Ramification Theoretic Methods in Algebraic Geometry (AM-43), Volume 43 9781400881390

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Annals of Mathematics Studies Number 43

ANNALS OF MATHEMATICS STUDIES Edited by Robert C. Gunning , John C. Moore, and Marston Morse 1. Algebraic Theory of Numbers, by H e r m a n n W e y l 3. Consistency of the Continuum Hypothesis, by K urt G odel 6. The Calculi of Lambda-Conversion, by A lonzo C hurch 10 .

Topics in Topology, by S o lom o n L efsch etz

11. Introduction to Nonlinear Mechanics, by N. K r y l o f f and N. B o go liu bo ff 15. Topological Methods in the Theory of Functions of a Complex Variable, by M a rsto n M orse 16. Transcendental Numbers, by C a r l L u d w ig S iegel 17. Probleme General de la Stabilite du Mouvement, by M. A. L ia p o u n o f f 19. Fourier Transforms, by S. B o chner and K. C h an d r ase k h a r an 20. Contributions to the Theory of Nonlinear Oscillations, Vol. I, edited by S. L efsc h etz 21. Functional Operators, Vol. I, by J ohn

von

22. Functional Operators, Vol. II, by J ohn

Ne u m an n

von

Neum an n

24.

Contributions to the Theory of Games, Vol. I, edited by H. W . K uhn and A. W . T u c k er

25.

Contributions to Fourier Analysis, edited by A. Zy g m u n d , W . T r a n su e , M. M o r se , A. P. C a l ­ deron , and S. B ochner

26. A Theory of Cross-Spaces, by R obert S ch atten 27. Isoperimetric Inequalities in Mathematical Physics, by G. P o l y a and G. S zego 28. Contributions to the Theory of Games, Vol. II, edited by H. W . K uhn and A. W . T u cker 29. Contributions to the Theory of Nonlinear Oscillations, Vol. II, edited by S. L efsc h etz 30.

Contributions to the Theory of Riemann Surfaces, edited by L. A h lfo rs et al.

31.

Order-Preserving Maps and Integration Processes, by E d w ar d J. M c S han e

32. Curvature and Betti Numbers, by K. Y ano and S. B ochner 33. Contributions to the Theory of Partial Differential Equations, edited by L. B e r s , S. B ochn er , and F. J ohn 34. Automata Studies, edited by C. E. S hannon and J. M c C ar t h y 35. Surface Area, by L a m b e r t o C e sar i 36. Contributions to the Theory of Nonlinear Oscillations, Vol. Ill, edited by S. L e fsch etz 37. Lectures on the Theory of Games, by H arold W . K uh n . In press 38. Linear Inequalities and Related Systems, edited by H. W . K uhn and A. W . T uc k e r 39. Contributions to the Theory of Games, Vol. I ll, edited by M. D re sh er , A. W . T u c k e r and P. W o lfe 40. Contributions to the Theory of Games, Vol. IV, edited by R. D un can L uce and A. W. T ucker 41. Contributions to the Theory of Nonlinear Oscillations, Vol. IV, edited by S. L efsch etz 42.

Lectures on Fourier Integrals, by S. B o chn er . In press

43.

Ramification Theoretic Methods in Algebraic Geometry, by S. A b h y a n k a r .

RAMIFICATION THEORETIC METHODS IN ALGEBRAIC GEOMETRY by Shreeram Abhyankar

P R IN C E T O N , P R IN C E T O N

NEW

JER SEY

U N IV E R S IT Y

1959

PRESS

Copyright © 1959 / by Princeton University Press All Rights Reserved L. C. Card 59 - 11 071

Printed in the United States of America

PREFACE These are the notes of a semester course that I gave at Columbia University in the winter session of 1955 -5 6 . There is little overlapping of the existing books on algebraic geometry with the contents of these notes. One novel aspect of this course is a simple proof of the local uniformization theorem on algebraic surfaces over algebraically closed ground fields of characteristic zero by a straightforward application of ramification theory (no continued fractions!). The contents of these notes is mainly based on the works of Abhyankar, Krull and Zariski re­ ferred to in the Bibliography. For an understanding of the main stream of discourse, Van der Waerden’s Modern Algebra (Chapters I to VII and parts of XII, XIII, XIV) and Northcott's Ideal Theory (Chapters 1 to k ) are enough prerequisites. The writing down of these notes was partly supported by a National Science Foundation project in the Department of Mathematics at Harvard University, to which I am grateful. The following is a novel aspect of the style of presentation. At several places proofs are omitted and are referred to the original sources, mainly Abhyankar and Zariski and at times to Krull. But this omission is done in well formed chunks and the statements of such results is made quite understandable so that even with these omissions the reader can travel a clearly visible pathway. Moreover partly because of these omissions, the supplying of many of which would have required a fair amount of preliminary material, this pathway is quick and short cutting across the subject carry­ ing the reader a fair distance. It is well known that exhaustive treat­ ments of algebraic geometry are liable to become rather formidable. The usual attempt of avoiding this formidability has mostly resulted in the production of a merely introductory exposition which leads the reader hardly to the gateway of the subject. Hence the author decided on the present method of exposition which in a semester would give the student a taste of at least some middle ground, no doubt very highly selective, of the subject. This can be achieved at the first reading of this monograph. While if during the second reading the student attempts to read up on the omissions, the general references for which are always given, that should v

PREFACE serve as a good guide and motivation for reading of the original memoirs. Undoubtedly when he tries to study some of these memoirs he may have to go prepare some preliminaries, but that will always be indicated by the references in these papers and thus the student will not feel lost. This method of chronologically going backwards and forwards, I have always found instructive. , The construction of this 1shortT path owes its shortness only to a smaller extent to the above mentioned omissions. In the main it is due to things like the elegant, ramification theoretic, proof of uniformization; and the point of view that varieties are collections of local rings etc., i.e., to the prominence given to local rings. The first portion of the monograph can serve as a unified treatment of the ele­ mentary aspects of ramification theories required in algebraic geometry as well as in algebraic number theory. A possible sequel could include the following: (1) more useful lemmas concerning the topic of Section 7 ; (2) completions of local rings; (3) higher ramification groups of local rings (Hilbert-Krull); and (t) ramification theory of local rings for in­ finite extensions (yet to be developed). Definitely there are many ways of approaching our subject and many angles to look at it from. This provides the reader with one.

Shreeram Abhyankar Columbia University Harvard University October, 1956

vi

CONTENTS Page PREFACE

.............

INTRODUCTION....... I.

II.

......

GENERAL RAMIFICATION THEORY........................ 1. Elmentary lemmas from ideal theory........ 2. Primary decomposition in non-noetherian rings................................... 3 • Integral dependence....................... k. Linear algebra............................ 5 * Discriminant of an algebra................ 6. The discriminant ideal.................... 7 « Galois theory of local rings..............

v 1 k 5

9 11 21 25 31 35

VALUATION THEORY . 8. Ordered abelian groups.................... 9 . Valuations................................ 10. Specialization and composition of valuations 11. Ramification theory of valuations......... 12. Valuations of algebraic function fields

M tl k7

III.

NOETHER IAN LOCAL RINGS............................. 13. The dimension of a noetherian local ring.... ................ 1t. Quadratic transforms

66 66 72

IV.

TWO-DIMENSIONAL LOCAL DOMAINS. . .................. 15 * Limits of quadratic sequences............. 16. Uniformization in a two-dimensional regular local domain............................ 17 - Uniformization for a two-dimensional algebraic functionfield........

75

V.

55 59

61

75

79 85

VARIETIES AND TRANSFORMATIONS...................... 18. Projective varieties...................... 1 9 * Resolution of singularities of algebraic surfaces................................

89 90

BIBLIOGRAPHY......................................

95

9k

RAMIFICATION THEORETIC METHODS IN ALGEBRAIC GEOMETRY

INTRODUCTION Let k be an algebraically closed field and let An be the affine n-space over k, i.e., An is the set of all n-tuples [called points of A ^ (a) = (a1, ..., aR ) with a^ in k. Let f^(X) = fi (X1, ..., X^) = o (i = 1, ..., h), be a finite number of polynomial equations with coefficients in k. The set of points (a) = (a1, ..., an ) of A^ which are common zeros of f 1, ..., f-^, i.e., for which f 1 (a) = f2(a) = ... = f^(a) = o is called an algebraic variety V. f 1 = o, ..., fft = o is a set of defining equations of V. If f is the union of two proper subvarieties, then V is said to be reducible, otherwise V is irreducible. To an irreducible (algebraic) variety V can be attached an integer called the dimension of V. There are many ways of doing this. The intuitive description of dimension can be given in two ways: (1) It gives the number of degrees of freedom of a point allowed to move freely on V; (2) It gives the number of independent parameters in terms of which the n coordinates of a variable point of V can be expressed. There are many ways of making these intuitive concepts precise, e.g., (3 ) Set d = - 1 if V = 0 . d = o if V consists of a single point, d = 1 if V consists of more than one point, and if any proper irreducible subvariety of V is either a point or the null set and so on, i.e., indutively having defined irreducible varieties of dimensions - 1, 0, 1, ..., s, we define an irreducible variety V of dimension s + 1 by the

Simple Singular

1

2

INTRODUCTION

condition that V has irreducible proper subvarieties of dimension s and that any irreducible proper subvariety of V is of dimension at most s . If d = 1, 2 or 3 then V is respectively called a curve, a surface or a threefold. If V Is a d-dimensional irreducible variety in then d < n and if d = n then V = AQ . A point P of V is said to be a simple point of V if in the neighborhood of P, V has the structure of an affine A^. If P is not simple, then P is singular. If V has no singular points either in or at infinity then V is said to be non­ singular. Let V and W be two irreducible varieties of dimension d. Let g be an algebraic correspondence (i.e., a correspondence defined by algebraic equations) between the points of V and W.

V —

> w

It can be shown that the equations of g are rational if and only if g is single-valued at a generic point of V, i.e., single-valued at almost all points of V. Hence if g and g“1 are both rational, i.e., if g is hirational, we get a correspondence between the points of V and W which is one-to-one almost everywhere. If such a birational correspondence between V and W exists, then V and W are said to be birationally equivalent. Projective equivalence is a very special case of birational equivalence (circle and ellipse). One of the central themes of algebraic geometry (which is the branch of mathematics studying algebraic varieties or equivalently "theory of equations in several variables") is to in­ vestigate the properties of an algebraic variety V which are invariant under birational transformations (e.g., genus of a curve V 1 = number of holes in V 1 if k = C the field of complex numbers). Many questions of birational geometry can be studied only on a nonsingular variety. Hence, given a d-dimensional variety V it is of utmost importance to transform V birationally into a nonsingular variety or rather to know whether V can be thus transformed. This is the so-called problem of resolution of singularities. For curves, the solution is classical (Max Noether — father of algebraic geometry — Mathematische Annalen, 1876—18 8 4 ) for k = C, and for arbitrary k was given around 1930). For surfaces, the (first complete) solution was given by Walker in 1935 (Annals) for k = C, by Zariski in 1939 [see Z3 , Z5 in Bibliography] for fields of zero charac­ teristic, and in 195^ see [A2 and A6] I completed the proof for fields of nonzero characteristic. In the case of zero characteristic, Zariski gave a solution for threefolds in 1 9 ^ [Annals]. The climax of this course will be a simplified solution of resolution of singularities for surface. point on

Let V be a d-dimensional Irreducible variety, and let P be a V. Let K be the set of rational functions on V. Then K is

INTRODUCTION

3

a finitely generated extension of k of transcendental degree d and is called the function field of V. Let R be the set of rational functions on V which do not have a pole at P, then R is a noetherian ring with a unique maximal ideal M and is called the local ring of P on V. It turns out that a whole lot of information about V is concentrated in the function field K/k of V and the set of the local rings of the various points of V. Hence we shall adopt the attitude that a variety is de­ fined by its local rings. To take care of finer aspects of V one has to consider branches (or, equivalently, sequences of points) on V; this will be taken care of by valuations (which represent the arithmetization of the idea of a branch) of K/k.

