduce ordering which is opposite to the inclusion relation. Then we see that the set becomes an inductive set, and we prove the assertion by Zorn's lemma. (2.6) COROLLARY: The radical of an ideal a is the intersection of all minimal prime divisors of a.
6
GENERAL COMMUTATIVE RINGS
(2.7) If ~I , ••• , ~n are prime ideals of a ring R and if a is an ideal of R wMch is not contained in any of the ~i , then there is an element a of a which is not contained in any of the Pi . Proof: If one of the pi, say ~n , is contained in some other pj , then we may omit Pn . Therefore we may assume that there is no inclusion relation among the ~i . Then there is an element ai of
PI ... Pi-IPi+l ... \lnu
L:
which is not in \li , for each i. Then a = ai is in a and is in none of the Pi . An ideal q of a ring R (q ~ R) is called primary if every zero divisor of Rlq is nilpotent, or equivalently, if ab E q, a Et q(a, b E R) imply the nilpotency of b modulo q. This definition can be expressed as follows: (2.8) Let P be the radical of q, then q is primary if and only if ab E q, b Ef \l (a, b E R) imply a E q. (2.9) The radical P of a primary ideal q is prime, hence it is the unique minimal prime divisor of q. Proof: ab E \l, a EE p( a, b E R) imply that a nb n E q for some natural number n and -that an IE p. Therefore, by (2.8), we have bn E q, hence b E p, which proves that P is a prime ideal. The last assertion is obvious. If q is a primary ideal with radical p, we say that q is a primary ideal belonging to P or that q is primary to ~, or that ~ is the prime divisor of q (the justification for "the" will be given in (7.6)). (2.10) If the radical ~ of an ideal a in a ring R is a maximal ideal, then a is primary. Proof: ~I a is the unique prime ideal of Ria. Therefore, by virtue of (2.2), we see that an element of Ria is- either a unit or nilpotent, which proves the assertion. (2.11) If qI, ... , qn are pr-imary ideals with the same prime divisor ~ in a ring R, then the intersection a of the qi is primary to p. Proof: Let x be an arbitrary element of ~. Then x mi E qi for some mi (for each i), hence xm E a with m = max mi , which shows that ~ is contained in the radical of a. This implies that the radical of a is p. Assume that ab E a, a Ef pea, b E R). Since the qi are primary to p, we have b E qi for every i, hence a E -.., which implies that a is a primary ideal belonging to ~. (2.12) Let cf> be a homomorphism from a ring R into a ring R'. If q'
CHAPTER I
is a primary ideal with prime divisor ~' in R', then q primary ideal with prime divisor p
=
7 cf>- I (q') is a
cf>- \ p') .
Proof: Let x be an element of R. x is nilpotent modulo q ~ xn E q for some n ~ cf>(xf E q' for some n p cf>(x) E ~'~ x E p. Thus we see that ~ is the radical of q. If ab E q, a Et ~,then cf>(a)cf> (b) E q', cf>(a) Et ~', hence cf>(b) E q', and therefore b E q. Thus we see that q is primary to p. (2.13) If q is a primary ideal of a ring R with prime divisor ~ and if a is an element of R which is not in q, then q: aR is primary to p. Proof: Since a EE q, we have q C (q: a) C ~, and p is the radical of (q:a). Now be E (q:a), bEEp imply ca E q because bca E q.
Hence cEq: a, proving our assertion.
3. Noetherian rings We say that a ring R is Noetherian if it satisfies the maximum condition for ideals, namely, any non-empty set of ideals of R has maximal members. We say that a module JJ1 over a ring R is Noetherian if it satisfies the maximum condition for (R- )submodules. Observe that a ring R is a Noetherian R-module if and only if R is a Noetherian ring. (3.1) THEOREM: Let JJ1 be a module over a ring R. Then the following three conditions are equivalent to each other: (1) JJ1 is Noetherian. (2) If NI , ... , N n , ••• are submodules of JJ1 such that Ni C NiH for any i = 1, 2, ... , then there is an n such that N j = N n for any
j
2:. n.
(3) Every submodule of JJ1 is a finite module. Proof: Equivalence between (1) and (2): If there are N i as in (2), but no such n, then the set of the N j has no maximal member. This shows that (1) implies (2). Assume, conversely, that (2) is true and let F be any non-empty set of submodules of JJ1. Let NI be any arbitrary member of F. When Ni is defined, we define N i+1 to be such that: (i) NiH = Ni if Ni is maximal in F, (ii) NiH is a member of F such that Ni C NiH if Ni js not maximal. Then, by the validity of (2) we see that some N n must be maximal in F. Thus (2) implies (1). Equivalence between (2) and (3): If an R-submodule N of M is not finite, then we see easily that (2) is not true. Thus (2) implies (3).
8
GENERAL COMMUTATIVE RINGS
Conversely, assume that (3) is true and let Ni be as in (2). The union N of the Ni is a submodule of M, hence N is generated by a finite number of elements, say Xl, ... , x m • Then they are in some Ni , say N n , whence N n = N, which implies that N j = N n for any j 2:: n. Thus the proof is complete. (3.2) COROLLARY: If an R-module M is Noetherian, then any (R- ) homomorphic image of M and any submodules of Mare Noetherian. (3.3) Let a be an ideal of a ring R and let b be an element of R. If both a + bR and a: bR have finite bases, then a has a finite basis. Proof: Let ai and cj(i = 1, ... , r; j = I, ... , s) be such that a + bR = L: aiR + bR and a:bR = L: cjR. We may assume that the ai are in a. Let a' be the ideal generated by the ai and the bCj . Then we have a' C a. Let a be an arbitrary element of a. Then, since a E a' + Rb, a == rb (modulo a') with an r E R. Since a E a, we have rb E a, hence r E a:bR = cjR. Therefore rb E a'. Thus a E a', and we have a = a' and the proof is complete.
>.:
(3.4) TH}<JOREM: A ring R is Noetherian if and only if every prime ideal of R has a finite basis. (THEOREM OJ!' COHEN) Proof: The only if part is a consequence of (3.1). Assume that every prime ideal of R has a finite basis and that R is not Noetherian, which means, by virtue of (3.1), that there are ideals which have no finite basis. Let F be the set of ideals of R which have no finite basis. Then F is an inductive set; for, if {Nxl is a well-ordered subset of F and if the union N of all the N x has a finite basis, say Xl, ... ,X r , then they are in some N j and we have N j = N, which contradicts the assumptipn that N j has no finite basis, hence we have N E F. Therefore there is a maximal member a in F. By our assumption, a is not a prime ideal, hence there are elements a, b of R which are not in a such that ab E a. Then both a bR and a:bR contain a properly, hence they ha\'e finite bases by the maximality of a in F. Therefore, by (:).3), we see that a has a finite basis, which contradicts the fact that a E F. Thus F must be empty, and R is Noetherian.
+
(3.5)
THEOREM:
A finite module M over a Noetherian ring R is a
X oetherian module. Proof: We prove the assertion by induction on the number of generators of M. If M i~ generated by the empty set, then M = 0 and the assertion is obvious. Assume now that M = RXI + ... + Rx, and that M' = L:~ RXi is Noetherian. Let N be an arbitrary R-sub-
9
CHAPTER I
module of M. Let a be the set of elements a of R such that aXl E N + M', i.e., a = (N + M') :RXl. Then a is an ideal of R; hence a has a finite basis, say al , ... , am. Let di be, for each i = 1, ... , m, such that d i E N and such that d i - aiXI E M', and let N' be the submodule L: Rdi of N. Then N = N' + (N n M'). Since lVI' is l'\oetherian by our assumption, N M' has a finite basis, hence N has a finite basis. Therefore M is Noetherian.
n
(3.6) THEOREM: If a ring R' is generated by a finite number of elements over a Noetherian ring R, then R' is Noetherian. (HILBERT BASIS THEOREM)
Proof: Using an induction argument on the number of generators, we may treat only the case where R' = R[x] with a single element x of R'. Let a' be an arbitrary ideal of R'. Let a be the set of elements a of R such that there is an element a' of a' of the form a' = ax' + S 1 CIX + ... + cs (for a suitable natural number s and elements Ci of R). Then a is an ideal of R. Since R is Noetherian, a has a finite basis, sayal , ... , am . Let a: be, for each i, an element of a' such that a: - aix' E RX'-l + ... + Rx + R for some s. Considering elements of the form a:x a , we may assume that s is common to all a: . Let a" be the ideal of R' generated by the Then, by our choice of a", we sec that every element of a' is congruent to an element of Rx 1 + ... + R.Y + R modulo a", namely,
a: .
S
-
a'
=
(a'
n ( L:~-l RXi)) + a". n
Since L:~-l RXi is a finite R-module, it is Noetherian. Hence a' (L:~-l RXi) has a finite basis. Therefore a' has a finite basis, which proves that R' is Noetherian. (3.7) THEOREM: Let 111 be a finite module over a Noetherian ring R and let N, N' be (R- )submodules of M. Let a be an ideal of R. Then there is a natural number r such that
for natural numbers n which are greater than r.
(LEMMA OF ARTIN-
REES)
Proof: By the principle of idealization and by the fact that R EEl M becomes a Noetherian ring by virtue of (:3.6) (Hilbert basis theorem), we may assume that M, N, N' are ideals of R. Let aI, ... , as be a basis for a and let Xl , ... , Xs be indeterminates. Let Sn be the set of
10
GENERAL COMMUTATIVE RINGS
homogeneous forms f(xl , ... , :1: of degree n in the Xi such that f(al, ... , a.) E anN N'. Let S be the union of all the Sn and let ~ be the ideal of R[x] generated by S. Since R[x] is Noetherian by the Hilbert basis theorem, ~ is generated by a finite subset fl , . " , ft of S. Let di be the degree of fi and set r = max d i • For n > r, let a be an element of anN N'. Since a E an, there is anf E Sn such that f( al , .. , ,as) = a. Sincef E S, f = L figi with gi E R[x]. Comparing the degrees, we may assume that gi is a homogeneous form of degree n - d i . Thus we have a = f(al, ... , as) = Lfi(al, ... ,as) X gi(al, ., . , as) E L an-di(adiN N') c an-T(aTN N'). Thus we see that anN N' C an-T(aTN N'). Since the converse inclusion is obvious, we have completed the proof. (3.8) Let a be an ideal of a Noetherian ring R and let M be a finite anM. Then aN = N. R-module. Set N = Proof: By (3.7) (the lemma of Artin-Rces), we have anM N = an-T(arM N) for n > r. Hence we have N = an-TN, which proves our assertion. (3.9) Let aij and tj(i, j = 1, ... , s) be elements of a ring R. If Lj aijt j = 0 for i = 1, ... , s, then, denoting by d the determinant I aij I , we have dtj = 0 for any j. Proof: Let dij be the cofactor of aij in the determinant I aij I . Then Li dijaij = d, Li dijaik = 0 (j ~ k). Therefore 8 )
n
n
n
n
n
n
nn
n
n
o=
Li dij(Lk aiktk) = dt j
•
(3.10) Let M be a finite module over a ring R such that aM = M for an ideal a of R. If no element a of R such that a - I E a is a zero divisor with respect to M, then M = O. Proof: Let tl , . . . , ts be elements of M such that M = Rti . Then the relation aM = M shows that there are elements aij E a such that ti = L: aijtj for every i = 1, ... , 8. Let d be the determinant I Oij - aij I (Oij being the Kronecker 0). Then d == 1 (mod a), hence d is not a zero divisor with respect to M. On the other hand, dti = 0 by (3.9), which implies that ti = 0 by our assumption. Thus M = O.
L
(3.11) THEOREM: Let M be a finite module over a Noetherian ring R and let a be an ideal of R. Then anM = 0 if and only if no element a of R such that a - I E a is a zero divisor with respect to M. (INTER-
nn
SECTION THEOREM OF KRULL)
Proof: Assume at first that such an element a is not a zero divisor with respect to M. Set N = anM. Then aN = N by (3.8). Since
nn
11
CHAPTER I
111 is X oetherian, N has a finite basis. Therefore the relation aN = N shows that N = 0 by (3.10). Conversely, assume that there is a zero divisor a with respect to jJ{ .such that a - 1 E a. Let m be a non-zero element of M such that am = O. Then we have (1 - a)m = m, hence (1 - a)nm = m for any natural number n. Therefore m E anM for any n, and anjJ{ :;t' O. (3.12) THEOREM: Let 111 be a finite module over a Noetherian ring R, let a be an ideal of R and let x be an element of R. Then there is an integer r such that anM:xR C [O:XR]M + an-rM for n > r. Proof: x(anM:xR) = anM xM = an-r(a™ xM) C xan-TjJ{ by (3.7) (the lemma of Artin-Rees). Therefore, if y E anM:xM, then there is an element y' E an-TM such that xy = xy', whence y - y' E [0: XR]M , which proves the assertion.
nn
n
n
(3.13) COROLLARY: With the same notation as above, if N is a submodule of M, then there is an integer r such that (N + anJl;I): xR C [N::l~R]M + an-rM for n > r. Proof: Applying (:3.12) to 111' = JJIIN, we prove the assertion. (3.14) COROLLARY: Let 111, N, a, and R be the same as above and let x be an element of M. Then there is an integer r such that (N + anM) :xR C (N:xR) + an- r for n > r. Proof: The proof of (3.12) can be applied to the case where N = 0 if we replace xM with xR, whence we prove the assertion in the same way as in (3.13). (8.15) If an ideal a £s generated by a finite number of nilpotent elements, then a is nilpotent. Consequently, if a' is the rad£cal of an ideal a in aN oetherian ring, then a' is nilpotent modulo a. Proof: Let aj , ... , am be a basis for a such that each of the ai is nilpotent. Then there is a natural number n such that a7 = 0 for every i. If r is greater than m( n - 1), then any monomial of degree r in the ai has a formal factor a7 for at least one i, which shows that such a monomial must be zero, and aT = O. The last assertion follows from (3.1) and what we have proved. We add here the following theorem and an application:
ni bl
(3.16)
bi
=
THEO REM: If a ring R has ideals Ih, . . . , bn such that 0, and such that each Rlb i is Noetherian, then R is Noetherian.
Proof: Using induction on n, we may assume that n = 2. The sum + b2 is a direct sum in that case. There are a finite number of ele-
1:2
GENERAL COMMUTATIVE RINGS
L
ments bi of OJ such that OJ + 02 = biR + 112 . From the properties of direct sums, we see that the b; generate l1 J . Thns l1J has a finite basis; similarly 02 has a finite basis. Let now p be an arbitrary prime ideal of R. Then p contains one Oi . Since lJ i and P/Oi have finite bases, we see that P has a finite basis, which proves the assertion by virtue of (3.4) (the theorem of Cohen). (3.17) COROLLARY: If M is a Noetherian module over a ring R, then R/(O:M) is a Noetherian ring and M is a finite R/(O:M)-module. Proof: The finiteness of M is obvious by (3.1). Let UJ, ... ,Un be a basis for M. Then RUi is Noetherian and is isomorphic to R/ (O:RUi). Hence R/(O:Rui) is Noetherian for each i. l\'ow, O:M = ni (O:RUi) by (1.2), and we have the proof by (3.16). We note that we have made use of the following in the above proof: (3.18) The structure of the R-module 111 is the same as the structure of the R/(O:M)-module M. EXERCISES: 1. Prove (3.7), (3.8), and (3.11) without assuming that the ring R is Noetherian but assuming that M is a Noetherian module. 2. Let a and bI , . . . , bn be ideals of a Noetherian ring R. Prove that there exists a natural number 1" such that n i nUbi = nn-r( n i nrb i ) for n > r. (Hint: Consider the direct sum of rn copies of R as an R-module.) 3. Let a and b be ideals of a Noetherian integral domain R. Prove that there exists a natural number r such that uu:b = an-r(ur:b) for n > /".
4. Jacobson radicals The intersection m of all maximal ideals of a ring R is called the ,Jacobson radical of R. It is obvious, by virtue of (2.2), that if a i :: m, then a is a unit in R. ( -4:.1) THEOREM: Let m be the ,Jacobson radical of a ring R and let JI be a finite R-module. If N is a submodule of M such that M = mJI N, then M = N. (LEMMA OF KRULL-AzUMAYA) Proof: Set M' = M/N. Then M' is a finite module and M' = m1l1'. Hence, by (3.10), we have 111' = 0, namely, 111 = N.
+
(-4:.2) THEOREM: If m is the ,Jacobson radical of a Noetherian ring R, and 1f 111 is a finite R-module, then nn mnM = 0; in particular,
nn mn =
O.
This is an immediate consequence of (3.11). -4:.3
J
THEOREM:
Let a be an element of the ,Jacobson radical of a
13
CHAPTEH I
Noetherian ring R and assume that a and 0 are ideals of R such that b C a and a CaR 0. If a:aR = a, then a = b. Proof: a modulo 0 is in the Jacobson radical of Rio and therefore we may assume that b = O. Then a eRa, which implies that a = a(a:a) = aa. Therefore we have a = 0 by (3.10), which proves our assertion.
+
5. The definition oj local rings Local rings and semi-local rings, which are defined below, are naturally topological rings. But the topology will be introduced in Chapter II, §16. We give here only ring-theoretic definitions of them. A ring R is called a quasi-semi-local ring if it has only a finite number of maximal ideals; it is called a quasi-local ring if it has only one maximal ideal. A Noetherian quasi-semi-Iocal ring is called a semi-local ring; a Noetherian quasi-local ring is called a local ring. When we say that (R, lh, '" , prj is a quasi-semi-Iocal ring (or semi-local ring, or quasi-local ring, or local ring), we mean that R is a quasi-semi-Iocal ring (or semi-local ring, etc.) and that the maximal ideals of R are the Pi . It is sometimes convenient to have a name for quasi-local rings (R, m) such that m = O. Note that, by virtue of (4.2), local rings are included in the family of such quasi-local rings. Therefore we call such quasi-local rings local rings which may not be Noetherian. Similarly, a quasi-semi-Iocall'ing R with the Jacobson radical m is called a semi-local ring which may not be Noetherian if mn = O. We give some applications of the lemma of Krull-Azumaya (4.1): (5.1) Let (R, m) be a quasi-local ring and let M be a finite R-module. Then a subset {u}J of M is a basis for M if and only if the set of residue classes {u~l of {u}.} is a basis for MlmM over Rim. Consequently, the u" form a minimal basis for M if and only if the u~ form a linearly independent basis for MlmM over the field Rim. Proof: Setting N = 1: RUi, the assertion follows immediately from the lemma of Krull-Azumaya. We note that the above proof shows that: (5.2) The first half of (5.1) is true for any ring R if m is the Jacobson radical of R. The following is an immediate corollary to (5.1) and shows a nice property of quasi-local rings:
nn n
nn
14
GE)[ERAL COMMUTATIVE RINGS
(5.3) If Mis a finite module over a quasi-local ring R, then any basis for NI contains a minimal basis for JJ;I as a subset; if UI , . . . , U m and 1'1 , .•• , Vn are minimal bases of M, then m = n and there is an invertible linear transformation T ouer R such that (Uj, ... , Un) T = (1\ , ... , Vn). Let R be a subring of a ring R'. We say that an ideal a' of R' lies Ol'CF an ideal a of R if a = a' R. We say that a ring R is dominated by another ring R' if: (I) R C R', (2) every ideal of R which is different from R generates an ideal in R' which is different from R', and (3) every maximal ideal of R' lies ove'.' a maximal ideal of R. In that case, we write R ::; R'; R < R' means that R ::; R' and R ~ R'. Therefore, a quasi-local ring (R', m') dominates a quasi-local ring (R, m) if and only if R C R' and m = m' R.
n
n
6. Rings oj quotients Let R be a ring and let S be a multiplicatively closed subset of R which does not contain zero. Let U be the set of non-zero divisors of R. Let n be the set of elements a of R such that there is an element s of S with as = O. Since S is mUltiplicatively closed, n becomes an ideal of R. Let cf> be the natural homomorphism from R onto R/n. Then we have the following lemma: (6.1) If a is an element of the multiplicatively closed set generated by e and S, then cf>(a) is not a zero divisor in R/n. Proof: Since U and S are multiplicatively closed, a = us with II E U, s E S. Assume that cf>(a)cf> (b) = 0 (b E R). Then ab = usb E n, hence there is an element s' of S such that usbs' = O. Since u is not a zero divisor, we have ss'b = O. Since ss' E S, we have bEn, hence cf>(b) = 0, which proves that cf>( a) is not a zero divisor in R/n. Now we define the notion of rings of quotients; we must treat three eases. In the set P = {( a, u) [ a E R, u E Ul, we introduce an equivalence relation such that Ca, u) is equivalent to (b, v) if and only if av = bu. We denote the equivalence class of (a, u) by a/u. The set Q of the equivalence classes becomes a ring under the operations such that the 8lill and the product of a/u and b/v are (av + bu)/uv and ab/uv, respectively. Q is called the total quotient ring of R. Elements a of R ran be identified with a/I of Q. Thus Q contains Rand Q is generated by R and the inverses l/u of the elements u of U.
15
CHAPTER I
Now we consider the general case: By (6.1), ef>(S) consists only of non-zero-divisors, hence the subring of the total quotient ring of ef>(R) generated by ef>(R) and the inverses of the elements of ef>(S) is well defined. This subring is called the ring of quotients of R with respect to S. Note that Rs = lef>(a)!cf>(s) I a E R, s E S} and thatef>(a)!cf>(s) = ef>(a) ·ef>(S)-I. The third case occurs when S is the complement of a prime ideal ~. Though this is a special case of the above, we use a different notation; Rs is called the ring of quotients of R with respect to ~, and is denoted by R~. A ring R* is called a ring of quotients of R if there is a multiplicatively closed subset S of R such that (0 ~ Sand) R* = Rs. We use the following notation: (a) If a is a subset of R, then the ideal of Rs generated by ef>( a) is denoted byaRs. (b) If a' is an ideal of Rs , we denote by a' R the ideal ef>-l( a') == ef>-l(a' ef>(R)). The following is a characteristic property of Rs : (6.2) If there are a ring R' and a homomorphism (f from R into R' such that for any s E S, (f( s) has an inverse in R', then there is a homomorphism r from Rs into R' such that (f = ref>. Proof: Let a be the kernel of (f. Since all elements of S are mapped to units, we see that (f(n) = 0; i.e., n C a, which means that there is a homomorphism r' from ef>(R) into R' such that (f = r'ef>. We define a mapping r by the relation r(ef>(a)!cf>(s)) = r'ef>(a)[r'ef>(s)r 1 and we see easily that this r is the required homomorphism. As a consequence, we have: (6.3) Let a be an ideal of R which does not meet S and let (f be the natural homomorphism from R onto Ria. Then (f(R)u(s) = RslaRs.
n
n
n
(6.4) THEOREM: If a' is an ideal of R s , then (a' R)R8 = a'. Proof: We have (a' R)Rs = ef>(ef>-l(a' ef>(R) )Rs = (a' ef>(R))R s ~ a'. If ef>( a) Ief>( s) E a', then ef>( a) is in a' ef>(R), and therefore ef>(a) !cf>(s) is in (a' R)Rs (because l!cf>(s) is in Rs), which proves that (a' R)Rs :::> a', and the assertion is proved.
n
(6.5)
n n
COROLLARY:
n
n
n
I; R is Noetherian, then Rs is Noetherian.
16
GENERAL COMMUTATIVE RINGS
'13,61 THEOREM: Assume that q is a primary ideal of R belonging to a primE ideal ~. Then: I'a) If p meets S, then ~Rs = qRs = Rs. Ib ,J If ~ does not meet S, then q contains n, ~Rs is a prime ideal, qRs is primary to ~Rs , ~Rs n R = ~,and qRs n R = q. Proof: If ~ meets S, then q meets S because any power of an elemen t of S is in S. Therefore we have (a). Assume tha t ~ does not meet S. If a is in n, then there is an s in S such that as = 0, hence as -:: q. Since s EE ~, it follows that a E q, hence n C q. Let b be an elementofqRsnR. Then <1>Cb) = <1>(q)!
( q). Since n C q, we have bs E q. Since sEEp, we have b -:: q, which proves that q contains qRs n R; since the converse inclusion is obvious, we have q = qRs n R. As a particular case, where P = q, we h&,ve ~ = ~Rs n R. Next we assume that <1>(ab)!(a)!(b)/<1>(t))" E qRs (for some r), which proves that qRs i;; a primary ideal. Now, applying this to the case where ~ = q, we see that ~Rs is a prime ideal because in that case r can be taken to be 1. Since elements of ~ are nilpotent modulo q, elements of ~Rs are nilpotent modulo qRs . Therefore qRs belongs to ~Rs . Thus the proof is complete. (6.7) COROLLARY: Let ~ be a prime ideal of R. Then mal ideal if and only if ~ is ma..Timal with respect to S. (6.8)
COROLLARY:
~Rs
is a maxi-
If an ideal a of R does not meet S then, aRs "s Rs .
(6.9) THE~REM: Let ql, ... , q" be primary ideals of R. Then n ... n qn)Rs = qlRs n ... n q"R s . If ql ~ i ;"2 qi and if aiRs "s R s , then qlRs ~ i ;"2 qiRs . Proof: Set a = ql n '" n qn. We renumber qi so that qi n Sis empty if and only if i ~ 1'. Since qiR s contains aRs , aRs is contained ill alR 8 n ... n qnRs (= qlR.~ n .. , n q,.R.~). Let <1>(a)/<1>(s) (a E R, .~ ,:: S) be an element of qlRs n ... n q,.R.~ . Since qiR s n R - qi for i ~ r, a is in ql n '" n qr. Take elements S"+I , . . . , Sn of S so that 8"_j is in qr+j' Then a' = as"+I'" snisina.Therefore<1>(a)!(ssr+1 ... Sn) is in aRs , whieh proves the converse inclusion. Thus we see that aRs = q1R s n qrRs. Now we assume that \11 ;R i ;"2 qi and that r ~ 1. Take an element a of ni ;"2 qi which is not in al . Then since qlRs n R = ql, <1>( a) is not in qlR s , which shows that aIRs ;f2 i ;"2 qlRs . 101
n
n
n ...
n
n
17
CHAPTER I
(G.lO) COROLLARY: Assume that the zero ideal of R can be e.rpressed as the intersection of primary ideals ql , ... , qn of R, where qi n S is empty if and only if i :::; r. Then the idwl n( =cp-I(O)) coincides with ql qr. Proof: n = cp-I(O) and therefore n = (O)Rs n R. By our assumption, (O)Rs = q1R s n ... n qrRs and therefore n = ql n ... n qr. (6.11) Let R be a ring and let S be a multiplicatively closed subset of R which does not contain zero. Let S' be a multiplicatively closed subset of Rs which does not contain zero. Let S" be the multiplicatively closed subset of R generated by S and all elements s" of R such that with a suitable element s of s, cp ( s") !cf> ( s) is in S', where cp is the natural homomorphism from R into Rs . Then R s" = (Rs)s' . Proof: Let e and 1r be the natural homomorphisms from R into R s" and from Rs into (Rs)S' , respectively. Then every clement of 1rCP(S") has inverse in (Rs)8' and (Rs)s' is generated by 1r¢(R) and inverses of clements of 1r¢( S"). Therefore there exists a homomorphism from R~" onto (Rs)s, . Let n" be the keruel of 1r¢. Then for every element a of n", there exists an element cp (s" ) / cp (s ) (s E S, s" E 8") of S' slH~h that cp( a )cp( s")!cf>( s) = O. Then as" is in the kernel of cp, whence there exists an element s' of S such that as"s' = o. Since s" s' is in S", a is in the kerncl of e. Therefore e = 1rCP, whence R,s" = (Rs)s'. (6.12) Let R be a ring and let S be a multiplicatively closed subset of R which contains no zero divisor. If a ring R' contains R and is contained in Rs , then Rs = R~. The proof is straightforward, and we omit it. We introduce the following notation: Let R be a ring and let:r be a transcendental element over R. Let S be the set of polynomials f E R[x] whose coefficients generate the unit ideal R in R. Then S is a multiplicatively closed subset of ReS contains no zero divisor because of the following lemma: (6.13) An element L~ aixi(ai E R) is a zero divisor in R[x], if and only if there is an element b ~ 0 of R such that bai = 0 for every i. Proof: The if part is obvious. Assume that L aix i is a zero divisor. j There is a non-zero element L~ bjx such that
n ... n
( L aixi) ( L bjx
j )
=
o.
We prove the existence of b as in the assertion by induction on the j degree s of bjx • If s = 0, it is obvious and we assume that s > o.
L
18
GENERAL COMMUTATIYEJ RINGS
If (L aixi)b s 0, then there is nothing to prove, and we assume i that aib s ~ 0 for some i, which means that aj( bix ) ~ 0 for some j; let t be the largest j such that the non-equality holds. Then
L
,( L~ aixi) ( L
j bjx )
0
=
L
j
and therefore atb. = o. Therefore f = ate bjx ) is different from zero and deg f < s. Furthermore, obviously (L aixi)f = 0, and we prove the assertion by induction. Now since S does not contain any zero divisor, the ring R[x]s contains R. This R[x]s is denoted by R(x). When Xl , ... , Xn are algebraically independent, we can define R(Xl)(X2) ... (Xn). This last ring is denoted by R(Xl' ... ,xn ), or simply by R(x). The following lemma is easily seen (cf. (6.17) below): (6.14) With R and the Xi as above, if R is quasi-local, or local, or quasi-semi-local, or semi-local or Noetherian then so is R (Xl , . . . , x n ), respectively.
As an immediate consequence of (6.13), we have: ( 6.15) If q is a primary ideal belonging to a prime ideal p in R, then in the polynomial ring R[x] = R[Xl, ... , x n], pR[x] is a prime ideal and qR[x] is primary to pR[x]; obviously qR[x] R = q.
n
We note by the way that the following is obvious from the uniqueness of the expression of polynomials. (6.16) For ideals ai, ... , an in R, we have, in the polynomial ring R[x], (a l an)R[x] = aiR [x] anR[x]. As for R(x), we have the following results: (6.17) (1) If a is an ideal of R, then R(x)/aR(x) = (R/a)(x); (2) "if q is a primary ideal with prime divisor p, then pR(x) is prime, qR(x) is primary to pR(x), qR(x) R = q, pR(x) R = p; (3) if ai, ... , an are ideals in R, then (al an)R(x) = aIR(x) anR(x); (4) an ideal m' of R(x) is a maximal ideal of R(x), if and only if there exists a maximal ideal m of R such that m' = mR (x) ; and (5) R < R (x) . Proof: (1), (2), and (3) are obvious by the case of R[x] ((6.16)). The if part of (4) is obvious. Let m' be a maximal ideal of R(x). The set m of coefficients of elements of m' R[x] forms an ideal of R. m ~ R by the construction of R (x). Since m' C mR (x) and since m' is maximal, we see that m' = mR(x) and that m is a maximal ideal of R, which completes the proof of (4). (5) follows from (4).
n ... n
n ... n n
... n
n ... n
n
We add here a remark on tensor product:
n
n
CHAPTER I
19
(6.18) Let S be a mu/tiplicatil1ely closed subset of a ring R such that
o~
S. Then ®R Rs is exact. Proof: Let M be an arbitrary R-module. In the set
P
=
{(m, s) 1m E M, s E S}
we introduce an equivalence relation such that (m, s) is equivalent to (m', s') if and only if there is an s" E S such that s"sm' = s"s'm. The set P' of equivalence classes of P becomes an Rs module by the operations (m, s) (m', s') == (s'm sm', ss'), cf>(a)/cP(s)(m, s') == (am, ss') (cf> and == being the natural homomorphism from R into Rs and the equivalence relation respectively). Then, as in the proof of (6.2), we see that there is a natural homomorphism from P' onto Jl![ ® R~ , and by the universal mapping property of the tensor product, we see that P' is naturally isomorphic to M ® Rs. Thus we may identify P' with IV[ ® R~ . If the same is applied to a submodule N of 11[, then we see that N ® Rs ~ 111 ® R~ , and the assertion is proved.
+
+
EXERCISES: 1. With R and the Xi as in (6.14), prove that R(x" ... ,x") = R[x, , ... ,x,,]s with the set S of polynomials whose coefficients generate R.
2. Let a, , '" ,a" be ideals in a ring R and let S be a multiplicatively closed subset of R such that 0 ~ S. Prove that IllRs
n '" n
Il n Rs
=
(Ill
n ... n Il n )Rs .
7. Prime divisors For a given ideal a of a ring R, let U be the set of elements of R which are not zero divisors modulo a. Then U is multiplicatively closed and does not meet a. A prime ideal ~ is called a maximal prime divisor of a if ~ is a maximal ideal with respect to U and if ~ contains a. A prime ideal q of R is called a prime divisor of a if there is a multiplicatively closed subset S of R which does not meet a such that qRs is a maximal prime divisor of aRs . (7.1) A prime divisor q of a contains a, and all elements of q are zero divisors modulo a. Proof: S being as above, qRs contains aRs by definition. (6.6) shows that q = qRs R, which proves that a c q. Applying (6.1) to Ria, we see that all elements of q are zero divisors modulo a. (7.2) A prime ideal ~ is a maximal prime divisor of a if and only if ~ is a maximal member of the set of prime divisors. Proof: By the definition, a maximal prime divisor is a prime di-
n
20
GENERAL COMMUTATIVE RINGS
visor. Let q be a prime divisor of a. Then q consists of zero divisors modulo a by (7.1), hence q is contained in a maximal prime divisor of a, which proves the assertion. (7.3) Let P be a minimal prime divisor of a. Then aRp n R is primary to p. This aR~ n R is called the primary component of a belonging to p. Proof: Since p is minimal among prime ideals containing a, the same is true for pRp and aR p . Therefore aR p is primary to pRp , ,vhich proves our assertion by (2.12). As an immediate consequence, we have: (7.4) A prime ideal p is a minimal prime divisor of a if and only if it is minimal among prime divisors of a. (7.5) Assume that an ideal a of R is the intersection of primary ideals ql , ... , qn and if the intersection nqi is irredundant, then the set of prime divisors of a coincides with the set of the prime divisors Pi oj the qi . Proof: (6.9) applied to Rp with \l = Pi shows that each Pi is a prime divisor of a. Conversely, assume that p is a prime divisor of a. Then for a mUltiplicatively closed set S which does not meet a, pRs is a maximal prime divisor of aRs = nqiRs . An element a of Rs is a zero divisor modulo aRs if and only if a is in some piRs such that piRs ~ Rs. It follows from this and (2.7) that the maximal prime divisors of aRs are some of piRs. Hfmee (6.6) shows that p = Pi for some i, which completes the proof. (7.6) An ideal is primary 4 and only if it has only one prime divisor. Proof: The only if part is a consequence of (7.5). Conversely, assume that an ideal a has only one prime divisor p. (7.4) shows that p is the unique minimal prime divisor of a, hence p is the radical of a. Assume that b is a zero divisor modulo a. Then a + bR consists of zero divisors modulo a, hence there is a maximal prime divisor of a containing b by (2.2). Hence b E p and b is nilpotent modulo a, which proves that a is primary to p. A prime divisor of a which is not minimal is called an imbedded prime divisor of a. For a prime ideal p, p is the unique minimal prime divisor of pr (for any natural number r). Hence the primary component qr of p" belonging to p is well defined. This qr is called the rth symbolic power of lJ and is denoted by p(,) . (7.7) If Rp is a local ring which may not be Noetherian, then the in-
21
CHAPTER I
tersection n of all the symbolic powers \JCr) of \J is the set of elements x of R such that xs = 0 for some s of R which is not in \J. Namely, n is the kernel of the natural homomorphism from R into Rp . Proof: Since Rp is a local ring which may not be 2'\ oetherian, the intersection of \JrR p is zero. Since pcr) = \JrR p R, the assertion follows. As a corollary, we have: (7.8) If R is a Noetherian integral domain, then the intersection of symbolic powers of a prime ideal is zero.
n
8. Primary decomposition of ideals vVe prove here a theorem on the primary decomposition of graded ideals in a graded Noetherian ring, which includes the usual decomposition theorem as a special case (because any ring is a graded ring with the trivial gradation, which will be defined below). We say that a ring R is a graded ring if R is the direct sum of additive subgroups R o , R r , .,. , R n , '" such that RiRj C Ri+j (then it follows that 1 E Ro). An R-module M is called a graded module if M is the direct sum of Ro-submodules lifo, ... , Jl;I n , • •• such that RiNI j ~ lIf i+j . Elements of Rn or M n are called homogeneous elements of degree n. When we say that R = L Rn is a graded ring or that M = L M n is a graded module, we mean that Rn or M n are as above. A submodule N of a graded module M = L: Mn is called graded if N = L (N M n). This definition can be applied to ideals. A gradation is called trivial if all homogeneous elements of positive degrees are zero. (8.1) A submodule N of a graded module M = L Mn over a graded ring R = L Rn is a graded sub module if and only if N is generated by homogeneous elements. Proof: The only if part is obvious. Assume that N is generated by homogeneous elements fniUni E kIn). Lfn E NUn E Mn) implies that Lfn = Lfij( Lk gkij) for some gkij E Rio, hence
n
which is in N. (8.2) If Ni are graded submodules of M and if Ui are graded ideals of R, then nN i , nUi, L:N i , L:Ui, Ur:U2, Nr:N2' Nr:ur, urNr are all graded. The proof is straightforward and we omit it.
22
GENERAL COMMUTATIVE RI""GS
L:
(S.3) THEOREM: A graded ideal q ( ;;t: R) of a graded ring R = Ri is a primary ideal with prime divisor ~ if and only if the following conditions are satisfied: ~ is the ideal generated by all the homogeneous elements which are nilpotent modulo q and, for homogeneous elements a and b, ab E q, a ~ ~ imply b E q. Proof: The only if part is trivial. We prove the if part. Since, by our assumption, every element of ~ is nilpotent modulo q, it is sufficient to show that (L:! ai) ('[~ bj ) E q, L: ai ~ ~ imply L: bj E q, where ai E R i , bj E R j • 'Ve prove it by double induction on m = t - sand n = v - u. If m = 0, we have 2: asb j E q, as ~ ~. Since q is graded, we have asb j E q, hence b j E q and L b j E q. The case where n = 0 is proved similarly. Now we consider the general case. Since asb" is the (s + u) th degree part of (2: ai) (2: b j) and since q is graded, we have asb" E q. We have (2: ai) (2: asb j ) E q. SInce a,b u == 0 modulo q, 2: asb j == L:+1 a,b j mod. q. Hence by our second induction, applied to ("L! ai) (2::+1 asb j ) E q (which has the same m and one less n), we have 2: asb j E q. If as ~ ~,then we apply the case where t - s = 0 and we have L b j E q. If as E ~,then
2:!+1 ai and
2: b
~ ~
(2:!H ai) ( 2: b j
E q. Hence by our first induction, we have j ) E q. Thus the proof is complete.
(S.4) COROLLARY: If q is a graded primary ideal of a graded ring, then the prime divisor of q is also graded. (8.5) If a graded ideal u of a graded ring R is not primary and if R is ~V oetherian, then there are graded ideals 0 and c such that u = 0 n c, u C 0, and such that u C c. Proof: There are homogeneous elements band c such that bc E u, b ,::: u and such that c n ~ u for any natural number n by virtue of 18.:3). Set bi = u:ciR. Since R is Noetherian, there is an n such that 1)'0-,-1 = bn • Set 0 = u bR and c = u cUR; they are graded because U and c are homogeneous. It is obvious that u c 0, u c c. Therefore it i" sufficient to show that 0 n c c u. Let x be an element of 0 n c. Then.r = a+ by = a' cnz(a, a' E u;y,z E R).cx = ac bcy E u, henee c.r = a'c c n+1z E u, which implies that cnHz E u and
+
+
+
+
z E u:c
Tiereiore cnz E u and x
+
nH
= a'
= bn+1 = bn = u:c n •
+ cnz is in u, which completes the proof.
CHAPTER I
23
(8.6) A graded ideal of a graded Noetherian ring R is the intersection of a finite number of graded primary ideals. Proof: Assume the contrary and let F be the set of graded ideals of R which are not the intersections of any finite nnmber of graded primary ideals. Since R is X oetherian, there is a maximal member a of F. a E F implies that a is not primary, hence there are graded ideals lJ and c such that a c lJ, l1 C c and such that a = lJ n c by (8.5). lJ and c are the intersections of graded primary ideals ql , ... , qm and qm+l , .. , , q" respectively by the maximality of a. Then a = nqi, which is a contradiction. Thus F is empty and the proof is complete. Now we have: (8.7) THEOREM: If a is a graded ideal of a graded Noetherian ring R, then (1) there are only a finite number of prime divisors, say PI , ... , . . . , pr, of a, (2) the Pi are necessarily graded, (3) there are graded primary ideals qi belonging to the Pi such that a = nqi and (4) if Pi is a minimal prime divisor of a, then the qj in (3) is necessarily the primary component of a belonging to Pj . Furthermore (5) a cannot be the intersection of primary ideals of number less than r. The primary decomposition given in this theorem is called a shortest primary decomposition of a.
Proof: (1), (2) and (5) follow from (8.6), (7.5) and (8.4). follows from (8.6) and (2.11). (4) follows from (6.6) and (6.9).
un
(8.8) COROLLARY: A prime ideal P of a Noetherian graded ring R is a prime divisor of a graded ideal a of R zj and only if there is a homogeneous element a of R such that a: aR = p. Proof: Let qi be as in (8.7) for a. Since a:a = n(qi:a), we prove the if part easily by virtue of (2.13). If P = Pi for some i, say 1, theri let b be a homogeneous element of q2 qr which is not in a. Then a:b = ql:b, which is primary to p by (2.13). Let c be a homogeneous element in (ql: b) : P which is not in ql: b. Then (ql: b) : c contains p. Hence (ql:b):c = P by (2.13). Thus, with a = bc, we have a:a = P by (l.2).
n .. , n
(8.9) THEOREM: Two graded ideals a and lJ in a graded ring R coincide with each other if and only if aRm = lJRm for every graded maximal ideal m of R. Proof; The only if part is obvious. Assume the validity of aRm = lJRm . Set c = a: lJ. cRm = aRm: lJRm = Rm , hence c is not contained
24
GENERAL COMMUTATIVE RICi'GS
in any graded maximal ideal of R. Since c is graded by (8.2), we have c = R, and 0 c a. Similarly a C 0, and a = b. (8.9) can be generalized to the case of graded modules by the principle of idealization, hence in particular we have: (8.10) COROLLARY: Let 1lI be a morhde over a ring R. If J11 ® Rm for every maximal ideal m of R, then 111 = o.
=
0
EXERCISES: 1. Generalize the above (8.7) to )Toetherian graded modules in the following way: Let lvl be a module over a ring R. A submodule P r" M of M is called primary if any zero divisor with respect to M/P is nilpotent with respect to M/P. In this case 0: (M/P) (= P:M) is a primary ideal. (Prove it.) The prime divisor of 0: (M /P) is called the associated prime ideal of P. If a submodule N of M is an irredundant intersection of primary submodules P, , ... , P n then the associated prime ideals of the Pi are called associated prime ideals of N. Assume that M = L: Mn is a graded module over a graded ring R = L: Rn . (a) Prove that the ring R EB M given in the principle of idealization is a graded ring with the structure L: (Rn EB Mn) and a submodule N of M is graded if and only if it is a graded ideal of the ring R EB M. (b) Prove that a submodule N of M is a primary submodule with associated prime ideallJ if M r" N and if there is a primary ideal q* of R EB M with prime divisor lJ EB M such that N = q* n M. (c) Then apply (8.7). 2. Prove the converse of (b) above in the following form: If N is a primary submodule of M with associated prime ideal lJ, then with the primary ideal q = N:,'vl, q* = q EB N is a primary ideal belonging to lJ EB ill, and, with this q*, N = q* n M. 3. Show that an irredundant primary decomposition (or the associated prime ideals) of the submodule N of M does not necessarily correspond to an irredundant 'primary decomposition (or the prime divisors) of the ideal (N: M). (Hint: Let q, , q2 be primary ideals of R, sct ill = R/q, EB R/q2 , and consider the decomposition of (0) in M.)
9. The notions oj height and altitude We say that a ring R is of altitude r if there is a chain of prime ideals Pi such that po :J P, :J ... :J pr but there is no such chain with more terms. If there is no such r, we say that R is of infinite altitude. For a prime ideal P of R, the altitude of Rp is called the height of p. Xamely, height p is the maximum of lengths of descending chains of prime ideals which begin with p (length of a chain is defined to be one less than the number of terms of the chain). For an ideal a of R, the minimum of heights of minimal prime diyisors of a is called the height of a; the maximum (or supremum) of
25
CHAPTER I
these heights is called the altitude of a. altitude R/a is called the depth of a. 'iVe shall prove a fundamental theorem of Krull on altitudes of ideals in a :x oetherian ring (( 9 .:3), below). In order to prove it, we need the if part of the following theorem on rings with minimum condition: (9.1) THEOREM: A ring R satisfies the minimum condition for ideals if and only if (a) R is Noetherian and (b) altitude R = 0 (namely, every prime ideal is maximal). (THEOREM OFAxIZUKI) Proof: Assume first that R satisfies the conditions (a) and (b). It follows that any maximal ideal is a minimal prime divisor of the zero ideal, hence we see by (8.7) that there are only a finite number of maximal ideals, say ~I , ••• , ~n, and that the intersection m of the Pi is the radical of O. Hence (:3.15) shows that m is nilpotent. Let l' be such r that m = O. Then R = R/ m", which implies by virtue of (1.4) that R is the direct sum of R/~: . Each ~;-I/~{ for j = 1, '" , l' is a finite module over the field R/~i , hence it has a composition series. Therefore R/~; , and hence also R, have composition series. Thus R satisfies the minimum condition. Conversely, assume that R satisfies the minimum condition. Let ~ be an arbitrary prime ideal of R. R/'p satisfies the minimum condition because R does. If R/p has a non-unit x ~ 0, then {xn(R/~)1 has no minimal member because R/p is an integral domain. Thus R/~ must be a field, which proves that ~ is maximal. Thus altitude R = O. We see also that R has only a finite number of prime ideals; for if lh, '" , ~n, ~n+l, .. , were different maximal ideals, then \\'e should have a descending chain ~I ::::J (PI n ~2) ::::J .•• ~ (~1 ~n) ::::J (~I ~n+1) ::::J ••.. Let the maximal ideals be ~I, ••. , ~n , and let m be the intersection of them. Among the powers of m, there is a minimal member; let it be n = m T • Assume for a moment that n ~ O. We have n 2 = n ~ O. Let b be a minimal ideal among those which are contained in n and whose product with n is different from 0 and set p = 0: bn. Since bn ~ 0, we have ~ ~ R. cd E ~,c tf ~ imply cdbn = 0 and con ~ 0, hence cO = b by the minimality of 0 and therefore don = 0, which implies that d E 'po Thus ~ is a prime ideal and therefore n ~ 'p, which implies that on = on2 ~ fmp = 0, which is a contradiction. Thus n = 0, i.e., m = O. Therefore (1.4) shows that R is the direct sum of R/~~ , each of which satisfies the minimum condition. Hence each ~{-I/~~ satisfies the minimum condition as an R/~i module, hence over the field R/~i. Therefore
n .. , n
n ... n
T
26 \.1~-I/\.1~ has a compo~ition ~E'riE's, and therefore R/\.1: , and R, too, have composition serie:". Henc·e R satisfies the maximum condition, too. We prO\~e ne:;;:t the following lemma of Krull: (9.2) LEt Cl bE a !lon-unit of a Noetherian integral domain R such that a ~ 0. If p is a minimal prime divisor of aR, then height \.1 = l. Proof: Considering Rp , we may assume that \.1 is the unique maximal ideal of R. Let q be a prime ideal of R such that q C \.1. Set ai = q(i: + aR. Since RlaR satisfies the minimum condition by (9.1), we see that there is an n such that ai = an for any i ::::: n. Then we have q,n) C q(i) + aR. By the minimality of \.1, a is not in q. Therefore q(n):aR = q(n). Hence (4.3) implies that q(n) = q(i). Thus i q(i) = q(n). On the other hand, (7.8) implies that i q(i) = 0, hence we have q(n) = 0, and q = 0. Thus height p = 1. Now we come to an important theorem of Krull:
n
n
(9.:3) THEOREM: If an ideal a of a Noetherian ring R is generated by r elements, then, for any minimal prime divisor p of a, height P is not greater than r, i.e., altitude a ::::; r. (ALTITUDE THEOREM OF KRULL) Proof: Let P = Po :::J \.1I:::J .. , :::J ps be a chain of prime ideals Pi . I t is sufficient to show that s ::::; r. If there are prime ideals between P and PI, then adding a maximal member of prime ideals between P and \.11 , we may assume that there is no such prime ideal. Considering Rp instead of R, we may assume that p is the unique maximal ideal of R. Let aI, ... , aT be a basis for a; we may assume that al ~ \.11 ifor, a Q; PI)' Then there is no prime ideal except \.1 which contains DI aIR, which shows that \.11 aIR is primary to p. Therefore there i;; a natural number t such that a; E PI aIR (for all i). We write 0: = alb i Ci with bi E Rand Ci E PI- Set a' = ciR. Let p' be a
+
+
+
+
E;
minimal prime divisor of a' contained in Pl. Since the radical of c' + aIR contains the ai, a' + aiR is primary to p, hence p' + aIR ~:" primary to p, which means that, in the ring RIp', pip' is a minimal prime divisor of a principal ideal, which shows that height pip' = 1 by (9.2). Since p' C Pi, we have p' = \.11' Since a' is generated by . - 1 elements, we have height p' ::::; r - 1, hence s ::::; r. Thus the ,. heorem is proved. [1..1) COROLLARY: The altitude of an ideal of a Noetherian ring is ~\.5
THEOREM: If a is an ideal of a Noetherian ring R and if
CH.-\..PTER I
27
height a = r, then there are elements ai, ... , ar of a such that I::~ aiR is of height s for any s ::; r. Proof: We construct ai inductively. Assume that there are elements ai, ... , a -1 of a such that height (I::i aiR) = t for any t ::; s - 1. Let ~l, • • . , ~n be all of the prime divisors of I::f- l aiR (= 0 if s = 1) such that heights of them are s - 1. If s - 1 < r, then a is not contained in any of the ~i , whence there is an element a. of a which is not contained in any of the ~i by (2.7). Then height I::~ aiR is at least s, whence it is s by (9.3). As a consequence of (9.5), we see that: (9.6) If (R, m) is a local ring of altitude r, then there are r elements ai, ... , ar of m which generate a primary ideal belonging to m but there is no primary ideal belonging to m which is generated by r - 1 elements. The set of such elements Oi as above is called a system of parameters of R. A system of parameters Xl, . . • , Xr of a local ring R is called a regular system of parameters if it generates the maximal ideal of R. A local ring which has regular system of parameters is called a regular local ring. The following are consequences of (9.6). (9.7) Let (R, m) be a local ring. For a given element X of m, there is a system of parameters of R containing X if and only if altitude R/xR = altitude R - 1; each of the follOWing conditions is sufficient: (a) altitude R/xR < altitude R, (b) there is no prime ideal of altitude o which contains x, (c) there is no prime ideal of depth equal to altitude R which contains x. Proof: Let Xl , ••• , Xs be elements of m. Then their residue classes modulo xR generate a primary ideal belonging to m/xR if and only if x, Xl , ••• , Xs generate a primary ideal belonging to m. From this fact, the assertion follows immediately. The same idea proves: (9.8 ) With the same (R, m) as above, if Xl , ... , Xs are elements of m such that height L xiR = s, then 8
altitude
R/I:: xiR
=
altitude R - s.
(9.9) If (R, m) is a local ring, then length R1l11 m/m2 ;:::: altitude R; the equality holds if and only if R is regular. The proof is straightforward by virtue of (5.1). (9.lO) If R is a Noetherian ring and if Xl, . . • , Xn are indeter-
28
GE"ERAL COMMUTATIVE RINGS
minates, then altitude R[XI , ... , xnl = n + altitude R, and altitude R(Xl, ... ,x n ) = altitude R. Proof; It is sufficient to consider the case where n = 1. 'Ve see that altitude R[xll 2: altitude R + 1 by (6.15). Let )TI' be a maximal ideal of R[xd and set m = m' n R. It is sufficient to show that height m' :s: height m + 1. In order to prove it, we may assume that R = Rm. R[xll/mR[:td = (R/m) [xll, which is a principal ideal ring, whence there is an element f of m' such that m' is generated by m and f. Let Yl , . . . , Yr be a system of parameters of R. Then we see that the Y i and f generate a primary ideal to m', whence height m' :s: height m + 1, and the first assertion is proved. Considering the case where 111' = 111R[xl + xR[x], we see that height 111 = height mR[x], whieh proves the second assertion. vVe make a remark, whieh is obvious from the definition of regularity. (9.11) Let R be a local ring. If there are r elements aI, ... , a r of R such that height, LaiR = r and such that R/ L aiR is regular, then R is regular. We note that (9.10) is not true in general if we do not aRSlimc that R is K oetherian. (ef. Seidenberg [1]) We note that though (9.4) implies that the altitude of a semi-local ring is finite, it is not true in general that altitndcR of Noetherian rings are finite; see Example 1 in the Appendix. Let 11'£ be a module over a ring R. Then altitude R/(O:M) iR called the operator altitude of M and is denoted by op. alt if/I. M is called a faithful R-module if O:M = O. Note that the structure of M as an R-module is substantially the same as the structure of 111 as an R/(O:M)-module.
10. Integral dependence Let R be a subring of a ring R'. An element a of R' is said to be integral over R if there are elements Cl , ••. , Cn of R such that an + n 1 Cla + '" + Cn = 0, i.e., a is a root of a monic polynomial over R. We say that R' is integral over R if every clement of R' is integral over R. (10.1) An element a E R' is integral over R if and only if there is a subring R" of R' such that R" is a finite R-module and such that
a E R". Proof: If a is integral over R, then R[al is a finite R-module. Conversely, assume the existence of R" = L~ RUi. aUi = L j aijUj
29
CHAPTER I
with aij E R, hence, denoting by d the determinant I Oija - aij I (Oij being Kronecker 0), we have dUi = 0, hence dR" = and d = 0. Since d is expressed as a monic polynomial in a with coefficients in R, we see that a is integral over R.
°
(10.2) COROLLARY: The set R" of elements of R' which are integral over R forms a ring. Proof: If a, b E R", then R[a, b] is finite over R, hence every element of R[a, b] is integral over R. The ring R" given above (10.2) is called the integral closure of R in R'. If R" = R, we say that R is integrally closed in R'. A ring is said to be integrally closed if it is integrally closed in its total quotient ring. (10.3) COROLLARY. If a E R' is integral over a subring R* and if R* is integral over R, then a is integral over R. Consequently the integral closure R" of R in R' is integrally closed in R'. Proof; Assume that a E R' is integral over R*:a n + clan-I + ... + Cn = with Ci E R*. Then R[cI , ... , Cn , a] is finite over R, whence a is integral over R. When R is a subring of a ring R', R' is an R-module. R:R' is called the conductor of R in R'. (10.4) A ssume that R is a Noetherian ring and that b is an element of a ring containing R'. (1) The conductor of R in R[b] contains elements a and ideals a of R such that ab n E R for any natural number n; ab C a. (2) If the conductor of R in R' contains an element a which is not a zero d~visor in R', then R' is a finite R-module, is integral over R and is contained in the total quotient ring of R. Proof: (1) is obvious, while (2) follows immediately from the fact that R is Koetherian and that R' eRa-I. (10.5) Let a be a non-zero-divisor of a ring R. If a- I is integral over R, then a-I E R. Consequently, if a field is integral over an integral domain I, then I is a field. Proof: There are elements CI , . . . , Cn of R such that (a-I) n n I CI(a-l)n-1 + .,. + Cn = 0, hence a-I = - (CI + C2a + ... + cna - ),
°
+
which is in R. On the other hand, the definition of integral dependence implies immediately the following lemma: (10.6) If a ring R' is integral over its subring R, then (1) for any homomorphism cf> defined on R', cf>(R') is integral over cf>(R) and (2)
30
GENERAL COMMUTATIVE RINGS
for any mUltiplicatively closed subset S of R which does not contain zero, R~ is integral over Rs . Now we have the following important result. THEOREM: Assume that a ring R' is integral over its subring be a prime ideal of R and let S be the complement of P in R. Then a prime ideal p' of R' lies over p if and only if p' is maximal with respect to S. Proof: Assume at first that a prime ideal p' of R' lies over p. Then p' does not meet S, therefore p'R~ is different from R~. Hence pRs C p'R~ Rs c Rs ,which proves that p'R~ Rs = pRs . Hence R~/p'R~ is integral over the field Rs/pR s , which shows that R~/p'Rs is a field, hence p'Rs is maximal, which means that p' is maximal with respect to S. Conversely, assume that p' is maximal with respect to S. Then p'R~ is a maximal ideal and the field R~/p'R.g is integral over RS/(plR~ Rs). (10.5) implies that this last integral domain is a field, which means that p'R~ Rs = pR s , which implies that p' R = p.
(10.7)
R. Let
~
n
n
n
n
n
(10.8) COROLLARY: With the same R, R ' , and p as above, there are prime ideals of R' which lie over p; there is no inclusion relation between any of these prime ideals of R'. (LYING-OVER THEOREM) (10.9) COROLLARY: With the same R, R' as above, if Po c '" c Pc is an ascending chain of prime ideals in R and if a prime ideal p~ of R' lying over Po is given, then there is an ascending chain of prime ideals lchich begins with p~ such thal p~ R = Pi for each i. If, in this case, there is no prime ideal between Pi and Pi+! , then there is no prime ideal between p; and Pi+l . (GOING-UP THEOREM). Proof: The existence is easy by induction on n, while the last assertion follows from (10.8) (the lying-over theorem).
n
p;
(10.10) COROLLARY: If a ring R' is integral over a ring R, then altitude R' = altitude R. This follows from (10.9). _~nother immediate consequence of (10.8) i,,:
'10.11) i.~
COROLLARY:
If a ring R' is integral over a ring R and if a
an ideal of R different from R, then aR' j1 1.
Let R be an integral domain. A ring R' containing R is called an ,"'d'gral extension of R if R' is an integral domain and if R' is integral ,~,l,-er R. If the field of quotients of R' is finite over that of R, we say :kt R' is almost finite over R.
31
CHAPTER I
An integrally closed integral domain is called a normal ring. The integral closure of an integral domain R in its field of quotients is called the derived normal ring of R. When R is a normal ring, an integral extension R' of R is called a Galois extension of R if R' is the integral closure of R in a Galois extension (not necessarily separable) of the field of quotients of R, in the sense of Galois theory: the Galois group of the field extension is called the Galois group of the integral extension. ( 10 .12) THEOREM: Let R be a normal ring and let R' be a Galois extension of R. Then for any prime ideal p of R, the prime ideals of R' which lie over p are conjugate to each other; that is, if prime ideals p~ and p~ lie over p, then there exists an automorphism 0" of R' over R such that p~u = p~. Proof: We consider at first the case where R' is almost finite over
R. Assuming the contrary, let p~, ... , p~ be all of the prime ideals of R' which are conjugate to p~ . Since there is no inclusion relation among the p; by the lying-over theorem «(10.8)), there is an element a of p~ which is not in any of the p~, ... , p~. Then no conjugate of a is in any of the p~, ... , p~ , whence the norm a* of a with respect to R is not in p = p~ R. This is a contradiction because a E p~ implies that a* E p~ R = p. Thus we have proved this case. Let us turn to the general case ..Consider the set F of pairs (S, 0") of Galois extensions S of R contained in R' and an automorphism 0" of S over R such that (p~ Sr = p~ s, for all possible Sand 0". Then we introduce an order in F as follows: (S, 0") ::; (S', 0"') if and only if S C S' and the restriction of 0"' to S coincides with 0". With this order, we see that F is an inductive set, hence there is a maximal member, say (S*, 0"*) of F. It is sufficient to show that S* = R'. Let S" be a Galois extension of R such that S* C S" C R' and such that S" is almost finite over S*. There is an automorphism of S" over R such that its restriction to S* is 0"*; we denote such an automorphism by the same letter 0"*. Then (p~ S"r* and (p~ S") lie S*), hence there is an automorphism over the same prime ideal (p~ 0"" of S" over S* such that (p~ S" = p~ S". By the maximality of S*, we have S" = S*, which implies that S* = R', and the proof is complete.
n
n
n
n
n
n
n r*u"
n
n
(10.13) THEOREM: Let R be a normal ring and let R' be a ring such that (1) R C R', (2) R' is integral over R and (3) no non-zero ele-
32
GENERAL COMMUTATIVE RINGS
ment of R is a zero divisor in R'. If a prime ideal ~~ of R' and a descending chain PI :J P2 :J .. , :J Pr of prime ideals in R are given such that PI = P; n R, then there is a descending chain of prime ideals P; in R' which begins with P; and such that p~ R = Pi; here if pr = 0 alld zj a prime ideal P' of R' which lies over 0 and such that p' C P; is pre-assigned, then there is such a chain with p~ = p'. (GOING-DOW~
n
THEOREM)
Proof: Since
P;
contains zero,
p;
contains a minimal prime divisor
q' of zero by (2.5). Since non-zero elements of R are not zero divisors
in R', q' lies over zero by (7.1). Thus it is sufficient to prove the last assertion. R'/p' is an integral extension of R; if we see the existence of such a chain in R'/p', then, considering the inverse image of the chain, we prove the assertion. Thus we may assume that R' is an integral extension of R. Let R" be a Galois extension of R containing R' and let p~ be a prime ideal of R" which lies over P; . Let q~ C ... C a; be a chain of prime ideals in R" such that q~ R = Pi by the going-up theorem (( 10.9)). Since p~ and q~ lie over the same prime ideal PI of R, there is an automorphism 0" of R" over R such that q~" = p~ . Then, obviously the chain of (q~U R:) is the required one.
n
n
dO.14) THEOREM: With the same Rand R' as in (10.13), let a' an ideal of R' and set a = a' R. Then height a = height a'. Proof : We first consider the case where a' is a prime ideal. For a c-hain a' = p~ :J p; :J ... :J p; of prime ideals in R', we have a chain II = lJ~ R :J p; R :J '" :J p; R by the lying-over theorem 10.8)), whence height a' ::; height a; similarly the converse int"'iuality follows from (10.13), and height a' = height a in this ":i"e. ~ ow we consider the general case. Let p' be a prime divisor of J' ",ueh that height a' = height p'. Since P' R contains a, we have height a ::; height p' n R = height P' = height a'. Conversely, let P ["E' a prime divisor of a such that height a = height p. Let p' be a prillle ideal of R' such that a' C p' and such that p' n R = p; the exi",tenee follows from the lying-over theorem applied to R'I a' and R II with the prime ideal pia. Then we have height a' ::; height p' = height b' n R = height a. Thus~eight a' = height a.
n
i)r,
n
n
n
n
10.15 I THEOREM: Let R be a normal ring and let f(x) be a monic .:'<;:/i'Jmial in an indeterminate x with coefficients in ./1. Set R' = R::: f z R[xl and let d be the discriminant of f(x). If R* {s the integral ~::'!'~ •.:. (~i R' in its total quotient ring Q, then dR* is contained in R'.
33
CHAPTER I
Proof: If d = 0, the assertion is obvious, and we assume that d ;;t' O. Let a be the residue class of x in R'. Let K be the field of quotients of R and let L be a field containing all roots of f( x). We note first that any element a ;;t' 0 of R is not a zero divisor in R', whence K is contained in Q and Q = K[a]. For each root ai of f( x ), there exists an R-homomorphism 1>i from R* into L such that 1>i(a) = ai. Kow let j b be an arbitrary element of R*. Then b = uja (Ui E K, n = degree of f( x ) ). Then 1>,( b) = U ja~ and we regard that these equalities as linear equations in the unknown Uj . The determinant D of the coefficients is IIi<j (ai - aj), hence D2 = d. Since 1>i(b) and ai are integral over R, the dUi are integral over R. Since dUi are in K and since R is a normal ring, we see that dUi are in R, hence db is in R', whence dR* is contained in R'.
L
2:;-1
(10.16) COROLL,\.RY: If R' is an almost finite separable integral extension of a Noetherian normal ring R, then R' is a finite R-module. Proof: Since R' is separable, there is an element a of R' such that R' and R[a] have same field of quotients. Let f be the irreducible monic: polynomial over R which has a as a root and let d be the discriminant of J(x). Because of the separability, d ;;t' O. (10.15) implies that dR' C R[a], whenee R' is a submodule of the finite Rmodnle L:;-I (aijd)R (n = deg f(x)). Since R is Noetherian, it follows that R' is a finite R-modnle. ( 10 .17) A ssume that f (x) is a monic polynomial over an integral domain R. Let the roots of f( x) be Ui (i = 1, ... , r) and let l' (x) be the derivative of f(x). Then the discriminant d of f(x) coincides with (_1)(1+2+'" +r-1) IId'(ui). Proof: f(x) = (x - Ul) .,. (x - u r ). Therefore, setting gi(X) = f(x)j(x - Ui), we have1'(x) = Lgi(X), hencef'(ui) = 9i(Ui) for each i. Therefore 1I1'(ui) = IIgi(Ui) = (_I)(I+2+···+r-1) IIi<j (Ui - Uj)2 = (_1)1+2+'" +r-Id.
By virtue of the above result, the following is a generalization of (10.15) in the case where R' is an integral domain. (10.18) ThEOREM: Let R be a normal ring and let f(x) be a monic polynomial over R. Let a be a root of f(x) (in an integral extension of R). Let l' (x) be the derivative of f (x) and let R* be the derived normal ring of R[a]. Thenf'(a)R* C R[a]. Proof: Let the roots of f(x) be a = 1&1 , U2 , . . . ,Ur and set gi(X) = f(:r) / (x - Ui). Then l' (a) = gl (a). Therefore it is sufficient to prove
34
GENERAL COMMUTATIVE RINGS
the assertion in the case where f(x) is irreducible. If a is inseparable, thenf'(a) = 0 and the assertion is obvious. Therefore we assume that a is separable over R. Let R" be an almost finite separable Galois extension of R containing a and with Galois group G. Let H be the subgroup of G which corresponds to R* and let 0"1 = 1, 0"2, ••• , O"r be elements of G such that a~i = Ui . Then it holds that G = L HO"i . Furthermore gi(X) = f(x)/(x - a Ui ) = g~i(X). Let Ci E R[a] be such that gl(x) = cr-d- I co. Then, for an arbitrary element b of R*, we have bf'(a) = bgl(a) = L bUigi(a) = Li,j b~ic?aj. Since G = L HO"i, Li bUicf' is invariant under any 0" E G, hence is in R. Thus we have bf'(a) E R[a], and the proof is complete.
+ .,. +
EXERCISES: Let R' be a ring and let R be a subring of R'. Let a be an ideal of R. An clement b E R' is said to be integral over a if there are elements a, , '" , an such that ai E ai for each i and such that bn a1b n - 1 an = O. 1. Prove that b E R' is integral over a if and only if there are elements 1/, , .•. , Un of R' such that bUi E L: aUi for any i and such that annihilators of L:R'Ui annihilate some powers of b. 2. Let b be another ideal of R. Prove that if a E R' is integral over a and if b c:: R' is integral over fl, then ab is integral over ab. 3. Define the integral closure a* of a in R' to be the set of all elements of R' which are integral over a. Prove that a* is an ideal of the integral closure R* of R in R'. Prove also that a* is integrally closed in R', namely, that the integral closure of a* in R' is a*. 4. Prove that when R is aN oetherian ring, an ideal b contained in a has the oame integral closure in R (or E') if and only if there is a natural number r ouch that bar = ar+l. 5. Assume that a ring R' is integral over its subring R and that x is an indeterminate. Prove that R' (x) is integral over R (x). (Hint: Let R" be the integral clo~ure of R(x) in R'(x). Use the fact that every maximal ideal of R" lies over il maximal ideal of R (x).)
+
+ .,. +
11. Valuation rings IVe say that a ring R is a valuation ring if R is an integral domain :,uch that for any two elements a, b of R, it holds that either aR C bR or bR CaR. When K is the field of quotients of a valuation ring R, we say that R is a valuation ring of K. ( 11.1) THEOREM: A ring R with a field of quotients K is a valuation ring of K if and only if one of the following conditions is satisfied: I I I Ifa E K,tlvneitheraora-1isinR. ,:2) Every finitely generated ideal of R is principal and R is quasi-
',xol.
CHAPTER I
Proof: (1) is nothing but a restatement of the definition. Assume that R is a valuation ring and let m be the set of non-units of R. Let ai, ... , aT be arbitrary elements of m. Then there is an element as (s ~ r) such that aj E a,R for any j. Hence the ideal generated by these elements is the principal ideal asR which is contained in m, which implies (2). Conversely, if (2) is true, then for any two elements b, c of R, the ideal bR + cR is principal, hence by (5.3), either bR + cR = bR or bR cR = cR, which shows that R is a valuation ring. Thus the theorem is proved. (11.2) Let R be a valuation ring. (1) If a is an ideal of R and if b E R is not in a, then a c bR. If \J is a prime ideal of R, then (2) R/p is a valuation ring and (3) p is set-theoretically equal to pRp . Proof: If a E a, then aR c bR (because bR Q; aR), which implies (1). (2) is therefore immediate from the definition. Let q be any element of pR p • Then there is an element a of R which is not in p and such that p = aq is in p. Since a ~ p, pR c aR by (1), and we have q E R. Since aq E p, we have q E p, which proves (3). (] 1.3) If R is a valuation ring of a field K, then an arbitrary ring R' such that R c R' c K is a valuation ring, and there is a prime ideal p of R such that R' = R~. Proof: The condition (1) in (11.1) is satisfied by R, hence by R', too. Therefore R' is a valuation ring, which implies R' is quasi-local by (11.1). Let p' be the maximal ideal of R' and set p = p' n R. Then R~ ~ R'. Since any element of R' which is not in R is the inverse of a non-unit of R, by (11.1), we see that R' c R~ . Thus R' = R~ . (11.4) Let R be a valuation ring of a field K, let p be the maximal ideal of R and let R* be a valuation ring of the field R/p. Then the set R' = {x I x E R, x modulo p E R*} is a valuation ring of K. R' p = R and R'/p = R* This R' is called the composite of R with R*. Proof: Let a be an arbitrary element of K. If a ~ R, then a-I E p and a-I E R'. Assume that a E R. Since R* is a valuation ring, either a modulo p E R* or a-I modulo p E R*, which shows that either a E R' or a-I E R'. Thus R' is a valuation ring. That R' /p = R* is obvious by the construction. That R'p = R follows from (11.3). (1l.5) Let R be a valuation ring of a field K and let k be a subfield of K. Then R k is a valuation ring of k. The proof is straightforward by (11.1). (1l.6) A valuation ring R is a normal ring.
+
n
36
GENERAL COMMUTATIVE RINGS
Proof: Let a be an element of the derived normal ring of R. If a ~ R, then a-l E R, hence we see that a-I has its inverse in R by (10.5), which is a contradiction to the assumption that a ~ R. Thus a E R, and R is normal. A mapping v from a field K onto a set G* is called an additive valuation, or merely a valuation, if the following conditions are satisfied: (1) G* is the union of a linearly ordered addithTe group G and an element CQ which is defined to be greater than any element of G, (2) v(a) = CQ if and only if a = 0, (:3) v(ab) = v(a) + v(b) (hence, v is a homomorphism from the multiplicative group of non-zero elements of K onto G), (4) v(a + b) ;::: min (v(a), v(b». G is called the value group of v. Two valuations v and v' of a field K are called equiva:lent to each other if there is an isomorphism (J" from the value group G onto G' such that (J"(v(a» = v'(a) for any a (>'" 0) E K. (11.7) Ifvisavaluation,thenv(l) = O,v(a- I ) = -v(a),v(-a) = v(a), and when v(a) < v(b) it follows that v(a b) = v(a). Proof: vel) = v(l·l) = vel) vel), whence v(l) = 0. Hence = v(1) = v(a) v(a- I ), and v(a- I ) = -v(a). = vel) = 2v(-1),andv(-1) = O,hencev(-a) = v(a). v(a) < v(b) implies v(a b) ;::: v(a) = v(a b - b) ;::: min (v(a b), v(b», which b) = v(a). shows that v(a If v is an additive valuation of a field K, then by the definition,
°
+
+
+
+
+
+
°
+
it is obvious that the set R = {x I x E K, vex) ;::: O} forms a ring. The formula v(a- I ) = -v(a) implies that either a or a-I is in R, hence R is a valuation ring. v( a) ;::: v (b) if and only if aR C bR. Thus it is easy to see that the value group is isomorphic to the multiplicative group {aR I a >'" O} with opposite order. This R is called the valuation ring of v. Conversely, assume that R is a valuation ring of a field K. Then the mapping v* such that v*(a) = aR is a homomorphism from the multiplicative group of non-zero elements a of K onto that of the aR. For two elements a, b E K, let c, a', b' be elements of R such that a = a'/c, b = b'/c. Then a'R + b'R = max (a'R, b'R), which shows that aR C bR or aR => bR and that aR + bR = max (aR, bR). Thus we see that {aRI is linearly ordered and that a + b E max (aR, bR). Therefore we see that v* can be modified to be a valuation v (by the opposite order and changing multiplication to addition), and the valuation ring of v is R. v is unique up to equivalence. Thus we have (1l.8) There is a one-one correspondence between valuation rings R
37
CHAPTER I
of a given field K and equivalence classes of additive valuations v of K in such a way that R corresponds to the class of v if and only if R is the valuation ring of v. We prove next the following existence theorem of valuation rings: (11.9) Let R be a subring of a field K and let 0 C \h C \.12 C .. , C \ls be an ascending chain of prime ideals in R. Then there exists a valuation ring V of K· such that V has prime ideals nl, . .. , n. which lie over \l1, . . . , P8 respectively. Proof : We prove the assertion by induction on s. When s = 1: Considering Rp with \l = \ll, we may assume that \ll is the unique maximal ideal of R. Let F be the set of sub rings S of K such that \lIS ;;z:: S and such that R c S. F is an inductive set, and therefore there is a maximal member S* of F. Then S* is quasi-local, for, otherwise, if m is a maximal ideal of S* containing \lIS*, then S: E F, which contradicts to the maximality of S*. Let x be an element of K which is not in S*. x ~ S* implies that \lIS*[X] contains 1, namely, PIX there are elements po, ... ,pn of \lIS* such that 1 Po Pnxn = O. Since S* is quasi-local, 1 + Po is a unit in S*, which means that X-I is integral over S*. Hence lhS*[X-I] does not contain 1 by (10.11). The maximality of S* implies that X - I E S*. Thus S* is a valuation ring, and is obviously the required one, because we assumed that \ll is the unique maximal ideal of R. Now, we assume that such a V, say V' with prime ideals nl , ... , nS-1 , exists for the chain \l1 C \l2 C ... C \l'-I. Considering V~'_l , we may assume that n,_1 is maximal. R* = R/\ls-I is a subring of the field V' /nH and has a prime ideal \ls/ps-I . Hence, by the case where s = 1, there is a valuation ring V* of V' /ns-I such that V* has a prime ideal which lies over \l,/\l8-1 . Then, as is easily seen, the composite of V' with V* is the required valuation ring. (11.10) Let RI , . . , , Rn be valuation rings of the same field K. For a given element a of K, there exists a natural number s such that both a/(l a as-I) and 1/(1 a as-I) are in the intersection D of the rings Ri . Proof: Let \li be the maximal ideal of Ri . We consider an arbitrarily given i (::;; n). If a ~ R i , then by a valuation Vi with the valuation ring R i , 0 = Vi(l) > vi(a) ;:::: vi(a 1 ) = vi(1 a as-I) for any s ;:::: 2, whence is;:::: 2, these elements are in the R, . Consider the case where a E Ri . Since 1 a as-I = (1 a')/(l - a), when a modulo \li is a primitive (ei)th root of 1 with
+ +
+ + ... +
+ ... +
+ + ... + S
-
+ + ... +
+ + .,. +
38
GENERAL COMMUTATIVE RINGS
ei ~ 2, then for s which are prime to ei , these two elements are in the Hi . When a - 1 E ~i, for s which are not multiple of the characteristic of R/~i , these two elements are in the Ri . In the other case, these two elements are in the Ri for any s. By the finiteness of the munber n of the Ri , there is surely an s which satisfies the above requirement for any i, and the assertion is proved.