CHAPTER I:

GENERAL RAMIFICATION THEORY

Summary. Let A be a domain integrally closed in its quotient field K, let K* be a finite separable algebraic extension of K and let A* be the integral closure of A in K*. Let [K* : K] = n. Let W and W* denote the sets of prime ideals in A and A*, respectively. Obviously P* e W* = > P = P* fl A e W, i.e., P* lies above the prime ideal P of A. We shall show that, conversely, for each P in W there is at least one member of W* lying above P, and there are at most n members of W* lying above P. In some sense, In general over P € W there lie n members (counted with their separable residue degrees) of W*, i.e., W* is an n-fold covering of W. Those P e W over which lie less than n "points" of W* are the "branch points" of this cover­ ing; they are given by a certain ideal in A called the discriminant ideal of A*/A. Now assume that K*/K is galois. Let P* e W* and P = P* fl A e W. Then with the pair P*/P we associate two subgroups of the galois group of K*/K called the splitting group and the inertia group; their fixed fields are the splitting field and the inertia field, the splitting field is contained in the inertia field. As far as P* is concerned, in passing from K to the splitting field, p splits, ac­ quires no residue field extension and stays unramified; in passing from the splitting field to the inertia field, P acquires the separable part of the residue field extension and stays unramified; and finally in passing from the inertia field to K*, there is pure ramification and the in­ separable part of the residue field extension if any. The above described situation of the n-fold covering W* of W is what we call "ramification theory" and it arises in two situations in algebraic geometry: (1) When A is the valuation ring of a valuation of K (in fact, for W we may take the set of all the prime ideals in all the various valuation rings of valuations of K. Then W is the Riemann mani­ fold of K and W* is the Riemann manifold of K*); (2) When A Is the affine coordinate ring of an algebraic variety V, in this case A is noetherian and W is in one-to-one correspondence with the set of ir­ reducible subvarieties of V; A* is an affine coordinate ring of a derived k

1. ELEMENTARY LEMMAS

5

normal model V* of V in K* and the map of ¥* to ¥ (whose inverse is n-valued) is the rational transformation from V* to V. Now the local rings on algebraic varieties do not satisfy the valuation condition (ex­ cept when they are one-dimensional) and the valuation rings do not satisfy the noetherian condition (unless they are real discrete). Hence, in this chaper, we develop a general ramification theory without assuming either of the two conditions; special aspects of the two cases will be dealt with in the following chapters. A breakdown of this chapter into various parts is clear from the contents. 1. Elementary lemmas from ideal theory. In this section we re­ call some lemmas from ideal theory; the proofs of most of them will be either left to the reader as exercises or referred to Northcott1s Ideal Theory cited as N. By a ring we shall always means a commutative ring with identity. For the ease of reading, we shall try to use the letters: A, B for rings; H, I, J for ideals; P, Q for prime ideals; Q, for primary ideals; R, S for semilocal rings and M, N for their maximal ideals. Let A be a ring and let P, Q, be ideals in A. Recall that Pis a prime ideal if P 4 A and one ofthe following three equivalent conditions is satisfied: (1) a, b e A, ab e P, a i P implies b e P; (2) H, I ideals in A, HI C P, H t P implies J C P; (3) H, I are ideals in A which properly contain P implies HI is not contained in P.. P is called a maximal ideal if P 4 A and if H is an ideal in A with P C H 4 A implies P = H. If P is prime and Q is primary, then Q is said to be primary for P if a, b e A, ab € Q, a £ Q implies b € P. By \Tq we denote the radical of Q. If P + Q = A then P and Q are said to be coprime. LEMMA 1.1. Suppose Q C be Q. Then either Q = P = A or [N, Lemma 1, p. 11.]

P C Iq and that ab € Q, a i p = > P is prime and Q is primary for

P.

LEMMA 1.2. Let P ^ P2 be prime ideals in A which are co­ prime and let Q 1, Q2 be ideals in A which are primary for P , P2, respectively. Then Q 1, Q2 are coprime. PROOF. Otherwise,there would exist a maximal ideal containing Q 1 + Q2 - Q. C M and M is prime implies Pj_ C M i = 1, 2, so that P 1 + P2 C M j A which is a contradiction. LEMMA 1.3.

Let

H 1, ..., P^

M in for

be pairwise coprime ideals in

Then n

n

i= 1

i= 1

n %=n % •

A

A.

6

I.

GENERAL RAMIFICATION THEORY

Let ..., an be given elements in such that a = a^ mod for i = 1,

A.

Then there exists

a

in

A

PROOF. For ±4 j, and Hj are coprime implies that there exists h^j € and hj^ e Hj such that h^j + hjj_ = 1 so that hj^ = 1 mod and 0 mod H .. Let

hi - II hji . jVi Then h^ = 1 mod and 0 mod Hj for all j 4 i* Let a = a1h 1 + ... an^n" Then a = a^ mod for i = 1, ..., n. To prove n = fl H^, we make induction on n; obvious for n = 1, so let n > 1 and assume true for n - 1. Then

n %cn Hj_=n %

i=1

i/1

i/l

and

n % n % n% c

i=i

since

H1 + H2 = A

=

i/2

i/2

we have n

n

« ! = (Hi + v

i=1

I n \ CH, n % \i^1 I

n

%

i=1

I n \ + h2 n Hi =h, \i?te I

=

h,

n

ii=1

%

n

+

%

i=1

I n n %

\

\i/l

/

/ n

,

\

+ h2 n % \i/2

n

= n % I

i=l

i.e., LEMMA 1.4. Let ..., Pn be prime ideals in A and Q, an ideal in A such thatQ i P^ for i = i, ...,n. Then there exists a cAsuch that a € Q and ai P^ for i = 1, ..., n. PROOF.A proof by induction on n is given in Proposition 6, p. 12 of [N]; a direct proof is as follows: omitting each Pj which is contained in some P^ with i / j we may assume that Pj £ P^ whenever j 4 i; then there exists a^. € ? • , ] [ Pi; fix b± e Q, with b^ i P^. Let

1 . ELEMENTARY LEMMAS ai = b±

Then Then

e Q, a € Q

n

a±j

7

•

i P^

and

and a^ e if j / i. Let a / P^ for i = 1, ..., n.

a = a 1 + ... + an «

LEMMA 1 .5 If the set M of all the nonunits of Ais an ideal then M is the unique maximal ideal in A. Conversely, if A contains a unique maximal ideal M then M is the set of all the nonunitsof If a ring M 1* Mn sayingthat also write

R

contains only a finite number of maximal ideals

then R ls called a semilocal ring; we shall express this by (R; M 1 , ..., Mn ) is a semilocal ring. If n = 1 we shall (R, M 1) for (R; M 1).

LEMMA 1.6. Let

B

be a

ring and

A

a subring of

B . Then we

have: (1 ) For an ideal I in A, I C IB fl A always, andI = IB fl A If and only ifthere exists an ideal J in B with J D A = I, i.e., if and only if I is a contracted ideal forthe ring extension B/A. (2)For an ideal J in B, J 1 (J fl A)B always, and J = (J flA)B if and only if there exists an ideal I In B with IB = J, I.e., if and only ifJ is an extended ideal for the ring extension B/A. J HA ~JJ 1 a one-to-one correspondence between the contracted and the extended ideals, and it preserves the ideal theoretic operations, +, fl, ., : . Now let S be a nonempty multiplicative subset of nonzero divisors in A; by As we denote the quotient ring of A with respect to 3 , i.e., the set of elements of the total quotient ring of A which are of the form a/b with a in A and b in S. If P is a prime ideal in A, we write Ap instead of A^_p . LEMMA 1.7. Let H the set of elements a/bwith k2]

be an ideal in A. Then HAg coincides with a in H and b in S. [N, Lemma 3, p.

.

LEMMA 1.8. Let P be a prime ideal the unique maximal ideal in Ap .

in a ring A.

Then

PAp

is

PROOF. This follows from the next lemma; a direct proof Is as follows: Let x e Ap, x / PAp . x e Ap implies x = a/b with a, b e A, b i P. Since x / PAp, a / P by Lemma 1 .7 and hence x“1 = b/a€ Ap, i.e., PAp contains all the nonunits in Ap . Since any element in PAp is of the form a/b with a e P , b / P so that a / b ; therefore PAp does not contain 1 and hence It does not contain any unit of Ap. Hence, by Lemma 1.5,

PAp

is the unique maximal ideal in

Ap .

A.

8

I.

GENERAL RAMIFICATION THEORY

LEMMA 1.9. Let P be a prime ideal in ring A, let A* = Ap P* = PA*. Then: (1 ) Every ideal in A* is extended. (2) If H is an ideal in A then HA* = A* if and only if H fl P / H . (3 ) Contraction and extension sets up a one-to-one correspond­ ence between all the prime ideals of A* and those prime ideals of A * which are contained in P. Let P 1 and P 1 be two such corresponding prime ideals in A* and A; then we also have a one-to-one correspondence between ideals in A* which are primary for P and ideals in A which are primary for P 1. (2 *) If Q is an ideal in A which is primary for a prime ideal not contained in P then QA* = A*, i.e., Q is lost in A*. (^) If H 1 , - • - > Hn are ideals in A then (H1 fl ... fl Hn )A* = H 1A* n ... n. HjA*. (5) A/P and A*/P* are canonically isomorphic. (6 ) If A is noetherian, then so is A*. and

PROOF.

ring

Most parts are in [N, Propositions 9 to 11, pp. M - 4 3 ].

LEMMA 1.10. Let W Then A = Hpe^- Ap.

A.

be the set of all the maximal ideals in a

Now let A be a ring and let A 1, ..., AQ be nonzero ideals in A such that A = A 1 © .•• © An (direct sum). Then Ap is a ring and this is a direct product; let ep be the identity in Ap, then Ap = epA and e^, ...,en are orthogonal idempotents (i.e., e? = ep and eiej =

0

i ^ j ) with

PROOF. For Ai fl Aj =(o), i.e., A = A^ © ... © Ah is a A. such that 1 = e- + 2 2 2 1 = 1 = ep +... + en, aie2 +

••• + aien

1 = e 1 + ... + en -

ap e Ap, ap / A , with i / j we have aj_aj e apaj = °i therefore the decomposition direct product. Let ep be the unique element of ... + e . Thenfor i / j, e.e. = 0 and hence 2 ^ i.e., ep = ep; for ap g Ap,ap =apl = ape^ +

= aiei

30 that

ej_

the identity inApandAp = epA.

Conversely, let e ^ ..., eR be orthogonal idempotents in A with 1=e1 + ... + en and letAp = epA. Then it is easily verified that we have the direct sum A = A 1 © .. . © A ^ One instance in which such a decomposition of the identity into orthogonal idempotents occurs is the following: Suppose we have {0} = H.jH2 ... where the Hp are pair­ wise coprime ideals in A. Since the Hp are pairwise coprime there ex­ ists apj in A such that apj = 0 mod R. if j 4 i a^d apj = 1 mod Hp. Let ei -

n aij j/i

2. Since by Lemma 1 .3 ,

PRIMARY DECOMPOSITION

{0} = R

H2 ...

= H 1 fl H2 fl ... fl

it it

follows that e? = e^, = 0 if* i ¥ j and 1 = e-j + ••• + en * Then we have the direct sum decomposition A = A 1 @ .. . © Ah where Ap = epA. Let now f . be the canonical homomorphism: A -- > A/H. = A. and let f * * be the homomorphism of A into A 1 © .•. © Ah given by: f(a) = (f^a), ..., fn (a)). Then fp maps Ap isomorphically onto Ap and for j 4 i it is zero on A.. Therefore f maps A isomorphically onto J ¥: ¥: A 1 © ... © Ah and hence we may identify A with A 1 + ... + An 2. Primary decomposition in nonnoetherian rings. The aim of this section is to prove the following proposition which gives a primary decomposition for certain types of ideals in an arbitrary (not necessarily noetherian) ring A. PROPOSITION 1.11. Let H be an ideal in A with H 4 A such that there are only a finite number of prime ideals P , ..., PR con­ taining H and each of thePp is a maximal ideal. Then H = Q 1fl ... fl (= 0,-iQo ••• Qn ) where Q.is primary for Pj_. Furthermore theQ. are # * unique, i.e., if A = Q 1 fl ... n ^ where Qp is primary for Pp, then %

= Qj_ for

1=1,

. . n.

The proof of this proposition will be preceded by several lemmas Let H and I be ideals in A and h an element of A. H is said to be prime to I if (I : H) = I; h is said to be prime to I if hA is prime to I, i.e., if ah g I with a A implies a g I. h is said to be a zero divisor mod I if for some a in A with a / Iwehave ah g I, i.e., if the residue class of h modulo I is a zero divisor in A/I; H is said to be a zero divisor mod I if each element ofHis a zero divisor mod I. Notethat h is a zero divisor mod I if and only if hA is a zero divisor mod I. Also observe that h is prime to I if and only if h is not a zero divisor mod I. Now assume I / A and let S(I) denote the set of elements a in A which are prime to I; then S(I) is a nonempty multiplicative subset of A and H (respective ly h) is a zero divisor mod I if and only if H fl S(A) =0 (respective ly h i S(I)). If S is any nonempty multiplicative set in A then by Ig we shall denote the set of elements a in A such that as € I for some s € S; it is clear that Ig is an ideal In A with I C Ig C A; Ig Is called an isolated component of I and also the S-component of I; it follows from Lemma 1 .7 that if S contains no zero divisors of A then Ig = A D IAg. If P is a prime ideal in A we shall write Ip for

I^-P

andcall it the P-component of

I.