(11.11) ThEOREM: Let RI , . . . , Rn be valuation rings of the same field K and assume that Ri R j for any (i ~ j). Let ~i be the maximal ideal of Ri , let D be the intersection of the Ri and set qi = ~i D. Then we have (1) an ideal m of D is maximal if and only if m = qi for some i and (2) Ri = D iq • (THEOREM OF INDEPEXDEXCE OF VALUATIOXS) Proof: We prove (2) at first. Let a be an arbitrary element of Ri . Let s be such that both 1/(1 a as-I) and a/(l a as-!) are in D, by virtue of (11.10). Since a E R i , we have 1 a as-I E R i , whence it is a unit in R i . Therefore 1/(1 a as-I) Et qi and therefore a is in D qi • Thus Ri ~ D qi • Since the converse inclusion is obvious, we have (2). (2) implies that q, qj for any (i ~ j). Therefore there is an element ei of nj,ei qj which is not in qi for each i. Let a be an ideal of D which is not contained in any of the qi . Then a contains an element ai which is not in qi , hence a contains aiei. aiei is not in any qi . It follows that aiei is a unit in any Ri , and therefore it is a unit in D, too. Therefore a = D. This implies that any maximal ideal m of D is one of the qi . Since qi Q; qj for any (i ~ j), we see the converse, and the proof is complete. 'iVe add here some remarks on normal rings.
%
n
+ + ... +
.. . +
+ + ...
+ + .. , +
+ + ... +
%
L
L
L
(11.12) THEOREM: An integral domain R is a normal ring if and ollly if R is an intersection of valuation rings of the field of quotients K of R. Proof: If R is the intersection of valuation rings VA , then, since each VA is normal, R is normal. Conversely, assume that R is normal and let b be an element of K which is not in R. Set R' = R[l/b]. If there is a relation (l/b)(L ai(l/b)i) = 1 (ai E R), then we 5ee that b is integral over R, which is not the case. Therefore l/b is not a unit in R', whence there is a valuation ring V of K such that 1b is a non-unit in V and such that R c R' c V by (11.9). Thus we 5ee that b is not in V, whence b is not in the intersection of valuation
CHAPTER I
39
rings of K which contain R. Therefore we see that R is the intersection of all valuation rings V of K such that R C V. (11.13) Assume that R is a normal ring and that b is an element of the field of quotients K of R. Then the kernel n of the R-homomorphism cf>, from R[x] (x being a transcendental element over R) onto R[b] such that cf>(x) = b, is generated by polynomials ex - d such that b = die (c, d E R). i Proof: Assume H:at L:~ a;x E n (ai E R) and let V be an arbii trary valuation ring of K such that R C V. Since L aib = 0, we n 1 have anbnV = (an_1b - + ... + ao) V C bn-1V, which implies that anbV C V, i.e., anb E V. Since V is arbitrary, we have anb E R by (11.12), whence (1;nX - dEn with d = a"b. Therefore we complete the proof by induction on n. EXERCISES: 1.
Willi the same notation as in (11.11), prove that if
0.1 , '"
... , an are ideals of R1 , . " , Rn respectively such that there is no inclusion
relations among the minimal prime divisors of them, then D / n (ai) is the direct sum of the Ri/ai . 2. Prove that a quasi-local integral domain with the field of quotients K is a valuation ring of K if and only if any ring R' such that R c R' k: K contains the inverse of some non-unit of R. 3. Assume that a ring R' dominates a valuation ring R of a field K. Prove that R' n K = R. 4. Let R be a valuation ring of a field K and let K' be an algebraic extension of K. Let R' be the integral closure of R in K'. Prove that valuation rings of K' which dominate R are just rings of quotients of R' with respect to maximal ideals, in the following way: (1) When K' is a finite Galois extension of K; let V be a valuation ring of K' dominating R and let D be the intersection of all the conjugates of V. Prove that D = R'. Then apply (11.11). (2) When K' is finite over K; apply the result in (1) to the smallest Galois extension of K containing K'. (3) Prove the general case, applying (2) to finite subextensions. 5. Let R be a valuation ring of a field K and let K' be an over-field of K. Prove that there is a valuation ring R' of K' such that R < R' and such that the residue class field of R' is algebraic over that of R, in the following way: Let {u~l be a transcendance base of K' over K and set K" = K({u~l). Prove the existence of such a valuation ring of K". Then apply 4 above.
12. Noetherian normal rings Let us begin with the following remark: (12.1) A local ring (R, m) which is not a field is a valuation ring if and only if height m ;:::: 1 and m is principal. In this case, height m = 1. Proof: The only if part is obvious. We prove the if part. height
40
GENERAL COMMUTATIVE RINGS
m = 1 by the altitude theorem of Krull. Let q be a minimal prime divisor of zero and let p be a basis for m. Since p ~ q, we have q:p = q. On the other hand, q cpR, hence q = 0 by (4.3). Thus R is an integral domain. Let a ~ 0 be an ideal of R, and let n be such that a C pnR, a Q; pn+lR. Then a = (a:pn)pn. If a:pn ~ R, then a:pn C pR and a C pn+lR which is not the case. Thus a = pnR. Therefore we see that every non-zero ideal of R is a power of pR, hence R is a ,~aluation ring. Next we have the following lemma: (12.2) Let a, b, c, d be elements of a ring R such that ad = bc. If a is not a zero divisor, then aR: cR C bR: dR. Consequently if both a and b are non-zero divisors, then we have aR:cR = bR:dR. Proof: x E aR: cR implies cx = ay with y E R, hence ayb = bcx = adx. Therefore by = dx, and x E bR: dR. (12.3) Let ~ be a prime ideal of a Noetherian ring R. Assume that Rp is not a valuation ring. Then for any non-zero divisor a E p and for any element b of aR: p, bI a is integral over R and the conductor of R in R[bla] contains ~. Proof: Let p be any arbitrary element of ~. Then bp = ar with an r E R. If r ~ p, then p C aR:bR C pR:rR C ~ (by virtue of (12.2)), which implies that ~ = pR:rR and we see that pRp is generated by p, hence Rp is a valuation ring by (12.1), which is a contradiction. Thus (bla)p C ~, and the assertion is proved by (10.4). ~~s an immediate consequence of (12.3), we have (12.4)
THEOREM:
e12.5)
THEOREM:
A normal local ring of altitude 1 is a valuation ring. Let p be a prime ideal of a Noetherian ring Rand assume that p contains an element a which is not a zero divisor. Then ~ is a prime diviso: of aR if and only if either height ~ = 1 and Rp is a 1'01 nation ring or there exist a, b E R such that bI a is integral over R and such that the conductor of R in R[bla] coincides with p. Proof: Assume at first that ~ is a prime divisor of aR and that Rp i" not a valuation ring. Let b be an element of R such that ~ = aR: bR Iby (8.8)). Then b E aR:p, hence, by (12.3), bla is integral over R and the conductor c of R in R[bla] contains p. If c E c, then cbla E R, i.e., cb EaR, hence c E aR: b = p. Thus the conductor c coincides ,yith p. Conversely assume at first that height p = 1. Then obviously :J is a minimal prime divisor of aR. Assume next that ~ is the conductor of R in R[bla]. Then pRp is the conductor of Rp in Rp[bla].
CHAPTER I
41
Since (bjd)'p C R, we have bp CaR, and 'p C aR:bR, and therefore 'pRp CaRp: bRp ~ R p , which implies that 'pRp = aRp: bRp ; hence 'pR p is a maximal prime divisor of aRp , and the assertion is proved. (12.6) THEOREM: If R is a Noetherian ring and if a prime ideal 'p is a prime divisor of aR with a non-zero-divisor a of R, then for any non-.zero-divisor b contained in 'p, 'p is a prime divisor of bR. Proof: By (8.8), there is an element c of R such that aR: cR = 'po Then, since b E 'p, there is an element d E R such that bc = ad. It follows from (12.2) that aR:cR = bR:dR, hence 'p = bR:dR, which implies that 'p is a prime divisor of bR by virtue of (8.8). (12.7) COROLLARY: Let R be a Noetherian ring and let a, b be elements of R such that a is not a zero divisor, b/a ~ R and such that b/a is integral over R. Then either there is a minimal prime divisor p of aR such that Rp is not a normal ring or there exists an imbedded prime divisor of aR. Proof: Assume that the prime divisors 'pi of aR are all minimal and that Rp are normal for any 'p = 'pi. Let qi be the primary component of aR belonging to 'pi . Since b/a is integral over R, it is integral over R Pi whieh is a normal ring, hence b/a E R Pi , and b E aRpi R = qi. Thus bEn qi = aR, and b/a E R. Thus we have proved the assertioll. (12.8) If R is a normal ring, then any ring of quotients of R is also a normal ring. Proof: If a is integral over R s , then there is an element s E S such that as is integral over R, which proves the assertion.
n
(12.9) THEOREM: A Noetherian integral domain R is a normal ring if and only if the following two conditions are satisfied: (1) If 'p is a prime ideal of height 1 in R, then Rp is a normal ring and (2) if 'p is a prime divisor of a principal ideal ~ 0, then height p = l. Proof: Assume that R is normal. Then (1) holds good by (12.8), while the validity of (2) follows from (12.5). The converse is an immediate eonsequence of (12.7). A Koetherian normal ring of altitude 1 is called a Dedekind domain. (12.4) and (1.4) imply that: (12.10) Any ideal a ~ of a Dedekind domain R is the product of maximal ideals of R, and such an expression is uniquely determined
°
bya. EXERCISES:
1. Let \l be a prime ideal of aN oetherian ring R and assume that
42
GENERAL COMMUTATIVE RINGS
height p;::: 1. Prove that Rp is a valuation ring if and only if there is no primary ideal q such that p(2) C q C p. 2. Prove that a Noetherian ring R is integrally closed if and only if R is the direct sum of finite number of Noetherian rings Ri such that either Ri is a normal ring or every maximal ideal of Ri is a prime divisor of zero (namely, every non-unit of Ri is a zerO divisor). 3. Let R be a Noetherian normal ring. Prove that every principal ideal aR of R is an intersection of symbolic powers of prime divisors of aR. 4. Let R be a Noetherian integral domain. Assume that every ideal of R is a product of prime ideals. Prove that R is either a Dedekind domain or a field. 5. Let R be a Dedekind domain with field of quotients K. Prove that the set of non-zero finite R-submodules of K forms a multiplicative group by the natural multiplication. 6. Assume that M is a finite module over a Noetherian ring R, that a E R is not a zero divisor with respect to M and that p is an associated prime ideal of aM (in M). Prove that if b E p is not a zero divisor with respect to M, then p is an associated prime ideal of bM.
13. Unique factorization rings ~~n element a of a ring R is called irreducible (in R) if a is not a product of any two non-units of R. We say that an integral domain R is a unique factorization ring if (1) every element a ~ 0 of R is the product of a finite number of irreducible elements and (2) if a = al . . . am = bl . . . bn with irreducible elements ai and b j , then Tn = n and there is a permutation 7r of the i such that a"(i)R = b,R. _~n element a ~ 0 of an integral domain R is called a prime element ii aR is a prime ideal of R. It is obvious that a prime element is an irreducible element, but not conversely. :\"ote that the condition (1) above is satisfied by any Noetherian integral domain.
(1:3.1) THEOREM: A Noetherian integral domain R is a unique factorization ring if and only if every prime ideal ~ of height 1 in R is principal. Proof: Assume that R is a unique factorization ring. For any prime ideal P of height 1, let a be an irreducible element of R contained in ~. Let b be an element of aR: p which is not in aR. Assume for a moment that aR ~ ~. Then let c E p such that c ~ aR. Then be EaR, and Thpreiore there is an element d E R such that be = ad. But a is not an irreducible factor of b or c, which is a contradiction. Thus ~ = aR and -hE' '1I!1y if part is proved. Assume conversely that every prime ideal 01 height 1 is principal and let a1 ••• am = bl • . . bn be factoriza-
CHAPTER I
43
tions of an element c as products of irreducible elements ai and bj • We prove the uniqueness by induction on n. \Vhen n = 1, c is irreducible and the assertion is obvious. By assumption, irreducible clements wh;ch are non-units are prime elements. Since al ... am E blR, which is prime, some ai E blR; we may assume that a l E blR. Since aiR is prime, we have aIR = blR, and therefore there is a unit u such that a2 ... am = (u~) ba • . • bn , whence, by induction, we have the proof of the uniqueness. (13.2) If R is a unique factorization ring, then for any multiplicatively closed set S which does not contain 0, Rs is a unique factorization ring. Proof; A prime element of R is either a unit or a prime element of Rs . Furthermore, ideals of Rs are generated by ideals of R, hence the assertion follows from the definition. (13.3) A unique factorization ring R is a normal ring. Proof: Assume that alb (a, b E R) is integral over R: (alb)n + cI(alb)n-1 + ... + Cn = (Ci E R). We may assume that a and b have no common prime factor. Assume that p is a prime factor of b. Then, since an + Clan-Ib + ... + cnb n = 0, we have an E pR and a E pR, which is a contradiction, and alb E R.
°
(1:3.4) COROLLARY; The polynomial ring in a finite n'tlmber of algebraically independent elements over a field is a normal ring. EXERCISES: 1 . Let R be aN oetherian normal ring and let S be a multiplicatively closed subset of R which does not contain O. Assume that (1) there is a natural number e such that q(') is principal, for any prime ideal q of height 1 in R such that q meets S and that (2) there is a natural number f such that p'(f) is principal for any prime ideal p' of height 1 in Rs . Prove that for any prime ideal p of height 1 in R, p(,j) is principal. 2. Assume that p, , ... , pr are prime elements of an integral domain R. Prove that an ideal a of R has prime divisors only among the piR if and only if a is a principal ideal generated by the product of some powers of Pi . 3. Prove that an integral domain R is a unique factorization ring if and only if (1) R satisfies the maximum condition for principal ideals and (2) every prime ideal of R different from zero contains a principal prime ideal different from zero. . 4. Give an example of a normal ring R, which is not a unique factorization ring, in which every prime ideal of height 1 is principal.
14. A normalization theorem When I is a subring of an integral domain R and when K and L are the fields of quotients of I and R respectively, then we know
GENERAL COMMUTATIVE RINGS
the transcendence degree of Lover K, which is defined to be t,he transcendence degree of R over I. (14.1) Let K be a field and let Xl , ... , Xn be algebraically independent elements over K. If YI is an element of K[x] = K[XI, ... , X"] Ichich is not in K, then there are elements Y2 , ... , Yn of K[x] such that (1) Yi = Xi x7 i for some natural numbers mi (i = 2, ... , n) and (2) K[x] is integral over K,[YI , ... , Yn] (and therefore YI , ... , Yn are algebraz'cally independent over K). Furthermore, (3) if a natural number s > 1 is given, all the mi can be chosen to be powers of s. Proof: We write YI as ai111i, where ai E K, ai ~ 0 and the kI; are monomials in the Xi . "\111 e define weights ml = 1, m2 , ... , mn of XI, X2 , ••• , Xn such that one Mi , say MI , has greater weight than
+
1:
the others; let t be a power of s such that t is greater than the degree i d of YI and set mi = t - l for each i. Then, since weight x~ < weight Xi+l for any i = 1, 2, ... , n - 1, we see that the mi satisfies the requirement on the weights (considering a lexicographical order of the monomials Mi)' Set Yi = Xi x7 i for i = 2, .. , , n. Then YI w l ran be written ±alxr + flx + fw where w = weight MI and thefi are polynomials in 1)2, ... ,Yn with coefficients in K. Therefore XI is integral over K[YI , ... , Yn], whence the Xi , which are in K[.Ih , ... , y" , xd, arc integral over K[YI , ... , Yn]. Thus the Yi are the required elements.
+ + ...
(14.2) ThEOREM: Let K be a field and let XI , . . . , Xn be algebraically independent elements over K. If a is an ideal of height r in the polynomial ring K[x] = K[xI, ... , Xn}, then there are elements YI , ... , Yn of K[.r] such that (1) K[x] is integral over K[y] = K[YI, ... , Yn], (2) L1 K[y] is generated by YI , ... , Yr and (3) Yr+j = Xr+j fj with polynomials fj in XI , . . . , Xr with coefficients in the prime integral domain 7r of K for each j = 1, 2, ... , n - r. If K is of characteristic p ~ 0, then the fj can be chos6n such that fj E 7r [xl, ... , x;'J. (1\ OR-
n
+
~L\..LIZATlON THEOREM FOR POLYNOMIAL RINGS)
Proof: We prove the assertion by induction on r. If r = 0, the asis obvious. Assume that I' 2:: 1 and let a' be an ideal of K[x] :ou('h that height a' = r - 1 and such that a' c a. Then, by induction, chere are elements YI, ... , Yr-I, Y~ , ... , y~ of K[x] which satisfy the o::onditions in our assertion with a' instead of a. Since height a = r, we han' by (10.14) that height (a K[YI, ... , Yr-I , y;,J) = ~. (In the other hand, we have Yi E a' k a(1 ~ i ~ r - 1) by con~-:n(·tion. Therefore there is an element Yr of a n K[y;. , ... , y~] ~ertion
n
y;., ... ,
45
CHAPTER I
'which is not zero. Then, applying (14.1) to Yr and K[y~ , . " ,Y~], we see the existence of Yr+l, ... , Yn of K[y~, ... , y~] such that (a) Yr+j = y;.+j + y~mi (if the characteristic p ~ 0, mj can be powers of p) and such that (b) K[y~, ... , y~] is integral over K[Yr, ... ,Yn]. The first condition (a) implies the validity of (3) (and the statement in parentheses implies the validity of the last statement in the theorem), while the condition (b) implies the integral dependence of K[x] over K[y]. Since YI , ... , v,. are in a, (t K[y] contains them. Since Lr yiK[y] is a prime ideal of height r and since height a n K[y] = r, the inclusion implies the equality. Thus the theorem is proved.
n
(14.3) COROLLARY: Let I be an integral domain and let XI, '" , Xn be algebraically independent elements over I. Let K be the field of quotients of I. If a is an ideal of I[x] = I[xI, .,. , Xn] such that a n I = 0, then there are elements YI, ... , Yn of I[x] and an element a( ~O) of I such that (1) I[a-I][x] is integral over I[a-I][y], (2) aI[a-I][x] I[a-I][y] is generated by YI, ... ,Yr with r = height aK[x] and (3) Yr+j = xr+j + fj with polynomials Ii in XI, ... , x,. with coefficients in the prime integral domain for each i = 1, '" , n - r. Proof: Set a' = aK[x], and let YI, ... , Yn be as in (14.2) applied to a' and K[x]. Then (3) is valid for them. Since YI, .,. , Yr are in a', there is an element ai ~ of I such that aiY E a for each i. Since K is a field, aIYI, ... , ary,. are as good as YI, ... , Yr . Therefore we may assume that YI, '" , V,. are in a. Since Xi is integral over K[y], there is an element CiC ~o) of I such that CiXi is integral over I[y] for each i. Let a be the product of all the Ci . Then we see easily that this a and the above Yi are the required elements.
n
°
(14.4) THEOREM: Assume that a ring R is generated by elements JI , . . , , bn over an integral domain I. Assume furthermore that no elenent a( ~o) of I is a zerO divisor in R. Then there are elements ZI , .. , • . • • , Z t of 1r[b l , ... , bn] (where 1r is the prime integral domain of 1) which are algebraically independent over I and an element a( ~o) of I such that R[a- I] is integral over I[a-\ ZI, ... , Zt]. (NORMALIZATION THEOREM FOR FINITELY GENERATED RINGS)
Proof: Let XI, ... , Xn be indeterminates. Then there is a uniquely determined homomorphism c/> from I[x] = I[xI, '" , Xn] onto I[b] such that c/>(Xi) = bi for each i. Let a be the kernel of C/>, and let YI, '" , Yn , and a be elements as in (14.3) applied to a and I[x]. Set Zi = c/>(Yr+i). Since Yi E a for i ~ r, I[a-I][b] is integral over
46
GENERAL COMMUTATIVE RINGS
n
I[a -I][zl. Since a I[Yr+l, ... , Ynl = 0, the Zi are algebraically independent over I. Since Yr+j E 1r[XI' ... , xn], we have Zi E 1r[b l , '" , bn ], and the assertion is proved. (14.5) THEOREM: Let R be an integral domain which is generated by a finite number of elements bl , ' " , bn over a field K. If po, '" , Pt are prime ideals such that Po = 0, PI is maximal, Po C PI C .. , C Pt and such that there is no prime ideal q such that Pi-I C q C Pi for any i, then t must be the transcendence degree of Rover K. Proof : We prove the assertion by induction on t. There are algebraically independent elements ZI, ... , Zu over K (Zi E R) such that R is integral over K[z], by (14.4). If t = 0, then K[z], is a field by (10.5) and the assertion is true. Assume that t > 0. Then apwhence u = plying (14.1) to PI K[z] and K[z], we may assume that ZI E PI. Then PI K[z] = zIK[z] by (10.14). Then, applying the induction assumption to R/rl1 , which is integral over K[zl/zIK[z], and to the prime ideals pi/PI, we see that t - 1 = u - 1, whence t = u, which proves the assertion.
°
n
n
(14.6) COROLLARY: If an integral domain R is finitely generated over a field K, then for any prime ideal P of R, height p depth p is equal to the transcendence degree of Rover K and depth P is equal to the t,.anscendence degree of R/p over K.
+
(14.7) THEOREM: If an integral domain R is generated by n elements a field K, and if m is a maximal ideal of R, then m is generated by n elements and R/m is algebraic over K. Proof: That R/m is algebraic over K follows from (14.6). Let bl , ' " , bn be such that R = K[b l , • . • , br.l. Let Ci be (b i modulo m) for each i and let f;(xi) be the irreducible monic polynomial over K[ci , ... , ci-d which has Ci as a root. Let fi be the monic polynomial in b; with coefficients in K[b l , ' " , bi-l] which is obtained from f: replacing CI, '" , Ci-I , Xi with bl , . . ' , bi- I , bi , respectively. Then we see that R/ ( L fiR) = K[CI' '" , cn ], and therefore m = L fiR, which proves the assertion. Ol'e!'
(14.8)
COROLLARY:
Let
Xl, ... ,
Xn be algebraically independent ele-
mJ?llts over an integral domain I. Let p be a prime ideal of I[x] I~Il'
=
'" , xnl. If IcpnI) is a regular local ring, then I[x]P is a regular local
,:,illg. Proof: We may assume that I = IcpnI) . Let h,
...
,fs be a regular
47
CHAPTER I
n
system of parameters of I. We set q = l-1 I, K = I/q, and Zi = Xi modulo qI[xl. Then, obviously the Zi are algebraically independent over K and I[xl/qI[xl = K[zl. We set ~' = ~/qI[xl, and apply (14.2) to ~' and K[zl; let Yl , ... , Yn be as in (14.2) applied to our case. Then K[zl = K[ZI' ... , Zr , YrH, ... , Ynl. Let L be the field of quotients of K[Yr+l, ... , Ynl. Then ~'L[ZI' ... , zrl is a maximal ideal of L[ZI, ... , zrl, and therefore it is generated by r elements, whence ~'K[z]P, is generated by r elements, which implies that pI[x]P is generated by r + s elements. Since height p' = r, we see that height ~ ~ r + s by (6.15), whence I[xlp is a regular local ring. (14.9) THEOREM: If a ring R is finitely generated over a field K, then the Jacobson radical of R is the radical of R. (HILBERT ZEROPOINTS THEOREM)
Proof: We first assume that R is an integral domain. Let f be an element of R which is not zero and consider R[l/fl. (14.6) implies that maximal ideals of R[l/fllie over maximal ideals of R, which implies that there are maximal ideals which do not contain f. Since f is arbitrary, it follows that the Jacobson radical of R is zero. Now we consider the general case. The above result shows that if ~ is a prime ideal of R, then the intersection of maximal ideals of R which contain ~ coincides with ~, and therefore we get the result. (14.10) Let Xl, ... , Xn be algebraically independent elements over an integral domain I. Then there is a maximal ideal m of I[xl such that m I = 0 if and only if there is an element a( ~O) of I such that I[ a-11 is the field of q11/Jtients of I. Proof: If there is such an element a, then a maximal ideal m of I[x], such that aXl - 1 E m, lies over zero of I. Assume conversely that there is a maximal ideal m of I[xl which lies over zero of I. Then, applying (14.4) to I[xl/m, which is a field, we see that the t in (14.4) must be zero in this case, and I[a -11 is a field by (10.5). Thus the proof is complete.
n
EXERCISE: With the same K, Xi and 11 as in (14.2), aswme that K contains infinitely many elements. Prove that there are elements Yl , ... , Yn of K[xj satisfying (1) and (2) in (14.2) and such that the Yr+i (j = 1,2, ... , n - r) are linear combinations of the Xi with coefficients in K.
CHAPTER II
Completions 15. Formal power series ring Let R be a ring and let Xl, '" , Xr be indeterminates. Let Fd be Lilli module of homogeneous forms of degree d in the Xi with coeffiI',inll('s in R for every d = 0, 1, ... ,n, .. , . The set F of infinite sums L: ai with ai E Fi forms a ring by the obvious operations ( L ai) + (Lb i) = (ai + bi ), (Lai)(Lb i ) = Ln (Li+j~naibj). This /1' is called the formal power series ring or merely the power series ring ill 1,ho Xi with coefficients R, and is denoted by R[[XI, ... , xrll or I'illlply by R[[xll. Elements of R[[x]] are called (formal) power series ill Lhexiwith coefficients in R.For an element at of R[[x]](ai E F i ), Llw Ilumber n, such that an ~ 0 and such that ai = 0 for i < n, is (lnllcd the leading degree of the element, and the an is called the leadin(J form of the element. The leading form of 0 is defined to be O. ao il' (:alled the constant term of the element. Note that R[x] is a sub ring of R[[x]] by the obvious identification "hut a finite sum L~ ai is identified with infinite sum L~ ai with !/'i = 0 for i > r. The above notation will be fixed throughout this section.
L
L
(15.1) THEOREM: There is a one-one correspondence between all ma.Timal ideals m of R and all maximal ideals m* of R[[xll in such a way that m* corresponds to m if and only if m* is generated by m and the Xi . Proof : We see obviously that, when m is a maximal ideal of R, (,he ideal mR[[xll + L xiR[[xll is a maximal ideal of R[[x]]. Let m* be a maximal ideal of R[[xll and let 111 be tho set of constant terms of elements of 111*. Then m is obviously an ideal of R. If we know that lit ~ R, then 111* is contained in mR + xiR[[x]], and we seo that m* = mR[[xll + xiR[[xll and that 111 is maximal. Therefore it is I'ufficient to show that 111 ~ R. Assume the contrary. Then there is ILIl eloment ai of 111* with ao = 1. Set b = - (L~ ai). Then we
L
L
L
[49]
J
~'I;'
()'
L (.,
10 ",IiI'I'I' ('11.1'11 ('" iH til' ~::' h' 1'111' Hii iii . II, 1It'III'(' 1'01' allY . eJ' ,., '~I/,{' I ('\",,(1 I (,,11 'i/l, 2:. n. ,-'IIH'n (~III)( ~II I) ( ))( L.,.II I) I , wn K(~(: Lhal, /:1) O~ 1+ (~.: (LI:I'IIIK 01' dl:gn:<: 1/,»), whil:lt ll?-eans that the nth degree pml, of (~L ai) ( C,.) it:: I or )\1:1'0 uI:I:ol'd(:11.11 ('OIlHidf'I'
lilu:d Lo Ill'
1.\It' II
wllil·1i iCI 1'111111.1
I. Ii tI(·J.!:I·('I· pal'l.
(L/1,i)(LZ
or
L
>
ing as n is zero or not. Thus we see that the element L a,. it:: unit, which contradicts to the assumption that [: ai E m*, and the proof is completed.
(15.2) COROLLARY: An element f E R[[x]] is a unit if and only if the constant term of f is a unit in R. (15.3) THEOREM: If R is Noetherian, then R[[x]] is also Noetherian· Proof: Let a be an arbitrary ideal of R[[x]]. It is sufficient to show that a has a finite basis. Let a* be the set of leading forms of elements of a. Then a* generates an ideal of the polynomial ring R[x], which has a finite basis (over R[x]) , say ii, ... ,ii, consisting of elements of a*. Let fi be an element of a which has f.i as its leading form and let a' be the ideal of R[[xll generated by the f1, '" ,ft. Let g be an arbitrary element of a. By the choice of a', there is an element [!ihi,o of a' which has the same leading form as g, and the same is applied to g - L fih i .O and so on. Thus we see that there is a sequonce of elements Lfih.,.,n(h.i,n E R[[x]]; n = 0, 1,2, ... ) such that (1) Lfihi,n has leading degree greater than that of Lfihi,n-l and (2) Lfihi,n and g - Lj
... ,xrll =
R[[xJll ... [[xrl].
16. An ideal-adic topology Let a be an ideal of a ring R and let M be an R-module. Let F be the family of anM. We introduce a topology on M, which may not be a To-topology, taking F to be a hase of neighborhoods of zero of R, which means that open sets of M me unions of an arbitrary number
1.1
1'11.\ 1"1'11111, II
I'dI' 01' II\(' 1'01'111 /J I 0"111 (I, I 11/). 'I 'Ii i,.., l.0pI,loJ.!:'v iH I~nllt'd 1.111' II IIdi,' lOjlulofJII III' /11. 'I'hnllll.ddil.ioll iH (',oltl,illlloIiK (i.".,.J'(.f, II) .f I· '!f iH a eOllLilillOllH 1'l\\I(',l.ioll 1'1'0111 M X M iliLo M) alld L1\1~ 1l1ldl.iplil~aLioll 01' elenwlIl.K o\' Nil' alKo mllLilillollS (i.e., f(;c, V) = :cy if,; a (:ontinuoliK 01'
fuuetion from R X Minto M, where the topology of R is assumed to be the a-adic topology of R). We retain the above meanings of R, M, and a throughout this section. (16.1) The a-adic topology of M is To if and only if anM = O. If the a-adic topology of M is To , then the following distance function r makes M a metric space: rex, x) = 0; r(x, y) = Tn if and only if x - y E anM and x - y ~ an+1M. The proof is straightforward and we omit it. (16.2) A submodule N of the R-module M is an open set of M in the a-adic topology, if and only if N: lJ;f contains some power (({ a. In that case, N is also a closed set. Proof: If N is an open set, then 0 E N implies anM eN for some n. Conversely, if anM ~ N, then since N is a submodule, b + anM eN for any bEN, and N is the union of the open sets b + lln]j,i, whi(:h proves that N is open. If N i.s open, then the complement M - N of N is the union of :c + N with x E M - N, and each x + N is open, hence jJJI - N is open. Thus N is closed.
n
(16.:3) TUEOHEM: The closure N* of a submodule N of Min M with the a-adic topology coincides with (N + anM). Proof: Since each N + anM is an open set, it is a closed set by (16.2). Hence N* eN + anM for any n, and N* c n(N + anM). Conversely, let x be an arbitrary element of n(N + anM). Then x = bn an with bn E N, an E anM for each n, which shows that, for any n, x + anM meets N. Since the x + anM form a basi.s for the neighborhoods of x, it follows that x is in the closure N* of N. Thus the assertion is proved.
nn
+
(16.4) COROLLARY: For a submodule N of M, the a-adic topology of MIN is To if and only if N is a closed subset in the a-adic topology ofM. Proof: The a-adic topology of MIN is To if and only if n(anM + N)IN = 0
by (16.1), or equivalently, n(anM sertion.
+ N)
=
N, which proves the as-
l'I'~II'ldl:'I'II'NI\
WI! 1\111", Lllltl, IVIII'II N iH n Hlillillodlll(' oi' III, 1111' IHtdi(! I,opology oj' N IlIaY hi! iliITI'I'I'IIi, 1'1'0111 Lhn Lopology or N :1.1' lI. HliI,Hjl:W.I' or III. III'IICI~ Lhn \'ollowill!l: thl!()I'cllI iH I'pally IloLnwol'Lhy:
(lG.5) THEOHEM: If M is a Noetherian module, then for any sul!module N of M, the a-adic topology of N coincides with the topology of N as a subspace of M with the a-adic topology. Proof: By virtue of (3.17), we may assume that R is Noetherian. It is obvious that anN C anM N. The lemma of Artin-Rees (3.7) implies that anM N = an-rearM N) C an-rN, thus we prove the assertion. On the other hand, it should be remarked that: (16.6) Let R* be the ring R EEl M in the principle of idealization. Then the a-adic topology of M coincides with the topology of M as a subspace of R* with (a EEl M) -adic topology (which is equivalent to aR* -adic topology) . The proof is straightforward and we omit it. Now we consider semi-local rings. On a semi-local ring which may not hc Noetherian, we define the Jaeobson-radic:al-adin topology to Iw UII! natuml topology. When lYI is a finite module over a semi-local I'illg N with Jacobson radical m, then the m-adic topology of M iR tll'iilll!d (,0 be the natural topology of M. ThiH definition iH jllHtified by ( IIUi).
n
n
n
THEOREM: Assume that M is a finite module over a semiring R with Jacobson radical nt. Then an arbitrary submodule N (~r M is a closed subspace of M and N = (N + mnM). i'mol': Since MIN is Noetherian, it is a To-space by (4.2), and we pl'Ovn the assertion by (16.3), (16.4) and (16.,5).
(10.7)
{lIl'ILl
nn
( I G.8) THEOREM: Assume that a semi-local ring R' is a finite module mwr a semi-local ring R. Then the topology of R' as a semi-local ring eoincides with that of R' as a finite R-module. If, furthermore, R' conLaing R (whence R' dominates R), then R is a closed subspace of R'. Proof: Let m and m' be the Jacobson radicals of Rand R' respecLivciy. R'/mR' is a finite R-modulc, hence is a finite Rim-module. Tb(~rcforc, there exists a natural number n such that m,n C mR', whieh shows that m,n, C m'R' em", and the firi:3t assertion is proved. The last assertion follows from the first one by virtue of (16.7). BXIJJHClSES:
1. We say that (R, a) is a Zariski ring if R is Noetherian ring
(1111\1"1'1'111, II
1IrIIIOl'illi,,,tI wH11 LI,,, 1\ lI.tlio I,lIpol0Il:Y (I h,.illll: 11,11 itil'lI.l or 10 Hilldl t.l1I1.I, (lVOI'Y id"n.l or U ill II. ... loH"d ,wI., 1','oVIl 1."".1. 11,11 n II.di,\ No"i,""rilt.ll rillll: H iH n /';ll.I'iHki rillll: if nllcl oilly if e'Ve\I''y ,,1,,111(,111, It or "~HII,\" (J, 1 ( a i8 a uuit in U, :1. I'rov" 1."11,1. il' (N, el) iH IL /';miHki I'ill!!: ILllel il' U i8 an iucal of R contained ill n, I,"nll (N, II) iH ILtHO II /';nriHki ring. :'1. (:"lInl'lLli7.(l (10.7) to the case of a Zariski ring R. 1. Adll.pl. (HUl) 1.0 j.he case of a Zariski ring and prove it. Ii. (,,,1, (U, a) be 11 /';ariski ring and let x, , ... , Xr be indeterminates. Prove 11111(, (NII~;II, allUxll + L xiR[[x]]) is a Zariski ring. II. (,uI, M be a module over a ring R with an ideal a. Assume that the a-adic I,,,polo!!:y of M is To. Set b = (a + (O:M))/(O:M). Prove that (1) the a-adic 1,lIpolo!!:y of M coincides with the b-adic topology of M and (2) the b-adic I,opolo!!:y of R/(O:M) is To. 7. Let M be a module over a ring R with an ideal a. Let N be a submodule "I' WI which is closed in the a-adic topology of M. Prove that N:M is closed in Lito (\-adic topology of R.
""II,I.
17. Completions This section is concerned with completions of semi-local rings and wit.h completions of finite modules over semi-local rings. But we begill with a case a bit more general so that the readers can see some g(~lIeral facts in the case of semi-local rings which may not be NoethnI'ian. Let R be a ring and let M be an R-module. Assume that M is a lIIotric space with a distance function rex, y) such that (1) rex, y) = '/'(:c - y, 0) for any x, y E M, (2) for any positive real number E, Lho set U, of x such that rex, 0) < E forms an R-submodule of M. We note the following fact which follows immediately from the above condition. (17.1) If rCa, b) > rCc, d), then rea, b) = rea c, b d). With this metric, we can discuss completions as usual. Namely: A ~equence (c n ) of elements cn(n = 1,2, ... ) of]}f is called a Cauchy ,WXjuence if for any given positive number E, there is a natural number N such that, for any m and n which are greater than N, r(c m ,cn ) < E. An element a i~ said to bo a limit of a Cauchy sequence {Cit} if
+
+
if such an a exists for a given (en), then a i~ unique and is denoted by lim Cn • M is said to be complete if every Cauchy sequence in M has a limit in M. An R-module M*, having the following properties, is naIled a completion of M: (1) M* is a metric space with a distance
(II 1M 1'1.1.\'1'(( INK
fllllnl.iolll'+ 1'1111,11 Lim!' I"~(fl', 1/+) 'f'+(,I'~' I/'~, 0) 1'01' "'lIy ,I"f, 1/>1< ( M>I< (:.l) 11/* iH I'.olllplnl.!', (:l) III iN II, tlnllHn HI I hHPH("l'. of 111*, HIHI Cl) if ,r* nlld HI'(, lillliLH of (:nlll'.hy H('00'1'(:1:,. , !in). WI) !!;ivc one mor'n definition couecrning Cauehy scquonees: A
,,*
(:a\l(,hy sequence {Cn} is called a regular sequence if r(cj ,cn ) < !n for allY n and for any j > n. Then as a general fact in metric spaces, we have (17.2) For any Cauchy sequence {Cn} in M, there is a regular sequence {dn} such that lim r(c n , d n ) = O. N ow we prove the existence of completions:
(17.3) THEOREM: M has a completion M* which is unique 11,p to isomorphisms (with regards to topologies). The set of elements x* of M* such that r*(x*, 0) < f fonns an R-submodule of M* and is the closure of U, in M*, M* /U; is naturally isomorphic to M /U, . If M is a ring and if the U, are ideals, then M* is naturally a ring whose multiplication is such that (lim c:) (lim d:) = lim(c!d:) for anJ! Cauchy sequences {c!} and {d!} in M*, and the are ideals of M*. Proof: Let C and C* be the set of Cauchy sequences and regular sequences in M respectively. In C, we introduce the following operations: {en} {c~l = {Cn c~}, a{cn} = {acn}(a E R), and when M is a ring (as above), {cnl{ c~} = {CnC~}, Thus C becomes an R-module; when M is a ring, C becomes a ring. Let n be the set of Cauchy sequences which have zero as limits. Then n is a submodule of C; when M is a ring, n is an ideal. We shall show that C/n is a completion of M, identifying an element a of M with the class of Cauchy sequences which have a as the limit. We define a function r*(x*, y*) (x*, y* E C In) to be limn~oor(:rn, y,,) with representatives {x,,}, {Yn} of x*, y*. By virtue of (17.1), we see that r*(x*, y*) = r(xn, Yn) for suffieiently large n except for the case where x* = y*. With this r*, C /n becomes a metric space as is easily seen. Furthermore, if :c* is a class in C/n containing {c n }, then x* = lim Cn in this topology. Thus M is dense in C/n. Let {x:} be a Cauehy seqnence in C/n. Let {x n ;} be regular sequences in M such that = lim Xn; • Then by the regularity of the {Xni}, the sequence {Xnn} becomes a Cauchy sequence, hence has a limit x* in C In. Then we see that x* is a limit of the sequence {x!}. Thus C/n is a completion of M. The uniqueness of M* is rather obvious, because elements of M* must be in one-one correspondence
U:
U:
+
+
x:
I II I ;11"1'11111, II
wiLli
!,II'III(1l1irl
1>1'1'01111' I''''H.\'
oj'
"'lid
('/11.
Till' oLlII'" "'H,'iI'i·iioll.'I 1.111' d!~LaiIN.
ill 0111'
LI 1l'1I !'! 'II I
Ilav!' III1IY
\1'1' ollli!.
'l'III':OIII'IM: ,,1 88 If, II 1.1: thal (1":8 11.1/, '£deaf OJ' H and that the metric III /8 /1£1 1('11 b!f thl' n-ruJ-£c topology (by ( I G.1)). If a has a finite basis, Ihl'lI Ihl' tnl!11fnu.IJ I~r tlw (;nml!felion M* is the a-adic topology. l'l'oo\': II. i:-; HufficiPllt, by virtue of (17.:3), to show that the closure N~ oj' n"M ill M* is a"M*. Let x* be an arbitrary element of N!. 'I'bnll :1:* = lim :r;i with (Xi) such that Xl E a"M and such that Xi n .1"; II ( (1'11 i'iJYI = a ( aiM). Let ai, ... , at be a basis for an. Then ,I'j .- :1:j'H = Lj ajb ji with b ji E aiM. For each j, the series Li bji iH I:()II vergent (i.e., { L~~l bj;} is a Cauchy sequence) and express an l'lnnwnt of M*. Then we see that x* = L ajb;, which is in a"M*. 'l'huH N! C anM*. Since the converse inclusion is obvious, the proof
(,,'.,1)
I~r
b;
iH (\omplete. (17.5) THEOREM: Assume that the a-adic topology oj' R is To and that a has a finite basis ai, ... , aT . Let Xl, .. . , Xr be indeterminates and consider the formal power series ring R[[x]]. Then the completion U* of R is (isomorphic to) the ring R[[xll/n*, where 11* is the closure oj' tlu! ideal n = L (x,; - ai)R[[x]] in R[[xll with the (n aR[[x]])-adic topology (= the (aR[[x]] LXi R[[x]])-adic topology). Proof: We consider the map cf> from R[[x]] into R* such that if J = Lfi(x) E R[[x]] (fieX) being homogeneous form of degree i) I.Illm cf>(f) = Lfi(a). Then we see that cf> is a homomorphism from UI[x]l onto R* and that the kernel n' of cf> contains x - a, hence n. f = L!i(X) is in 11' if and only if Lf;(a) = O. Let f,,(x) be the Imding form of f. Then Lf;(a) = 0 implies that fuCa) E a"H, and Lh ere is a homogeneous form hUH (x) of degree u 1 such that .fAa) = hUH(a). f - f,,(x) huH(x) has leading degree greater Lhan u and f == f - fu(x) hU+JCx) modulo 11. Thus, repeating the Hame, we see that if fEn', then fEn (L xiR[[x]]) " for any n. Conversely, if f E n( n L xiR[[x]]) "), then we see that L fie a) E an = O. Thus we have n' = nn (n + (L xiR[[x]])n) = nn (n + n"R[[x]]) = n*.
+
+
+
n"
+(
+
+
+
(17.6) COROLLARY: Assume that R is a semi-local ring with J acobNon radical m and assume that elements ai, ... , aT generate an ideal which has 111 as its radical. Then the completion R* of R is a semi-local ring and is (isomorphic to) the ring
1111~11'1.lli'I'III~n\
NII.I'I, .. , , .1',11/( ~
(.1',
where the Xi are indetc'I'm'inllJi:8. '/'/11' IO/lo{o(1!I ol /tl. 118 1/1.1' t:IIllIllll'lio/l. IJ' /?, coincides with that of /t* as a semi-fo('(d 1'/11(1. I'roof: Set a = L aiR. Then since tlw rail il'al or n iK Ill, Lhe n-adie Lopology is equivalent to the m-adic topology. TIII,rdo\'(, WI) eall apply (17.4) and (17 ..5). Since R[[x]] is a semi-local ring by (L5.4), every idnal of R[[x]] is closed, whence we have R* = R[[x]l/(L (Xi lI,i)R[[xlJ) by (17.5). The coincidence of topology follows from (17.4).
Another important fact concerning the completions of semi-local dngs is: (17.7) THEOREM: Let (R, ~1, ••• , ~r) be a semi-local ring. Then the completion R* of R is the direct sum of completions Ri of local rings Ri = Rpi. Proof: Let m be the Jacobson radical of R, i.e., m = i ~i • Then, for each natural number n, Rim" is the direct sum of R/~~ and there n are ei.n E R such that 1 - Li ei,n E m , ei,n E ~j if i ~ j, hence 1 - ei,n E ~~ . Then for each i, ei,n - ei,n+! E ~~ ll~ = mn and therefore the sequence {ei,,,l has a limit ei in R*. Since
n
n .,. n
.n,
we have Lei = 1. Since ei,nej,n E mn (i ~ we have eiej = 0 if i ~ j. Hence we have e~ = ei. Thus R* is the direct sum of ideals R*ei, which are rings with identities ei. We shall prove now that R*ei is (isomorphic to) the completion of Ri . Let {c n} be a Cauchy sequence in Ri such that Cn - Cn+l E ll~ Ri . Since R/~~ = Ri/~~ Ri for any n, there is a sequence {c~} of elements c~ of R such that (c n modulo ~~Ri) = (c~ modulo ~n for every n. Then the sequence ' } 'IS a C auc h y sequence In . R and l'1m Cnei,n I l'1m Cnei,n I 2 {cnei,n = = ei(lim C~ei,n)' Hence the limit of {c~ei,nl is in R*ei . It is easy to see that the map lim Cn --'> lim C~ei,n gives a one-one correspondence between the completion Ri and R*ei. Then we see easily that is isomorphic to R*ei . Now we go back to the completions of modules:
R.;
( 17.8) THEOREM: A ssume that R is a semi-local ring and that M is a finite R-module. Let R* be the completion of R. Then M ® R R* is the completion of M, whose topology as a finite R*-module coincides with its topology as the completion of M.
('II;\I"I"'iI(, II l'I'IIl,I': 1,1,1.1111,1' UII'.JHI·(tI,~iI)lll'lI,dil'lI.llIr N, '1'111'11 IIIN'" is UII!.l:I.(~lIh fl I'ntiil'allll' HI' I,y (17,(j). 1"IIl'LiI('I'IIIIlI'n N*/III"N* ~~ N/llt l{ Now, 11 (III (,"'! N*)/IIII/(M C9 U*) = (M /1II J111) ® (/l*/m"R*) = M/mnM. Hill(:I~ III" III = 0, WI) s(~() that M ® R* contains M (by the identifica!.illil 'In 'fn ® I for any m EM), and furthermore that mn(M ® Hilil
n
N*) n M = mnM. This shows that M is a subspace of,M ® R*. An 1I.1'i>iLra,I'y clement c* of M ® R* can be expressed as L mi ® wiLh mi E M and r; = limn n,n with ri,n E R. Then c* is the limit of I L miri,n} as is easily seen. Thus M is dense in M ® R*. Let {c!} be II. mgular sequence in M ® R*. Let Ul, . , . , Ut be a basis for M. Since (M ® R*)/mn(M ® R*) = M/mnM, there is a sequence {c n} in M Hlldl that c! - Cn E mn(M ® R*). cn+! - Cn is in mnM, hence is expnlHHed as L Uimin(min E mn). For each i, L min has limit r; in U*. Furthermore, we see that ~ Ui ® is the limit of the sequence I (:';}. Thus M ® R* is complete.
r;
r;
(17.9)
COROLLARY:
Let a be an ideal of a semi-local ring R. Let R*
III: the completion of R. Then the completion of a is aR* and aR* is iso-
n
'/lwrphic to a ® R*. Furthermore aR* R = a, and R* / aR* is the completion of R/a. Proof: The first assertion follows from (16.5), hence the second HHHertion follows from (17.8), Since a is a closed set by (16.7) and :-;ince aR* is its closure in R*, we see that aR* R = a. The completion of Ria is (Ria) ® R*, which is obviously R*laR*. Similarly, we have for modules:
n
(17.10) COROLLARY: Let N be a submodule of a finite module M IIvcr a semi-local ring R. Then the closure N* of N in the completion M* (~r M is the completion of N and we have N* M = N. Furthermore, M* I N* is the completion of MIN. If we identify M* with M ® R*, then N* is surely identified with N ® R* in M ® R* as is easily seen. That N* is the completion of N implies that the usual tensor product N ® R* is isomorphic to the Hubmodule N ® R* in M ® R*. Hence we have
n
(17.11) COROLLARY: If R is a semi-local ring, then ®n R* is exact. (17.12) Let (R, lh, ... , prj be a semi-local ring and let R* be the completions of R. Then we have altitude R* = altit.ude R. Proof: altitude R = max {altitude R~J. On the other hand, R* is Lhe dircet sum of completions of R~i . Therefore it is sufficient to prove the assertion in the case where R is a local ring with maximal
.'I'MI'I,I'I'I'I.INII l ill. '11,1 III ).11. Wn PI'O\'I' Ih .. IIHlll'l'Lillll II,\' illdlH'lioll Oil II.ll,il.lldn It '. l II' nll,il,ll 0, IPI. ,I'I , ... , ./:, I)(~ a .'1,VHI.PIII of pal'II,III('I.I'I·.'1 of N. Tilnll U\!~ .r; g;clwl'a(,p all ideal whi.~h iH primary Co lIlU* ill U*. 'I'hel'dol'e we H(\(~ iJmt altitude R* :::; altitude R and that thero iH un element Xi , say
a
such that R* jxlR* has altitude less than altitude R*, whence altitude R*/X1R* = altitude R* - 1 by (9.7). Since R*/x 1R* is the eompletion of R/x!R, we see that altitude R*/x!R* = altitude R/xJR by our induction assumption. Thus we have altitude R* = altitude R*/XlR* + 1 = altitude R/XlR + 1 2:: altitude R, whence we have altitude R* = altitude R. :1:1,
(17.13) COROLLARY: Let R be a local ring and let R* be the completion of R. Then any system of parameters of R is a system of parameters of R*. EXERCISES: 1. Prove that the completion of a Zariski ring is a Zal'iski ring. 2. Confirm that (17.9), (17.10) and (17.11) can be generalized to the case of Zariski rings without any substantial change of proofs. 3. Let R be a ring with an ideal a and let M be an R-module. Assume that the Il-adic topologies of M and R are To , hence metric. Prove that the completion M* is naturally a module over the completion R* of R. Prove furthermore that if a hasa finite basis, then the topology of M* is the aR*-adic topology.
18. Exact tensor products We saw in Section 17 that if R* is the completion of a semi-local ring R, then ® R R* is exact. Therefore, we prove here some general properties of exact tensor products, which can be applied to the case of semi-local rings immediately. (IS.1) THEOREM: Let R be a ring and let R* be a ring which is an R-module, and such that ® R R* is exact. Let M be an R-module. Then we have the following, where Ni are sub modules of M and a is an ideal of R: (1) (N! N r ) ® R* = (N l ® R*) (Nr ® R*). (2) (N!:N 2) ® R* = ((Nl ® R*): (N2 ® R*)), provided that N2 has a finite basis. (3) (N!:a) ® R* = (N! ® R*) :aR*, provided that a has a finite basis. (4) If an element a of R is not a zero divisor with respect to Nl , then a is not a zero divisor with respect to N! ® R*.
n ... n
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lill('(ll'll! inr/IIJllllu/(lnl II/WI' N*. 1"'1101': III 111'£1('" 1,11 provn ( I ), wn haVil oilly 1,0 pt'()v(~ the ease where I' ~. 1,"'0111 (N I 1- NJ/N, =~ N 2/(N l N 2 ), we have (N 2 ® R*)/ (N, n N,,) ® N* , ((NI + N 2 ) ® R*)/(N l ® R*) = (N2 ® R*)/ ((N I ® W) (N 2 ® R*)), which proves (1). Since N2 has a finite hHNiN, ill 0 nkr to prove (2), we may assume that N 2 = bR with an (l!nIlWIlL (, C N 2 , by virtue of (1) above and (1.2). The map cf> delillnd by cf>(:c modulo Nl:bR) = (bx modulo Nl bR) is an isomor"biellil from R/(Nl:bR) onto bR/(N l bR) by (1.5). Therefore h~*/(Nl:{)R)R*
/1" (x)
n
n
n
n
n
((N I ® R*)
n (b-®
R*))
(/1* is defined similarly ascf>. Hence we have (2). (3) is proved similarly /l,N (:2), using M instead of R in the above proof. That a is not a zero divisor with respect to NI implies [0:aR]N 1 = 0, hence by(3) applied Lo Nl = M we have [0:aR*]N1 ® R* = 0, which proves (4). As for (fi), Ui ® 1 in (L uiR) ® R* are linearly independent, he nee the n:mdness implies (5). (18.2) COROLLARY: Let M be a finite module over a semi-local ring N and let M * and R* be the completions of M and R respectively. Then 'lOr: have op. altitude M* = op. altitude M. Proof: O:M* = (O:M)R* by (2) in (18.1), and R*/(O:M*) is the eompletion of R/ (0 :M) by (17.9). Therefore we have the assertion hy (17.12).
(18.3) THEOREM: With the same Rand R* as in (18.1), assume furthermore that there is no ideal a of R such that a ~ Rand aR* = R*. Then R C R* and, for any ideal a of R, it holds that aR* R = a.
n
Proof: Let a and IJ be ideals of R. Assume that IJ has a finite basis. If IJR* C aR* then R* = aR*:IJR* = (a:IJ)R*, whence IJ Ca. Applying this to a = 0 and IJ = bR with b EO: R* (as an R-module), we have IJ = 0 and R C R*. Applying the same to IJ C aR* R, we have aR* R C a, whence aR* R = a.
n
n
n
(18.4) With the same Rand R* as in (18.1), assume furthermore that ((, C R* and that aR* R = a for any ideal a of R. Then (1) the total quotient ring K of R is naturally a subring of the total quotient ring [(* (If R*, and (2) for any ideal a of R, K aR* = a, hence in particular J( R* = R.
n
n
n
110
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R = va, and we have (2).
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(lS.5) With the same R, R* and M as in (VU), a.wu'ffW .r1J,/'lI/'(~I'IIWI'(! the same assumptions as in (lS.4) above. Then the mal) ¢ 8'11t:h that ¢(m) = m ® 1 (m E M) givesanisomorphismfromM into M ® U*. Proof: For a fixed mE M, mr _mr ® 1 (r E R) gives an isomorphism from mR into mR ® R*, hence the exactness proves the assertion. N ext we give some sufficient conditions for a tensor product to be exact. (lS.G) Let R be a ring and let R* be a ring which is an R-module. If, for any ideal a of R, the natural map from II ® R' onto aR* is an isomorphism, then ® R R* is exact. Proof: Let M be a finite R-module and let N be a finite R-module contained in M. We shall prove that N ® R* c M ® R*. Let u l , . . . , U r be a basis for M, and we set Ni = (L~--I ujR) + N. If N i - I ® R* C Ni ® R* for any i, then tho assertion is proved. Thus we may assume that UI, . , . , Ur-I E N. Let F be a free Rmodule with free base U 1 , •• , , U r , and let ¢ be the homomorphism from F onto M such that ¢( U i ) = Ui. Set G = L~-l RUi , H = cp-l(N) , K = cp--l(O). Then, with a = N:M = N:urR, H = G EEl aUr • Therefore H ® R* = (G ® R*) EEl (aU r ® R*), F ® R* = (G ® R*) EEl (UrR ® R*). Since, by our assumption, aUr ® R* C UrR ® R*, we have H ® R* C F ® R*. Hence K ® R* in H ® R* coincides with K ® R* in F ® R*, which proves that N ® R* = (H ® R*)j (K ® R*) C (F ® R*)j(K ® R*) = M ® R*. Thus we complete the proof. (IS.7) THEOREM: Assume that R and R* are Noetherian rings such that R* is an R-module. Let ¢ be the homomorphism from R into R* such that ¢ ( a) = a· 1 (in R*). Let ill?:* be the set of maximal ideals of R* and let ill?: be the set of prime ideals m of R such that m = ¢-I( m*) with m* E ill?:*. Then ® R R* is exact if and only if the following is true: If q is a primary ideal with prime divisor m E ill?: and if b is an element of R such that q: bR = m, then qR*: bR* = mR*. Proof: The only if part is obvious by (IS.I). We assume that the condition holds good. We remark first that the condition is carried
01
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llilpoCnllL. I II ol'(kl' to prOY(, ChaC II. 0, WII IlIa.v tJmt NiH an iiI"al and Lhal; 111 =cc N by (Is.n). WI' lIKfI 1.111' iii Ihlfll.ioll olllenp.;Lh R. J[' blgth U '''' I, UWIi U iH a lildd, allli 1.\111 aHcH'I' Lioll iA obvious. AHHurnillp.; that lobp.;Llt R > I, IPI, b hI' all 1111'1111'111. or N Hileh that length bR = J, i.e., O:bR = 111. \{y Olll' illd\l(',Lioli a,H:-;\i1I1P /.iOIl, ®RlbR (R*/bR*) is exaet, which mCl1IlH that i,hl, naLul'al III:1p I'l'f)tn (N/bR) ® (R*/bR) into R*/bR* is an iflomorpbiHIII. '1'111'1'1'1'01'1' I{ iH contained in b ® R*. Since 0 :bR* = mIt*, we }mvil LhaL l)ol.Ii II ® R* and bR* are faithful R*/mR*-modules, hcw,!) J( = O. 'I'I\II.'-i lVI, hnve settled this case. (2) Now we prove the genoml (:aH(I. By (I) I\,l>0ve, we see that ®R/mn R*/mnR* is exact for any n, whidl iIIlPlil'~{ Chat (N/(mnM N)) ® (R*/mnR*) is naturally cOlltaill(\d ill 11,(' !irHL that 111 iN tl.HHillIl
n
(M/mnM) ® (R*/mnR*). n
n
Therefore we see that K is contained in (m M N) ® /(,* ill N (~ HI. Hince mnM N c mn~rN for an r and for any n ::::: r by the 1101111)\:1, oj' Artin-Rees, we see that K c nn mn(N ® R*), which mw.;/; lin :1,1'1'11 because R* is a local ring. Thus K = 0, which completes the PI·I)O/'.
n
(18.8) COROLLARY: Assume that a Noetherian ring R is dorn:ina/('d liy a Noetherian ring R* and that for any maximal ideal m (~r N till' 'ideal mR* is a maximal ideal of R*. Then ® R R* is exact if and oniil If qR* R = q Jor any primary ideal q of R such that the radical III q is a maximal ideal. In particular, ~r a semi-local ring R is a dense subspace oJ a ,w:llli local r'ing R', then ® Tl R' is exact. We notc that thi:-; (18.8) gives another proof of (17.11). }I'lli'Ll II 'I' more, as an immediate eonsequence of (18.8), we have tho f()lI()w~
n
ing corollary hy virtue of (G.l7).
(18.9)
COROLI,AHY:
Let
Xl,
..• , Xn
lie algebraically independ,I'IIJ
I !()MI'I,III'I'II INII
clc'f/wnts liI'I"/' 1/ N (}I 'Ihl ,/,11/11 /'111(/ H. "'!tl'll (.)),' N( ,I'I iN 1',1'1/1'1. ( 'f. II:,~I'I' cise 2 below). (18.10) Let Rand U* 1)(: 'l'inus slU:h thllJ h~" 'i,~ lUi '1~-I/I(}111/11' I/lid slt('h that ® R R* is exact. Let cf> be the nat'uTal !J,o'/l/'(l'llIo'l'jlh£sln Jl'lrill NI:/lto R* (cf>(r) = '1'·1 with 1 E R*). Let a be an ideal I({ U and lel 8 and 8* be multiplicatively closed subsets of Rand R* respectively 8uch that cf>(S) C S* and such that S a = S* aR* = empty. Set R' = U"/llU 8 and R" = R~* I aR~* . Then ® R' R" is exact. Proof: If M is an Ria-module, then M is an R-module and
n
M
®Rln
(R*laR*)
n
=
M
®R
R*.
Therefore we see easily that ® Rln R* I aR* is exact. Hence, we may assume that a = O. Since ®R* R" is exact by (6.18), we see that ®R R" is exact. If M is an R'-module, then M is an R-module and M ® R' R" = M ® R R", and therefore the assert.ion is proved easily. (18.11) THEOREM: Let Rand R* be Noetherian rings such that R* is an R-module and such that ® R R* is exact. Let a be an ideal of Rand let p* be a prime ideal of R*. Then p* is a p'l'ime diviso'l' of aR* if and only if there is a p'l'ime divisor p of a such that p* i8 a p'l'ime divisor oj pR*. p* is a minimal prime divisor of aR* if and only if there i8 a minimal prime divisor p of a such that p* is a minimal prime divisor of pR*. Proof: Assume that. p* is a prime divisor of 'pR* wit.h a prime divisor p of a. Then t.here are element.s b E Rand c* E R* such t.hat. p = a:bR, p* = pR*:c*R* by (8.8). Since pR* = aR*:bR* by (18.1), we have p* = (aR*:bR*) :c*R* = aR*:bc*R*, and p* is a prime divisor of aR*. Conversely, assume t.hat. p* is a prime divisor of aR*.
Let. ql n ... n qr be a short.est primary decomposition of a. Then aR* = nqiR*, whence p* is a prime divisor of some qiR*, say, qIR*. Let p be the prime divisor of q = ql. We shall prove that p* is a prime divisor of pR* by induction on length Rpl qRp . We may a88ume that q = 0 by (18.10). If the length is 1, then the aS8ertion is obvioutl beeause p = q. Let a be an element. of p such t.hat p = 0: aR, and let q' = aR)) n R. Then there is an element d of R which is not in p 8ueh that dq' CaR. Let qi n ... n q; be a shorte8t primary decomposition of zero in R* such that the prime divisor pi of q; i8 a prime divisor of pR* if and only if i ::::; t. Since p = 0: aR, pR* = n (q;: aR*) by (18.1), whence by our assumption on t.he prime divisors pR* qi : aR*) n ... n (qi: aR*). Let b* be an arbitrary element of
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0, WIIl'III'n b>l>d . O,III'III:n b* ( O:dN*' (O:dfnu+ 0, alld s .. I, whinb PI'OVI'H the HKKm(,ioll, Now we l'OIlHidl'l' t.lH~ InHt IIHYl'rtioli. AHHllrne thaL ~* il'l a lI\illimall)l'iml~ divil'lo]'o!' IlN*. 'l'III'1l p* il'l a pl'imc diviHot' of ~/(,* with a prime tiivitlo]' ).I of 11. 'I'lw llIillillIItiity or p* impliel'l Chat p* i.'l a miJlimal prime divisor of ).IU* alld that ~l iH /I. minimal prime diviso], of a. Conversely, assume that ~ iH H, minilll:d primo divisor of a and that ).1* is a minimal prime divisor of ).1n*, L.. t. ).1** be a minimal prime divisor of aRt such that p** C ).1*. Let, q) II11 Lbo natural homomorphism fromR into R* (cf>(a) = a·l with 1 ( U*), and let p' be cf>-l(p**). Then ).1' is a prime ideal by (2,12) and p' iN ('.011,' I.nined in p because p = cf>-l(p*) (for, non-zero elements of H/~l :IXI' 1I0!; zero divisors with respect to R*/pR* by (18,1), By the millillIality of ).1, we have ).1' = p, whence p** is a prime divisor of ).IN*, whieh implies that p** = p* by the minimality of p* and by (,haL p** C pt. Thus p* is a minimal prime divisor of aRt. We say that a module M over a ring R is torsion-free if every 11011;!,pro-divisor in R is not a zero divisor with respect to M, '1'111'1'101'01'1' Ilt*
(18.12)
THEOREM:
Let Rand R' be Noetherian rings such that H'
is an R-module and such that ® R R' is exact, If M is a torsion-free Hmodule and if every prime divisor of zero in R is of height at most I I then M ® R R' is a torsion-free R' -module. Proof: Let R* and R'* be the rings R EB M and R' EB (M ® R') I'nspectively in the principle of idealization. If N is an R*-module , Llten N is an R-module, and N ® R R' is naturally identified with N 0 R* (R' ®R R*) = N ®R' R'*. Therefore we see that ®R* R'* is exact. In order to prove (18.12), we may assume that M is a finite module. Assume that a E R' is a zero divisor with respect to M ® R'. Then a is a zero divisor in R'*, and there is a prime divisor p'* of ;!'ero in R'* such that a E p'*. (18.11) implies that there is a prime divisor p* of zero in R* such that p'* is a prime divisor of p*R'*, rlince M2 = 0, we have M C ).1*, and therefore p* = p EB M with a prime ideal p of R. Similarly, ).1'* = p' EB (M ® R') with a prime ideal ~' of R'. Since M is torsion-free, we see that ).1 consists merely of zero
til tlil'iHIII'H, 111'111'1' i:1 II. prillll' tlil'illlll' IIi' '/,111'11 II,\' 0111' 1I:1rllllllpLiIIIl oil
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iH a l',('I'O diviNOl' ill N', wllil'll 1',OIlIPld(~H (.1)(' pi'ool'. 1';XI,JIWlf;JC:-:: I. AHHll1nc l.lmL U, U*, U** are riJl~H KJleh thaI. ®u U* and ® u* fl*" are exact" Prove that ® II H.** iH exact, 2, Let x, , .,. , x" be algebntically independent elements over a ring R. PI'ove that ® R R[x] and ® R R (x) are exact. (Hint: Prove first the exactness of ®R R[x] using the fact that R[x] has a free base over R.) 3. Let R* be a ring which is a module over a Noetherian ring R. Prove that ®R R* is exact if (a:bR)R* = aR*:bR* for any ideal a and any element b of R, 4. Let (R, a) be a Zariski ring and let M be a finite R-module. Let S be the set of non-zero-divisors with respect to M and let N* be the completion of a submodule N of M. Prove that (M ® Rs) n N* = N, where, M ® Rs and N* are naturally imbedded in M ® R~ (R* being the completion of R). 5. Let (R, a) be a Zariski ring and let R* be the completion of R. J~et b be an ideal of R. Prove that if bR* is principal then b is principal. (Hint: b/ba = b/ba ® Ria = b/ba ® R* /aR* = bR* IbaR*.) 6. Prove that if the completion R* of a Zariski ring (R, a) is a unique factorization ring, then R is also a unique factorization ring. (Hint: Let lJ be an arbitrary prime ideal of height 1 in R lind let S be the complement of lJ in R. Then, show that lJR~ is principal and that every maximal prime divisor of \JR* does not meet S, Then conclude that \JR* has only prime divisors of height] and then that \JR* is principal.) 7. Let R* be a ring containing a ring R such that (a) ®R R* is exact and (b) aR* n R = 11 holds for any ideal a of R. Let M be an R-module. Prove that if M ® R* is a finite R*-module then M is a finite R-module.
19. The theorem oj transition We say that the theorem of transition holds for rings Rand R' if R is dominated by il' and (2) if q is a primary ideal in R such that the prime divisor of q is a maximal ideal, say m, then ~ 1)
lengthR' R' / qR' is finite and it holds that lengthR' R' / qR' (19.1)
THEOREM:
=
(lengthR' R' / mR') (length R R / q) .
Let Rand R' be Noetherian rings such that
R :s; R'. Assume that, for any maximal ideal m of R, the length of R' / mR' is finite (i.e., the prime divisors of mR' are all maximal). Then
(II L\ 1"1'11111
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(\:)
,~j' l/i" follol/lill(/i{ i8 " II"",'IIS"I'!/ 1/1/(1 iill.lli,.;,.1I1 ,.o/It/ifio// Jor III,. I'Irlitlilli ,(/' Iii,' Ih"o/'t'1II qj' 1/'IIIINil;oIIIo/' l/i,' /'i//flN It allrl N': (a) II q /S 1/ /lrilll((I'!1 itll'lll hdo//I/illf/ 10 (('1/11/,,/,/11111'/ ir/I'II./ 111 in It I(.I/.({ 1/ q: hh~· 111 Iol' an d,:II/,'II'/' h ,~r H, the//. qU': hl{" = III /{,'. ( I») M I,: N' /:8 (',!>ad. ((~) 11'111' IJ://'!I '/II.(u'£'ma.1 ideal 111' (if Il', the theorem of transition holds JOI' Hlm'n,,: and N,~" . (d) If Il and 0 arc idcals of R, then (a:o)R' = aR':oR', ('(/I'll
1'1'001': We show first that the validity of the theorem of transition illiplins (a). Set q' = q + bR. Then length R/q = length R/q' + 1, !Il'III:n length R'/qR' = length R'/q'R' + length R'/mR', which impline; that length R'/mR' = length q'R'/qR' = length bR'/(bR' q/(,') = length R'/(qR':bR'), which implies that mR' = qR':bR'. Thie; proof being reversible by induction on length R/q, we see the eqllivalence with (a). (a) implies (b) by (lS.7). (b) implies (d) by (IR.l), while (d) implies (a) obviously. Thus (a), (b) and Cd) HI'P equivalent. It is obvious that (c) implies the validity of the theorem of transition; the converse is also obvious except for the i'lwt that R* = R(lIlnR) C R** = Rill" which is proved as follows: Q<),,:*R** is exact by (b) and (lS.1O), whence (1S.3) implies the indllsion relation.
n
(19.2) COROLLARY: Assume that the theorem of transition holds for Noetherian rings Rand R'. (1) For any ideal a of R, the theorem of transition holds for the rings /f,/a and R'/aR'. (2) A ssume that 8 and 8' are multiplicatively closed subsets of R nnd R' respectively such that every maximal ideal p' with respect to 8' '£n R' is a minimal prime divisor of pR' for some maximal ideal p 'with respect to 8 in R and conversely that if p is maximal ideal with reNPt;ct to 8 in R, then pR' does not meet 8'. Then the theorem of transition holds for the rings Rs and R~, . (3) If Rs and R~, in (2) are semi-local rings, then Rs is a subspace of R~,. Proof: We note first that ® R R' is exact by (19.1). Since aR' R = a by (1S.3), the validity of (1) is obvious. ® RS R~, is exact by (IS.1O). Therefore, by our assumption and hy (lS.:3), Rs is a suhI'ing of R~, , hence Rs is dominated by R~, . Therefore (2) is proved
n
00
, '( 1M 1'1 dl)'I'IIIN I \
I!I.IIIY (). (:\) iH l'HHY III"'HIIHI'
N,Y,
i1I'III'!!
ill
n"h',~j'
n Ii','j
p:t,l'l.it:IlIaI' 1'01' 1.I\1~ .I:1.I'01i:-1011
n"
1'01' Hlly ili":!i
n
III'
I'adi,'al n,
II;X,,]/\l:I;\I';: LoL N alld H' ho Nodh,:l"ia,/I l"i/lgH HIII:h l.lIal. /t,' iH all N /I\odll!". ASHlImc OwL Lhe following Ht.aComonj,H a,',) CnlO: (a) 1.1)(,"0 iH 110 IrInxilll:d id,:nl of R which generaieH il' in if,', (b) the natural imagc U* or It, ill a' (N" = R/(O:R ' )) is dominated by R' and (c) the condition (2) in the uoiiniLioll of the theorem of transition holds for Rand R'. Prove that R r;;;;, R'.
C II A PT II~ I{, 11 [
Multiplicities 20. Homogeneous rings i\ ring R is called a homogeneous ring over a ring A if R is a graded "illgL Ri such that A = Ro and such that R is generated by RI over A. II' furthermore Rl is generated by algebraically independent elements over A, then R is called a homogeneous polynomial ring. A graded ideal or a homogeneous ring is called a homogeneous ideal. The above definition 8how8 that homogeneous rings arc nothing hut homomorphic image8 of homogeneous polynomial ring8 who He IW['Jwls are homogeneous ideal8. (20.1) A homogeneous rIng R = L RII is Noetherian ~f and only z{ Nil and RI are Noetherian (i.e., Ro is a Noetherian ring and RI has a .finite bas'is over Ro). Proof: The if part is obvious. Convertlely, assume that R itl NonCherian. If a and b are ideals of R o , then that a r" b implies that nR r" bR (because a = aR R o , b = bR R o), and we see that Ro iN Noetherian. If a and bare sub modules of R I , then that a r" b implies that aR r" bR, and we sec that Rl is K oetherian. Let R = L It" he a graded ring and let M = L Mit itl a graded module over R. We retain these deiinitiolltl of Rand M throughout Chis seetion. The length of M" (as an Ro-module) is a function of n ['or n = 0, 1, 2, .... This function is called the K-function of M (over Ro) and is denoted by KRo (M; n) or simply by K( M; n). If there is a polynomial f( x) in one indeterminate x whose coefficients are rational numbers such thatf(n) = K(M; i) for sufficiently large n, then the polynomial f(x) is called the O"-polynomial of M and is denoted by O"R(M; n) or by O"(M; n). It is obvious that M has a O"-polynomial if and only if each K( M; n) is finite and furthermore K( M; n) is a polynomial in n for sufficiently large n. When a is a graded ideal of R, K (R / a; n) is called the IIilbert characteristic function of a and is denoted by X(ll; n). (20.2) If Nand N' are graded s'ubmodules of 1.11, then
n
n
L;
[67]
os
M, 11/I'II'ld( '1'1'1 WI
KeN IN';
11.)
I /dN
n N';
1,(
II)
N;
1/)
I (;( N';
11.).
Similarly if a and U 0/1"(: {fradol id('als (~( N, 1/11'1/, X(il I b; 1/.) -In 0; n) = xCa; n) + xCo; n). Proof: CNn + N~)/Nn = N~/CNn N:,), whidl implicH Lho assertion. C20.3) Let N = L N n be a graded submodule of M and let f be a homogeneous element of degree d in M. Then KCN + fR; n) = KCN ; n) + x(N:fR; n - d). Similarly, if M = R, then
xCa
n
xCN
+ fR; n)
xCN; n) - xCN:fR; n - d).
=
Proof: By C1.5), CN + fR)/N is isomorphic to R/CN:fR). By the nature of the isomorphism, we see that CN + fR) n/N n is isomorphic to Rn-d/CN:fR)n-d, which implies the assertion. C20.4) Assume that R is a homogeneous polynomial ring in indeterminates Xl, ... , Xs over a ring A which satisfies the minimum conditionfor ideals. Then KCR; n)
=
(n;-
~~
I)-length A.