LEMMA 1.12. Let S be a nonempty multiplicative subset which is disjoint from a given ideal I / A (in particular we may I = {o)). Let W be the set of all ideals in A which contain are disjoint from S. Then W contains maximal elements and they

of A take I and are

10

I.

GENERAL RAMIFICATION THEORY

necessarily prime. Furthermore, let I1be an Ideal contained in I and let the set of all ideals in A which contain I1 and are disjoint from Then any maximal element of is also a maximal element of W.

¥ 1 be S.

PROOF. Obviously W has the Zorn property. Let P be amaximal element of W. Let H 1, H2 be ideals in A which properly contain P; then for i = 1, 2 there exists h^ € with h^ € S so that h^h^ G ^1^2 and h 1h2 € S so that i^hg i P; hence I^Hg t P. Therefore P is prime. The last assertion is obvious. In particular, zero divisor ideal of I

if S = 3 (1 ), we shall say that P is a maximal and that Ip is a principal component of I.

If P is a prime ideal in other prime ideal between I and P mal prime ideal of I.

A such that I C P and there is no then we shall say that P is a mini­

LEMMA 1 .13 . Let H and I be Ideals in A different from A such that H is a zero divisor mod I. Then there exists a maximal zero divisor P mod I such that P D H. PROOF. H is a zero divisor mod I implies that H + Iis a zero divisor mod I and hence H + i n S ( I ) = 0 . By Lemma 1.12, there ists a maximal zero divisor P mod I such that P D H + I 0 H. LEMMA 1.1I+. Let I be an ideal in A different from I = the intersection of all Its principal components.

A.

ex­

Then

PROOF. Clearly I is contained in the intersection. Now let u be an element In the intersection and let H = (I : uA). Then [P is a maximal zero divisor ideal of I] = > [u € Ip] = > [there exists v in A, v i P such that uv e I so that v e H] = > [H t Pi • Hence, by Lemma 1.13 > H is a nonzero divisor ideal mod I, i.e.,H contains a nonzero divisor h mod I. Now h e H implies hu € I and hence u e I.

P P

be a is a

LEMMA 1.15. Let I be an ideal in A different from A and let minimal prime ideal of I . Then (1 )Ip Is primary forPand (2) zero divisor mod I.

PROOF. Toprove (1) we have to show that (i) Ip C P C vTlp, and that (ii) u, v e A with uv e Ip and u i P implies v e Ip . Nowu € Ip implies there exists v e A, v i P such that uv € I so that uv e P and hence u e P. Therefore Ip C P. Next, given u e P let S be the smallest multiplicative set con­ taining u and A - P. If Iwere disjoint from S then byLemma 1.12 there would exist a prime ideal Q containing Iand disjoint from S. But Q, Is disjoint from S implies that Q, is disjoint from A - P, i.e., Q CPwhich in view of the minimality of P implies that Q = P so that

3.

INTEGRAL DEPENDENCE

u i P

which is a contradiction. Therefore I is not disjoint from S, there exists v e A - P and a nonnegative integer n such that unv e I. Now unv g I and v i P imply that un € Ip, i.e., u € vTlp. Therefore P C */lp. This proves (i). I.e.,

Next, let u, v g A with uv g Ip andu i P. Since uv g Ip, there exists w € A, w i P such that w(uv) is in I. Since u i P we have wu i P; since (wu)v = w(uv) is in I we conclude that v g Ip. This proves (ii) and hence (1).

w / P

Finally, let ug P. Then, in view of (i), un g Ip for some n whence unv g I for some v in A with v i P so that v i I. There­ fore un is a zero divisor mod I and hence u is also a zero divisor mod I. Thus P is a zero divisor mod I which proves (2).

actly Qp

the

PROOF OF PROPOSITION 1.11. By 1.15 (2); P], ..., PR are ex­ maximal zero divisor ideals of H. Let Qp = Ip . By 1.15 Cl),

is primary for

Pp

and by 1.1k , n

h

=

n

%

•

i= 1 Since the Pp are pairwise coprime, uniqueness follows from [N], Theorem

by 1.2 and 1.3 we have DQp= nQp. 2 on p. 1 5 -

The

3. Integral dependence. Let A be a ring and B an overring of A (with the same identity). Let P and Q be prime ideals inA and B respectively; if Q D A = P then we shall say that Q lies above P or P lies below Q. An element u of B said to be integral over A if there exist aQ, a1, ..., an_1 in A such that un + an 1'an~1 + ... + aQ = o. If every element of B is integral over A we shall say that B is integral over A and express this by saying thatB/A is integral. If A is a domain such that A contains every element of its quotient field which is integral over A, then A will be said to be normal. LEMMA 1.16 (Kronecker). Let A be a normal domain with quotient field K. Let f(X), g(X) be monic polynomials in K[X] and h(X) = f(X)g(X). Then h(X) € A[X] implies f(X), g(X) g A[X] . Let K* be an overfield of K. Then u g K* with u/A In­ tegral implies that the minimal monic polynomial of u/K is in A[X]. PROOF. The second part follows from the first. To prove the first part, let L be a root field of h(X) over K. Let n f(X )

= £

i=o

n fn .iX 1 =

n

1=1

(x - u p

and

12

I. GENERAL RAMIFICATION THEORY

with f\ e K, fQ = 1 and Up e L. Let P q ^ , • • • > Yn ) (_1 times the i-th elementary symmetric function in , ..., Y , so that fi = Pp(u.j, ..., un ). Now f(u.) = o implies h(Up) = o which implies that Up/A is integral. Since the integral closure of A in L is a domain, u ], *••> ^ ape integral over A implies that fp = pp(u.j, un ) is integral over A. Since A is normal and fp e K we must have fp e A for i = 1, ..., n, i.e., f(X) e A[X] . Similarly, g(X) e A[X] . Another proof of the above lemma using valuations will be given in Chapter II, see Corollary 2.27.

A.

Then

LEMMA 1.17. Let Ap is normal.

PROOF. is integral over

A

be a normal domain and

a prime ideal in

Let u be an element of the quotient field of Ap. Then

un + a ^ 11”1 + ... + an = 0 a^ e

P

Ap implies that ap = t>p/Cp

,

with

with bp, c. e A

and

Awhich

a^ e Ap Cp / P.

Let

n e=

n

ci

^ 1 = bi

and

i=i Then

= d^/e

II

ci •

jVi

with d^, e € A

and

e i

P.

Let

v = eu

and a| =

d^e1-1.

Then vn - a'v11-1 + ... + ♦ ...

= e1^ 11 + d ^ 11"1^ " 1 ♦ d2een-2un-2

+ dnen-’ = en[un + ( d p e ) ^ " 1 + (d2/e)nn-2

+ ... + (dn/e)J = o and

that

a|€ A.

Since A

is normal,

v €

, A.

Since e i P, u = (v/e) € Ap.

LEMMA 1.18. Let B be a domain and A a subdomain of B such B/A is integral. Then A is a field if and only if B is a field.

PROOF. Assume A is a field and integraland hence u/Ais algebraic so that B is a field. and hence

let u-1

0 / u e B. Then u/A is e A[u] C B. Therefore

Now assume that B is a field and let o / u e A. Then u“1 e B u”1/A is integral, i.e., (u~1 )n + an_1(u“1 )n_1 + ... + aQ = o

withap e A so that fore A is a field.

u”1 = - (a

1 + an_2u + ••• + a0un-1) e

There­

3.

INTEGRAL DEPENDENCE

13

LEMMA 1.19. Let B be a ring and A a subring of B such that B/A is integral. Let P be a prime ideal in A lying below a prime ideal Q, in B. Then P is maximal if and only if Q, ismaximal. PROOF. ¥e may canonically assume that (B/Q)/(A/P) is integral. Now apply Lemma 1.18.

B/Q, 0 A/P.

Then

LEMMA 1 .20. Let B be a ring and A a subring of B such that B/A is integral and let P be a prime ideal in A. Then there exists a prime ideal Q in B lying above P. PROOF. Let ¥ be the set of ideals J in B for which J fl A C P. Then ¥ has the Zorn property and hence has a maximal element Q. ¥e assert that Q is prime. For u, v e B; u, v / Q and uv € Q = > (Q + uB) H A P; (Q + vB) fl A / P = > there exist u*, v* e A, i P such that u* = au (mod Q) and v* = bv (mod Q) with a, b e B = > u*v* = abuv (mod Q) = > u*v* e P which is a contradiction. Hence Q is prime. Now assume if possible that Q fl A / P. Then there exists u e P, / Q so that Q + uB properly contains Q and hence (Q + uB) fl A / P, i.e., there exists v e A, / P and b e B such that v = bu (mod Q). Since B/A is integral, we have bn + an_lt)n"1 + ••• + aQ,

with

a. e A

Therefore (bu )n + uan-1 (bu)n_1 + u2an__2(bu)n"2 + ... + unaQ = o Since

v = bu (mod Q,)

.

we have

vn + (uan_1)vn_1 + (u2an_2 )vn_2 + ••• + (unaQ ) = 0 (mod Q) Since the left-hand side is in A, it must be = 0 (mod P). we have vn e P and hence v e Pwhich is a contradiction. Q flA = P. LEMMA 1.21. Let B be a ring P be a prime ideal in A. Then there lying above P < = > PB fl A = P.

.

Since u e P Therefore

and A a subring of B, and let exists a prime ideal Q,in B

PROOF. (==>) Obvious. ( P* fl (B - Q*) = 0 = > P* C Q*. Also p* n s =0 = > (p* n a) n (a - p) = 0 = > p* n a c p. since P* fl A D PB fl A 3 P, we have: P* PI A = P. PROPOSITION 1.2 5 . Let A be a normal domain withquotient field K and let A* be the integral closure ofA in a finite normal extension K* of Kwith [K* : K] = n. Let P be a prime ideal in -g* A. Then there are only a finite number of prime ideals P^, . P in A* lying above P, with m < n; and P*, ..., P* form a complete set of K-conjugates. * PROOF. By 1.20, there exists a prime ideal P. in A* lying -H* -Xabove P. Let P2, ..., Pm be the other K-conjugates ofP1; we must then have m < n and P . fl A = P for i = 1,..., m . Suppose if possible that Q* is a prime ideal inA* lying above P with Q* 4 Pj_ for i = 1, ..., m. By 1 .24 , Q* 4- P^ for i = 1, .. ., m, and hence by 1A there exists u e Q* such that u £ P. for i = 1, ..., m. Then 1 * none of the K-conjugates of u canbe in any P^ and hence Nk*/t^-(u) i P? for i = 1, ..., m. But g Q* ^ A C P* which is a contradiction. LEMMA 1.26. Let A be a normal domain with quotient field K, let K* be a finite algebraic extension of K, let A* be a domain such that A C A* C K* and such that A*/A is integral, and let P be a prime ideal in A. Then there exist only a finite number of prime ideals in A* lying above P. PROOF. Pass to the integral closure of extension of K containing K* and invoke 1.25.

A

in a finite normal

DEFINITION 1.27. Let (R, M) be a normal local domain with quotient field K and let R* be the integral closure of R in an ex­ tension field K* of K. Let M*, M*, ..., be the prime ideals in R* lying above M. It follows by 1.1 9 that M 1, Mg, ..., are exactly the maximal ideals in R*. Let S^ = R^* and N^ = M^S^. We shall say that (S.,, N1 ), (S2, N2 ), are the local rings in K* lying above (R, M). If [K* : Kj is finite it follows by 1.26 that there are only a finite number of local rings in K* lying above (R, M); let M 1,.. •, be the maximal ideals in R*: then we shall use the expression: Let [the semilocal ring] (R*; M 1, ••., Mn ) be the integral closure of R [or of (R, M)] in K*; if furthermore K*/K is normal, then it follows by 1.25 that (S^ N1), ..., (S , Nn ) form a complete K-conjugate set.