Proof: The number of monomials of degree n is equal to (
n
+s
-
s -
1
1)
and the assertion is proved. C20.5) THEOREM: Assume that R is a Noetherian homogeneous ring over a ring of altitude zero. If a graded R-module M has a finite basis, then M has a O"-polynomial. Proof: At first, we consider tho case where M = R/a with a graded ideal a. Assuming the contrary, let F be the set of graded ideals a such that R/a has no O"-polynomial. Since R is Noetherian, F has a maximal member, say a*. If depth a* = 0, then R/a* has a finite length, say l, and R/a* has the constant l as its O"-polynomial, which is a contradiction. Thus depth a* ~ 1, thorefore there is a homogeneous element f of degree 1 which is not in a*. C20.3) implies that xC a* + fR; n) = xCa*; n) - xCa*:fR; n - 1). By the maximality of a*, xC a* + fR; n) is a polynomial in n for large n, say, for n ~ m. If a* :fR 7"" a*, then xC a* :fR; n - 1) is a polynomial in n for large nand we see that xC a*; n) is a polynomial for such n, which is a contradiction. If a*:fR = a*, then summing up the equalities xCa* + fR; r) = xCa*; r) - xCa*; r - 1) for r = m, m + 1, ... , n, we see that
1'11111"1'1'111, III
1111
L:,',
I\ln+; 1/) x(n" I .IN; i) 1 (I~OIIHLJI,IIL), HIIlI x(n*; 11.) iHn poly, lIolliial 1'111' 1n1'J.!:I· II, whil~h iH HINO II. 1~III1Ll'll.!lil',Lioli. "'hUN '" iH (~lIlpLy "'1111 LhiN ('HNn iH HI'LLind. Now wn pl'ovn Lh(~ gnllcml (~aN(~ hy ilidudiOl; 1111 ('111' 1111 II Ii 11'1' III' hOlliognlll'ollN gCllcmLol'N of M, Let Ul, . , . ,us be 11 IIHNiH i'ol' AI NiI(',1i LhaL ('adl 'lti iN homogeneous, If s = 1, then M is iHOIIIOI'pliil' Lo Il/(O:M), and therefore the assertion is true in this 1'11.141'. Hd N = I uR. Then, by our induction, we assume that N 1""14 n IT-polYliomial f(x). M/N = usR/(usR N), and therefore M/N IinH a IT-polynomial g(x), Since length Mn = length N n 11'11J.!:Lh (M.,jN n ), we see that f(x) g(x) is a IT-polynomial of M, II,lId tlw atlsertion is proved. (~(l.(i) If Ro/ (0: M) is the direct sum of local rings A i( i = 1, , .. ,r), Ihen denoting by ei the identity of Ai, M is the direct sum of Mei and /(/.,,(M; n) = KAi(Mei ; n). I 'l'Oof: IT i( m) = mei defines a homomorphism from M onto M ei , h"II(',e into M. Obviously ITi is the identity map, hence M is gen('I'aLnd by the M ei. mi = 0 (mi E M ei) means that, since miei = IIli, 0 = ej( mj) = mj. Thus we see that M is the direct sum of I.hn Mei , and the assertion is proved. (:20.7) Assume that Ro is a local ring with maximal ideal 111. If ( II, 11) is a local ring which is dominated by Ro and if K = R o/111 is a ,finite algebraic extension of K' = A/n, then
L';
n
+
+
Li
L
L
L
KAM, n)
=
[K:K'J'KRo(M; n),
I 'I'oof: Let M n = a(O) :::J a(l) :::J ... :::J aCr) = 0 be a composition /'Im'ies of M n as an Ro-moduk Then each aCi) /a(i+1) is isomorphic to /{, whence it has length [K:K'] as an A-module. Thus we prove the II. H.'-IPl'tion. A polynomial f(x) in one indeterminate x, whose coefficients are l'II.Lional numbers, is called a numerical polynomial if there exists an lIlL(~ger N such that fen) is an integer for any integer n such that /I ~ N. IT-polynomials are numericaL (20.8) If f(x) is a numerical polynomial of degree d, then there are
integers co, ... ,
,,. +
Cd-J
(x
Cd
such that f( x)
i 1) +
Cd,
=
Co (x
1
d)
+
Cl
(x
1~ ~ 1) +
where the (:) are binomial coefficients. Con-
Iwqnently, f( n) is an integer for any integer n and if a is a coefficient of
I(:c) then a· (d!) is an integer.
70
1\1111 il'll'l ,1('1'1'11<111 PI'!)D!':
W(\
pl'lIV!! CIi!! /I,HH(II'CiOIl
aKK(~I'LiDIi iK ol,vioIiK. AHKIIII\(\
hy
illilll!'l.ioll (III
I.hll.l. d I) ~·c I:d.r"
>
O. 1,('1. ('
d, II' d "II
0, 1.11('11 I,hll I,hll IIOl'lJinil'll1. or
xd • Thon I(x) - I(:I: 1-+ (1.1'1'1111' or lowI'!' dllJ,!;I·('('H). Since f(x) - I(:c - 1) it:: aiHo II\UXWl'i(lal, ('0 c'c (:(d!) iN :1.11 illl.l'g(,I·.
Therefore f(
x) - (x ! d) is a numeriwl polYliomial or dngn'(' 1(,1'1' Co
than d, and we prove the assertion hy our indtwtion.
21. A-Polynomials Let a he an ideal of a ring R and let M he an R-module. Set F n an/an+! and Gn = anM/an+!M, for n = 0,1,2, ., .. When a E F m , b E F n , we define ab as follows: let a' and b' be elements of am and an respectively such that a = (a' modulo am+1 ), b = (b'modulo an+!) and then we define ab = (a'b' modulo am+ n+1 ) ( E Fn+m). This multiplication defines a ring structure in the direct sum F of all the F n , and F becomes a graded ring. Furthermore, since Fn = an/a n +\ it follows that Fn = (F1)n, which implies that F is a homogeneous ring. This homogeneous ring F is called the form ring of R with respect to a. When a E an and a ~ an+!, n is called the degree of a with respect to a and (a mod an+!) is called the a-form of a. The direct sum G of all the Gi becomes similarly a graded module over F. This G is called the form module of M with respect to a. It is obvious that length M/a n +1M = length ai M/ai+ 1l\1 1:~ K(G; i). Therefore we see by virtue of (20.5) that (21.1) A ssume that Rand Mare Noetherian and that depth a = O. Then there is a numerical polynomialf(x) such that length M/an+1M = f( n) for all n that are greater than a fixed integer; f( x) is nothing but the O"-polynomial of the form module G. Now, applying (20.6) and (20.7) to the above result, we see that:
1:;
(21.2) THEOREM: Let M be ajinite module over a Noetherian ring R and le,t a be an ideal of R such that depth a = O. Furthermore let R' be a Noetherian ring such that (1) R is an R' -module, (2) with a'
=
(a I a E R', aR
c al,
R' / a' is dominated by R / a and (3) [R / mi: R' / m: 1 are finite, where mi (i = 1, ... , s) are the prime divisors of a and m; are the maximal ideals of R' such that m;/a' = (mi/a) (R'/a'). Then: (a) there is a numerical polynomial f(x) such that fen) = lengthR' M/anN\.for
n
71
('11111"1'1')11, III
if (/ iN thl! fOI'/11 I//lIdlll,' I~j' IlIu'illil'l'.Y/u'('I/IIIl,I/t"II./'(n I I) 1111'1",((,';//) • ./'(1/ I I) /(1/) h/,"III'( 0; 1/.) JII/' iiI/JII('':I'/lII!! 11/('(/1' //IIIIII'//.( //11I1I.{/('/,8 n I/l/.tI (('.) J( 1/) LIN/III i: N' /111; I· (1'lIlIlh J,'i • III (>0 10' Njn" AI C>() II Ni , I/Ihl '/'1 , Hi Nil, ( , Wn dnlloCn LlIl'/(.r) J.!:iVI~II.iIll-d, ahovn hy IIIIJI NII:f!ir'/I'ltllllirll'{/I' //1111/./'111 /llllI/hr'I' II. (I,)
AIo:'(O; Ai; .r), whidl i:-; ('alled a A-]!II(.l/II,(lfIl:ia( or M. A/,:(n; AI; .r) Illay "I~ dn1ioj,I'd Hilllply by A(11; M; .r) HIHj 111 may 1)(: omiCCI~d ir 111 = H. By our defillition, we Nne iIIlln<~diaLely Chai.:
(:2I.:n When R iii a l!erni-loC(J,t ring with cornlJtclilin I~*, tlu:1/
Aw(a; 111; ;;;)
=
Aw(all*; 111 ® R*; x).
We see also the following by virtue of (lS.!)), (19.1). (21.4) With the same R, R', a and M as in (21.2), if X
·i.~
((.
11'11.11.8
eendental element over R, then Aw(x)(aR(X); M ® R(X); .r) All' ( a; M; x).
(21.5) THEOREM: With the same R, R', a and lJ!l as in (21.:2), 11'1 N be a subrnodule of M. Set g(x) = AR,(a; N; x) + Aw(n; M/N; :1:) - Aw(a; M; x). Then g(n) = lengthR'(anM N)/ll"N :S length R'. an-rN /anN with a fixed natural number r and JOT a 8UJ/i-
n
eiently large natural number n, and therefore g( x) is of lower degree thaI/, Aw(a; N; x), except for the case where Aw(a; N; n) is of degree 0 and g(x) = O.
Proof: Since (M/N)/an(M/N) = M/(anM + N) and Hilll'l~ (anM + N)/anM = N /(anM N), we have the first equality. 8ilJ(~n anZVJ n N = an-rearM n N) by the lemma of Artin-Rees, we havli anN C anM N C an-rN, which shows the inequality. If Aw(a; N; :r:) is a constant, then amN = 0 for sufficiently large m and we see thaI, g(x) = 0 by the inequality just proved. Assume that AR,(a; N; :1:) is not COIlHtant. Then g(x) .::;: Aw(a; N; x) - Aw(a; N; x - 'r), which is of lower degree than AR(a; N; x), and thus we complete thn proof.
n
n
22. Superficial elements Let It = L Rn be a Noetherian homogeneous ring. An element .f of R is called a superficial element of R if f E Rl and if there is an m such that (O:fR) n Rn = 0 for any n ;:::: m. (22.1) Assume that Ro is a semi-local ring with maximal ideal.'!
7'2
M111/1'11'1,11 lI'I'llI\rl
1111 , " . , III", Ilia.!. ('1'1'1'11 U/illi l'OIlIf/InN il/jillilt'I!1 Illf/II!! l'it'IIII'II.IN f/,//(!
Ihl/,I "~I
iii
I,d III , ... , II j III' tWO/II'"/' 8Ii.1l/1wdli.lI'N oI 8upcljir:'/af I'f(!'II/,(!nl f lif U whit:!l, '£8 'not '£11, any of lit.(:
/8 a FII:I:II' Nw·l/I.odnit'.
. 'l'lum, t/wn; '£8
(J,
II, .
Proof: Consider all of the prime diviHors of )1;01'0 wh ieh do not eontain R 1 ; let them be ~l, .•• , \Ju. Set nt+i = \Ji n R 1 • Then, by the lemma of Krull-Azumaya, we see that nj 5tR I ~ RI for any j = 1, ... , t u,5t being the Jacobson radical of Ro (5t = ni mi). Hence N j = (nj 5tR 1 ) /.£R I is a proper submodule of M = R1/.£R I . We shall show that there is an element g of M which is not in any of the N j • M is the direct sum of Mi = RdmiRl ,which have free bases mi,l, ... , mi,n(i) over the fields RO/mi . We prove the existence of g by induction on Ln(i). SetN; = ml,lR N j • We may assume that N; = M if and only if j > v. Applying the induction assumption to /ml,lR in M/ml,lR, we see that there is an element g* of M/ml,lR which is not in any of the /ml,lR for j = 1, ... , v. Let g' and g" be two different elements of M whose residue classes modulo ml,lR are the same g*. If both g' and gil are in the same N j , then N j contains ml,l , which cannot happen for j > v. Therefore, if we take different gW, ... , l') of M whose residue classes are all the same g* and if r > t u v, then at least one of l i ) is not in any of the N j for j > v. By the choice of g*, we see that the element is not in any of the N j for j = 1, ... , t u. Thus the existence of g is proved. Now letfbe an element of R such thatfmodulo5tR = g. Sincef ~ \Ji for any i by our choice, 0 :fR coincides with 0 up to primary components whose prime divisors contain Rl ; namely, there is an ideal b which contains a power R7 of RI such that b n (O:fR) = 0, whence (O:fR) n Rn C (O:fR) n b = 0 for n ~ m, and f is superficial. When a is an ideal of a Noetherian ring R, an element a of a is called a superficial element of a if there exists a natural number c such that (an: aR) n aC = a,,-l for any natural number n > c. (22.2 ) With the notation as above, a E a is a superficial element of a 2 if a modulo a is a superficial element of the form ring F = L F" of R with respect to a. Proof: Set a' = a modulo a2 • Let a natural number c be tlllchLhat (O:a'F) n Fn = 0 for n ~ c. Let n be an arbitrary natural number such that n > c. It is obvious that a n - 1 C (an:aR) n aC • Let b be an arbitrary element of (an:aR) n aC • Then ab E an. Let m be such that m l b E a - and that b ~ am. Set b' = b modulo am. Then b' E F m-l .
+
+
+
+
N;
N;
+
+
7:1
.'11,11"1'1111(, III
n
Hilll'.' II I 1\", \1'1\ 1111,1'.' III . 1', ",lil'lll'(\ (.0:0'/,') I,'", I O. II' III • II, Ihl'll oh ( n" illl"lil'~j Lhnl. 1//1/ 0, whidl iN n I·OIlI.J'wlidioli. 'l'hnl'l'foJ'l'/II, . 1/., HIIII h ( nil I. ,I'hll:-l I.hn pmol' iH I~OII1J1ltd,n. (:.l:.l.:\) '1'''Wllt,I')M: 1"ll (\ Iw ((,/1. 'I:t!(/(/,l I~r Illljill!. z(/ro 1:'1/. a Nor;lIW1"£an '/''://(/ N, and lel I./u: 7!1'irrw d£/r£,w)'/'S I!/, (1 he 1111, ... , 1\l". Let 8 he lhe inler.wlci'ion I~r til.!: co'fnpl(mwnls I~r 11\.[. I,et 01, ... , 0 t be ideals oj R wlf.ch lho,l aa" is nol contained in any oiRs . If R/mi contains infinitely IWIny elements Jor every i, then there is a superficial element a of a which .,:.~ not in any oj the OJ . I)roof: Let F = F n be the form ring of R with respect to a. Set 2 2 2 Iii = FI ((OJ )/a ). If nj = F I , then a C OJ a , hence a = 2 (Uj a) a , whence aRs = (Uj a)R8 by the lemma of KrullJ\,mmaya, and it is a contradiction. Thus nj ~ FI for any j. Therefore there is a superficial element a' of F which is not in any of the IIj by (22.1). An element a of a such that a' = a modulo a is a superfieial element of a by (22.2). It is ohvious that a ~ OJ, and the [I$sor-
n n +
L +(
+
n
!,ion is proved. (22.4) With the same R and a as ahove, if (t = aR, then a is a .mper.ficial element of a. Proof: It itl sufficient to show that, in the form ring F = Fn of
S
Il with respect to a, a' = a modulo a2 is a superficial element of F. fi' is generated by a' over Fo = R/aR. Therefore, if \.1 is a prime divisor 0[' zero in F which does not contain F I , then a' Et \.1, hence 0: a' F (:oincides with 0 up to primary components whose prime divisors nontain Fl' Therefore, a' is a superficial element of F, and a IS a Kuperficial element of a. (22.5') With the same notation as in (21.2), iJ b E R, then length w R/(a n
+ bR)
= length w
R/a n - length1/' R/(an:bR).
This follows from (1.5). (22.6) THEOREM: With the same notation as above, assume that a E a. Set a' = a/aR. Then A1/'(a'; n) ~ Aw(a; n) - A1/'(a; n - 1) for large n. If a is a superficial element of a, then, with the intersection S of the complements of the prime divisors of a, kngthR' (O:aRs) is finite and AR,(a'; x) = Aw(a; x) - Aw(a; x - 1) length1/' (O:aRs). Proof: The firtlt assertion is immediate from (22.5). Let c be a
+
sufficiently large natural number which is fixed and let n be,greater Lhan c. Then (an:aR) n a = an-I. By (22.5), we have C
7,1
MIII!I'll'ld('I'I'II'ill
'\"" (n;
/I )
'\10" (n;
I)
1/
ill Hill", I.hi:-; length i:-; a eOlli:iLan(" flay C:, for large n. We have only to prove that C = length w c (O:aRs). Since Rla = RslacRs we may assume that R = Rs. Then an:aR is contained in (O:aR) + a by (3.12) for large n, whence C = length w (O:aR)/(O:aR) a Since this C is independent of c C (when c is large), (O:aR) a must be zero, and the assertion is proved.
Situ:<: !.Iw minimum <'.ollditioll lLold:-;
C
n
n
C
•
(22.7) THEOREM: With the same notation as in (21.2), the degree OJ AR' (a; M; x) is equal to altitude (a + (O:M))/(O:M) (in the ring RI(O:M)). Proof: By virtue of (21.2), we may assume that R = R' and furthermore that R is a local ring, with maximal ideal m = Jnl . We may assume obviously that 0: M = O. Let Ul , . . . , Ut be a basis for lVI, We prove the assertion by indnc:tion on t. If t = 1, then M is isomorphic to R, and A(a; M; n) = A(a; n). We prove thi:-; ease by indudion on altitude R. If altitude R = 0, then the assertion iN ohvious. Assume that alt.itude R > O. Conflidering R(X) if neceRsary (with a t.ranscendental element X over R), we may assume that R/1I1 eontains infinitely many elements by virtue of (21.4) and (9.10). (22.:3) and (22.5) imply that there is a superficial element. a of a such that. alt.it.ude RlaR = alt.itude R - ] (by virtue of (9.7)). Then 1..( al aR; x) is of degree (altitude R - 1) by our induction assumption. Therefore, we see that A(a; x) is of degree (altitude R) by (22.6), which proves t.he assertion in this case. We consider now the general case. Since 0 = 0: J\!J = ni (0: uiR), there is an Ui , say Ul, such that depth (0:u1R) = altitude R. Then, applying (21.5) with N = u1R, deg (A(a; M; x)) = max {deg A(a; N; x), deg A(a; MIN; x)}. deg A(a; N; x) = alt.itude R by the case where t = 1 and deg A(a; MIN; x) :::; altitude R by induction, and thus we prove the assertion. (22,8)
COROLLARY.
With the same a and R as above, if altitude
a 2:: 1
and if a is a superficial clement of a, then altitude alaR
=
altitude a - 1.
• '11.\1"1'1'111 III
(:'.:'..11) 'l'III':"lll':~'I: .'I.~NIIIII(' Ihai II/(' 11t('OI'r'11I
N (wllt('l'illll 'l'iIlIlN U ((lIrI Itl'. If 111/111:111111 WillI(! dill/NO'I'
((f'
pN*,
).1
((I'
is (( pl'illl(' irll'rli
I/I,('n hniJ.!:hL
/I'IIIINilili/l IwldN fol'
rd' Nand '1/
p '" heighL
~)Jf<
iN (/,
~l*. (,'on..<;equcn,Uy,
nlLiLlitin U HII,il,".h~ U*. VII,'I'!'/w'/'m,(m~, 1/ a i8 an ideal of il, then hniglll, 4l Iwii!:ld, nJ?,* and altitude II = altitude aR*. 1'1'001': 'I'he L!ICorem of transition holds for Rp and by (19.2), WII(HI(~(lA(pR:* ;:c) and A(pRp ;x) have the same degree, and we have height p = height p'. Therefore, it is obvious that altitude R = nlLitude R~ because R* dominates R. The last assertion follows from ( 18.11) and the above result (by virtue of (19.1)).
R:.
II~XERCISES: 1. Prove the converse of (22.2). 2. Let M be a finite module over a semi-local ring R and let u be an ideal of N Hllch that the radical of u is the Jacobson radical of R. Prove that altitude R iH equal to the altitude of the form ring F of R with respect to u and that op. alt M is equal to the operator altitude of the form module of M with respect to a (as an F-module). 3. With R, M and u as above, let b be an element of R. Prove that
}.(u; 1'01'
M/bM; n)
=
}'(u;
M; n) - length (M/(anM:bR»
sufficiently large n.
23. Multiplicities Let R, R ' , a, M, etc. be as in (21.2) throughout this section. Set altitude a. Then the degree of Aw(a; M; x) is at most equal to r by (22.7). Let a be the coefficient of xr in this A-polynomial. Then (r!) a is an integer by (20.8). This integer is called the multiplicity of a with respect to Mover R' and is denoted by fLR'(a; M). R' may be dropped if R' = R. M may be dropped if M = R. Note that J.!R'(a; M) = 0 if and only if altitude (a + (O:M) )/(O:M) < altitude (l. Note also that if M is a ring, then J.!R' (a; M) is either J.!R' (aM) or lIero according as altitude aM is equal to altitude a or not. (21.2) implies that (23.1) J.!R'(a; M) = 2::::> J.!(aR i ; M ® R i ) [R/mi:RI/m'], where i runs over those indices such that height mi = altitude a. 'I' =
(23.2) COROLLARY: Let S be the intersection of the complements of the mi such that height lni = altitude a. Then J.!w(a; M) = J.l-R'(aRs ; M ® Rs). Note that cine can reduce multiplicities to those of the case where R = R' and where R is a local ring by (2:3.1). As an immediate consequence of (21.5), we have an important
711
MIII.'I'II'I,II I I'I'IIoirI
(~l:l.:t) "'IIWllt.i'IM:
(/.IIIi, ,in llll,I'i'il'll/tI.!',
If N
it op.
iN 1/ NI//lll/otill/I'
ali. AI /N
<
Id 11/, 1/1t'11
nll.il.udn U,
t//./'I/. '1/1(: halN:
i-!1I'(ll; N) = i-!w(n; IY/). (2:3.4) COnOI,],A I~Y: If N has a free base al , ... , af and if op. alt 111 /N is less than altitude R, then we have i-!w(a; M) = r}tw(a). In particular, if the finite module M is a ring contained in the total quotient ring of R, then i-!w(aM) = }tw(a). N ext we prove
(23.5) THEOREM: Letting p run over all prime divisors of zero in R such that depth p = altitude R and such that a + p ~ R, we have }tn-(a; M)
=
Lpi-!w((a
+
p)/p).lengthRp(M ® Rp).
Proof: We note first that the vanishing of the right hand side implies that op. aIt M ® Ri < altitude R, and therefore that }tw(a; M) = O. Now we prove the assertion by induction on t = length (M ® Rp). Since the case where t = 0 is proved already, we assume that t > O. Let m be an element of M such that 0: mRp = pRp for one p, say q. Then there is an element c of It which is not in q such that O:cmR = q. Set N = cmR. Then N is isomorphic to R/q, and }tll'(aj N) = }tR'(a; R/q) = }tw((a + q)/q). This equality and our induction applied to M/N proves the assertion by virtue of (2:3.:3). The following lemma is immediate from the definition: (2:1.()) If 0 is an ideal of R such that 0 c a and if a and 0 have the same radical (or, moregeneralty, if altitude 0 = altitude a), then}t",(a; M) :::; }tR'(o; M).
L.
(2:1.7) THEOREM: With 0 as above, assume that J.!(a) }tR'(aj M) = }tR'(o; M). Proof: By the pin (23.5), we have }tea)
=
Lp
}tea
=
}teo). Then
+ p/p) . length R p , }teo) =
Lp }t(o + p/p) . length Rp.
SinceJ.!(o + p/p);:::: }tea + p/p) by (23.6),}t(a) = }teo) implies that J.!(o + p/p) = }tea p/p) for every p. By (23.1) we see, in the same way, that J.!/?'(o + p/p) = i-!w(a + p/p). Therefore, again by (23.5), we h:we the equality J.!R'(aj "~f) = }tR'(o; M).
+
77
1'11,11"1'1111[ III
WI' ndd 111'1'1' Llil' !'oll:;wili/o!: l'I'IIIIU'I,: If '" ~ /I'" 1I11t/ (: 111'1' lit ( fU1'11I '1'111(/ I~r N /1111/1/11 flll'lll //Ilit/llit' 4 II! ,1'1'81/1'l'Iil'I'/,I/, '11)illt '1'1'8/11'1'1 III n, I/II'//. 1l1",(Il; AI) Il""( /1'1"'; (il, op. all, III op. all, (/, 1I,lId ,ill. 'l)((,l'ti(IIi./U/I', aiLiLIIII<' n
L (/"
( ~:\.H)
nlLiLlldn FIV "'lin PI'OO!' iH illlllll'dia(.1' I'; x HllI:h
(lJ
n
l'1'01n
(21.2) and (22.7).
I. LnC !J nllli e h() ideals of R such that depth b > depth e and R. Prove Lh,tC MR,((a &)/0; M/oM) = MW((a (& e)/ e); M/\b elM) = MR,CCa bel/be; M/beM). 1m!)! 1-11-;;;:
LhnL a
+ u 'i"n
+
+
2. ASHume that Mi are finite R-modules and that 0 - t M, M" - t 0 is exact (with suitable homomorphisms). Prove that
+
-t
M2
n
-t ... -t
8. Let R* be the ring REEl M in the principle of idealization. PrOve that MI,:,(a; M) = MwCaR*) - MwCa).
24. System oj parameters We first generalize the notion of system of parameters: Let a be nil ideal of a Noetherian ring R such that depth a = O. Set r = altiLude a. If there are r elements al , ... , ar of a such that a and LaiR Imve the same radical, then we call the set of the ai a system of paramdeI'S of a. The set of r elements bl , ., . , br is called a system of param(~lers of R if depth L biR = 0 and if altitude L biR = r. (9.6) shows that if R is a semi-local ring then the Jacobson radical of R has a H'y8tem of parameters. On the other hand, (9.3) implies that if Lf biR (hi E R) has the same radical with a, then s ::::: r. Therefore, by virtue of the case of system of parameters of a local ring, if Xl , '" , Xr is a H,Ystem of parameters of a, then altitude al(xlR + .,. + xtR) = '/' - t. In this section, we mainly treat the case where a is generated by a system of parameters of a. One should note that the general I,use may be reduced to such a case by virtue of (21.4), (23.2), (23.7) II,nd the following: (24.1) THEOREM: If (R, ml, ... , ms) is a semi-local ring of altitude r, if every Rlmi contains infinitely many elements and if the radical (!f a is the Jacobson radical '5t = mi of R, then there is a system of 7)arameters al , '" , aT of a such that Il (a) = Il ( LaiR). Before proving (24.1), we prove a lemma. (24.2) With the same R, R', S, and a as in (22.6), we have the following:
n
7H
Mill ,'1'11'1,1/ '1'I'IIIH1
11.ll.i(.IIlI(l n/oN (~)
If (( 'is a
11.1 i,i 1.111 In 11- I 811'f/1'/:/ir/o,{ ('1t·11/1·11.!, (~!'
II'lIgL"!!, N/II.U
(:l) II a
n alld U /l.1(.iCll!h~ n
I, 111,'/1,
J.!w(n) -I- InllgLh/I,(O:oH,,).
·i8 a. 8I1.J1I·I:/ic·ia.l demcnt 01 n (f,nd J.!w(a/aR)
·tJ
all.i Clitin n
>
I, lIwn
= J.!/",(ll).
Proof: If a E a\ then an:aR contains an- t for n > t, and "w(a/aR; n) ;::: "R'(a; n) - AR' (a; n - t) for sufficiently large n by (22.6), and we prove (1). (2) and (3) follow from the last assertion in (22.6). We note, by the way, that if altitude alaR = altitude a, then J.!w (a/ aR) :::; J.!R' (a) by the definition of multiplicities (or by (23.3». Now we prove (24.1). We prove the assertion by induction on r. If r = 0, then the empty set is the required system of parameters. Assume that r ~ 1. Let al be a superficial element of a; the existence follows from (22.3). Assume at first that r = 1. Applying (2) ill (24.2), we have lengthR'R/aIR = J.!w(a/aIR) = J.!w(a) + lengthw(O: aIR). Since al is a superficial element of aIR by (22.4), we have by (22.6) that lengthR'R/aIR = J.!w(aIR) + lengthR'(O:aIR), and we have J.!w(a) = J.!w(a;R). Assume now thatr > 1. 'rhenJ.!/t,(a/aIR) = J.!w(a) hy (24.2). By induction, there is a system of parameters I '" ' a2, ' .. , arI of a/aiR such that J.!R,(a/aIR) = MR'( "L.. ai(R/adl». Let ai be elements of a such that a; = ai modulo aIR. We are to prove that al , ... , ar are the required elements. Since a; E a, we have MR'( LaiR) ;::: J.!w(a). On the other hand, by (1) in (24.2) we have J.!1l'( (L aiR)/aIR) ;::: J.!R'( L aiR), whence J.!w(a) = J.!R,(a/aIR) = J.!w(La~(R/aIR» = J.!wC(LaiR)/alR) ;::: J.!R,(LaiR), which implies that J.!w( L a;R) = J.!w(a), and the proof is complete. (24.3) We use the same notation as in (21.2). If a E at and if altitude a/aR = altitude a-I, then Mw(a/aR; M/aM) ;::: t·,uu,(a; M). If al , ... , ar form a system of parameters of a and if ai E ani, then lengthwM/( L aiM) ;::: nl ... n r • J.!R,(a; M). Proof: (22.5) can be generalized to the case of modules by virtue of (1.6), and we have lengthwM/(an + aR)M = lengthR'M/anM lengthwM/(anM:aR). Since a E at, we have length M/(anM:aR) :::; length M/an-tM, and Aw(a/aR; M/aM; n) 2:: AR,(a; M; n) Aw(a; M; n - t), and we have the first result. By a repeated application of the first result, we see that lengthR'M/( aiM) = Mw(O; M/C L aiM» ~ nl ... nr·Mw(a; M), which proves the last assertion.
L
71l
( 'II" I "1'11111, III
(:JI.,I) lI'ill! Ih" N!I,~/I'III
of
Nil 1111'
UN O,bOl'(', ,1/ n 'iN Ihl'!/' J,!",,(Il; AI)
N, U' , 11, (/,nd M
/)(//,(/,1111'/1"1'1-/ 01, '"
/I:
,
(/'",
I/I'/I.("/'(/'[I'd
bl/ (/'
lilllll';I1(",:) .,"
a;'iil! )/(111'" '/I.,,). (LIOMMA 01" LW:II) 1'!"Ool': 'l'hnC 1011I1:CII/,.,/I1 a;'iM ;:::: 'ILl'" nr'J,!W(!l; M) follow/,; 1'1'01\1 (:J,I.:n, alld wn hav~ lim inl" (lengthwM/L aiM)/nl ... nr ;:::: It"" (11; M ). Therefore, we have only to show that (IClII/l:UI1,,,M
/L
lim Nllp (IengthwM/( L afM) )/nl ... nr ::; J,!w(a; M).
L
IJd /I' = /1'" be the form ring of R with respect to a and let G = 0" be the form module of M with respect to a. Set Xi = 2 ((,l modulo a • Then the form module of M/( L a;iM) is a homolIIorphic: image of G/(L xtiG), whence longthwG/(L x;iG) ;:::: 1(~lIgt.hIl'M/( L a;iM), whence it is sufficient to show that
L
lim sup (lengthwG/( L x;iG) )/nl ... nr ::; Itw(F I ·/?; G), I)(~(:ause Itw(a; M) = ItR,(FI·F; G) by (23.8). Thus we may HNsume that F = R, M = G, and ai = Xi. R is then a homomorphic image of the polynomial ring in r indeterminates Xl , . " , XI' over Fo. We prove the ease where M = mR (m E Go) by induction on Inngth Fo. M is isomorphic to R/(O:M). If r = 0, then the assertion il'l obvious, and we assume that r > 0. Assume that length Fo = 1, 'i.e., a is maximal. Since Fo is a field in this case, the Xi must be algebraically independent over Fo because altitude a = r. If O:M = 0, Chen obviously M is isomorphic to the polynomial ring, Itw(a; M) length w Fo , and
longth w M/( =
L
x;iM)
=
lengthRI R/( L xfiR)
length w (Fo[xd/x?lFo[x])
(9 Po ••• (9 Po
=
(Fo[xrl/xrnrFo[xrD nl ... n r • length R, F o ,
und we have the required equality in thiN case. Assume that 0: M ~ 0, and let .r be a homogeneous form of degree, say 8, which is ill O:M (f T" 0). Thell M is a homomorphic image of R/JR. Then Icngthll' M / (L x;'iM) ::; length w R/(Lxi'iR) - lengthw(.fR + L xtiR)/( L Xl'iR). lcngth w R/ ( L Xfi R) = n l . . . nr length R, Fo as was shown above, and length w (fR
+L
x;'iR)/(L x;iR)
=
length w R/((
L xi'iR) :JR)
by (1.5). Since Fo is a field, (L xtiR) :!R C L X;'i- R and therefore length R, R/ ( ( LXi" R) :!R) is not smaller than 8
HO
MIII,'I'll'ldlll'I'lllili
N) • ,. (II, TItIIH
Iilll
HlIll
(1(~llgLh M / (
L
a<;'i M ) ) /UI
' ..
1/",
:::; lim sup (n! ... nr - (n! - s) .. ' (n.r - 8))' ]ollgLh u , FO/'fh ... nr =
and thus we prove this case. Now assume that length Fo let a E Fo be such that length aFo = 1. By induction,
+
lim (lengthIl'M/(aM
L xfiM»/nl ... nr
>
0,
1, and
J.!R,(a; M/aM) ,
=
and
J.!R,(a; aM). Now lim sup (lengthIl' M/( L xfiM) )/nl " . nr =
lim sup [(lengthIl' M/(aM
+
L Xi'iM)
+ lengthIl' aM/CaM n L xfiM)]/n! ... nr :::; J.!Il'(a; M/aM) + lim sup (lengthR,aM/a( L =
x~iM)/n! '"
J.!Il'(a; M/aM)
+ J.!Il'(a;
nr
aM),
this last sum is equal to J.!Il' (a; M) by (23.3). Thus the case where M = mR is proved. Now we prove the general case. Let m!, .. . , ms be a basis for M which are homogeneous. We prove the assertion by induction on s. Sot N = L ~-! mR. lim sup (lengthw M/( L xfiM) )/nl ... nr lim sup (length w M/(N
+
longth w (N
+
:::; J.!w(a; M/N)
L XfiM)
L x~iM)/( L xi'iM»/n! .. , nr
+ lim =
+
sup (lengthR'N/(Lx~'N»ln!'"
J.!R,(a; MIN)
+
J.!R,(a; N)
and we have proved the assertion completely. (24.5)
=
J.!w(a; M), ~
With the same notation as above, we havd nl ... nr·J.!R,(a; M).
COROLLARY:
J.!R,(L ai'R;M)
nr
HI
I'll 11"1' IIilt III
;\lIoi,lll'I' l'OI'OIlIl,I',Y 11'1'
Hllliliid
1'1'11111,1'1\ iN Cill' roilowillJ.(
(:J.I.(\) (!IIItIlI,J,A Ill': lI'ilh llil' I/lila/illll ((8 ((.1111/'1',1)' '/1'1' dn /III/' ((,881(.11/('
llirtl !l /8 {11'IIITUlt'd (1/1 1/ 8!Jslnll I~r 11((./,(/:/l1('/1'/,8 1m!' 1I.8S/WU: lIwl n I'm./I'd bll III , ••• , ((." 'W';II!. S > 'I' al(.iLllIln n. 'I'lwn
lilll (11~1\p;j,h"" M/(La;"M))/nl'" Proof: Hilll~n
n8
'';8 {/('I/,.-
= O.
have only (,0 "how that lim sup (length M / ... nr ~ 0, in tho same way as in the proof of (24.4), WI' lIl:1y tLHHUme that R is the form ring F = L Fn of R with respect Co (1. Tlwll lJ;f is a modulo over the polynomial ring Fo[X! , ... , X,] ill illdeterminates Xi with operation such that Xim = aim for allY /II. ( 111, and in this sense, the multiplicity of M is zero, and thm; we PI'OVO tho assertion. N ow we prove an important result.
(L at';M) )/nl
Wp
(24.7) THEOREM: With the same R, R', M, and u as above (which are Ihe same as in (21.2)), assume that u is generated hV a system oj param('tors a!, .. , , a r • For an arbitrarily fixed integer s such that 0 ~ s ~ '1', ,w:/, 0 = L~ aB. Then, letting ~ run over all min1:mat prime divisurs of b ::;uch that height ~ = s and such that depth ~ = r - s, we have the following formula: J.!II'(U; M) = J.!R'(u + ~/~) ·J.!(oRp ; M ® Rp).
Lp
(ASSOCIATIVITY FORMULA)
Proof: For all arbitrary natural numher t, set Ot = L~ a:R, ai,t = Ot and ci = Liai,t(R/ot). Then we see by (n.5) that f..!1l'(ci; M/otM) = L: J.!ll'(ci + ~*/p*) . length (lVi/o t) ® (R/ot)p*, where p* nms over all prime divisors of zero in R/o t sueh that depth ~* = r - s. Therefore, letting ~' run over all minimal prime divisors of Ot of depth r - s, we have ni modulo
J.!wCc*t ; M/otM)
=
Lp'J.!w(U+ p'/p') . length M ® (Rp.jotRp')'
On the other hand by the lemma of Lech (24.4), we have limn~"" (length w M/Cot + L;+IafR)M)/nr-s = J.!w(c*; M/11 t M). Therefore, we have that J.!w(u; M)
=
limt,n~oo length w (M/Cot
+
L;+! afR)M)/t"'n
rs
lim t~oo (limn~oo (length w M/ ( 0 t + L; +I afR) M) / t n s = limt~oo Lp'J.!w(U p'/p') (length JJ;[ ® Rp,/otRp,)/t . S
=
r
S )
+
By (24.4) and (24.6), we see that lim (length
1~1
® Rpoj 0 tR p' ) / t
S
MIII,'I'II'I,II'1'I'1111I1
i~ 1'1/11/1,1
1.0 ('iLlH'I' p(
~I' iN ('(l'tHl 1.11
[,Ull' ; II/ C·) H\I'I
N III' 1I0!., 1I,lId i.lH'I,,,j'1I1'1' WI'
III'
~,I'I'II 1I,1'1',oI'tiillll:
"l'lIvn
II.I·j IIl,jll:lil.
(.ill' nHr-lI'I'j,illll.
25. Macaulay rings We Nay thaI. all ideal (1 oj' a l'illg iN 1:,~(}l!alh!! if nVl'l'y pi'ill\(~ diviNoI' P of a hm.; d(~pth equal to d(~pth a. It iN obviouN I.hal. all i:'iOlmL!ty ideal cannot have any imbedded prime divisor. We Nay thuL a Nemi-I()(~al ring R is unmixed if the zero ideal in the completion of R i" iNobathy .. It is known that there are local integral domains which arc not unmixed; an example is given in the Appendix, Example 2. (25.1) Let R be a semi-local ring with Jacobson radical m. If thcre is an ideal a of R such that the radical of a is m and such that the zero ideal in the form ring F = L Fn of R with respect to a is isobathu, then R is unmixed. Proof: Let R* be the completion of R. Then F is the form ring of R* with respect to aR*. Assume that there is a prime divisor p* of zero in R* such that depth p* < altitude R = altitude R* (by (17.12)). Set p' = L (p* an)/(p* an+I). Then p' is an ideal of F and F/p' is the form ring of R*/p* with respect to (ItR* + p*)/p*. Therefore depth p* = altitude (aR* + p*)/p* = altitude (FIF + p')/p' by (2:1.8), and this altitude is equal t.o depth p' by (14.5). Since p* is a prime divisor of zero, there iN an element a ~ 0 of R* Nllch that. ap* = 0. Let at be the a-form of a. Then we have p'a' = 0, whence p' consists only of zero divisors, whence p' is contained in a prime divisor q' of zero in F, whence depth q' :::::. depth p' = depth p* < altitude R, and the zero ideal of F is not isobathy. Therefore we complete the proof. A system of parameters al , ., . , aT of a Noetherian ring R is called a distinct system of parameters of R if Jl- ( LaiR) = length R / ( LaiR). (Note the inequality }t(L aiR) ::::; length R/L aiR, which was proved in (24.3).) A Macaulay local ring is a local ring which has a distinct system of parameters. A Noetherian ring R is called a locally Macaulay ring if Rm is a Macaulay local ring for every maximal ideal m of R. A N oethorian ring R is called a Macaulay ring if it is a locally Macaulay ring and if height m = altitude R for every maximal ideal m of R.
n
n
(25.2) THEOREM: A system of parameters al , ... ,aT of a Noetherian ring R is distinct if and only if, in the form ring F = L F,. of R with
('11,11"1'11)1(, III
I'/'NIH'I'I 10
L
n
fI,li',
.1'/
hl'll/('rr/ill /1Ir!r'fI/'IIr/r'lIl Ol'r'/, 1"11
1'1'001': 1,1'1. I'
L
(II,' 1IIIiIlilio 0")
(i
I, ... ,
I') //1'/' 111(/1'
"~/(1.
I'" lin I.lin liOIlIOJ,!;I'III'III1H polYllolllial I'illg ill X" OV!"I' N/n. Tlinll p.(I',/') C~ 11'lIg1.II N/Il hy (:JO.·ll. 1,1'1. (i' 1)(' 1.I1I' hOlllolllol'pli iH11I 1'1'0111 I) Oil 1.0 F Hlleh I.hal. ril( .\';) .1'; , :l.Ild lid, II I.n I.III~ k(,I'llId or e/>. W(~ lIav(~ 11'llgth /(/11 = /I({',/') C~ p.(l',/'; II) p.(I',/'; F) by (:~:~.:)). Obviow-;Iy, P.(J'IJ>; F) iH 1'1111:1,1 to M( .r/P), whidl iH eqlml to 101(0.) by (2:3.8). Therefore It'IIJ,!;1.11 N/n 101(11) if and ollly if p.(J\J>; n) = O. Assume that n 7"" o tl.lld Id J be a homogeneous element of n. Multiplying a suitable llil'lnclI!; or Po if necessary, we may assume that fm = 0 with a IIIlI,ximaJ ideal l1l of Po. Then O:fP = mP, and fP is isomorphic to 1'/1111>, whenceM(PIP; 11) 2: M(PIP; fP) = 101 (PIP; (Po/m) [XI, ... , X,.I) = 1. Therefore, M(PIP; n) = 0 if and only if n = 0, and the tl.H,~I~l'tion is proved. We have by (25.1) and (25.2) the following: illti!"j,I'I·llIillal.l'K .\', , .,. ,
L
+
7=
(:25.3) COROLLARY: If a semi-local ring R has a distinct system of 7111:mmeters al , ... , aT of the Jacobson radical of R, then R is unmixed.
(21).4) THEOREM: A system of parameters al , ... , aT of the J acoblion r'adical of a semi-local ring R is distinct if and only if ai is not a
~I'I'O divisor modulo L~-I ajR for each i = 1, 2, ... , r. Proof: Set 0. = L~ aiR. We prove the assertion by induction on r.
II' r = 0, then the assertion is obvious because the empty set is a diHLinct system of parameters of a ring of altitude zero. Assume that 'I' ::::: 1. Assume at first that the ai form a distinct system of parameters. Thon, length R/a 2: M(a/aIR) 2: p.(a) by (24.3), whence the equaliCieH hold, and the ai modulo aIR (i = 2, ... , r) form a distinct H'yHtem of parameters of R/aIR. Therefore, by induction, ai is not a v,nl'O divisor modulo L~-I ajR for each i = 2, ... , r. Thus it is suffiI'ient to show that al is not a zero divisor, which is obvious by virtue of (25.2). Conve~sely, assume that each ai is not a zero divisor modulo I ajR. Then al is not a zero divisor and, by induction, the ai modulo aIR (i 2: 2) form a distinct system of parameters of R/aIR. L(~t F = L F n be the form ring of R with respect to 0., let XI, ... ,Xr 1)(\ indeterminates and lete/> bethe homomorphism from P = (R/a) [Xl 2 Ollto F such that e/>(X i ) = ai modulo 0. • Since the ai modulo aIR U 2: 2) form a distinct system of parameters of R/ aIR, the form ring /1'* of R/aIR with respect to a/aIR may be identified with P/XIP,
Li-
HI
~1111
11'11'1 ,\1 '1'1'1 WI
wllil'lI HiIlIWH 1,1111.1. 1.11(' 111'1'11(,1 II III' (I' in 1'!llIin.illl,d ill .\' 1/' /1,11(1 Llia!. /1"1' idl'lliili('d wilit /I'/
When bi is defined, let g E P be a homogeneous form of degree d + i i + 1 such that ¢(g) = alb i modulo ad + +2. Since the kernel of tho natural map from F onto F* is¢(XIP), we see that g E XIP, whence there is a homogeneous form g' of degree d + i such that g = Xlg'. Let c be an element of ad +i such that ¢(g') = c modulo ad+i+l. Then d i 2 alb i == alC modulo a + + • Set bi+l = bi - c. Then bi - bi +1 = c E d d a +\ alb i +l = alb i - alc E a +i+2, and bi +l is defined. Now, let b* be the limit of the sequence (b i ) in the completion R* of R. Then we have alb* = O. Since al is not a zero divisor in R hence in R* by (lS.1), d we have b* = 0, which is a contradiction to bi ~ a +\ and the assertion that n = 0 is proved, whence the ai form a distinct system of parameters.
(25.5) THEOREM: If ai, ... , arform a distinct system of parameters of a N oethcrian ring R and if a system of parameters b] , . . . , br of R generate an ideal 0 which has the same radical as a = aiR, then the b i form a distinct system of parameters. Proof: By the nature of multiplicities, we may assume that R is a semi-local ring, and that the radical m of a is the .Jacobson radil:al of R. Note that every maximal ideal of R has height r by (253). We note also that the distinctness is not the property of the members of system of parameters but is the property of the ideals generated by them. We prove the assertion by induction on r. If r = 0, then a = 0 = 0 and the assertion is obvious; if r = 1, then, since R is unmixed, bl is not a zero divisor, which proves the assertion by (25.4). Assume that r > L By the above remark, we may assume that aiR blR has height 2. By (25.4), the ai modulo alR(i ::::: 2) form a distinct system of parameters, whence by induction, every system of parameters of m/alR is distinct, which means that every system of parameters of m which has al as.a member is distinct. Since height aiR blR = 2, there is a system of parameters of m having al and bl as members, and it is distinct by the above result. Hence, applying
L
+
+
1'11.\ 1"1' 1'1 It III
111(' IIIHII'I' 10 {" 1IIId 10 liliH IHHj, H,VHj,I'1I1
01'
or 1'11.1'11 II 1('1.1 '1'H,
WI' HI'I' I.hll,1, Hlly
iH diHj,illl'l., :l.lld 1.11(' h, 1'01'111 H di,..,l.illld, HYHLcll1 01' pal'alllni.('I'H. WI' HHy LIm!. Lill' I/.I/.III:I:./'(,r/I/.(!SS f/t,(!(}l'cm holdN ill a Nootherian ring R il' III(' I'ollowing iH L"np: II' all ideal n of R it:: generated by s elements and il' hl~ighL n ~~, S (N I:all bo ally non-negative integer) then every prime diviNol' 01' It iH oj' i\(light s. (intilw Lhit:: terminology, we can state the following characterization or 11)(:ally Macaulay rings.
H,YHII'1I1
PII.I'/l.IIIl'i.I'I'H
or
III whi('h IlnH
h,
HH II. 1111'1111>1'1'
ill 1'1I,l'l.il'IIII1,I'
(:,),:).0)
THIDOREM:
A Noetherian ring R is a locally Macaulay ring
U(lnd only if the unmixedness theorem holds in R. Pl'oof: Assume first that the unmixedness theorem holds in R, let 111 be an arbitrary maximal ideal of R. Let aI, ... , aT (r = lil'ight 111) be elements of 111 such that height I:~ aiR = s for any N ::; r; the existence follows from (D.5). The validity of the unmixodIII'HH theorem implies that each ai is not a zero divisor modulo ajR, ",hellee modulo ajRm , too. Therefore the ai form a distinct sysLI'IlI of parameters of Rm , and R is a locally l\1:aeaulay ring. Con1'1'I'Hdy, assume that R is a locally Macaulay ring, and assume that 11011 ideal a is generated by s elements aj, .. , , as and that height n = s. Let 111 be an arbitrary maximal ideal of R such that a C 111. 'l'lwn there is a system of parameters of Rm which contains the ai as a /'wiJt::et. By (25.5), we see that snch a system of parameters is necesHHl'ily distinct, whence Rill/aRm has a distinct system of parameters hy (25.4). It follows that R»,! aRIIl is unmixed by (25.3), whenco nN,,, has no imbedded prime divisors. Sinee this is true for any 111 I:ontaining a, we see that every prime divisor of a is of height s by virtue of (9.3), and the unmixedness theorem holds in R; and the pl'oof is eomplete. N ext we prove a eharacterization of Macaulay rings. IIolld
I:{
I:;
(25.7) THEOREM: A Noetherian ring R is a Macaulay ring if and only if one of the following conditions is satisfied: (1) Every system of parameters a1 , ••• , ar of R is distinct. (2) If maximal ideals 111, 111' (111 may be equal to 111' if R is a local 'I"/ng) of R are given, then there is a distinct system of parameters al , ... , ... , ar of R such that clIcry ai is in 111 n m'. Proof: Assume at first that R is a Macaulay ring. Let aI, ..• , a r be a system of parameters of R and let 1111, •. , 1118 be all of the prime
HO
M111i/'II'I.lI'I'I'llllIl
L
diviHOI'H III' n t.lld!'
U,
H,li! ,
WIII'III'n
1I,h',
Hilll'I'
\VIII'III'!' !.Ill' II, 1'111'111
hy
(~:1.1
h'
I\,
iH II.
1\1 11,1'/1,1 Ii 11..1'
diHt.illl'l. ~I,YHklll
I'ill/l:, IH'i/l:ill. III,
or
lI.il.i
,,11,1'1I,/lI('I.I'I'H ill l'il.I'1i
)
and (I) iH provnd. II, iK oi>viollK thuJ, (I) implinK (:2). AKHlllllP (,haL (~)
holds good. H iN i'Hlff1(:inllt (,0 show that R is a Mal:au'lay ring. Ln(, rat, ... , ar \ be a distinet system of parameters and set It = a,;R. Theil !-l(a) = length R/a = L length Rn/aR n , where 11 rum, through all prime divisors of a. On the other hand, by (23.1), )L( a) = L !-l( aR n,), where 11' runs through all prime divisors of a such that height 11' = r. Rince length Rn./aR n, 2: !-l(aR n,) by (24.3), we see that height 11 = r for any 11 and that the ai form a distinct system of parameters of R" . Hence (2) implies that height 111 = height 111' and that Rm , R"" are Macaulay local rings for any two maximal ideals m and m', whence R is a Macaulay ring. Thus the proof is completed. The following remark is immediate from the definition: (25.8) A semi-local ring R is a Macaulay ring or a locally Macaulay ring if and only if so is the completion of R. On the other hand: (25.9) If R is a locally Macaulay ring, then every ring of quotients of R is also a locally Macaulay ring. Proof: By the definition, it is sufficient to prove that Rp is a Macaulay ring for any prime ideal p of R. Let r be the height of ).l and let al , ... , a,. be elements of ).l such that Lf ajR is of height s for any s :::;; r (by (9.5». Since the unmixedness theorem holds in R, ai is not a zero divisor modulo L~-l ajR for each i, and therefore the ai form a distinct system of parameters of Rp by virtue of (25.4), which proves the assertion. The following is a generalization of the classical unmixedness theorem, because a field is obviously a Macaulay ring:
L
(25.10) THEOREM: Let Xl, '" , Xn be algebraically independent elements over a Noetherian ring R. If R is a locally Macaulay ring then so is the polynomial ring R[x). Proof: The general case follows easily from the case where n = ], and we assume that n = 1. Xl is denoted by x. It is sufficient to prove that if m is a maximal ideal of R[x), then R[xl m is a Macaulay loeal ring. Set \l = m R. Then Rp is a Macaulay loeal ring. In order to
n
'/J,7
I'll ;\1"1' 1<111 '"
P"OI'I' 1,IiI'IIHH('I'j,ioll 1'01' 141'111' , wn IIllly IIHI~I"lll! Lilli!, N· Nv , Ill/pNI,r! If'( /I. IIl/I.~jlll/l.l idl'lloi of (N/p)I,/'/' WIWIII'1! 11I/~J/4rj iH I!;nlH1l':tLnn 11011 1~11~lllnlll. of 111 HIWh Chat f' = J modulo 1.lNI,I'I, I,('/, 1/( , '" , (1", he a diHLiliet HYHtmn of parameters of R = R~, By ((i.!:l) alii! (2GA), we H()I~ that aI, ", , a r , f form a distinct sys1.('111 of PHI':t.IlWCOl'H of Il[:t:]", , and the assertion is proved, (2r), II) (JOItOJ,LAHY: Let Xl, , ' , , Xn be algebraically independent ('/I"lIw'nls over a Macaulay ring R, Then R[Xl, "" Xn] is again a Maennla,Jj nng if and only if there is no prime ideal 13 of R such that (I) P is not maximal and (2) there is an element a EE 13 such that
(Rlp)[l/(a modulo 13)]
'ill lhe .field of quotients of Rip, I 'I'oof: The non-existence of 13 as above is equivalent to the statement 1.11/1,(; every maximal ideal m of the polynomial ring R[x] lies over a maxi11111.1 ideal n of R, by virtue of (14.10). The proof of the if part: Since height n = n lieil!;ht min = n by (14.5), we see that height m 2: n ",ll.iLude R. Since altitude R[x] = n altitude R by (9.10), we see 1,linL height m = altitude R[x] for every maximal ideal m of R[x], ",,,d the assertion is proved by (25.10). The proof of the only if part:
+
+
+
AHHume the existence of such a p. Hence if follows that there is a Illltximal ideal m of R[x] such that n = m R is not maximal. Applyilll!; the above observation to m(Rn)[x], we see that height m = linight n + n. Therefore height m < altitude R + n = altitude R(x], whence R[x] is not a Macaulay ring. The following two remarks are obvious by (25.4) and (12.9). (25.12) A Notherian ring R of altitude 1 is a Macaulay ring if and only if no maximal ideal is a prime divisor of zero. Any Noetherian ring of altitude zero is a Macaulay ring. (25.13) A normal Noetherian ring R is a Macaulay ring if every 'maximal ideal of R is of height 2, We prove furthermore that;
n
(25.14) THEOREM: A regular local ring is a normal Macaulay ring. Proof: Let (R, m) be a regular local ring and let Xl, .. , , Xr be a I'ogular system of parameters of R. Then the form ring F = L:: Fn is lI:onerated by r elements over the field Fo = Rim, and altitude F = r by (23.8). Therefore the r generators must be algebraically independent, whence R is a Macaulay ring. That R is a normal ring follows from the following lemma:
~IIII,'I'II'I'/'
(:.l{,.1 r,) /'1'1 ( h~. III I II(' f/ /I/f'ft! I'IIIf!,
mil/!.
'/'f'S/JI'I'i
/1/ III
1:8 III/,
il/lf'r/I'II.!
'1'1'11':11
/f
Ihf' /01'11/ I'il/f! '"
dl/I/Illill, I//('I/
II'
~ /1'" f~r Ii.
IN fl/I il/lf'(/I'I/.( dl/I/llt'://..
If /I' '1;8 (J, 'fI,o'I"rnnl
)',:",U, I,hf'''' Ii. i8 n 1/,11/,1/1/1.1 1'11/(/, Proof: ab = 0 (a, II ( U) illlpiiPH 11,'1>' () il' ft' nlld // aI'(' /11·1'01'1111' of a and b. Therefore we have pl'OV(~U !.Iw find, :LHH('I'Lioli. NnxL wn HKsume that F is normal. Assume that a/h (a, !J ( N) iN illkgmi ,)V(\J' R and let c 'F 0 be an element of the conductor c of R ill Ula/hl, Wn are to prove that a E bR + mn for any n, by inductioIl on n. W(~ may assume that a, b are in m, Then the above is true for n = 1, Assume that a E bR + mn, i.e" a = bq + I' withq E R, I' E 111n. ItiH n sufficient to show that a E bR + 111n+1. If I' E m +\ then there iH n 1 nothing to prove, and we assume t.hat I' EE m + , Since alb = q + (r/b), we have c(r/b)m E R for any m, i.e., cr = b"'d m wit.hd m E R, m Let c', 1", b' and d~ be m-forms of c, 1', b, dm rei:lpeetively. Then cr = bmd m implies t.hat c'r,rn = b,md~ or c'r,m = 0 or b,md~ = O. Since II' is an integral domain, we see that c'r,m = b,md~ , whence c'(r'/b')'" E F. Since this is true for any natural number m and since F it.; tt normal Noetherian ring, we have 1" /b' E F. Then theI'e ii:l all element e of R such that the Til-form of e it.; 1" /b', which means that eb modulo m n+1 = 1", whence I' - eb E 111n+1, and, siIlf'e a = b(q + e) + (I' eb), we get a E bR + llln+l. Thus a E bR + mn for any n, and a C nn (bR + 1Il") = bR (by lG.7)), whence alb E R. ThuH the normality ifl proved. Another interest.ing re:mlt on regular loeal ring is the following rtl
(25.16) THEOREM: Assume that (R, lll) is a regular local ring and that a ring R' is a finite R-module and furthermore that R is a subring of R'. Then R' is a free module over R if and only if R' is a Macaulay ring. Proof: Let Yl, ... , Ys be clements of R' whose residue classes modulo mR' form a linearly independent base of R' /lllR' over R/m. Then R' = RYi by the lemma of Krull-Azumaya. Let Xl , ... , Xr be a regular system of parameters of R. R' is a Macaulay ring if and only if t.he Xi form a distinct system of parameters of R', or equivalently J.l( mR') = length R' /mR' namely
L:
J.lR(mR')
=
length n R'/mR' = s.
Let F be a free R-module with free base Y 1 , ••• , Y s and let 11 be the kernel of the homomorphism ¢ such that ¢( Y i ) = Yi. If n 'F 0, n contains a free module. Therefore
(I"
,11"1' 1111/, III
/1.1111 11)(· (''Illldily IilildH if /I,IIIIIIIIi.l' il' II 0, i.f'., h~' iH 1'1'('(\. 'I'IIIIH I.lill Pl'olll' i~4 "Olllpll'l,('. ()II !.Ii(' 1I1,1i('1' 111L11d, (.JI(' rollowill/!; iH oilvio\lH I' 1'0 III I.hn dd;llil.;oll: (~;).17) ;1.~SIlIl/(· llinl, Ihl' 111.1'/1/,11111 ItJ' 1/'(/'/I..~ill;/in holds/orllu! N/ll'/hl'r /11/1 /'':n(f,~ N n/l.d U'. '1'//./'/1. U is /l, Jl/llmu/nll riJ/,1/ ,,;r nnd (In/lI,;r .~/)':.~' h~'. Wn :t,dd 1\t'1'(~ I.ltn following; LI\(~()I'nllJ, wltidt iH a (!ol.'oll:t,I''y 1.0 (~!i.I;)): (~!i.I~) 'l'III<:violl:->. A:->:->lIme that R/n iH regular. Theil I.IWI'(~ iH :t. I'cglliar system of parameter!:) Xl , ... , Xr of R such that
Xl, .,. , Xs
Ea
nlld :-meh that the residue dasses of Xs+I , ••• I'(~glliar system of parameters of Ria. Since
1).1' (:25.14) or (25.15), we see that
II =
, Xr
modulo
a form
;1,
I:i xiR is a prim(~ ili('n.l I:f x.Jl, which eomplnLPH
I."(~ proof. /I;XERCISE: Assume that an ideal a of a Macaulay ring R is generated by (lluments and that height a = '1'. Prove that Ria is a Macaulay ring.
l'
26. Definition of syzygies Lnt R be a NooLlwriall rillg, lIHlmlly a 10enl ring, thl'OlIgholll; LlliH ~(i(:Li()n.
Ld Ul , •.. ,Un be elements of an R-module M and let U1 , ... , fJ" I>n indeterminates. The set N of elements L aiU i (ai E R) of (,\W free module generated by the U i such that L aiUi = 0 is an R-modulp, whence it is a finite R-module. This last R-module N is ealled t1w 1'(dation module of the elements Ui. Note that (L RUi)/N iH nal;lIl'HlIy isomorphic to RUi' When the Ui form a minimal baHiH fol' M, the relation module N of the Ui is called a relation module of M. Nol;e that, in this last case, if R is a local ring with maximal ideal Ill, I,hull N C L mU i as is easily seen by virtue of (5.3).
L
(26.1) THEOREM: If R is a local ring, then a relation module of (~ II£'/Jcn finite R-modtde M is unique up to isomorphism. If VI, '" , 71" ,,;,~ ~ basis for M, then the relation module N of the Vi is the direct sum oJ IIIl! relation module of M and a free module. Proof: The uniqueness of the relation module of M is an immediate consequence of (5.3). As for the last assertion, we may assume by virtue of (5.3) that VI , " . , Vm is a minimal basis for M. Let VI, . . . , ... , Vn be indeterminates. For each i > m, there is a relation II.j = Lf=1 aijVj , whence the relation module N of the Vi contains ILII element fi = Vi aij V j . Since the V,: are free, we see that )'11/+1, '" ,fn generate a free submodule N* of N. On the other hand, N contains the relation module N** of VI, •.. , Vm , and N** is the I'olation module of lYI. We are to prove that N is the direct sum of N** and N*. It is obvious that N* N** = O. Therefore it is sufficient 1,0 show that N is generated by N* and N**. Let g = L biV i be 1111 arbitrary element of N. Then 9 - L;':+1 b,ji is in RV1 RVm , whence the element is in N**. Thus N = N* + N** and the /tssertion is proved.
L:7'
n
+ ... +
[91]
'1'111'1 '1'1111111111' 1)/1'
By Vil'Llln of (~(;'I), WIII'Ii ,,~ iH
/I,
1\,,~,\'IIIil:r\
IOI'/I,II'ill/l: lVI' 1~/l.1I "1'lilll' 1.111' lIoLioli
of lJyzygie8 of fini(.c modl/II~ Hi"! I'ollowi"!: Let R be a local ring and let M hl~ a filliLn N-IIIOd"II'. Tlan mil i"!Y:'.yJ.!;Y of M is M itself; when the ith syzygy Zi of M iH ddillnd, Lito (i + I) Lia syzygy of M is the relation module of Z i. The iLh Hy:'.ygy of 111 i,.., denoted by syzi M; if we want to express that M is an R-moduIC', WI' write syz~ M. Even if R is not a local ring, we can give the following notion of a weak syzygy sequence of a finite R-module JJ![: A sequence of finitl~ R-modules Zi (i = 0, 1, 2, '" ) is a weak syzygy sequence of M if Zo = M and if each Zi( i 2:: 1) is the relation module of a finite basis for Zi-l . Zi is called a weak ith syzygy of M. (26.2) If M = Zo, ZI, ... , Zn, .. is a weak syzygy sequence of a finite module M, then, with an arbitrary, multiplicatively closed subset S of R such that ~ S, Zo ® Rs , ZI ® R s , .. , , Zn ® Rs, ... is a weak syzygy sequence of M ® Rs over Rs . Proof: Let 11,1, •• , , Un be a basis for Zm such that the relation module of the Uj is Zm+l . Let N be the relation module of the
°
Ui
® 1
(1 E R,s).
It is obvious that Zm+l ® Rs is contained in N. In order to prove the converse inclusion, let La~ Ui be an el(ement of N, i.e.,
Since the elements of S are mapped to units, we may assume that a; are images of elements ai of R. Then L a:(u ® 1) = implies that L aiUi is in the kernel K of the natural map from Minto M ® Rs , whence there is an element s of S such that L a;sui = 0, and L Ui E Zm+l ® Rs . Thus N = Z",+1 ® Rs , and the assertion is proved. (26.3) When R is a local ring, any weak itk syzygy Z of a finite module M is the direct sum of the ith syzygy of M and a free module, hence the relation module of Z coincides with syzi+l M. The proof is immediate from (26.1). From now on, let R be a local ring. A finite module M is said to be of homological dimension n (over R) if n + 1 is the smallest i such that syzi M = 0. In symbols, we write hd M. = n; when we want to express "over R" explicitly, we write hd R • The definition implies that: (26.4) Let M be a finite module over the local ring R. Then: (1)
°
a;
if fll/d (llIluif III
1111 III
0,
(~)
lid M
() '(/' (t'Iul only if M
(:l) if M ~ 0 and if F is a free //lOr/II/I', 1/11'11 II(I M = hd(M E9 F), (4) hd(syzi M) = max(-l, lid III - ',;) anti (Ii) ~f Z is a weak ith syzygy of M then hd Z is equal /(1 hd M - i for i ::; hd M; it is equal to either or -1 for other i. III
II ,1'1'1'1' 1I1I1//ltll' tll)il'/'('II/ fl'/I/II, Z(:/'II,
°
(:W.Ii) THEOHEM: If a submodule N of a finite module M is given, Ihl'//', for each n, there is a weak nth syzygy Zn of M and an isomorphism If from syzn N into Zn such that ZnlCT(syzn N) is isomorphic to
syzn (MIN). I'roof: By the definition of syzygies, we have only to prove the case wltnrc n = 1. Let nl , ... ,nr be a minimal base of N and let m~ , ... , , .. , m: be a minimal base of MIN. Let mi be an clement of M such l,hnL m: = mi modulo N. Let Zl be the relation module of the base nl, ... , n r , ml, .. , , m., of M. L aiMi E syzl (MIN) if and only II' L a,m: = 0, that is, L aim, E N, or equivalently, L aimi + bjnj = with some lij E ii, whence syzl (MIN) is the image of :11 by the map ¢ such that ¢( aiMi + L bjNj) = L a;lYf; . The Iwmel of ¢ is the set of clements of Zl of the form L bjNj , which is obviously the relation module of N, i.e., syzl N. Thus the proof is I'omplete.
2:
°
L:
(26.6) COHOLLAHY: Among the homological dimensions of N, M, and M IN, one of the following must hold: (1) hd N ::; hd M = hd( MIN), (~) hd 111 ::; hd N = hd(MIN) - 1, or (3) hd(MIN) < hd M lid N. Proof: Let Z" be a weak syzygy of lYI sueh that Zn/syz n N ~'yzn (MIN). Case I : ASimmethat hd N is minimal among homologieal dimensions of M, N, MIN, and let n be hd N. Then Zn+l is isomorphic to syzn+l (MIN). Therefore, either hd M = hd (MIN) 01' ~'yzn+1 AI = 0. In the latter ease Zn+l is a free module, and therefore we see that, in ease I, either (1) or (2) holds. Case II: Assume that hd M is minimal among the homological dimensions of M, N, MIN, nnd set m = hd M. Since syzm M is free, syzm N is the relation module oj' a base of syzm (M)N), henee either both syzm (MIN) and syzm N nrcfree and different from zero or hd (syzm N) = hd (syzm+l (MIN)), nnd therci'ore either (1) or (2) holds. Case III: The remaining case iH where hd (MIN) is smaller than both hd Nand hd M. Set n' = hd (MIN). Then Zn'+l/syzn'+l N = 0, i.e., Zn'+l = syzn"I'1 N. By
III
'1'11111 '1'1111111111' 1111' IIH,I'IIIIIIII
11111' II,HHI II II pl,ioll
wn ItIl.V(~
!.ImL hd (II/ IN)
iH 1111111,\\1'1'1,\11\.11\1111,,,
lid N /1.1111 ltd nI,
(:1). '/'ItII14 I,hl' prolll' iH 1·lIllIpll'I",.
11;~(II:ItI'INI':H: LI1L
U 111,11, N"dhnrill,lI l'illJl; II.lld 1'''"Hid,,1' Iillil.n 11i",IiIl"H OVIll' ,,~. modliin Ai iH II, '1II'IIjIW/'il'l: 'lllIIt/III" il' M Iiii N", iH 1'1'111, 1'01' IUlY maximal idlJILl 111 or u. Th" hOl//,o[o!/,£I:/J,1 tlhll{;II,,~io/(, of II IIIOdlll" M iH d"fillnd 1.0 lie Lhe maximum (or tho HII\lrmnllm) Ill' 1i0IllOlogil::d U '" over R", . 1. Prove that hd M :::; n (n ::::: 0) if and only if an arbU,l"lIl'Y wcn,k nt,h HyZygy of M is projective, or equivalently, if and only if a suitable weak nth HyZygy of M is projective. 2. Prove that if Z and Z' are weak ith syzygies of the same module M, then there are free modules F and F' such that Z Ell F ~ Z' Ell F'. 3. Generalize (26.5) and (26,6) to our case. 4. In (26.2), assume that R' is a ring which is an R-module and such that 18>R' is exact. Prove that Zo 18> R', Z, 0 R', ... ,Zn 0 R', '" is a weak syzygy sequence of M 0 R'. .
W"
HHy LIla" 1J,
27. Change oj Rings (27.1) Tm]OREM: Let M be a finite module over a local ring (R, m) and let x be an element oj m. Assume that x is not a zero divisor with respect to both Rand M. Then for every natural number n, there is a natural isomorphism between (syz n M) I x (syz n M) and SYZ~/XR (MlxM).
Proof: Using the induction on n, we have only to prove the caSe where n = 1. Let Ul, ... , Ur be a minimal base of M. Then u' = Ui modulo xM form a minimal base of MlxM. L a:U i E syzl (MlxM) if and only if L a:u~ = 0, i.e., L aiUi E xM with ai such that a: = ai modulo xR, namely L aiUi = L xbiui. Thus the map ¢ such that ¢( a,U i ) = (ai modulo xR) U i is a homomorphism from syzl M onto SYZ1/XR MlxM. The kernel of ¢ is the set of elements xaiUi such that xaiUi = 0, whence aiUi = because x is not a zero divisor with respect to M. Th us the kernel of ¢ is x (syz 1 lrf), and the assertion is proved. We say that a sequence of elements Xl, ... ,X r of the maximal ideal m of a local ring R is an M -sequence, M being an R-module, if Xi is not a zero divisor with respect to M/L~-l xjM for each i = 1, '" , r. Then we have:
L
L
L
L
L
°
(27.2) COROLLAIty: Let M be aflnite module over a local ring (R, m). If Xl, '" , Xs is an I11-sequence and at the same time an R-sequence, then hd R }J![ = hdR/oM I aM with II = L xiR.
('III 1"1'1'111 I V
(~7,:11 'l'1111:IIIII':M:
/,rl ill III' (/ ./illill' IIlIIdll/I'
(/1/1//1'1,/'111' II//I'!I'IIII'III
Id
III /I'hil'" iN
/101 ill
III'/'I' /I /lIl'Itll'ill(/
(/t', 1111
III", ,lssllI/lI'.I'III'I!lI'l'lIllil'I'I!Jn{
/11111/ '::/'I'II dipislIl' ':/1. /i', Ihol. M -i8 l'onlrt.i/II'r!':1I I/, .1'/'1'/1 U/.oduk F ,,~{ Ii ( {I" ' , , , (I,. Iwi/l(/ I/, fl'(,(, 111/,81 1 I~( /1') IUUf, thai, ,rF ~ JI,l ~ 111/1'. '/'lil' II , fol' I'lli'll. I/I/./I/i/,(/./ 1/.1/./11111"1'1/., HYY." ill m'fl, hi: imlwddr:d in a Iter: II/Udllk /1'" '£1/. SI/.(lh (!, 'W(/,I! Uw,/. :r/I'n C HYII" A;j C lIl!?n and such that (HYY." JI/) /.d l '" '/:s nalu,raUy isomorphic to SYII{;lxR (M / xF). Pl'oof: lJHill1!; illdudion on n, we have only to prove the case where I/. "= I. Lei; a~, ... ,a; be a minimal basis for M/xF and let aI, ... ,as
,/' iN
L
1)(1 elements of M sueh that a~ = ai modulo xF. Then aI, .. , , as, .rl1, , .. , , xUr is a basis for M. They really form a minimal basis for M, since otherwise there would be a relation Clal + .,. + csa s + d,:tU l + ... + drxU r = 0 with some Ci EE m or some d j EE m. If some (:,: ~: )11, then this contradicts the fact that the a~ form a minimal basis for M/xF; if some d j EE m, then the coefficient of U j in L Ciai is not ill )112, whence some Ci EE m (because M emF) which is not the case. Thus the ai and the xU j form a minimal basis for M. syzl M is therefore the relation module of aI, .. , , a r , x UI , .. , , X Us. L ciA i + L bjVj E syzl M if and only if L Ciai + L bjxU j = O. Let (J be Lhe map from syzl Minto FI = L RA i such that (J( L CiAi + L bj V j) = L ciA i . This (J is obviously a homomorphism. Element of the kernel of (J is bjVj with bjxUj = O. Since x is not a zero divisor and since Uj are free, we have bj = 0 and therefore (J is an isomorphism. Since the ai and the xU j form a minimal basis, we have (J(syzl M) C mFl . Since xai E xF, we have xAi E (J(syzl M). Thus :lPI ~ (J(syzl M) C mF I . L c;A i E syzlAxR (M/xF) if and only if L cA i E xF with Ci such that c; = Ci modulo xR, or equivalently L ciAi E (J(syzl M). Thus we see that
L:
syzAlxR (M/xF)
L:
=
(J(syzl M)/xFl,
and the assertion is proved completely. (27.4) COROLLARY : With the same notation and assumptions as in (27.3) above, we have hd R M = 1 hd RlxR M/xF, except for the case
+
where F = O. Proof: If both hd M and hd M / xF are infinite, then the assertion is obvious. Therefore, we assume that one of them is finite and we prove Lhe corollary by induction on the finite homological dimension. hd RlxR 111 /xF = -1 if and only if M /xF = 0, i.e., M = :rF, or equivnlently hd M = 0 because, in the notation in the proof of (27.3) the
II(I (/, Hlld
'1'1111\ '1'111,1(111\'
.I'l'/
1'111'111 II. lllildlilid 11I1./liI1 1'111'
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III.
Hilll'I'
1.I'1i
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ill
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WI'
" I1(\ a,~:-:I'I'1.111I1 . II,V .1111 I11('1.11111. . 11I'OVI'J
ca.;;) COI((lLLA H,\': lVillt. /.II.(: Nf/./l/l' U, III, (/./I{I .1' n.~ 1/.11111'1', Id III Iii' fI, .finite H-'Ynotin/f: ,~uch Olat .r!l1 = O. 'I'h('/I. lid/,' !11 I + Iid/,'/J'{': Ill) except for the case where M ,= O. Proof: Let UI , '" , U r be a minimal bake of J\1. L c J 1; ( HYII:I M if and only if L CiUi = O. Then with F = L R Ui , we have :tF ~ syz1 M emF, and SYZ1/XR M = (syz1 M)/xF. Therefore, we can prove the assertion, using (27.4). We give, by the way, some results on M-sequences .. Let (R, m) be a local ring and let M be a finite R-module. For an M-sequence Xl, . . . , Xr the number r is called the length of the Msequence. An M-sequence Xl, '" , Xr is called a maximal M-sequence if there is no element y E 111 such that Xl , .. , ,X r , y is an M-sequence. (27.6) With the notation as above, zf Xl, .. , , Xr is an M -sequence, then any permutation of the Xi is an M -sequence. Proof: If we show that Xl and X2 permute with each other, then the general case follows immediately, using the induction on r from the fact that X2, .. , , Xr is an (M/xd\1)-sequence. Thus we arc to prove the permutability of Xl and X2. Assume that X2m = 0 (m EM). Then, since X2m E x 11l1, we have m E x 1N1:x 2R = xllJ;[, and m = xlm' with m' E 111. Thus xlx2m' = O. Since Xl i" not a zero divisor with respect to M, X2m' = O. Thus we have O:x2R (in M) is contained in Xl(O:X2R), whence by the lemma of Krull-Azumaya, we have
i.e., X2 is not a zero divisor with respect to N1. ASHume that m E X2M:XlR. Then Xlm = X2m' with m' E M. Then m' E x 11l1:x2R = xlM, i.e., m' = xlm" (m" EM). Then Xlm = xlx2m". Since Xl is not a zero divisor with respect to M, we have m = X2m" , and x2111:xIR C x2M. Thus X2 , Xl , X3, . , . , Xr is an M-sequence. (27.7) With the same notation as above, if there is an M-sequence Xl , . . . ,X r , then there is an N1 -sequence YI , . . . , Yr such that their residue classes modulo m2 are linearly independent over the field Rim; if the Xi form a maximal1l1-seqllCnce, then so do the Yi . Proof: Assume that Xl, . , . , Xs (s may be zero) modulo m2 are linearly independent. In that case, let YI, •.. ,Ys be the Xl, . , . , Xs .
1'11.\ 1"1'11111 I V WI' PI'II\'I' 1111' 1I1111111'Cioll II.\' illlllll'Lilll1 Oil fllll'll /1,11 8,
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IS 1101, II. 1:1'\'11
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8/',1111'11 1.\11'1'('
III In III , I.III'I'P n I· III" (III' J'nUII~J', (nM: 111)+ wi Lit 1'(':-IPI'I'1. 1,11 III IIlM, \lIW:LII:-II'
,I'" 1~4 11111. /I. 1,1'1'11 dIVIHIII' wIl.li I'I'HPI'I~L 1.11
III' 111 wllll'lI i:-l 11111. III
II'
I'. HI'L il
2.:'; iH
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nil I'lpllIl~IIL !I
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LII(~ :-11'1, III' 1:('1'(1 dIVi~II/'~ wiLh I'('~PP('I, 1,0 1111111111'1' IIJ' p!'ill\(~ id('al~ (:-,,~o 1';x('I'(li:-le
M luM i~ Ul(~ union of a finite I ill §8), Then :£1, '" , Xr-l , Y
iH 11.1\ JI1-Hoq\WIt('(~, and w(~ arc reduced to the case of greater s (by (:,)7,(;)). ThlH it remains only to prove that if the Xi form a maximal III -Hcquence, then so do Xl, ,., , Xr-l, y. Considering M luM and U/(uM:M), we may assume that r = 1 and that M is faithful. That ,1'1 is a maximal M-sequence implies that every element of m is a zero divi80r with respect to MlxlM, whence there is an element m of M wll ich is not in xlM but is in xIM: m (cf. Exercise 1 in §8). Then 11m = xlm' for an m' E lJ!!. If m' = ym* for an m* E lt1, then m = xm* which is not the case. Therefore m' ~ yM, Let z be an arbitrary eleIllent of m. Then zm = xm" for an m" E M, whence xym" = yzm = ,I'zm', and ym" = zm'. Thus z E yM:m'R. Thus every element of 111 is a zero divisor with respect to AII yM, and the maximality of 1/ is proved. Thus (27.7) is proved completely. (27,8) CORtlLLARY: Maximal M -sequences have the same length. Proof: Assume that Xl, ,., , Xr and 1/1, ." , Ys are maximal MH(~quences. We prove that r = s by induction on r, If r = 0, then the HHsertion is obvious; if r = ], this has been proved in the last step of the proof of (27.7). Assume now that r > 1. Then the union of the sets (If zero divisors with respect to M I xlM and M I ylM does not cover 111, whence there is an element z such that both Xl, Z and Yl, z form MHnquences. Let Xl, Z, ZR, " . , Z t and Yl, Z, W3, ,." w" be maximal M-sequences. Considering M IxdYI, we have r = t by induction, ConHidering M/zM, we have t = u, and, considering Mly1M, we have 'I/, = 8. Thus r = s, which completes the proof. EXERCISES: Let M be a finite module over a Noetherian ring and let a be an illeal of R. Then the notion of M-sequence in a is defined simibrly using a inHtead of m in the previous definition. 1. Confirm that, under the assumption that M laM T" 0, (27.6) and (27.8) (Jan be genemlized to such a case. 2. Prove (27.2) when R is a semi-local ring and when the Xi are in the Jacob~on mdical. 3. Adapt (27.4) to the case of semi-local rings.
11K
'1'111'1 'I'll 1':01/ \
11/" i1 \ 'f,\ 1111'11 \
2M. U('lllllm' IO('t11 1'I1I1l.\' Wn Il('gili wiLli /I, 11'111111/1,: (::lKI) I,d. (N, 11/) 1)(: a {,I/(:(/,( 'rill!!. If 0: III / 0, /Ju:n hd M ,D fll/' eveTY finite mod'ute M which i8 'not II'I'(:. Proof: Assume that hd M = n < 00 amI Chat n ?:. I. 'I'lwll :..;yll /I III is free and different from zero. Since syz n M i::; j,he l'elaLiol\ modltin or syzn-l ].II, there is a free module F such that syzn M ~ lllf?, wh(~II(',n (O:m)(syzn M) = 0, which is a contradiction. Now we prove: (28.2) THEOREM: Let (R, m) be a local Ting. If hd m is finite, then R is TegulaT. ConveTsely, if R is a TegulaT local Ting, then faT eveTY finite R-module M diffeTent fTom zeTa, we have hd M =c altitude R - s, wheTe s is the length of a maximal M -sequence. Proof: Assume that hd m is finite. We prove the regularity of R by induction on r = altitude R. We begin with the following remark: If O:m ,L 0, then m must be free by (28.1), whence m = 0 because 0: m ,L O. Therefore, R is a field in this case. Now, if l' = 0, then 0: m ,L 0, and the assertion is proved already. Assume that l' > O. Then by the above remark, we have 0: m = O. Therefore, there is an element x of m which is not in m2 such that x is not a zero divisor. Then hd m = hd R / xR m/xR 1 by (27.4), whence RfxR is regular by induction. Therefore we see that R is also a regular local ring by (9.11). Conversely, assume that R is a regular local ring and let M be a finite R-module different from zero. Let Xl , ... , Xs be a maximal M-sequence. We may assume, by (27.7), that Xs ~ m2 if 8 ?:. 1. We prove the equality hd M = l' - S (1' = altitude R) by induction on 1'. If l' = 0, then R is a field, and M is free (8 is obviously zero). Thus this case is obvious. Assume that l' ?:. 1. If s ?:. 1, then hd M = hd R / xsR M/::c M by (27.2), whence by induction hd M = (1' - 1) (8 - 1) = l' - s. If s = 0, then M is not free, whence syzI M ,L 0 and hd M = 1 hd syzl M. Since syzl M is contained in a free module, y E m (y ,L 0) forms an (syzl M)-sequence; if it is maximal, then by the above proof, we have hd (syzl M) = l' - 1, and hd M = r. Thus, it is sufficient to prove that every element z of m is a zero divisor with respect to (syzl M)/y(syzl M). Let F be a free module such that F/(syzl M) "" M. Since z is a zero divisor with respect to M, there is an element a of F which is not in syzl M such that za E syzI M, whence yza E y(syzl M). But ya ~ y(syzl M) because y is
+
8
+
('11111"1'11111, IV 1111(, II, Y,PI'I)
divir4ol' II.lld
111'1'/1,111'1(11/
II HY If, , II!,
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(:.lH,:I) (!O/l.Old,AI1I': IJ' p /8 (/,fII'/I/I/' il!('(/l /~r /I /'I'(/lIlu/, lul'/l.!, '/'/11(/ h', h\, ':8 //, 'I'I'l/n{(I./' lo/'/r!, '1"1:11(/, "I'OO\': lid p iH 1;lIil.(', wlH,tll:n 1111/"" ~>Iip iH lillil.n I,.v (~o,:.n, wllidl
IIM'/I.
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Lnl. (/(" Ill) h(: a I'n~I\III,J' IO(Ial J'illp; :l,lullnL 71 h(: 1.11(: cilal':I.(,i,(!I·iHI.iI' oi' iH (:allnd an 'Il'f/,r'(un'!j,:/,r/ '/'c(J'I1.lllr local r"';n.(j if nil.l\(\l· f{, ('OI\(,II,iIIH II,
U/lli.
n
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2
• Namely, H iH 1I1lmmilicd if H.l\d only if H/p!?' iH I'<'P;IINow we Imvn: (28.4) If P is a prime ideal oj an unrarrtified regular loca,l rinfJ H, Ihen Il~ is an unramified regular local ring. Proof: Let p be the characteristic of the residue class field or U. II' pll Q;; lJ, then R~/lJR~ is of characteristic zero, whence R~ contailiH I,ll<: rational number field, and R~ is unramified in this case. ASRlIlll(', I.hat pR C lJ. Since R/pR is regular, we see that (R/pR)~!l'''' oc Nu/pR~ is regular by (28,3), and the assertion is proved, (28,5) Let (R, m) be a local ring and let a be an ideal of R. Assunw that a = xR:yR (x E a, y E R) and that x ~ am. If hd a S 1, then 1\ = xR. Proof: Assume the contrary. Since x if am, there is a minimal basiH a" aI, ... ) an for a. Let b be such that xb = yal . By the assumption Hyzl a = {cX + L CiAi I CX + L Ciai = 01 is a free module, Let i" ... ,fm be a free basis for syzl a. syzl a contains elements a1X a:Al and bX - yAl' whence there are elements dij such that a1X :1:A 1 = Lj dlifj and bX - yAl = Lj d2ifj , y(a1X - xA 1) = xbX :1:yA = x(bX - yA), whence we have yd1j = xd2j for any j, which implies that d1j E xR: y = a, whence the coefficients of did; are in am, which implies that x is in am because a1X - XAI = L d1dj, whieh iR a contradietion and we prove the assertion. (28.6) Let:1: and !J be elements oj a ring R. If J: is not a zero divisor, awn either J:R + yR is principal or hd (xR + yR) = ] + hd (xR:yR). Proof: A:osume that .rR + yR is not. principal, t.hon, since hd xR = 0, we have by (26.6) that hd (xR +yR) = hd (xR + yR)/xR, whence it is equal to hd R/(xR:yR) by (1.5), and the a:o:oertion is proved by the fact that xR:yR = syzl R/(xR:yR).
fidd
OJ'
m
111,1'.
(28.7) THEOREM: An arbitrary regular local ring (R, factorization ring.
111)
is a unique
10(1
'1'111,) '1'1111)01'"
Oil' t\\'~,"llIll)t\
!'I'OOi': 1.1'1. " hI' III(' nILiI,lIdl' 01' N 11.11(11(11. II 1.(1 11,11 II.I'hiLI'/I,I'Y pl'illl(l id(,ld (II' 1\(~i/!;liI, I ill Ii, W(I 11,\,\1(' oilly 1,(1 P"(lV(' Lllld, ~I iH pl'ill('ipnl. II' I' •.~ I, LlI('11 L11(\ HHHI'ltioli iH ol,vioIlH, AHH(lIII(\ 1.1111.1, .,.. :!, II' ;~ ( III i:-: 1101. ill p, 1.11('11 Z rOI'IIIN all U/).I-HI\qll(\III~(\, wllil~1I illlplinN Lh:I,(, hd N/p <: 'I' I, 1\(,1\(,1, 1\(1 P ~ 'I' - ~ "nl:II.IIH
induction on r. Set. a = xR + yR. Since x is irreducible, if a is prineipal, t.hen yR = R, and p is principal. Therefore, we· assume t.hat. a is not principal. Let. t be so large that a' : m = a' with a' = a: m t. Let z be an element of m which is not in m2 such that a':zR = a' . R' = R/zR is regular, hence it is a unique factorization ring by induction. Therefore hd R , (a + zR)/zR ~ 1 by (28,6), whence hd R , (R/a + zR) ~ 2 < altitude R ' , which implies in particular that (a + zR)/zR:m/zR = (a + zR)/zR, t
by (28.2), whence (a + zR):m = a + zR. Therefore a' = a:m C (a + zR): mt = a + zR, Since a':zR = ai, a' C a + zR implies that a' = a by (4.3). Therefore, there is a maximal R / a-sequence Xo, ., • . . . , Xs such that Xo = z. Then Xl, '" , Xs is a maximal R/ (a + zR)sequence. Since hd R , R/(a zR) ::; 2, we have s :::: r - 1 - 2 = r - 3 ,whence hd R R/a ~ 2, and hd a ~ 1. Since we assumed that a is not principal, we have hd a = 1, whence hd p = 0 by (28.6) and p is principal. Thus the proof is complete. We say that a Noetherian ring R is a regular ring if R Itt is a regular local ring for every maximal ideal m of R. Under this definition, one can assert the following corollary:
+
(28,8) COROLLARY: A regular semi-local integral domain is a unique factorization ring. This follows immediately from the following lemma: (28,9) Let (R, 1111, ... , 111,.) be a semi-local integral domain. If every Ri = R Itti (1; = 1, '" , n) iii a unique factorization ring, then R is a unique factorization ring. Proof: Let p be an arbitrary prime ideal of height 1 in R. Then pR i is generated by an element Pi of p. Let ai be an element of nll mi-l mi+l mn which is not in mi. Then pRi is generated by Piai . Set P = L Piai . Then since pjaj E pmiRi for i 'F j,
... n
n
n ... n
n
1() I
I'll \ 1"1'1'111 I V
II
WII tll'I' 1,11111.
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add,
111'1'1', HOIIII' l'I'lIlnl'l(H Oil IIOII··)'('glilll,t· 1(lI:al l'ill!2;H.
(:.lS.IOJ /,1'1. (It', III) IIf'
(I, /II(:(/'I, 'I"';"'r/. /1 881/./lU: that an .1: ( 111 ,is not a n.finiil' 1////(i'II.I(: 111 'F 0, it hold& that heL" M = 1 + Itd{";,'1 (HY1.' 1I1)/'r(HY1.' M). I '!'Ool': Itd ll M = 1 + hd", Hyzl M. Since x is not a zero divisor and Hi 11(,(: HY1.' 111 iN a Nubmodulc of a free module, we see that hd R syzl M = lid ",/,.", (Hyy,l M)/X(syzl M) by (27.2), and the assertion is proved.
:('/'/1
dill/mI'. '1'11.1'//./111'
(28.11) Tn.MORIDM: Let (R, nt) be a local ring and let M( 'FO) be n .Ii.nite R-module. Let s be the length of a maximal M-sequence. If hd M iH .finite, then hd M + s is equal to the length t of a maximal R-seq'uence. Proof: We prove the assertion by induction on t. If t = 0, then 0: III 'F 0, and the assertion is true by (28.1). Assqme that t > 0, /1,11<1 let Xl , ... , :0t be a maximal R-sequence. If s 'F Q, then we may /I,HHllme that Xl i" not a zero divi"or with respect to ],';[ by virtue of (:!7.8). Then hd" M = hd R / xlTl Mj.TlM. (X2 modulo xIR), ... , (Xt Illoclulo xlR) is a maximal R/xIR-sequence, and if Xl , Y2, ... , Ys is a lIIaximal M-sequenc:e, then (Y2 modulo xIR), ... , (Ys modulo xIR) is fI, maximal M/xIM-sequence by definition. Thus we have settled this (:11,He. Assume that s = 0. We can prove that Xl form::; a maximal (Hyzl M)-sequence, in the same way as in the last step of the proof oj' (28.2) (with y = Xl)' Therefore, by the case where s > 0, we see Chat hd syzl M + 1 = t, and hd M = t, which completes the proof. 1'~XERCISES:
1. Let a be an ideal of a regular local ring R. Prove that if
p, , .. , , lJn are all of the prime divisors of a, then hd a is not greater t.han the
IIItLximum of the depth lJi + 1. 2. Let R be a Noetherian ring. Prove the equivaJence of the following condiI,ions: (1) hd M :::; n for every finit.e module M, (2) R is a regular ring such that IIHitude R ;;; n. 3. Assume that M is a finite Ria-module (a being an ideal of a local ring Il). I't'Ove that hd Rla M + hd R Ria = hd R M, provided that each of them is tillite.
29. Syzygies of graded modules Since there i8 a very close relation between graded ring::; and semilocal rings or between graded rings over loc:al ring" and local rings, we Itdd some remarks on syzygies of graded aodule". Let R = L Rn he a graded Noetherian ring sueh that Ro is a 10c:a1
III',',
'11111\ '1'111 11(111\' 1111'
~\\1,\1111'1I\
L"
I'illg willi 11111.\illlll.l idl'/I,I 1111), HI'I III 1111) I I h'", III iN oill'iollNly 11, Ill/I,xil1\1d ilil'/I,\. 1.1'1 M 1)(' H lillill' /l;I'HIIi'd 11111111111', WI~ d('(illl' Ii, NIP/lllli ,~/'ql//'I//'/' or III /1,,'4 rollnw,'4: TII(~ lil'HL 1111'1111)('1', wllidl IIlay 1)/· d/'liol.l~d I,y HYY,I) M, iN 11/ il.NI'IL WIII'II Clip ",1,11 1111'111111'1' NYX" I 11/, whidl iN a gl'/Likd [illil.l~ m()dllll~, iH ddilll'd, WI' ddill(\ !.Iw
+
('II. I )NC Il\mnl>I~1' NYX" 11/ aN folioWN: Ldo 'UI , '" , U r IH~ a Illillimal IIaNiN for NyxlI I M (',onNi,,('ing only of homogeneollN nli~In<~III.N, of dl~
gl'ce, NUY, d1 , , . ' , dr. Let U1 , . , ' , Ur be ind()t()rminat(~N and <:011Nider the relation module N = (L aiU i I L aiUi = 01. If we regard U i as an element of degree d i , then N is a graded module, ThiN graded module N is defined to be syzn M. Then: (29,1) Every syzn M is unique up to isomorphism. Proof: If VI , ••• ,Vs is another similar basis for lYI and if d~ = deg Vi , then we see, first, that there is a linear transformation from the members of Ui with smallest d i to such ones of the Vi, hence the same can be generalized to be the case of the Ui with d i at most the second smallest and such ones of the Vi , and so on, and we see that r = sand there is a linear transformation which maps the Ui onto the Vi , whence syzn M is independent of the choice of bases. Thus "syzygy sequence" is well defined, and therefore the same treat.ment as in the case of local rings can be applied. But we need not repeat. the same again, because of the following theorem. (29.2) THEOREM: With the same notation as above, we set R* = Rm. 'Then syz;. (M ® R*) is naturally isomorphic to (syz; M) ® R*. Proof: Since we are considering graded modules, any element a of R which is not in m is not a zero divisor with respect to modules, whence M is naturally contained in M ® R*. One can see easily that a minimal basis for M consisting of homogeneous elements becomes a minimal basis for M ® R*, whence we see easily that syz~. (M ® R*) is naturally isomorphic to (syz~ M) ® R*. Applying the same to syz; M instead of M, we prove the assertion. By virtue of the above results, one can state a generalization of the classical syzygy theorem of Hilbert as follows: (29.3) THEOREM: Let R be the homogeneous polynomial ring in algebraically independent elements Xl, . , . , Xn over a regular local ring Ro . Then jor any finite graded module 11{ over R, hd M is at most n altitude Ro , and with t = hd M, we have syzt+l M = o.
+
l-'heory of Complete Local Rings and Its Application 30. Some properties of complete local rings (:30.1) THEOREM: Let R be a complete semi-local ring with Jacobson 'I'adical 111. If un ( n = 1, 2, 3, . ,. ) are ideals of R such that lln C Un+l for any n and such that nn.Un = 0, then for any given natural n1lmber n, llime exists a natural nnmber m(n) such that llm(n) C 111n. Proof: Assume the contrary, namely, assume that there is a natural r lIl/mber r sueh that U 'II 111 for any m. Then, for any n :2: r, U m mn. Hillce altitude Rlmn = 0, the minimum condition for ideals holds in N/m n , whence there is a natural number t(n) such that at(n) + mn = n,n + mn for any m :2: t(n). We may assume that t(n) < t(n + 1) n 1'01' any n = r, r + 1, '" . Then Ut(n) C Ut(n) + mn = Ut(n+l) + m , Hiid therefore for any given element Xn of Ut(n) there is an element n ;t~"+l of Ut(n+l) such that Xn - Xn+l E m , Starting with an Xr E Ut(r) r which is not in m , we have a sequence of Xn as above. Then the Xn rorm a Cauchy sequence, which has a limit x* in R. Since Xn , Xn+l, ••• 1I,I'e in Ut(n) and since x* is the limit of the sequence, we have x* E Ut(n) I)'y the closedness of ideals. I-Ience x* E nn Ut(n) = 0, and x* = 0. On Lite other hand, since Xn - Xr E mT for any n, we see that x* - Xr E mr , whence Xr E nt'", which is a contradiction.
%
It
%
(30.2) COROLLARY: If a complete semi-local ring R is dominated by semi-local ring R' which may not be Noetherian, then R is a subspace
of R'. Proof: Let m and m' be the Jacobson radicals of Rand R', respecLively. Then obviously m" C m,n n R. By (30.1), we have m,m(n) n Il ~ m n , and the assertion is proved, We say that a quasi-local ring (R, m) is a Henselian ring if the following i8 true: If a monic polynomialf(x) over R is such thatf(x) == (J1I(x)ho(x) modulo mR[x] with monic polynomials go and ho with the l103]
1(11
'1'111111111\
PI'IIPI'I'I.I'
III" I'IIMPI,I'I'I'N I,III'!\!, 11111;(/11
111111 (l1Ih'1,1'1 I hllh'I,1'1 I IIINI,I'I
m"t
POIYlllllllill,le1 (1(,1') 1i,lId Ii(,,') Hlll'il IIl1i.1,/(,I')
O( ,I')
(III ( ,"), h( ,")
"11(.1') 11.1'1' ill
(I( ,"
1111'11 111,'1',' 111'1' 1IIIIIIi"
III( ,")
I\.lId HIII'it
CIlIioL
IIIUI,I'I,
Ill\dl'l' CitiH LI'I'llIiIlOloll:.Y, WI' l'Ii,1I 11.14141'1'1. 1,1,11.1,:
(:m,:l) '1'lIlmll,I'IM: If (U, 1Il) is (J. mll/pit:!,(! IU(:fI'/' ri/l.!1 'whi(:h,l/.fI,l! II.O!, 11f: N Of' lIwT'/a:n, !'hen R ":,,, a 11 r:fi8!il inn r"inll, We prove thiH Uwol'cm in the following form, whieh iH JllO],(, l!;(,IH,ral ill appent'n1H:e: (:10.4) If (R, m) is a complete local rino which may not be Noetherian, f(x), ho(x) aTe polynomials in an indeterminate x over R and if Oo(x) is a monic polynomial in x over R such thatf(,y) - Oo(x)ho(x) E mR[x] (Lnd such that Oo(x)R[x] + ho(x)R[x] + mR[x] = R[x], then there are polynomials O(x) and h(x)( ER[xJ) as follows: f(::c) = o(x)h(x), o(x) - Oo(x) E tnR[x] , hex) - ho(x) E tnR[x] and O(x) is a monic polynomial, Proof: We may aSimme that deg ho + deg 00 :::; deg f. Starting with 00 and ho , we construct sequences of polynomials On (:r;) and h" (:1:) such that f - O"h" E m,,+1R[x], 0" - 0,,-1 E m"R[x], hn - h,,-l E mnR[x], deg hn + deg On :::; deg f, and sueh that On is a monic polynomial (for every n): Namely, when an and h n are already defined, then we define 0,,+1 and h,,+l as follows: Since 00 - On and 110 - hn are in mR[x], we sec that On , h" , and m generate R[:r], whence there are polynomials ai( x), bi ( x) and mi( x) such that Xi = Onai + hnb, + mi with mi E mR [x] , Since On is monic, we may assume that deg bi < deg 0" = deg 00 ' By the existence of the term mi , we may assume that the coefficients of ai and bi are units or zero, Then we sec that deg ai :::; deg.f - deg On if i :::; deg f. Now we write f - Onh" = L CiXi(Ci E 111n+1). This last sum is taken up to the term whose degree is equal to d = deg f· Therefore, f - O"hn = On( Cia'i) + hn ( Cibi) + L Cimi, and deg (L Ciai) :::; deg f - deg On , deg (L cib i ) < n 2 deg (/n, Cimi E m + R[x]. Set 0,,+1 = On + cib i , hn+1 = hn + Ln C,ai. Then f - On+1hn+1 = L Cimi - (L cib i ) ( L Ciai) E m +2R[x], and as is easily verified, On+1 and hn+1 satisfy the requirements. Thus the existence of the sequences is proved. Since R is complete, we can consider the limits of (On(X)) and (hn(x)); let them be g(x) and h(,y). Then we see that f - Oh E mnR[x] for any n, and f = gh, and \vt' prove easily that these 0 and h are the rrquired elcments. Since some general properties of Henselian rings will be observed ~r
L
L
L
L
('11.\1"1'1,111
lOll
I'
IIIL('/' ill ( 'lill,pll'l' VII, illl'IIIdillg IIII' 1'11.1'1. ClmL ('lil' 11t'IIKI,1 1('11111111, or UII' 1'01'111 ill (:10,11 i1, 1\, gl'III'I'II,1 propl'I'Ly or 111'111-I1,linll I'illgl-l, \VI' t41111.11 IIO\. I/l'l'i\'I' HII.\' of 11i1'111 HI. IH'I'~I'ltI" I'XI'I'Pi. 1'01' 1.111' followill/.!:: (:IO,r,) If Ii is I/, /l1'lIsl'tinnillle!lJ'nl dOl/III,,://. (/,//.110' It' 'I:,~ ITn ":/1[1'(11'111 1',I'II'II,~i/i1l 'I~r Ii, Ili/'/l N'/:8 If /I,/I,~'/:-I/lcal, 1'1'001': 1.1'1 III hi' III(' IlIa:-:illl:l.1 ideal of N, A:otlumc that H' hU8 maxi111/1.1 idl'al~ '"', '""(111' ~ 1/1"), alld let a be all element of m' which is 1101, ill 111", ~"kL N" = Hlaj and let
!Iii ILII irreducible monic polynomial over R which has a as a root, i"\illce a Em', we have Cn E m. Since a ~ m", an ~ mR[a], and there is /I, I:,: which is not in m. Let j be such that Cj ~ m, Cj+s E m (for Ii> 0). Then 1 ~ j ::::; n - I, andf(x) := (;7;j + CIX i - 1 + .. , + Cj)x n - j IllOdulo mR[xJ. Since R is Henselian, f(x) must be reducible over R, which is a contradiction, and the assertion is proved. (:~0.6) THlDoREM: Let R be a complete semi-local ring with Jacobson 'I'((li'ical m. Let M be an R-module. If MjmM is a finite R-module and ':I'the m-adic topology of M is To , then M is a finite module: Let Ul , ... , , , , Us be elements of lJ;[ such that lJ;[ j mM is generated by their residue l'Iasses, then lJf = L RUi. Proof: Set N = L~ RUi. Let a be all arbitrary element of M. We W/I,lIt to show that there is a sequence of elements al , .... ,an, .. , of N :ouch that an = L mniUi with mni E m n- 1 and such that
a -
L~ aj E l11nlJ1.
We use induction on n. The case "vhere n = 1 is immediate from the ILHI-Illmption. Assume that ai, ... , an are already defmed. Then n a. - L~ aj = mib i with mi E m , bi E M. Let Ci be elements of N 1'11(,h that b i - Ci E: mlJr, and set an+l = L miCi . Then an+l is the I'equired element, and the sequence is well defined. Set = mni /Ll\d a* = L m:ui' Then a - a* E mnM for any n, whence a = a*, which implies that M eN, and M = N.
L
m;
Ln
EXERCISES: 1. Assume that a local ring R, with principal maximal ideal Till, is dominated by It semi-local ring R' which may not be Noetherian. Prove I,/Int R is a subspace of R'.
2. Generalize (30.1) to the case of finite module Over a complete semi-local
ring.
100
1,(,1, N
'1'11111(11/\
III'
/I,
III" (1IIMI'ldl\'I'I': 1,(11',\1, IIIN(III
H(\lI\i .. lo('all'illl!; wil,it ,JH('OIIC~((1I l'II,lIil'lI,1 111 11,111111'1. ,1'1, ., . ,
... "I'" III' nlnlll('lli,H 01' III. 1,('1, I III' /I, HliI)I'illg 01' N. '1'11(\11 pownl' H(,I'inH ill I.Iw ,/'1 wiLli ('o('(Ji('inIII,H ill I hns 1ll('IIollilig ill ('\1(, (',olllpl(\l,iolJ U* 01' U alld I,h(, :·wl, 01' all SlId\ POW(,I' s(\ries 1)(\(~OIl\(\H a sllIll·jllg or U*. This sllbrill!!; is d(ll\o\.<-<1 by 111:1'1 , ... , :(;,,11. It'llI'1.1WI'IllO[,(', if X, , '" , X" :\1'(. illde1;(,rminai.(\,'i, I.h(\1\ tiwr() is a homomorphism ¢ from JIIX, , ... ,X" II onto Il[x1, .,. , x"Jj over I such that ¢( Xi) = Xi' If (,h(, homomorphism ¢ is an isomorphism, we say that :r1, ... , Xn are anal!Jtically independent over I. We note that if I is Noetherian then I[[X1, ... , ... , xnll is Noetherian. If I is a semi-local ring or a .local ring, then so is I[[X1, ... , x n ]]. (The proof is immediate from (15.3) and (15.4).) The main result in this section is the following:
(31.1) THEOREM: If (R, m) is a complete local ring which may not be Noetherian, then R contains a ring I which satisfies the following condition: Let p be the characteristic of Rim. Then m I is generated by p (i.e., p-fold of the identity), I is a complete local ring and Rim is naturally equal to I I (m n I). Consequently, if Ia A} form a basis for m, then every element of R is expressed as a power series in the a A with coefficients in I. (STRUCTURE THEOREM OF COMPLETE I,OCAI, RINGS) Such a ring I, as above, is called a coefficient ring of R; if I is a field, then I is called a coefficient field of R. It is obvious that I is a field if
n
and only if R contains some field. Proof: We begin with the case where p = O. Let I be a maximal subfield of R; the existence follows from Zorn's lemma. We want to show that this I is a coefficient field. Since m I = 0, I is regarded as a subfield of Rim. If x' E Rim is transcendental over I, then with an x such that x' = x modulo m, I (x) is a subfield of R, which contradicts the maximality of I. Hence Rim is algebraic over I. Assume that Rim contains an clement a' which is not in I and let f(x) be the irreducible monic polynomial over I which has a' as a root. The same f(x) is regarded as a polynomial over R. Since p = 0, a' is a simple root of f(x) modulo mR[x], whence f(x) = (x - a)g(x) with an a E R such that a' = a modulo m, by virtue of (30.3) or (30.4). This implies that I (a) is a subfield of R, which contradicts the maximality of I, and Rim = I. Thus this case is proved. We consider the case p -,:6. O. Before proceeding with the proof, we give some preliminaries. When K is a field of characteristic p -,:6. 0, a subset B of K is called a
n
107 I'
'InSI'
oi' /1 ii' /I Hllli,dil'H I.hl' i'ollowillg I.wo l'oll(liLiollH: ( I ) 1\
=~
1\ I' ( U) ,
(~)
ii' /'1, ... , {,,. u]'(\ IlIutually di"tinct clements of B, then /),,):1\".1 = p". 1IIIIkl' Lhis terminology, we have the following lemma:
111111
1/1 "(11
1 ,
(:{I.~) /U1'lI
."
,
An arbitrary field K of characteristic p -,L 0 has a p-base,
n. For any natural number n and for any mutually distinct elements
"I, .. , , b
r
of B, we have that K [Kpn(b 1
,
•••
=
Kpn(B) and that pn ,br ) : K ] = pnr.
I It'oof: The existence of B is easy by virtue of Zorn's lemma, considerillg subsets of K satisfying the second condition in the definition. We pi'ove the other assertions by induction on n. Since the map ¢ such that pn c/)(n) = aP is an isomorphism from K onto K P , we have K P = K (B P ) P hy induction and therefore K = [(1ln(B). pnr 2: [K \b 1 , ••• ,
pn ... ,br ) : Kpn] = [K (b 1 , . . . , br ) :Kpn(bi , ... , bn]· [KP\bi , ... , pn 1 P ) 'K]!n] > r • • . , br ' _ p "[Kpn--l(1 . II, . . . , b) r''K - ] -_ p' p (n-1)·,. . 1'1lllS'
WI~
prove (31.~). We note that (:n.2) implies that every element of K is expressed as pn II. polynomial in elements of B with coefficients in K in such a way Chat the degree of the expression is less than pn for each member of B /l,lId that such an expression is uniquely determined by the element or K. Next we prove another lemma: (31.3) Let p be a prime number and assume that an ideal m of a ring
pn - b Em, then a Il" E m n+1. Conseq1lently, if, furthermore, m t = 0 and if m is a maximal 'I:deal, then the map ¢ such that ¢( a) = a P' with s 1 2: t induces a one-one map from the field Rim into R. q q 1 Proof: Set c = b - a, q = pn. Then bq = a qa - c q (;) aq-"c r c . If r = ptr', (p, r') = 1, then a simple calcula/?, contains p (i.e., the p-fold of the identity).
If a
+
+
+ '" +
+ .. , +
Lion shows that (;) is a multiple of pn-t, whence (;) E m,,-t. Since
> t, c" E mt+\ and therefore the assertion is proved. Now we proeeed with the proof of (31.1). Let B* be a p-base of f( = Rim and fix a set B of representatives of B* in R (i.e., we take only one b E R for each b* E B* such that b* = b modulo In, and B 7)1
IOH
'1'1111:(11/1
1111'
I'II~II'I"'I'I'II:
1,111'\1, IIINIIII
iN 1.111' Hi'1. 01' 1-1111,1111),1 1'01'1'11.1'11 I 111,1.11 I'HI 1111111111'1' 1/ 11'1. 'i'" III' 1.111' IIlnp 1'1'11111 NIIII" illl.o il.HI,11' giVI'1i I,y 'i'" (0. 1 (/,I!''', '1'111'11 'i'" illillll'I'H I/. lilli' IllIn Il\ap 1'/'11111 NIIII illl.o h~/III" h.v (;\1.;\); I.Ili:-: lilli' olin Iliap iN ill'III1LI'iI hy ITn' Wn (kIlIlLI~ by II n 1.111' illlngl' III' H/llI by (/" ' 1.1'1, 8" III~ Lill' NI'I, 01' polynomialN ill einnll'nLN III' /) I.aknll \l\lIdulo III" wiLli codli('.i(,IIL:-: ill /1" such that the degree of 1,he polynomial ill n11<:11 dnllwlIl. 01' /) i:.; IeN:-:
than p2n. Since CPn induces on K = Rim an isomorphiNlll from 1\ ollLo IJ2n KP2n, there is a natural one-one map between elemenLt-: of K alld 2n An (such that (a modulo m)P corresponds to (a modulo mn)p2n). Therefore, we have a one-one map from K onto Sn, and Sn beeomeN a complete set of representatives of Rim in Rlmn. Set I n = Sn + pSn + ... + pn-1Sn (this notation means that I n is the set of elements of the form L~~~ aipi with ai E Sn. We want to show that .1n is a ring. If it is known that the sum of two elements of So. is in .1n , then we see that the sum of two element8 of .1n is in .1n , and furthermore, since the product, of two elements of Sn which are monomialR in the elements of B are in Sn , we see easily that the product of two elementR of Sn is in J n , whence the product of two elementH of .In iR in I n • Thus, in order to prove that .1 n is a ring, we have only to show that. the Rum of two element" L ak,1'VI and L liMM (aM, liM E An ; M = b;) '" Ii~r with Iii E B, 0 ::; Ci < p211) is in .1n . We prove it in the form that piC L aMM -I- L Ii Ml'Vl) iN in pi.1." by inullei,ion on i starting with i = n (until i = 0). ThiN last asseri.iolt i:-: obvious for i = n. Assume that 0 ::; i < n. Set q = p2n. For each M, aM and bM are in An and therefore they are qth powers of element" C M and d M of Rim". Since aM and b M are uniquely determined by the residlle elasses of CM and d M modulo m, we may assume that CM and riM arc
. S n. N ow, aM m
+
bM
=
( CM -I- d M) q -
(;) arc multiples of p, we see that piC
(Q) CM q-rdM.
1 ",aL."l r
r
o' "mce
I>.M (;}~rd~ )1\11 i8in pHlJn
by induction because CM, d M are in Sn. Therefore L pi(aM
+ h)M =
L pi(CM
+
dM)aJV[
+
piC Lr,M
(;)C~rri~)M
is in piJn. Thus, the above assertion ii:> proved, and we see that I n is a ring. Obviously J n ml 111 n = pJ" , whence pJn is a maximal ideal of I n • The one-one correspondence between Sn and Rim now induces the natural isomorphism between JnlpJ n and Rim. We want to show
n
I'IIA 1"1'1'111 V
lOll
II(\~L LlIII.I, LlII' 1l/I,i,III'n1110IIIOllllll'pltitHIl II" 1'1'0111 U/III" Olli,o "~/I"" I ill 11,1111,1.111'11,1 hOlllOlllOl'pliiHl1i 1'1'0111 .I" 0111.0.1" I, II, iH ohvioliH UmL 1r,,( III ) l'I'I'ol\ll'H /1.11 ('11'11\('111, 01' ,'-,'" I ,111'111"(' 7r,,( 8,,) C:.I" I, whieh implil'H LIm!. '71'" iH H h01\1011IOl'phiH1l1 1'1'0111 .1/1 illto .1/1-1. On the other hnlld, iL iH ol,violiH hy Lh" nOIlHLI'llId,ioli !.ImL (~adl element of 8 n - 1 is in 71",,(,'-1,,), whidl implinH Lh:1.L 71",,(.1,,) l,onLaintl '/n-I' Therefore 7I"n in"III'('M a lIaLlIl'al hOIllOIlIIlI'phiHIlI 1'1'0111./" onto '/n-l' Now, let {an} be a M(\II'II'III'(: of <:I(:III<:IILH Hildl Lfiai;(l" E I n and such that 7I"n(an ) = an-I' 1,'01' nHeh Hileh "eqll<:lIec {an}, let {b n } be a sequence of elements of R Hili'll Lhat bn modulo mil = an. Then it is obvious that Ibn} is a regular H(~qILCllec, hence there is lim bn • This last limit depends only on the Hnqllcnee {an} as is easily seen. Let I be the set of such limits. Since nHeh .Tn is a ring, I is a ring. Furthermore, each regular sequence in I with pI-adic topology comes from such a sequence {an} as above, whence I is complete in its pI-adic topology. Therefore we see easily lI.H in the proof of (15.1) that I is a quasi-local ring with maximal ideal 'liT. Hence, in order to prove (31.1), it is sufficient to prove that I is Noetherian, which follows from the following two lemmas: "II"PH
(31.4) If a quasi-local ring (R, m) is dominated by a local ring ( /l', m') which may not be Noetherian, then R is a local ring which may not be Noetherian. Proof: This is obvious because m em'. (31.5) If a local ring R which may not be Noetherian has principal maximal ideal pR, then R is Noetherian. Proof: Let II be an arbitrary ideal of R such that R 'F II 'F O. Then there is a natural number n such that II C pnR, a ~ pn+lR. Then n:pnR is not contained in pR, whence a:pnR = R, which implies that 11 =:J pnR and a = pnR.
(31.6) COlWLLARY: If R is a complete local integral domain, then R contains a complete regular ring 8 such that R is a finite 8-module and such that S = I[[:CI, , .. , x r ]] with a coefficient ring I (if R and analytically independent clements Xl, ••• , Xr over I. Proof: When I if) a field, let Xl, ••• , Xc be a Rystcm of parameters of R; when I is llot a field and if the maximal ideal of I is pI, let XI, .•. , Xr be such that p, XI, . . . ,Xr is a system of parameters of R. Set 8 = I[[xI, .,. , Xr]]. (30.6) implies that R is a finite 8-module, whence altitude R = altitude 8. Let Xl, ... , Xr be indeterminates, Then since Xl, ... , Xr or p, XI, ... , Xr form a regular system of parameters of I[[X I , . . . , Xr]], we have altitude I[[XI , ••• , Xrll =
110
1I,IUl,lIdn
h~. '1'111'1'1'/'01'(\
I)(! 11.11 iHOIIlOI'plliHIII,
!'lin "'lid
1111,1,\/1'/1,1 II III,ppi IIj,I; 1'1'0111 I,hl' Il.rlHIII'l.ioli
lei
III.\' 110111,0 11\.1'1\
AH II.IIOLII(!I· l:o/'OIIHI''y 1.0 (:11.1), WI' 1111.,,11 !.III' followillJ,!; "il'l,lII~ of
I II II HI,
pl·o"l.d. I'I'HIlII. I),V
(11i.:l):
(:lI.7) COI(()LI,AHY: If ,,~ is a CO'tI/,7)/dn /I/(;al 'I"I:'/I.!I w":ichmay 'not /)(: Noetherian and if the maximal ideal 1Il of R It,a8 a .finitc ba81:8, then H 'I;.~ N octherian. As an application of (31.7), we prove the following: (:-n.8) THEOREM: A semi-local ring R which may not be Noetherian is really Noetherian if and only if: (1) every finitely generated ideal of R is a closed subset of R, and (2) the maximal ideals of R have finite bases. Proof: The only if part is obvious by (16.7). Assume that R is not Noetherian and that the maximal ideals have finite bases. Then there is an ideal which has no finite basis, whence there exists an ascending sequence a] C U2 C ... C an C ... of ideals an of R such that each of the an has a finite basis. Let R* be the completion of R. Then R* is Noetherian by (31.7) and (17.7). Therefore there is an n such that umR* = unR* for any m ~ n, which implies that an is not a closed subset of R, because llnR* R, which contains Um , is contained in the closure of Un • Thus (31.8) is proved. Next we give some remarks on the choice of coefficient rings. When (R, m) is a complete local ring which may not be Noetherian, such that Rim is of characteristic p -F 0, let K be the maximal perfect subfield of Rim. For each element a' of K, let bn be a representative of a,p-n in R and set Cn = b~n. Then since b~+l - bn E m, we see by (31.3) that the Cn form a Cauchy sequence, whose limit a is uniquely determined by a' independently of the choice of the bn • This a is called the multipl1:cative representative of a'. Note that a is a representativc of a' and that if [) is the multiplicativc repreHentative of an elemcnt b' of K, then ab is the multiplicative roprmlentativc of a'b', as is obviously 8cen by our definitioll; these are the reason:-1 for (·.alling a the mult,iplicative reprcsentatives of a'. Now we have:
n
(31.9) THl
III
('11.11"1'1,111 V
('//'1111'111(/
I I~r h'
1/l11I1'1t I'I>IIIII/IIN
If Ill1til'III'I'/lI'III:{,/it'il'lI./
,.':11(/ 1(1'
U l'IIII./I/,;/1.,~
IIII' IIIlIlli"lil'lllil'I' l'I'JiI'I'NI'llllIlifll'8 I(/' HIII'I"IIII'n/'8 I~r 1\. 1'1'0111': ( I) iH (lhv iOIiH hy I.itn wool' (II' (:ll.I). 'l'hn nxinj,()II(:e of 1 ('(llll.nillillg I.it(' giv('11 U wan I'mlly IlI'ov()d in Ow proof of (31.1). We
Ibn Ia,;!, hall' or (2). lJc;ing the notation employed just !.lin /)" (:an h() dlOl:len from a given coefficient rillg l' (II' N, wll('lI('() {,/w 1ll11ll,iplieaLive representative a of a' E K is in I'. /I./'n 1.(1
P"OV('
1)('1'01'(' (:: I.D),
(::1.10): COltOLI,AHY: Let (R, TIl) be a complete local ring which may 'lllIt 1)(; Nodhcr-ian such that m -F 0. (1) If Rim is oj chamcter-ist'ic zer-o, !/Wf!, U has only one coefficient field when and only when Rim is algel/miG over- the pr-ime field. (2) If Rim is of chamcteristic p -F 0, then U has only one coefficient r-ing when and only when Rim is per-fect. Lastly, we prove a structure theorem of ramified regular local rings. We note, first, that the classical theorem of Eisenstein on the irredlteibility of polynomials can be stated as follows: (31.11) Let ~ be a pr-ime ideal in a r-ing R and let f(x) = xn n 1 (LIX + ... + an (ai E R) be a monic polynomial in an indeter-minate :c over- R. Assume that all the ai ar-e in ~ and that an \f ~2, then f(x) is 'irred'ucible over R. Hence, if furthermore R is a normal ring, then f(x) is irreducible over the field of quotients of R. Proof: If f(x) is reducible, say f(x) = g(x)h(x) with monic poly/lomials g(x) and h(x), then, since f(x) == xn modulo ~, we have {/(x) == x", h( x) == x n- r modulo ~ and an E ~2 which is a contradiction, which proves the first assertion, from which the last assertion follows. When (R, m) is a local ring, a polynomial f(x) as above with 1.1 = m is ealled an Eisenstein polynomial over Rand R[xlIJ(x)R[xl iH ealled an it'iscnstein extension of R. Now we can state:
+
(31.12) THEOREM: Every complete regular local ring (R, m) is an fCise1Lstein extension of a complete unr-amijied r-egular local r-ing (Ro , 1110)' which is necessar-ily the power series r-ing in a finite number- of analytically independent elements over a coefficient ring of Ro . Ever-y Eiscn.~tein extension of a regular local ring is again a regular local r-ing. Proof: As for the first assertion, we have only to prove the case where R does not contain any field. Let I be a coefficient ring of R and let p be the characteristic of Rim. Let Xl, ... , Xn be a regular 2 Hystem of parameters of R. If R is unramified, i.e., if p \f m , then we may assume that Xl = p, whence R = I[[x2, ... , xnlL which proves the last statement of the first assertion. Returning to the general
I I :J
'1'111':0111
III!'
l'I)~II'ld,:'I'I'1
!,I)I',II, IIINIIII
I'.II,HI', WI! IIll1.y ItH,YIIIIII' 1.111\.1, I', .1'.,: , " . "I'" ill II, H,VHI.I'III or PII.I'HIII(
L;
°
EXERCISES: 1. Give a direct proof of (31.7) using the fact that the form ring of R with respect to the maximal ideal is aN oetherian ring. 2. Let CR, m) and (R', m') be complete local rings which may not be Noetherian, such that R' is integral over R (R ::; R'). Prove that if R'/m' is separable over Rim, then, for any coefficient ring I of R, there is a coefficient ring I' of R' which contains I. Prove furthernlore that such an l' is uniquely determined by I. 3. Let CR, m) be a complete local ring which may not be Noetherian. Assume that R contains a field of characteristic p r" O. Assume that (R', m') is a local ring which may not be Noetherian such that R'P eRe R'. Prove that there are coefficient fields I and l' ofR and R', respectively, s~h that I r;;; 1'.
32. Finiteness oj derived normal rings vVe first prove the following:
(32.1) THEOREM: If R is a complete local integral domain, then an arbitrary almost finite integral extension of R is a finite module. As for the proof, since R is a finite integral extension of a complete regular local ring by (31.6), we have only to prove the following: Let R be a complete regular local ring and let K be the field of quotients
I~J'
N, /,t'l /, III' 1/ .Ii II ill' IIIW/I/'IIII' 1',I'll'lIllioll I~J' /\' (/Ild II'! HilII' IIII' illll'fll'lIl h' ill I" '1'111'/1. /a" in 1/ .Ii II ii,' Ii' 1110'/1111',
1'108111'1' ,~r
"!'ool': 1.('1./1 '11'1.111' 1:IIII,I'I1oI'I,I'I'i cil,i(' 01' /1, II' /1 n, 1.11('11" iH HI:pal'ahln OVI'I' f\ , /Iolld II~I iH fi II i1.1', 11'y ( In, I() ), '1'11('1'1,1'01'1' WI' :I,:-::-:IIIlW Lhal. p -F 0, WIiI'IH'.I' U i:-: I.lil' pOWI'I' HI'I'il'H !'illg ill lumlyLieally illdependent eleIIWIII.:-: ,I'I , , , ' , ,I',. !lVIH' a finld /, 'I'11('l'c is a finite purely inseparable I'XI.(:II,yioll !/ of 1\ HlII'II Lila I. /;( 1/) iH separable over D, If we know that Llle illtngml dOHul'n U" oj' It, ill (/ is finite over R, then we see that the ill(,(:gml dosul'e oj' It, ill L(f/) it> finite because L(V) is separable over 1/, and therci'ore R' it> finite over R, by (3,1), Thus we may assume Lhat L = L', Let q = [L:Kl (q is a power of p) and set I* = Ilia, q i/.,: = x~lq, R* = I*[[Yl, '" , Yrll, If a E R', then a E R, whence U' C R*, which implies that every element a of R' is uniquely expressed as a power series in the Yi with coefficients in I*, and in that Hense we define the leading form of an element of R', We want to show first that when aI, ". , as are elements of R' such that the leading forms /I, .,. , I" or these elements are linearly independent over R, then aI, ... , as are linearly independent over R. Indeed, if ~ aio.i (Ii i E R) is a non-trivial linear combination, then the leading form of L aib i it:: a non-trivial linear combination of the Ii which cannot be lIero because of their linear independency. Thus aI, ... ,as are linearly independent. Since L is finite over K, we see that there are elements aI, ". , as of R' with leading forms iI, .,. , is such that if a is an dement of R', then the leading form i of a is linearly dependent on II, .. , , is . Let Cl, . , ' , Ct be the set of coefficients of 11, .. , , Is . Then the eoefficients of 1 are in I(cl, ... , Ct). Therefore, if do = 1, dl , . , . , du is a linear base of I (Cl' . , . , C t) over I and if mo = 1, ml, .. , , mv is the set of monomials in the Yi of degree less than rq, then 1 is in the module L Rmid j , whence the module M generated by leading forms of elements of R' is a finite R-module; let gl , ... , gw be a base of M such that each gj is the leading form of an element bj of R' and let R" = R[b 1 , • , • , bwl. R" is a finite R-module, whence it is complete. Therefore, R" is a subspace of both R* and R' by (30.2). Let d be an element of R', We want to show by induction on n that there is a sequence I dn} of clements of R" such that d - dn has leading degree not less than n, We may start with do = O. If do, ... , dn are already defined, let f be the leading form of d - d71 . We can write I = L hjg j with hj E R, and we may assume that h j are homogeneous forms, whence d* = L h)}j has the same leading form f, and Lhe1'e-
III
'1'111'1111/\ 01" l'IIMI'",q'I'I'! 1,111' II, IIINIIi!
tI" I t/'" ill LI", l'I''IlIil'l'd ,,1, ' 111,'111, '1'111111, 1,111' 1'~ild"'III'I' IIi' 1.111' 111'11\'1'11. ~illl'(' It'" ill /I, IlltlllIPHI'(' IIi' 11"" 1.111' HI'IIiII'III'I' iH if, (!/I,I/('hy HI'ql/('III',y ill It'''. HilWI' U" ill ('111111'1('1.1', LlII' H('(II/I'III'I' 1111.:-1 lI. liJlli!. tI" ill /(,". tI 1/" IW('II.lIHI' U" iH II. HI IItHIIII.I',I' III' h~', Whi('11 illllllil'H !.Im!. U' ~ U", :1.111\ N' iH:I, lilli!.n 1(lIllIillIll'. Wn NlI.y LIm!. a HI'mi-lo(',al rill/!: R iN wnn{lIli(:u,Uy '11:n;/,(/;It/,'/jil'd ir (,\II~ ('.olllpll'Lioll N* IIi' N linK ])0 nilpoLcl\L e1mncl\i; (,xenpL )ICI'O; all i(i
1'111'1' 1/" ! I
~"'IJlII'III'I' i~1
(;{2.2) TH1Wm~M: If a semi-local integral domain R is analytically then the derived normal ring R' of R is a finite R-module. I lr()oj': Assuming the contrary, let R = Ro C Rl C R2 C ... C //'" C '" be an infinite ascending chain of subrings of R' such that m(:h Rn is a finite R-module. Let R* be the completion of R. Then the (.ompletion R: of Rn is generated by R* and Rn (and is identified with /(,1/ ® R*) by (17.8), which implies that R: is a finite module over R* and is contained in the integral closure Q* of R* in its total quotient ring. Let K be the field of quotients of R. Then R: K = Rn by (18.4), which implies that Q* is not finite over R*. On the other hand, let be all of the prime divisors of zero of R*. Then Q* is the direct sum of derived normal rings of R* In; , whence Q* must be finite over R* by (32.1). Thus R' must be finite over R nnram~!ied,
n
n;
33. Derived normal rings oj Noetherian integral domains (33.1) Let R be a Noetherian integral domain with field of quotients K and let R' be a ring such that R C R' c K. If R is integrally closed in R' and if, for any given prime ideal ~ of R, there is a prime ideal ~' of R' such that ~' n R = ~, then it holds that R = R'. Proof: Assume the contrary and let b' be an element of R' which is not in R. Let a be the set of elements a of R such that ab' E R. Let ~ be a prime divisor of a and set c = ll:~. Then a ~ c, whence there is an element c of c such that cb ~ R. Let ~' be a prime ideal of R' sueh that ~' R = ~. (cb')~ C R and furthermore (cb')~ C ~', whence (cb')~ C ~' R = ~, which implies that cb' is integral over R, by (10.4). Since R itl integrally dosed in R', we have cb' ( R, which is a contradiction, and we complete the proof.
n
n
1'11,\ 1"1'1,:11 V
(;l:I.:~
I
'1'111,1( IIII'I~I:
I,d Ii' /11'
II
I Ifl
N III,/III'/,"Ull i/l(I'(//'(/.I dlil/I,ain wl:!,h Jif'id
11',11'1 " /w II, jill ill' II.ludi'm":l: 1'.rtC/I,8'1:()1I, I(/, I\. and let H' be a '/'ill(/ 81/('/1. Ihlll U C N' C I,. II al Li Lwin H = 1, then Ior any ideal a' of Id 1///11("1'1118
/{' 811ch 11i1/,(
0'
/0, /t,'/n' i8 a module of finite length over RICa'
II R).
111,1)(1.1'/";('1/,10:1', /{,' iii a Noetherian ring of altitude at most one. (THEOREM Ill>'
(\InJI,lrAKlZUICl)
Proof; Taking a finite integral extension of R we may assume that " = K. Set II = a' R and let ~l, ... , ~8 be all of the prime divisors of (I. Since a' 'F 0, we have a 'F 0, and therefore the ~i are maximal. Ld S be the intersection of complements of the ~i in R. Then every nkment of S is a unit module a, whence R'la' = R~/ll'R~ and Ria = Us/aR s . Therefore we may assume that R is a semi-local ring, that II iH eontained in the Jacobson radical m of R and that R is dominated by R'. Let R" be the integral closure of R in Ii/. We first prove the :tHsertion for R" instead of R'. Let x be an clemcnt of a which is different from zero and let YI , ... , Y t be arbitrary elements of R". Then Nand R[YI, ... , Ytl are IVlaeaulay rings, and therefore length R "~/xR = f..l(xR), length R R[Yl/l:R[yj = f..lR(xR[y]). By (2:3.4), we have f..l(:cR) = f..lHCxR[y]), whence we have
n
length R/rR
=
length R R[yJ/:l:R[y].
Hince the Yi are arbitrary, we see that length R R" IxR" .:::; length RlxR, hence the assertion is proved for RI! instead of R'. Therefore, we may l'8place R with RI!. Thus, we assume that R = R". Then the assumptions in (33.1) are satisfied by Rand R ' , whence R = R', and the asHertion is proved in the general case. We note that the derived normal ring R' of a local integral domain of altitude 1 is not necessarily a finite module. See Example 3 in the Appendix; cf. Exercise 1 below. An integral domain R is called a Krull ring if the following two conditions are satisfied; (1) If ~ is a prime ideal of height 1 in R, then R~ is a Noetherian valuation ring. (2) An arbitrary principal ideal aR (a 'F 0) of R is the intersection of a finite number of primary ideals of height 1. (33.8) The condition (2) above is equivalent to the following two conditions: (2.a) Every principal ideal of R has only a finite number of prime divisors ~ such that height ~ = 1.
lin (~~,I,)
U
!'dllllll
\1 1'/1/1 111'1'1'
1111 1"'1//1/'
1f/('tI/ii
u,llll'iJ,';hl. I ill 11',
/1 ' (' /IIII'!'
nIl 11'11'
1'1'11111': AHHl/illI' Ihal. (:!) l,oldH, (:2,11.) IluldH II.\' (7':,), ~('I. /) nUll 1('1. ('Id ((', (I, ( It,) II(' 11.11 n.l'hiLI'HI'y ('11'111I'liI, III' n, 'I'lli'll ('Up ~ (lNI' 1'01' allY pl'illln idl'al p or III~igh L I, WIl(~III'(\ (: ( n((LU p n h~), alld (,/, iN 111.~t. illI,n(,~I~I'liOIl l'oillc:iilnN wiLh (IN by I:OlldiLiOlI (2) by virLllI: of ((Ui), WIWIlI:I~ (:lrI C n, and we RCO Chat R = f), whidl prOWH Lhl: validil.y of (2,1», Conversely, assume that (2.a) and (2,1» hold alld /1:1. (!, 1)(: :til clement of R which is not zero. Set II = n(aR p' n R) whl:rn pi rUll::; over all prime divisors of aR such that height pi = 1. Wn havI: only to prove that a = aR, hence that a C aU. Let b be all arbitrary clement of ll, If Pis a prime ideal of height 1 and if p -F pi for ally pi, then a ~ p, whence aRp = Rp. Therefore, we see that b E a/t,p for any p (height p = 1), whence bla E nR" = R, and beaR. ThuR, the proof is complete. We note that: (:~:3.4) A Krull ring is a normal ring. A Nocthcrian normal ring 1:8 a Kr'ull ring. Proof: The first assertion followA from I'olldition (1) and (2,h), while the last assertion follows from (12,9). (:~3.5) Let R be an integral domain with field oJ qu.otients K and assume that a set F of Noetherian valuati()n rings V of K 8ati.~fie8 the Jollowing two conditions: (1) R is the intersection of all V E F. (2) If a (E R) is not zero, then there are only a finite number oj V in F such that a is non-unit in V. Let S be a multiplicatively closed subset of R such that 0 ~ S and let F' be the subset of F consisting of those V in which every element of S is a unit. Then we have the equality Rs = v EF' V. Proof: Let D be the last intersection. Since Rs C D obviously, we have only to prove that D C Rs. Let cld (c, d E R) be an arbitrary element of D. Let VI, ... , Vr be the set of V E F in which d is a non-unit; we renumber them so that Vi E F' if and only if i :; s. For each i > s, there is an clement Si of S which is non-unit in Vi. If m is sufficiently large, we have sci d E V j for every j with s = (ss+l ... . . . 8r)'n. TheIl scld E V for any V E F, whence seld E R. Therefore eld E Rs. 11.111/
n
(33.6)
THEOREM:
Let R be an integral domain. Then there is an F
(1111\ 1"I'lill! V
117
fin ill (:I:\.f)) 1/ (/1/(1 II///ll if N in (/ /1/'1/1/ 1'///(/. /1/ IIII/i f'ltW, Nil is 1/, ///1'/// lu'l' 4 /1' IIII' /'/'/'/'ll Ill'i/l/(' itlm.! ).1 /d' IU'i(/hi li/l. U. hoof: If N iH /I, 1\/'1111 "illg, Lhnll I.II(~ Hd of U~ , P (willg JlI'illw id(~al" of \wigliL I, HII.Li"fil'H I.hn (',olldi(.ioIlH 1'01' ]I'. Co II v
(33,7) If V is a Noetherian valuation ring of a field L and if K is a K is also a Noetherian valuation ring. subfield of L, then V Proof: Let a be an element of K. Then either a E V or a-I E V,
n
IIH
r)
n
n
\' 1\ 01' II I , I' 11, IVllil'll HIIOWN 1,11/1.1. I' II ir~ /I, valllll,l.ioll I'illg, 1"01' (,I('III('liI,H 1/, I, of 1', (/1' c: Id' if nlld oilly if 1//11 iN ill thl! Ilinxilllll.l iil('II,1 111 of I', WIHIII('(' II/II ( Iii 1\'). 'l'hllH a.V C IN iH ('l{lIivnll!lIl. 10 Nayillg 1.11:1.1. fI.( \' 1\) C Ii( V 'l'll(m~ I'm!!, I.hl! lII:txillllllll nOll(liLioll ill V illlpli!!H I.ImL ill II 1\, wllinh pmV!!H i.he :LHHeI'Lioli.
WIH'III'I' 1/ (
n
n ('" n n /\). n
(3;).8) Let R be an integral domain and lot J!' be a family oJ prime ide(ll8 of R such that, for any element a of R, and fur any prim(: dVl!isor ~ oJ aR, there is a q E F such that ~ C q. Then R = n qE F Rq . Proof: The inclusion R C nRq is obvious. Let alb (a, b E R) be an arbitrary element of nRq and set c = bR:aR. alb E:: Rq implietl that bRq:aR q = R q, hence that cR q = Rq. Assume for a moment that C -F R and let ~ be a prime divisor of c. Let p be an arbitrary element of ~. Then, there is an element y ~ C such that py E c. ,Then pya E bR. But, since y ~ c, we have ya ~ bR, and we see that p itl a zero divisor modulo bR. Therefore, there is a prime divisor ~' of bR such that ~ C ~', hence there is a q E F such that ~ C q, whence cR q C ~Rq -F Rq , which is a contradiction. Therefore c = R, namely, alb E R. Thus, the assertion is proved.
(33.9) COROLLARY: If R is an integral domain, then R m runs over all maximal ideals of R.
=
nR
11t
if
(33.10) THEOREM: Let R be a Noetherian integral domain and let R' be the derived normal ring of R. Then: (1) R' is a Krull ring and (2) if ~ is a prime ideal of R, then there are only a finite number of prime ideals ~' of R' such that ~ = ~'n R, and, for any such ~', R' I~' is an almost finite integral extension of R/~. Proof : We consider, first, the case where R is a local ring and ~ is the maximal ideal of R. Let R* be the eompletion of R and let n* he the radical of R*. Set 8 = R* In*. Sinee n* n R = 0, R is regarded as a subring of S. Since any nOll-zero element a of R is not a zero divisor of R* by (18.1), a is not ill any minimal prime diviflor of zero of R*, whence a is not a zero divisor in S. Thus, the field of quotienttl K of R is a subring of the total quotient ring L of 8. Let 8' be the integral closure of Sin L. We want to show first that R' = S' K. The inclusion R' C S' K is obvious. Assume that alb (a, b E R) is in S' K. Then, since alb is integral over S, there are elements c~, '" ,c~ of S such that (alb)n + c~(alb)n-l + ... + c~ = O. Let E R* be such that c~ = c7 modulo n*. Then we have an +
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I I \I III 1 " . 1 I':~II" I 11'1', ~ill('(\ 11+ if! lIilpol'('IIi" 1-101110 I)I'IV/'I' of UdH III.HL HIIIII 1)I'I'OIlI('H 11('1'0, 1V1i('III'(' 1.11('1'(' 11.1'11 ('I(,IIII'IIi,H d'i , ... , tI:~, or Itl. HII('1i l,lill.L fI.'" 1 d'i'II'" III 1 . . , 1 d:~Ji'" O. WII('III:O n'" ( ( L't (/'" 't,'lt n U, 11.1111 IliiH III.,d, idnal iN 2:)" a'" ii/a by (17.\)), wliil'il ililply Lllal. 1,11('1'(' 111'1' (,I('III/'lti,H til, .• , , Ii,,, of R such that (/."'·1 diU'" III + .,. 11",/1''' 0, alld alb iH integral over R, hence II/II ( NI :l.lld 1.I1(~ <'qllality N' '" 8' K is proved. Let pi , ... , p; lin \.Iw pl'imn diviHOI'H or iI*. Theil L iH the direct sum of the fields of I(lIoLiclltH of U* /jJi lw,d 8' is the direct sum of the derived normal l'illl!;S of R* /jJi . Since R* /jJi is a complete local ring, S: is a finite (W/jJi)-module, by (32.1). Let ei be the identity of K i • For an 1I,l'hitrary prime ideal qi of height 1 in S:, (S:)q*i is a Noetherian vltluation ring, because is a Noetherian normal ring. Set o(q;) = /(1 + .. ' + K i - 1 + (S:)q*i + K i +1 + ... + K r • Then we see that ehe intersection of all possible o(qi) becomes SI. Therefore, we have It = n(K n o(q":)). It is obvious that a E K is in K n o(qi) if and only if aei E S;;, hence K n o( qi) is isomorphic to Kei n (S:) q: , which implies that K n 0 (q i) is a Noetherian valuation ring, by (33.7). Thus R' is the intersection of Noetherian valuation rings 'J( o(qi) of K. If a E RI (a -F 0) is a non-unit in o(qi), then it rollows that a E S~ + .. , + S~-l + q; + S~+l + ... + S;, and therefore there are only a finite number of such 0 (qi ). Thus R' is a Krull ring in this case. Let Y1 , ... , y t be arbitrary elements of R' and consider R[y] = R[Y1, ... , y t]. R[y] is a semi-local ring, whose completion is generated by R* and the Yi . Therefore SI is the integral closure in L of R*[y] modulo its radical. Since this is true for any y, and since the completion of a semi-local ring having u maximal ideals is the direct sum of u local rings, we see that the number of maximal ideals of R' is at most r, and furthermore, the maximal ideals of S' lie over those of R'. Whence the assertion is proved for the case of a local ring R and its maximal ideal jJ. Now we consider the general case. Let S be the c:omplement of jJ in R. Then we see that R: is the derived normal ring of Rs = R~. Therefore, by what was proved above, R~ is a Krull ring and furthermore R~ has only a finite number of maximal ideals p' and R:/jJl is finite over Rs/pR s , which implies the last assertion in our theorem. Furthermore, since jJ is arbitrary, for any given prime ideal ql of height 1 in R I , applying the above for p = ql R, we see that R~, is a Krull ring, which shows that R~, is a Noetherian valuation ring. Furthermore, we have R' = where 1';11"
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p' I'IIIIH (1\'('1' 11.11 11I1I."iIIiHI idl\/dll of It", 11,\' (:;;\.\) I, \VIII'III'I' "" rlU:I, \Vitl'I'I' q' I'IIIIH 0\'1'1' 11,11 Pl'illll' iill'/i.l1I of IlI'i/l;itl, I ill h~', !.y LlII' fwd, I.it:!'!' 1111' "';,,11,1'1' 1\1'It1ll'iIlI!;N. 'J'itl'I'dol'I', hy (:1:1.:1), il, l'I'lllniIiH oilly 1,0 PI'OVI' LI,,('/, /l,1I,V pl'illnipal iil('al flH' 01' H' linN 01l1,V II. lillil.l\ 111IIH!.I'1' 01' IlIillilll:l.l Pl'illll' i1iviHol'H. Nillnl~ N\n\ iN Nodhl\/'in,1I HIIiI Hilll',I~ h~' iH i,J1I~ i1(,l'ivl'd 1101'11111,1 I'illl!; 01' U\n\, WI~ limy aHNlIflW LiIa,1. a ( U. aN has only a lillil.(\ IIIIIII!.I'I' 01' prilll(: diviNOI'N, hellee by LIte lillil.eneHH of lyillg-ov(~1' I>l'illw iill'nli.; PI'OVI\i1 abov(~, the lillit(\]WNI'> of minimal prime diviNon; of aR' l'oliowH 1'1'(l1\l Lit() following lemma. (:1:1.11) With the same R, a, and R' as above, if ~' is a minimal I)/'il//,I: d£m:,w/, (d aR', then ~ = ~' R is a prime divisor of aR. 1'1'001': Lot /) be the eomplcment of ~ in R. Then ~'R~ is a minimal pt'illln diviHOI' of aR~ , and ~Rs = \)'R~ Rs . As is easily seen by the pI'illlnry (k(~omposition of aR, ~ is a prime divisor of aR if and only if ~IUp iH a prime divisor of aRp . Therefore, by the fact that R~ is the i1Pl'iv()d lIol'malring of Rs , we may assume that R is a local ring with IllH.ximal ideal ~. Therefore R' is a Krull ring, in this case, whence it('ighl. \)' = 1. Since ~ = lJ' R is maximal, we see that ~' is also a lIIaximal ideal by (10.7). Let x be an element of ~' which is not in lI,ny other maximal ideal of R'; x exists because R' has only a finite 1lllIlIhm' of maximal ideals by our assumption. Set R" = R[x] and p" = \)' R". Then ~' is a unique prime ideal of R' which lies over ~l". Therefore height ~" = height ~' = 1 and ~" is maximal by (10.8). COIlHidcr the completion R"* of R". R"* is the direct sum of compldn local rings Ri where RI is the completion of R;, . Let e be the id(~lltity of RI . Let R* be the completion of R. Then R"* = R*[xJ. There is an elcment b 7"" 0 of R such that bR" ~ R, whence bR"* C /t,*[xJ. Since height ~" = 1, there is an n such that (~RI)n C bRI' Therefore \)ne C bR I C R*, whence ~tne C btRI C bt-IR*. Assume for a moment that ~ is not a prime divisor of bR. Then ~ is not a prime divisor of btR for any t, by (12.6), hence \J.- contains an element p which is not a zero divisor modulo btR (in R), whence modulo btR* in R*. Therefore, the fact that ~tn(~ne) = ~(t+l)ne C btR* implies that ~ne C btR* for any t, which is impossible because nt btR* = O. Thus ~ is a prime divisor of bR, hence of aR by (12.6). Thus the lemma is proved, and the proof of (33.10) is complete.
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(33.12) THEOREM: The derived normal ring R' of a Noetherian integral domain R of altitude 2 is again Noetherian. Proof: We have only to prove that if ~' is a prime ideal of R ' , then
1'11,\ l"I'I'il!
I~ll
V
pi lilili IIlillill' lill.IIiH, II.\' \il'lll(I III' (:\.1) (1./11'111'1'111 III' (~oll('II). WI! Iin-iL 1'1'(1\'1' 11)(' I'II,H(' \VIiI'I'I' )\' IH lilli, IIIn.xilllll,l, II.HHlllllillf!: I.hal. I~V(~I'y maximal iill'II,1 IIar1 /I. lillill' 1,11,,"1', WI' Iliay aHHIIIIII~ I.hal. pi -F O. Set P = pi R. '1'1i1'11 ~) iH 11111. III/I,xillial by (10,7), wlwlI<:e height p = 1. Let a be an 1,llHlIl'lti, oi' pi which iH 1I0t ill any other prime ideal of R' which lies IIVI~I' p, Nillcn Hlal is Noetherian, we may replace R with R[aJ. Thus w(~ may aHHlime that pi is the unique prime ideal lying over p, Since N~, iN a Noetherian valuation ring, there is an element b of pi such I.hat bit;, = p'R;,. Considering R[b], we may assume that b E R, whence that p'R;, = pR;,. Set q' = pRI:p', q' contains bRI:p'. Since H' is a Krull ring, bR ' = pi n q~ n . , . n q~ with primary ideals q~ (~ pi) of height 1. Since p' is the unique prime ideal lying over p, q; n R is not contained in p, whence q' n R, which contains (q~ n R) ... (q~ R), is not contained in p. Since q' R contains p, we R) = 0, Let q'* be the radical see that q' R ::J P and depth (q' of q'. Since depth (q' R) = 0, we see that q'* is the product of a finite number of maximal ideals, say m~ , '" , m~ . Since each m~ has a finite basis, we sec that q,*n has a finite basis and that R/q,*n ii:< Noetherian by the theorem of Cohen, and furthermore that q' contains some q,*n. It follows that q' has a finite basis. q'/p'q' is an R'/p'module, Since q' has a finite basis, q' / p' q' is a finite module, whence its 8ubmodule (p' q')/p'q' is a finite module because R' /p' is Noetherian by the theorem of Cohen. Since p' q' cpR' c p' q', we can consider (p' q')/pR', and it is a finite R'/p'-module. Since p has a finite basis, we see that p' q' has a finite basis. Since R' / p' and R' / q' are Noetherian, we see that R'/(p' n q') is Noetherian by (3.16), and we see that p' / (p' n q') has a finite basis, and therefore p' has a finite basis because p' n q' has a finite basis. Thus, it remains only to prove that every maximal ideal of R' has a finite basis. Let m' be an arbitrary maximal ideal of R'. By the same reason as stated above for p', we may assume that,m' is a, unique prime ideal lying over m = m' R. Then R'/mR' = Rm,/mRm" Therefore we may assume that Rand R' are local rings. If height m = 1, then the assertion is true by the theorem of Krull-Akizuki (33,2), and we assume that height m = 2. Let a, b be a system of parameters of R and let x be a transcendental element over R, If we know that m'R' (x) has a finite basis, say II, , .. , ., , ,Is (F I{[.rD, then we see that 1l1' is generated by the coefficients of I, , and therefore 111' has a finite basis. Therefore it is sufficient to show that m'R' (x) has a finite basis. Since aR and bR have no common
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'l'III'HIII\ III" I'IIMI'I.!!)'I'I>! 1,111'\1, 1111\1111\ Illillilllni 111'illll' dil'illlll'H, 11'1' HI'I' LilaL
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l',wl,H Cilll,j, It" ill I/. 1\1'111' rill/!; I\.l\d illIl,\. III' iH Lill' IllIiqlll' 1I111,xil\\H1 illl'HI or U'. WI' \1'/1.1\1, 10 :dlll\V 1,11/1,1. II,/' I I, gl'III'I'li,I,('cl a primp i
Ii'[xJlp* = R'[li/al (:J: IIlOdlllo p* . , ,- II/It). 'l'hl'll l.l* i~4 gl'llnl·l\.LI'" by elements cx + d with c(b/a) = d by (11.1:l). Nilll:n aN' alld 'IU'lta'/n no common prime divisors, we see that c EaR', wlteIl(:(: p* i:-: g(:Il(:ml.(·1 I by ax b. Therefore, we see that ax b generates a primo idett! ill Ii'ex). Consider R* = R(x)/( (ax + b )Ii'(x) R(x)). This ring i:-: Noetherian and of altitude 1. Therefore R' (x) / (ax + b )Ii' (x) i:-: Noetherian by the theorem of Krull-Akizuki, which implies th"L m'R'(x)/(ax + b )R'(x), hence m'R'(x), hence m', too, have finite bases, which completes the proof. We note that there is a local integral domain Ii of altitude 2 such that it has non-Noetherian integral extensions contained in its derived normal ring and that there is a local ring of altitude 3 whose derived normal ring is not Noetherian; see Examples 4 and 5 in the Appendix.
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EXERCISES: 1. Let R be a semi-local integral domain of altitude 1, and let R* be the completion of R. Prove that R is analytically unramified if and only if the derived normal ring R' of R is a finite R-module. Prove also that the number of the prime divisors of zero in R* coincides with that of maximal ideals in R'. (N ote that the above assertions are not true in general if we do not assume that altitude R = 1; see Examples 6 and 7 in the Appendix.) 2. Let R be aN oetherian integral domain of altitude 1 and with field of quotients K. Prove that if L is a finite algebraic extension of K and if R' is a ring such that R k: R' k: L, then there is an integral extension R" of R such that R' is a ring of quotients of R".