3•

INTEGRAL DEPENDENCE

17

LEMMA 1 .28. Let A be a domain with quotient field K and let P be a prime ideal in A. Let R = Ap and M = PR. Let K* be a field extension of K, let R* be the integral closure of R in K* and let A* be a subdomain of K* integral over A. Then it is clear that R* DA*. Let M*, M*, ..., be the prime ideals in R* lying above M and let P? = M? 0 A*. Then (1) P*, P*, ..., are exactly the prime ideals in A* lying above P (although they need not all be dis­ tinct). Now assume that A* is the integral closure of A in K*. Then (2) Pg, ..., are all distinct and R^* = Ap* for i = 1, 2, ... PROOF. (1) Obviously P*, P*, ..., lie above P. Now let P* be any prime ideal in A* lying above P. Let S = A**, N = P*S and S* = the integral closure of S in K*. Let N* be a maximal ideal in S*. Then N* fl S = N. Now A* D A

and

P* fl A = P

.

SDR

and

N PI R = M

S* D R*

and

N* fl R* = M?

Therefore

Therefore for some

i

Therefore P* = N* PI A* = Mp fl A* = Pi (2)

Obviously

1

normal. Therefore Ap* D R*

RM

and

D Ap*. 1

Also

P?Ap* fl R* = Mj

Ap* D R i

. and by 1 .17,

for some

Ap* i

j

Therefore

-^M* ^ ^P* ^ ^M* j i i Therefore M . - R* fl

Therefore *

Rm *,

M jRj^* - R* fl PpAp* - R* fl R^* - Mp j i i

** * R^* = Ap*. In the above argument we have shown that RM*> i i 1 * * are all distinct. Therefore P1, Pg, ..., are all distinct.

is

I.

18

GENERAL RAMIFICATION THEORY

Another proof of part (1) of the above lemma, using valuations, will be given in Chapter II, see Corollary 2.28. LEMMA 1.29. Let (R, M) be a normal local domain with quotient field K and let (S, N) be a local ring in a finite algebraic extension K ’ of K lying above R.Then S fl K = R and N flK = M. PROOF. Let K* be a finite normal extension of K containing K !. Let R 1 and R* be the integral closures of R in K* and K* * respectively. Let M l = N D R f and let M be a prime ideal in R* lying above M 1. Then M. is above M. Let Mp,..., M. be the other maximal ideals in R*. Let Sp= R^*, Np = and let G be the galois group of K*/K. Now M 1 = M* n R 1. ^Therefore S* D S and hence * 1 1 S 1 fl K 0 S flK. Hence toshow that S fl K = R it is enough to show that S* H k =R. Let u e S* H K.Given S? there exists g e G such that 1* * ' * 1 * gs 1 = Sp and hence gu e S.. But u e K implies gu = u. Hence u e Sp for i, ..., t, i.e.,

- (n si) ■r*> by 1.10. Therefore u € R* fl K = R andhence S fl K = R and hence N f l K = N flR = M.

S*

fl K = R . Therefore

DEFINITION 1.30. Let K be a field, K*a finite algebraic extension of K and S a normal local domain with quotient field K*. It follows from 1.29 that if there exists a normal local domain R in K such that S lies above R, then R is unique. Hence if R exists, we are justified in calling R "the local ring in K lying below S." LEMMA 1 .3 1 - Let A be a normal domain with quotient field K, P a prime ideal in A, K* a finite algebraic extension of K, K ! a field between K and K*, A* and A* the integral closures of A in K f and K*, respectively. Then A* is the integral closure of A f in K*. Let PJ, ..., P* be the prime ideals in A 1 lying above P. Let * * * Pjj > • • • > be the prime ideals in A* lying above P^. Then the u, + ... + u_^ ideals P?_. are all distinct and they are exactly the prime 1 s * * ideals in A* lying above P. Let = A£, and T^j = Ap* .Let Sp be the

integral closure of Sp in K* and let N ^ , ..., N£y be the * maximal ideals in S^. Then v^_ = u^ and after a suitable relabelling we have T± . = (S?)N# for j = 1 , . u± . ij PROOF.

Exercise in view of what we have proved up to now.

3. In particular, if

INTEGRAL DEPENDENCE (A, P) = (R, M)

19

is a local domain, then we

have

Level of K*(T11,0 11

1 i N* = > N* fl A* / A*. Let P 1 be a maximal ideal in A* containing N* fl A* and let R* = Ap* and M* = P*R** Then 3 * D R* and R* is a local ring in K* lying above R. Let R* be a local ring in K* lying above R*. Then R* is above R and hence by 1.25 there exists a K-automorphism g of K* such that

flR=

20

I.

GENERAL RAMIFICATION THEORY

g(R*) = R*Let g(S1 ) = S* and let S ’ = S* fl K 1. Then S' Is a local ring in K 1lying above S and S* 3 R* = = > S* 3 R* = > S ’ 3 R T. LEMMA 1. 32B. Let (R, M) (S, N) , K, K f be as in 1 . 3 2 A. Let (S’, N ? ) be a local ring in K 1 lying above S. Then there exists a unique local ring (Rl, M* ) in K 1 lying above R such that S ’ has center M 1 in R 1. PROOF. Let A f be the integral closure of R in K f. Then S’ 3 A '. Let P* = N 1 fl A f. Then P 1 fl R = ( N ! fl A 1 ) fl R = N 1 fl R = (N* n S ) f l R = N f l R = M and hence P* is a maximal ideal in A* . Hence R* = A £ f is the required unique local ring. h . Linear algebra.Let

A be a ring. An element a of A is called nilpotent if a11 = 0 for some positive integer n. An ideal H in A is called a nil ideal if all its elements are nilpotent; H is called a nilpotent ideal if Hn = {0} for some positive integer n; ob­ viously a nilpotent ideal is a nil idea. Observe that the set of all the nilpotents in A is an ideal, in fact it is equal to nT(o}. If J { 0} = {0 }, i.e., if A has no nonzero nilpotents, then A is said to be semisimple. The intersection of all the maximal ideals of A is called the radical of A and is denoted by Rad (A). H is said to be an Artin ring if the descending chain condition for ideals holds in A. A is said to be a primary ring if A contains only one prime ideal. LEMMA 1 .3 3 - Let S(A) denote the setof all the elements ring A for which (1 + xa) is a unit for all x in a. Then S(A) = Rad (A).

a

in

PROOF, a i S(A) = > there exists x in A such that (1 + ax)A / A = > there exists a maximal ideal M in A such that 1 + ax e M = > ax i M = = > ax i Rad (A) = > a £ Rad (A). Therefore Rad (A) C S(A). a i Rad (A) = > there exists a maximal, ideal M in A such that a i M = > M + aA = A = > there exist m e M andx e A such that m + ax = 1, i.e., l - a x = m € M = > 1 - ax is a nonunit = > a i S(A). Therefore S(A) C Rad (A). LEMMA 1.3^.

Let

A

be an Artin ring.

Then

PROOF, a e sT{0 } = > a11 = 0 for some n = > (1 - ax)(l + ax + a 2x 2 + ... + anxn ) = 1- an+ 1xn+1 = 1 a e Rad (A). Hence nT{0 } C Rad (A).

Rad (A) = nT{o }. for all x in A, and hence by 1 . 33,

Now let a e Rad (A). Since A is an Artin ring, there exists n such that anA = an+ 1A, i.e., there exists x in A such that a11 = an+ 1x; also by 1.33 there exists y in A such that (1 - ax)(1 + y) = 1, i.e., ax(l + y) - y = 0, i.e.,

k.

c = an (ax)(l + y) - a^V = Rad (A) C VCO}. LEMMA 1 .3 5 -

21

LINEAR ALGEBRA +y)

- s P j = a11;

In an Artin ring

A

therefore

every nilideal

a e nT{o}.

H

Hence

is nilpotent.

PROOF. Since A is Artin, there exists n such that Hn = Hn+1 = Hn+2 = ... = I, say. Suppose, if possible, that I 4 0 and let ¥ be the set of all ideals J in A such that JI 4 o. Since A is Artin, W contains a minimal element K. Then there exists k in K such that kl 4 0so that kA e V. Since kA e ¥ and kA C K we must have kA = K. Now KI2 = KI / o and hence KI e ¥. Since KI C K we must have kl = KI = K = kA. Therefore there exists i in I such that 2 n k = ki and hence k = ki = ki= ... = ki = ... . Since i is nilpotent, we must have k = o,i.e., KI = o,which is a contradiction. Therefore 1=0. LEMMA 1.36. Let A be a primary Artin ring. Then every zero divisor in A is nilpotent, i.e., belongs to nT { o ) = Rad (A), and Rad (A) is the unique prime ideal in A. If Rad (A) = {0} then A is a field and conversely. PROOF.

Straightforward.

PROPOSITION 1 .3 7 * Let (1)

A

A

be an Artin ring.

Then

contains only a finite number of prime ideals

P 2 ’ •••' Ptn' (2 ) Each P^ is maximal. (3 ) A has a unique direct sum decomposition

(*0

A = B 1 © — © Bn ¥here are ideals each being a primary ring and we have n = m. Let e^ be the identity in B^. Then after a suitable rearrangement we get ( 1 mod P^

ei =

|

( 0 mod P .

(5)

(6) (7)

(8)

if

i 4 j

.

Let N and N^ be the radicals of A and B^ respectively. Then N^ = Ne^ = Pj_ajA N = N, © ... © N m . Every chain of ideals in A is finite and hence in particular A is noetherian. If A is semisimple, then each B^ is a field and B 1, •.•, Bm are exactly the nonzero minimal ideals of A. If A is an algebra over a field K and if ta^.},

I.

22

GENERAL RAMIFICATION THEORY

j = 1, ..., is a K(= Ke^)-basis of the alge­ bra B^, then [a^ •} (i = 1 , m; j = 1 , ..., n^) is a K-basis of A and hence [A : K] = z[B^ : K(= Ke^)]. Finally, if A is semisimple, then is an over­ field of K(= Ke^) [Theorem of Dedekind]. PROOF. Let N be theradical of A.By 1 .3 ^ and 1 .35 ^ the set of allnilpotents in A and there existss such that Ns Since A is Artin, A has only a finite number of maximal ideals ..., Pm and by 1.2 and 1.3 ve have m

s

s

For

i

4 j,

Pj

1=1

t Pj_

s

n8- n p±- n p± •

n=n pi =n i= 1

N is = o.

1=1

1=1

exists x^je Pj,i

and hence there

P^.

xi =n x±j • jVi Then xi i P1 and exists c^ in A Then

xi € P. if j 4 i. Since A/P^ is a field, there such that Cj_x^ = UP^). Let e^ = 1- [1 - (c^x^)3]3 .

1

mod P3

o mod

J[

P^

in

For

i / j,

eiej €n Pk=n3=o , m

k= 1

i.e.,

ej_ej = °*

Also m e±(l - e± ) € I] Pk - Ns - 0 k=l

i.e.,

e^l - e^) = o,

i.e.,

,

e^ = 1 . Finally

m (e, + ... + em - 1 ) 6 P| k=i

= Ns = o

,

Let

i.e.,

e 1 + ... + em = 1 • Let

A = B 1©

Bj_ = e^A.

... ©

Bm

Then we have the direct sum decomposition

which is also a direct product decomposition.

j / 1, e 1 e Pjj e iP j

contains

e^ = e 1

Since for

which is the identity in

B1

and

hence m

n

pke i = pi ei

•

k=i

4

Since for

j

f.j

homomorphism

of

N 1 = P 1e 1

is a maximal

ideal in

nilpotents

of

is the

1>

e • €P 1,

B1

we conclude that prime ideal in

and since N1

B1

which implies

then

Nj_ =

of all nilpotents

P] D B2 ©

A onto A

B1

B 1.

.. . ©

Bm = kernel f 1

given by: f ^ a ) = a e 1 . Since

N 1 =N e 1

is Artin implies B 1

is the only maximal ideal of

P = N 1.

Similarly

we have

is the set of all

is Artin, by 1.3^

B 1.

Now if

N 1 = (the set of all nilpotents in Therefore

N1

in B^.

P

is a

B1) C P

is the only prime ideal in

isthe only prime ideal in Therefore B^

where

Therefore

B^

and

is primary.

B 1.

N^ is the set P

Again

is a

prime i.e.,

ideal in A = > N C P = > P-jPg**• Pm ^ P = = '> Pi ^ ^ Por some P = Pj_. That N = N - , © • • • © ^ is now obvious, in fact for any

ideal

H

we have H = H 1 Now let

A = D1 ©

©

.. . @

—

©

where

= He^.

Dn be a direct sum decomposition

of

A

into nonzero ideals D . . Then 1 = d 1 +... + d_ with d .€ D . , p l 1 n l i7 d i = d i' d id j = 0 Por 1 ^ J % = d iA * Also = Nd^ = radical of D^. Assume D 1, . .., Dn are primary. Then is the unique maximal ideal in

D^.