34. Chains oj prime ideals Let R be a ring and let p and q be prime ideals of R such that p C q. A chain of prime ideals po C Pl C . " C pr in R is called a maximal chain of prime ideals between p and q if po = p, Pr = q and if there is no prime ideal p' of R such that Pi-l C p' C Pi for any i = 1, ... , r. A maximal chain of prime ideals bet'('1een a minimal prime divisor of zero and a maximal ideal is called a maximal chain of prime ideals in R. We say that Ii satisfies the chain condition for prime ideals if, for any prime ideals p C q, and for any integral extension R' of Rq/pR q , every maximal chain of prime ideals in R' has length equal to altitude R'. Note that this is equivalent to saying that if R* is integral over
I'll·' 1"1'11111
V
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illl'al:-1 Ill'tlVl'l'll p' Hlld q' iN 1'!j11/l,1 1.0 III'ighl. q/\). II. iH oll\'ioll,y 11.1' 0111' 1I1'lillil,illll Llml.: (:\.1.1) If (/. 1'/1/.(/ U 8(/J,.,~/il'8 IIII' dUI/n I'II/I·di/.illl/. fli/' 1J1"I:nu: 1:d('(!,l.~ ll!.I'//. 111.1' I·III/.d//.-ill!/, i" 8/1,1";.~/i/'(1 hI 1111/1 U,~/n' N.~ , 'Wh.ere U' £s (zn Ol)(;r-r1:nu I~r h~ 8/1.th. aU/,/, N' ·i8 '/;nl('!lm,{, (J/lI"I' N, II' £8 (/'1/. 'irie(zt of R', and S £s a mulli-
pl-iw,/,';,)(:l!/ dosed 811.1)8(:/ (~r h~', III O"del' Co diHt:IlHH Lhe condition, we introduce the following i,()!'IllillOlogy for II. ring it. We say that R satisfies the first chain condition for prime ideals if nvery maximal chain of prime ideals in R has length equal to altitude I( We say that R satisfies the second chain condition for prime ideals iI', for every minir.:ml prime divisor ~ of zero in R, and for any integral ()x(,ension R' of R/~, the length of any maximal chain of prime ideals in R/ is equal to altit.ude R . .1'\01.8 that the validity of the second chain condition for prime ideals in II. ring R whose altit.ude is finite is equivalent to (a) every maximal ideal m of R has height equal to altitude f(, and (b) R satisfies the chain condition for prime ideals. We note that, though the validity of the second chain condition for prime ideals implies the validity of the first one, the converse is not true. See Example 2 in the Appendix. We note furthermore that a ring R whose altitude is finite satisfies the chain condition for prime ideals if and only if for every minimal prime divisor ~ of zero in R and for every maximal ideal m of R eontaining ~, the ring R nt/~R '" satisfies the second chain condition for prime ideals. Therefore, it is of great importance to discuss the second chain condition for prime ideals in local integral domains. (34.2) Assume that R is a subring of a ring R' and that R' is integral over R. If R' satisfies the chain condition, or the first or the seccmd chain condition, for prime ideals, then R does, too. The proof is immediate from (10.9). (34.3) A Noetherian integral domain R satisfies the second chain ccmdition for prime ideals if and cmly if the first chain condition is satisfied by every finite integral e:ctension R" of R. Proof: The only if part is obvious by definition. Assume the validity of the first, ehain condition in any R" as above and let R' be an arbitrary integral extension of R. Let 0 = ~~ C ~~ C ... C ~~ be a maximal ehain of prime ideals in R'. Set ~i = \.1: R. Then there are only
n
1:.'.1
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III I'
I iii' I '(
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1,( 11'1 I, III NIIII
a lillil(~ 1IIIIIdll'I' or Pl'illl(' iill'!!.lrl ill It" ",itil'it Iii' 01'('1' )I, , II,V (:1:1.10), whencn UWI'(~ iN a lillil.p ('.\I,(~IINioll h'" or Ii' (/i" c: Ii') Nlll'it Llilii. ('VI'I'V ~~ is the unique prillw ilkal or h~' l,Villl!: OVI'I' -~I;: ~: nu", Nilll:(' ~~' c ~~ c '" c ~; i:s a maximal (·ltaill or pl'illIP idnalN ill h~" II,Y LlII' going-up theorem, we see that r = all.il.ll 0, We consider at first the case where I is a field. Let Xl be an arbitrary nonzero element of ~l • Then there exists a system of parameters Xl , ••• , · .. , Xr of R, and R is a finite module over the complete regular local ring I[[xl] = I[[xl, ' . , , x r]] by (31.5). Since a regular local ring is a normal ring by (25.14), height (~l I[[x]]) = 1 by (10.14), hence ~l I[[xll is generated by Xl, Thus R/~l is a finite module over I[[x]J/xd[[xJ], which is of altitude r - 1. Since 0 = ~d~l C ~2/~1 C ... · .. C ~8/~1 is a maximal chain of prime ideals in R/~l , we see that s - 1 = r - 1 by induction, which proves the assertion in this case. Assume that I is not a field and let p be a prime element of I. If p E ~l, we see that R/~l is a finite integral extension of I[[xl , ... , · , . , xr-lJ]/pI[[XI , ... , xr-lll with Xi such that p, Xl , •• , , Xr-l is a :SYNt.em of parameters of R, and we prove this ease 8imilurly. If p ~ ~, let Xl , . , . , Xr--l be sllch that Xl ( ~ and :such that p, :[;1 , ... , Xr -1 is a system of parameters of R. Then we prove the ease similarly. Thus we prove the a8Hertion. We say that a semi-local ring R is quasi-unmixed if every minimal
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l'II,II"I'IIiI/ V pl'illl(' di\'illlll' III' ~,I'I'II ill 11,1' "lIlllpll'Lilili IIi' Ir' I,nrl dl'pl" 1'11111'" III Hili 1.111\1' U. (:\,U,) If It' i8 II I//II/si IIII/lIi,/'I'tI 81'II/i III/'Irl ,.ill(/ /lilt! (f' )l iN /I 1/1'/11/( itll'lIl I~r H, Iltl'/I. Nip i,~ III/It.~i IIIIII/i,/'I't! 1111/1 11"igltl, )1 I I\I'pll, )l ",ll.il.lldn N. 1'1'001': 1,1,1, It~* III' 1.11(' 1'lIlIlpll'Lillli III' U nlld 11'1. )1" III' HII tl,d,iLI'II,I',I' Inillilllni pl'illll' diviHol' or pN*. ~ill(,(' LI'I' 1.111'11/'/'111 or 1.1·:/,II.'iil.ioll 1IIIIdii 1'01' II~ alld /{I', WI! H('(: 1.":1.(, II('ig"(' P III'ig"l. p'I' by (:J:J.!I). ~illi'" It'+ HII.LiHlil:H t}1(: lil'HI. dlaill l'olidiLioll I,y (;;,1.,1), WI' HI'I: 1.11:1.1. d.,pl.l, polII.ILitudo R - Iwight p* = :dLil.lltin U - II(:ighl. p. ~ill':I' i.I,iH iH 1.1'111' 1'01' any minimal primo diviHorH p* of pU* and Hill(:(: u'I,/~"rl' iii I./rl' I:ompletion of R/~, by (17.D), wo H(:(: (,haL tlnp!.h p* 11,ILiLlldl' N*/~R* = altitude R/~ = depth ~ and thai. LlH: :tHHW'i,ioli iN 1.1'111"
(34.6) THEOREM: Every quasi-unmixed semi-local .,.ill!! U {he second chain condition for prime ideals.
.~lIlii';.fil'N
Proof: We see easily the validity of the first ehaill I',olldi!.ioll ill "'. by (34.5) and by induction on altitude R. Let. p be all nl'hil.l'HI'.\' minimal prime divisor of zero in R. Then, sinec height p = 0, ii, 111I1d:1 that depth ~ = altitude R. Therefore, by the remark givoll abovI' HII" by (34.3), we have only to prove that every finito int(:gl':l.l I'X/.l'III-lioll R' of R/~ is quasi-unmixed, which is easy because Hlp iH IIIIH.Hi 1111 mixed by (34.5).
(34.7) COROLLARY: Assume that R is an unmixed semi-lu('(f} illll'lIl'IIl domain and that R' is an integral extension of R. A chw:n I({ 111'11111' ideals ~~ C ~~ C .,. C ~; in R' is maximal if and only if (~l:' n N) I (~~ n R) c ... c (~; n R) is a maximal chain of prime id(,(),f,,~ ill N. Furthermore, if a' is an ideal of R', then height a' = heigh!. (il' n h'), Proof: The first is an easy consequence of the validity of (,11/: Hl't'IIll(l chain condition for prime ideals. Consequently, the last aHi-\(:I'i.ilili i," true if u' is a prime ideal, whence the proof of (10.14) can hi: applil'd, which proves the last assertion.
(34.8) COROLLARY: If R is a homomoTphic image of a locally J\1 (/,{'/II/ lay ring, then R satisfies the chain condition fOT prime ideals. If II/I'ilil'/' more R is an integral domain and if R' is a finite z'ntegral exlen,~':/I11 I~r R, then for an ideal a' of R', it holds that height a' = height (n' n fa'). We add here the following result on unmixedness of 10call'illgN:
(34.9) THEOREM: A local z'ntegral domaz'n z's unmz'xed homomorphic image of a Macaulay local ring.
~r
£l i8
II
I~(I
'1'111'111111
IIII' 1'()MI'I,I'i'I'H 1,1)('11, IIINIIII
1'1'001': AHHllinn LliaL p iN /I. WillI(' idllll.1 or II, I\III,I'II.IIIII..Y 101'/1.1 I'illg 11'. LeI, l' 1><: til(: I,('iglll, or ~I Hlld 11:1, .1'1, .. , ,.1', III' 1,11'IiII'IIiH or )l HilI'" LilaL a = xj( is or hnighL r (b'y (lUi)). Ld N* 1)(1 Lhll 1·.OlllpII'LiOI\ or u. Then R* is a Macaulay ring, hCII('(1 (.Iw IUllnixl'dlll'"" (.1i<:w·(lIl\ hold" in R* by (25.6), whieh implic:,; HUt(, (lV('I'.; Pl'illll' divi"o\' or nN* i" a minimal prime divisor, and R* I aR* is lUunix(ld (d. 1';x(ln:i,,(1 ill *21)). Since ~ is a minimal prime divisor of a, every prime divisor of pH* is a prime divisor of aR*, by (18.11), and therefore it follow:,; 1.Ila(, R* I~R* if) unmixed. Thus the assertion is proved.
L
(34.10) COROLLARY: Assume that R is a Macaulay semi-local ring. For an ideal a of R, Ria is an unmixed semi-local ring if and only if every prime divisor of a is of height which is equal to height a.
35. Localities We say thai, a rilll!; It iH or fin'dely (J(!/I'{!'ml(!d l//'l)(; OWl' It rill/!: h' if is a ring of qIlOti(:lItH of a filliLniy l!;ellcmLnd I'illl!; N* OV(,I' N; N' iH said to be of ,ft:r~itc type over H, if fUI'L1Wl'ffiOI'C U* (',all III: <"iIOe'('11 III be integral over R. A ring R is called an affine ring over It rilll!; I il' H iH :1.11 illli'/!~I'al domain and if Ii is finitely generated over J, w\I(HI(:n I IIIIIHI, 11(' 1111 integral domain in this case, A quasi-local illi.q!;I'al dOIl\Hili which ill of finitely generated type over a ring 1 is ealbl a {/lm.!il/! OV('I" I, Such an I as above is called a grrnmd ring of (.it(: :dlilln rill/!: ot 1"1' locality, Note that local rings which are UH(:d ill IIHII:I.I : t.i/!:('\I l"iI,i(: /!:I' ometry are localities over fields, which we (::1.11 (l,/'ye/n'w';(:-IJI'/lIII('/I'i('(r/ local rings, In ring-theoretic formulations of all!;rlil'ai(: /!:(\OIli('I.I''y, i L iH sometimes convenient not to restrict oneself to I(H:aliLinH hilL Co ('Oil sider local rings which are of finitely generated (.ypn ovnl' fi(.JdH (II' some kinds of rings, But, most of the import.ant rCHtilLH Oil 1.11<'111 11.1'(' conveniently formulated in the case of integral domaillH 0/' HI'(: (1('I'i V('( I easily from that case, and this is the reason we defined n HIH'(',ial 1.('1'111 "locality," Of course, we do not have any good results wiLllollL allY restriction on the ring I, Since the case where I is a field iH I'aLil<'l' 1.011 specialfrom the ring-theoretic point of view, we shall Pl'OV(: i/II/1Ol'I all I reslllts coneerniIlg algebrltie-geomei,rieal loeal rings for (lIMn /!:('II<'I'nl CaRt'S in thiN ehapter, vVe say that an integral domain I satisfies the finitcn(',~8 ('/III/iii ill II for integral e:ctensions if every almost finite integral exkllHiol1 or Ii!: a finite extension, We note that: (35,1) If an integral domain I satisfies the finiteness c/il/,d,:li/i1/. f/ll' integral extensions, then so does every ring of quotients of I, The proof is straightforward and we omit it, A field L is called a function field over a ground ring 1 if 1.1 iN Lllo
a'
[127]
llliII'~"IJ'I'/lII'
I,IH',\/' IIINdll
(inl" or 'il/nLi('IILH or nil 11,1Ii11(1 rill!!; 0\'1'/' /, II' /, iH Lhll 1il'Id or '1l/oLjI'III.H II. I'illg It: wllil'h iH 11,11 11,llilll' I'illg 01' /I, IIII'II,liLy 0\'1'1' I, Lhl'll WI' Hay Lilal. I.; iH I.h(~Jl/,lId'';/iI/..lidd 4 N Hlld thaI. h~ iH /1,11 (/./Iill/' ,.illi/ 01' a 1/lI'IIIil!1 of L. We begin with the following lillil.clIPHH I.hnol'm'" 1'01' polYllollli:d l'illgH:
or
(35.2) THEOREM. Let Xl , ... , X" be algebraically indepenti/:nt cl/:ments over a Noetherian integral domain I. If I sati.'!jies the .linitenr:N'~ condition Jor integral extensions, then so does I[x] = /[;1::1, ... , XII]' Proof. Let L be an arbitrary finite extension of the field of quotient.H K of I[xl. We have only to prove that an arbitrary integral extension R of I[xl contained in L is a finite I[xl-module. By our assumption for I, we may assume that I is normal. If L is separable over K, then the assertion is obvious by (10.16). When L is inseparable over K, take elements al , ... , a r of I and a power q of the characteristic p of I such that L' = L( ai/q, ... , a~/q, xi/q, ... , x;;q) is separable over K' = K(ai/ a, .,. , a~/q, xi/q, .,. ,x~/q). Let l' be the derived normal ring of I[ai, q , ••• , a~/ql. Then l' is finite over I by our assumption, q and 1'[xi, , ... , x~/ql is finite over I[xl. Since L' is separable over K', q q the integral closure of 1'[xi/ , ... ,x~/a] in L' is finite over I'[xi, , ... , ... , x~/ql, whence it is finite over I[x]. Therefore R is finite over I[xl by (3.1). (35.3) ThEOREM. Let A be an affine ring over a Noetherian integral domain I. Assume that I satisfies the finiteness condition for integral extensions and that the derived normal ring of Ap is a finite A~-module for every prime ideal ~ of A. Then the derived normal ring A' of A is a finite A-module. Proof. There are elements ZI , ••• , Zr of A which are algebraically independent over I and an element a (~O) of I such that A[l/a] is integral over I[l/a, zj = I[l/a, ZI, ... , zrl by (14.4). Let L be the field of quotients of A. Then the int.egral closure of 1[1/ a, z] in L iF; A'[l/a] and is a finite I[l/a, z]-modulc by O~5.2). Thus we see that A'[l/al is a finite A [l/a)-module, wh(m(:e Lhere are a finite number of elementH b1 , •• ' , bs of A' fiuch that A'[l/a] = A[1/a, bl , '" , u.,j. Set Al = 11 [b 1 , ••• , b.,). If a iH a unit in 11, then the assertion is obvious and we assume that a is a non-unit in A. Let the prime divisors of aAI be ~1 , ••• , ~t. Then there are a finite number of clements Cl, ... , Cn of A' such that (Al)~JCl , ... , cu ] is normal for every ~i. We set A2 = A [b 1 , ••• , Us, Cl , ••• , cu]. A2 is a finite A-module. We prove here the following:
I ~.,) 1,1'1,
/I I II'
II. NOI'IIII'I'ilill I'illl!; 11111,1, 11,111, ,I~,
c:
/I
c:
,I'. '1'111'11 (
I
I
ii'
P il-l 110 I'1'illll' illl'lI.l or l'I'il1)lI, I ill II, 1.1\('11 11,\ i:1 H 1101'1111'" 1'11if/: 11,1111 ('J) if 1.111' illtl,,'lIdl'" prilll(' di\'i,YOI'~1 or (//1 /1.1'1' lj, , ,., , q" /l.llt! ii' til , " . , , .. , dot, 1101"1' 1'11'11\('111,1-1 or .1' HIll'h Lhll,L tl",ltI, , .. , , tI",1 iM Il0fllm\ fol' I'VI'I'Y q,', U\('II for /1.\1 IIoI·biLI'II.I'.\' illll)('d,kd prillII' dil'iNol' q' or flt1ltt, , ... , II",\, q' n H 1'011 Laille; NOIIII~ q i properly (q i C q' n H). IlId(',(~d, if a (I p, Ullm H~ 1'(IIIL:l.ill,Y I/It, alld fill ii':I. I'ill/!: or qlloLil'IILH oi' A'll/al, which implin,., LhaL HlJ il' 1I01'JIII'" ill Lhi,Y mSI'; ir I/. I p, ,'11'1, t)' = P n A, , pI! ='" P n /I~ . HiJ\(:': Iwi/!:h!' p = I, we Hnl~ LilaL p' iN H. prime divisol' of aA, by (:::l.ll) (by vit'LlI(~ of Lite goillg-Ilp LItI'OI'I'1I1 applied to 13 and A' with () c p). Tlwl'doJ'(', (A 2 )P" ie: lIonnal, !1,lld (A2)~" contain::; A', whence 13~ i::; a rillg of quotiClli,H or II' alld iN /I, lIormal ring, and (1) is proved. A::; for (2), sct q = q' H. Hilll:I' q' ii' an imbedded prime divisor of a13[d 1 , ••• , dwl, R[d[, .. , , tI'II!.,' ie: not a normal ring, by (12.9). By the same reaNOIl aN roJ' /I" above, we see that 13 q is not a normal ring, whence, hy vil'tt In 1'1' ( 1 ), we see that a13q has an imbedded prime divisor q" B q , wltil,1t must be some qi13q . Since 13q;[dl , ' " , tis] is a normal rillg, q 1',:\,\11101, be the qi , whence qi C q, and (2) is proved. Now we proceed with the proof of (35.3). Starting with Eo = .I,! , we eonstruct a sequence 13 i of rings as follows: \Vhen Bi ie: ddi I wd, let dl , ••• ,dw be as in (2) above, applied for 13 = 13 i , and Ret 8; II ~, Bi[dI , • , • ,dw]. Then, by the finiteneAs of the ascending ehain of prillll' ideals in A2 , we see that. there is a 13n such that a13n has no imbnddl'd prime divisor. Then the assertion is proved, beeause En iR nOJ'Ill:d, whence 13n = A', by the following lemma: (35.4) Let R be a Noetherian integral domain and let a (7""0) be an element oj R. Assume that R[l/a] is a normal ring and that Rp is a nor-
n
mal ring for every prime divisor p of aR. Then R is a normal ring. Proof. Let q be a prime ideal of height 1 in R. If a ~ q, then Rq ie: a ring of quotients of R[I/a] and is a normal ring. If a E q, then q is a minimal prime divisor of aR and R q is a normal ring. Let b ( 7""0) be an element of R and assume that bR has an imbedded prime divisor q. Since R[I/a] is a normal ring, bR[l/a] has no imbedded prime divisor, whence qR[l/a] = R[I/a], that is, a E q, which implies that q is an embedded prime divisor of aR by (12.6), which is a contradiction, and (35.4) is proved by virtue of (12.9). If I is a subring of an integral domain R, the trHnscendence degree of R over I is denoted by trans. degr R. We say that the altitude formula holds for an integral domain I if,
(It', III) 0\'('1' I, nll.jLlld(. Ii' I LI'HIIH, 11,"1111 I LI'/I,IIH. II('KI N.
1'111' lUI,\' illI'n1iL.v /l,ILil"ld(,
dl'glll,"I1/) U/III
(:lii.ii) "'III,;OIU':M. II (/1/. i",It'(/1'fI1 fili//llli", 1 is fl. hli///Ii//IIIJ'jihi/: ',:/1/(/.(//' III('ally M IU'/wlll.l/ '/"':n(/ III, 1.//1'1/, 1;'/: IIllill/d/' IIII'IIi'III(/, hllld,~ fill' I. jJ"urtlwrrnOTc, any !o(:(J,(,:ly N ()JWJ' 1 ',:,0.; a /i.O'I/I,/!II/III'II/ii/: iI/WI/(' /~r (/, )14 am,'//.lay ring, and c()ns(JqU/~ntly, the chain condition Jo'/' prinu: ,,:dnu.ls and t/U' altitude formula hold in R. /~r II.
Proof. There are algebraically independent elcmcntH ;/:1 , .• , , XII over M and prime ideals q C ~ of M[x] = M [Xl, ... , Xn] such that R = M[xh/qM[xh . Since M[x] is a locally Macaulay ring by (25.10), M[xh is a Macaulay ring, and R is a homomorphic image of a Macaulay ring, whence the last assertion follows from (34.8) and the first assertion. In order to prove the first assertion, we have only to prove the formula stated above for I and R, because R is arbitrary. We prove it by induction on n as follows: If n = 1, then the assertion is straightforward by the validity of the chain condition for prime ideals in any affine ring over I and, in particular, in M[x]. Assume that n > 1, and set B = I[xd/(q I[Xl]) , ~' = 111 B. Then altitude B~, + trans. degI/wnI) B/~' = altitude I(~'nI) + trans. deg[ B by the case n = 1, and altitude R + trans. degB/p' R/m = altitude B p' + trans. degB R. Since m 1 = ~' I, we prove the formula. Thus (35.5) is proved completely. As an application of (35.5) to general Noetherian integral domains, we can prove the following theorem:
n
n
n
n
(35.6) THEOREM. Let A be an affine ring over a Noetherian integral domain I and set r = trans. degI A. If ~ is a prime ideal of I and if ~' is a minimal prime divisor of ~A such that ~ = ~' n I, then trans. degI/~ A/~'
2:
r.
Proof. Let S be the complement of ~ in 1. Then considering Is and As, we may assume that (I, ~) is a local ring. We prove the assertion by double induction on height ~ and the number of generators of A over I. If height ~ :::; 1, then by our assumption, I is a Macaulay ring, whence by the altitude formula we prove the assertion in this case. Therefore, we assume that height ~ ~ 2. If height ~' T'" 1, then let q' be a prime ideal of A sueh that q' c ~', q' I -F 0 and such that height q' = 1. Sinc:e ~' is a minimal prime divisor of ~A, we see that q' n I -F ~. Sinec height q' = 1, q' ii:l a minimal prime divisor of (q I)A, whence by our induction on height ~, applied to q' I,
n
n
n
1'1111"1'1111/ 1'1
It'/q'
by 0111' illlilll',l.illll H,c,HI'I'l,illll iH 1.1'111' ill 1,liiN l'nNI', 11'1' 1I~lclllllll' LlmL IlI'iglil, ~l' I, LI'C ,1'1, , , , , ,I'", !In a H(~L of gl'III'l'al.ol'~ oi' ,I OVI'I' /, Ii' ,Ihl' iH 1101. algl'l>l'ai(~ OV(H' /Ip, Hay if ,(;, IIloilltio p' iH IIO!, algidH'ail" OVI~I' I Ip, U)(~II we eonHider A" = I[xI) Hlld ~I" ~I' n /1", Hilwe :1:1 modulo p" is transcendental over lip, W(~ HI~n I.haL ~l" ~c, pll" and Chat height p" :::;: height p by the altitude Lllnol'(~rn of Krllll :llld by the fact that a system of parameters of I~ gnncml.cH an ideal which is primary to p"A"~" in A"~", whence by our indudion on n applied to A over A" with the prime ideal p", we Hne i1wt trans, degA"w Alp' 2: trans, degA" A, Since trans, degrr~ II" Ip" = 1 and since trans, degr A" :::;: 1, we prove the assertion in Clii::; case. Therefore, we assume that Alp' is algebraic over lip. It Huffices to prove now that A is algebraic over I. Let A * be the derived Ilormal ring of A and let p* be a prime ideal of A * which lies over p', Then height p* = 1 by the going-up theorem, Since A * is a Krull I'ing by (;33,10), A:, i:,; a Noetherian valuation ring. Let K be the field of quotients of I and set B = A:, K, B is a Noetherian valuation ring by (33.7), Set C = B[A); this last ring is an affine ring over B, Set furthermore p** = p*A;. n Band p" = p*A;. n C, and let C* be the derived normal ring of C. Then obviously A * c C* C A;. , whence is a ring of quotients of C*, that is, = C: with r = p* A;, C*. It follows that r is a minimal prime divisor of p**C* and height r = 1. Since B contains I, Blp** contains lip, whence A;,/p*A;, is algebraic over Blp**. Therefore C*/r is algebraic over Blp**, whence C* is algebraic over B by (35,5) and by the fact that B is a Macaulay ring, which proves that A is algebraic over I. Thus the proof is complete, 1\'1' 111'(' 11111,1, 11'11.1111. dI'IJ:'il'l'III)
npplil'd III 'l'III'I'I'i'III'I',
P,'(
q' () /),
11'1' ~II'I' /1111.1
I'. '1'111'1'1·1'111'1',
/111'
n
A:.
n
A:.
36. Pseudo-geometric rings We say that a ring R is a pseudo-geometric ring if R is Noetherian and if, for every prime ideal p of R, Rip satisfies the finiteness condition for integral extensions. We note first that: (36,1) If R is a pseudo-gcometric ring, then every homomorphic image of R, every ring of quotients of R, and every ring which is a finite module over R arc pscwio-geometric rings, The proof i:,; :,;tmightfol'ward. (::)6.2) Let R be a semi-local integral domain and assume that p is a
I :I:~ Will/I' i"I'II/ I!!, /iI'i(/h/ I ill II' NIII'1i 111111 /i'p ill /I /11/11/1/1/111/ I'illfl· If P /N al/.oJy/.im.lly 1/1/.1'/('lIlIjil'" 1/.1/." if pl' i.~ II IIli/li/l/1l1 11,.//1/1' dillisor I!f' pH'f, R* being tlw (;mnJltl'l'ill/l, I!!, N, thl'//' h~I;' /8 1/ /111///(/,/.//11/. rin!/. Proof. Let w be an dement or \J whidl iH IIO!, iii p~N~\ . 'I'h(~11 pNp
wR~ , whence p*R;. = wR:• . Since w i" not a J1pr() diviHOt' ill N* hy (18.1), we see that is a valuation ring by (12.1). (36.3) Let R be a semi-local integral domain and let x (~O) be an element oj the Jacobson radical of R. Assume that xR has no imbedded prime divisor and that, for every prime divisor P of xR, P is analytically unramified and R~ is a valuation ring. Then R is analytically unramified. Proof. Let R* be the completion of R and let the prime divisors of xR be PI , ... , pr . By our assumption, and by (36.2), if the prime divisors of p;1l* are p7j (j = 1, ... ,n(i)), then eachp7j contains a prime divisor lj3;j of zero which is the kernel of the natural homomorphism from R* into the valuation ring R:. i i . Let n be the intersection of all the l.l3;j . Sinee the Pi are all the prime divisors of xli, we sec that the p7j are all the prime divisors of xR* by (18.11). Since 1j37j is contained in any primary ideal belonging to p7j , we see that n is contained in xR*. Since x is in the .Jacobson radical of R* and since n:xR*= n, we sec that 11 = by (4.3), which proves that the zero ideal of R* is semi-prime, i.e., R is analytically unramified.
R:.
°
(36.4) THEOREM. A pseudo-geometric semi-local integral domain R is analytically unramified. Proof. We prove the assertion by induction on I' = altitude R. Let R' be the derived normal ring of R. Then R' is a finite R-module by our assumption, whence R is a subspace of R' by (16.8). Therefore, it is sufficient to show that R' is analytically unramified. Since R' is pseudo-geometric, we may assume that R is normaL If r = 0, then the assertion is obvious, and we assume that r ;::: 1. Let x (~o) be an element of the .Jacobson radical of R. Then, by (36.3) applied to this x, we see that R is analytically unramified, which proves the assertion. (36.5) THEOREM. If R is a pseudo-geometric ring, then every ring which is of finitely generated type over R is a p8e1ldo-geometric ring. Proof. By the definition, and by (:36. J ), we have only to prove that if R is a pseudo-geometric integral domain, then every affine ring A over R is a pseudo-geometric ring. Using induction on the
1'11.11"1'1':11 1'1 111I/lillI'l' of J!;I'IIPI'II,i,ol'f: of ,I, WI' 1111.\'1' 11111,1' 1,0 /11'(11'1' 1.111' l'II.fIl' ",111'1'1' ,I /"1,1'1 IVilll 1111 1'11'1111'111 ,/, of ,I, II Ilidlil'I'H III "I,ow Lila!. if q iH :f, /l1'illl(' iill'I'" or ,I, 1.111'11 ,I /q ,~/I,Li:4Iil',cl 1111' liIIiLI'III'HH l'olliliLillll fill' illll' j.t;1'1I.1 I'XI.I'IIt-!ioIlH, ~illl'I' /,'1 ( q n N) it-! /lH(,lIlIo gI'IlIIII'LI'il', WI' Ilal'I' Olily (,(1 /11'111'(' Lllal. ,I II.H 11.11111'1' HII.liHlil',Y 1.111' lilliLI'III','iH I'ollilil.illil 1'111' ilil,I'gl'al I'XI.I'IiHioIlH, If ,I' iH 1,l'aIlNI'I'IIIII'IIL:I,1 01'1'1' N, 1,111'11 I.II(~ aHHI'I'l.ioll follolV,e; fl'lllli (;i!'i,~), 11,1111 WI' aNHtlllw Ihat. ,I' iH aigillll'ail: ov('l' U. 'I'hl'lI, 1'11/1Hiilcrill!!: a HltiLablc filii\'(' illl.q.!;I'nl l.xLl'lif'doll of fl, whidt iH PHI'tlilo gl:ome',l'ie by (:1(LI ), WI' may aHHllme that x is in the field of lIl/olil'l 11,'; J( of R. Let L be an arbitrary finite algebraic extensioll (If /\', III order to prove the finiteness of the integral closure A' of A ill I), Hilll'I' the integral closure of R in L is a finite R-module, we may :I,,'iHIIIIII' that L = K and that R is a normal ring. Thus it is sufficient to PI'II"I' the finiteness of the derived normal ring of A (assuming that U iH normal, and that x E K). By virtue of (35.3), it suffices to show t.hal. if lJ is a prime ideal of A, then the derived normall'ing of A~ is a finite module over A~. Obviously, we have only to prove it in the case where ~ is maximal. Since Rlpn R) is a pseudo-geometric local ring, we may assume furthermore that R is a local ring with maximal ideal R. Since ~ is maximal, x modulo ~ is integral over Rim, m = ~ If x E R, then there is nothing to prove. Assume that x EE R. Let f(X) be an irreducible monic polynomial over R which is irreducible modulo m and such that f( x) E ~. Considering a suitable, finite integral extension of R, we may assume furthermore that f(x) is of degree 1, hence that x E ~. Let F be the set of pairs (a, b) of elements of R Ruch that ax = b. Then the set of aX - b generates a prime ideal q of the polynomial ring R[X] such that R[X]/q is naturally isomorphic to R[x] by (11.13). Then the set b of the b has no imbedded prime divisor by (11.13), and by its proof, and AI xA is isomorphic to Rib, which implies that xA, hence xAp , both have no imbedded prime divisors. If ~' is a prime divisor of xA~ , then ~' n R is a prime divisor of b, hence we see that RWnR) is a valuation ring and x is in the valuation ring. Thus we see that A~ is analytically unramified by virtue of (36.3), whence the derived normal ring of Ap is a finite A~-modtlle by (32.2), which completes the proof of (36.5).
n
(36,6) COROLLARY. If A is an affine ring over a pseudo-geometric integral domain, then the derived normal ring of A is a finite A -module.
I: II
(;1(1.',') .1 1II'IIIi 111('(1/ 1'/1/(/ h' /.4 /11111/11111'/1111/ /I/lI'/I//lljil'tI If 1 III /S (1111111/1/('(11//1 IW I'll IIIlji/'(1 fill' ('1 ('1'/1 111(1.1'/111111 /(1('(/1 III (~r 'l'lin PI"OO!' is illllll('dill.l.n hy (17.7).
i/ U
/11111 11111.1/
u.
(;W.H) '/'I1I':()I{,II:M. I,d .1'1, ..• , :1'1/ II(' 111(/1'111'11/('(11//1 JIlt/(·II(·lliI('III. I'!(' ments O'I!(~r (t sem:i-iow/ 'I"inl/ N. 11881/,/11(' /,//,(/,1, N -/8 (Ulldlli/mlll! 11I/1'(/'llIlji('d. If a semi-local rino H' is a rino (~r (Jlwl'il'n/'8 I(/' 11,(, jllltY/WIII/ill.! I'ill!! R[x] = R[XI , ... , Xn], then Il' is analytically unmll/:~/it:d. In PW'I'I:('It.iIll', 1j p is a prime ideal of R, then R~ is analytically unram'~/i('d. Proof. By virtue of (36.7), we may assume that R' is a loeal rillg. Let q be the prime ideal of R[xj such that R[x]q = R' and set p = q R. Let R* be the completion of R and let p* be a minimal priffi(~ divisor of pR*. Then the theorem of transition holds for R~ and R:. and R~ is a subspace of R:•. Since R* is a pseudo-geometric ring by (32.1), we see that is a pseudo-geometric ring by (36.1), which implies that the completion of has no nilpotent elements except zero, whence the same is true for its subspace R~ . Thus Rp is analytically unramified and we may assume that R is a local ring with maximal ideal p. Since R*[x]/pnR*[x] is isomorphic to R[x]/p"R[x], qR*[x] is a prime ideal of R*[x] (because R*[xJlqR*[x] = (R*[xJl pR*[x]) / (qR*[xJlpR*[;r])), and furthermore the theorem of transition holds for the local rings R' and R*[X]qR'[x] by virtue of (IS.S) and (19.1), whence R' is a subspace of R*[X]qR*[X] . Therefore we may assume that R = R*, whence R' is pseudo-geometric by (32.1) and (36.5), and the assertion is proved by (36.4). We add, here, a result on the normality of pseudo-geometrir: semilocal integral domains. We first prove it in a more geneml case:
n
R:.
R:.
(36.9) THEOREM. Let R be a Noetherian integral domain and assume that a ( ~ 0) is an element of the Jacobson radical of R. A 8sume f urthermore that: (1) aR has only one minimal prime divisor p, (2) aR~ = pR~ , and (3) R/p is a normal ring. Let R' be the derived normal ring of R. If R' is a finite R-module and if every (minimal) prime divisor p' of aR' lies over p, then R itself is normal and p = aR. Proof. R' /p' is integral over R/p. Since R~ is a valuation ring by our assumption (2) and by (12.1), we see that R' is contained in R~. Therefore R'/p' C R~/pR~, which implies that R'/p' and R/p have a common field of quotients. Therefore that R/p is normal implies that R'/p' = R/p. On the other hand, R' C R~ implies that p' R' pR~, whence p' is unique, and that R;, = Rp , whence aR;, =
n
I!II A l"I'llllt V I
,
p'h~""
fL roliowH LlmL flH' )1'. 'l'IH'I'I'/'OI'(I WlaN' Nip, whidl illlpli"H LlI/I.L h~1 (/'I~' I N. Nilll~n H' j~ /I. IilliLn 1~-lI\odlll(~ alld !:lineo ((. iH ill LlII' .1n""ilHoll 1'11."i':/l.1 or u, w(~ H('(l LllaL U = "~' by LIw lemma, of 1\I'ldl,AY-IIIIIII.yll., wialllH:n (J."~' "" pi implieH that all = p. Thus the pl'ool' iH I:oll\plnl,n. Now Wp apply j,}w above result to the case of pseudo-geometric illLegml domaiuH. (36.10) TllIWRlDM. Assume that a psewlo-geometric integral domain U is a homomorphic image of a locally Macaulay ring, and that aI, ... , ... , ar are elements of R such that height (L aiR) = r and such that they are in the Jacobson radical of R. If L aiR has only one minimal 71rime divisor p, if the ai form a regular system of parameters of Rp and 'zf Rip is a. normal ring, then R is norma.l and p = LaiR, and furthermore every RIL{ aiR is a normal ring. (LEMMA OF HmONAKA) Proof. We prove the assertion by induction on r. If r = 1, then (36.9) a,nd (34.9) imply the validity of the assertion, and we assume Chat r > 1. Let q be a minimal prime divisor of arR. If r is a minimal L aiR, then height r/q :::;: r - I, whence, by prime divisor of q the validity of the chain condition for prime ideals, height r ::; r, and p = r. Therefore q c p. Since the ai form a regular system of parameters of Rp we see that q is unique and that arR q = qRq. Consider R" = Rlq. Then, applying the induction to R" with elements ai modulo q, we see that p/q = (q L aiR)/q and that RI(q L{ aiR) is normal for each j = 0, I, ... , r - 1. In particular, Rlq is normal. This, the uniqueness of q and the equality arR q = qRq imply that R is normal and that q = arR by the case where r = 1. Therefore the proof is complete.
+
+
+
EXERCISE. Prove that a pseudo-geometric semi-local integral domain R is unmixod if and only if R satisfies the second chain condition for prime ideals. GenernJi7.o (36.10) to the case whor~ R it! a pseudo-geometric unmixed local illtOgl'al dOIll~dn.
37. Analytical normality We I:la,y that ul:lemi-local integral domain R is analytically irreducible if the completion of R is an integral domain. R is called analytically normal if the completion of R is a normal ring; in this case, R itself must be a normal ring by (18.4). We say that a prime ideal p of a semi-local ring R is analytically irreducible if Rip is analytically irreducible.
1:\11
(lIiiIlMII)'I'I/.1(1 I.(II'ill. IIINIlII
"1'1
(;i7.I) U III' 1/ 'I'il/(/ (I//d (/IlIlI//I1f' Iho! I, ." ( fa: (//'(' I. ·,:s '/I./I! It Z(·'/'o···dilli8U'1' ,:'" N /UU{ (:!) '''~:I/U IH.!J /' /8
N/II'II
IIII/l ( I )
(/./1. 1'//'/1/('//1 /(/
the total quotient r-i'/l.(f 4 H 8U,(:/i. UUI.l ll! ((.nd ·W) nrc -i", N, !hl'''' ·0 '/N ":/1. fa'. Proof. Since tuv E tU, we have tv C tH:'uN = lU, wl\l~I\(:c LlII'I'I~ i.'j an element v' of R such that tv = tv'. Since t is noL a ;!'(:I'O divi:-iOl', W(' have v = v' E R. (37.2) Let R be a normal semi-local-ring and let R* be its compl/:/i/ll/.. A SSUme that t is an element of R which is neither zero nor unit in N such that every prime divisor of tR is analytically unramified. If v ·is an element of the total quotient -ring of R* such that v is integral over HI and such that tv E R*, then v is in R*, Proof. Let the prime divisor of tR be ~1 , • • • , ~r • They are of heigh (, 1 by (12.9). Let S be the intersection of the complements of ~i in U. Then Us is a semi-local Dedekind domain with maximal ideals ~iR,~ , whence ~iRs is a principal ideal for every i by (28.9). Let Xi be an element of ~i such that ~iRs = xiRs for each i, and let ei be natural numbers such that IRs = X~l ... x;r Rs . Since tR*: sR* = tR* for any s of S by (18.1), it is sufficient by virtue of (37.1) to show that there is an s of S such that tvs E x?··· x;rR* (observing that X~l ... x~r ( tR). Let ~jj (j = 1, .. ' ,n(i)) be all the prime divisors of ~iR* and let Wij be the valuation of the field of quotients of R:' ii with R~ij as valuation ring and such that Wij(Xi) = l. Let q,ij be the natural homomorphism from R* into R:. i i . Let fl , ... ,fr be non-negative integers satisfying the following condition: tvs is in X{l .. . X;rR* for some s of S but for any s of S and for any i, tvs is not in X{l .. ·x~r. xiR*. Then it is sufficient to show that fi 2: ei. Assume the contrary, for instance that fl < el. We take an element s of S such that tvs is in x{l .. . X;rR* and let z be an element of R* such that tvs = x{l ... x{rz. We may regard q,ij as a homomorphism from the total quotient ring R** of R* into the field of quotients of R:. i j • Then, :,lince v is integral over R*, q,ij( v) is in R:. i; , whence Wi/( q,ij( v )) 2: 0, Since WJj(q,lj(t)) = el > 11 = Wlj(q,lj(X{l .. 'x~r)), we have Wlj(q,lj(Z)) 2: 1. This shows that q,lj(Z) iH in q,lj( ~~j). Since the kernel of q,l; is contained in ~~j, it follows that Z is in ~ij . Since x 1U; = j ~ij R,~ , Z is in x1R; and therefore there is an element s' of S such that ZS' E xlR*. Thus, with s" = ss' which is in S, tvs" E Xl·x{l .. 'x~rR*, which is a contradiction and we have ei :::; fi for every i, Thus the proof is complete. (37.3) A ssume that a normal local -ring R is analytically normal. Let
n
I'll
11"1' 1'111 1'1
1',rll'll,ql"l/ I~( 1/11' ,1il'ld I~r 1IIIIIIil'lIl,q II I~r Ii' 111111 11'1 //1/1'(/1'111 I'lmilll'l' 1(1' Ii' ill /" ,\ ,%'/11111' l/il/I 1'1'1'1'/1 111'ill/I' itll'lll )I 1(1' /it'il/hi I ill I iN 111I1I1,l/lil'llll,ll 1I11I'1//llIjil,tI, '1'/11'11 1/11' 1'11111/1/1'110 /1 I" I~I' I IN illll'tll'lIlI!! 1'11I,o.;I'd, IItIII 1,0.;, fill' 1'/'1'1'11 11I1I,rllllll! itll'l/I III IIf I, I", is 111111 1/11//'((,11/1 /l.III'IIIIIJ,
" III'
1/
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l/il'
./il/III' 81'llltl'l/llll'
(,11'1\\('111, or I cHlI'h LhaL I, 1\(1/.) alld 1('(, dill' il'l'('dlll'ibll~ IlIOlli(, (loIYllollli:t.i OV('I' N \Vliil,li bn,.; /1, HH a root.. Ld Ii"' lin Lhl~ (lolllpidioll of It nlld ltd. 1** I )I~ (,he ill" Lnp;m[ e/OHlll'l, of /* ill i(.,y Lotal quotillld, rillp;, Hi II I'.i\ / iH a {illiLlr u~ I\lOdlde hy (10, J(;), t.he ill (,egral I'lmmre of n*[aJ in itH LoL:d (jlloLi('11 L I'illg I'oineide::; wiLh 1** by (J7,8), Therefore we see that dJ** C 1(,*1111
P('OO\'. L(,(,
II, 1)(' :1.\1
(.1\1' di:-;('l'illlill:l.IIL or Lill'
hy (10,15), whieh implies that 1** = J* by (37,2), and therdOl'(I 1* iH illtegrally closed, If b is a nilpotent element of J* and if x iH lIoi, :\, y.W'O divisor, then bj:r; is integral over 1* and is in J**, and b E :r1** ~,~ :r/*, which implies that b E mnJ* with Jacobson radieal 111 of R alld with an arbitrary natural number n, Therefore we have b = 0 by (4,2), whieh proves that 1* has no nilpotent elements except J1(1J'(), i,I)" the zero ideal of J* ilS tlw intersection of a finite number of prim(r ideals, say pi , ' , , , The idempotents of t.he total quot.ient rillg of J* are integral over J*, and t.herefore they are in J*, and we HCO that J* is the direct. sum of 1*/p; . Sinee 1* is integrally closed, c[l,(Ih /* / p; mURt, be normal, and the proof if; (~omplcte,
P: '
(:i7,4) THEom;J\L Assume that a normal local ring (R, nt) is analytically irreducible, Let K be the field of quotients oj R, Assume that a local ring (R', m') satisfies the following three conditions: (1) R ::s; H' c K, (2) RI/mR' is a finite R/111~mod1lle, and (:~) altit.ude R' = altitude R. Then R' coincides with R, Proof. Let R* and R ' * be the completions of Rand R ' , reRpectively. R = m, we see t.hat 111' C m,i R for every i, Therefore Since m' there exists a natural homomorphism ¢ from R* into R ' * and ¢(R*) becomes the elosure of R in R ' *, By assumption (2), we see that R '* is a finite ¢(R*)~module by (30,6), whenc:c altitude R ' * = altitude ¢(R*) by (10,10), Since altitude R ' * = altitude R' = alt.itude R = altitude R* by (17,12), we have :t11.iLllilc R* = altitude ¢(R*), Since R* is an integral domain, it. follow:-: that ¢ is an isomorphism, Thus we sec that. R is a subspace of W :,,"1 !t,'* is integral over R*, Let alb (a, b E R) be an arbitrary nl(rlll('I\1, or R ' , Since alb is iutegral over R*, there are elements ci , ", , c~ of Hoi' such that (a/b)n + ci(a/b)n-l c! = 0 :tlld Lhnl'dol'c an ban-lei
n
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+ '" +
+
+ ,., +
(1I1\()~IIII'I'111i
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II" ('~: 11ft," I, ' " ,II". Nil\(·n (L;' 1)'1/" 'Itl') we "n(~ LhaC UWl'n a/'(~ nl('III('IIL" ('I , ..
+ ... + {),'c"
= 0, wh(~I\(~(1 (a/II)"
nH I
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'u hy
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(17.\1),
I 1)(/" I ('"
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and alb is integral over U. Sil\(~e His IIOI'III:t.i, (1/1) i" ill h~, whil'll pt'III'I':' the assertion. Now we prove a theorem on loealitie" over field" 01' Ikd(~kilid dll mains which satisfies the finiteness condition for intnl!;ml (~XL(~IINioll::. Note that a Dedekind domain I is pseudo-geometric ring if alld old.\' if it satisfies the finiteness condition for integral exten:sions. A pal'L of the following theorem (analytical unramifiedness and the fillill' ness of the derived normal rings) follows from (36.4) and (::n.;)). But we shall give a direct proof of this part in our special case: (37.5) THEOREM. If I is either a jield or a pseudo-geometric /)ed(' kind domain and if R is a locality over I, then the derived normal rill!! R' of R is a finite module, and R is analytically unramijied. If .fu.rl/ln more R is normal, then R is analytically normal. Proof. Let r be the altitude of R. We prove the assertion by ill duction on r. If r = 0, then the assertion is obviollS, and we aSSlllll(' that r > O. Our induction assumption means by virtue of (36.6) thaI, if S is a semi-local ring of altitude ~ r - ] sllch that, for any fila:; imal ideal m of S, Sm is a locality over I, then Sis analytic:ally lIll ramified. Let m be the maximal ideal of Rand snt ~ = I m. Tlu'll I~ is either a field or a Noetherian valuation ring and is pseudo-g(~o metric. Therefore, we may assume that I = I~. Let Xl , . . . , Xn b(' elements of R such that their residue classes modulo m form a traIl, scendence base of R/m over I/~. Since I is a field or a valuation riIlf!:, we see that the Xi are algebraically independent over I. Set B = I (Xl, .. , , xn). Then B is either a field or a Noetherian valuatio II ring and satisfies the finiteness condition for integral extensions by (35.2), whence we may assume that I = B, i.e., that R/m is algebraic over I/~, Let YI , ... , Yr be a system of parameters of R, whe)'(~ if I is not a field, we choose YI to be a prime element of I. There is a chain of prime ideals 0 C ql C ... C qr = m in R sllch that Yi E qi if and only if i ~ j, and therefore, in the ring I[y] = I[YI, ... , Yri, we have a chain of prime ideals 0 C ql I[y] C .,. C qr I[yl, whence height qr I[y] 2: r, whic:h implies that lJI, ... , YT (01', Y2, ... , Yr if I is not a field) are algebraically independent over I and, by virtue of the validity of the altitude formula for I[y], thaI.
n
n
n
n
1'1111"1'1':11 \'1
It' iH 11,lgI,I'I'II,i(' 01'1'1' 1111:. 1,1'1. " III' 1111' flllll'Lili1i 1il,IiI (If Ii' IIlid 11'1. ./ Lill' illl.l'I1;I·II,1 I·IOHIII'I' III' Illll III L. 11'1I1'I,hl'I'IIIIII'I', HI'1. U" "'1./ I. HiIlI'I' .I i." /I. lillil.l' 11!f11l1(1I11I11', WI' HI'I' I.hal. H" iH II. lillil.l' UIIIO,,"II'. (11.) WIII'II /, iH ,",I'PHI'II,hll' OVI'I' /1//1. II'!. III" 111'11.11 :l.l'hil.l·IU·'y 11111,,111111.1 idl'al of U" 11.1111 HI'1. " I"" H'y (:17.:\) iLllli Il'y 0111' IlIilllI:l.ioll I!.HHllIlIpLIIIII, WI' NCI: LII:i.!, 1.111' I:OIllpldilili of .lit iH a 110 rllla I I'illl-(. Hilll:l: Lilli :tJ,: 1'111'111 a H.YH!,I:1I1 III' pamllll:!,(lI'H III' N:~" and of .lit, we HC!) that U::,,, OC~ b'y (:n.-1). ThcJ'(:fol'c N" iH :illulytieally unrumified, whence iIH Nli bHp:we H (by (Hi.8)) iH allalytieally unramified. On the other 11"'1111, Hiuee R:~" iH normal for arbitrary m", we see that R" is' normal b'y (:\:~.n), whener: H" = H', and the first assertion is proved in this (:II,HI:. If R iH normal, then R = R", and R = .Tit , which is analytically IlIlrmal. (b) Next we eonsider the case where It is not separable over I[y]. Take elements al , ... , a" of I and a power q of the eharactcristic p q of [ such that L' = L( a~/q, ... , a;/\ yi/ , ... , y~/q) is H8parable over q q Iln:/ , ... , a;/q, yi/ , ... , y;/q]. Let I' be the derived normal ring of Iial/", ... , a;/"] and let J' be the integral closure of I[y] in L'. Sinee 1/ is separable over I'[yi/ a, ' .. ' , y~iaJ, for every maximal ideal r of I) = R[J'], the completion of P, is a normal ring and P, = .T'crnJ') h'y our observation in (a) above. Hence P is normal. r iH a finite U-module because .T' is a finite I[y]-module by (:35.2), whenee R' is fI, (illite R-module. P is analytically unramified as in (a) above, whence i I.H subspaee R is analytieally unramified. Thus the first assertion is proved. Assume now that R is normal. Let b1 , ••. , bt be elements of .J which form a linearly independent basis for L over the field of quoLim1ts of I[y]; let Cl , ••• , Cu be element of JI which form a linearly illdependent basis for L' over L. Since J' is a finite I[y]-module, there it-! an element d (~O) of I[y] such that dJ' C L I[y]bicj. Let P*, U* and S* be the completions of P, Rand I[Y]cmnI[Y]) , respectively, /Liid let T be the integral closure of R* in its total quotient ring. By the dlOice of d, we have dP* C L S*bicj. Since P* is integrally closed, we see that T is contained in P*, whence dT ~ L S*bicj. Since the I:,; are linearly independent over S*[b 1 , ••• , btl by (18.1), we sec that tiT C S*[b J , . , • ,btl, and dT C R*. Therefore T = R* by (37.2), and therefore we see that R* is a normal ring because R* is a local ring (d. the last step of the proof of (37.3)). Thus the proof is complete. II(I
n ./.
./"
(37.6) COROLLARY. If R is a locality over a field or over a pseudoveometric Dedekind domain, then the number of prime divisors of zero
I HI /1/ 1111' 1'1i1//11/1'lillll
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11'. 1'1'001', 'I'll(' l'Olllp/pl.ioll or It" iH U;'lh"l, IVllil'l, It/I.H 1./,1' HII.III(' l.ol.lI.l qIIO(.i(,IIL I'illg wi!'h N*, W\\('III'.I' till' 1I1111t1)(,1' 01' Pl'illll"/iviHol','-\ or 1,1'1'0 ill R* e()ill<:id(~H with I.hal. ill U*llt,'!. Nilll:(~ N*IU'I iH 1.111' dil'l'I'!, HIIIII or
11/1' 1/1'1'/1'1'1/ 11/11'1/11/1 /'i//(/
I(/,
normal local rings, we prove the aHH(~l'(,ioll. The following is another corollary to (;n.!i), alld iH I'('ally l:ol\Ln,illl,d in the proof of (37,5), (37.7) If R is a normal locality over a ring I which i8 either a .lil'lt! or a pseudo-geometric Dedekind domain, then there are a .li:nite n"IJ!m/I/'/' of algebraically independent elements Xl , . . . , X", , YI, ., . , Un /IIwr I such that R is of finite type over I(XI , .,. ,Xm)[YI , ... 'Y1I1~ with pr'ifll.I' ideal ~ generated by a prime ideal of I and the y i . We want to add here an applIcation of (37.::n t.o pseudo-gromei.l'ic rings.
(37.8) THEOREM. If a normal pseudo-geometric local ring R is 1(1' finite type over an analytically normal ring S, then R is analyticall!/ irreducible. If furthermore R is separable over S, then R is anal!Jticall!/ normal. Proof'. The separable case follows immediately from (:n .:i) alld (;:)G.4). Let. R' be, ill the general casc, the Heparable algebraic clOHUl'l' of S in R. Then R' iH analytieally normal, whenee it is analytically irreducible, Since R is purely inseparable over R ' , it follows t.hat tlw completion R* of R has only one minimal prime diviHor of zero. Since R is analytically unramified by (36.4), it follows t.hat R is analytically irreducible. (37.9) COROLLARY. Assume that a pseudo-geometric semi-local integral domain R is of finite type over an analytically normal ring, and let R' be the derived normal ring of R. Let r be the number of maximal ideals of R'. Then the zero ideal of the completion oj R is the intersection of r prime ideals. We note that (37.10) Let I be a complete local integral domain and let Xl , . . . , x" be algebraically independent elements over I. If a local integral domain R is of finite type over I[xl , ... , Xn], then R satisfies the conditions in (37.9). Proof'. I is a finite integral extension of a complete regular local ring by (31.6), whence we may assume that I is regular. Since every
('11.\1"1'1'111 VI
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38. Some types of ring extensions LnL (U, lll) hi' a 10eal illl.ql;l'al domain and let (V, b) be a valuation I'illl!: Hitch thai, R :::;; V, Let n be an ideal of R such that a ~ Rand Ii'.L aJ , ... , ar be a ba"is for a. Then there is an ai , say al such that n V = al V. Set bi = ai/al, B = R[b l , . , . , brl and ~ = b B. Then nH will be shown below, the ring B~ is uniquely determined by R, a, nlld V, and this B~ is called the dilatation of R by the ideal 11 with 1'(,Hpect to V. When a = m, we call B~ the quadratic dilatation of R wi th respect to V. The proof of the uniqueness: Let a~ , . , . , a~, be another basis for a. We assume first that ai = a~ for i = 1, .. , , r. Let a; be such that (/,~V = aVand set b; = a;/a~, B' = R[b~ , . , . , b~,], ~' = \.) B ', I ' n C.,: = ai/aj, C = R[Cl, ... , Cr ']' q = \.) ' C. Smce alI V = ajI V, we /-I(\e that Cl is not in q, whence Cl is a unit in C q , which implies that e q contains B ' , and B~, :::;; Cq . Similarly, since b; is not in ~/, we see Lhat C q :::;; B;, , Thus C q = B;, . On the other hand, since the are linear combinations of the ai with coefficients in R, we see that B' C H, and B = B ' , whence B~ = B;, , Thus we have proved the uniquelIess in this ease. The general case can be proved easily, considering L' I I 1,118 baSIS aI, ' . . , a r , al , . . . , a r , • The main interest of dilatations lies in geometric applications, which should be stated with regard to varieties or something similar. Therefore, we shall deal only with very special results concerning quadratic dilatations which are of ring-theoretic interest. Let (R, 111) be a regular local ring and let ~ be a prime ideal of R KlICh that ~ ~ 111 and let V be a valuation ring which dominates R ILild ha" a prime ideal q sueh that V q dominate" R~ ; j,he existence of V follows from (1 UJ). Starting with Ro = R, let R i +l be, if Ri is alI'cady defined, the quadratic dilatation of Ri with respec:t to V, Then we have the following:
n
n
a;,
(38,1) imal and
n
Each Ri is a regular local ring, q Ri is not max(Ri)(qnRi)' If furthermore depth ~ = 1 and if the de-
THEOREM.
R~ =
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,~lIl'h lim/'
n
l'!'Ool'. We pl'OV(~ 1.I1(~ lin,(' aHH(~I'I.iOIl hy ilililinLiol1 011 I. TII(~ II,HHnI'Lioll is obvioUt-i for 'i = 0, alld W(~ HHHIIIll(\ Lha(, ,,: > O. Lnt. .1', , .,. , .1',. II(' /I, regular system of parame(;el'::i of U i - l , wltnrn w(\ may aHHlIllW LiIHL x j E Xl V for every j. Then Ri is a ring of quotien(,H or Hi d';):2j.1:, , ... , ... , Xr/XI]' Since q n R i - l is not the maximal ideal, :1:1 q: q n H, , , which proves that (Ri-I)(qnRi_l) = R~ contains Xl as a unit. Thet'nforl' (Ri)(qnRi) is a ring of quotients of R i- l [l/XI], and coincides wiLh Nil' Xl is a non-unit in Ri , which implies that q Ri is not maximal. Applying this to the case where height ~ = r - 1, we see that all.i tude Ri = altitude R i- l (= r) in the special case. This fact impli(,,'1 in the general case that X2/XI , ... , Xr/XI modulo xIRi-l[xdxI , .. , . '" , xr/xd are algebraically independent over Ri-I!( L xjR., ,), whence R;/XIRi is a regular local ring by (14.8), and therefore we H('" that Ri is a regular local ring. Now we shall prove the last assertioll, Let R' be the derived normal ring of R/~ and let a be an arbitrary element of R'. Let L' be the field of quotients of R/~ and set V' (V/q) n L'. Since depth ~ = 1, R' is a Dedekind domain by tltl' theorem of Krull-Akizuki, whence V' is a ring of quotients of H'. Let v' be a valuation of L' whose valuation ring is V'. We want 1,0 show that there is an i such that Ri/ (q n R i ) contains a. Let a III' bi-dCi-l. with bi - l , Ci-l E R i- l . If sand t are the degree of bi - l and R i - l ), th(~11 Ci-l with respect to the maximal ideal L XjRi-I! (q bi- l and Ci-l are divisible by x; modulo (q n R i ) in Ri/ (q n R i ) with u = min (s,t).Therefore,ifa EE Ri_l,thena = bi/ciwithbi,ci E H, such that V'(Ci) < V'(Ci-I). Since V' is a Noetherian valuation ring, there is an n such that v(cn) = 0, i.e., a E Rn. Since R' is finite over R/~, it follows now that there is an n such that Rn/ (q Rn) contaillH R'. Since V' is a ring of quotients of R', it follows that Rn/(q n Rn) = V', and the last assertion is proved by virtue of (25.18).
n
n
n
(38.2) COROLLAHY. Every quadratic dilatation of a reg·ular lOClli ring is again a regular local ring. Proof. If Xl , ... , Xi' is a regular syst8m of parameters of a reguhl' local ring R, then XdXI , ... , Xr/XI are algebraically independent ovel' R/( xiR) by the proof of (38.1), and the assertion is proved similarly. As an application of (38.1), we prove the following:
L
II:I 1'('/ f ( "'0) II/' (//1 ('11'//11'//1 (~f' (/ 1'('Olllu/' 1(1('(111'/1/(1 /fII1I II'! P III' (( Ilri//II' /1/1'1/1 /~J' II'. '/'Ii/'/1 ,!I/' 1//'(/1'/'/' 1/ /~J' f II'/Ih /'/' 111)/'('110 III is //oll/','iN IIi((/I. II,,' 1/('(/1'/'/' 1/' 1(1'/ lI,ilh"('N/w('llo pN\, (':/1. Np), /'1'1101'. II' WI' klllllV 1./11' vHlidil,y 01' 1./1(' II.H,'iI'I'Lioll ill L111' I':l.HI , Whl'l'l' dl'pl./I ~I I, I.hl'II, 1'llIlHidl'l'ilig a I 1111 .., iII Ill. I I'haill 01' pl'illl!' it/I':!,IH 1111,· I.WI'I'II P alld III, Hay, ~). ~10 C " , C P,. III, alld applyillg (,lin valid· il.y to l'al:h Np ,; wiLh Willii' iill'al Pi IUp; , WI' pl'ove t.he aHHI,l'tiol1 by vil'l.lIn of (~S,;n, ThllH 1Vi' llJay aHHllllle that. dl,pth p = I. AHHIIIII!' (,hal. It,' iH a I'I,gular loeall'ing dominating R such that (1) tuR' iH L1w IIlHximal ideal, (2) there is a prime ideal p' in R' such that p = p' n n N, and (3) 111 R' n R = m n , Then d is the degree of f with respect 1.0 mR', Since p c p', the degree of f with respect to p'R;, is not less Lhan d'. Therefore Rand p may be replaced by R' and p'. Hence, ill particular, we may assume that R is complete. Let I be a coefficient dllg. We treat from now on the case where I is not a field, because Lhe case where I is a field can be treated similarly but in a simpler IV tty. Let p be the characteristic of the residue class field, Then there iH a Noetherian valuation ring J* snch that I* Ipl* is the algebraic 1,Iosure of llpl; the existence is proved easily by virtue of Zorn's l!Hnma. Let Xl, . , . , Xr be a regular system of parameters of R. Then l,fll1re is a homomorphism ¢ from the formal power series ring in indeterminates Xl , ... , Xr over J such that ¢(X i ) = Xi. The kernel oi' ¢ is generated by a Eisenstein polynomial say F by (31.12), ConHider R* = l*[[XJ , .. , ,Xrll/FI*[[XI , ., , ,Xrll. Since F is an EisenHLein polynomial, we see that R* is regular. By the construction, we HI:e that R* satisfies the conditions for R' above and has a prime ideal P' as stated there. Thus we may assume that Rim is algebraically dosed. Let (V, n) be a valuation ring which dominates R and which has a prime ideal q such that R~ ::::; V q and such that V 111 is algebraic OVer Rim, hence, Vln = Rim. The proof of the existence of such a V is similar to the proof of (11. 9), considering a maximal chain of prime ideals of R which goes through p. Then there is a regular sysLem of parameters Xl, ' .. , Xr such that XiV C Xl V for every i = 2, ... , r. Then, obviously, Xl, XdXI, . . . , xrlxl is a regular system of parameters of the quadratic dilatation RI of R with respect to V. Hince f is of degree d with respect to 111, f is a homogeneous form in Lhe Xi of degree d with coefficients in R, and not all the coefficients ILre in m. ThereforeiI = flx~ is in RI and the degree of iI with respect (;0 the maximal ideal of RI is not greater than d, Since Xl EE p, the
(:11-1,;\) 'l'III'IIIIII'I~I.
(/I~, III)
III
dq!; I'l'l , 01' II wiLli I'I'HPI'I'L
(II
1,0 ph'l'
n h'll! 11'11 1'111/0'1 1 , i~1 1'111111.1 10 tI',
I iH I'I'plll.l'I'd by i,l, ~illl~I' WI' hnvl', nI'LI'I' II. lillil.,' 1I(IIItill'I' III' 1.111' above replaeemellLH, tlw (,aHn wli"I'n ~I iN gl'"I'I':l.LI,'1 hy :I. Nllb,'-:I'L 01' :I. regular system of parameterH of H by (:{~, I), nlill (,Iw aNHI'I'Lioll iH easily proved in this case, Thus the proof iN eomplni,I', Secondly, we consi.der a special t.ype of inseparalM eXkllHioJlH of local integral domains, (:38.4) Let (R, m) be a local integral domain and let a be an element of an integral extension of R, Assume that a is not in the field of quotients of R, that the characteristic p 01 R is different from zero and that a P E R. Then R[aJ is a local ring. Let m' be the maximal ideal of R[aJ. ,2 Then either length Rlm m/m2 = lengthR[a] I til' m' /m , or (length R/lil 2 m/m ) + 1 = lengthR[a]/III' m'/m'2. The first equality holds if and only if either the irreducible polynomial X P - a P over R is irreducible modulo m or there exists an element b E R such that (a - b) P E m, Ef m2. Proof. Since a is purely inseparable over R, R[aJ is a local ring. Let Xl, ... , Xr be a minimal basis for m. If X P - aP is irreducible modulo m, then it is easy to see that m' is generated by m. Furthermore, the Xi form a minimal basis for m'. For, otherwise, there is a linear combination y = L aiXi of the Xi such that a'iR[a] = R[a], and such that y E m,2 = m2R[a], which is impossible because 1, a , . , . , aP - l are linearly independent over R, and this case is settled. Assume now that X P - aP is reducible modulo m. Then there is an element b E R such that a P - bP E m. Therefore, considering a - b instead of a, we may assume that b = 0, that is, a E mi. Then 111' = mR[a] + aR[a] = m + aR[a]. Assume that. an element y E m, which is not in m 2, is in m,2. Then, since m' = m + aR[a], t.here is a relation i 2 1 dixi)a + eia )a (Cij , d i , of the form y = L CijXiXj + P l ej E R). Since 1, a, ... , a - are linearly independent over R, we have y = L CijXiXj + ep _2 aP • This shows that the residue class of y modulo m 2 is of the form unit X (a P modulo 1112 ). Therefore lengthR[a] 1111' m' /111,2 is either r or r + 1 according as a P Ef m 2 or a P E 2 m • Thus the proof is complete. '1'111'1'1'1'01"(' Ii IlIlI.y III' l'I'pIHI'l'd II.\' HI (/1,1111
1.11(' OI'igillal R waN
1'.111 II plnt.n,
2::
(2::
(2::6-
(38.5) COROLLARY. With the notation as in (38.4), if R[aJ is regular, then R is regular, too. Lastly, we consider unramified extensions. A quasi-local ring R' dominating another quasi-local ring R is said to be unramified over R if the maximal ideal of R' is generated by that of R and if the resi-
1'11,11"1'1,:11 1'1 III' Ii" illlll'plln"'''I' 11\'1'1' 111111111' Ii': III I I('I'\\'ilil', 11" ill 1'/1,111'" Ii'. .\ Pl'illll' ilil'HI p' III' H' irl Iinid III III' /1I//'IIi/lljit'tI III' /'I/lllIjil't! (1\'1'1' 11', 11/' 11\'1'1' p' /i', il' h';'. ifI 1IIII'HIllilil'd III' I'Hlllilil'd, 1'1' Njll'I'Livl,I,V, 111'1'/' Hql'fl',,) , [INillf,!; 1.111' il.11I)VI~ Ci'l'llIiJllllllf,!;,V, lVI' Imvl' I.!HI 1'1I1111Willf,!; LlII'III'I'III: iiii('
I,I/lrlil lil'lli
l'I/I//lji"tI ()I'!'I'
n
(:{K,O) 'l'IIII)OIUI!M, l1.~N/f,/I/'(· I.lUI/. /T. III/,/I,,~i-I(/('(/,(
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III)
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( I) iI H' i8 a 'l'i'flO (~r Il'lwliwlltg oI Rlu.] with an element v. IIf I{ II/I,d II t/wn! is a polynomial I(.r) O1}(:'r Ie sv.ch that f( v.) = 0 and 811.111/ 1/11/ I, denoting by!'(x) the derivative of j(x), f'(v.) is not in lit', thl'//. H' /8 uuramijied over R, (2) Conversely, assume that R' is v.nramified over R, and lel H" III' II .finitely generated sv.bring of R' whz'ch is integral over Rand 8'Wlh l/ill./ It is a ring of quotients of R", Let = be the ma,I":/IIII.I 'ideals of R", where m" is chosen to be R' = R;~", Let v. be an element IIf N" such that v. modulo mil generates R'I m' over R I nt and such t/w,{, denoting by fi( x) a monic polynomial over R such that fi( x) modnlo 1lI ";8 the irreducible monic polynomial jor modulo over Rim, Jl ('Ii. ) 'i8 not in for every j 'F 1. Then R' is a ring oj quotients of R[u] and 'Il isa root of a monic polynomialf(x) suchthatf - fd~2" E mR[x] lor some nat'ural numbers nj ; hence, in particular, if 1'(x) is the derivative of f above, then l' (u) is not in m', Proof. If R[u] is isomorphic to R[xl/fR[x], then (1) is obvious, and Lhe general case of (1) follows from the following obvious remark: (38,7) If a quasi-local ring R' is unramified over a quasi-local ring R and if a' is an ideal of R ' , then R' I a' is unramijied over R I (a' R), We shall prove (2), Set Q = R[u], n = mil Q and Q' = Qn ' Since fl ( u) is in m" but not in any other m; , mil is the unique prime ideal of R" which lies over n, Therefore R' = R~" is integral over Q', Since R' is of finite type, it fo11owR thtt R' is a finite module over Q', Since R' is unramified over R alld Nill(~(~ N :::; (2' :::; R', R' is unramified over Q', whence nt' = nR', FIII'I.lWl'ilIOI'I', by 0111' dlOi(:e of u, R'/m' = Ql/nQ', Therefore R' = Q' 1lN', wlii!'!1 ill\f1iil~R I.ha(, N' = Q' by the lemma of Krull-Azumaya, WiH'III'(: Lill' lil',-d, aRHI'I'Lio/i ill (2) is proved, Sinee Q' = R ' , there al'n 11:1.(,111'111 1I11111hl'l'H /I RIII'Ii 1,11:1.(" with g(x) = fd~2" -j;r, g(n) E= mUll/I, 'l'lil'I'!'I'()I''' /I i,e;:I. !'IIoL of a polynomial f*(x) such that 1* - il I 1111t'I,!'I, III III'dl'l' 10 Nhow that we can choose sHch an 1* to be lIIollil', ii, ill Illil\iI'il'lIl III Hllow Chat 'U. is a root of a monic polynomiaiI*'i' III' "1"';1'1'1' III II III: ii, 1111' dngl'ee of g, Set
m"
m7 , " , , m:
u
m;
m7
,pr
n
n
+
J
/1/ U I UII I h'l/d I. 'l'iI('ll, Hilll'I' /11"'1111 illlplil'H 111111. /1'1111 111 I IIIit'll/I. '1'1t1'1'I' fol'(~ by I.Iw 11~1\l1I1:1, or 1\ I'll 11-;\:;'1 II I I:I.,\'/I. lVI' /11'1' 1.1111.1, UIIII M, /1.1111 u is a root of a mOlli(~ polYllomial or liIW'PI' tI, IVllil'il ('.()lIIpidl'H 1.111' proof. The following three assertions are eorollarinN (,0 (:{S.O). (38.8) Assume that a ring R' is of finite type OWl' a ring R. 11 a prirnl~ ideal ~' of R' is unramified over R, then every prime ideal q' of R' such that q' C ~' is unramified over R. The proof is straightforward. (38.9) Let f(x) be a monic polynomial over R such that the discriminant d of f( x) is a unit in R. If u is a root of f( x), then every prime ideal of R[u] is unramified over R. In particular, if furthermore R~ is a regular local ring for a prime ideal ~, then for every prime ideal ~' of H,' which lies over ~, R~, is a regular local ring. The proof is straightforward by virtue of (10.17). If
U iN
dl'/.!; fl, II.lld l'IIII:lidn
I\lOllil', f/( 1/.)
I
(88.10) THEOREM. Assume that an integral domain R' is of finite type over a normal ring R. If every maximal ideal m' of R' is unramifled over R, then R' is a normal ring. Proof. With the notation as in (38.6), R:n' is a ring of quotients of R[u] such that f( u) = 0, f' (u) EE m', whence R~l' is a ring of quotients of R[u, l/f'(u)], which is a normal ring by (10.18). Therefore R~l' is a normal ring. Since R' is the intersection of all the R~, by (33.9), we sec that R' is a normal ring. EXERCISE. Let ~ be a prime ideal of a regular local ring R such that R/~ is regular and let V be a valuation ring which dominates R. Let R' be the dilatation of R by the ideal ~ with respect to V. Prove that R' is a regular local ring.
39. Separably generated extensions We say that an integral domain R is separably generated over its subring I if either R is of characteristic 0 or, denoting by Land K the fields of quotients of R and I respectively, and by p the characteristic by R, the tensor product L ® K K 1/ p is an integral domain. In order to prove a characterization and some properties of separably generated extensions, we introduce the notion of derivations of a ring (we need in this section only those of field, but we need the general (' ase later). A derivation D of a ring R iN an additive endomorphism of the total quotient ring L of R which satisfies the following condition: (1)
1'11.1 I "I'ldi/ 1'1
II'/,
/)('1',1/) ,"lI!! I 1/11,1' 1'111' ,/', II I "lllleI I:!) 1111'1'1' iii !III c,II'III('III, dill' U whil,li iH 1101. II, !I,c'I'O eli I' i,'lOI' Hill'll 1.1111.1, dl)1 h') <:: II~, II' tI c'/1.I1 1)(' dlOHc'/1 I.() 1)(' I, I.ltC'1I /) iH c'allc'd :1.11 //I!/'O/'u'/ tI/,/'i/I//.fiol/ of' h~. Lc'l, /) hn:r. dl'l'ival.ioll 01' Ii alld Icd. I j,l'. ii, Hlil)l'jllf,!; of N. II" til n, LI\(~II we Hay Ulal. /) iN a ti('I"1;/ial-iofl, otJ(:'/' I; '~f' liN = 0, I.ltnll /) iN (~all('d I.ltc~ Z/i/'II d(I'I"i/ml1;on and iN dnlloj,od by O. Wn 1I()l.c~ 1.1IaL: (;m. I) Jf J) is a derivation of a ring R with total quotient rinu ", then Dl = 0, D(x/y) = (yDx - xDy)/y2 (x, y E L, y not beinll a
zero divisor) . Proof. Since 1 = 12 , we have D1 = 2D1 and D1 = 0. Since x = y(x/y), we have Dx = yD(x/y) + (x/y)Dy, and D(x/y) = (yDx xDy)/y2. The set of derivations of a ring R over its subring I forms an Rmodule, which is denoted by '£Jer(R/I). Linear dependence of derivations always means dependence in this module, hcmce over R. (39.2) I is a subring of a ring R and if R is of finitely generated type over I, then, denoting by L the total q1wtient ring of R, we have '£Jer(R/I) = '£Jer(L/I). Proof. It is obvious that '£Jer(R/I) c '£Jer(L/I) by the definition of derivations. Let D be an arbitrary derivation of L over I and let Cl , ••• ,C n be elements of R such that R is a ring of quotients of I[cl , ... ,crJ Since DCi E L, there is an element d of I[cl , ., . , cn ] which is not a zero divisor such that dDci E I[cl, ... ,c,,]. Since every element a of I[cl , ... ,cn ] is a polynomial in the Ci with coefficients in I, we sec that dDa E I[cl, " . ,cn ]. Let S be a multiplicatively closed set such that R = I[cl, ... ,cn]s . If b E R, then b = a/s with a E I[cl, ... ,Cn ], S E S. Then dDb = d(sDa - aDs)/i by (39.1). Since dDa and dDs are in I[cl , . " ,Cn ], we see that dDb E R, whence D is a derivation of R. Thus '£Jer(L/I) = '£Jer(R/I). Let R be a ring and let Xl, ... ,Xn be indeterminates. Then there are derivations Di (i = 1, ... ,n) of R[[Xl , ... ,Xn]] such that Di(Lair ... jnX{l···X~') = Ljiah ... jnX{l ... Xinn/Xi. These D i are called the partial derivations and are denoted by a/ax i ; if f is in the total quotient ring Q of R[[XI , ... ,Xn ]], thon D;f m:1y be denoted by aj/ax i . When fl , ... ,1m arc elements of Q, then the matrix (af/aX j ) is called the Jacobian matrix of II, .,. ,!m and is denoted by JUl, ... ,fm ; Xl, '" ,Xn ) or by J(b, ... ,fm)' If D is a derivation of R, then there is a uniquely determined derivation D'
rf
111'1
(111:01\1111'1'1111' 1.111',1/1 IIINlill
or h'II.\,
L
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.. ,
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(/)(/,11,,·;,,).\':1 ...
hy
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Wn IIOLI~ L!taL Lhn parLi:t.i dl'I'ivaLiol\l--i :\.1111 I.Iln aill,vI' 0' ;'01' ill Iq2;r1 I. I the polYllolllial rillg NIX, , ... ,X"I. If iel a !tomolllOl"phiNm (ldillcd Oil /t,IIX, , ... , X"II alld if' :/', :\,1'1'. NII/·I,
/) :\Yn illLq!;l':d dnl'ivaLiollN oi"
I.ImL :1'.: = cp(X i ), i.lWlI, [or Hni ( RlIX" ... , X"II,
n
0-
(39.4) COROLLARY. Assum.e that a field L is generated by elements Xl , ... , Xn over a field K and let .f1 , ... ,.fs be a basis .for the kernel of the hom.omorphism cp over- K from K[X1 , ... , Xn] onto K[X1 , ... ,xn ] such that ¢ (X i) = Xi . I.f r- is the rank of the Jacobian matrix J (.fl , . . . , ... ,fs), then every maximal linearly independent set of derivations of Lover- K consists of n - r elements. On the other hand, ~f L is separably algebraic over K, then every derivation of K can be extended uniquely to a derivation of L.