Now

d 1 = d ie i + ••• + d iem = die i

= di

Therefore

c^e^

iseither zero or

every nonnilpotent is a unit,d ^ ^ or B 1©

B. fl

= {o}.

••• ©

Therefore P i = ^i

Bg .

••• + d ? e m

o

or

Similarly after a suitable Hence

p =

e^.

B

reordering:

Mn = o.

M

4

{0},

Consider the B-modules

of them is annulled by

M,

B^,

and

of

B.

let

B/M, M/M2,

n

B^

B^ C D 1

D1 = . . . © D^..

B 1= D 1 ©

we have:

(1 ) to (5 ).

i.e., to prove that

every chain of ideals is finite.

the radical and hence the only prime ideal So assume

Since in

n = m and after a suitable reordering

..., m. Thus we have proved statements

in a primary Artin ring

and

B^.

Therefore either

reordering we have:

To prove (6), it is enough to prove it for each

nothing to show.

•

is an idempotent in =

Hence after a suitable

B 1 = P 1. Por

+

If

M =0

Let

M

be

there is

be such that Mn - 1/Mn .

Mn“ 1

40

Since each

they may be considered to be modules over the

I.

2k

GENERAL RAMIFICATION THEORY

field B/M. They are Artin rings and vector spaces and hence they are finite dimensional vector spaces. Therefore there are only a finite number of B-modules, i.e., B-ideals between and for i = 1, ..., n. Therefore by the Jordan-Hoelder theorem every chain of ideals in B is finite. This proves (6).Now let us prove (7 )- N^ = o implies B^ is a field which implies that B^ is a (nonzero) minimal Ideal. Now let P be any other (nonzero) minimal ideal in A. Then Pe^ = 0 or B^ and hence P = B^ for some i. Finally, now (8) is obvious. 5. Discriminant of an algebra. Let A be a ring, and K a subfield of A such that A is a finite-dimensional vector space over K, say [A : K] = n. Let (w) = (w1, ..., wn ) be a basis of A/K. Given a in A let TQ d be the linear transformation of the vector space A into itself given by Ta (b) = ab for b e A. Let n

awi =I kifj J =1

with e K so that is matrix of Ta with respect to the basis (w^, ..., wn ). Define thecharacteristic polynomial of a with respect to A/K by: Cha (X) = ChAyK , a (X) = det (XE - ((k^ .))), [E = unit matrix] = Xn + c^Xn ~ ] + ... + cn

.

Also define: n S(a) = SA^K (a) = (trace of a relative to A/K) = - c1 = ^ k ^ i=1

,

N(a) = NA ^ ( a ) = (norm of a relative to A/K) = (- 1 )ncn = det ((k^j )) . Let (w*) = (w*, ..., w*) be any other basis of A/K and let L be the matrix: (w*) = L(w). Then the matrix of Ta relative to the basis (w*) is: ((k?j )) = L ”1((k^ .))L and hence det (XE - ((IC-lj ))) = det (L“1(XE - ((k±j))L) = det (XE - ((k*^))) Thus ChQ(X) and hence S(a) Cl choice of the basis of A/K.

and

N(a)

•

do not depend on the particular

Let

a, b

e

5.

DISCRIMINANT OP AN ALGEBRA

A.

Let Ta : wi — Tb

:

> awi

Wi —

25

= I k i j wj

> bW± = Z ^ j W j

*

Then

Therefore

N(ab) = det(((kij. ) ) ( ( i 1j ) ) ) = det ((k^Jtoet ((•‘ ■g)) (1 ) = N(a)N(b)

S(a + b) = ^ ( k i± + i±1) = ^ k i± + £ i ±j (2 )

= S(a) + S(b) Let

k e K.

Then kav. = £ (kkij)wj

•

Therefore S(ka) = kS(a) T1 :

-- > w^.

Hence

and

S(l) = n

and

S(k) = nk Let

(a^

.

and

(3 )

N(ka) = knN(a) N(1) = 1.

Therefore by ( 3 ),

N(k) = kn

.

(k)

an ) be an arbitrary set of elements of

A.

We define:

D(a1 , — , an ) = DA ^K (a1 , — , an ) = (discriminant of (a^, • • • , aR ) relative to A/K) = det ((S(a^aj))) Let

(b1, • • • , bn ) be another set of elements of e K. Then bybu = | ) k„4a< £ kviai) ( l kujaj) 1

.1

A

with

b^ = Ek^a^,

= I. kviaiajk]u ij .T

'

26

I.

where

kju =

^uy

GENERAL RAMIFICATION THEORY

Therefore

K S ft v ,)) )'-

( ( 3( £

,:« sl a j kju j j

■ ( ( ^ W R ^ u ) )

= ((ki j ))((S(aia.)))((k^j ))

.

Therefore DCb,, Let

bn ) = tdet ((k±j))]2 D(a.,,

now (w*) = L(w)

be any other basis of

w*)= [det (L)]2D(w ). though

D

Since

e^

A = A.j (+)... © A^

be the unit of A/K.

(5 )

.

D(w*) =D(w^, D(w) =

...,

o.Hence,

depends on the basis, it makes sense to say that

A^,

For

A/K

is zero (or nonzero)"

Kj_ = Ke^,

and

(i = 1, ••.,

a € A, a =

=

m; j = 1,...,

n^)

ae1+ ... + aem -

£ k jJ)eiwiu u= 1

(with

k ju} e K)

= I kj i )wlu • u= 1

Hence ni awi j = ( I aei) wij = aeiw ±j = I k j i )wiu u=1 Therefore

SA/K^a ^ -

A,

(w^ •) (j = 1, ..., n^)

ni (ae±

(6)

be a direct product decomposition of

let

basis of A^/K^. Then of

Then

det (L) / o, D(w*) = o < = >

"the discriminant of Now let

A/K.

a^)

+ ••• + 3Am/Km (aem ) »

let be a

is a basis

5.

DISCRIMINATE OF AN ALGEBRA

27

m NA / K ^ =

n

>

NA /K:laeP

1=1

and

i / u = >

'Wj_j^r uv = 0 = = >

DA/K^w 1 1 ’ w1 2 ’

= °*

Therefore

•••■’ Wm 1 ' W21 ’

•••»

wtnn1J

“ det ((SA/K(wijwu v )))

=

m IJ det ((SA /K ((wijWiv)))

[j, v = 1 ,

1=1

n± ]

,

and

\ / K ± {wU >

wln1 ) = det ((SA1/K1 (wijwiv)}) = det ((SA/K((wljw lv)))e1

.

Therefore The discriminant of A/K is zero < = > of A^/K^ is zero for i = 1, ..., m

the discriminant (7 )

Now suppose that the radical N of A is nonzero, i.e., A has nonzero nilpotents, and let s be such that Ns_1 4 0 and Ns = o (s > 1, N° = A). Then A = N° > N 1 > ... > Ns = {0}. N1 is a vector space over K and hence we can find elements , ..., p^ in N^" whose residue classes

modulo

and then fPjj) (i = °> A/K. Now a £ N = >

N^+1 —

N^/I (1+1

form a K-basis of the vector space > s “ 1j j =

•••>

Q.j_)

is a basis of

aPlj. = Y kijvuPvu (kljvu € K) = = > S(a) = 0 v>i

N 4 0 = > there exists a basis w 1, ..., wn of with wn e N = > discriminant of A/K is zero

.

(8)

A/K (9 )

Recall from field theory that in the special case when A is a field we have: For a given a e A let a = a^1 \ ..., a^n ^ be the con­ jugates of a (counted with proper multiplicities). Then

28

I.

GENERAL RAMIFICATION THEORY

S(a) = a^1 ^ + ... + a ^ , N(a) = a ^ a ^ ... a ^ , D(a.j, ..., an ) = det ((a|^))2 and discriminant of A/K is zero if and only if A/K is inseparable.

(10)

Now assume that A 1, ..., A^ are primary rings, i.e., A =A 1 © ...© Am is the unique decomposition of Areferred to in 1 .3 7 Then N^ is theunique prime ideal in A^ andA^/N^ is afield. We define [A^ : K]s = (the separable degree of = If

(A^/N^)/K^

over

K)

[(Ai/Ni ) : K±]a

is separable for A

Now (7 ),

A^

i = 1, ..., m

is separable over

(11 ) then we shall say that K

(12)

(9 ) and (10) yield

PROPOSITION 1.38. is semisimple and separable

A/K

has a nonzero discriminant

< = = > A/K

m [A : K] = Y [Ai : Kl i=1 Observe that any element a e A satisfies a polynomial equation f(a) = 0 where f(X) is a nonzero monic polynomial in KtX] of degree at most n. If asatisfies no equation of degree less than n,i.e., if 1, a, ..., a11”1 is a basis of A/K, then we say that a is a primitive element of A/K. Observe that if a is a primitive element of A/k and if f (X) = Xn +k-jX11”1 + .. . + kn is the unique monic polynomial of degree in

KtX]such that

f (a) = 0,

then we have

Da/k^) = -^A/K^19

*••, a*

) = En (k1, •••, ^n ) 9

-H -

where Dn (X1, ..., Xn ) is an element of ZtXj, ..., XQ] completely de­ termined by n where Z is the ring of integers (in fact D* is the discriminant of f in the elementary sense). Lemma 1 .3 9 * In the above notation, suppose that A/K is semi­ simple separable and assume that K is infinite. Then A/K has primitive elements. In view of 1.38, the proof of this lemma follows from the next more general lemma.

5.

29

DISCRIMINANT OF AN ALGEBRA

LEMMA 1 .Lo. Let K overring of K. Let A 1, ...,

be an infinite field and let A be an be nonzero ideals in A such that

A = A-| 0 ... © Let e^ be the identity in A^ and let Kj_ = Ke^. Assume that A^ isa primary ring in which the unique prime ideal N^ consists entirely of nilpotents (i.e., N^ is the set of all the nilpotents of A^). Let q1, .••, qm be positive integers such that qj_ < [(A-j/KjJ : Then (1 ) there exists a in A such that a does not satisfy anyequation in KtX] of degree < q 1 +••• + qm ; (2) assume furthermore that for some j either Nj / (o) or that [(A./N.) : K.]. > 1. Then there exists a in A such that a does not J

J

J

satisfy any equation in K[X] of degree < q1 + — + qm - [If [A : K] = n < «> then we can state this as: (1 ) Lemma 1 .3 9 ; and (2) if A/K is not semisimple separable, then there exists a in A such that a does not satisfy any equation in PROOF. Cj_.

class of

For

K[X]

c^

in

of degree

< 2i=i *-Ai :

A^ we shall denote by

the N^-residue

Observe that since any zero divisor in Nj is necessarily a nilpotent we have the following: For aj in Aj let H be the ideal of polynomials f(X) in K*(X) such that f(a-) = 0, then H = u(X) K.[X] r J J # J where u(X) is the monic minimal polynomial of aj/^j it is a power r of u(X) which is irreducible and is the minimal monic polynomial of a. over K.. J

J

We are going to choose an element Fix bj in Aj such that the monic minimal over K. is of degree s . ^ q.. If N. = 0 J * J J J set a . = bn * and g n*(X) = the minimal monic J *

J

J

-x-

a. in Aj for j polynomial kj(X) and [A. : K-L = j j 1 polynomial of a. j

= 1, ..., m. °T ^j 1, then we over K-. J

Then = and deg g j ^ = = sj - ^ j ' Now assume that either N* 4 0 or [(A./N.) : K . L > 1. We want to show then that there exists J J J J 1* / \ a. inA. whose minimal monic polynomial g.:(X) over K. is of degree / \ // 0 we may set a- = b., so assume that h •(b / .)\ = 0. t. > q«. If h.(b.) J

J

J

J

J

J

J J

CASE 1. [(A./N.) : K*]. > 1. Then by the theorem of primitive J J J elements [W1, p. 126] there exists a^ in A . such that the minimal monic polynomial of a,j over Kj is of degree > Sj + p > q^ (where p is the characteristic of K.). J

aj Of

CASE 2. [(Aj/Nj) : = bj + Cj. Then hj(bj)p 4 0 hj(X)),

hj(bj) t Nj-

i.e., W

= 1. Then there exists 0 4 Cj € Nj. Let for all r (here hj(X) t s t h e derivative Now

= hj(bj + cj } = cj[bj(V

+ cT (bj )]

with f(X) e K,tX], hl(b.) i N. and c.f(b.) e N. = > h i ( b . ) + c.f(b-j) i / X J .J J J J J J J J J J N- = > hub,*) + c.f(b-) is a nonzero divisor = > h.(a.) 4 0. Since J

J

d

J

J

J

J

50

I.