1'1'" 1"1' I,ll!, V I
IIII
III' I(. lil'ld L ill I'nll('d I(. NI'IUII'II/I//(/ III' Ii 11\'(11' illl Hili die,leI II il' LIllI .1', 1/.1'(' li.lg(·III·II,i('l/.lly illcil'llI'llcic'liI, IIVI'I' 1\ nllci if L iH H('PII.I'II,hl.v I/.IJ1;nill'nil" IIvnl' Iq.I·, , ., . , 1\ 111'1. IIi' 1'11'/111'111.11 .1'1 , . . . • . 1'1
/1'1111.'11'1 'iltli '/I/'t. /)/(.~I'
, .. ,.1'1.
A HI'L III'
1'II'III1~ltI,H .1'1 , ••• ,.1'1
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III' 11.11 illl.l~gml domain R i" called a III' N ovcl' iLH Huhl'ing I if it isa separating lildel of qllotients of R over the field of
(:{D.I'i) 'l'IIIGoIUoJM. '"e{ L lye a Junction field over a field K, and let p 11/: the characteristic of K. (1) If t = trans. degK L, then length L :vel' (fl/K) ~ t and the equality holds if and only if L has a separating transcendence base over K. (2) Assume that p -F 0 and that a is an d(:ment of L such that a P E K, a ~ K. Then there is a derivation D of I\(a) over K such that Da = 1 and :ver(K(a)/K) = K(a) ·D. Proof. We prove (2) first. The existence of D is obvious by (39.:)). Hince J(XP - .aP) = 0, we see that length :ver(K(a)/K) = 1 by (;{9.4), which proves (2). Next we prove a particular ease of (1) as follows: (39.6) With K and L as above, L is separably algebraic over K iJ and only if :ver(L/K) = O. Proof. If L is separable over K, then the zero derivation is uniquely extended to a derivation of L by (39.4), which shows that :ver(L/K) = O. Conversely, assume that :ver(L/K) = O. Let Xl, . . . ,Xn be dements of L which generate L. We prove that L is separable over K by induction on n. Set K' = K(Xl)' Since :ver(L/K') C :ver(L/K), we see that :ver(L/K') = 0, whence L is separable over K' by induction. Therefore (39.4) and :ver(L/K) = 0 imply that :ver(K'/K) = O. (Thus we have reduced the problem to the case where n = 1.) If :(;1 is transcendental over K, then there is a derivation a/ aXI which i:-; not zero, whence Xl is algebraic over K. If Xl is not separable over K, then K is of characteristic p -F 0 and K (xi) -F K', whenee 'IJer(K'/K) :::2 :ver(K'/K(x{')), which is not zcro by (2) in (39.1'», proved above, and there i" a eontradietion. Thus Xl i" "cparable, whence: L is ,,('parable over K. Thw; (39.6) is proved. Now we proceed with the proof of (39 ..5). AS8ume fir"t that L has a separating transcendence base Zl , .•. ,Zt over K. Since L is separable over K(z) = K(ZI, ... ,Zt), the partial derivations iJ/iJz i are uniquely extended to derivations Di of L by (39.4). Assume that ciD, = 0 (Ci E L). Then 0 = (L, ciDi)zj = cjDjz j = Cj, which
L
IfI()
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PI'OI'('H 111/1.1, 1.111' /), 11.1'(' lill('II,"'.\' illdl'p('lIdl'III,. 1.(,(, /) Iii' lUI n!'ilil.l'lu·.y dw·il'al.ioil or "0\'(,1' /\ 11,11" H(,(' 1/, /);:, . '1'111'11 (/) 'I/,,/), );:j . 0
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1\1'1'0 dnl·iva·0 11I'I'HIIHn /; iH f.,'pnl'ahle OV(,I' K(z), whieh prove::; thai, ';Dcr(/.,/J() iH !!;('I\('I·al.nd by (,}W f).;, whene(\ length'tJer(L/K) = t. Now we comiiuer the !!;(\llnml (,at-:(,. We usc thn 1'01'
notation in (39.4). Reordering thefi and Xj if lleces:::;ury, we may aHsume that the determinant I af;j ax j I for 1 :s; i :::; r, 1 :::; j :::; r is noi. zero. If we consider a similar Jacobian matrix for Lover K (Xr+l , ... , ... ,xn ) with generating element~ Xl , . , . , Xr , then the rank of the new Jacobian matrix is r, which implies that :fJer(L/K(xr+1 , ... , ... , x,,)) = 0.. Therefore, we see by (39.6) that L is separably algebraic over K(Xr+l' ... ,xn ), whence t :::; n ~ r = length'tJer(L/K). If t = n - r, then it is obvious that Xr+l, •.. , Xn is a separating transcendence base of Lover K. Thus (39.5) is proved completely. (39.7) If Lis a function field over afield K, if L' is,afield containing K, and if L' is purely inseparable over K, then L ® K L' is a local ring of altitude zero. Proof. Since L is a ring of finitely generated type over K, L ® L' is of finitely generated type over L', which implies that L ® L' is Noetherian. Let p be the characteristic of K. If p = 0., then the assertion is obvious because L' = K in this case. Therefore we assume that p ~ 0.. Let f = L ai ® bi (ai E L, bi E L') be a non-unit of L ® L'. Let q be a power of p such that bi E K for every i. Then = L a~ ® b'f = L aibi E L. Since f is a non-unit, f q is a non-unit, whence.r = 0.. Thus every non-unit of L ® L' is nilpotent, from which the assertion follows. (39.8) A field L is separably generated over its subfield K if and only if every finitely generated sub extension of Lover K is a separably generated extension of K. Proof. The if part is obvious by the definition, while the only ~r part is easy because ® K L is exaet (for K is a field). By virtue of the above lemma, we consider finitely generated extensions:
r
(39.9) Tm]oREM. Assume that L is a function field over its subfield K. Then the following three conditions are equivalent to each other: (1) L is separably generated over K. (2) L has a separating transcendence base over K. (3) If a field L' contains K and if every element of L' which is
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Prool'. Wn :-:liow nl. li!',YC CliaL (I) ill\pli(~:-: (2). Thi::; i::; obvious if K is or ('lial'H(~(,(~"isCi(: 0, wll(~I\e(~ we assume that K is of characteristic 11 ,;L' O. Ld, :1', , ... , :1:", be elements of L which generate L, and we pl'Ovn !.Iw a:-::-:m"i,ioll by induction on n. Let ~ be the kernel of the IIOIHomol'phi::;m ¢ over K from the polynomial ring K[X I , " . -, XnJ onto K[xJ , ... ,xn ] such that ¢(X i ) = Xi, and let i -F 0 be a polylIomial in ~ which is of the smallest degree among those in p (if ~ = 0, Chen the assertion is obvious and we assume that ~ -F 0). Since l1p is a field, hence is the field of quotients of K[XI , ... , Xn] ® 1.1 ® K f(llp '" Kl1p[X l , ... , Xnl/~Kllp[Xl , ... ,Xn], it follows that l1p ~Kllp[Xl , ... ,Xn ] is a prime ideal, and i is irreducible over K . It I'ollows that there is one i such that ai/aX i -F O. We may assume that ai/aXn -F O. Then we see that Xn is separable over K(XI, ... , Xn-l). Hince K (Xl, . . . , Xn-l) is separably generated by (39.8), it has a :-:eparating transcendence base, which becomes a separating transcendence base of Lover K by the separability of x n . Thus we have proved that (1) implies (2). Next we prove that (2) implies (3). Assume that (2) is true but (3) is not true. We want to show a (:ontradiction. Let ¢ be a homomorphism over L' from L ® L' into tt field such that trans. deg£, ¢(L ® L') = trans. degK L. Let Xl, . . . , ... ,Xt be a separating transcendence base of Lover K and let Yi = ¢(Xi). Then Yl, ... , Yt are algebraically independent over L'. Among non-zero elements ai ® bi (ai E L, bi E L') of ¢-I(O) (which is not zero by our assumption), let i = ® be one which has the smallest number of terms. We may assume that = 1. Let K* be the field generated by the Then, by our assumption, K* is not separably algebraic over K, whence there is a non-zero derivation D of K* over K by (39.6). Since the Yi are algebraically independent over K*, D can be extended to a derivation of K*(Yl, ... ,Yt) so that DYi = 0 for every i by virtue of (39.3). Since every element of ¢(L) is separably algebraic over K (YI , ... , ... , Yt), hence over K*(Yl, ... , Yt), it follows that the extended D can be extended uniquely to a derivation of K*(¢(L) ) by (39.4). Since D(K(Yl' ... , Yt)) = 0 and since ¢(L) is separably algebraic over K(Yl, ... ,Yt), D(¢(L)) = 0 by (39.4). Thus the extended derivation D is a derivation over ¢(L). Since L:¢(a*)b* = 0, we have o = D( L: ¢(a;)b;) = L: ¢(a;)(Db;), which implies that L: a; ®
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(39.10) THEOREM. If L is a fnnction field Oll('r (J, .field 1\. '1'11,1'11, 1./1r'1'I' is a finite. purely inseparable extension K' of K stu;h that /,(1(') ,i" separably generated over K'. Proof. Let K* be the smallest perfect field containing K. Then 1\* is purely inseparable over K, whence L ® K* is a local ring of altitu
Proof. Let p be the characteristic of I. If p = 0, then the assertion is nothing but a special case of the normalization theorem for finitely generated rings. Therefore we assume that p ~ 0. Let Xl , . . . ,Xt be a separating transcendence base of A over I and let Xt+l , . . . ,X n be such that A is generated by Xl, . . . , Xn over I. Let Xl , ., . , Xn be indeterminates and let cp be the homomorphism over I from I[X] = I[X I , ••• , Xnl onto A such that cp(X i ) = Xi. Let a be the kernel of cp and apply the normalization theorem for polynomial rings and the proof of (14.3). We see that there are elements Y I , . . . , Y n of I[X] and an element a of I such that: (1) I[X][I/ a] is integral over I[I/a][Y I , . . . , Y n ], (2) aI[X][I/a] I[I/a, Y I , . . . , Y,,] is generated by Y t + l , • • . , Y n , and (3) Y i = Xi + Fi for i = 1, ... , t with F; E 1r [X f+l , ... ,X:] where 1r is the prime integral domain. Set Zi = cp(Y i ) for i = 1, '" ,t. Then (1) implies that A[I/a] is integral over I[I/a, Zl, ... ,ztl, and (2) shows that the Zi are algebraically independent, as we have done in the proof of (14.4). What we have to prove here is that A is separably algebraic over I[zJ, ., . ,zd.
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IGXI')IWI:-l)')H, 1, Lot (N, "I) lin l\, 1I111'lIlld loealiLy over a ring 1 whidl iN nil,I,,'!' fiold 01' tt Dedekind dOllll\,i I) I\,lid nHHlIml~ that R is separably g(\I\())':i.1,nti IIVI'!' I, Prove that there is a Hopnmt.illg transcendence basis z, , ' " , z(, til' II: (IV('!' I such that n = m n l[z, , ' " , z,J is generated by m n I and It l-llIliHI'1, III' 1.11" Zi and such that R is of finite type over l[z, , ' " , ztln ' 2, Let K and L be fields of characteristic p ;;£ 0 such that K ~ L, l'rovt: that L is separably generated over K if and only if a p-base of K is a subsot of a suitable p-base of L, 3, Let (R, m) be a complete local ring which may not be Noetherian and let p be the characteristic of Rim, Assume that a local ring I is dominated by R, that pI is the maximal ideal of I, and that Rim is separably generated over IlpI, Prove that there is a coefficient ring of R which contains I, (Hint: Making use of Exercise 2, above, adapt the proof of the structure theorem of complete local rings,) I\,
40. Multiplicity oj a local ring When (R, rn) is a local ring, the multiplicity f.I(m) is called the multiplicity of R and is denoted by m(R), (40,1) THEOREM. Let ~ be a prime ideal of a local ring R, If height p = altitude R and if p is analytically unmmijied, then the multiplicity of Rp is not greater than that of R. Proof. Let 111 be the maximal ideal of R. If R/111 cont.aim; only a finite number of elements, then, taking a transcendental element x over R, we consider R(x) and R~(x), By the facts that R~(x) = R(xhR(x) , pR(x) is analytically unramified by (36,8), meR) = m(R(x)) and m(R~(x)) = m(R~), we may assume that R/m contains infinit.ely many clements, Let R* be the completion of R and let ~* be a minimal prime divisor of pR*, Then the t.heorem of t.ransition holds for Rp and R: * , whence m( R~) = m( R:.) by the analytical unramifiedness of ~, Therefore we may assume that R = R* (because m (R) = m (R* ) ). Let. 1)31, ." , I)3r be all the prime divisors 1)3 of zero such that depth 1)3 = altitude R, where we assume that. l)3.i k: ~ if and only if i ~ s, Then (2:3,5) implies that meR) = J.t(m) = L~ f.I( 111/l)3i) ,length R'lJi and that m(R~) = L~ J.t( ~Rp/Ij.LR~) ,length ~
+ depth
/i'111, . 'l'h(,I'I,I'III'(', il' WI' 1111o\V j,lJ(' l'II,lillil,.\' III' 0111' /I,HNI' I' j.jll II ill LlI(1 ('/Irl(' wlil'l'n U iN lUI illl,('glni dllllmill, LlII'11 \1'(1 PI'II\'(' Lil(' gl'II('I'II,1 I'/I.~('. ThllH WI: may aMNII 1111', 1'1I1'L!II'I'IIIOI'I', I.hal. N iN 11.11 illi.l'gl'll,1 d1l11l1l.ill. 1.('1. I Iii' a coefficient rill!!; 0[' N alld Id ./'1 , ... , :1'/, hl~ II, NYHl.nll1 01' p:I,l'nllll'l.nn-l III' R such that m(R) = J.teL: xJt,); Ow l~xiNLI~III',(~ or I rlillowH 1'1'0111 (31.1), and that of the Xi follows from (24.1). HeL 8 = 111./'" .. , , · .. , Xt]], n = L XiS and q = ~ s. We denote by [N: 8J OlP deg"('I' of extension of the field of quotients of R over that of S. Then m( H) J.t(nR) = J.ts(n; R) = [R:S]·J.t(n). Similarly, if we denote by T 1.111' complement of q in S, then m(R~) :s; J.t(qR~) :s; J.t(qR T ) = J.tsq(q8 q ; R T ) = [R:S]·J.t(qSq). Hence it suffices to show that J.t(n) ~ J.t(q8 q). Therefore, it suffices to prove the assertion in the case where R = 8. If R is regular, then Rq is regular by (28.3), whence J.t(qR q ) = 1 alld the assertion is true. Therefore we assume that I is not a field. Lid. Xl, ... , X t be indeterminates and consider l)t = I[[X I , ••. , Xtll. Let c:p be the homomorphism over I from ill onto R such that c:p(X i ) = Xi . The kernel f of c:p is a prime ideal of height 1, whence f is principal by (28.7); let J E l)t be such that f = fln. (40.2) below, implies tha.I. m(R) is the degree of f with respect to c:p-l (m) and that m(R~) is the degree of fwith respect to c:p-l(~)m
n
element oj m, then m(R/JR) is equal to the degree of J with respect to m. Proof. Let d be the degree of J with respect to m and let Xl , . .. , be a regular system of parameters of R. Then there is a homogeneous form h( Xl , ... ,Xr ) of degree dover R such that h( x) = f and h(X) ~ mR[X]. We may assume, in the same way as in the proof of (40.1), that R/m contains infinitely many elements. Then, considering a sufficiently general linear transformation of the Xi , we may assume that h(X) has a term uxf with unit u in R. ThenJ, X2, •• , , · .. ,Xr is a system of parameters of R and length R/ (fR xiR) = d, whence the multiplicity of the ideal generated by the (Xi modulo fR) is equal to d because R/JR is a Macaulay ring, and m(R/fR) :s; d. Let Y2, ... ,Yr be such that J.t(fR L yiR/JR) = m(R/fR), whence m(R/JR) = length R/(fR L YiR) d= J.t(fR LyiR) which is not less than d by (24.3) becauseJ E l11 • Thus we prove that m(R/JR) = d. We shall prove a generalization of a well known result in algebraic • •• ,X r
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1'1/,11"1'1"" 1'1 ",;1'0111('11'.1' II'I,il'l, 1111111'i'lrl 111111. III(' 111'1. eli' Ilill/1:11I1I1" pOillll1 Oilltil /l.1/1:I,III·/lil' 1'liril'I.I' i~1 II l'lttlll'ci ~II'I ill 1,"l'i,I" i Ittjlolog,I', III Ill'cil'I' III 1'01'1111111111' 1111 ielC'al Ilil'OI'I'lil' 11:1'1I1'1'11,liv"lI,ioll or 1.111' Id'OI'I' l'I'fHIII" \\'1' illl,mellll'I' I \Vel I~Ollllil.i/)II,,, 1111 NOI'I.I'I'l'ill.lI l'illgH: (*) I I)('illg ii, NI)('I.II1'I'iall ill i.c'g rll. I dOlllll.ill, lVI' l'OIiHicil'l' 1.111' I'olilli I,ioll I. Ii: I. I. : 1.111'1"1' i:-: all iell':I,1 II whil'll iN clirTI'I'I~II1. 1'1'0111 ;/'1'1'0 II.lIci :-:111'11 1.1111.1. if a pl'iIliC~ idc'al p or I 111)(':-: 1101. 1',011 !.:!.i II n L111~1I I p i:-: II. rq!;1I111.1" I()I~II.I rillg. (**) N hpill!!; II. Nol'I,hnrill,1I rill!!;, WI~ (:oll:-;idl~r the c,olldiLioll 1.11:1.1,: i,lwre me idnal:-: 13, 13[ , ... , 13" , .. , of It, :-IIIC,h LilaI'., fot, a prime icll~:l.1 p of R, (I) Rp iH lIot reglllHl' if alld only if p c~ontaill:-: 13 and (:.n N iN of multiplic:ity !!;l'eatel' than a given natural number n if and oilly" if p contains 13" . Then our generalization is formulated as follows: (40.3) THEOREM. Assume that I is a pseUdo-geometric ring 8uch that if p is a prime ideal of I and if I' is a finite purely insepara/J[o integral extension of 1/ 'p, then I' satisfies the condition introduced in (*) above. Then every ring R which is of finitely generated type over I satisjies the condition introduced in (**) above. In order to prove the above assertion, we first prove the following lemma. ( 40.4) Let R be a Noetherian ring and let M be the set of prime ideals of R. For a subset N of M, there is an ideal a of R such that pEN (p EM) is equivalent to a ~ 'p if and only if N satisfies the following two conditions: (1) if 'p E N, then every prime ideal which contains p ,is in Nand (2) if a prime ideal 'p is not in N, then there is an ideal b such that b %p and such that if 'p C q E N then q contains b. Proof. Assume first the existence of a. Then the validity of (1) is obvious. As for (2), it is easy to see that a satisfies the requirement for b. Thus we settle the only if part. Assume that N satisfies the conditions (1) and (2). Let c be the intersection of all the prime ideals in N. It suffices to show that c satisfies the requirement for a, and for that purpose, it is sufficient to show that if p E M is not in N, then p does not contain c. Assume the cont.rary, and let q be a minimal prime divisor of c which is contained ill p. Since 'p ~ N, and since q C p, it follows by (1) that q (I N. Let the other minimal prime divisors of c be ql , " . , qr and Ic:!, b hl~ an ideal, as in (2), applied to q. Then it is obvious that all r ( N I'olll.ain (6 + q) ql nqr and therefore c ;;;;;2 (6 + q) q, qr and qRq = cR q =
n n .. , n
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s/l.I':,~/il'g tll/' l'I//I,r/'I:l'I:lInin (**) a/ili/W, 1'1'001'. LnL N, N I , , " , N i , . " l)(~ Ul(~ :-:cL:-: of willie ideal:-: p of Ii' :-HI,'II Lllal,: (1) U~ i:-: noL 1'(Wtial' if and Oldy if P i:-: in N and (~) '111,( NI» > , : if and only if p iH ill N i . II, follow:-: from (2~.:n and (40.1 )
'1'/11'/1. H
IhaL Ihe :-:d:-: Nand Ni :-:ai,isfy tho eondition (1) in (40.4), hmH,(1 il. HIllJi("(':-: Co :-:how that they satisfy the condition (2) in (40.4). AsslI 1111 , I.hal, a pr'ime ideal,p is not in N. Then R~ is regular. Let Xl , . . . , .Tr I II' (,Jnllwld~:-: of p ::lueh that their natural images in Rp form a reglli:l.r :-:y:-:Lem of parameter::; of R~ . Then there is an element a of R which i:-: IIOi, ill P sneh that, with 8 = I an I n = I, 2, ... }, xiRs = pHs. Lnt a be the ideal of R such that pea and such that alp satsfies tI\I' (,(Hldition for a in (*) above applied to Rip. Set 6 = aa. If a prill\!' ideal q is such that q ~ p and such that 6 q, then Rq is a ring 01' qlLOtients of Rs and furthermore Rq/pRq is regular, which implies th:'" H q is regular, by our choice of a and by (9.11). Therefore we see till 1 existence of 13 by (40.4). Assume that a prime ideal p is not in N,. Then m(R~) ::; i. Let Xl, ..• , Xr be a system of parameters of R~ sueh that m(R~) = J.I.( xiRp) and such that every Xi is in p. Then ther(' is an element a of R which is not in p such that, with 8 = (ani n = I, 2, ... }, xiRs is a primary ideal belonging to pRs . Set 6 = aa with an ideal a such that pea and such that alp satisfies the condition fol' a in (*) applied to RIp. If a prime ideal q contains p and if 6 q, then Rq is a ring of quotients of Rs and furthermore Rq/pRq is regular. Let Xr+I, •.• , Xs be elements of q such that their residue classes modulo p form a regular system of parameters of Rq/pRq . Then, it is obviouH that the natural images x~, ... , x~ of Xl, . . . , Xs in Rq form a system of parameters of Rq. It follows from the associativity formula (24.7) that J.I.( x~Rq) = J.I.( x;R qmodulo pR q ) · J.I.( L~ x:R~) = fJ,( L~ x:R~), which is meR\»~ by our choice of the Xi. Therefore m(Rq) = J.I.(qR q) ::; J.I.( x:Rq) = m(R~) ::; i. Therefore we see the existence of l3 i by virtue of (40.4), which completes the proof. Now we want to prove (40.3). Sinee t.he validity of the condition stated in (**) for an R is carried over any ring of quotients of R, it. suffices to prove the case where R is finitely generated over I. It is
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1',1' \'ili.III' 01' (IOJ,) III H"II\V I.lial. il' p iH II. pl'illl(' id('nl 01' U, 111('11 N/p 1",liHlil'~i llil' I'ollllil.ioll HIII.I.I'd ill (*), linlw(', :tHHllIllillg LliaC 1 iH nil ild,I'/.!;1'1I.1 dOlllnill Hilli 1.1,11.1. N iH all a(JiIl(~ I'illg over 1, we have only /'0 Hliow 1.1i1l.1. U HaLiHlioH L1tn (:olldiLioll Htuted in (*). There is a finite PIII'I'i.Y inHnpamblc integral extension I' of I such that I'[R] is separ"'illy gCllnmi.nd over I' by (39.10). We prove the assertion by induc/,ion Oil I.he degree of the extension of the field of quotients K' of I' OV(:I' the field of quotients K of I. If [K':K] = 1, then R is separably g(:rwrated over I, whence there is a separating transcendence base Z" ... , Zt of R over I and an element a (~O) of I such that R[I/a] iH integral over I[I/a, Zl, ... ,Zt]. Let c be an element of R which generates the field of quotients L of Rover K (Zl , ... ,Zt). Let f( x) bo the irreducible monic polynomial over K(Zl , ... ,Zt) which has c HH a root. Crmsidering elements of type cs (s E I[Zl , ... ,ztl) instead of c if necessary, we may assume that f(x) is a polynomial over [[Zl , .. , ,Zt]. Let d be the discriminant of f( x). Let 6 be such an ideal as a described in (*) applied to I. Then we sec that adb satisfies the requirement for a in (*) by virtue of (38.9) and (14.8). Assume now that [K':K] > 1. Then the characteristic p of K is different from zero. Let c be an element of I' which is not in K and such that cP E I. If eEL, then c = y/x (x,y E R) and we may replace R with R[I/x]; then we have c E R. Then, applying our induction to Rover I[c], we prove this case. Thus we assume that c ~ L. By induction applied to R[c] over I[c], we see that there is an ideal a of R[c] as described in (*) applied to R[c]. Let a' be the ideal a R. Noting that a' contains every pth power of elements of cr, we see that a' satisfies the requirement for a in (*) applied to R by virtue of (38.5). Thus the proof of (40.3) is complete. As is obvious, if R~ is a regular local ring, then m(R~) = 1, but not conversely. If we replace 13, 131, ... , 13", .. , in (40.3) by their radicals, then we see the inclusion 13 C 13 1 C ... C 13" C .... Concerning the relationship between 13 and 13 1 , we add the following result: Hllllil'il'liI
n
(40.6) THEOREM. A local ring (R, m) is regular if and only if R satisfies the two conditions that R is unmixed and that meR) = 1. Proof. The only if part is obvious, and we prove the converse. Assume that R is unmixed and that meR) = 1. Let R* be the completion of R. Then R* is unmixed and m(R*) = meR) = 1. It follows from (23.5) that 1 = j.t(mR*) = L~. j.t(mR*/~*) . length R:*,
wlin!'n p'" I'II1IH (lVI'!' all Willll' divirll))'r1 III' 111'1'0 (HIWh 1.!tIl.L dl'pl.11 p"" H*). 'l'11I~!'I'I'OI'I~ WI: :-l1:1~ 1.111101, h~* 11111;;1, hi' :\.11 ilil'I'/.!;I'lI.1 dOIlIlI.ill Since it suffices to :,;how LbaL U* iH I'(~glllar (bcI'all"n or Lhn 1'lIllaliLy 2 m/m2 = mR*/m R*), we may aHHllIrlO LhaL It, iH (·clllpln!,l:. II' N/III contains only a finite number of clement:,;, then Lake a tl':l.IIHI·1:11I 11:11 1.:1.1 element x over R and consider R(x). Since R is a homomorphil', im:lgI' of a Macaulay ring by our assumption, the same if) tnw for N (.1' ), whence R( x) is unmixed by (34.9). Then, since length ml 1Il~ length mR(x)/m2R(x) by (18.9) (cf. Exercise 2 in §18), we sec thai R is regular if and only if R(x) is regular. Furthermore, since thl' theorem of transition holds for Rand R(x), we see that m(R(x)) = I. Therefore R may be replaced by R (x), whence, considering the COlli pletion of R(x), we may assume that Rim contains infinitely many elements. We now prove the assertion by induction on altitude R. I I' altitude R = 0, then R is a field and the regularity is obvious. I I' altitude R = 1, then there is a superficial element a of m and p,(aR) = p,(m) = 1 by (24.1), while by (24.2) or by the fact that R is a Macaulay ring in this case, we see that p,( aR) = length RI aN, whence length RI aR = 1 and m = aR. Therefore R is a regular local ring. Assume that altitude R = r > 1. Let a be a superficial element of m. Then 1 = p,(m) = p,(mlaR) by (24.2), which implies thai, m(RlaR) = 1. Since every minimal prime divisor p of aR is of height ] by (9.2) and since the first chain condition is satisfied by every complete local integral domain by (34.4), we see that depth p = r 1, whence, applying (23.5) to mlaR, we see that aR has only one minimal prime divisor p, that aR~ = pR~ and that m(Rlp) = 1. By induction, we see that Rip is regular, whence Rip is normal by (25.14). Therefore we see that aR = p (and that R is normal) by the lemma of Hironaka (36.10). Therefore RlaR is regular, whence R is regular by (9.11), which completes the proof. altit\ldl~
41. Purity oj branch loci The purity of branch loci at a simple point, which is known in algebraic geometry, can be formulated in a very general form as follows: (41.1) THEOREM. Let CR, m) be a regular local ring and let (P, n) be a normal local ring which dominates R and which is a ring of quotients of a finite separable integral extension of R. Assume that every
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IlI'gili wil,,, HOIIII' I'I'HIIII,H ill (;aloiH i.I\('oI'Y. Lnl. U 11(\ II, lIol'llIal I'illg :1.1)(1 InC H' !In a HI'pamhlc (;aloiH nxLnllHioll or H. wil.h (;aloiH gl'OIiP a. Lui, p' bl~ a pI'ime ideal in U'. Then the Hid, lJ of 0' ill U :meh I.lml. ~t"T = p' iH a Hubgroup of G. This H is called the splitting grtYup of p'. The Hct 1 of 0' in G such that aU - a E p' for every a E R' is a ~:iUbgroup of H. This I is called the inertia group of p'. The invariant subrings Sand T of H and I are called the splitting and inertia rings of p', respectively.
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(41.2) THEOREM. With the above notation, set p = p' R, q = s, q' = p' T. Then: (1) p' is the unique prime ideal of R' that lies over q, (2) Sq/qSq = R~/pR~, (3) q is unramified over R, (4) q' is unramified over R, and (5) I is a normal subgroup of H. Furthermore, (6) H/I is the Galois group of R'p,/p'R'p' and of Tdq'T q, over p'
n
Sq/qSq
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Proof. Considering R~ , S(R-~) , etc., we may assume that p is the unique maximal ideal of R. The splitting group of p' over S is the Galois group H, which shows that p' is the unique prime ideal of R' which lies over q. In order to prove (2), it suffices to show that if a E S, then there is an element b E R such that a - b E q, hence, considering an almost finite Galois extension containing a, we may assume that G is finite. Let a be the intersection of maximal ideals of S other than q. Then, a + q = S because of the finiteness of G, whence there is an element a' of a such that a - a' E q. If a' E q, then a E q and b can be O. Therefore we assume that a' ~ q. Let 0'1, '" ,O'n (0'1 = 1) be such that G is the disjoint union of the Hl'i. Then pfUi are all distinct from each other, whence a,Ui E p' except for i = 1, whence if we set b = a,Ui then b E R and a' - b E p' n S = q, which proves (2). Since (2) is settled, it is sufficient to show that pSq = qSq in order to prove (3), henee, considering an arbitrary clement of q, we ean reduce to the almost finite case, as in (2), and we assume again that G is finite. Let e be an element of q whieh is not in any other maximal ideal of S. Then, with the 0' i as above, CUi is not in p' for i -F 1, whence the product e' = cu • ... eUn is not in p'. Since ee' E Rand e' is integral over R, we see that e' E S. Since e' ~ p', we see that e' ~ q, whence e E pSq because ee' E p. Let e* be an arbitrary element of q a (= Jacobson radical of S).
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'i'hen c c* Ha(,jHfil~H ChI', Hallin 1',olldiCioll 1'01' (~ II.bOVI~, wlll'III'I' (' I ('+ I ~Rq and therefore c* ( ~Uq. Nilll~n (q Il)U q qU", wn HI'I' 1.11111. ~Rq = qR q , which proves (3). (5) is eaHily He(~II. Wn HI~(', by ill(' definitions of unramifiedness and Galois groups, that if (II) alld (n I are proved for the almost finite case, then the geneml l~aHC I·o!lowcl. Thus we may assume again that G is finite. In order 1,0 prove CII, it suffices to show that q'is unramified over S, and we may aHHlllIII' that R = S. Since I is the Galois group of R' over I, R;, I~' R~, i~~ purely inseparable over T q , I q' T q , , hence we may assume that il' T (for both (4) and (6)). Let 1, a, .,. , a m - l be such that tlll,il' residue classes modulo q' form a linearly independent base of a maxi mal simple sub extension L of Tlq' over Slq and set T' = 2:;r:- l Ha'. Since I = II} by our assumption, every element ( 7'" 1) of G = II induces a non-trivial automorphism of Tlq' over S/q, whence m ~' order of G. Since 1, a, ... ,am - l must be linearly independent OVI'I' R, we see that m ::; order of G, whence m = order of G. Furthermore, we see at the same time that L must be separable over Slq, whence Tlq'must be separable over Slq, and Tlq' = L, whieh proves (6). Furthermore, we see that the degree of irreducible moni l' polynomial f( x) over R which has a as a root is equal to m, whenc(\ T' is a ring. Since Tlq' is separable over Slq,f(x) modulo q is separable, whence the discriminant d of f(x) is a unit in R, whence T' iN a normal ring, and therefore T' = T. Since f(x) is irreducible modulo q, we see that qT is prime, which proves (4). Thus the proof of (41.2) is completed. Let al , ... ,an be linearly independent elements of R' over R such that L Rai is a ring. Let G* be the subgroup of G which is the sei. of (J" such that (J"( ai) = ai for every i, and let (J"l, . , • , (J" m ((J"l = 1) be such that G is the disjoint union of the G*(J"i' Since L Rai is free, we have m = n. Then the square of the determinant I ati I is called the discriminant of the clements al , ... ,an. Note that if the ai arc n l 2 1, a, a , ... ,a - for some a, then the discriminant of these elements is t.he discriminant of the irreducible monic polynomial which has a as a root. If A is a matrix of linear transformation over R acting OIl Rai and if A ( Rai) is a ring which has the same field of quotients as L Rai, then we see obviously that (discriminant of A(al), ... ,A(an )) = (det A)2 (discriminant of al, .,. ,an). In particular, the discriminant of a linearly independent base of L Rai is unique up to units, whence it is called the discriminant of L Rai .
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whieh limi nv(\!' tl. Thl'llJ'(I/,) iH 1101. ill ~\*, Nill('.I' ~\* iN II.l'hil.l'II.l',V, \VI' ill'I' that any conjugate of f'(a) it-; ]loL ill p* h'y (IO,I~), WIIi'III~I' I.hl' di:1 criminant d of f(x) is not in p*, whcnec 1I0C ill p, b'y (10,17), (41.5) THEOREM, Let R' be a finite separable inte(!l'o'/' (:,f/,O/l,"':OI/. I~r /I Krull ring R, If R' is a free R-module and if every prime 1;r/eal /(( /i.1'iylti 1 in R' is unramified over R, then every prime ideal p' !if H' is 'unmnujil'll over R, Proof. Set p = pi R. Then, considering R~ and R~ R-~) , we mH.,V assume that p is the unique maximal ideal of R. If height p = I, then the assertion is obvious, and we assume that height p > 1. Ld d be the discriminant of RI. Assume that d is a non-unit. Then th(~I'I' is a prime ideal q of height 1 in R which contains d. Let a be sueh ilil element as given by (41.4) applied to q. Then the discriminant d', of the irreducible monic polynomial f( x) such that f( a) = 0, is not ill q, As was remarked before (41.3), d' is the discriminant of R[aJ and iN in dR, which is a contradiction, whence d is a unit in R, which prOV(~H the assertion by virtue of (41.3). ( 41.6) Let (R, m) be a regular local ring with a regular system /~r parameter s Xl , . , , ,X r and let (V, n) be a valuation ring which dom'inates R. Set a = xlR xsR (2 ::::; s ::::; r), If XdXl , . , , ,xS/rl modulo n are algebraically independent over R/m and if R* is the dilatation of R by the ideal a with respect to V, then R is a subspace of R*. Proof. Set RI = R[xdxl' . , . ,Xs/XlJ. Then xlR' is a prime ideal and xlR I = aR'. Set R" = R:w . Then R" is a Noetherian valuation ring. Obviously R* is dominated by R" and RI! is a quadratic dilatation of both Rand R*. For an element f of R, f E mn if and only if f E x~ R", and, if we denote by m* the maximal ideal of R*, then fol' an element f of R*, f E m*n if and only if f E x~ RI!. Therefore m*n n R = m n and we complete the proof. We need one more preliminary:
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(41.7) THEOREM. Let (R, m) be a complete regular local ring with a regular system of parameters Xl , ' •. ,Xr with r 2: 3, Let P be a normal local ring which is a finite separable integral extension of R. Then there is a finite number of elements al , . .. ,as of R such that, letting y be a transcendental element over R, and c an element of R, the local ring Pc = P(Y)/(X3Y - Xl - CX2)P(y) is analytically irreducible, whenever c is such that II (c - ai) ~ m. Proof. Let b be an element of P such that P is the derived normal
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I'ill~'; or 11'1111111111 11'1/1 .\' I III' III(' II'I'I'IIIII'illll' 111111111' pol,I'llolllinl 11\'1'1' Ii' 1<11i'1i 111111. ,f'( 'I) (), ~I'I, II', It'l //1, 1,1':1/1 ,/'1 1',/", 1h'1 /11. 'l'hl'll il, iH ohl'iolJ,'i 11:111. I'" iH /.1:1'111'1':1,1.1'" 11'y N" II.lld I', ",1"'111'1' I'" iN H (illil,11 ill 1.1'11:1'11.1 1'\I.I'IIHioll oi' U, alld haH 1.111' HII.IIII~ 1(IIOIil'III. (il'ld IIH "',1111. TI\(~rdol'I' 1.11(' ('OIIl\lll'l.ioIlH oi' I'"~ 1\,lId /(',·11" 1mI'I' I.hn H:t.Il\I' l.oLal qllol.il'lll, rill!!:, WIt('III'(~ 1.1t!: allalyl,il~ irrndlll:iliilil,y (Ii' I'" iH nqllivalnllL 1,0 1.1\11.1, of /[',,1111. '1'(wl'l'forn iL iH Hldli(:icIlL (,() Hhow LlmL .r(X) iH il'rcdlll:illi!: 01'1'1' Lhe eoml'leLioll of No: if II ((; - (li) q 11l, AHHlIllIC Lhe (~oIlLI"II,I'y, llamely, aHHllm(~ Lhat there are in(iniLely many c, Hay CI , ('~ , whose residue da"N(~N modulo 11l arc difTerent from each ()t1\(~I', Hili'll l.hatf(X) is reducible over Before proceeding with the proof, we make some remnrkH. (1'111" a C E R, we set Ue = (Xl + cX2)/xs . For two c, d E R such Lhal, I: d ~ 11l, set R e.d = R[ue , 'Udlm' with m' = mR['Ue , 'Udl; note iJlaL /(',.,'/ is a Jilatation of both Re and Rd of the type in (41.6) (and 1.lwl·(,fol·(' m' is a prime ideal). Therefore Re and Rd are subspaces of Re,d . AHsume that e E R is such that (e - c) (e - d) ~ m. Then we (Iall consider R e •e and R d •e • Set a = (d - e)/(d - c), (3 = (e - c) --;(d - c). Then we have 'U e = at~e + (3'Ud. Therefore 'U e E Re,d and, 'U c , 'U e modulo mRe,d are algebraically independent over Rim. Thus Re,d dominates Re,e' Symmetrically, Re,e dominates Re,d and we have Re,e = Re,d' Therefore Re,d = Rd,e = Re,e. We apply this fact to ReI' RC2 , ... and we have that the Rei are subspaces of R" = R Cl ,e2 . Let the completions of R", Rc; be R"*, Rc:, respectively. Then Rc~ ::;
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R"*. Therefore we can consider factorization of f( X) in the algebraic closure of R"*. Since there is only a finite number of ways in which the polynomial f( X) splits into two monic factors, there are at least three mutually distinct elements among the Ci , say d1 , dz , d s , such that f(X) has the same factorization f(X) = g(X)h(X) over all R:i . For simplicity of notation, we denote 'Udi by 'Ui. Let Q be a complete set of representatives of the residue class field of R. Q['Ul , 'U21 denotes the set of all polynomials in 'UI , 'U2 with coefficients in Q. Q( 'Ul , 'U2) denotes the set of all FIG such that F, G E Q['UI , U2], whose residue classes modulo m have no common factor, and such that the coefficient of the lexicographical highest term in G is 1. Then every element of R"* is uniquely expressed as a power series in Xs , '" ,Xr with coefficient in Q(UI ,U2)' Thus we may write (formally)R"* = Rd! ,d2 = Q( UI, U2) [[xs, ... ,xrJl. Q( UI) and Q( U2) being defined similarly,
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"'" a n2 "' nr ( U2X3 )"2 Xa" n" · h I. I In 1'(1,' I' · SInce a . IS .In R*d1' a = L..., " Xrn r WIt ·· t . Q( ) H ""( ""n3 £) n'l " fJClen -s In Ul. ence a = L..., L...,£~O ai(n,_i),q- .. n,U2 Xa' '" .r,' From this expression, we derive an expression for a as a pow(~r :-wrjl'I' in Xa , . . . ,X r with coefficients in Q( Ul, U2). By the uniqueli('N,c; III the expression, we see inductively on n3 that ans"'n, = F'c,,)/U,oI" G(n) E Q[ud and F(n) is a polynomial in U2 of degree at most n3 wil.h ('11 efficients in Q[u!l. Considering R:2 , we see that G(n) E Q[1/,21 :111,1 F(n) is a polynomial in U! of degree at most no with coeffic:iellLM ii, (J[u2l. Thus G(n) = 1 and an""'nr is a polynomial in u! and 1./,2 IV iI" coefficients in Q and its degree on each u, is at most n3. Set (V (d3 - d2 )/(d a - d1 ) and (J = (d 2 - d!)/(d 3 - d1 ). Then a, IJ :\1'1' units in Rand U2 = au! (JU3. If we substitute for U2 the exp"I'I' sion du! CU" in an""'n, , we obtain a polynomial a!"".nr in 1£, nlld 1./,3 with coefficients in R; we can choose d3 so that the degree of a;'"."", in Ul is equal to the total degree d(na, ... ,nr ) of an"".", for a giv('11 . * n,X"n3 .. , Xrn,'IS th en a POW,'I ' an"."", . Th e expreSSIOn a = "" L..., a"3". series expansion of a with coefficients in R[u!, ual. Assume LlI:1I there is a d(n3, ... ,nr ) which is great.er than n3 , and let am3 ".,,,,,. \)(' onc which has the lexicographically smallest suffix ma, ... , mr am (I IIf', those a m3 "'m, such that d(ma, ... , mr) > m3' We choose da so thlll a!3."m, is of degree d(ma, ... , mr) in U!. Let the expression of (J, i,l Q(UI ,U3) [[X3 , '" , xr ]] be a: 3 .. • n ,x;3 '" x;'. Then, :111 1 ,d3 = is obvious, each a: 3"'n, is the coefficient of X;3 ••• x;' in the re-I'~ pression of LSi
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UII' 1"IIIIIIn, or l\ndl.AJlIIIIIHya.
Now w(' :t.l'n to Pl'ovn CII.I ). ( ~ollKidnring pail'S (il, P) of lo(~al I·illg" ('11.1), W(' "tl,y ill till' 1)]'I'''(~llL proof that (R, P) is eqt~ivalelll III W', I)') if it; bold" ihat P iK unmmified over R if and only if 1'/ iH 1IIII'amili(~d over R' (henee, we see, when (41.1) is proved, thai. all pni I'" are equivalent to each other). The first step of our proof is to Hllow that: (*) For any given pair (R, P), there is an equivalent pair (R*, P**) 81/.ch that R* is the completion of R. We denote in general by c an element of P which is integral over N and which generates the field of quotients L of P over the field of qllotients K of R, and by f( x; c) the irreducible monic polynomial in nil indeterminate x over R which has c as a root. Furthermore, we (knote by gi(X; c) (i = 1, .. , ,n(c)) the irreducible monic fadors of .f(x; c) over the completion R* of R. Let P* be the completion of fI and let qi , .,. , q~ be the prime divisors of zero in P*. Let P' be Lite integral closure of R in P. Then, since P is separable over R, P' ii, a finite R-module by (10.16), whence the completion P'* of P' has Lile tmme total quotient ring as R*[c], whence P'* has no nilpotent dements exeept zero and the prime divisors of zero in P'* corresponds ill a one to one way to gi(X; c). Since P is a ring of quotients of P' with respect to a maximal ideal, P* is a direct summand of r*. Therefore, we see that 0 = qi n ... n q~ and, after a suitable renum11I',ring of the gi, gi(C;C) E qi for i ~ m, and gj(c; c) EE q; if j 7"" i. We shall show that m = 1. Assume for a moment that m > 1. Set (It = q; + (nj,,<'i qj) and a* = Assume that there is a prime ideal ~* of P* containing 11* such that height (~* n P) ~ 1. Since ~* (lontains at least two of the qi ,f(x; c) modulo ~* has a multiple root for any possible c, whence, if f' (x; c) denotes the derivative of f( x; c), Lhen f'(c; c) E ~* for any c. Thus 1'(c; c) E ~* P, and ~* n Pis ramified over R by (38.6), which is a contradiction. Thus there is no l;uch ~*. Let d be the discriminant of f(x; c) for a fixed c, and let S be the set of elements s of P such that dP:sP = dP. Since P is normal, every prime divisor of dP is of height 1. Therefore the nonoxistence of ~* above shows that S meets every prime ideal of P* II" ill
nat .
n
I(H\
(j1
(~()lIl.aillillg n*, 11('11('.(' I',~ ('OIILniIIH nil idl'lIll)(ti,(,IIL 1,1('III('liI, I' ",IIiI'11 1:1 not the iUellLiLy, Hill(',n c iH illl.('gl'ni OV('j' U'l<, wn lilw(! til' l I d ' 1),1' (10.15). Sincc e ( 1).~ , there iH all ninIlI(!lIL s or /'I. HI II ,.I I I.hal. I',~ 1 1"'. Therefore we see that e E P* by (;n.I), which iH a ('(,liI,l'l\,didioll Ill' cause P* is a local ring. Thus m = 1, and P* iH all ill(.q!;nd dOllmil1. Let P** be the derived normal ring of P*. AHHum(, t.hal. I.h(!/"(! i,; II prime ideal ~* of height 1 in P** which is ramifi.ed over U*. 'l'hl'll f'(c; c) E ~* for any c by (38.6), and ~* P is ramified over Il, wliil,h is a contradiction. Thus R*, P** satisfy the conditions in (41.1). I I' Pis unramified over R, then P is regular, and P* = P** by (21).11), hence P** is unramified over R*. Assume that P is ramified OV(iI' II' and that P** is unramified over R*. Then P**/mP** 7'" Pin I)), (41.8). Let a' be an element which generates P**/mP** over R/l11, and let h(x) be a monic polynomial over R such that h modulo 11/ iH the irreducible monic polynomial for a' . Let n' be the maximal idn:d of P[x]jh(x)P[xl which corresponds to the irreducible factor h*(x) or (h(x) modulo n) over Pin of which a' is a root, Then we see thaI. h*(x) has a linear factor x - a' over P**/mP**. The completion (JI of Q = (P[x]jh(x)P[x])n' coincides with P*[x]lhl!(x)P*[x] with :I factor hl!(x) of h(x) over P*. Since P** is Henselian, and since h(,I") is separable, the factor hI! of h( x) has a linear factor x - a over yl'l with an a E ai, which shows t.hat Q is analytically reducible becauNI' a ~ P* and deg hI! > 1. Since the discriminant of h(x) is a unit, it. iN obvious that Q satisfies the conditions in (41.1) with respect to N. Then, as we have proved above, Q must be analytically irreducibk, which is a contradiction. Thus the pair (R*, P**) is equivalent to (R, P) and the statement (*) is proved. We remark here that: (**) For a given pair (R, P), if x is a transcendental element over 1>,
n
then (R(x), P(x)) is equivalent to (R, P). The proof of this statement is straightforward and we omit it. By virtue of (*) and (**), we may assume, in order to prove (41.1 ), that R is a complete regular local ring and that Rim contains illfinitely many elements. Let L be the maximal separable subextensioll of Pin over Rim and let a' be an element which generates Lover Rim. Let h(x) be a monic polynomial over R such that h modulo 111 is the irreducible polynomial for a ' . Then, since P is Henselian by (30.3), h(x) has a root a such that a E a' . Then (R[a], P) is obviously equivalent to (R, P). Therefore we may assume furthermore that Pin is purely inseparable over Rim.
107
"IIA 1"1'11111. VI
WI' 111i/1.I1 :/1'(1\'(1 (11,1) I,'y illdlll'lioli Oil II.II,i(.lIdll I{ II' II.I(.il.lIdn N -:: I, LI"'II 1,111'1'(' iH 1I0Lllillll; (.0 III'OVI', If U :.J, LI 11'11 , Hilll:n I' iH 1I0i'l1lHI, I' iH n IVIII.I'Hllln,V I'ill/.!:, WI"'I1I'" I' i,'\ II. 1'1'1'11 1('llIodllln hy (:Ui, J(j), and !.lin HHcll'rl.ioll l'ollowH I'I'OIl\ (/II,ii), '1'''nl'(~fo1'O we tM:l:,;ume that r 2: 3, 2 LnL :/: !In nil nlmllel\(, of lit wlii(:h i:,; 1I0t in l1l , and let ql , ' , , , qs be pl'iuw diviHOnl of xJ>. By the assumption on P, we have xP = nqi' Hct (Ji = J> I qi and let Q: be the derived normal ring of Qi ' Since Qi i!'i complete, Q; is a normal local ring, By the induction assumption, if r is a prime ideal of P different from 11, then P r is unramified over R(rnR) , Applying this fact to those r containing qi, we have: (1) r/qi is unramified over RlxR; hence (Qi)r/qi is a regular local ring, and consequently (2) the conductor of Qi in Q; contains a power of the maximal ideal; and (3) is unramified over RlxR, In particular, the residue class field of is separable over RI m, whence Pin = Rim by our assumption made above. If QI 'F Q~, then L~ 'F Pin by ( 41. 7). Then, taking an element a' which generates over Pin = Rim, we extend both P and R so that their residue class fields become L~ by the method we used above. Namely, let f(x) be a monic polynomial over R such that f modulo m is the irreducible monic polynomial for a'. Set PI = P[x]lf(x)P[x], RI = R[xllf(x)R[x]. Since the discriminant of f is unit in R, it follows that PI and RI are normal and are unramified over P and R, respectively. mP 'F n if and only if mPI 'F nPl and therefore (RI, PI) is equivalent to (R, P), Furthermore, f(:1:) modulo ql is reducible over QI , and we see that qi splits into several prime ideals. Since the total number 8 of the qi does not exceed the degree of extension of the field of quotients L of P over the field of quotients K of R, we see that, after a finite number of steps, we come to the case where QI = Q~, whence QI = RlxR. Thus we may assume that there is an element x of m which is not in 2 m such that xP has a prime divisor ql with the property that Qlql = RlxR. Now let c, y and Xl, •.. ,X r be as in (41.6) and consider (R(y), P(y)), which is equivalent to (R, P). Set z = X3Y - Xl CX2. Then zP(y) is prime, and our observation for X can be applied to z and we have: (1') if r' is a prime ideal of P(y) such that z E r' c nP(y), then rlzp(y) is unramified over R(y)/zR(y) and consequently (2') the conductor of P(y)/zP(y) ill its derived normal ring P" contains a power of the maximal ideal nP(Y)/zP(y). Since P(y)/zP(y) is analytically irreducible, the completion of P" is an integral domain, which implies that P" is a local ring. By our induction assumption, we have (3') P" is unramified over R(y)lzR(y).
Q:
L; Q;
L;
10i'{
(j1'l(IMlli'I'IJ,I(! 1,11i'AI, IIINilil
WI: h:w(~ 0Ii1'y 1,0 Hhow I,!IHI, I'" 1'( /I )/'1'( 1/), 1,(11, 1.1\1' 11Il\,\illlid iil('al of I'" hn II". Ld r" hn a lIIillimal pl'illl(' di"iHo(' oi' //I/'(!/) I ,,'I'(!/), Then the COl 1(1 u ei,o I' 01' I'(U)/zl'(!!) ill I'" iH 1101, I'olll.nilll't! ill r"/zP(y), whence the derived lIol',"al rillg oi' 1'(1/)/1''' haH I\. ('I'Hidlll' class field which contains P"/n". On iJw oU\(~1' halld, :-:il\('(\ 1'/<11 R/xR, P(y)/qIP(y) = R(y)/xR(y), and tlH~ref()re F(y)/r" iK a 1'('/,';11 lar local ring, whence P(y)/r" is normal. Therefore IW/n" U/III, whence P(y)/zP(y) = R(y)/zR(y) by (41.8) and F iK 11l1I':I,lIIili('d over R. Thus the proof of (41.1) is complete. EXERCISES. 1. Show by an example that the normality of P in (41.1) iH illl portant. 2. Show by an example that the regulari ty of R in (41.1) is important. (II i III Consider an extension R' of the following type. Let R be a normal loc:d I'i II v, in which there is a prime ideal p of height 1 such that p itself is not prilll:i p:,J. bat there is an e such tha.t pi,) is principal. Take the smallest e. Let R' btl 1.i1,. derived normal ring of R[al with a such tha.t a'I~ = p('),)
42. Tensor products (42.1) Let Land L' be fields containing an integral domain I whir!i has field of quotients K. If L is finitely generated over K, then L ® / is Noetherian, L ® r L' = L ® K L', and the zero ideal in L ® L' /UI,'; no imbedded prime divisor. Proof. L is of finitely generated type over K, whence L ® L' is oi' finitely generated type over L', which proves that L ® L' is No!' t.herian. Since Land L' contain K, we see that L ® r L' = L ® I, (K ®rL') = L ®KL'. Let (x) = (Xl, '" ,xn ) be a transcendml('I' base of L' over K. Then L ® L' = (L ®KK(x» ®K(x) L'. 1,('1, K' (=L(x» be the field of quotients of L ®KK(x). Since K(x) iM a field, L ® L' is a subring of K' ®K(X) L'. Therefore the tojal quotient ring of K' ® L' contains K' ®K(X) II Since L' is a fini!(' algebraic extension of K (x), we see that K' ® K(x) L' iH a finite ]\.' module, whence K' ® K(x) 1/ satisfieH the minimum condition [or ideals, which proves the last aHsertion. Let (R, ~l, •• , ,~m) and (R', ~:, ... , ~~) he semi-local ringM whieh are modllleH over a ring I. Let ¢ and ¢' he the natural mappingH from I into Rand R', respeetively (¢( a) = a·l for a E I and tIll' identity 1 of R; similarly for¢'). Set qi = ¢-l (~i)' q~ = ¢,-I (~~) and T = R ® r R'. Then we have the following lemma. (42.2) ~iT +~;T 'F T if and only if qi = q~. In this case, if one IIf R/~i and R'/~~ is of finitely generated type over I/qi, then: (1)
I"
1(\11
I'll \1"1'1':11 \'1
'1'/1 , )l,'I' I ):>1') /,'1 ;VIII'II/I'/'il(l/, I:!) I'I'I'I'!I /ll'illll' dil'islI/' Il,~ I!f' \1,'1' p,'I' iN (/ lIIillil//(// l'I'illll' lfil'/soi', (///11 1:\) 11,( I/(/,~ II ./i//ill' 11f/,,~':,~, "I'(lo!'. II' (j, ~~ 111('11 ~\,'I' I p;'I' l'olli:l,iIIH~\;"1' '1/1' N''/' = '1', Hlld \VI' PI'O\'I' 1./11' fin,:!, aHHI'I'Lioll I'HHily. Hil\(~<: 'I'/(P i 'l' p;T) (N/~\i) 1Xl1/1J, (N'/P;), W(~ pl'llve (I) alld (2) by (42.1), Since PiT ~1~'1' ImH a fil\it(~ bnHis, (:l) follows from (1). With Lh(~ H1tmC llotation as above, we assume that every R/Pi is of fi'lii.eiy I!;cnerated type over I/qi' Let S be the intersection of complcmentH of prime divisors of PiT p;T for all pairs (Pi ,p;) such that PiT P; T 'F T. Then T s is called the local tensor product of U. and R' over I and is denoted by R XI R' or by R X R'. By this definition and by (42.2), R X R' is a quasi-semi-Ioeal ring and the maximal ideals of R X R' have finite baRes. We note that if Rand R' are of finitely generated type over I, then so is R X R ' , If R is of finitely generated type over I, then R X IL' is of finitely g81lPrat.ed type over R ' , whence ill thiH (~afle R X R' iH a semi-Ioeal ring. Now we go hack to the general ('aHe where R X R' is dC'fillccL Let m'C he the JaeobHon radical of Il X R', Set n = n" Then R X R'/n ifl a semi-local ring which may not he Noetherian. Then the ('omplction of R X R ' /ll is called the completc tensor product of R alld R' over I and ie; denoted by R ®I R' or by R ® R'.
q; ,
+
+
+
+
+
men.
(42.3) THBOREM. R ® R' is a semi-local ring (which is Noetherian). Proof. EVNY maximal ideal of R X R' has a finite basis, hence every maximal ideal of R ® R' has a finite basis. Therefore the asflertion is proved by (31.7). (42.4) With the same notation as abovc, if R* and R ' * are semilocal rings which contain Rand R' respectivcly, as dense subspaccs, then R* ® R ' * = R ® R'. Proof. Let m and 111' be the Jacobson radicals of Il and R', respectively, and let a(n) be the ideal of R ® R' generated by 111 n and m'n. Then a( 1) contains a power of the Jacobson radical a of R ® R' by (42.1). It is obvious that a(1) is eontained in u, Sinee u(1)2n C a(n) C 0(1), we see that the system {a(n)} is a base of neighborhoodi:l of zero of R ® R'. The same is true for R* ® R'*, i.e., the system of a*(n), which are the ideals of R* ® R'* generated by mnR* and m,nR*, forms a basis for the neighborhoodi:l of zero. Therefore the aRi:lcrtion follows from the fact that R* ® R'*/a*(n) = (R*/ mnR*) X (R'*/m,nR ' *) = (R/mn) X (R'/m,n) = R ® R'/a(n).
1'/'0 \l'illi llil' 1IIIIIIIill/i. /IN 1/1111/ 1', if N, iN IIII' 1'11111 U1" //.Iut 'tJ U~ i.~ aU! ('IIIII/l/dill/l, I~/ i Jill' I'/II'I'!I (I, /), Ihl'lI H ® /t '£8 l/.atl/./'(/Jt!1 '':Slrtlt(/'/,l)!I.'';(: til Iltl! (11:1'1'1'1 8/1/1t I~r N, (I) N, fll/' (/1/ ('I:, .i) .,11,(://, that ~l/I' + p~'I' 7'" '1', and ('(I.r://, H, ®, N'; ('I/inl'idcs '1Ililli !f" ®Ii h~'; '~r i (lmwt(:.~ tlw (;O'Ynplct'I:WI, (~( / qi • We say thaI, n fidel f( iR a ba8ic .field or a N(,111 i-local I'i III!: (N, PI , . , . , .. , , ~n) if K is a subfield of R and if every N/p,;, iN:L fillil,(1 nlgnhl'ai,' extension of K. (,I~.r)) (~()II.o1",'\lI,\'.
1
N;,
J){l'Iill/l Id
/
(42.6) THEOREM. Assume that Rand R' are semi-lowl rino" /llId that a field K is a basic field oj both Rand R'. Set R* = R ®f( R'. '/'11.1'11 altitude R* = altitude R + altitude R' and, Jar any ideals a anll n' of Rand R', respectively, such that depth a = depth a' = 0, we ha/'I J.i,K(aR* + a'R*) = J.i,K(l1) ·)1.[((a'). Proof. Set j(n) = lengthK an/an+! and g(n) = length K a'u/ a,,, I I. If we consider R/a n and R'/a,n as K-modules, then they are dil·(II·1. sums Li
0, then the assertion is obvious. Set f*( n) g*(n)
=
(n;:.
~
1).
Thenf(n)
=
af*(n)
=
(n
+ r r - 1
1)
and
+ (polynomial of 10w(~1'
degree), g(n) = bg*(n) + (polynomial of lower degree), whencn, by induction, sen) = Li+i
I'll ,II "I' 11:11 \' I
J.!:I'I'I' /1,1, lI\I)r1i, /'
f
pI>
I
/,1
I ';' I
I), 'I'IIIIN WI' 1111,\'1' 1'l'dlll'I'd 1,0 I,hl' 1'11,1'1' Wltl'I'I'
h,Y Lhn I'HHI' WIIl'I'I' U alill ill 'I' Hlld r' I(~L(,I~I'H, I'cHpe(',tiveiy, ill whil'" 1':tHI' U* iN obviollNly Uw Jormal power Heries ring ill r + r' 1n(,(,(~I'N, which p,'oveN Ute aNHcrtion, A Nodhel'ian valuation ring I is called a basic valuation ring of a local ring R if R dominates I and if the residue class field of R is a finite algebraic extension of that of I, (42,7) If Rand R' are local rings which have a common basic valuation ring I with a prime clement p, if altitude R/pR = altitude R - 1, and if altitude H' / pR' = altitude R' - 1, then altitude (R ®I H') = altitude R + altitude R' - 1, Proof. Let rand r' be the altitudos of Rand R', rospectively, Then there are systems of parameters p, X2, ' , , , Xr and p, X2, , " , , X,r ' 0 f' Rand R', respectively, We may assume that Rand R' are complete, whence thoy are finite modules over S = I[[x2, ' " , xrll and S' = I[[x~ , ' , , ,x~,]L respectively. Then, as is easily soen, R ® R' c:ontains the formal power series ring S* = I[[x2, ... , Xr , x~ , ... , x~, II and is a finite S*-module. Thus we prove the assertion. 11,1111 (/
(/+, ThiH I/I,NI, l'aH(' iH !'I'alit-I'd
Ii~' al'l~ Lill' l'ol'lll/l,l POWI'" Hl'I'il~H l'iligH
( 42.8) THEOREM. If a and h are ideals in an unramified regular local ring R, then height (a + h) ~ height a + height h and altitude (a + h) :s; altitude a + altitude 6. Proof. We prove the first formula first. Let a' and h' be minimal prime divisors of a and h, respectively, such that height a' = height a, height h' = height h and let ~ be a minimal prime divisor of a' + h'. Then it suffices to show that height ~ ~ height a' + height h'. Since R~ is an unramified regular local ring by (28.4), we may replace R with Rp , and we may assume that ~ is the maximal ideal of R. Furthermore, we may replace R with its completion, whence R is the power series ring in analytically independent elements, say Xl , . . . , ••• ,Xr over a coefficient ring 1. Let Yl , .. , , Yr be indeterminates and let CJ be the isomorphism from R = I[[xl , ... ,xrll onto I[[Yl , ... , YrlJ such that CJ(Xi) = Yi. Sot R* = I[[xl, ... ,Xr , Yl , ... ,YrJJ. Then R*/(a'R* + CJ(h')R*) is identified with the complete tensor product of R/a' and R/h' over I. Let b be the ideal of R* generated by Xi - Yi (i = 1, .. , ,r). Then, since a' + h' is primary to ~, since R* / (a'R* + CJ(h')R* + b) is isomorphic to R/(a' + h'), and sillc:e b is generated by r elements, we soo that altitude R*/(a'R* + CJ(h')R*) :s; r. If [is
I ~" /
IIldl )1111':'1'1/11'
d
NIII\
1,111' \I, 1/1
a lil'ld, tlll'll nlt.it.Ili1(' /i+/IIl'lt l' I o( II')""') ill'ptll II' I iI"plll Ii' by (/1~.(i) alld it. i:-: (''llial Lo (r hl'i/1;lit. n') I (I' hl'i/1;11i II') II.\' the validity of ihe ('haill l'Olldit.ioll for prillII' idl'al.-.;, wllil'll illiplil'il height a' height h' 2: r, alld Uw a:-::-:cI'Lioll i:-: prowd ill t.hi,; I·a,'il'. Assume now that I is not a field, and In1, Ji 1)(: a prillt(> ('II'llIl'liI. 01' I. If p is in none of a' and h', then we "ee that aitit.IL<\<: N*/(a'li* I a(h')R*) = depth a' depth h' - 1 by (42.7), alld we :-:1'(: L1H::I:-: sertion similarly. If p is in both a' and h', then eOlll-lid(:l'illg U/7i1I: instead of R, we prove the assertion easily and we sec ill i.hi:-: (·:l.NI' that height ~ 1 :::;; height a' height h'. Assume that p ( n' alld t.hat p f h'. Then, set h" = pR h'. Then height h" = uli.iLII(I<' h" = height h' 1 by (9.2), and height ~ 1 ::::; height a' h('igltl. h" by the case stated just above, which proves this casco Thut\ 01(' formula for heights is proved completely. Now we prove the h~t. formula. Let ~ be a minimal prime divisor of a h such that height. ~ = altitude (a h) and let a* and h* be minimal prime divisol',~ of a and h, respect.ively, which are contained in ~. Since ~ is a minimal prime divisor of a h, ~ i8 a minimal prime divisor of a* h*, whence altitude (n h) = height ~ :::;; height a* height h* ::::; altitude a altitude h, and t.he assertion is proved. Kext we consider tensor prodlH't.s of normal ring8. We begin witlt the following lemma: (42.9) Let R be a normal ring which contains a .field K. If a field L is separably generated over K, and ~f L ® H is an integral domain, then L ® R is a normal ring. Proof. L ® R is the union of all L' ® R wit.h finitely generated subfields L', so we may assume that L is finitely generated over K. Then L has a separating transcendence base Xl , ... ,Xn over K. Sci. Il' = R[Xl' ... , xn], K' = K (Xl , ... , Xn) (with regard to the fact that K and Rare sub rings of L ® R). Since the Xi are algebraically independent over R and since R is a normal ring, R' is a normal ring. Since K'[R] is a ring of quotients of R', we see that K'[R] is a normal ring. Since every clement of L is a root of a monic polynomial over K' (hence over K'[R]) whose discriminant i" a unit in K' (hence in K'[R]) , we see that L[R] = L ® R iR a llormal ring by (10.1 [)). (42.10) THEOREM. Let Rand R' be normal rings which contain a field K. If Rand R' are separably generated over K and if R ® R' is an integral domain, then R ® R' is a normal ring. Proof. Let Land L' be the ficld:-: of qnotients of Rand R', rm;pec-
+
+
+
+
+
+
+
+
+
+
+
+
+
J(
J(
J(
+
+
1'11,\ 1"1'11111. V I
liv('I.\'. " (0) Ii" IIlliI /i (il //111'1' 1101'IIII1II'illf.':H I,.y ( I~.\I). '1'1"'1'1'1'111'1' jj, Hllllil'I':4 10 rdllll\' 111111. ( /, (.) U' ) () (Ii' (.) //) /i' (.) h". I ,1'1. III~I Hlld II/~'I I,I' lilll'nl'ly illlll'PI'lIdl'llj, I,JI,HI'H III' Ii' Il.Ild /1" 11\'1'1' /\', nlill 11'1. 11',,1 :l.Ild 11';.,1 II(' lilll'al'iy illil"pl'lIdl'lIl. l'iI,HI',., III' /, nlld // 111'1'1' /1 witil'it (,1I11j,aill II(,~I alld 111~'I, I'nHpl'I'l.ivl'ly. Till'li nVI'I'y 1,11'1111'111. I, III' /, (.) // iN (,XI)I'('NH('d lilliqlldy ill I it I' I'III'IlI (/',I,"P" C<) I'~, ((/."", I /\'). I I' II i,'1
L
ill L ® /(1, tli(,11 ill thiN (,XllI'('NHilill I'~' iN ill 11I~'1 1'111' I'VI'I'y 1/ HIII'II 1.11111, aMP! ~ 0 (for Home /1): il' b iN ill Ii, (8) //, Lh('11 I'M iN ill luxl 1'111' ('I'I'I'.\' /1 such that a MI"7"" 0 (fIJI' NOIlIn /1' ). TIII'I'I'I'III'I', il' /) iH ill (/) ~") /t) (, (R ® L'), thCll 1.lw ()XPl'()NHioll JrlllHI. h(, I.lre 1'1I1'1lI (rx~,fll. (.) II~' , which implies that Ii ( R ® N'. ThuN (I, ® N.') (H IXi 1/) fa' (.) R ' , and the proof is complete. The analogue of (42.1 0) does noL hold ill g('II('ral 1'111' II'IINIII' pl'od uets over a ring which is not a field, even if the I'illg iH :I. NIII'lltl'I'inll valuation ring. A generalization in that ease eaH h(, Nl.aLI,d aH 1'1I11111VH:
or
n
L
( 42.11) THEOREM. Let Rand R' be normal N()dh('l'ill.nril/.fI'~ Il'ft I('I! contain a Noetherian valuation ring I with a prime o/eulI'nf, .1'. ,188111//(' that R* = R ® I R' is a Noetherian integral domain IJ.nd Ihll/ 11111/1 It' and H' are separably generated over I. Then H* is a ,(W'l'uutl n://!I if II II Ii only if R:. is a normal ring for every prime divisor \)* Id .rli*. Proof. The only if part is obvious and we proV() LlI(, 'if IIJ1,I·I.. 1,1,1 K be the field of quotients of I. Then K = 1[1 j.rl alld {('*II j.rl R[l/x] ®I R'[l/x] = R[l/x] ®K R'[l/x]. Thus R*ll/.rl i:-l :I. 11111'111111 ring by (42.10), which implies that R* is a normal rillg by vil'llIl' III'
(35.4). (42.12) COROLLARY. Let Rand R' be as above and aNNUli I.!' flll'll/I'/' more that for every prime divisor \) of xR, R/~ is separably !/I'II('/'l/ll'd over I/xI and that xRp = ~Rp. Then R ®I R' is a norma/rillfl. This follows immediately from (42.11) and the followill/!:: (42.13) Let K' and L bejields which contain afield K. If /, is 81'/1 arably generated over K, then K' ® K L has no nilpotent (:/(:/III'II! 1'.1'1'1'111 zero. Proof. By the definition of tensor products, we may
:1.,-;:-1111111' lilll!.
L is finitely generated OV8r K, whence L has a scpamlill/!: 11':1,11:-11'1'11 denee base Ul, . . . , Un over K. Then K' ® K L ha:-: Lil() NII.IIII' 1,111.11.1 quotient ring with K'(u) ®K(U) L, and we may ItNNIIlIII' LlmL /, i~1 algebraic over K. Let a be such that L = K(a). TII('II II iN iI. 1'11111, III'
171 /I.
(11'11 1M IWI'III(' 1,1 II ',\ I, III NI III
HI'PII,I'II,hll' IIIOllil' POIYllOlilill.1
i( ,I')
01'1'1'
Ii,
Il.Ild
Iii
Co) " iH iHIIIIIOI'plii"
1,0 1\' '[,1'[/'/( ,I') 1\'[,1'[, '1'111'1'('1'01'1' WI' PI'OI'I' IIII' II.:-:HI'l'l.ioll, LII"YLly, I'll' illl,l'od'llll' I,hn lIol.ioll of lUI ol'dl'l' of ill~('plI.l'lI.hiliL.v, whil'li iH of gl'ollll'l,l'il' ill I 1'1'1',,1., WI: bl'/.!;ill wiLli ChI' followillg 11'11I1I1n:
,I) I,('{ Hand U ' IIf' 'I"'I;nos 81U:h thai H' 1;8 (1,1/, N-Ul,/It/II./(, , If, flli' U' , tlwn: 'is a III:I: N-'Ifwdn!n (:ontai",illrl I'fI"fI'I,I'n/'s and contwi;u:d ,in N', tlwn ® /1 H' ,z8 e,w.ct. In pa'l"ticll.tll.l',
(,I~,I (/./1.,1/
./il/.';'(' I/./I,'/II./If"r I~r ('11:'IIWI/,/'8 Id
t!U:8('
I)' N (:onlain8 a N odhnrian vahtation riny I, if l' i8 a r'ing containinfl I /T:Iul 8'1U:/i. that (:/Jery non-zero element of I is not a zero divisor in f', ani I '1/ /t,' = H ® I f', then ® R R' is exact. The proof iH Htraightforward and we omit it. A:-::-:ume that fieldfl K' and L have a eommOll tmbfield K and I.h:ll OlIO of K' and L iH a funetion field over K. Let ~* be a prime ideal of the local tensor produet L* = L XK K' and set m = length L;;•. Then, obviously m is uniquely determined by the fields K, K', /, Hnd L */ ~*. This m is ealled the order of inseparability of the pair (K', L) over K with respect to L*/~*, and is denoted by iK(L, K'; L* /~*). Note that if L' is a field eomposed of Land K' and if eith('l" trans. degK L = trans. degK' L' < 00 or trans. degK K' = trall:-:. deg L 1,/ < 00, then L' is naturally isomorphic to L */ ~* (with regard to the eomposition) with a suitable ~* and the order of inseparabilil.y iK(L, K'; L') is well defined. Note also that the above eondition 011 the transeendenee degrees is equivalent to the two eonditions thai trans. degK L = trans. degK' L' and trans. degK K' = trans. degL 1/. (42.15) (1) If one of K', L is separably generated over K, then iK(L, K'; L*/~*) = 1. (2) If K is oj characteristic p 'F 0, then iK(L, K'; L*/~*) is a power of p. (3) If K is of characteristic p 'F 0, if K' conl tains K / p , and if iK(L, K'; L*/~*) = 1, then L is separably generated over K. Proof. (1) iH immediate from (42.13). AH for (2), we may aSHume by symmetry that. L itl a funetion field over K. Let L" be a sub field of L Hueh that. L" itl separably generated over K and such that L i" purely inseparable over L". Then L" X K' has no nilpotent eicment", (L" X K'). whenee (L" X K')V is a field K'(L"), where~" = ~* Obviously, K'(L") ® L" L iH a loeal ring and coineides with L : •. Therefore iK(L, K'; L*/~*)· [L*/~*:K'(L")l = lengthK'(v') L:. = [L:L"], and we prove (2). As for (3), we see that L ® K 1 / p has no nilpotent elements, and is a local ring, whenee it is a field, hence the assertion is proved.
n
1'1 I.
1'11,11"1'1':11 VI
(·I:J.IO) 'I'm:OII.ldM. /'1'1 /.;, Ie, I), /,+ III/.d ~)~. III' UN U'/WI'I'. ,'18.~1/./l11' {!till ( /, q) il'; II N /I/'I//('/'i(/./I. 110./11.((./.io/l. 'l'i/l.(/ 81/.('11, /.Iud /\ C~ /Iq (/ ,//W/J mi/l.('itll' 'II'ilh /1). /1.~SIIIlI.I· ((.t.~1/ thal (N, m) mul (1', q') are local rings 'IPhil'li tll/lili//nll' / SIU:h that /) = UI llt, f(' = 1'1 q', q'l q1' is nilpotent 1/,Ild 811,eh t!tnl nonon-zuro clement oj I is a zero divisor in I'. Let 1]3* be the prim.I' -idwl of R* = R ®I I' such that 1]3*/(mR* + q'R*) = \.1*. Then, fl/I' I'vcr// primary ideal n of R belonging to m, we have
/.t(nRi*)
(length 1'lqI') ·iK(L, K'; L*/\.1*)· /.t(n),
=
length (R;*/nR:*) =
(length 1'lq!') ·iK(L, K'; L*/\.1*) . length (Rln)
If R XI I' is Noetherian, hence in particular if one of R and I' is of finitely generated type over I, then R* may be replaced by R X I 1'. Proof. R** = R X I !' is a ring of quotients of R*** = R ® I I'. ® R R*** is exact by (42.14), whence ® R R** is exact. Let 1]3** be the prime ideal of R** such that 1]3**R* = 1]3*. Then we see that ®R (R;::') is exact, whence ®R/n (R;::'lnR;:.) is exaet by (18.10). Since R;::'/nR;:* = R;./nR;** and since the above is true for any n, we see that the theorem of transition holds for Rand by virtue of (19.1), hence /.t(nR;.) = (length R;./mR;.)·/.t(n) and
R;.
length (R~./nR~*)
=
(length R~./mR~.) . length (Rln).
If the same is applied to K, I'/qI' and L instead of I, R and I', respectively, then we have length R;./mR;. = /.t(O·R;./mR;.) length (Ii;./mR;.)/q'(R;*/mR;.) ·/.t(O·1'lqI') = iK(L, K'; L*/\.1*)· length 1'1 qI', and the equalities are proved. The last remark is obvious. (42.17) THEOREM. Assume that U = K'(L) is the field descl'ibed just before (42.15). If K" is afield between K and K', then iK(L, K"; K" (L» and iK"(K"(L), K'; L') are well defined, and we have iK(L, K'; L') = iK(L, K"; K"(L» ·iKH(K"(L), K'; U). Proof. Since trans. deg K L 2: trans. degK" K" (L) 2: trans. degK' L', we have trans. degK L = trans. degK" K" (L) = trans. degK' L'. Since trans. degKK" 2: trans. degL K" (L) and since trans. degK" K' 2: trans. degKH(L) L', we see that the above inequalities are equalities. Therefore the first assertion is proved. Let \.1* be the prime ideal of L* = L X K' such that L' = L*/\.1* and set L** = LX K", \.1** =
I 'ill
III':II~III;'I'IIII'
1,111'\1, IIINllli
II;:"'~ , 11 () 1IIId I' 1(" lil'l'l', '1'111'11 h'+ Ilil'I'(' 111'1~IIIIII'c; h~ (0) I';', \Vliil'li j iH II. I'illg ill' 1lii0Lil'IilH 01' I, ·, \VIII'III'I' N~', Ilil'I'p l'OiIWidl'H wii,ll lIill' I,~\ , '1'111'1'1'1'01'1' i/,(!" I';'; 1/) 11'lIgl.h !,;~, (ldL .. halid Hidl' 01' llil' 1'(jllll,1 iCy ill (,I~,lli)) i""(I,, 1\'; 1/),II'llglll /{i""(/,, 1('; 1/),/"(/,, I":"; II"( I,)), wllil'h l'olllpll'l,I'H ChI' Pl'oof.
11\+ (1 1'++, '1'111'11 \\'1' npply (1:l,ln) Iii '"
II', II
Iii 1'1'1' Il('illf..'~
":8 II. Onto'iN (',I' I{. IA't I{" Iw a ma:eirnal 7J'wrely ,ins/:1J1LraiJl/: sulw,l'tcn.~ion /~( I\' U1W/, f{. 'l'hen iK(L, K'; ]('(1,» is independent (d the choice (~r Oil' fO'If/.Jiol·;r:d jield Ie (L) and is equal to iKe 1" K"; K" (1,». III i.hi;.; ease, iK(L, K'; K'(L» may be denoted by iK(L, K'). Proof. Since K' is a Galois extension of K, we see that K' is sepal'able over K", whence the equality iK(L, K'; K'(L» = iK(L, K"; 1\"(1,» follows from (42.17). Since K" is purely inseparable, then: i;.; only one composed field K" (L) of K" and L, and the proof is complete. Assume that I, is a function field over the field K and let K* bl: the algebraic closure of K. Then iK(L, K*) is well defined; it is calleel the order of inseparability of I, over K, and is denoted by [1,:K]o . (42.19) Let I, be a ftlnction field over a field K of characteristic p ~ O. If[L:K]. = pm ~ 1, then iK(L, K llp ) > 1 (hence::::: p) and L(K P-"') m m is separably generated over K P - , i.e., iK(L, K P- ) = pm. Proof. Since I, is not separably generated over K, 1, ® K llp is not llp an integral domain, and iK(L, K ) > 1, hence the order of in;.;epllp llp arability of L(K ) over K is p" with n < rn, henee by indllctioll we complete the proof. (42.20) Assume furtherrnore that I, = K (XI, ... ,:rr+l) and that trans. degK I, = r. Let m be such that [L:KJ, = pm. Let F, G, H be the irreducible polynornialsinindeterminates Xl, ... ,Xr+dor (XI, ... , m ... , Xr+l) over K, K P- , the algebraic closure K* of K respectively. Then we have, except for constant factors, F = GP'" and G = HH' with a polynornial H' over K* such that H'(xI , ... ,Xr+l) ~ O. Proof. Set A = K[XI , ... ,Xr+l], A' = KP-m[X I , . . • ,Xr+l], and A* = K*[X I , '" ,Xr+IJ. length I, ® Kp-m= pm by (42.19) and m I, ® K P- is a ring of quotients of A'/FA'. Therefore we see that FA' = GpmA' because KP-m is purely inseparable over K, which m proves the first factorization. Since L(K P - ) is separably generated m P over K - by (42.19), GA * is semi~prime, and we prove the last factorization. (T..l.IX) (~Oll()""AI(Y. /\.~gl/,nw fu.rlli./'J''fII,Of'(' that I{'
!/'I/..w:o!/, /~r
('III I"I'II;I!
\'1
1'/,/
I,: \ I'll(' 'I: li'II',. /. (;"11"1'111 i1,,, (I~_(i) 1.0 1.111' 1'11.111' I\' 111'1'" /1 iH II ''''111111011 Hllhlil'ld "I' t.' :1.lld Ii" 1111.,11111111. 1"'llidll" .'lntlH li .. ldH "I' Ii' 11.1111 W 11.1'1' lillil.('ly l!;"I)(~I':l.Ll'd OV(~I' /\ . :~. (:(,III'I·l1.li~., (1~.7) 1,0 1.1111 ('.:tHI'
wbnl'l' 1 iN a VtL[UtLtiOtl ring dominated by III iN defined. :'1. AHHlIIlll' I,haL L, U are function fields over a field K such that L C L', 1\.1\,1101, 1\* bo the algebraic closure of K. Prove that [L':Kj./[L:Kj, = [L':Lj.+ [K*(U):K*(L)j. = h(L', K*(L); K*(L')). 4. Goneralize the last half of (42.14) to the case where I is a Dedekind do,,~ :Iolld
main.
W
:l.11I1·HlI(,h Lhu.L /{, @I
CHAPTER VII
IIenselian Rings and Weierstrass Rings 43. Henselization We begin with a supplementary remark to Galois theory, discussed in §41. (43.1) Let R be a normal ring, let R' be an almost finite separable Galois extension of R with Galois group G, let ~ be a prime ideal of R and let ~~ , .. , , ~:, be all of the prime ideals in R' which lie over ~. Let R" be the splitting ring of ~~ and set ~" = ~~ n R". Then: (1) if a E R" is in ~~ and is not in any of ~~ , .. , , ~~ , then a is a root of an irreducible monic polynomial f( x) = x" + CIX r - I + .,. + Cr such that c,. E ~,Cr-l EE ~ and R;" is a ring of quotients of R[a] and (2) if b E R" is not in ~~ and is in all of ~~ , ., . , ~~ , then b is a root of an irreducible monic polynomial g(x) = xr + d1x r- 1 + .. , + d r such that d I EE ~,d2, ... , dr E ~ and R;, is a ring of quotients of R[b]. Proof. ~" is unramified over R by (41.2), whence the assertions follow" from (38.6). By virtue of the above result, we have the following lemma which will be generalized later: (4:3.2) Let ~ be a prime ideal of a nor'mal ring R. Then the following four conditions are equivalent to each other: ( 1) Rp is a II enselian ring. (2) Every integral extension of R has only one prime ideal which lies over ~. r 1 (:3) Every monic polynomial f( x) = x + CIX - + ... + Cr over R, such that Cr E ~ and Cr-l EE ~, has a linear factor x + a with a E ~. (4) Every monic polynomial g( x) = x' + d1x r- 1 + .,. + dr over R, stlCh that d I EE ~, and d2 , ... ,dr E ~, has a linear factor x + b with b such that b - d1 E ~. Proof. We note that, denoting by S the complement of ~ in R, the "et of integral extensions of Rp coincides with the set of R~ where [179]
It 1'11111111\'1'1' nil illl.q~I'II.II'\II'IiHiIIlIH III' II'. '1'111'1'1'1'01'1' (:! I flllluwll 1'1'11111 (II 11.1' "irllll' III' (:IIl}II, (!oll\'I'rHI'ly, il' N p it! 11111. 111'IIHI,liall, 1.lil'II 1111'1'1' ii, 11.11 il'l'I'dlll'ihll' 1IIIIIIil' PIlIYIlIlIlIi:i.I /1'(.1' I 0\'1'1' 11')1 '-III1'1i llilll II'(,/, I 11111111110 pH1,\.r1 ,,",pliI,,; illl.o "WO 1':1,1,1,01''; IVliil,1i liav(' 110 1'111111111111 l'Old" '1'111'11 HII I,I'I/( II'(,/')U~I./'I) i,; all illl.q.!;l':d l'xLI'IiNioll III' U~ wliil'll iN 1101, I(lIaNi-loI'.:d, WIiI,II(:I' Lhl' ill Lq!;ml I'.IONIII'I'. 01' U ill Lhal. iIlLq!;I'al I'XLI'IINinll of Up haN al. II,aNL Lwo Pl'illW idealN whidl Iii' OVIT p, 'l'hliH (I) alld (2) arn l'qllivlIlcliL 1,0 (,ach other. By Llw Nallln 1''':lNOII, WI' NI'I' Lhal. IIOII-validii,y of one of U~), (4) implicH t.he lIoll-validiLy 01' (2), CO])VCrHe1y, afifiume that (2) iH not true; let R* bl' :t Hepara!,!I' illLI'gml cxtenHion of R which has at leatlt two prime idealN pi , ~I'; whi(:h lie over p. Let a be an element of pi which is not in pi alld l'OIlHider an almoHt finite Galois extension R' of R which containN (/. Then (4::U) implies the non-validity of (;)) and (4). Therefore 1.111' ltHtlertion is proved completely. When R is an integral domain, the separable integral closure of It iH the set of separably algebraic elements which are integral over Ii (in a fixed algebraic closure). Now we shall define the notion of H enselization. vVe consider first the caHe of normal rings: Let R be a quasi-Ioeal normal ring. Let R' be the separable integral closure of R and let p' be a maximal ideal of R'. Let R" be the splitting ring of p' and sl'!. pI! = R" p'. Then R;" is uniquely determined up to isomorphismN over R, for, if H is the splitting group of p' and if p~ is another maximal ideal of R', then p~ = p/lf with an element (J of the Galois group G of R' by (10.12) and the splitting group of is (J-1H(J. This R;" is called the Henselization of R. Now we consider the general case. Let R be a quasi-local ring. Then there is a normal quasi-local ring S which has an ideal a such that Sia is isomorphic to R. Let S* be the Henselization of S. Then S* I as* is uniquely determined up to isomorphism over S / a as will be shown below, hence, identifying Sia with R, we call S* laS* the Henselization of R. Before proving the uniqueness, we note that: (43.;3) If R* is a Henselization of a quasi-local ring (R, m), then R* is a Henselian ring. Furthermore Rim = R*/mR* and R* is unramified over R. Proof. The laHt assertion follows from (41.2) in the normal case, whence in the general case by the definition. If R is normal, then
n
p;
1'11.'\ I"I'IIII!
VII
IHI
\VI' PI'O\'I' llil' lind, Ilrlrll'l'l.illll II,V Liw dl'lillil.illil /1.1111 II,V (,I;I.~ J, lVill'lll'I' Lill' gl'IH'I'II,1 l'aKI' rllllllw,., 1'1'0111 LlII~ rollowillg O"ViOIiK rad: (,1:1.-1) 1\'1 'I '/'!/ 1t1l1l11l1l1111'pltiil: i/lUl(jll I!!, a flensdian ring is a Hcn8/'11:/11/. .,.i/l.l/. III o/'{kl' to prove the uniqueness of a Henselization of R, we prove the following j,heorem:
(4:'LIi) THEOREM. If a Henselian ring H dominates a quasi-local ring R, then there is one and only one R-homomorphism ¢ from a given Henselization R* of R into H such that ¢(R*) ::; H. If R is a normal ring, then ¢ is an isomorphism. Proof. Let S be the normal ring which is employed in order to define R* and let a be the ideal such that Ii = Sla. Let S* be the Henselization of S. Let F be the set of pairs (T, (]') of sub rings T of S* and homomorphisms (]' such that: (1) T is a quasi-local normal ring dominated by S* and dominating Sand (2) (]' is a homomorphism from T into H whose restridion on S coincides with the natural homomorphism from S onto R and such that (]'( T) ::; H. Let F' be the subset of F consisting of such (T, (]') that (T, (]') E F and (T, (]") E F imply (]' = (]". Defining (T, (]') ;::: (T', (]") if T ::::) T' and if the restriction of (]' on T' coincides with (]", we make F' an ordered set. Then it is easily seen that F' is an inductive set, whence there is a maximal member (T*, (]'*) of F'. Assume that S* 7"" T*. Then, by the definition of the Henselization of a normal quasi-local ring, we see that the separable integral closure of T* has at least two maximal ideals, whence T* is not Henselian by (43.2). Therefore there is an irreducible monic polynomial f( x) = xr + a1x r - 1 + ... + ar over T* such that aT is in the maximal ideal WC* of T* and ar-1 is not. Since H is Henselian, there is a root a' of (]'*(f(x)) such that a' is in the maximal ideal m of H, whence (]'*(f(x)) = (x - a')g*(x) with a monic polynomial g*(x) such that g*(O) ~ m. By the existence of a', we can extend (]'* to a homomorphism (]'** from T** = T*[akUl*+aT*[aJ) into H so that (]'**(a) = a', a being a root of f(x) which is in the maximal ideal of S*. Thus (T**, (]'**) E F. By the maximalityof (T*, (]'*), there is a (T**, (]''') E F such that (]''' 7"" (]'**. Since (T*, (]'*) E F', the restriction of (]''' on T* must be (]'*. Therefore (]''' 7"" (]'** implies (]''' ( a) 7"" a'. Since a is in the maximal ideal of T**, (]''' ( a) must be in m, hence g*(rT"(a)) is a unit in H. Since f(a) = 0, it follows that (]''' ( a) - a' = 0, which is a contradiction. Thus T* = S*, and the existence and the uniqueness of ¢ are proved completely.
Nil\\' \VI' II.KHIIIIH' LlIII,L N iH IInl'lIll1.1. WI' l'OII:,idl'I' 1.11(' ('11.:,1' \VI 1('1'1 , It' 8. '1'111'11 8~' iH 11.11 illl,I'glnl dOlllnill wltil'll i,"i II.lgI'III'II,il'. nVI'I' h~. 'J'111'1'''l'oi'l' 11.11 idl'II.1 II'" or 8* iH iWI'o il' II.lId oilly il' 0* It~ O. Applyillg thiH I'II.I·! 10 1111' 1"'1'111'1 or (/), WI' Hnl' IImL '/> iH n.11 iHIlIIiOIVhiHIIi. '1'1111." 1.111' proo\ i:1 l'olllpll'Ll'd hy 1.I11~ I'nllowillg pl'Oo\' of Uw IllIiqlll'III'NH of 111'IINI'li1.a!ioll: /''/'/11(/' I~r 1h.1'11,1I//:1/1/I·/1.I'8" Id 1/1'II.,w:I'izal'ion: Ld U* a.lld U** hn I [I'IINI,I iv,n.LioIIH of It. '1'111'11, applyillg (4:Ui) wiLh U** == II, we NI'n Lhat. ('\11'1'1' iH :I. hllll\OIlIOl'pltiHIll ¢ from R* into N** and, NYlllmdrieally, Lila! !h('i'l~ iH a hOIllOIllOl'pitiHlll ¢' from R** into R*. Then ¢ .¢' iH a holtlo 1I101'phiHIIi I'rom N** into it.Helf. Siuee there is the identity map, (,\)(' IIItiqlll'IWHN ill (4:~.!i) implie~ that¢·¢' = 1, and symmetri('ally ¢'.¢ I. 'l'lwl'c\'ore we Hee that. ¢ and ¢' are isomorphisms. 1\ (,xl, WI' lIote that: (1:Ui) If f(x) is a monic polynomial over a ring R and if R' 'U; 1/ ,.inu (~r ql)'otients of R[xl/f( x )R[x], then ® R R' is exact. Proof. R[:tl/f(x )R[x] is a free R-module, whence ® R (R[xl/f(x )Rl·rl) iN (,xaet. Since ® R[x]/ fex)R[x] R' is exact, we prove the assertion. OIL the other hand, it is obvious that: (4:{.7) Let a ring R' be a module over a ring R. If, for any finill' f/.'wrni)cr oj elements aI, ... ,an of R', there is a sub module R" whit:ll, contains aI, '" , an such that ® R R" is exact, then ® R R' is exact. Now we come to an interesting result:
n
( 43.8) THEOREM. Let R be a quasi-local ring and let R* be 1:1" 1/ enselization. Then ® R R* is exact. Proof. If R is normal, then the assertion is obvious by (43. J ), (4:1.6), and (43.7), whence the general case follows from the definitioll alld (18.10). (4:3.9) Let (R, m) be a quasi-local ring and let R* be its Henseliza lion. For a finite number oj elements bi of R*, there is an element a I~r r 1 mR* which is a root of a monic polynomialJ(x) = xr CIX -t Cr over R with Cr-l ~ m, Cr E m, such that the ring R' = R[a]ent+aR[al) contains bi . R* is the Henselization oJ R'. Proof. Let (S, WC) be a normal quasi-local ring which has an ideal a such that R = S/o.. Let T be the separable integral dosure of Sand let ~ be a maximal ideal of T. Let T' be the splitting ring of ~ all( I set ~' = ~ T'. Then S* = T~, is the Henselization of S, and R* = S* /o.S*. Let b: be an element of S* whose residue class modulo 0.8* is bi for each i and let S' be an almost finite separable Galois extensioll of S containing all the b; . Let SIt be the splitting ring of ~ n s' and
+
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11'1, (/' 111'11111·11'11'1'111, III' Ii n 8" wllidl it, 11111, ill Hlly 111,111'1' IlIlI.xilllll.l illl'al III' 8", '1'111'11 1/ /I' 1II11i1l1lo il8'~ iH 1.111' l'I'(IIIiI'I'd I'II'IIII~III. h,Y viI'LII(~ of (·1:1.1 ), AHHIIIIII' 1I1'~1. 1,1111,1. !.III'I'n iH all n II.H Hl,atnd ill Llw HN8crLioll. Let (,'(,1') 1)(':1. 1I1l)llil~ PIII.YIIOIl\iall)v(~I':-; NlIdl that /1'(:1:) modulo a = f(x). 1,1'1, ((,' hn a I'oot IIf P (,/,) NI1(:11 that a' E [f(S* (the existence follows I'I'IlIll 1.11(: I'a('t that 8* it-! Hent-!elian). Then U = S[a'](Wl+"'S[a']) is a II OI'lIla I rillg by virtue of (38.10) and is dominated by S*. Let U* h(: I.h<: II<:n8elization of U. Since U and S are normal, we see that (f* = S* by (43.5). Therefore R* = S*/aS* = U*/aU* is the Henselization of U/(aS* n U) = R[a](1lt+aR[a]) .
( 43.10) THEOREM. If (R, m) is a local ring, then its H enselizativn R* is a local ring and contains R as a dense subspace. Proof. R*/mR* = Rim by the construction. Since ®R R* is exact, we have mnR* n R = mn by (18.3). Assume that b E nmnR*. Let a be as in (43.9) applied to this b. Then R' = R[a](m+aR[a]) is Noetherian, whence nm nR' = O. Since R* is the Henselization of R', ® R' R* is exact, which implies that nl nR* n R' = mnR', whence b E nmnR' = o. Thus b = 0 and R* is a loeal ring whieh may not be ~T oetherian. Therefore the facts remarked at the beginning imply that R is a dense subspace of R*. Now we want to prove that R* is Noetherian. Since the maximal ideal of R* is mR*, it has a finite basis. Let R** be the completion of R*. Then R** is the completion of R by what we have proved above. Let b* be an ideal of R* which has a finite basis, say b1 , .•• , b" and let c be an arbitrary element of b*R** R*. Let a be as in (43.9), applied to c, b1 , .•• , bs • Then, sinee R* is the Henselization of R' = R[a](m+allra]) , R** it-! the completion of R', and since R' is Noetherian, c E b*R** R' = (L biR**) R' = L biR', whence c E L biR*. Thus b* i8 a closed subset of R*, and therefore R* is Noetherian by (31.8). N ow we derive some properties of Henselian rings. (4:3.11) Let R be a Henselian ring. Then the Henselization of R coincides with R, whence R is a homomorphic image of a Henselian normal ring. Proof. Let R* be the Henselization of R. Then there is a unique homomorphism ¢ from R* into R by (43.5). ¢ is a homomorphism from R* into R*, whence by the uniqueness, ¢ must be the identity, and R* = R. (43.12) THEOREM. A q1.1asi-local integral domain R is H enselian if and only if every integral extension of R is quasi-local.
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hU*. Conversely, if R* is the di'f'ect sum oj ideals II and 6, then then' = gR*, 6 = hR*. Proof. We note first that since R* is integral over R, every maximal i(kal m* of R* lies over m, hence the Jacobson radical of R* contaills mN*. Now, since g, h modulo m have no common root, we see thaI. I.hey generate R* I mR*, which proves by virtue of the above remal'k I.hat gR* + hR* = R*, hence gR* n hR* = ghR* = jR* = by (l.:{). 1(, follows that R* if; the direct sum of gR* and hR*. Conversely, aSHllme that R * is the direct sum of ideals II and 6. Set :c* = :,. modulo fR[:c]. Let ¢ be the natural homomorphism from R* on1,l1 R* I mR*. Since R* I mR* is a homomorphic image of the Euclideall ring (Rlm)[:c], there are monic polynomials g", h" in ¢(:c*) over Rim such that ¢(a) = g"(R*/mR*), ¢(6) = h"(R*/mR*). Let 'f' and s be the degrees of g" and h", respectively, and let g* and h* be polynomials in :x:* over R such that ¢(g*) = g", ¢(h*) = h", g* E u and such that h* E 6. Set A = L ~-l :x:*ig*R*, B = .L~-l :c*ih*R*. Since the 'f' + 8 polynomials 0.", ¢(:x:*)g", '" ,¢(X*)"-l g", h", ... , ... ,¢(:x:*)"-lh" form a linearly independent base of ¢(R*) over Rim, it follows that A + B + mR* = R*. Since R* is a finite R-module, we see that A + B = R* by the lemma of Krull-Azumaya, whenc(~ i A = ll, B = 6, and x*Sg* = L~-l ai:c*ig*. Set h = :c L~-l ai:c . Then h(:c*)g*R* = 0, and h(:c*)R* C o:u = 6. Since ¢(:c*)ih" (i = 0, ... ,'f' - 1) form a linearly independent base of ¢( 6) over Rim and since they are monic polynomials of mutually distinct degrees, we see that ¢(h(x*)) = h". Thus we may employ h(x*) instead of h* above. Similarly, we may assume that g(:c*) = g*. Since g(:c*)h(:c*)R* = 0, i.e., gh E jR[x] and since j is of degree r + s, we
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(·I:Ur;) TII)i)()i(J·JM. If (j'uasi-lvcal ring (R, m) is Henselian if and only 'Ij" e/!(:ry ring H,', which contains R and is a finite module over R, ":8 Ow direct sum of a finite number of quasi-local rings. Proof. We want to prove the if part first: Let f( x) be a monic polynomial over R such that f( x) modulo 111 splits into a produet of two polynomials g", h" which have no common root. R* = R[xl/ f(x)R[xlis the direct sum of quasi-local rings (R I , ml), .,. , (Rn , m n ) by our assumption. Let the maximal ideals of R* be mi , ... , m~ (1117Ri = 111i). We may assume that g" E m7/mR* if and only if i ::; r. Then, as is easily seen, a = RI + .. , + Rr and 6 = Rr+1 + ... + Rn are such that a modulo mR* and 6 modulo mR* are generated by g" and h", whence we see the factorization f = gh such that g modulo m = g", h modulo m = h" by virtue of (43.14). We shall prove the converse. Assume therefore that R is Hensclian. Let m~ , ... , 111:, be the maximal ideals of R'. Considering direct summands of R ' , we may assume that R' is not a dired sum of two rings. We have only to prove that n = 1. Assume for a moment that n 2: 2. Let. a be an dement of m; which is not in m~ . Set R" = R[a]. Since R' is integral over R", m; R" is maximal, whence R" is not quasilocal. Let f( x) be a monic polynomial over R which has a as a root. Then t.he maximal ideals of R* = R[xl/f(x) corresponds in a one to one way to the mutually distind irreducible fact.ors of f modulo In, it follows that R* is the direct sum of quasi-local rings by (4:3.14). Since R" is a homomorphic image of R*, we see that H" is the direct sum of quasi-local rings, whence R" has an idempotent e which it.; not the identity, and e E R ' , which implies that R' is the direct sum of eR' and (1 - e)R' and we have a contradiction. Thus n = 1, and the assertion is proved completely.
n
(43.16) COROLLARY. If R is a Henselian ring and if a quasi-local ring R' is integral over R, then R' is Henselian. (4:3.17) THEOREM. Assume that a quasi-local ring (R', m') is integral over a quasi-local ring (R, m). If R* is the Henselization oj R, the R' ®n R* is the Henselization of R'. Proof. Every maximal ideal of R' ® R* contains m(R' ® R*) because of integral dependence. Since (R' ® R*)/m(R' ® R*) =
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(,I:l.IX) 'i'111';()/U<:M. IfS8U,11'W that a 1j1w8i-local ring (H', lit') dmr/:/;/I. fI, 1/l/fI,8£-lor:al r'ing (IL, 111) and that H' is of finite type over R. 1'11,1'1/, Ihl' I/I'//.,w:{-£za{·ion H'* (d a' iii a finite module over the Henselization It* I~( U. 1'1'001'. H'* dominal.eN H', whence it dominates R, too. Therefol'l' 1,1l('/'I~ iN a lilliquely dctermined R-homomorphism ¢ from R* into H'* I,y (, I:l.!i). Ld; al, ... ,an be clements of R' such that they are int.egral OVI'I' U alld N\l('h that R' is a ring of quotients of R[al , ... ,an], LI'1. N** I)(~ Uw Hubring of R'*generated by the ai over ¢(R*). Then R** iN IlnllNI~liali by (48.16), and we see that R** = R'* by virt.ue or CI:l.!i). Thu8 R'* isa finite module over ¢(R*), and the assertion iN (/It,,~
pl·oved. III o/'der t.o investigate Henselizat.ions of quasi-local integral doIlinillN, we prove the following auxiliary result: (1:1.I D) Let (R, m) be a qllasi-local normal ring and let p be a prime it/I'll.! 1(( R. Let R' be an almost finite separable Galois extension of H '/Iiith Galois group G and let m' be a maximal ideal of R'. Let R" be the 8plining ring of m', set m" = m' R" and set R* = R:~". Let p* be nn arbitrary prime divisor of pR* and let S be the complement of p in U. Then: (1) p* n R = p, (2) p* is unramijied over R, (3) R,~/pR: i.e; Noetherian, and (4) ~R* is semi-prime. Proof. Let a be an element of m" which is not in any maximal ideal of R" other than m" and let f( x) be the irreducible monic polynomial for a over R. R* is a ring of quotients of R[a] by (43.1), whence ® Il R* is exact by (43.6). Therefore no element of S is a zero divisor modulo pR* by (18.1), which proves (1). By the choice of a, a (modulo p*) is a simple root of f( x) modulo p, hence (2) is true by (:l8.6) and by the fact that R* is a ring of qnotientf.; of R[a]. Furthermore, pR: is a ring of quotient.R of a Noetherian ring
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NOW WI' l'OIIl\! 1,0 Lhn following; illtCl'c::;ting result: (/1;t20) TIIl~Om]M. Let (R, m) be a quasi-local integral domain and lel U* Ilc the H enselization oj R. Then: (1) a prime ideal ~* oj R* is a prime divisor oj zero iJ and only iJ ~* R = 0, (2) the zero ideal oj R* is semi-prime, and (3) there is a one to one correspondence between maximal ideals oj the derived normal ring R' oj R and prime divisors oj zero oj R*. Proof. Since Q9 R R* is exact by (43.7), every prime divisor of zero in R* lies over zero of R by (18.11). Since R* is a ring of quotients of a ring which is integral over R, every prime ideal of R* which lies over zero of R is a minimal prime divisor of zero, and therefore (1) is true. Let T he a normal quasi-local ring which has a prime ideal ~ such that T j~ = R. Let U be thc separable integral closure of T and let T* be the Hel1i:lelization of '1' which is defined by employing a maximal idealu. Since R* = T* j~T* by definition, we want to prove (2) and (:3) in terms of T* and ~. Assume that an element a of T* is nilpotent modulo ~T*. Let T' be an almost finite separahle Galoi::; extension of T containing a, set ?m' = un T' and let T" be the splitting ring of ?m'. Then TZWl'nT") :s: T*. Since ~TZwl'nT") is semi-prime by (43.19), we have a E ~T7m'nT")' whence a E ~T*. Thus ~T* is semi-prime, which proves (2). Let G be the Galois group of U over T, let J[ be the splitting group of u, and let U' be the splitting ring of u, whence '1'* = U;unu')' On the other hand, let q be a prime ideal of U which lies over ~ and let K and I he the splitting group and the inertia group respectively, of q. Then the Galois group of Ujq over R' is KjI by (41.2). Therefore we see that: (*) The set of maximal ideals 0 of U which contain q and such that ojq contains a given maximal ideal m' of R' is the set of u" with rT E HrT(m')K with an element rT(m') of G. On the other hand, let ~* be a prime divisor of ~T* and set ~" ~* nu'. Then it is easy to see that: (**) The set of prime ideals of U which are contained in the maximal ideal u and lie over ~" is the set of q
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II: x 1'111,1 '11-\1111-\. I. 1.1'1. "', Ii', U", iJ, alld lJ~ hI' aH ill (.1:1.I) Il,lld lid. 8 hn /I, 1l0rllll\,1 rill!!; Hilidl 1.1111,1. '" S 8 S U'. 1.101, q, = pi n 8,11", .. ' , qUi lin all 1.11(, rnaxilll:ll idl'nl.y or ,'{ wllil'lI lin OVI'I·lJ. I'rovn 1.11",1,: (I) if (J, iH all nlnllHlIlL of il, n W' HIlI·I, 1,1111,1. {(, i,y 1101, ill ally 01''1'' , _.. , '1"" , UlIIIl a i,c, a rooL or a monie polynomial J;I'-I 1',.1"- , ,I, '" + ('r 01'111' /{, Hlieh Lhai, Cr E lJ, Cr - l Ef lJ, and (2) if Ii is :m cleIlll'lll, or q" n ... n '1", n U" and if b is not in '11 , then Ii is a root of a monic poly Ilolllini :1'" 1i,:D" j d" over R such that d j Ef lJ, d 2 , .,. , d, E lJ. 2. WiLl. I.he not:li,ion in (43.20), assume that a maximal ideal m' of U' I,OI'l'IIHPOlllIH 1,0 a prime divisor lJ* of zero in Ft* by the correspondence givIIIl ill Lhl] prool'. Prove th"t the derived normal ring of Ft* /lJ* is the Henselizatioll
-+
-+ .. , -+
of U[n'. 3. Aswme th"t R is a quasi-local integral domain and that the derived I101'm1lJ ring of R is q\lasi-local. Prove that if a Henselian ring H dominates a, i,hen H dominates the Henselization of R. 4. Let V be a valuation ring and let V* be the Henselization of V. ProVl' Lhat V* is a valuation ring, and that every principal ideal of V* is generated hy an element of V (or equivalent,ly, that the value group of a valuation de fined by V* is naturally identical with that defined by V). 5. Let lJ be a prime ideal of a Henselian valuation ring V. Prove that V p is Henselian.
44 . Hensel lemma We begin with the following c:orollary to (37.9): (44.1) THEOREM. If R is a Henselian pseudo-geometric analytically normal ring, then every finite integral extension R' of R is analytically irreducible and is algebraically closed in its completion R ' * (i.e., every element of R'* which is algebraic over R' is already in R'). Proof. Analytic irreducibility is an immediate consequence of (37.8). Let a be an element of R'* whic:h is algebraic: over R'. Let b ~ 0 be an element of R' suc:h that the element ab = c is integral over R'. Then the c:ompletion of R'[c] is R'[C] Q9w R'* by (17.8). The first assertion, applied to R'[C], implies that R'[c] is analytic:ally irreduc:ible, and R'[c] Q9 R'* is an integral domain, which implies that c E R'. Therefore a is in the field of quotients K of R' and a E K n R ' * = R' by (18.4), which proves the assertion. (44.2) THI~OR]<]M. If R is a pseudo-geometric local ring, then the Henselization R* of R is pseudo-geometric.
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1'1'001', LI'I p'" 111'11.11 1/,l'ilill'HI',Y Pl'illll' idl'illol' U+ IllId HI'I. ~I ~I+ K II, iN 1'/IH'y 1.0 rH'I' 11,11.1, I'VI'I',Y lilli!.I' illl,I'gl'HI I'XI,PlIHiOII N** or 1t,*/\J* if-: gl'llI'I'Hkd OVI'I' It'hl* hy a 1111 ',al ,.illg I( whinh dominates Rip and which iH or lillitu type ovcr if)p. Hinee R is pseudo-geometric, the (kl'iv(~d /Ionnal rillg of R' is of finite type over Rip. Let I be the derived normal ring of R**. Then I is a Henselian normal ring and I dominates a ring of quotients J of the derived uormal ring of R'. Thell there is a uniquely determined J-homomorphism from the Henselization J* of J into I by (43.S). Sinc:e J* is a normal ring, our construction of J and I implies that J* = I. Sinc:e J is of finite type over Rip, we see that J* is a finite module over R* Ip* by (43.18), whic:h proves that R* is pseudo-geometric.
(44.3) COROLLARY. Let R be a pseudo-geometric analytically normal ring and let R* be the Henselization of R. Then every integral extension R' oj R* is analytically irreducible and is algebraically closed in its completion. We note that normal localities, over a field or a Dedekind domain which satisfies the finiteness c:ondition for integral extensions, are pseudo-geometric: analytic:ally normal rings, as was proved in §37. This fact and (44.3) will be used in the proof of the following theorem whic:h may be c:alled the Hensel lemma: (44.4) THEOREM. Let (R, m) be a Henselian ring and let J(x) be a polynomial over R in an indeterminate x. Assume that there are polynomials ho(x) and go(x) over R such that: (1) f(x) - go(x)ho(x) E mR[x], (2) go modulo m and ho modulo m have no common root (i.e., goR[x] + hoR[x] + mR[x] = R[x]) , and (3) go i15 a monic polynomial. Thcnf(x) has a monic factor g(x) such that g - go E mR[x], and! = gh with h(x) such that h - ho E mR[x]. Proof. R is a homomorphic: image of a HerUlelian normal ring by (43.11), henc:e we may assume that R is a normal ring. Let I be the ring generated by the coeffic:ients of f, go, and ho over the prime integral domain and let I' be the derived normal ring of I. Then since R is normal, I' is contained in R. On the other hand, sinc:e I is pseudogeometric ring, so is I'. Set P = I~mnI') . Then P is a normal locality over the prime integral domain, whence it is analytically normal by (37.5). Let P* be the Henselization of P and let P** be the completion of p, Then t.here are g and h as in the assertion but with coefficients in P** by (30.4). Since g is monic, the coefficients of g and h
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45. Convergent power series rings Wn Hay t,lmt a ring R is a Weierstrass ring if R is a pseudo-geomd.l'il' ring sueh that, for every prime ideal ~ of R, R/~ is a finil.(· ill Legl'al extension of a regular local ring. We note first the following general property of Weierstrass rings: 11(~IIHelian
(41).1 )IJ ~ is a prime ideal of a Weierstrass ring R, then ~ is anaLyli (:aLLy irreducible and R/~ is algebraically closed in its completion. This follows immediately from (44.3).
On the other hand, it is obvious that: (45.2) IJ R is a Weierstrass ring, then every ring which is a finilr' R-module is a Weierstrass ring.
In order to introduce the notion of convergent power series, WI' introduce the notion of multiplicative valuations. Let K be a field. A map v from K into the set of nonnegative real numbers is called a multiplicative valuation of K if it satisfies tlw following three conditions: (1) v(a) = 0 if and only if a = O. (2) v(ab) = v(a)v(b) for any a, b E K. (3) v(a + b) S v(a) + v(b). We note that if there is an isomorphism ¢ from K into the field of complex numhers, then v such that v( a) is the absolute value of ¢( a) is a multiplicative valuation of K. We note also that if v* is all additive valuation of K whose value group can be imbedded in the additive group of real numbers, then, with a real number c such thai, o < c < 1, the map v such that v( a) = cv*(a) becomes a multiplicative valuation of K. Let K be a field with a multiplicative valuation v, and let Xl, ••• , Xr be indeterminates. A formal power series L a'lj" .nrx;l ... x;'·
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JJ1 Hueh that AI 1'01' {W{~I''y (nt, ... ,nr ). 'I'IIC~ Kt'/. or C'.OIIVCI'f,!;CIl!, power scrios, in indeterminates XI, •.. ,Xt , fOl'lnK a CilliJl'illf,!; of /(l[:X:I , ' .. ,xr]L is called the convergent power series ring ill the variables XI, ... ,Xr over K, and is denoted by ](<<:/:1, ... ,xr». Note that if v is such that v(a) = 1 for every a (rOO) of K, then K«XI, .. , ,xr» = K[[XI, '" , xrlJ. We note that, as is easily seen, an invertible linear transformation of the variables defines an automorphism of K «XI, ... ,xn ». 10 I')
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(45.3) THEOREM. Let K and XI, . . . , Xr be as above. Let f = L adl".drX~l ... x~r be an element oj K[[xi , ... ,xrlJ such that aO"'Oi = Ojori = O,l, ... ,n - 1 (n? 1) andsuehthatao".on ~ 0. Thenjor every element g oj K[[XI, , .. , xr]L there is a uniquely determined element q oj K[[xi , ... ,xrlJ such that g - qf is a polynomial in Xr oj degree at most n - 1 with coefficients in K[[xi , " . , Xl'-I]]' Ij f and g are convergent power series, then q E K «XI, ... ,Xl'» and g - qf is polynomial (oj degree at most n - 1) in Xr with coefficients in K «XI, ... ,Xl'-I»' (WEIERSTRASS PREPARA'l'ION THJ~ORlTIM) Proof. Set R = K[[XI, ... ,xrlJ. Since Rj(L~-1 xiR + jR) is a K-module generated by 1, Xr , ... ,X;-I modulo L~-I xiR + fR, it follows from (30.6) that R is a module over K[[xi , ... , Xr-l , IlJ generated by 1, Xr , ... ,x;-\ which proves the existence of q. The uniqueness may be proved easily by induction on r (considering RjXIR if r ? 2). But for convenience in proving the last assertion, we give a more complicated proof of the uniqueness. (Also the existence may be proved in this fashion.) We may assume that aO"'On = 1. Set g = L bd,,,.d,x~' , .. x~r, q = L Qdl"·drX~J ... x~r. Then, since g - qf has no term of degree greatrr than n - 1 in Xr , we see that
We prove the uniquenestl of qdl".d., by induction on W(qdJ"'d,) d r + (n + 1)· L~-I d i . If W(qdl."dJ = 0, then all d i are zero, and the ahove condition implies that bd, ,, ·dr_l (d,+n) = qdl" .d, hecause ao".o = .. , = aO·"O(n-l) = 0, aO"'On = 1. Assume that qd1·"d, such that W(q,zl·"d,) < t are unique and consider a qdl."d, such that W(qdl"'d,) = t. Then, the condition
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(IL[ , ' , , ,dr ), Let N, y*, z* he pOl:litive real number::; sudl (:{M/z") 1, zz* ~ (3M/zn) 1, l/yy* ::;; 1 (I (zz*) -(r.-l,,»)-I/(r-1), We want to prove that V(qdl".d r ) ::;; NI/*",·I· .. ·,{r-'z*d r by induction on w = W(qd,,,.d r ) = dr (n 1) X L~I d,;. If w = 0, then we have V(qd,,,.dr) = V(bd1"'(dr+r») < At /Z" < N, and we consider the case where w ~ 1. By the equality hd,,,.dr_ddr+n) L a(d,-itl"·(dr_,-ir_,)(dr+n-ii)qi,,,·ir' we sec that 1'01'
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Thus we have V(qdl".d r ) ::;; Ny*dl+".+dr-lz*dr, and the assertion IS proved. As a corollary to (45.3), we have the following lemma: (45.4) LetJ(y) and g(y) be monic polynomials in an indeterminate y over K «Xl, ... ,xr». A ssume that f( y) = yn + Cly"-l + .,. + Cn with Cl, ... ,Cn E 2:xiK«Xl, ... ,xr».lfJ(y) is afactm> ofg(y) in K«Xl , .,. ,:cr , y», then fCy) is a factor of g(y) in K«:Cl, .. , , ... ,Xr»[Y]· Proof. Let h E K «Xl, ... ,X,. , y» be such that g = fh. On the other hand, let q, s E K «Xl, ... ,xr»[y] be such that g = qf + s and sueh that s is' of lower degree than J (in y). Then this q is the one in (45.3), applied to our J and g in K «Xl, ... , Xr , y». By the uniqueness in (45.3), we have h = q, whence r = and the assertion is proved.
°
(4S.5) THEOREM. With the same K and the Xi as above, we have: ( 1) K «Xl, ... , x r » is a H enselian regular local ring of altitude r, whence the completion of K «Xl, ... ,xr» is K[[XI , ... , x r]], and (2) if a is an ideal of K «Xl , ., . ,xr », then K «Xl, . . . ,Xr» / a contains K «Yl , ... ,Ys» with suitable variables Yi and is a finite module over K «Yl , ... ,Ys». Here, iJ K contains infinitely many clements, then the Yi can be linear combinations oj the Xi with coefficients in Ii.
(N ORMALIZATION THEOREM FOR CONVERGENT POWER SERIES RINGS) Proof. If K contains only a finite number of elements, then v( a) = 1 for any a T" 0, and K «Xl, ... , Xc» = K[[XI , ... ,xr]] in this ease. Therefore the assertion in this ease follows easily from the theory of complete local rings. Thus we assume that K contains infinitely many elements. Since K «Xl, ... , Xc»/ 2:;+1 XiK «Xl, ... ,Xr» K «Xl, ... ,Xt», we see that altitude K «Xl, ... ,Xr» ~ r. There-
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(45.6) COROLLARY. Assume that I( is a perfect field. Then any riuy which is a finite module over a convergent power series ring in a finil(' number of variables with coefficients in I( is a Weierstrass ring. EXERCISES. 1. With the notation of (45.3), prove that there are a mOllie polynomial f* = x; + CIX;-l + .,. + Cn with Ci E L:~-l xiK«XI , ... , xr-J» and a unit q in K«XI , ... , x,.» such that f* = qf. 2. Let a local ring R be a ring of quotients of a complete local ring. Prov\, that the Henselization of R is a Weierstrass ring. 3. Let K be a field of characteristic p ,c 0 and let x, , .,. , Xr be indeterminates. Prove that K«XI , '" ,xr» is a Weierstrass ring if [K:Kl'l is finite. 4. Let I be a field or a pseudo-geometric Dedekind domain and let R be u, locality over I. Prove that the Henselization of R is a Weierstrass ringo 5. Let R be a Weierstrass ring which has no nilpotent elements, and let R* be the completion of R. Let a be a given element of R*. Assume that ·for every minimal prime divisor p* of zero inll*, a modulo p* is algebraic over Rj (p* n R). Prove that a is an element of R.
46. Jacobian criterion oj simple points We begin with a remark on derivations of a local ring:
(46.1) Let (R, 111) be a Local ring and let R' be a subring oj R. As-
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M (~f' h' III'w'mll's II I'il/[/ N" /Jill'/, N' NI/I'''' I./UlJ U/Iil = N" I ( III n h~"), I,I'! fl , ... , fJ' 1)(' n 1)(1,81:,~ fol' III. If I), J)' arc derivations of !l,)/J!'/' N' 8111'h tha/' Ihn o~, /)'JII, for aLL 'fYI, ( M and such that DJi = n'l,: fol' I'})(T!/ 'i, then Wi: haveD = D'. P/,ool'. By Uw ddinition of derivations, we may assume that D and NII/I/I' llilll (/ 8,(1),'01'1
0' are integral derivations of R, whence Dm n ~ mn-r, D'mn ~ mn - 1 • Since every element of R is expressed as a power series in the Ji with coefficients in R", the above result implies that Da = D' a for any a E R. Let a be an ideal of a ring R and let D be a derivation of R. Assume that there is an element d of R, which is not a zero divisor modulo a, such that dD is an integral derivation of R and such that dDa C a. Then, denoting by ¢ the natural homomorphism from R onto Ria, we can define a uniquely determined derivation D' of Ria such that D'(¢(x)) = ¢(dDx)/¢(d) for every x E R (and D' is independent of the particular choice of d). The derivation obtained in this manner is called the derivation induced in Ria by D. When K is a field of characteristic p T" 0 and if 1(* is a subfield of K, then elements Zl, ... ,Z" of K are said to be p-independent over K* if [K*(K1') (Zl, . . . ,zn) :K*(K1')] = pn. (46.2) Let Xl, ... , Xc be indeterminates and let K be a field with a multiplicative valuation v. Let A be anyone oj K[Xl' ... , Xr], K[[X I , . . • , Xr]L and K «Xl, .. , ,Xr IJ K* is a subfield oj K such that [K: K*] is finite and iJ a is an ideal ( T" A) oj A, then every derivation oj Ala over K* is induced by a derivation oj A over K*. Proof. Assume first that K is of characteristic p T" O. Then every derivation of A or of Ala is a derivation over K1', and therefore we may aSSllme that K* containi:l K1'. Let Zl , ... , Zn be p-independent elements of Kover K* such that K = K*(ZI' .. , ,Zn). Let Di be the derivation of A over K* such that DiX j = 0 for any j and such that Diz i = 1, Diz j = 0 if i T" j. Let D' be an integral derivation of Ria over K*. Set = D'Zi and v~ = D'xj , where Xj = X j modulo a. Let Ui and j be representatives of and in A and set D = UiDi vjalaX j • Then D induces D' by (46.1). Thus every integral derivation of Ria over K* is induced by an integral deriva-
».
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u;
v;,
tion of Rover 1(*, which proves the assertion in thii:l case. Assume now that K i::; of charaeteristic O. Then K is separably algebraic over K*, whence we may assume that K = K*, and we call prove the assertion in the same way as above using ollly the partial derivations aI aX i .
Willi 1111'11/11111' Illlinlil'll Ili\ IIIHII'I', il'II' .,' '/'" 111'1' 1,11'1111'lillllIl' ,I, 111'111.111' IlInll'i\ (1),/, ,I, n,\',,) Ii 11/111 k 1'111' 1·IIII'H:J.IIlI.i 1'111' I'lilillllllll) '\ /'11.111'.1 II. lIIi,I'I''{ ,/11/'11/1/'1111 11I1I11'i.1' /11111 iN 1I1'lllIkll by ./1'1.1'1 , ... ,I"" .;+), NIIII' I Ii II I ,/'I'lfl, '" ,f", ; /\'+) i,'1 IIllilPIP lip III Ii I 11'/1.1' 11'11111/ 1I1'lllnl,illl1C4 111111 11111,1. ./"1'(/1 , ... , Jill; /\) iN II .I:l.I·III,ill.ll 1II:I.l.l'i~. WI' 1'1 II Il-liill'l', 1'1'0/11 III1W 1111, 1.1i(' 1·:l.HI~ Wlll'l'p UII'Ii 1'111'111 II, l'II,HiN 1'111' 1,1'1, pili' II, Pl'illll' diviHol' III' n alill II'!. q Iw II, pl'illw idpal I'lllll.aillili/,: I. HI'I, h' ,1'1' Ily I.lil' 1IIII'Ill:Liiy,aLioll LIIcorcmH, WI~ HI'(', 1.1111,1, LiIl'i'I' 11,1'1' '111'i:d,II'H fll , '" ,fl" Hili'll LIIaL I1/q iH a filliLe iIlL('I!:r:Li (~XLI'IlHioll of 1.111' illg .1' wllil'li iH III' Llil~ HallW j,yP(~ aK 11 11IlL wiLh val'iablnK /Ii. NIIW WI' ,eli'I'I'I, I. Ii:l I. :
(Hi.:I) TIII':IIIU':M. IVlwn A' can be chosen so that A/q is 8wpam/d(' ,I', Ih('11 h'/nH '£8 re(fular ~f and only if rank (JUI, ... ,I",) II/nill/o q) IlI'il!:li I, p. When j( is (~f characteristic p ~ 0, then UlaN 8 N'UII/II/, if rli/d onl!! ~f there is a subJield [(* of K such that [K:[(*\ is illil,' (/1/(/ ,'uwlt. Ihat rank (.1*(/1, ... ,fm ; K*) modulo q) = height. p. 1'/'11111'. WI' 111~l!:ill with a very special case where a = p = q. In Lliil' 11/·11' h'/llN iK I\, field, whence it. is regular. Assnme first that A/q i:: I/'('/,
1'l'lIl'lIl"I' IIVI'I' /1'. (46.1) and the last half of (:'~9.4) imply t.hal A', WhClll'I' 1'1 LillI!: .1'; X; modulo q, we see that the set of vectors (DXI , ... , .. ,n.I',) with j) c:: \[Jer((Alq)/K) is a vector space of dimen::;ioll , /11'1'1' 1{/qN. Hillee .lUI, .. , ,jm) modulo q is the matrix of I:oefli· 'il'III,'4 III' liIH'a!' eqnations of (DXI, , .. ,Dx,.) by virtue of (46.~) I'i'. (;W.:I)), we ::;rethat rank (JUi, ... ,jm) modulo q) = r - s = II'il!:liL q, which Hettles this case. Assume next that K is of character,Iii' Ji / O. We need the following lemma: (1(;.1) Set U = RlqR and let L be a field such that A' C L ~ I/. ,r'/ 1\* /)(' II 81tliJield oj K such that [K:K*] < 00 and such that K P c K*. ~'('r( (,l/q)ll\) iH I!:cllerated by the partial derivations of
,is a subfield K** oj K* such that [K: K**] < 00 and such \[Jer(L'IL**) = length \[Jer(LIL**), where L** is thejielri :!' (I"olieut:; (~f A ** which is the ring of the same type as A with vari1/1/"8 IIi' , ... , y7: over K**. 1'/'1'01'. Wn use induction on [L':L]. If [J'/:L] = 1, then the assertion H ol,violiK. Assume that L' ~ L. If a is an element of L' whieh is not
"!t('11
Ih(T(~
//(1/ 1I'III!:I.h
II I"~ tll<'ll, by our induction, there exists a field '\: 1\**\ < (0) such that
K** (J(**
c K*,
kngth \[Jer( L' I L**) = length \[Jer( L( a) I L**).
I'
rI
IH Hcparable over
L, then length \[Jer( LI L**)
length
('11111"1'1':11
'fl('1'( /,(0
1//'++) II,\' (:;1).·1 I, !Iolld 1.11('
('11,,'1\'
aHHllllIl' LlIII.I. // iH pll),l'ly iIIHI'P:!.I·::.hl(' OVI'I' I"~
I \)'i
V II
iH HI'I.CII'ti. 'I'IIIIH WI' IllII.y
I,. LI'i.
/I., :I.,"
ahovp, he Hill',h
(I U'. Wp lIlay aHHllUW that a E Ajq. Ld .'1 * :Llld 1/1' h(1 HU(',11 aH ;1 ** and L**, reHpeetively, in the case where /\** = 1\*. 1£ a" c: £*(£1') = LP(K*), then, since a P EE L P, we P C K** C K*, [K:K**] < (0) HI,C Lhat there itl a field K** (K P P Bueh that a EE L**(L ) = LP(K**). Replacing then K* with K**, we may assume that a P EE L*(L P). As was noted above, there is a field K** (K** C K*, [K:K**] < (0) such that length 'IJer(£1jL**) = length 'IJer(L(a)jL**). Since a P E L, a derivation D of L has an extension D" to L(a) if and only if D(a P) = 0, and when that is so, D" a can be assigned arbitrarily in L( a) by (39.:3). Hence I hal. (/."
I I,.
Hilll'l' (/,
II
/1,"
length'IJer(L(a)jL**) = 1 +length'IJer(LjL**(a P)). Since a P EF L*(L P ), we have aP EE L**(Lp) and length'IJer(LjL**)
= 1
+ length 'IJer(LjL**(a
P
)).
Thus length 'IJer(L(a)jL**) = length 'JJer(LjL**). Therefore length 'IJer(l/jL**) = length 'IJer(L(a)jL**), which implies the required equality. Thus (46.4) is proved completely. Now we go hack to the proof of (46.3). Let K**, A** and L** be as in (46.4), applied to the case where L is the field of quotienti:l of A' and K* = K. Let Z1, ... ,Zt be p-independent elements of K over K** such that K = K** (Zl , ... ,Zt), and consider the set of vectors (DXl, ... , Dx r , Dz 1 , •• , ,Dz t ) with D E 'IJer(L'jL**). Since length 'IJer(Lj L**) = s + t hy the proof of (4().2), we have length'IJer(L'jL**) = s + t, and therefore the rank of J*(fl, .,. , . .. ,fm ; K**) = (r + t) - (s + t) = r - s = height q. Thus this case is settled. l'\ext we consider the general case. Let gl , . " ,gn be elements of q such that a + L giA = q. Then the above reHult shows that J (f , ... ,fm, gl,... ,gn) or J* (JI , .. . , 1m, gl,... ,gn; K**) modulo q is of rank equal to height q. We treat here the second case, because the first case can be treated as a special case where K = K**. ASi:lume first that RjaR is regular. Then there is a regular Hystem of parameters Ul, ... ,Ub of R such that U1 , • • • ,Ua generate aR by virtue of (25.18). On the other hand, since ~ is a prime divisor of a contained in q, we have aR = ~R. That rank (J*(h , ... ,1m, gl , ... , ... ,gn ; K**) modulo q) = b implies that rank (J*( Ul, . . . ,Ub ; K**) modulo q) = height q = b, hence rank (J*(Ul' '" ,ua ; K**) mod-
1II':NIII'lIoi IN IIINIIII .II~II 11'1'111>11111'1'11.11111 HINIIII
lWj Ido II I
II
L';
II('if~liI,)1. Hilll'('!l1i' 1I,h', \\,(\ 11111'11 1'lI.ld, (./i'U', , III0dido q) IlI'il1:1 II, p. (~IIIIV('I',"I'ly, IJ.Ht1111111· I.IIHI. 1'/1,111-
'" ,I,,, ; 1\++)
(./+1/" ... ,J",; /{H) lillitililo q) lI('i~hl.).I, WI' Illay IJ.H,'IIIIIII' I.IIHI 1'I111k (./+1/" .,. ,f" ; /I.H') lillitildo q) 1/ IIl'iglll.).I, AHHIIIII(' 1.1111.1. 1'1, . , , ,I'" 11.1'(' (·I(·III(·III.H Ill' H HIII'Ii 1.1111.1. ('J, ( q::u, '1'111'11 fill' (,"(,I'y illl('gl'lI,l dl'l'ivn.l.ioll /) or ,I, /)( ('J;) ( q/( 1~1I1, f)( ('.I,) J,/>I'; I (';I~/'; I:;f~j'.i (Illotildil qU), Hill('(' .I'~UI , '" , , " ,f" ; 1\ H) Ill()dlilo q iH of milk nqllal j,o ft, WI' llav(~ (:,: I qU 1'01' ('v('t'y i. 'l'Il<'I'nl'ol'(~ 1.11(,1'1' ('xiHI.H a ]'(~glllar HYHl.nm or ImmIlW(.('I'H or H wlli('11 ('Olll.n,iIlH/I, , .. ,f" aH It HnbHd., whell('.e HIL~I,R iH I'l'gulal'. height.).I, we have ).IH = I;R whmwe all ).IH, Hlld Hill('p 11. It'/nN iH n·gulal', ThuH the proof it; completed,
L
L
L
L
L';
L
L't
(In,r;) COI{()LLA tty. f"el 1 be a Noetherian valuation ring with 1/ clenwnl p and let Xl , . , , ,Xr be indeterminates. Let A be eith.1'1' /l,rl , ' ., ,;(:rl or I[[XI , ' , . , x r]], and let II , ' ., ,Jm be elements oj ii, 8d a = L JiJl, let ).I be a prime divisor oj a and let q be a prime id(:a! I(j' ;1 r:onlaining ).I pA. Set R = A q • Assume that ).I n I = 0, IJ mil k (./ U, , ... ,f",) modulo q) = height ).I, then RlaR is a regular loca/' 1'1:t/.U in which p is a member oj a regular system of parameters. 'l'hl' WfW(TSC is true if AI q is a finite separable integral extension of a rin(/ /\' wh.ich is either the polynomial ring or the power series ring in soml' /lil.riablcs over IlpI. On the other hand, applying (46,:3) to the case where a = ).I and ;1 1).1 it:) a purely inseparable extension of a ring A" which is of th(' Harne type as A, we sec, at first that rank (J*(fI, .. , ,Jm ; K**) ]l'r£1I/.1:
+
lIlodulo ).I) = height ).I, whence:
( 4().6) COROLLARY, If JI is as in (46.3) and if A is pseudo-geometric, then A satisfies the condition Jor I in (40.3), Note that A may not he pseudo-geometric only when A ii:l a COllvergent power series ring over a field K of characteristie p ~ 0 slwh t.ha 1, [K: K P ] = 00 (cf, Exercise :) in §4!)) , (46,7) COROLLARY. A complete local ring satisfies the condition for [ in (40.:3). EXERCISES. 1. Let XI , ... , x, be indeterminates and let R be a ring. Prove that the partial derivations ajaxi generate mer(R[[xI , ... ,xrll/R), 2, Let D be a derivation of a Zariski ring R. Prove that D can be extended uniquely to a derivation of the completion of R. 3. Let K be a field of characteristic p r" 0, K' a finite algebraic extension of
11111
1 'II A l"I'lillC, V II
Ii, 11,111111'1,1\+ hll II. illiI"il.ld or Ii KIII"1 111111. Iii :/i'I'1 iH filli!.I', I'I'OVI~ 1.1111.1. (,here iH a KliI,fi.,ld 1\ 'H oi' Ii" HII"I> I.ltnl. IIi :/i""1 < en Hlld HIII,h Llml. Inllgth ':Der:(K/K**) ,~ 11'111':11, 'fil'l'(/i' /Ii""'),
47. Analytic tensor product Let K be a field with a multiplicative valuation v. We say that a ring R is an analytic ring over K if there are indeterminates Xl , ... ,Xn and an ideal a of K «Xl, ... ,xn » such that R r--..J K«XI , .' . ,xn»/a. The normalization theorem for convergent power series ring implies that a local ring R with coefficient field K is an analytic ring over K if and only if it is a finite module over a subring which is a convergent power series ring in a finite number of analytically independent elements over K. An analytic integral domain Rover K is said to be analytically separably generated over K if there is a system of paramet.ers Xl , ... , ... ,Xr of R sueh that R is separable over K «Xl, ... , X,». This definition is applied also t.o complet.e local int.egral domains wit.h a basic field considering K[[XI , ... , Xr II inst.ead of K« Xl , ... ,x,.». Let. Rand R' be analytic rings over K. Then there are indet.erminat.eB Xl, .. , ,X n , YI , ... ,Ym such t.hat R r-v K «Xl, ... ,xn»/a, R' r-v K «YI , ... , Ym»/b wit.h suitable ideals a and b. The analyt.ic ring K «Xl, ... , Xn , YI , . . . , Ym» / C, wit.h t.he ideal C generat.ed by a and b, is uniquely determined within isomorphisms by Rand R'. The new ring is called t.he analytic tensor product of Rand R' (over K) and is denot.ed by R c§ R'. One sees immediat.ely t.hat: ( 47.1) The complete tensor product R ®K R' is the completion oj
R c§ R'. In order t.o investigat.e analytic t.ensor products, we prove some result.s on complet.e t.ensor product.s over a field.
(47.2) THEOREM. Let (R, m) be a complete local ring with a basic field K and let (R', m') be a complete local ring which contains K. Let A be the totaL quotient ring oj R'. Then R' ®K R is naturally identified with a subring of A I§ K R. Proof. Since K itl a field, Q9 K R is exact, and R' Q9 R k:: A Q9 R. Let n he the Jaeobson radical of A. Since n nn = 0, we see that, for eac:h nat.ural number m, there is an n(m) such t.hat. nn(m) R' c m,m by (30.1). It. follows that. t.he t.opology on R' Q9 R induced by A ® R is st.ronger t.han or equal t.o t.hat. induced by R' ~ R. Therefore there is a natural homomorphism from t.he closure B, of the
n
1';1'111'1'11,1 I,d Ily /r" IIlld 11'111,1 (") H, 111111 Ii" ("j /r', Hilll'l' II Ifill. IllI.Hil' lil,ld IrI' N, h' III" iH II lillik I":' III()dlil(', \1'111'111'1' Ii" (") (1r'!11I") ii, II, ('lllllpll''''' ,:I'llli 1111'11,1 I'ill).';, Hlld, ::illl'(' /{/III"U H' (") (h~/III"), I! III"/{ iN l'lllllpll'll', Ii iH 1'lIlllpll'/'(' IlIldl'l' 11i/llldi(' /'lIpIIIII/-!:.Y, alld 1111'1'1'1'111'1' f( iH a ('lllllpll'/'(' HI'lili-lol:al I'ill).';. 'l'h('l'l'r(II'I', by IIII' d<'lilliCilil1 111':1, 1'lllIlpl('/'(' II'IIHII}' III'IIIIIII'/', Chl:l'!' IIlUH/' I){' a lIa/'III'alhlllllOIIlOI-phiHIIl 1'1'11111 /r" <Xl Ii Oli/'O 11. 'l'lrNdol'<: WI: Hnn Lhaf, n iH lIaLlIl'ally idl'lILified wi/'lr h" (x) H, whi('h 1',liIllplnf,('H L1w Pl'I)oi'. IlIiI 11'11 II',
(,17.:n 'I'm;OI(lolM. A ""ume lhat a complete local integral domain H is (/'//,Il'/lIhcnl/!/ 81']lara/II// {/cncmled over its basic field K. II Il' is a com111('fr' locaL inte{!ml domain such that K is separa/Ily aLgebraically closed ill Ilu, jield ((/' (IUolicnls L (~l Il', then il' ~ K R is a complete local inteural t/uuUliin. Pl'oof. We may assume, by virtue of (47.2), that R' = L. Let ,I'I , ' .. ,x" he a system of parameters of R and let c be an element or H hueh that c is separable over K[[Xl, ... ,xrll and such thaI, 1\11.1:1 , ... , xr]][cl has the same field of quotients as R. Let f(X) be tlw il'redueible monic polynomial for cover K[[XI , .. , ,xrll. We have oilly to show that f(X) is irreducible over L[[XI , ... ,xrll. Assume Lhe eontrary and let g(X) and h(X) be monic factors of f(X) such ihat f = gh. Coefficients of g and h are integral over K[[XI, '" , xrll (beeause so is c), whence they are in the integral closure of K[[XI, ., . , ... , xrll in L[[Xl , ... ,xrll. By our assumption on L, we see that the eoefficients of g and h are in a suitable purely inseparable extension of K[[XI, ... ,xrll. Since c is separable over K[[XI, .. , ,xrl], f( X) is irreducible over any purely inseparable extension of K[[XI, ... , xrll by (39.9), which is a contradiction, and the proof is complete.
rr
(47.4) COROLLARY. Ii is an algebraically closed jield, if R is a complete local domain which has K as a basic field, and ~f R' is a complete local integral domain containing K, then R' i§ K R is a complete local integral domain. The above result yields the following by virtue of (47.1) and (45,6): ( 47.5) TUEORlDM. Let K be an algebraically closed field. If Rand R' are analytic integral domains over K, then R ~ R' is an analytic integral domain. Next we prove some lemmas. (47.6) II R is a Noetherian normal ring and if Xl , , X" are indeterminates, then R[[XI , ... ,xnll is also a normal ring.
~)()
l'I'OIIi', WI'
1111\'1' Illdy ell
pl'll\'1'
1,111'
1'1I~H'
11'11('1'1'
II
I,
()II
(.111'
I
oL\Wl'
(lI,h'I' \\,111'1'1' P1'11111' 111'('1' all WillI(' iil('aIN Ill' Jwight L, whence n"h\,II,I'III. Nilll'n It~ iN a vallla( iOIl ring, IIp[[xill is a regular
111I.1ld, Ii
Nil,)"" I
11l1'al I'illg, 111'11<'(' iN a lIorma! ring by (25.14). Therefore R[[Xlll is II o I'IIm I, wllieh proves the aSBertion. (47.7) Let il and il' be semi-locaL rings containing a common .field K and such that R* = R ig) K R' is well de.fined. If fER and if.f' E R', then fR* n f'R* = ff'R*. Proof. There is a linearly independent base of Rover K cont.aining such one of fR and the same is true for R' and fR'. Therefore we sec easily that f(R Q9 R') n f'(R Q9 R') = fJ'(R Q9 R'). Let rn and m' be the ,Jacobson radioalB of Rand R', respectively, and apply the above faet to R*/(mnR* + m,nR*). We obtain th:1t (fR* + m"R* + m,nR*) (.f'R* + mnR* + 111,nR*) = ff'R* + mnR* + 111,nR* for every natural number n. Therefore f R* n J'R* k n,,(jf'R* + 111 n R* + m'''R*). This laHt intersection coineidcs with If'R* by (16.7), hence JR* n f'R* k .f.f'R*, whic:h provetl the aBBertioll. Thl' tlame proof giveB alBo the following lemma: ( 47.8) Let Rand R' be anaLytic rings over a field K. If J E Rand 4 f' E R', then, setting R* = R @ R', we have fR* n f'R* = fJ'll*. N ow we shall prove the following theorem:
n
(47.9) THEOREM. A ssume that Rand R' are analytic rings over a fieLd K. If Rand R' are normal rings and are analyticaLLy separa.'JLy generated over K and if R @ R' is an integral domain, then R ~ il' is a normal ring. Proof. Let b be an arbitrary clement of the d(~rived normal ring of R @ R'. Let Xl , ... ,Xr , C E R be such that the Xi form a ~3Ystem of parameters of R, such that R is separable over K «Xl, ... ,xr » and such that R has the same field of quotients as K«XI , .. , ,xr»[c]. Let. d be the discriminant. of the irreducible monic polynomial for c over K «Xl, ... , x r ». Since K «Xl, ... ,Xr » @ R' is a dense subspace of R'[[XI , ... ,xr]L and since this last ring iB normal by (47.G), we sec that K«XI, .. , ,xr » Q9 R' is normal by (18.4). Since b iH in the derived normal ring of (K«XI, '" ,xr » @ R')[cj, we see that db is in (K«Xl, .. , ,xc» @ R')[c] by (10.1[)), whence db E R @ R'. Similarly, there is an clement d' ~ 0 of R' Hueh that d'b E R @ R'. Then dd'b E d(R @ R') d'(R @ R'). This last intersection coincides with dd'(R ~ R') by (47.8), and t.hercf'ore we sec that b E R @ R', which completes the proof.
n
III"NIII,:IdIN
IIINIIII
INfI
( l'i,IO I (' 1"1,,1,1,11/\, I/' Ii' olltl I." IIIW'/II'lIimlll! I'ItJ"Wti.fil'ir/
I~,
1/11'11
11'1':11':1111'1'1111111 IIINII!I
11/'/' 1I11/,lllItill/lltillli,' /'II/(/N III'/'/'
II' (.) N' iN olNo
Oil
1//11'/1/(;1,
I':\I':I/,I'IHI'III, I. 1"'1. h' hI' II 1'''"lpII'I,I' 11)('111 illl'!'1':1'1I1 dOll"Lili willi II, hll.:ri .. Iil'ld II, 1'l'IIvI' 1,1111.1, 1,1,,' followilil': fOil I' I'IlIldil,ioIIH 11.1'., !'llIliv:dl'lli, 1,,, I'II.I'1i III I",,': (i) '" i,'4 II.II:dyl,i"ally H"pII.I'lI.bly 1':1'111'1'11.1,1'<1 OVI'I' II, Ui) 1,'01' Hlly lil'ld II II'lii,'1I iH I\. pOI'!'ly ill,'4.'pal'lI.bl<- 1',j,I'IiHioli of I';, Nw" I';' I
iH ,1.11 illLI'I':I·1I.1 dOlllai II , (iii) 'I'll" 1i'lIgt.1l of 'TIl'l'(N/li) OVi'1' Chi' fi"ld of IIIIOI,il'lIl,H 0)' '" ('Oilll'idI'H wiLli 1I.1j,iLlldl' N. (iv I 11'01' allY ('olllplPi.1' IO"ld iliLngral dmn:LiIi H' I'olll,aillillg I';, 1,111' I'illg Il 0" Il' ImH 110 lIi1pot.'~IIL elemclll. eXI',,]>L zel'o. :~. Lot. I~ be ILIl 1Lna.lyt.ie rillg (lver a field K. AH~lInle t.hat. U iH a.Il,L1yt,ieaJly il'l'.~dil.:ihl(~ (:I,H it, iH when j( is pe]'l'ect.). Prove t,hai, Il iH Illlalyt.ieidly Hopambly gl'lll'l'aL(~d over K if and only if l.hc completion of R i~. :1. LoL f~ bc 11 complete lo(:nl ini,egml domain with a basit: field Ie Prove t.lla!. Lile following two conditions are equivalent t.o each other: (i) fl is l1nalytically separably generated over K and K is algcbmicaJly (:1088d in the field of quotients of R. (ii) leor any complete local integral domain R' containing K, the ring R 0K Il' is a complete local integral domain. 4, Let R be a complete local integral domain with a basic field K, (i) Prove that if R is analytically separably generated over K, then R is separably generated over K. Prove also that the converse is true if [K :K"] is finite, where p is the characteristic of K, Show by an example that the above converse is not true in general. (ii) Prove that if R satisfies the conditions in Exercise 3 and if K' is a field containing K, then R 0K K' satisfies the same conditions with K' instead of K. 5. Prove (47,6) without assuming that R is Noetherian.
Appendix AI. Examples oj bad Noetherian rings 1. A Noetherian ring whose altitude is infinite. Let K be a field and let Xl , ... ,X n , '" be infinitely many algebraically independent elements over K. Let ml , ... , mi, '" be a sequence of natural numbers sueh that 0 < mi - mi-l < mi+l - mi for every i. Let lJi be the prime ideal of K[XI , ... , Xn , ... J generated by all the Xj such that mi: ~ j < mi+l , and let S be the intert;ec:tioll of complements of lJi in K[XI , ... ,x" , ... J. Then R = [([Xl, ... , ..• ,X n , ••• Js it; the required example. Proof. R~il? it-l a Noetherian rillg of alt.itude mi+l - m.; , whenee it is obvious that altitude R = 00. That R is Noetherian follow'=! from the following lemma: (ELl) Let R be a ring. Assume that: (1) if 111 is a maximal ideal of R, then Rut is Noetherian and (2) if f is an element oj R which is different from zero, then there is only a finite number oj maximal ideals of R which contain f. Then R is Noetherian. Proof. Let a be an arbitrary ideal of R such that a ~ O. By our assumption, there is only a finite number of maximal ideals which eontain a; let them be 1111 , •.. , mr . There are finite number of elements al , ... ,as of a such that there is no maximal ideal of R, other than the 111£ , whieh eontains all the aj. Sinee eaeh R1I1i is Noetherian, there is a finite number of elements a,+l , ... ,at of a which generate aRmi for every i. Then L aiRm = aRm for every maximal ideal 111 of R. Therefore LaiR = a by (8.9). Thus a has a finite basit; and (EU) is proved. We note by the way that the above result shows the following fac:t: (E1.2) Let (R, 1111, .•. ,111r) be a quasi-semi-local ring. If Rmi is Noetherian for every i, then R is Noetherian. EXAMPLE
EXAMPLE 2. A local integral domain oj multiplicity 1 which is not regular, which is at the same time an example of the chain condition Jor prime ideals. We first prove a lemma:
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8.
I'roo/'. II. iH ol,viollH IhaL 8 iH a rillg. 11'11/ ( j, 1.11('11 1/( I I 11/) I 1/1' 1'01' HOIII(' /II.' or j, which illtplicH UtaL j iH t./II' IIl1il/lI(' 11I1I,xilllnl jllt'al or 8, J.d, ('I , '" ,t( bc oI(~nwIII,H or U WhON(~ I'CHi 0) be algebraically independent elements over K(x). NI'I, ZiJ = (z.; - Lk<j aikxk)/Xj-I, Furthermore let YI, ., . , Ym (t.heI'(~ lIIay be \lOIIe) be algebraically independent elements over K[:c, ZI, , , . , ... , zl'l, Let RI be t.he ring generated by all t.he Zij and x over K and Nn(. N~ = ]l1[YI, , .. , Ym]. TI\{~II :dl l is a maximal ideal of RI , because XZi,j+1 = Z'j - aijX, 1t(,1I1'1~ Z'J C .TR I , Therefore (R1)XR, is dominated hy K[[x]], hence itl :t. 10m/ rill{l; which may not be Noetherian, Therefore (R 1 )X{fl is a I, lI':d rillg by (:)1.5), Let. m be the ideal of H2 generat.ed by x, Yl, ' , . , , , . , !/I" ' 11t i" a maximal ideal of height m + 1 and V = (R 2 )1lt is a 1'l'gli/ILl' lo<:alrillg of alt.itude m + 1, becanse V is a ring of qnotients 01' Ihn r-.;()(~therian ring (RI)XR,[YI, . , , , UrnJ. R1[L/r] = K[.r, l/r, ZI, , , , , 1
.\ l'I'lilNIII.\
,.:',1,111'111'1' IIII' idl'lI.llIlIl' N~: 11:1'1 II 'l'ti.l.1 'II ".v.l' -- 1,.:'1, , .. ,,:',., /11, ••• , .. , ,/1", iN /I, 1111I);illill-l illl'alor N" alill 1111: rillg IV (N~)II i:-; a J'(:gul:l,I' IllI'al rillg or :t,ij,il,IlIlp.,. 1-///, I. Ld.,'-\ lip Lhp iIlLm':-;c<:LiOIl or Lhe <,otnpl<'1I1I,1I1:-; or III alld II ill Hz alld:-;et il' = (R 2 )s. The maximal ideals of
+
H' arc 1ll/~' alld nU', and it holds that V = R~R" W = R~ R' • Therefore R' il:) Noetherian by (E1.2). Let i be the Jacobson radical of R' and set R = K i, Since R'/mR' = R'/nR' = K, we see that II is Noetherian and R' is the derived normal ring of R by (E2.1). Thus the ring R is a local ring of altitude r m 1 with maximal ideal i. The derived normal ring R' of R is a finite R-module, has two maximal ideals mR' and nR', and i = mR' nR', Furthermore R~tR' and R~R' are regular local rings of altitudes m 1 and r m 1, respectively. Thus the chain condition for prime ideals is not satisfied by R, whence R is not unmixed. Furthermore j.t(i) = j.tOR') = j.t( iR~R') = j.t(nR~R') = 1. Thus R is an example of a non-regular local integral domain of multiplicity one. We note here that: (1) if m = 0, then the first chain condition for prime ideals is satisfied by R, and (2) if m > 0, then R does not satisfy the first chain condition for prime ideals. Proof. Assume that m = 0. Then mR' is of height 1, whence there is a one to one correspondence between prime ideals of R~R' and those R = q, Since of R such that q' corresponds to l:l if and only if q' R:tR' is regular, the chain condition is satisfied by R~R' , whence (1) is proved, Assume that m > 0, Set q = xR' R. Then obviously q = xR' n nR'. Therefore qR~R' = nR~R' (because x Ef nR'), which shows that there is no prime ideal of R' contained in nR' which lies over q (because, since m > 0, xR' is not a maximal ideal of R', hence q is not a maximal ideal of R). Therefore we see that R~ll' is the derived normal ring of Rq , hence height q = height xR' = 1. Now, let C ql C q2 C .,. C qv be a maximal chain of prime ideals in R such that ql = q. Let C q~ C q~ C .,. C q~ he a chain of prime ideals of R' such that q: R = qi for every i. Then q; = xR' by what was proved above, and therefore each is contained in mR', whence v ~ height mR' = m 1 < m 'I' 1 = altitude R', which proves (2) .