GENERAL RAMIFICATION THEORY

h.(a-) = 0 for some r and since h.(X) is prime, for the minimal J J * J * -p monic polynomial g-(X) of a./K. we must have: g,-(X) = h.(X) , with J * J J J J r > 1, i.e., tj = deg g j(X) > q •. Thus the existence of aj has been established. g,(X) = XtJ + k-.xh"1 + *•• + kjt . be the monic ^ o l j -

Now let nomial in

KLX]

such that e / j

*

* •••

■

Multiplying the a^. by ouitable elements of the infinite fields may assume that g1 (X), ..., g^X) are pairwise coprime. Let a = a1 + ... + am . Then f(X) = kQXq + ... + kq e K[X] with fO'\a.) = o, J

where

divides

f (j)(X)

g-(X)

divides

f(X)

J

Therefore for the degree K we have:

d

f(a) = 0 = >

in

in

K-tX]

for

J

KtX]

for

j = 1, ..., m = >

j = 1, ..., m

of the minimal monic polynomial of

=^

we

f^'lx) = e.k X^ + ... + e^k = = > J ^ J

g tx) J

K.

a

over

+ ••• +

d ^ ti + ... + tm > q + ... + N. ^ 0 or J

if for some j we have [(A-/N.) : K-] > 1 J J J

6. The discriminant ideal. Let A be a normal domain with quotient field K. Let K* be a finite algebraic extension of K and let A* be a subdomain of K* with K* as quotient field such thatA* con­ tains A and is integral over A, let[K* : K] = n. Let (w*, ..., w^) be a basis of K*/K. Since K* is the quotient fieldof A*, we can write wi = ui^vi

with

u ±>

vi 6 A*v. 4 o.

Let n

a

= wi n

vj

•

j= i Then

(w1, ..., wn )

Then

w^w^ e A*

Therefore

is a basis of

and hence >

K*/K

and

SK*/K^w iw j^ € A

***> wn^ €

define

w^ e A* since

A*/A

for

i = 1, ..., n.

is integral.

6.

THE DISCRIMINANT IDEAL

31

D(A*/A) = (the discriminant ideal of A* over A) = the ideal in A generated by the discriminants of all the bases of K*/K which are in A*. Now let

A fbe the integral closure of D(K*/A)

= (thediscriminant of

A

in

K*.

K* over

We define A) = D(Af/A) .

Now let P be a prime ideal in A , then by 1 .26 there are only * a finitenumber of prime ideals P1, ..., Pmin A* lying above P. Let s. = Ap* and N, = pts.. Then A*/P< = S,/N, 3 A/P. If for a given J j « J J J J J J element of A* we reduce modulo P the equation of integral dependence over A, we obtain an equation of integral dependence for the corresponding residue class modulo P. over A/P. Therefore A*/P. = S«/N. is an J J J J algebraic extension of A/P. Now assume that A is a local ring with P as its maximal -ft-*■ ideal. Then(A*;P1, ..., Pm ) is a semilocal ring. Let P* be any primeideal in A* such that P* 3 PA*. Since P*, ..., P* are all the maximal ideals of A*, we must have P* C pt for some j. Now P C P* and hence P C P * f l A C P ^ n A = P , so that P* fl A = P and hence * J * * P* = Pj by1.19. Hence P ^ ..., Pm are the only prime ideals in A* which contain PA*. Therefore by 1.11 we can uniquely write PA* = Q* fl•-• fl Q^, where Qj is primaryfor Pj. By 1 .9 ; PSj = QjSj and Qt =q!s. n A* = PS. n A* and PS. = N. if and only if Q/t = ?*-. Since J J J J J J J J PS is primary for Nj, is a primary ring whose unique prime ideal Nj/PS coincides with its ideal of nilpotents and it is semisimple if and only if PS = Nj. Thus the following three conditions are equivalent: (1)

PSj = Nj and(Sj/Nj)/(A/P)

is

separable;

(2)

(S./PS.)is semisimple separable over(A/P); J J (3 ) = P^ and (A*/pt)/(A/P) is separable. J J J If one, hence all three, conditions are satisfied, we shall say that Sj (respectively Nj or Pj) is unramified over A (respectively over P). If S. is notunramified over A then we shall saythat S. (respectively J * . J Nj orPj) is ramified over A (respectively over P). If S1, Sg, ..., S are all unramified over A, we shall say that A* is unramified over A; otherwise we shall say that A* is ramified over A. Now let us get back to the general case when A is not nec­ essarily local and assume that A* is the integral closure of A in K*. Let R = Ap and M = PR. Let R* be the integral closure of R in K*. Let Mj = R* fl Nj. Then by 1.28, M*, ..., are exactly the maximal

32

ideals in

I.

GENERAL RAMIFICATION THEORY

R*,

S . = R™* and N. = M*js.. We shall say that P^ (or N. «j j j j j j j or Sj) is ramified or unramified over P (or A) according as S. is ramified or unramified over R. We shall say that P is a branch ideal or a nonbranch ideal of A-for the extension A*/A (or for the ex­ tension K*/K, or for K > K * ) accordingas R* is ramified or un­ ramified over R . LEMMA l.M. Let (R, M) be a normal local domain with quotient field K, let K* be a finite algebraic extension of K and let K* be a finite algebraic extension of K 1. Let (R1, M 1) be a local ring in K' lying above Rand let (R*, M*) be a local ring in K* lying above R 1. (1) If MR' = M f then MR* = M* < = > M*R* = M*; (2) If R* is unramified over R then R* is unramified over R = > R* is unramified over R 1. PROOF. Obvious. Lemma 1 .1+1A. Let the notation be as in 1.1+1. Assume that R containsa subfield k and a finite number of elements a1, ..., a^ such that R = Ap where A = k[a^ ..., a^] and P = A PI M. Then (2a) If R* is unramified over R, then R*is unramified over R. The proof of this lemma is rather involved, and hence we refer to the considerations of Section 2 of [A2] from which it follows directly. Now we are ready to prove the discriminant theorem in its local form. THEOREM 1 .1+2 . Let (R, M) be a normal local domain with quotient field K, let K* be a finite algebraic extension of K, let R* be a domain with quotient field K* such thatR* contains and is integral *#■ over R. Let M ^ .. ., M^ be the maximal ideals in R*. Then t V

[(R*/M^) : (R/M)l

< [K* : K]

,

k and the equality holds if and only if D(R*/R) = R [i.e., if and only if there exists a basis (w1, ..., w ) of K*/K in R* such that %*/K^w l' *•*' wn^ nonunits of R)].

is a

111 R

(since M

is the ideal of all the

PROOF. We shall give a proof under the assumption that R/M is infinite. For an extensionof the proof for finite R/M see [Kl ]. Let [K* : K] = n and [(R*/Mj) : (R/M)]g = g y Let N*= MR* = N* fl ... n N* where N^ is primary for M^. Let A = R*/N* and A. = R*/N?. Then A J J J J * and A. can be supposed to contain the field k = R/M. Let Q. = N./N* * J J and P. = M./N*. Then in A the zero ideal has the decomposition J J

6.

THE DISCRIMINANT IDEAL

33

Co} = Q 1 fl ... n Qj. into pairwise coprime primary ideals Q .. Hence A = A 1 @ ... © A ^ . Since Qj is primary for P., it follows that the ■unique prime ideal P. in A.coincides with the ideal of nilpotents of J J A.. It follows hy 1.16 that every element of A satisfies an equation J of degree at most n over k. Hence hy 1Ao, t £

t [(R*/Mj) : (R/M)]s = J

J =1

Let us agree to denote for N*

[Aj : k]g < n

.

J =1

a in

R* the residue class

modulo

by a.

Now assume that g 1 + ... + g^ = n. Then by 1.^0 and 1.16, A. has no nonzero nilpotents [hence N. = M^l, i.e., A. is a field and it tJ J tJ is separable over k, i.e., [A. : k] = [A. : k] = g*. Hence A. is a J J J J finite-dimensional algebra over k for j = 1, ..., t. Hence A is a finite-dimensional algebra over k and hence by 1 .38 A/k has anonzero discriminant. Therefore by 1 .39 we can find a primitive element u of A/k. Then 0 / D( 1, u, ..., u11-1 ) = the residue class, modulo M of D(1, u, ..., un"1) andhence D(1, u, ..., un“1) / M, i.e., it is a unit in R andhence D(R*/R) = R. Now assume that D(R*/R) = R. Then there exists a basis (w^, ..., wn ) of K*/K in R* such that D(w1, ..., wR ) is a unit in R. Certainly R* D w.jR + ... + w r R. We now show that WjR + ... + wnR = the integral closure of R in K*. Given w in K* we have w = a1w 1 + ... + anwn with &• e K. Also w / w^R + ... + wnRimplies some a. i R, say a1 i R; then D(w, w2,..., wn ) = a^Dfw^ w2, ..., wR ) R = > w/R not integral. Therefore R* = w^R + ... + wnR = the integral closure of R in K*. Therefore (w^ ..., wn ) is a k-basis of A and N* = w^M + ... + wnM* Let n (wiW j )wP = I a± j p q V q=1

Wlth

ai j p q € R

•

Then n (*

i

W

I 5ijpq\ q=1

•

Therefore D(w1, ..., wn ) = the residue class of

D(w1, ..., wn ) modulo M.

I.

3^ Hence

D(w.j, •.«, wR ) 4 °*

GENERAL RAMIFICATION THEORY Therefore, by 1 .38 ,

g 1 + ... + g^ = n.

The above proof yields the following two results. PROPOSITION 1 .4 3 . Let the notation be as in 1 .4 2 . Then D(R*/R) = R = > R* is normal (i.e., R* is the integral closure of R in K * ) and has an R-module basis of [K* : K] elements (which are there­ fore linearly independent over K), i.e., R* is a free R-module.

R = >

THEOREM 1 .4 4 . Let the notation be as in 1 .4 2 . Then is unramified over R.

R*

D(R*/R) =

We remark that the converse of 1.44 is not true in general (see [K1 ] ); however, it holds under certain assumptions, namely we have THEOREM 1 .44A. Let A be a finitely generated domain over a field and P a prime ideal in A. Let R = Ap, M = PR, K = the quotient field of R, K* = a finite algebraic extension of K and R* = the in­ tegral closure of R in K*. Then R* is unramified over R = >

D(R*/R) = R.

For the proof of this theorem, we refer to its original source [A2 , Section 2]. We can globalize Theorem 1.42 as follows: THEOREM 1.45. Let A be a normal domain with quotient field let K* be a finite algebraic extension of K and let P be a prime ideal in A. Let A* be the integral closure of A in K* and let *

K,

*

P 1' ***'

Ptbe the Prime ideals in A*

k

lying above

P. Then

t

V


D(R*/R) = R where R = Ap, M = PR and R* = the in­ tegral closure of R in K*. ( = > ) D(A*/A) 4- P implies that there ex­ ists a basis (w1, ..., wR ) of K*/K in A* such thatD(w1, ..., wn ) i P. Then Wj e R* and D(w1, ..., wn ) is a unit in R and hence D(R*/R) = R. ( < = ) D(R*/R) = R implies that there exists a basis (w1, ..., wn ) of K*/K in R* suchthat D(w1, ..., wn ) is a unit inR. Fix a^. e R such that m.

Wj Let

a^j =

with

m.-1

+ aj1Wj

+

cj_j € A

+ ajmj = ° ^

*

c^j 4 p * Let

- GALOIS THEORY OP LOCAL RINGS

cj - n

35

•

1=1

Then

w.c . e A* J

and

J

D(w1c1,

n cjj

f

D [g(Nj) = N± with 1 4 j = > N. U K ' = g(N- (1 K 1) = L (1 K 1 which is a contradiction] = > g e = G(K*/Kj). Therefore .

(so that

G(K*/K') C G(K*/ld) and hence J

THEOREM 1 .14-7 - We have Q Rj is unramified over

(1) R).

id C K'. J

R^/M^ = R/M J

and

J

(2)

MR^ = M5? J

J

PROOF. Let Tbe the integral closure of R in K*. Let Tj = Tn Kj, N± = M* fl T, N^ = M? fl Tj for i = 1, ..., u. ... Given a € Tj, by 1.46 and 1 .3 there exists b in Tj such that b = a mod Nj and 1mod N? for all i / j, so that b = a mod Nj and 1 mod N^ for all i4 j. Let Gj = G 1, G2,. . . , G be the left cosets of Gj in G. Fix g^ in G^. Then g^b), ..., gq(b) are the K j/K conjugates of b (counted properly) and hence q ° = V / K ( b ) = 11 k j /k t=i

Now g1(b) = b = a that g-fc(Nj_) = Nj for all t 4 “I, gj. (b) = 1 that c = a mod N^j. R^/M^ J

To prove maximal ideals in Tj,

J

J

g i (b)

•

mod N.and for any t4 1 so thatg^(b) = 1modN y mod N .. Hence c € R and = T^/N^ = R/M. This proves J

J

there exists such i Thus g.j(b) =a c = a mod Nj so (1 ).