+
+ +
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+
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+
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EXAMPLE 3. A normal local integral domain whose completion is integral over it and a local integral domain of altitude 1 whose derived normal ring is not a finite module.
I I';:\. I) Lei I':' In' 1/ .lilM 1'''I/I'lIdl'l'i,'llil' /' ... I) tI/lt! /I'! ,", , ' .. "I"" III' illtil'/I'I'lIlillll/t·8. 8d h'1' /\11.1"" ... ,.1",,11 111/1/ N 1\'/'11.1", , ... , ... ,.1"" III/I. I. '1'111'11.' ( I ) 1/11 ('h·IIH'"./' h I~r U+ /8 ill h~ if 1/1/1/ 111/1/10' I,h,' 1',n:l/i('il'lI/S I~l
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1'1'111'1'. ( I ) i:-1 "l.l'aigl1t.i'ol'w:t.l'd alld (:~) j" all ill\ll\(~diaLn (',OIl"CqIWIH'C "I' ( I ). ;\" 1'01' (:l), il' we kllow LlmL fl it-) N oethcriulI, thclI we provn IIII' nt-)HI'I·t.ioll (':l.Hily. 'l'lwl'dol'c W(~ shall prove that R it-) Noctherian. :-iilll'p Ii iH a 10('al I'illg wili(',h may not be Noetherian who8e maximal idl,:d it-) gl'lH'raf.tod hy LI1(~ :ri, it iH Hllffic:ient to 8how that every ideal n of N whil,l1 hnH a rillite baHiH iH a dOHed subset of R by virtue of 1:\1.:-:), 111'111'1' LhaL nH* U = n. Let. aI, .. , , am be a basi8 for a and 11'11/ I'I~ all al'hitl'ary nl('llwlli, of nR* it. Let K' be the field generated hy t 11(' l·ol'flil·i t 'Il!.:-1 of a,. alld (lover Ie. Then K' i1:3 finite over K P by I I ). LId, I dr.l (!I, = I) be a linear bal:3e of Kover K'. Since b E aR* 11', Ihl'/'(' an~ demnntt-l.f~ , ... , J! of R* slIch that b = a,f'i . Each .1':" iH t It(' HIlIft of two elements gi and hi of R* such that gi E K'[[x]] nlld HII(·.11 t.flU!' the coefficients of hi are linear combinations of the dr. oilll'l' Ihall I. Rince b, a,. E K'[[x]], we see that b = L aig'i , which illlpli('H Chat bEL a'iK'[[xll C a (because K'[[x]] C R by virtue of ( I ) ), am} the assertion is proved. TIIIIH, if, for instance, K is the field generated by infinitely many 1"lgI'hrai(,ally independent elements over a field of characteristic p ~ 0, I.hl'll H is a regular local ring which is not complete and such that the
n
n
L
n
l'oll/pLdion R* oj R is a purely inseparable integral extension oj R. ( 1<;:1.2) JI ssume that [K: K P] = 00 and let b1 , b2 , ••• , bi , . •. be jl-'independent elements oj K. Set c = L bixi, and consider the case 'IIl/WTC n = 1. Then R[c] is the required exampLe. PI·!)O£. Snt d = cPo Then dEll and Xv - d is irreducible over Il. NII:I '"'-' R[Xl/(X P - d)R[X], whence the eomplction of R[c] is isolIlorphi(, to R*[Xl/(XP - d)R*[X], in which the ret::idue dass of X - c iH a non-trivial nilpotent clement. Let R' he the derived normal rillg of R[c]. Then R' is a K oetherian valuation ring by the theorem of 1\l'lill-Akizuki and by the fact that R ::; R[c] ::; R' ::; R*. ThereforI' /l' it:: analytically irreducible. If R' it:: a finite R[c]-module, then Uk! it-) a subspace of R' and R[c] must be analytically irreducible, which it:: a contradiction. Thus R' cannot be a finite R[c]-module, and 1,111' aH"ertion is proved.
,11'1'lilNI.1 :\
l,lill.l.: NIII'!!tI'l'iul/ 1'It./IIII./itll/. 'l'illr/ I' I~r 1/ ,/il'ld II' SlIl'h 1/i.1/.1 llil' ,/il'lrll~r 1II/IIIil'lI/" 11.* 4 Ihl' l'OIII}i/l'/iol/, V* I~r V 'I;" (J, }III.'I'I'//l 'in81'1)(t.I'II./iI1' l'.rit'1I8iol/. of II. sw:h that \11.*: 11.\ = the eharacterist'ic I~r K. Proof. 1,('1. V* lw il* in the ea::;e where n = 1 and let C be a::; above. WI' Ilok 11111.1, 1\'1' 1'11,11 lIIodify 1111'11.10111'1' I'Xll.llipll' HII
( 1,;;\,;;) '/'lil'/'I' is
1/
Let (cl\l be a maximal set of p-independent elements of K* over the field of quotients L of R such that C E Icl\l. Let K be the field generated by all the CI\ except cover L. Then [K*: K] = p. Set V = V* K. Then V is a Noetherian valuation ring with maximal ideal Xl V, whence the assertion is proved.
n
EXAMPLE 4. A local integral domain 7' oj altitude 2 with derived normal ring 7" such that there is a non-Noetherian ring 7''' between 7' and '1". Consider the ring R in (E3.1) in the case where n = 2. Let us denote Xl and X2 by X and y. Let bl , ••. , bi , . . . be p-independent i elements of K and set C = Y L bix , Cn = (c - Li
+
+
+
+
EXAMPLE 5. A local integral domain oj altitude 3 whose derived normal ring is not Noetherian. Let R be the ring in (E3.1) in the case where n = 3. We denote by x, y, Z the variables Xl, X2, Xa. We change the notation bi for
,\ I'I'II:NI'I \ /, illdl'pl'IIIII'111. 1'11'1111'111.11; \\'1' dl'IIIIII' 1111'111 I,y III. ,'1, II", ("', ' , , , /I L, 1",1" I ," L, 1',,1", 11'01' 1.111' ~111,kl' 01' , , , ,II, , ( ' 1 ' ' , , , :-41'1. tI IlIlIqilil'ily, \\,1' 11~1c1111111' Ihal/I :.l, '1'1.1'11: ( I':r" I) Nitil i8 II,,' /'/',/"il'I'tI",/'Il/lIjlll', 1'1'1101', II iH Hldlil'il'lll. 1.0 HhllIV I.II:i.I 1,111' dl'l'ivl'd 1I0I'Iuai l'iu/2: 'I' of h'ltll i,'i 1i01. NllI'lllI'l'iHII. Hilll"~ 'I' iH II0I'lual, .IR*' n 'I' /1', WII<'III'I' ,1"/' iH a pI'illll' id('ii.1. WI' l'IIIIHidl'l' valll:i.l.illll l'ill/2:H H' N"I' and N" .. '1'", . . Hilll'I' If iH ill Ihol'Olllpl<'l,ioll of /~I, W(~ H('(~ t.hat. H' iH a d('IIH(' Hld,HPII.I'I' of W', WII('III,(' N'j.I'W = H" j.eN". 8ill(:(~ '1'///, iH illl'!'/2:ral ov!'!' Nj.I'U II.lId Hilll,n Hj.rN iH lIol'mal, it. followH i.lmt '1'/:r1' = ilj.!'N. Tlwl'\'1'111'1' 1111. maximal id(~al oj' '1' iH generated by x, y, z. AtiHUmC for a IIIIIIIII'II\. I.lIa!. 'I' iH No!'!.hnl'iall. Then '1' must be regular. Therefore i '/'/,//' iH I'q~lIlaJ', which implim.; that T contains L bix + zj for some i f I /i*. Theil we can write L bi:C + zj = (ao + ald)/eI, where 1111 , (J,I , 1'1 at'(~ dements of R sneh that t.hey have no conm10n factor. i i '1'111'11 W(~ have el L bix + elzj = ao + aIY L bix + aiZ L CiXi. i Hilll'(' I, L lii:c arc linearly independent. over KP[[x, yJl[K] (= R/zR), WI' hav\' ao C zR and el - alY E zR. Therefore we write ao = za~, i "I illY + ez (a~, e E R). Then (aIY + ez)( L bix + jz) = za~ + i 1/1/1 L bi.!:i + alz L CiX\ and t.herefore aIyzj + ez L bix + eji = ,~(/.:, + (J.,z L CiXi. We writ.e al , a~ , e and j as power series in z with l'oJdlil'.i!'nts in K[[x, y]], say, al = L aliz\ a~ = L aOiz" e = L e;z\ J L Jii. We want to show that e~ , ao r , and aIr are in yK[[x, yll by illdud.ion on r. Comparing the coefficients of z in the above equali it,y, we have aloyjo + e~ L bix = aOO + aIO L CiXi. Since 1, L bix\ L (:i,/:i are linearly independent over KP[[x]][Kl, we see that e~ , aOO , 1/.111 at'c in yK[[x, yll, which settles the ease where r = O. Comparing t.l1(~ ('ocfficients of zr+I in the above equality, we have y( L~ aldr-;) + '\' b i + L.. ,\,1'-1 '\' CiX i . S ·, , ('",. L...;,'r 0 ei'jr-l-i = aO r + ab' L.. k 1nce eo, ... ,er-I a!'c ill yK[[l:, yll by induction, we see t.hat e~, ao,· and all' are in yK[[x, Yll. Thus we see that. el , ao and al are in yR* R = yR, which contradiets our choice of t.hem. Thus T cannot. be regular, and T is not ]\;octherian, which proves the assert.ion.
n
6. A normal local ring which is analytically ramified. We make use again of the ring R in (E3.1) in t.he case where n ~; we denot.e Xl and X2 by X and y. As in Example 5, let bi ,C] , ••• , ... , bi , Ci, ..• be infinitely many p-independent elements of K. For i I.ht) sake of simplicity, we assume that p = 2. Set d = L(bix + Ciyi) alld T = R[d]. Then: EXAMPLE
,\ I'I'I':N III ,\ ( 1';lj, I) 'I' I." Iii,. l'I'I/lIilnl ,..I'lllIljll,·. !'I'oof. Thai 'I' iN aIIHI.vlil'all.v 1';l.lIlilil'd (i.e., 1.11<' 1'{)lIlpldioll
()r T 1m" Iloll-l.rivi:d IlilpoLmd, elnnwlIl,,) (~all 1)(' pl'{)vnll Himilarly a" (I~;).2). TllPl'dol'(' if, I'PlllailiH ()lIly to prove that '1' is a normal ring. Set en = 2:~ /In'l_i./ J , .f'" = Cn+iY\ and U = R[el, il , ... , en , in , ... ], and let U' be the derived normal ring of U. We want to show first that U = U'. Let g be an arbitrary element of U'. Then
2:;
with p, q, r, s, t E R, because R[el , il , ... , en , in] = R[e" , in]. Since p, q, r, s, t are in R, there is an integer N such that the coefficients of these elements are in K\b l , CI , ••• , bN- I , CN-I)' Since en = b" + bn+lx + .,. + bN_IXN-n-1 + xN-neN and in = Cn + Cn+IY + .. , + cN_IyN-n-1 + yN-niN, we see that g is in the derived normal ring of 2 V = K2[[X, ymb l , CI , .•. , bN- I , CN-I , eN , iN]' K [[x, ymb l , CI , •.• , ... , bN- I , cN-Il] is a complete regular local ring. Since the leading forms bN , CN of eN, iN are p-indcpendent over K\bl,CI, ... ,bN-I,cN-I),
we see that the ring V is a regular loeal ring, hence is a normal ring. Thus g E V C U, and U = U f • Since d = el + ii, T is contained in U, and therefore the derived normal ring T' of T is contained in U. This implies that if h is an element of T', then there ii:l an integer N such that xNyNh E R[e l ,id. Since h is in the field of quotients of T = R[d], have xN:t/h = a + a'd = ao + aIel + adl + a3edl with ai E Ii and with a, a' in the field of quotients of R. Since 1, el , .h , edl are linearly independent over Ie and since d = el + il , we have a = a = ao, a' = al = a2 and a3 = O. Thus XNyNh E T. We want to snow that xNyNh E T implies that h E T. For that purpose, we may assume that N = l. If h 1 T, then we have, by (12.7), one of the following: (1) xyT has an imbedded prime divisor, (2) there is at least one minimal prime divisor p of xyT such that Tp is not normal. Both are impossible because xT and yT are prime ideals, as can be seen as follows: T/xT is isomorphic to K2[[y]][K, il], which is an integral domain, and xT is prime; similarly, yT is a prime ideal. Thus the proof is complete. .
7. A normal local ring which is analytically reducible. Let K be a field of characteristic not equal to 2. Let x, y be indeterminates and let w = aixi (ao = 0, ai E K) be an element of EXAMPLE
2:
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H,e;ril'l'l, llial.: ,1,;7.1) NIXI/' X" z)NIXI":8 the 1'(1/f"i':/'I'r! e.I'II,lIlple. III 1I1'111'!' Lo prol'(' LlII~ :t.HHmtioll, WI' HLudy Horrw pl'opl'l'LieH of R. I L iH 1I1,\'i1l11H IliaL If. iH domillnt.cd by ]([[x, yJJ. 011 t.he other halld, LIi(' dl'filliLioll of z implieH thaI, Lhere it) a polynomial Ii (:X:, y) ( ](\x, yl 1'111' ('n('h .,: HII('h LllaL :rZ;I_l = Z; Ii(x, Y) (fi(O, 0) = 0). Therefol'(~ I'V('I'Y Zi iN ill .r!?' !Ill. ThuH th(' maximal ideal of R is generated by ./' :1.11(1 //. 1"ul'!.lH'rmol'c, it iH eaHY to sl'e that for any element a of R alld 1'01' allY giv()Jl llnL\lmlnumher n, t.here is a polynomial g(x, y) E 1\1.1', !II HIl('h that a - g(x, y) E (xU yR)n by virtue of the relali(lIl .I'Zill = Zi .fi(X, y), Thus we tlee that K[[x, y]] is the compleI iOIl of N ill view of the faet that R dominates K[x, Y]IXK[X.Yl+lIK[X.Jj]) • Let ~l bn a prime ideal of height 1 in R. If x E p, then it is obvioui:' 11111.1 P = .rR. Assume that x Ef p. Since every Zi is in K[x, y, z, l/x], ill(' I'illg R[l/xJ is a ring of qnotients of K[x, y, x]. Since x, y, Z are algnbraieally independent over K, K[x, y, z] is a unique factorization I'illg, whence R[l/x] is a unique factorization ring, which shows that pNII/:r] is principal. Let pEp be a generator of pR[I/x]. Since xR iH a prime ideal of height 1, RXR is a Noetherian valuation ring and nil .r"U = 0. Therefore we may assume that p Ef xR. Let a be an nl'bitl'ary clement of p and let r be an integer such that ax" E pR, r hell('n ax = pb with b E R. If r > 0, then pb E xR which is a prime idenl, and therefore b E xR because p Ef xR. Therefore we prove that (J. CpR, and p = pRo Thus every prime ideal of height 1 in R is prin('ipal. Let q be a prime ideal of R which is not of height 1. We want to show that q is maximal. Assume the contrary. Since depth xR = 1, aH is easily seen, we see that J; Ef q. Therefore qR[1/x] n K[x, y, z] is lL prime ideal of height 2 in K[x, y, z]. Therefore the transcendence dngree of R/q over K is one. Let x', y', z' be the residue dasses of :/:, y, z, respeetively, modulo q. Since the maximal ideal of R is generated by x and y and since x Ef q any polynomial (7"'0) in x with coefficients in K cannot be in q, which implies that :e' is transcendental over K. Therefore R/q is algebraic over ]([x'], whem:e R/q it) a 10I,ality of altitude lover K. But, in the completion of R/q, z' = 1'11.11
+
+
+
+
I
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L
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1/ ,r" I" Ni'll',' /I "I''' III 11'1111:11'1'111 /I 'I Ii./\, I 111'1'1' Ii (.1,1 ), 11'1' "ill"'I'.' Ol' 1/' 11111:iI. I", 11'II.1\111'l'lIdl'III.:'" 01'1'1' /I.(,,.'), IVlli,'h l'oli\,I':ulil'I:4 II", 1'1I.1'i Ihal "'/1\ i:: lIil';I,I'I'lI.i,' 0\'1'1' /,'1,,.'1, '/'111'" q IllIlN\' I)(~ 111:I,xillllll. 'l'III'I'l'forl' \VI' hal'I' PI',,\,,'d Ih:l.l. 1'1'I'I'y prill\('. id,'al or U hat) n fillil,I' 1':l.N", ",hil'h illlpli,',-,: 1/,:/.(. Ii' iN No('l.l1priall by t.he theorem of CollI'lI, ThliN IVI' H('(, I hal. Ii i,e; a I'('gl'\al' lo(:nl ring, Furthermore we :LHH('rl. Lhal. zH i" '" pl'illll' it/I':,! of Ii. J Ildc(~d, t;inee R[l/x] is a ring of quoLi(:nLN of 1\1,1', i/, 021. ;:; i" :I. prime dement of R[l/x]. Since zR and xR have 110 eomllloll pl'inw diviNor, we see that zR is a prime ideal. Now we al'(: 10 prove (E7.1). Since R is Noetherian,
(II'
HI'I'
11,:i.!
R[Xl/(X
2 -
z)R[X]
iA Noetherian. Sinee z is a prime element of the regular local ring R, and since the eharacterifltic: of K is different from 2, we see easily 2
that. R[Xl/(X - z)R[X] it; a normal local ring. The completion of 2 the ring ii:i [([r.1:, y]][Xl/(X - z)K[[:I:, y]][X]. Since z = (y + W)2, 2 X - Z = (X - (y + w)) (X + (y + w)), whence the zero ideal of t.he completion of the local ring has two prime divisors. EXAMPLE 8. A Noetherian integral domain T whose derived normal ring is not a finite T-module, such that, if ~ is a prime ideal oj T then the derived normal ring of Tp is a finite T~-module. Let. K, :r, y, bl , . . , , bn , ••. ,R be as in Example 4. Let. PI, ... , ... ,pn, , .. be iufinitely many prime elements of R (PiR ~ pjR if i ~ j). For eaeh natural number n, we set qn = PI ... pn. Set c = biqi. On t.he other hand, set I = R[l/x]. Then: (E8.1) T = I[c] is the required example. Proof. I is obviously a Dedekind domain, hence T is a Noetherian integral domain of altitude 1. Set Cn = (c - L~-! b.iq.i) / q". Then the derived normal ring T' of T contains all the Cn (for c~ E R). Sinee KP[[:r, ymb l , ., . , bn-lJ[cnJ is a regular local ring, we see easily that R[c!, '" ,C n , ••. ] is a normal ring, hence T' = I[cl, '" ,C n , ••• J. Let ~ be an arbitrary maximal ideal of T. If pn ~ ~ for any n, then Ci E 'Tp for every i, whence Tp is normal. If Pn E ~,then 'Tp does not contain Cn , whence 'Tp is not normal. The derived normal ring of 'Tp in thii:i case is TpfcI, ... ,cn , ••• ] = Tp[cnl. ThllH, in any case, the derived normal ring of 'Tp is a finite Tp-modllle. On t.he other hand, we saw t.hat. if Pi E ~,theu Tp is not normal, whieh implies that there are infinitely many prime ideals ~ of height 1 such that '1'p is not
L
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Hili, WI' Khall 1I0i, ('0111'·1')'11 OUI'HniVCK wit.h t.hOHC hist.orical f:L(~I.H which I'HII b<' HI'I'II ill Knill'K hook l4]. ('fUlfl/.I'!'
*
I,'
I. '1'1 W t.opicH in this seet.ion arc ralhnr daRRical except for the I'1'i III' i pll' of ideal iila fion and t.he exaetness of tensor products. The 1'0 I'll II 'I' waH liJ'Ht. lloticed by :;\Jagata [IG] and t.hen by Nagata-Akizuki III. TIi(' ('xal'!.!H'HS of a tensor product i8 a spec:ial case of exactne:-;H or fUllI'lo),H which was discussed in Cartan-Eilpnberg [1]. *:8. '1'11(\ topics in this section are all elaBsieal. *;;' (:~.:{) and the theorem of Cohen (3.4) were given by Oka [1] alld C{)hCfl [2], respect.ively. We note by the way t.hat t.he original 1'01'111 of the Hilbert basis theorem (:3.()) is in Hilbert [1]. The lemma of ;\r'l,ill-Rees (:3.7) was orally communicated to us by E. Artin in "iK I('dllfe at Kyot.o University in 1955. On t.he other hand, a spec:ial I':I.H(, of the lemma (the case where N = Rand N' is an ideal of R) waK published by D. Reetl [2]. The intersection theorem of Krull (:u I) is a generalization by Chevalley [1] of a theorem (our (4.2)) givell by Krull [9]. (:3.12) is a generalization of a result of Rees [2] whieh asserts only the case where x is not a zero divisor and JJI[ = R (I.he case where x may be a zero divisor was noted by Lech [1]), hence (;U:n and (:3.14) are new. (3.16) was given by Y. Mori and was omlly communicated to the writer hy him in 1952. Exereises 2 and :~ were remarked by H. Mat:mmura. §4. The history of the lemma of Krull-Azumaya (4.1) is somewhat (,omplieated. Namely, the ease where N = 0 and where Jl;1 is an ideal \yaK given and used effectively by Krull. But the one who effectively lIK(,d the module case is really Azumaya (d. Azumaya [1]), hence Llie writer once named this lemma "Azumaya's lemma" in Nagata
,\ l'I'I'INIII'
'21:1
/10/. lIlli, :Iill('(' ( I, I I iH 1111 1'11:1.\' f',"IIt'l'nli'I,II,lillll III' tlli' I':leil' gi\'(,11 hy i\nill, IIII' II'l'ill'I' ('hall!!;I'd Iii:: lllilill 11'111'11 hi' \\'1'1111' Nagala 1:..l:\I. ('1'111' Ii 1'1.;1, lil.l'I'II,IIII'I' wlii"Ii "11111 II illl-I I hi' 1"llIlIm ill 1111' PI'I'''I~III. flll'lll i" Nagala II/, 11111. 1.1((' Wl'il.('I· 11'lIl'IlI'd ilH 1'1I1'llIill:d,ioll from T. Nakayama ~Llld U. }\zllmaYlL ol'a,lly :1.1, NlI,goyn Illlivl'l':-;iLy when the writer was all undergmduat.e ,,1.11<1('111.) 1\lI'allwliilp, the writer saw that some mathematieians eall Lili" II'lllllla "Naknyama's lemma" and therefore the writer asked N:tkayallln, wilo had this formulation first, and what would be the best name for Lhis lemma? Then, Nakayama kindly answered the writer that he did not remember whether Nakayama or Azumaya was the first person and t.hat t.he name of Krull-Azumaya for t.he commut.ative case and the name of Jacobson-Azumaya for the non-commutat.ive ease would be t.he best names for the lemma. Thus the writer employs t.he name of Krull-Azumaya in this book. (4.2) was substant.ially given by Krull [9] (cf. the history of (3,11)). (4.3) was given by Nagata [14]. §5. The results in this section are contained substant.ially in Krull [9].
§6. The notion of rings of quot.ients was first st.udied by Grell [1], who t.reat.ed only the case where S consists merely of non-zerodivisors. The case was clarified by Chevalley [1]. Chevalley [2] defined the ring of quotients with respect. to a prime ideal, t.hen Uzkov [1] generalized eompletely. The method of Chevalley [1] can be applied t.o the generalized case and we obtain (6.4)-(6.9). The writer docs not. know any existing literature which cont.ains (6.13). The not.at.ion R(x) was introduced (in a more general case) by Nagat.a [13]. §7. Prime divisors (or associated prime ideals) of an ideal in a N oet.herian ring had been defined by virtue of primary decomposition, which is not. applicable to t.he general case; Nagat.a [10] generalized t.he notion and obt.ained t.he results in this scction, except. for (7.8) which we owe to Krull [1]. §8. The re"uits in this ::3e(,t.ion for non-graded Noetherian rings :m~ proved by J'{ octher [1] and expounded by Van del' Waerden [1]. The llotion of graded rings and graded modules arc immediat.e generalizatiom; of the notions of homogeneous rings and homogeneous ideals, hence one may say that these notions are classical. By virtue of our (8.3), t.he known method (cf. Van derWaerden [1]) can be adapted to the graded ease and we obt.ain t.he results. Primary decomposition in l'\ oetherian modules has been t.reated as an adapt.ion of the case of ideals, and t.he idea in the exercises is new,
"I I
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*11. (H.II 1\,11/1 p,il'I'11 1'.1' ,\I,i'l.IIl" 1:11 III :!I /lllIllhl'/IIIIIIIIII' Ih"I,I'I'1I1 "I' 1\ 1'1111 I \1.:1 I \\"'1'" p,II"'11 II.\' 1\ 1'1111 III, 111111 I \1. II, (lUll, 1111.1 (11.(i I /11'1' illllll"lIilll,' "IIIHII'IjIII'II"I','11Ii' (11.:\ I. 'l'It" 11111 ilill Id' II. 11.\'1i"" III III' 1'111'11.111 ,'I,I'I'H ,d' II. 1111'11.1 I'illl!: 11'/1,'1 iI1l,l'lIdlll,,'d 1I,lld IltI,'d ,'IT",'I,il'l'l.\' I,.\' ( !llI'l'nlil',V III. (11.10) 11:1.:-1 111'1'11 1'1'1'::1.1'111'.1 II.N /1.11 1I1,,,illl!." 1'111'1, I,.\' IlIal'.\' 1)1'111'11'. ~/(/. '1'111' 11111 ilill III' illll').!;I':d dl'P('lIdl'III'I' iN 1·!aHHil':I.I, 1111(. 1.111' olJjl'1'! Ila,01 111'1'11 J.':I'I 11'1'11.1 iz('d ; (,hi' 1':1. '-;1' III' :t.lJ.':I'''I'ail~ illLI'I!;I'I'H :i.I, lil',,,L, 1.1t('11 1.111' il1ll'J.':ral cil'I)I'lldl'III'I' OV(,I' a l'illJ.': (d. NonLIi(,1' 1::1), :1,lId Lh('11 IIII' Oil(, 111'1'1' :111 idl'nl (1'1'. 1\1'1111 1.')1). (IO.G), (10.7), t.lw lyill!!;-OV('1' 1111'01'1'111 (IO.H), IIII' I!;Oilll!;-lIP iJWOl'l'1ll (IO.!)), (10.11), (IO.I~), Lhe 1!;0illJ.':dllIVIl IIIi'III'I'III (10.1::), alld (10.14-) WnI'n givell hy Krull 17J alld J.':I'III'I·alizpt! a liLIII) liy CoiwlI-8cid(,llhnrg [IJ. nrnll [I] iH Lh(, lirHL IWI'HIIII who HLlldind !.ltp lloLioll or a eOlldlldor ill t.he gClleral WHIO. (10.1), (I()'~), (IO.:n, (10.1), (IO.I!»), :wcl (10.1()) arc either dasHieal III' illlllH'iliaLn gCll(OmliJlaLiollH of datl,'IinLi reHltlttl and shollid be re1"'I'I'I'd 1.0 N()(,t,hcr [;-~j (t.he validit.y of (10.15) in thiH form (i.e., thn ,':lH(' WIIl"'P.r may be reducible) was not.ed by Zariski [6]). (10.18) iH :I,IHII ('i:t.HHic:d and has been known t.o algebraic geometers. u.;xel'l:ise" I :; al'l) adapt.ions of t.he case of integral dependence over a ring. II;x()l'I~iHn 4- waH given partly by Northcott-ReI's [1], then by Nagata 11:-: I ill tltiH form. Exercise 5 is new. II. We owe the theory of valuation rings mostly to Krull [2]; (11.1) -(11.9), and (11.12) are either Krull's result.s or immediat.e gl'llI)mliJlations of Krull's results. (11.10) and the theorem of indlO[)()llIienee of valuations (11.11) were given by Nagata [4]. Exercise I waH given by l\'agata [4] and is an adapt ion of the classical approximaLion theorem (cf. Krull [2]). Exercises 2 and 4 were given by Krull
*
I~I· §12. (12.1) and (12.4) are classical. (12.2) was given by Sato [2J. (I~.:)), (12.5), (12.6), and (12.7) were given by Nagata [10], [11]. ( 12.9) was given by Krull [:)]. (12.10) ii:l clasl-lical. The Exerciso:o 1-5 : LI'(, dassieal. §13. All the results in this section (exeept for Exercises) are classical. §14. The classical llormalization theorem (for finitely generated illtegral domain) is the ease where the ground field contains infinitely many elements and was given by Noether [2]. A 1l0rmaliJlatioll theorem for polynomial rings over a field (containing infinitely many dements) was given by Chevalley [4]. A generalization of Noether':o
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(11.1) (11.·1) alld P"OO/'H o/' t.lWIll 1(\1 alid 11;\1. (1,/'1)) alld (14.G) arc daHi:iical. (1·1.7) waH l2:ivCII hy Y,ariHki [4]. (1'1.8) ii:i an immediate consequence o/' I.lw I'ILHI: wiw)'c 1 is a field and the case was proved by Zariski [4] (d. ]\:/LgaLa [1:3]). The Hilbert zero-points theorem was given by Hilbort [2] in a slightly different form, and, as is well known, there are many proofs of it by many authors. (14.10) was partly given by Artin-Tate [1] (in the Noetherian case) and then by Nagata: [13] in the general case. 1,:I.I'i,'lk i
1:21.
01'
fillill' 11111111'1' II/' I'II'IIII'IIIH wa,", l2:iVI'1i II'y
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Chapter II §15. (15.3) was given by Chevalley [1]. §16. (16.2)-( 16.4) are special cases of some elementary results in the 'general theory of topological groups. (16.5) was given by Serre [3]. As for (16.7) we should give a reference to Krull [9]. (16,8) was given by Chevalley [1]. Zariski rings were treated first by Zariski [:3]. §17. (17.7) was given by Chevalley [1]. (17,8) was partly given by Chevalley [1] and then given by Serre [3], (17.9) was given by Krull [9] and by Chevalley [1]. (17.11) was given by Serre [3]. (17.12) was substantially given by Krull [9]. §18. (18.1) was noted by J-P. Serre as was communicated to the writer by P. Samuel in a letter (in 1955-the writer believes). (18.3)(18.12) are new. On the other hand, the special ease of the results in this Beetion whC're R is a semi-local ring or a Zariski ring and R* is the completion of R were given by the following authorB: (18.1 I, (1) by Kagata [6], (18.1), (3) Zariski [5], (18.1), (4)-(1)) by Chevalley [1]. (18.4) by Y.Mori (orally; in 1952). (18.11), partly by Nagata [16], then by Sato [1]. Exercise .5 was given at firBt in the caBe of a local ring by Y. Mori (orally; in 1949), then by Yoshida [1] in the general case. Exercise 6 was given at first in the case of a local ring by Y. Mori (orally; in 1949), and Sato [1] remarked this faet. Exercise 7 is new. §19. AD results in this section are new, though (3) in (19.2) was Bubstantially proved by Nagata [16]. The origin of the "theorem of transition" is a theorem of transition of a multiplicity for geometric local rings given by Chevalley [4], which asserts that if R is a geometric local ring (in Chevalley's sense), if R* is the completion of
" I II
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('It(/I'II'I' III
'I'1t(' Illlti('iplil'i('y or a "y"LplD or paranwLcr" ill n lo(',Ll rilll!; witi,,1l ('lIldaill" a /il'ld WIL" illLl'odll<'e
)-(22.5) are easy generalizations of results by Samuel [16]). (22.5) is a non-trivial generalization of a result Ily Narnlld [I] and of a result by Serre [2]. (22.7) and (22.8) are easy /-(('1 II q'a,l iimLions of results by Samuel [1].
IiI
((:1'. Nagata
*:'1;1. Our definition of a multiplicity is a generalization of a relative 1lIlIlLiplieit,y defined by Nagata [16]. (Cf. Auslander-Buchsbaum [2]) (2:L I) was substantially noted by Nagata [16]. (23.:3) was first noted "llbH(alltially by Serre l2], (23.4) by Samuel [1], (23,5) by Nagata [I(i[ and Serre [2] independently, and (23.7) by Serre [2]. *2.1. (24.1) and (24.2) are generalizations of results by Samuel [I i, l\agata [16], Serre [2]. (24.4), (24.5), and (24.6) are generalizaI,ioll" of results by Lech [1] (cf. Sakuma [1]). The assoeiativity forIIlUIa (24.7) was first proved by Chevalley [4] in geometric local rings alld was generalized to the case of an arbitrary local ring by K agata [I ni, Serre [2], and Lech [1] independently. §25. Though the notion of a distinct system of parameters was il\Ll'Oduced by Samuel [1], the theory of Macaulay rings was given hy Xagata [16]; (2.5.1)-(25.13) were substantially given by Nagata [I () J. The normality of a regular local ring was given by Krull [9] whose proof was adapted by us for (25.15). That a regular local ring
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wII,-;givl'III,'y (~lil'vlI,lIl~'y III.
('ltupic/' / I' §2U, '!'lin f'(':-;UltH are elementary. §27, (~7.t;) and (27.8) were given by Serre [1] making use of ho-
mological algebra; other results except for Exercises can be regarded as new. §28. (28.2) was given by Auslander, Buchsbaum, and Serre; at first the converse part was given by Auslander-Buchsbaum (published later in [1]) then completed by Serre [1]; they used, of course, homological algebra. (28.3) was given by Serre [1], but some special cases had been proved by Cohen [1] and Nagata [16]. (28.4) was given by Serre [1] and Nagata [16] independently (partly by Cohen [1]). (28.5) was given by Auslander-Buchsbaum [:3]. (28.6) was given by Nagata [14], (28,7) has rather long history. The case of power series over an infinite field is seemingly dassical. Krull [8] proved the theorem for power scries rings over a complete Noetherian valuation ring which is not a field, and the result waH generalized by Cohen [1] to complete unramified regular local ringH. The algebraicgeometrical ease was proved by Zariski [4]. Then Y. Mori proved the theorem for the case of altitude not greater than 2 and also for unramified regular local rings (announced at the spring meeting of the Ma thema tical Society of Japan in 1949) and the same was given indepcndently by the writer in March of 1960 (unpublished); later in 1(;154, the sume was published by Krull [10] (written in 1952), As for the general case, the reduction to the case of ultitude :3 wus made by O. Zariski (unpublished) and ~agata [14] independently and the case of altitude 3 was proved by Auslander-Buchsbaum [:3]. §29, The classical result by Hilbert [1] is the one for the case of homogeneous polynomial rings over a field. Chapter V §30. (:=:0,1) was given by Chevalley [1], (30.2) was given by Cohen [I]. (::1O.4) is an eusy aduption of the well known Hensel lemma in complete valuation ringH and was noted by Cohen [1] (in the Noetherian catle, and the general case waF; noted by Nagata [1]). (:30,5) is rather classical and we I::iho\lld refer to Azumaya [1]. (30.6) was
"IS
Illilllillllillldl,l' P,il'j'll Ilv ('IH'I'll,llj'V III 1':,j'r"IIII' I 1\'1111 f!:II'j'll II.\' ('1\1\('11 III ~,iI, 1:11.11 1I'11111',il"'11 II.\' ('OIH'11 III. ",IHlllj' Pl'oo!' 1\'11;1 ::illq,Iili,·" 11,1' [\;lIl'illi II III lid (:j'''''j';: III, I:!I. 1 NIIJ-':IIII:III ::illq,)ilij'd 1111'1'1'001' ill ;-IIIIIH' :qH'j'inl j'n,'lj':: nil" hie< proof ill Ilw ~j'lwl'ld I'n;-Ij' 1"llilllilWd HOIII,' ,c;j'l·ioll.'; j'I'I'ol·H. (:I'tldl'H III :-4illlplilij'd Clw pmol' 1'01·!J. IO"nl I'ill/!; whij'h mlilll,illC; IJ, lil,ld. Nnl·il.n III I-!:III'" a proof whil,li WI' I'XIHllllldj'd ill I.Ili:-; Ilook, 111111 Ilij'll (:j,dt!I'N 1:.]1 I-!:nV<' allollll'l', bilL Nillliial' \)1'001'.) (:ll.:.n alit! 1:\1.:\) 11'1'1'1' ~il'l'll IIY '1'('illlllniillcl' III. (:\I.!i) waN l1:il'('11 hy :\'al-!::J.I.:I. I:~I. 1:\1.1;) alld (:11.7) wel'c l-!:ivcll hy CohcnllJ. (:IUn waN ~iVl'II by NlI~lI.(.a III. 'l'lw lIoLioli of mlllLipli(lative repl'eseni',t1LivcN waN giv(~11 hy 'I'j,idlllliillcl' III. (:~I.!)), (:~I.I()), and (:)1.12) were given by COlll'll III· 1';:\I'i'I'iN(~ ~ waN lIoL('d by ColwlI [11 and ah;o by Chevallcy III. 1':XI'I·j,i,.;1' :\ wm.; lloLed by Nal2:aLa l~2J. *;1.3. (:\~.I) wa:.; given by Nagata [7J and ~\1()\'i [2J. (:12.2) waH givell IIY [Vlol'i III. (:tl.l) wat3 given by Nagata [11J. At3 for (::1:-1.2): it wm; proved 11,V 1\ 1'1 ill (Math. Ann. 10;3 (1930); cf. [6]) that the derived normal rill!!; of a IllI'a[ integral domain of altitude 1 is K oetherian (and also Exl'I'(',iNC I), Lhen Akizuki [1] generalized the result of Krull ill Olll' fOI'IlI, but atlt3uming that R' is integral over R. The present form i,e; a Nli!!;ht generalization of Akizuki's result and was given by (!IIIIPII 12]. The theory of Krull rings was originated by Krull [3] ((:;;L:l), (:3;3.4), (33.5), (33.6); ef. Nagata [11]). (33.9) is easy and is 11'(,11 known; (33.8) is its generalization and was noted by Nagata 1111. (;~;UO) was given by Mori [1] for local rings and then by Nagat.a 1111 in the general ease; (2) in (33.10) was first explicitly stated by ( ~Iwvallcy [5], though it had been proved by Y. Mori (unpublished). (:{;{.II) was given by Nagata [11]; the present proof was given by II. Matsumura and was published by Akizuki-Nagata [1]. (3;:;.12) waH given by Mori [2] for local rings and was generalized by Nagata 1111· §84. (34.2) and (34.3) were given by N aga ta [17] (in which our (;{4.::;) was misstated). (34.4) was substantially given by Cohen [1]. (:~4.5) and (34.6) were given by Nagata [17]. (34.7) and (34.8) may "(~ :mid to be new. CI4.9) and (:34.10) were given by Xagata [17].
*.U.
('/wpter VI
§35. (35.2) was given by Nagata [13] and is a slight generalization of a classical result. (;35.3) was substantially given by K agata [1;:;].
\ I'I'I':N III \
I
;;!,,:,
I 11'11:1 f',III'11 ill II 111111'1' 1':"111'1'1111"11'111 11,1' 1\11-/2;/1.111,
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II
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§:iU. (:\(j.~) alld (;W.:n wm'() Hub:,;tantially given by Zariski [5] (who 1.1'<':1.1('<1 Ihl) m;.;e where Ii is normal). (36.4) was given by Nagata
II) I ati:Lpl.illg Zari:oki's proof (Zariski U5]) of the analytical unramif-iedIU)HH of algebraic-geometrical local rings. The analytical unramif-iedneHS of algebraic-geometrical local rings was first given by Chevalley [4]. (:36,[5), (36.6), and (36.8) are new (( 36.8) was partly given by Nagata [16]). (36.9) was substantially given by Hironaka [IJ. (:36.10) was also given by Hironaka. §S7. (::\7.1) was formulated in this form by Nagata [24] but was substantially given by Zariski [5]. (:37.2) was given by Zariski [6]. (37.3) was substantially given by Zariski [OJ. (37.4) is a ring-theoretic formulation of the so-called Zariski's main theorem on birational tramoformatiom; and the prespnt proof was given by Chevalley in his lecture at Kyoto University in 195:3. (37.5) was partly givell by Zariski [6] (the case where 1 is a field and R is separably generated over I), then by ~ngata [6] (the case where I is a field), then by Nagata [7] (nearly the present ease) and then by Kagata [1:3] (the present form). (37.6) is an immediate consequence of on.5). (37.7) was noted by Nagata [1:)]. (37.8), (37.9), (:37.10), and the exercise appear for the first time in this book. §38. (:38.1) and (38.2) are adaptions of the algebraic-geometrical case (d. Zariski [2]). 018.::;), (:38.4) and (:38 ..'l) were given by :"-J agata [14J. (38.6) was partly given by Zariski [1] (the ease where R is a normal locality over a field), whose proof can be applied to the case where Rand Rt are normal, then by Chevalley [6J (though Chevalley aSimmed that R is a locality over a field, he proved substantially the general case; he proved a little bit less than we have) and then by Nagata [24]. We make here a remark on the literature "Chevalley [6]." Expose 5 in t.he seminar was given by A. Grothendieek and no name was given in its Appendix 1. But the writer was told by A. Grothendieek that the appendix wat:; written by Chevalley. (:,18.9) is rather classical. (>18.10) wa:5 noted by Nagata [;2:3]. The exercise is an adaption of the algebraic-geometrical case (cf. Zariski [2]). §89. MacLane [1] proved that a function field L over a field K is separably generated over K if and only if L has a separating tran-
I
l'I'I'1N III \
111'1'1Ii11'11I'4' 1,11111'. (hll' l'I'11i1i111 ill l!rill 111'1'111111 111'1' 111111111,1' 1'1,1'111'111111111 ill II II fl.11'1'1i 1,,\' NlIf':1I11I [11[1,1' WI'il':1 In'IIII1II'1I1 I \V1'il [1[1, I:I!I,II I II'II,H f',il'I'11 h,l' Nllf':lIla [11[, 1';\I'I"'i,cII' I WII:j 11.11111 f':il"'11 1',1' N/I.~al:r.[1 1[, 1';\1'1' l'iHI':1 11'11:: lil','d, j.';il'I'1i ill !.IIi", 1",lIk, ~ iii, ( 10,1) 111111 ( 10,~) W"I'I' !2;ivI'1i I,y NlIglIllI, [1(\[, (10,:: I IVa,,", 1'111'11.1' gil'l'li 1,.1' N:rgn.la [:1[ IIl1d [~r)[, (111.1) iH a l'illg·Lill'ol'4'Cil' 1'01'11111 1111 ill II or II Li It '0 1'1 'II I (l'I'OPIlHiCilili Ii) ill NagaCa [1:1[, (,10':») waH Hili, ::III-IIi-ially giVI~11 11.1' N:lg:r.la [;l[, It, iH 110(, yeL kllOWIl LII III(' IYl'iC<'I",,", kllllwll'dgl' wlll,ther 01' lIoL (iIO,I) iH Lt'Il(' wiLhouL aHHllmill!2; thaI, p iH Hllal,vCil'ally lllll'alllified. II' it. iH tl'll<~, t.hell (40.:)) (hcrwp, itH Hp<~ci:t.I I'II,HI' (,IO.r;), 1(0) iH 1!'IIe wit.hout aHHuming that 1 or R iH PHCU
I l'I'I'iN III' lil'lill''illldil,l' III 1':"'I'I'iHI':\ \Va~1 g:il'I'1i II,\' WI,jllll. 1,111'11 a 111111'1' gCIIPml 1'01'11111111 11'1111 f':il'I'1i 11'y Nal-(lI,i,n
Ilrd.
('Imp/I'!' 1'/1
*.1:;. 'I'llI' 1IIII,ioli or HC\ll-lclian rings was introduced by Azumaya III alld Ill<' lIotion of Henselization was introduced by Nagata [4], I!)/' 12:~J. (4:-U) and (43.2) were given by Nagata [4]. (43 ..5) was givclI by Nagata [23]. (43.8) was newly given in this book. (43.9) wal-ll-lubstantially given by Nagata [.5]. (4:3.10) was given by Nagata [5], [23]. (43.12) was noted by Nagata [,I)] (cf. (43.1.5)). (43.14) was pointed out by T. Nakayama and was published by Azumaya [1]. (43.15) was given by Azumaya [1]. (43.17) was first given in this book. (43.18) was given by Nagata [23]. (43.19), (43.20), and Exercises 1 and 2 were given by Nagata [,I)]. Exercise 3 was first given in this book. Exercises 4 and 5 were given by Nagata [4]. §44. (44.1) was substantially given by Nagata [7]. (44.2) is new. (44.4) was partly given by Nagata [4] (the case where R is a valuation ring), and this general ease is new. The exercise was given by Nagata [4]. §45. Our notion of a Weierstrass ring was modified from that which was given by Nagata [7]. (4,1).1) was substantially given by Nagata [7]. (4,1).3) is classical. (Note that the original form of the theorem given by Weierstrass is Exereise 1, and the present form was noted by H. Spath in 1929 (J. Reine Angew. Math. Vol. 162), but the present form is substantially equivalent to that of Weierstrass as is easily seen.) (45.4) and (4.5.,1)) are also classical. (45.6) was given by Nagata [7]. Exercises 2 and 3 were newly given in this book. Exercise 4 was substantially given by Nagata [7]. §46. The notion of a mixed Jacobian matrix was introduced by Zariski [4], who proved (46.3) in the case where A is a polynomial ring. The generalized (46.3) was given by Nagata [19]. (46.4) was also given by ~agata [19]. (46.7) was noted by Nagata [25]. Exercise 2 was noted by Kagata [14]. Exercise 3 was given by Nagata [19]. §47. (47.2) and (47.3) were given by Nagata [9], and (47.4) and (47.5) are immediate consequences of (47.3). J\Tote that (47.,1)) says, in the case where K is the complex number field, that the product of two irreducible analytic varieties is again irreducible. (47.6) and Exercise 5 were published in Sugaku, Vol. 9 No.1 (19.57), p. 61
I Hollil illil IIi' 1'l'old"111 S I.!!, I; IIII' pi'old III 11111 1'llIllpll'II' 1IIId 1\ :lIlp pl"IIII'liI ill 1"III'I'II,tllli IIPPI'II,I' 11111111. I 1'('/1, I I'/.S i, 11,lld (I'/,.!li /1,1'1' 111'1'111 illf',ly 111'\1', '1'1I011/(1i 11111111' pl'l)pll' kllO\l' I 1'/,1(1) ill IIII' 1'1\.:11' 11'111'1'1' II. i:1 I,lli' l'l)ll\pll'~ 11111111,1'1' lil,ltI (1\11', 1\I'Ii/I:I.:l!li 0(' N:l.J.(o,\'1I. 1IIIil'l'I'Hily Iliid llil' II'l'ill'l' 1.1111.1, II" klll'lI' 1.111' 1'1',''11 Ii, I,11I lIi~1 Pl'olli' II':I~: diilil'Ii!I.), I,lli' \I'I'ill'l' l'llidd 110(. lilld :l.lIy lill'I':I,I,III'I' I'olililillilll~ 1111' I'I'olli!I, 1,;,1'1' l'i111'11 I, ::, :l.llti I IVI'I'I' l!:il'l'll hy N:lI!:II,(.a 1!lI,
,I }1}I/'I11/ i,l' ,II ( 1';1.1 ) wac; 1101.('<\ by Na)!;:Ll.a 12!iJ, (1 1;1.2), [I:xamplc 2, and (1':2,1) IVI'I'I' )!;iVI'11 by !\:a)!;al.a 141, All cxmnplc of a local integral domain of 11,ll.i (.lIlil' I wlioc;(~ d(~l'ivnd normal ring is not finite watl firtl(, givcn by Aki;l,llki III in i.Iw (:ac;c of <:hanwl.(~rist.ic zero, Exampletl 4 and [) w(~J'(' f!,'iVI'11 hy Nagata I~J, I';xample (i waH giv()ll by Nagata [12] and ExI\.lllpl(~ 7 waH giv(m by l'\agata [20], Example 8 was given by Nagata 12!i1·
References Y. [11 I';illige Bcmerkungen liber primare Integritatsbereiche mit TeilerkettensaLz, 1'roc. 1'hys.-Math. Soc. Japan, 17 (1935), pp. 327-336. [2J Teilerkettensatz und Vielfachenkettensatz, Proc. Phys.-Math. Soc. Japan 17 (1935), pp. 337-345.
AKI:.I!lKI,
AKIZUKI, Y., and NAGA'l'A, M. III Modern algebra (in Japanese). Kyoritsu, Tokyo, 1957. AUSLANDER, M., and BUCHSBAUM, D. A. [1] Homological dimension in local rings, Trans. Am. Math. Soc. 85 (1957), pp. 390-405. [2] Codimension and multiplicity, Ann. Math., 68 (1958), pp. 625-657; Errata Ann. Math. 70 (1959), pp. 395-397. [3] Unique factorization in regular local rings, Proc. Nat. Acad. Sci. U. S. 45 (1959), pp. 733-734. [4] On ramification theory in Noetherian rings, Am. J. Math. 81 (1959), pp. 749-765. ARTIN, E., and TATE, J. T. [1] A note on finite ring extensions, J. Math. Soc. Japan 3 (1951), pp. 74-77. AZUMAYA, G. [1] On maximally central algebras, Nagoya Math. J. 2 (1950), pp. 119-150. BOURBAKI, N. [1] Algebre multilineaire (Algebre, Chapitre 3), Hermann, Paris, 1948. CARTAN, H., AND ElLENBERG S. [lJ Homological algebra, Princeton University Press, Princeton, 1956. CIIEVALLEY, C. [1] On the theory of local rings, Ann. Math. 44 (1943), pp. 690-708. [2] On the notion of the ring of quotients of a prime ideal, Bull. Am. Math. Soc. 50 (1944), pp. 93-97. [8] Some propert,ies of ideals in rings of power series, Trans. Am. Math. Soc. 55 (1944), pp. 68-84. [4] Intersections of algebraic and algebroid varieties, Trans. Am. Math. Soc. 57 (1945), pp. 1-85. [5] La notion d'anneau do decomposition, Nagoya Math. J. 7 (1954), pp. 21-33. [6] Seminaire C. Chevalley 1956-1958; Appendix I to Expose 5. Ecole Normal Superieure, Paris, 1958. [223]
I H
('jHII11N,
III ()II I,\,,, 111"',,1'1,,,1'1' IIl1d ili"1I1 1,1,,'ol'.\' 01 "Oil' pi,,,,, 10"111 I'IIIV,I', '1'1'111111 ;\ III 1\llIl,h, Ho,' 1M IIlI,III), pp, hi IIII\. I'.~I ("IlIlIlIlII,II-I,iv(I l'illV,II \\'il,h ""lIl,l'i"I,,',1 IlIilllllllllll "Ollllil,joll, 1'111". ~IIII.II, ./, 1'/ (10['0), I'P, 'J'l I'J,
III I',·j,lln idl'lIlH ~lr.'J
II
lid illl,"!,:1'1I-1 <1111''''1111'11'''',111111. 11,11. M",I,h, Ho",
r,:J
(I!IIII), 1'1"
'J1i I ,
('''ow, W. L, II ()II th" Lhl1ol'lIlII .,1' Ikl'Lilii 1'01' 10,:,1.1 domaillH, 1'1'0<:. N ",t., lI,md, H"i, ( IilliH), I'P, IiHO IiH'1. (:lfllHII'lH,
(J,
H, ,II
A.
III ;\ "hol'l, 1'1'1101' of t.ite (\,XiHt.nllee .,1' ,:ot:llieiell\', lieldH for cOIDplef,() eqlli,:hal'an L"I'iHLi,: 10':',,1 I'i IIP;H, .1. LOlldon MaCh. Hoe. 2\J (1H54), pp. :3:~4--:~11. 1'21 (),I I.h" (o1II1)(,ddillP; t.lll,ol'em for eompleLo loe,d rinp;H, 1'l'oe, 1,01111011 MaLh. H,,('. Ii (I!llili), liP, :;1:3-:-154. (:,/,1>,1,("
II.
III I ~"1.i(Oh IIIlp;en zwischen del' Idealen verschiedener Ringe, Math. AHll. !17 (I D27), PI" 490-523. "",("')11,'1', Il. III (111(01' dioj,itcorie del' algebraischen Formen, Math, Ann, 36 (1890), pp, 471 1i:!,1. 1'21 0(,1\1' die vollen Invariantensysteme, Math. Ann. 42 (1893), pp. 313-37:3,
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n.
III II 1I0t.e on algebraic geometry over ground rings-The invariance of Hilbert. d'll,meLeristic function under the specialization process, Illinois J. Math. '2 (11158), pp. 355-366. 1\11,111,[" W. III I'l'imidealketten in allgemeinen Ringbereichen, S.-B. Heiderberg Akad. Wiss, 7 (1928). 121 IIllgemeine Bewertungstheorie, J. Reine Angew. Math. 167 (1931), pp. 160-196. 1:\1 Ober die Zerlegung der Hauptideale in algemeinen Ringen, Math. Ann. 105 (1931), pp. 1-14. 1,11 Idealtheorie, Ergeb. der Math, 4, No.3, Jlllius Springer, Berlin, 1935. llil Beitrage zur Arithmet.ik kommutativer Integritatsbereiche, Math. Z. 41 (1936), pp. 545-577. 16] Beitrage zur Arithmetik kommutativer Integritatsbereiche, IT, Math. Z. 41 (1936), pp. 665-679. [7] Beitrage zur Arithmetik kommutativer Integritatsbereiche, III, Math. Z. 42 (1937), pp. 745-766. IHI Beitrage zur Arithmetik kommutativer Integritatsbereiche V, Math. Z. 43 (1938), pp. 768-782.
,,'1 -
.,
IIII 1111".11111011111111'''111' III ~-\I'i'''I'''l'ill~.I·",.1 H,·illl' !\IIV,,'w.I\111I11. 1',illlll:IHI, I'll. ~O I ~~~~I \ 1101 1.111' 'l'III'lIl'il' cI,,,' 1'''III1I1I1LnLivl'lI 11I1'''I':I·iLii.LHI)(!I'ni,dl'', .1. ((,·i"" i\1I1':"W. MaUl. 1\1~l 110[,1), PI'. ~:\O ~Ii~.
L I'll 'II,
(~.
III ()II I.III~ :l.HHOeiILi,ivi Ly formula for multiplicities, Arkiv. Math. 3 (1956), pp. :\I)I:~I
(I.
F. S. [1] Algebraic theory of modular systems, Cambridge Tract.s Mat.h., 19 Cambridge University Press, Cambridge, 1916.
MA':AlILAY,
S. [1] Modular fields. I, Duke Math. J. 5 (1939), pp. 372-393.
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MORI,
Y.
[1] On t.he integral closure of an integral domain, Mem. Coil. Sci., Univ. Kyoto
27 (1952-53), pp. 249-256; Errat.a, Mom. Coli. Sci., Univ. Kyot.o 28 (19531954), pp. 327-328. [2] On the int.egral closure of an int.egral domain, II, Bnll. Kyot.o Gakugei U niv. B7 (1955), pp. 19-30. NAGA'rA,
M.
[11 On the struct.ure of complete local rings, Nagoya Math . .T. 1 (1950), pp.
12] [3] [4] [51 [6] [7J [8] [9] [10] [11] [12] [13] [14] [15]
63-70; Enata, Nagoya Math. J. 5 (1953), pp. 145-147. On the theory of semi-local rings, Proc. Japan Acad. 26 (1950) pp. 131-140. Local rings (in Japanese), Sugaku 5 (1953-54) pp. 104-114 and pp. 229--238. On the theory of Henselian rings, Nagoya Math. J. 5 (1953), pp. 45-57. On the theory of Henselian rings, II, Nagoya Math. J. 7 (1954), pp. 1-19. Some remarks on local rings, Nagoya Math. J. 6 (1953), pp. 53-58. Some remarks on local rings, II, Mem. Coll. Sci., Univ. Kyoto 28 (1953-54), pp.109-120. Note on integral closures of Noet.herian domains, Mem. Coli. Sci., Univ. Kyoto 28 (1953-54), pp. 121-124. Note on complete local integrity domains, Mom. Coll. Sci., Univ. Kyoto 28 (1953-54), pp. 271-278. Basic theorems on general commutative rings, Mem. Coll. Sci., Univ. Kyoto 29 (1955), pp. 59-77. On the derived normal rings of Noetherian integral domains, Mem. Coll Sci., Univ. Kyoto 29 (1955), pp. 293-303. An example of normal ring which is analytically ramified, Nagoya Math. J. 9 (1955), pp. 111-113. A general theory of algebraic geometry over Dedekind domains, I, Am. J. Math. 78 (1956), pp. 78-116. A general theory of algebraic geometry over Dedekind domains, II, Am. J. Math. 80 (1958), pp. 382-420. A general theory of algebraic geometry over Dedekind domains, III, Am. J. Math., 81 (1959), pp. 401-435.
'_" 'II
11111 '1'111' 1111''''1' "I III,IIllpl"'II.I' III ~y,"q,,1 1""111 111I~,iI, l'I',"'""dll'~,,, "I IIII' '"II'rlllIl,I""1I1 H,I'"I'''''IiIIlI, '1',,1'111 NII,I", 1ill,II ~1"'''"III,,' (""III"Ii "I ,1111'1111, '1'" I, .1''', 111[,1 I, PI' IIII '!:,~II 11','1 1111 1111' "llIdll 1'1',,1"""' "I' I'rillll, id,'" III, N "~,"I'II 1\11111, .J IIIII!I[,III, 1'1'. id II I I L~ I N OJ,l' 011 II, pll pl'I' 1)1' HIIIIIIWI ('OIIC'cq'lIi IIJ-l. JlIl,\' 1111 II 01 iI' 1I"qpPI'CJC'li III' iI/I'll III ~1I'1i1. (!"II.H,'i., (1IIiv.I\.I'"I,,;I() 11%111,',1, 1'". III[) 1'/1" 11111 1\ .JII,'ol,illll ""il,I'ri"" "I' {,ill""" poi II 1,". '"illl,i:,.1. 1\111111. I I 1%'iI. PI', 1:2',' 1:1:2_ 1:2111 1\11 mll-lIl1>lI' "I' II, 11111'11111.1 101.1..1 rill/-!: lVilil'l, i,y II.llltiyl,i"II.IIy !'I·dll",i!>I,·. ~I,,"I. (~oll, H"i,. (1IIiv, 1',YIII,II:\1 (IOIiX), PI>, X:I Xli, I~lll NoL,' Oil '" "I,,,,ill (lolidiLioli 1'01' I>rillll' idnalH, M"IlI. Coli. HI',i., lI11iv. 1\.1'''1,,) :1:2 (lilliO 1!lIiO) , PI>. Xli !IO. I~l~ll N"I,,, Oil l.o"f1i .. i(lIiL fiddH or ('lImplet(. llle:i.! I'ill/-!:H, M()Il'. Coli. H"i .• lilliv. l\yoLo ;1:2 (IHIiB WHO), I>p. !11 H2. 1:1;11 (III 1,11(1 L111'Ol'y or """Holiall I'ill/-,:H, "I, MOIll. ColI. Hei., lilliv. 1\'yoLil ;I:.l (IOIiO lOtiO), PI>. H:I 101. I~II ()II 1,1111 p'lriLy or !>mlleh loei ill t'('/-!:lllal' I(H,:d l'illgH, [Ililloi" '/. MaLII. :\ (IDW), PI>. :128 :\:1:1. I~lil Oil t.h« el()Hedll()~H of "ingular lOlli, Puhi". MIlII!. IIIHt.. II alit.. ji;l.lId. Hl'i. 2 (IHGH), pp. 29--36. NA"AI, Y. alld
NAGATA,
IV!:.
III I\I/-!:i'brnie geometry (in Japanese). Ky6ritsu, Tokyo, 1957. NAIII'I'A, M. III (III Lhe "il'ucture of complete local rings, J. Math. Soc. Japan 7 (1955), Pl'· ,1:11i,14:1. I~ I (Ill lhe unique factorization theorem in regular local rings, Proc. JaptLiI I\WI!. :\5 (1959), pp. 329-331. NI~III,
III
M, the dimension of local rings, Mem. ColI. Sci., Univ. Kyoto 29 (1955), pp. 7-9. ()Il
N OI,n'IlER, E, III Idealtheorie in Ringbereichen, Math. Ann. 83 (Hl21), pp. 24-66. 121 Der Endlichkeits~atz del' Invarianten endlicher linearer Gruppen der Charakteristik p, N achr. Ges. Wiss. Giittingen, 1926, pp. 28-35. [:\1 Abstrakter Aufbau del' Idealtheorie in algebraischen Zahl- und Funktionenkarpel'll. Math. Ann. 96 (1926) pp. 26-61. NOR'rHCOTT,
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III Hilbert function in a loeal ring, Quart .•T. Math. Oxford 4 (Hl53), pp. 67-80. D, G. and Rl!lES, D. III Heduction of ideals in local rings, Proc. Cambridge Phil. Soc. 50 (1954), pp. 145-158. 121 A note on reduct.ions of ideals wiLh an application to the goneralized Hilbert function, Proc. Cambridge Phil. Soc. 50 (1954), pp. 353-359.
NORTHCOTT,
( ) h '\
III
I
~._
,-~III' 1"11 1,1101'1101111
IIlIlilyl,i'lII""
.1 "1'1111 :1 I Iilid I. JlI', I( 1':1':11.
~~III
.I,.
1'II/IIi'''"'H vlIrillhle'H, VIII, .1, Malli, N"I',
:.l1,1,
I),
III 1\ lIIIII' 1111 VlllIIILi,ioIiH aHHoeiat.ed with a local domain, Proc. Cambridge Phil. HilI', 1)1 (1!11i1i), Pl'. 252-253. [21 'l'wl) I:In.HHieal theorems of ideal theory, Proc. Cambridge Phil. Soc. 52 (I!lSI;), pp. 155-157.
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1,11""1"'111 Oil
Y,""i"ki ri'I~JI, ('1111, .1. ~IIIIII II lillliIIl, 1'1' :11.
Y,,\IIHlI(t, (I.
III /\ 1",,1 "'II,in vnl'iC'Li"H OV,'I' I!:rolilid 1i1,leI of "Ii II rlll'II'l'iHI ie' II:! 11\1,10), PI'. IX7 :!:.ll.
1:21 1:\1 III
liil
11\1
171
1,'''lIlIclIlLilll'H of
II.
1.1'1'0, 11111 .• 1.
M /I I,ll,
/,:elll1l'nl Lheol'Y of ],iI'lLl,iolial "OITI1HPOllcllilll'I\H, TI'II,IIH. 11111.
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AND SAMUEL,
P.
III COllllllutative algebra, 1, van Nostrand, New York, 1!J58.
TAB L 1<; 0 F NOT A T ION
:1Jcr( /) module of derivations, p. 147 ii e./2:., ajUa:, partial derivation, p. 147 NotatiollH like jJJ, p. 148 hd homologieal dimension, pp. 92, 94 i() e.g., iK(K', L; L'), order of inseparability, pp. 174, 176 [ ]. order of inseparability, p. 176 J() Jacobian matrix, p. 147 J*() mixed Jacobian matrix, p. 196 K() K-function, p. 67 A() A-polynomial, p. 71
J.I.()
multiplicity, p. 75 multiplicity, p. 1.'>3 op. alt operator altitude, p. 28 Notations like p(r) symbolic power, p. 20 Notations like Rp , Rs rings of quotients, p. 15 O"() O"-polynomial, p. 67 syzn the nth syzygy, p. 92 trans. deg transcendence degree, p. 44 x() Hilbert characteristic function, p. 67 () e.g., R(x), p. 18 [[ II e.g., R[[x]], power series ring, pp. 49, 106 «» e.g., K«x», convergent power series ring, p. 191 X local tensor product, p. 169 Q9 complete tensor product, p. 169 ® analytic tensor product, p. 199 ~, <, etc. domination, p. 14 e.g., [a:bl c , a:b, p. 2
m()
229
A AddiLivll vaitkation, §11, p. 36 -atiil' t.opology, §](i, p. 51 Aflioe ring, §:35, pp. 127, 128 Akil,uki, Lheorem of, §9, p. 25
Algebraic-geometrical local ring, §35, p.127
Almost finite, §1O, p. 30 Altitude - - (of a ring), §9, p. 24 - - (of an ideal), §9, p. 25 - - formula, §35, p. 129 - - theorem of Krull, §9, p. 26 Analytic - - ring, §47, p. 199 - - tensor product, §47, p. 199 Analytically - - independent, §31, p. 106 - - irreducible, §37, p. 135 - - normal, §37, p. 135 - - separably generated, §47, p.
Coefficient field, §31, p. 106 - - ring, §31, p. 106 Cohen, theorem of, §3, p. 8 Complete, §17, p. 53 - - tensor product, §42, p. 169 Completion, §17, p. 53 Composite, §11, p. 35 Conductor, §1O, p. 29 Constant term, §15, p. 49 Convergent power series, §45, p. 191 - - ring, §45, p. 191 D
Dedekind domain, §12, p. 146 Degree, §8,21, §21, 70 Depth, §9, p. 25 Derivation, §39, pp. 146, 147 integral - - , §39, p. 147 linear dependence of --s, §39, p. 147
partial--, §39, p. 147 zero - - , §39, p. 147 Derived normal ring, §1O, p. 31 Dilatation, §38, p. 141 Discriminant, §41, p. 160 Distinct system of parameters, §25,
199
- - unramified, §32, p. 114 Annihilator, §1, p. 1 Artin-Rees, lemma of, §3, p. 9 Associated prime ideal, §8, p. 24 Associativity formula, §24, p. 81
p.82
B
Dominate, §5, p. 14
-base (p-base), §31, p. 107 Basis, §1, p. 1 Basis theorem, Hilbert, §3, p. 9 Basic - - field, §42, p. 170 - - valuation ring, §42, p. 171
E Eisenstein --extension, §31, p. 111 - - polynomial, §31, p. 111 Equivalent (valuation), §11, p. 36 Exact - - sequence, §1, p. 4 --tensor product, §1, p. 4
C
Cauchy sequence, §17, p. 53 Chain condition for prime ideals, §34, p.122
F
the first - , §34, p. 123 the second - - , §34, p. 123
Faithful, §9, p. 28
231
1"IIi/I(1
l'IiHiH, §I, p. ~l 11I0,(ltI", ~I, p. ~ j,.I'P" , §:!I), p. 1:.l'7 1"illil."I,I' /J,(\IIIII'/l.I.,'d j,,I'PII, ~;\Ii, p. 127 1"illil.(lIIIIHH ''IllIdij,illll 1'111' illl.I1I/;I·II.1 (IX 1.I'IiHioIlH, §ar;, p. 127 FIII'III
(0.1/;.,11 1'01'111), *21, p. 70 I'illl/;, ~21, p. 70 IIIOdld", §21, p. 70 I,'ol'lunl pOWll1' H"l'inH, §11i, p. 4!) rillJ.!:, §15, p. 4H 1"IIIII,j,ioll Jield, §:\5, ]lp. 127, 128
11I1j,(\"'((\d PI'IIIII' diviM"I', A7, I' ~o il"I"I"",,"'"I, (I' ), §·IH, p. lUll IlIdll""'( (d(lI'iVlIj,illll), ~III, p, Illri IIlMj,i", /{I'IIIIP, ~·II, p. Ir;o
rill/{, ~·II, p. 1m Illl,{\J.!:I'H.I ((lvl1r 11. l'ill/{) , ~\t), p. 2k (oVllr nil idonl), ~IO, p. :\·1 - (·'\ot-\I\l'!\, §IO, PI'. ~\), :1·1 -- derivat.ioll, §aH, p. '·17 ._- llx1,ensioll, §IO, p. ;10 Inj,eJ.!:rally <:losed, ~l(), p, 2H Intersection theorem of Krull, ~:\, p. 10
( lll.loiH
cx1,eIlHion, §1O, p. 31 - group, §1O, p. 31 (:oin).!;-uown theorem, §10, p. 32 (:oinl!;-op theorem, §1O, p. 30 ( :"nd"d -- ideal, §8, p. 21 -- module, §8, p. 21 --- ring, §8, p. 21 -- submodule, §8, p. 21 (: I'Olllld ring, §35, p. 127 H I [,\iJ.!:ht, §9, p. 24 lIensel lemma, §44, p. 189 lienselian ring, §30, p. 103 Henselization, §43, p. 180 II ilbert - - characteristic function, §20, p. 57 - - zero-point theorem, §14, p. 47 - - basis theorem, §3, p. 9 Hironaka, lemma of, §36, p. 135 Homogeneous - - element, §8, p. 21 - - ideal, §20, p. 67 - - polynomial ring, §20, p. 67 - - ring, §20, p. 57 Homological dimension, §25, pp. 92, 94
Irreducible elemen1,H, §1:1, Jl. ·12 Irredundant, §1, p. 1 Isobathy, §25, p. 82 Isomorphism theorem, §1, p. 2
J Jacobian matrix, §39, p. 147 mixed - - , §46, p. 196 Jacobson radical, §4, p. 12 K,K
K-function, §20, p. 67 Krull - - ring, §33, p. 115 altitude theorem of - - , §9, p. 26 intersection theorem of - - , §3, p. 10 Krull-Akizuki, theorem of, §33, p. 115 Krull-Azumaya, lemma of, §4, p. 12 L, A
A-polynomial, §21, p. 71 leading - - degree, §15, p. 49 - - form, §15, p. 49 Lech, lemma of, §24, p. 79 Length (of a module), §1, p. 4 - - (of an M-sequence), §27, p. 96 Lie over, §5, p. 14 Limit, §17, p. 53
1,'14'111 I'i 1111. ,
,n, p
la
II'liil'll IIIII.\' 1101 III' N""I,III'I'illll,
§fl, p. 1:1 L""IIII,''1trIOI' 11I'"dlll'l., ~·I~, p. 11111 ",wlllil\" ~:IfI, pp. 1~7, 1'.lK I '''''nll\,' Mn,"lIld",1' rinK, §~ft, p. ~t, Lyilil/:'''VI'I' 1,111'01:11111, *10, p. :\(I
1\'. MIII'/winy l'illP;, ~:lft, 11 . .'l~
IOl'lIl1y
, *:lli, p. K:l
Ma):illl:d
1,lIl1.ill of pt'illll, idl'IILy, *:1-1, p. 122 ide:t.i, §2, p. Ii . idnal wil.lt l'e~poel.1,o, §2, p. 4 (M)-Hequonee, §27, p. 96 -.- j)1'ime divisor, §7, p. 19
Minimal - - basis, §1, p. 2 - - prime divisor, §2, p. 5 Mixed Jacobian matrix, §46, p. 196 Multiplicative - - representative, §31, p. 110 - - valuation, §45, p. 190 Mu[t.iplicity, §23, p. 75 - - of a local ring, §40, p. 153 N
Natural topology, §16, p. 52 Nilpotent, §1, p. 1 Noetherian, §3, p. 7 Normal ring, §1O, p. 31 Normalization theorem for - - convergent power series rings, §45, p. 193 - - finitely generated rings, §14, p.45 - - polynomial rings, §14, p. 44 - - separably generated affine rings, §39, p. 152 Numerical polynomial, §20, p. 69
o Operator altitude, §9, p. 28 Order of inseparability, §42, p. 176 - - with respect to, §42, p. 174
I' I'
Il\\~", ~:\I,
p. 10'/
I' 1I1"1'IIIIIld'''II., ~l(\, p. IlIrl
",,,.j
1'111'1 illl VIII j,,", §:lll, p. 11'/ 1'''11'1'1' ""l'it'H, § I rl, p. ·111 I'ilil/:, §If" p. ·111 I'd\\m!',\', §2, p. Ii (\Olllpoll'"I.,
*7, p. 20
i.l"nl, *2, p. Ii HIIIIIIIIlIlid" *.'l, p. 2·1 ~II01't."Ht. dIHwllqIlIHit.ioll, §K, p. 2:1 !'rilile (ide:d), §2, pp. 4, (i .--- clement, §13, p. 42 Prime divisor, §7, p. 19 - - of a primary ideal, §2, p. 6 imbedded - - , §7, p. 20 maximal - - , §7, p. 19 minimal - - , §2, p. 5 Principle of idealization, §1, p. 2 Product, §1, p. 2 Projective module, §26, p. 94 Pseudo-geometric ring, §3G, p. 131
Q Quadratic dilatation, §38, p. 141 Quasi. - - -local ring, §5, p. 13 - - -semi-local ring, §5, p. 13 - - -unmixed, §34, p. 124 R Radical, §2, p. 5 Jacobson - - , §4, p. 12 Ramified, §38, p. 145 Regular --local ring, §9, p. 27 - - ring, §28, p. 100 - - sequence, §17, p. 54 - - system of parameters, §9, p. 27 Relation module, §26, p. 91 Ring of quotients, §6, p. 15 - - with respect to, §6, p. 15 S,II II-polynomial, §20, p. 67 Semi-local ring, §5, p. 13
NI\IIJi \I I'i 111(\, ~~, p. Ii Ne\lnmhle illk.p;I'II.I ('IOHllm, §·I:I, p. IHI) NDparahly gOIlOI'lti,O(I, §:l!J, p. I·((i analytically - - , §47, p. 1!J~)
Separating t.ranscendence base, §30, p.149
-sequence (e.g., M-sequence), §27, p. 94
- - in, §27, p. 97 Shortest primary decomposition, §8, p.23 Splitting - - group, §41, p. 159 - - ring, §41, p. 159 Structure theorem of complete local rings, §31, p. 106 Sum, §1, p. 1 Superficial element (of a homogeneous ring), §22, p. 71 - - (of an ideal), §22, p. 72 Surject.ive, §1, p. 1 Symbolic power, §7, p. 20 System of parameters, §24, p. 77 - - (of a local ring), §9, p. 27 distinct - - , §25, p. 82 Syzygy, §26, p. 92, §29, p. 102 T
Tensor product complete - - , §42, p. 169 local - - , §42, p. 169 Theorem of transition, §19, p. fi4
1'1'1 1" , §IH, p. li:1 '1'"1.,,,1 "'I,,LinIiL l'illP;, *0, 1'. II 'i'1·1I.IIH"I'IIII,mn{) deg'l'(le, § 1·1, p. II 'i'1·ivi:1.i p;r:ulaLioll, §X, p. ~I 'I'OI'Hioll
lJ
Unique factol'i~at,ion ring, §13, p. '12 Unmixed, §25, p. 82 Unmixedness theorem, §25, p. 85 Unramified, §38, pp. 144, 145 - - regular local ring, §28, p. 99 analytically - - , §32, p. 114 V
Valuation, §11, p. 36 - - ring, §11, pp. 34, 36 additive - - , §11, p. 36 multiplicative - - , §45, p. 190 theorem of independence of --s, §U, p. 38 Value group, §11, p. 36
w Weak syzygy, §26, p. 92 Weierstrass preparation theorem, §45, p. 191 Weierstmss ring, §45, p. 190
z Zariski ring, §16, p. 52 Zero divisor, §1, p. 1