(2), let X 1 = Njandlet X2, ..., Xy be the other so that giveni / j, N| = X^. for some t 4 1

4 j mod

Njand

7.

37

GALOIS THEORY OF LOCAL RINGS

and given t 4 1, Xt = N? for some i 4 j. Then MTj = Y 1 n ••• PI Yy = Y 1 ... Yv where Y^ is primary for X^. Let Z = X 1 fl ... fl Xy = X 1 ... Xy . Observe that [a e X 1, i X^. for i = 2, ..., v] = = > [there

,

exists a* g Tj, 4 X 1 such that aa* e M ( Y^ for instance let a* = n^=2g^(a)] = > [a g Y 1]. By 1.3 there exists e g Tj such that e = 0 mod X 1 and 1 mod X^ for all t 4 1 • Then by the above observation, e € Y^ . Now [f e Z] = = > [f + e € X.j, i X^ for all t 4 1 J s= > [f + e e Y 1] = = > [f e Y^ ]. Therefore Z C Y ] and hence X 1 C Y 1 so that X 1 = Y 1 . Therefore MR^j = X.R^! = MR*.

J

J

J

This completes the proof of 1 .4 7 Now let Ij = the set of g for all u € A*;

in

G(K*/K)

for which

it = the set of g J for all u e R ..

in

GS(Rt/R) J

for which

gu = u mod pt

gu = u mod Mt J

J

Then clearly ! • and I. are subgroups of G and Gj respectively; in fact, it Is obvious that g e l . = > g(M*t) = Mt = > g e G^; thus I. * J o J j J * J and I . are both subgroups of G ••Now let g e l . and v e R *. Then 4 J * J J and v2 i P.. Hence v v1/v2 with v^ v2 g A*

=

,

, , g(v, ) V g (V ) - V = --= g(v2 ) Vg

g( V

1

g(v )v2 - v.g(v ) =g(v2 )v2

)v2 - v ^ g + Vlv2 - vlS(v2 ) g(v2 )v2

V 2 (g(v1 ) - V, )) - V^gCVg) - Vg)) g(v2 )v2

e Therefore g e l . = > * J fore I. = I.. J

g

g

J

J

it.

v? i M?

[since That

J

g

g

J

it = >

and g e G? = = > g(v2 ) i M"t] J

J

gel.

is obvious.

j

We define G^(Rt/R) = Gi (Mt/M) = Gi (pt/P) = (the inertia group of J

R,* ■ h

J

over - ij-

R

or

J

MJ

over

^

M

or

P. j

over

P)

J

There

38

I. fRrJ/R)

=

R • over

= F 1 (P*/P) =

F1(Mj/M)

R

J

GENERAL RAMIFICATION THEORY

*

or

M . over

J

= the fixed field of Let R^ = J

fl r !

J

J

M

J

(the inertia field of

or

*

P . over

G^(R^/R). J

Gj = Gi(Rj/R), Kj = F1(r!/R), R^ = K^ and

P)

fl

By

Mj = Kj

fl

= ri fl M?.

J

J

J

Observe that since Rj is the only local ring in K* lying R^|, wehave: R? = the integral closure of R? in K*, GS(R^/R^) = G(K*/Kj ),FS(Rj/Rj) = Kj and G1 (Rj/R) = G^Rj/R?). * M •. J

*

For

a g R., J

Let

_

by

H] - R j / H - , For g gGs(Rj/R) itself by the

m!,

awe shall denote the residue class

Hj - R j / M j ,

above

of a mod

. Rj/M ] - R/M - H .

let us define a transformation f(g) equation: f(g)a = g(a). Then

of Hj

into

(1 ) f(g) is single-valued: PROOF, a = 5 = = > a - b g Mj = > g(a) - g(b) = g(a - b) g Mj [since . g g G s (Rj /R)] = = > g (a ) = g(b ). (2) f(g) is an automorphism of Obvious, since g is an automorphism of R./R. J

THEOREM 1.^8. h j /His normal and f is a homomorphism of Gs(Ri/R) onto G(Hi/H) and g R r F r ) = kernel of f, K^/k ! is galois. H'V J J J * J J J is the separable algebraic closure of H in Hj and f canonically in­ duces an isomorphism of G(Kj/Kj.) onto G(Hj/H) = G(Hj/H). Also wehave = M^R^ = MR^ (so that R^ is unramifiedover R as well as over R^). J

J

J

J

J

J

PROOF. Since all the groups in question are subgroups of Gs(Rj/R) and all the subfields of K* in question contain and since H^j = H, s it is clear that in the proof we may replace K by K- , or in other words we may assume that (Rj, Mj ) is the only local ring in K* lying above R. Hence we may drop the subscript j all the way through and observe that now

GS(R*/R) = G(K*/K) For

a g H*

and

K3 = K.

let

q(X) = the minimal monic polynomial q(X) = the minimal monic polynomial

of a/H; of a/K.

and

a g R* = > a/R integral = = > q(X) g R[X]. Let q*(X) be the (monic) polynomial gotten by reducing the coefficients of q(X) modulo M*. Then q(a) = 0 = > q*(a) = 0 = > q(X) divides q*(X). Now K*/K is galois

7.

GALOIS THEORY OF LOCAL RINGS

= >

with a^_ e K*. Now all t . Therefore

q(X) =

39

deg q n (X - at ) t=1

q(X) e R[X] C R*[X]

and hence by 1.16

a^ e R*

for

deg q q(X) =

n u= 1

(X - 5t ^ u

wlth

S1-6 H* u

•

Therefore H*/H is normal. It is obvious that f is a homomorphism into. Let H ’ = the separable algebraic closure of H in H*. Then by 1 . k 2 , H ’/H is finite algebraic and hence a galois extension. Assume that a is a primitive element of H t/H. Then there exists gu in G(K*/K) such that gu (a) = a^ .Then fCg ^ a = a^ . Therefore f is onto. For g eG(K*/K), [f(g) = 1 ] < = = > [for all b inR* we have f(g)b = b, i.e., g(’b ) = b, i.e., g(b) = b mod M*] < = > [g e Gi (R*/R)]. Therefore kernel f = G^(R*/R) = G(K*/K^-). Therefore G(K*/K^-) is a normal subgroup of G(K*/K). Therefore K^/K is galois. Also [K1 : K] = [index of G(H*/H)] = [H* : H].

GiK' /K1 )

in

G(K*/K)J = [order of

Obviously G ^ M * ^ 1 ) = G ^ M ^ M ) = GCK^/K1 ). Therefore GCH^/H1 ) is isomorphic to GCK-fr/K1 ) / G ( K * / K ^ ) = 1. Therefore H*/H^ is purely in­ separable. Hence H' C H1 . Therefore [H1 : H]s = [H1 : H]= [K1 : K] . Hence by ^ , H1 = H l and MR1 = M1 .

G(Hi/H).

Let Let

q denote the canonical isomorphism of F = qf. Then

G(H*/H)

onto

homo onto . P : G(K*/K) -------- > GClT/H) kernel P = GCK^/K1 ) For

g e G(K*/K)

set

p(g) = g/K1 . Then

homo onto . p : G (K*/K) -------- > GOr/K) kernel p = GCk V K 1 ) Let

f* = Fp~1 . Then As

f

f*

is an isomorphism of

was defined for

R*/R

we define

GfK^/K)

onto for

GCH^/H).

R^/R.

Then

^0

I.

for g e G(K*/K) we have completes the proof.

field R, K*

GENERAL RAMIFICATION THEORY f^^(p(g)) = f.

Therefore

g = f*.

This

PROPOSITION 1 . k 9. Let R be a normal local domain with quotient K, K*/Ka galois extension, R*a local ring in K* lying above afield between K and K* and R ! = K Tfl R*. Then GS(R*/R') G 1 (R*/R') FS(R*/R') F^(R*/Rl) PROOF.

= = = =

GS (R*/R) n Gi (R*/R) 0 compositum compositum

G(K*/K!), G(K*/Kf), of FS (R*/R) and K l, of F^(R*/R) and K T.

Straightforward application of galois theory.

PROPOSITION 1.50. Let (R, M) and (S, N) be normal local do­ mains with a common quotient field K; let K* be a galois extension of K and let (R*, M*), (S*, N*) be local rings in K* lying above R and S respectively. Assume that R C S and R* C S*. Then (1 ) (2)

G 1 (S*/S) C Gi (R*/R), and If S has center M in R, then

Gs(S*/S) C GS (R*/R).

PROOF. Let R' and S'be the integral closuresof R and S in K* respectively. Then R' C S'. Let N' = S' fl N* and M' = R'DM*. Let P 1 = R' n N'. Then N* J S* = > 1 i N* = > P U R * / R* = > N* fl R* C M* « = > P' = (N* 0 R*) O R 1 C M * OR' = M'. Now g e G 1 (S*/S) = > [a e R' = > a - g(a) e P 1 C M 1] = > g £ G^CRVR). Now assume that N 0 R = M. Then P 1 fl R =(R1 fl N' ) fl R = N' n R = (N* fl S' ) flR = N* fl R = (N* fl S) fl R = N fl R = M, and hence P' = M 1. Therefore g e G 3 (S*/S) = > [a e M' = > a £ N' = > g(a) £N' = > g(a) e N 1 flR' = pi = M'] = >

g £ G 3 (R*/R).

CHAPTER II:

VALUATION THEORY

8. Ordered abelian groups. In this section Z, Q and stand for the additive groups of integers, rational numbers and real numbers, respectively. In this section we shall consider ordered abelian groups a little more thoroughly than necessary, In order to give the reader a feeling about these groups.

A simply ordered set S Is a collection of objects with a binary relation ^ with the following properties: (1 ) (2)

a a^

(3)

a * b and b * a — a, b e S = > a ^ b

(* 0

We write

a > b

if

a b

a, b, ...,

for each a (reflexive) and b ^ c = > a ^ c (transitive)

a ^ b

and

> a = b , (t r i c h o t o t n y ) or b ^ a

a/b;

also we write

b < a

If

a > b.

An ordered abelian group S Is an abelian group which is simply ordered such that a, b e S with a ^ b = > a + c b + c for all c e S (i.e., the group operation is compatible with the ordering or vice versa). When talking about ordered abelian groups, by an ordered isomorphism (or homomorphism) is meant an isomorphism (respectively homomorphism) which preserves order. Let a e S. a is said to be positive or negative according as a > 0 or a < 0 respectively. Observe that a > 0 - a < 0. We define a

If a > 0

- a

if a < 0

For a subset T of S, by T+ we denote the positive part of T, i.e., the set of positive elements in T; the complement of T in S will be called the positive complement of T (in S). Observe that S+ is a subsemigroup of Ssuch that a e S = = > oneand only one of the following three conditions holds: (1 ) a € S+, (2) - a €S+, (3) a = 0. Con­ versely, given an abelian group S and a subsemigroup S+ with the above property, S can be (uniquely) converted into an ordered abelian group by setting a > b < = > a - b € S+ . in

R

will

II.

k2

VALUATION THEORY

Examples of ordered abelian groups. Z, Q, R with the usual ordering and any subgroup of an ordered abelian group (in particular, of Q or R ) with the induced ordering are all ordered abelian groups. Now let S1, ..., Sp be a finite number of ordered abelian groups and consider their direct sum S 1 © ... © Sp, the elements of which are of the form(s^ ..., sp ) with s^ e S^; let (t^ ..., tp ) with ti e S. beanother element of this direct sum different from (s^ ..., sp ) then we can choose q such that s^ = t^ for all i < q and s^ / tp . We set (s^ sp ) > (t,, tp ) or (t,, tp ) < (s^ sp ), according as sp > t^ or t^ < s^. This converts the direct sum into an ordered abelian group which we shall call the lexicographically (i.e., as in a dictionary) ordered direct sum of S1, ..., Sp, and we shall de­ note it by S 1 © ... © S p . If S 1 = ... = Sp = S then we shall write Sn for S 1 © ... © Sn * It will be seen that Rn and its subgroups are enough for algebro-geometric purposes to serve as value groups of valuations.

of

S

Now let S be an ordered abelian group. A nonempty subset T will be called a segment if t e T, s e S with |s| < t ===> s e T.

Let T i ^ T 2 t¥ 0 s e S ^ en^ s > then either thereexists t 1 € T 1, i T2 or t2 e T2, i T-j; if t 1 £ T 1, i T2 then t e T2 = > [|t| > |tj = > t 1 e T which is a contradiction] = > |t| < 111 | = > t € T 1, i.e., T 2C T 1; similarly, if t2 e T2, i T 1 then T 1 C T2 . Thus the set of all the segments of S is simply ordered by inclusion. An isolated subgroup of S is a subgroup which is a segment; it can easily be shown that a segment T is an isolated subgroup if and only if any one of the following four conditions is satisfied: (1

) T+ is a semigroup; (2) t g T, n e Z = > nt e T; (3) t e T, n € Z+ = > nt € T; (it) t 6 T+, n e Z+ = > nt e T+ . Since the set of all the segments is simply ordered by inclusion, hence so is the set of all the isolated subgroups of S; the order type of the set of the nonzero isolated subgroups of S is called the rank of S and is denoted by p(S). If there are only a finite number n of isolated sub­ groups, then p(S) is the order type of the sequence 1, 2, ..., n and we set p(S) = n. A subsetT*of S is called an uppersegment if T* C S+ and if t eT*, s e S with s > t = > s e T*. An upper seg­ ment T* is said to be isolated if s, t e S+, i T* = > s+ t i T*.Ob­ serve that a subset T* of Sis an upper segment if and only if it is the positive complement of a segment T and that T* is an isolated upper segment if and only if it is the positivecomplement of an isolated subgroup. Hence there Is a one-to-one inclusion reversing correspondence

8.

ORDERED ABELIAN GROUPS

I I I

1 I I 0 * * — *— *-------------------------*

ip

S

----- ( i l l

^

0

p*

7

*— *— *

between the set of all the segments and the set of all the upper segments and it induces a one-to-one inclusion reversing correspondence between the set of all the isolated subgroups and the set of all the isolated upper segments, and hence in particular the set of all the upper segments and the set of all the isolated upper segments are simply ordered by inclusion. Let h and k be two order types (an order type can be defined to be an equivalence class of simply ordered sets, the equivalence being the existence of an order preserving one-to-one map) and let H and K be two disjoint simply ordered sets of order types h and k respectively. Let L = H U K and order L as follows: Given a, b e L; if a, b € H then a ^ b if and only if a ^ b in H; if a, b g K then a ^ b if and only if a > b in K; if a € H and b € K, then a < b. Then L is simply ordered and the order type Z of L depends only on h and k and we set Z = h + k. Observe that if H and K are both finite, say consisting of H* and K* elements, respectively, then H, K and L are order isomorphic to the sequences 1, 2, ..., H*; 1, 2, ..., K*; and 1, 2, ... H* + K*, respectively; so that in this case the equation Z = h + k can be interpreted as giving the ordinary sum of two integers. EXERCISE.

Is

h + k = k + h

always true?

Now let S and S* be ordered abelian groups and let f be an order homomorphism of S into S*. Let s, t e S. Then |s| < |t| = > |f (s)| < |f(t)| and hence |s| < |t| and f(t) = o = > |f(s)| = o = = > f(s) = o.Therefore f“1(o) is an isolated subgroup of S. Now let f“1(o) = T and replacing S* by f(S) assume that f is onto. Then we have p(S) = p(T) + p(S*). [PROOF. It is enough to show that f induces a one-to-one correspondence between those isolated subgroups of S which properly contain T and the isolated subgroups of S*, i.e., to show that if H is a subgroup of S properly containing T then H is isolated if and only if f(H) is isolated. This follows from the fact that a, b g S, a > b = > f(a) ^ f ( b ).] Now, conversely, let S* be an isolated subgroup of an ordered abelian group S. Given A / B in S/T, fix a, b e S such that a e A and b g B. Set A > B or B > A according as a > b or b > a. This turns S/T intoan ordered abelian group. [PROOF. The only thing to be verified is that > is well defined. Say a > b and let a !, b* g S such that a 1 g A and b 1 g B. Then a ’= a + and b T = b + t2 with t ^ t2 g T. a* < b* = > [t.j b)J = > [a - b = a* - t 0, t2 “ ti > °> t2 — 1 1 g T, T is isolated] = > [a - b g T] = = > A = B

kk

II.

VALUATION THEORY

which is a contradiction. Therefore a ’ > b 1] and we shall denote it by S//T (if it is clear from the context, we shall write S/T for S//T). Observe that if S^, ..., Sp are ordered abelian groups and < p, then Sm+1® ... (x) S [= o ® . . . . . ® (o) ® S ® ^ (m times) *•• ® is an isolated subgroup of S 1 0 ••• (x) S and S1 0 ... 0 Sp //Sm+1 0 ••. ® Sp is canonically order isomorphic with S 1 ® ... ® Sm so that p(S1 ® ... (x) Sp ) = p(Sm+1 ® ... ® Sp) + P(S1 ® ... ® Sm ) and hence hy induction: p(S1 (x) ... ® S )= o< m

p (S p ) +

p CS^,

) + ...

+ P ( s 1 ).

EXERCISE. Let T be an isolated subgroup of an ordered abelian group S.Is S order isomorphic to (S//T) (x) T? IsS group iso­ morphic to (S/T) @ T? Now let S be a torsion free (i.e., S has no nonzero ele­ ments of finite order) abelian group. If for s e S and integers m and n with n / o there exists s’ in S such that ns 1 = ms, then s’ is unique and n*s 1 = m*s’whenever m/n= m*/n* and hence we may denote s' by (m/n)s. Let B = Cbj}, j 6 J, be a family of elements bj of S. An element s of S is said to depend rationally on B if there exist m-, .. ., m e Z, o / n e Z and j-, ..., j e J such that ns = nub. + ■ q • q • J1 ... +

m pbj

(e.g.,

if there exist n 1, ..., n^ e Q

and

...,

eJ

such that ^Ljb. exists and s = n-b. + ... + n b. ) then s is said 1 Ji 1 J 1 q Jq to depend rationally on B. If some b. depends rationally on the reJ maining b - fs, then B is said to be rationally dependent, otherwise B J is rationally independent. If B Is rationally independent, and if every element of S depends rationally on B, then B is said to be a rational basis of S. There exist rational bases of S and the cardinality of any two rational bases is the same. [PROOF: (The proof is entirely analogous to the corresponding theorems for the various kinds of bases, e.g., trans­ cendence bases of a field, vector space basis of a vector space — indeed, S can be considered to be a vector space or rather a module over Z. ) Let W be the set of all the rationally independent subsets of S par­ tially ordered by Inclusion, then W has the Zorn property, etc.] and this cardinality is called the rational rank of S and is denoted by r(S). If for every s e S and 0 / n e Z there exists s ! In S with ns 1 = s, then S is said to be rationally complete. LEMMA 2 .1 . Let S be a torsion free abelian group. Then there exists a rationally complete abelian overgroup S* of S such that s* e S* = = > there exists o / n e Z such that ns* e S. If S ! is any other such group, then there exists a unique S-isomorphism of S* onto S T. (We call S* a rational completion of S.)

8.

ORDERED ABELIAN GROUPS

PROOF. Consider the set of triples (m, n, s) with m e Z, 0 / n g Z and s g S. Identify (m, n, s) with (m’, n 1, s 1) if and only if n !ms = n m ^ 1 and then identify (1, 1, s)with s; this gives S* (just like getting Q from Z). LEMMA 2.2. Let S bea torsionfree abelian group and S* a rational completion of S. Then r(S) = r(S*) and every rational basis of S is a rational basis of S*. If S is ordered, thenthe ordering of S can be uniquely extended to S* (when this is done, S* is called an ordered rational completion of S) and p(S) = p(S*). PROOF.

Straightforward.

LEMMA 2.3.

An ordered abelian group

S

PROOF. Let 0 / s g S and 0 / n g Z+ s by - s, if necessary, we may assume that s > s + ... + s > 0. which is a contradiction.

is torsion free. with ns = 0. Replacing 0. Then s > 0 = > ns =

DEFINITION 2 . k . An abelian group S will be called an integral direct sum (or finitely generated free abelian) if there exists a finite number of elements s ^ ..., sn of S such that any element s of S can uniquely be expressed as s = m^s^ + ... + with m^ g Z; if this is so, then Sis torsionfree and r(S) = n; (s^ ..., sn ) is called an integral basis of S. LEMMA integral direct is the sequence direct sum for S^+ 1/S^ [hence tegral basis of

2.5. Let S be an ordered abelian group. Then S is an sum < = > p(S) = n < 00 and if 0 = SQ < S 1 < . . . < Sn = S of isolated subgroups of S, then 1 is an integral i = 1, ..., n and S = T 1 (x)... (x) Tn with = integral bases of T 1, •••, TR put together give an in­ S].

PROOF. ( < = ) Obvious. ( = > ) Let S be free with r(S) = m. Let S 1 be the smallest (nonzero) isolated subgroup of S. It is well known that a subgroup of a free abelian group with a finite number of generators is again such. Let (f^ ..., fn ) and (g1, ..., g^) be in­ tegral bases of S and S1, respectively. It is well known (see [¥2], section 108 or Selfert-Threllfall: Topologie, Chapter XII, section 86 , or Eilenberg-Steenrod: Foundations of Algebraic Topology, Chapter V, section 7 ) that we may arrange matters so as to have g^ = m^f^ with 0 ^ m^ e Z+ (mi divides m i+1 ) for i = 1, ..., q. Then gi g S 1 = > fi g S 1 = > m^ = 1 for i = 1, ..., q. Thus S = T (x) , where T is-the subgroup of S generated by f^+1, •••> Now aPPly induction to n. DEFINITION 2.6. Let S be an ordered abelian group. S is said to be archimedian if p(S) = 1 .S is said to be discrete archimedian if S is archimedian and if S does not contain any strictly descending

II.

VALUATION THEORY

infinite sequence of positive elements. S is said to be discrete if p(S) = n < oo and if o = SQ < S1 < ... < Sn = 3 is the sequence of isolated subgroups of S, then is discrete (archimedian) for i = 1, ..., n. [ o / s eS

LEMMA 2 .7 . An ordered abelian group S is archimedian and t g S ===> there exists n in Z such that ns > t].

PROOF. ( = > ) Otherwise there exist s, t g S, s 4 0 such that ns< t for all n e Z. Let H be the set of all elements h in S for which |nh| < t for all n e Z. Then H is an isolated subgroup of S with o / H / S so that p(S) > 1 which is a contradiction. ( o in S+ with s ^ H and let o 4 h g H. Then for all n in Z, nh g H so that nh < s which is a contradiction.

S

LEMMA 2.8. An archimedian ordered abelian group is order isomorphic to Z.

S

is discrete

PROOF. ( < = ) Obvious. (==>) Thenonexistence of a smallest positive element in S would imply the existence of a strictly descending infinite sequence of positive elements which would be a contradiction. There­ fore there exists a smallest positive element e in S. Now s e S and s 4 ne for all n g Z = > |s | / ne for all n € Z = = > (by 2 .7 ) there exists n g Z such that (n - 1)e < |s| < ne so that o < ne - |s| < e which is a contradiction. Therefore every element of S is of the form ne with n g Z. Since e is of infinite order ne -- > n gives the (order) isomorphism of S with Z.

S

LEMMA 2.9.An ordered abelian group is order isomorphic to a subgroup of R.

S

is archimedian < = >

PROOF. ( < = ) Obvious. ( = > ) Let us define a map f of S into R as follows: f(o) = 0; fix s > 0 in S and let f(s) = 1. Given t g S, let At = {m/n g Q | ms < nt) and Bt = {m/n g Q | ms > nt}. Then A^, Bt is a dedekind cut. Let f(t) = the real number defined this cut. Now it can be shown that f is an order isomorphism.

p(S) =

PROPOSITION 2.10. Let S be an ordered abelian group with n < 00. Then S is order isomorphic to a subgroup of Rn .

If S is rationally complete, then we have the stronger con­ clusion: there existsubgroups S^ of R such that S is order iso­ morphic to S1 (x) ... (x)Sn PROOF. In view of 2.2, it is enough to prove the second as­ sertion. We make induction on n, the case n = 1 is covered by 2.9, so let n > 1 and assume true for n - 1 . Let H be the maximal iso­ latedsubgroup of

S

other than

S.

Then

H

and S/H

are rationally

by

9.

bl

VALUATIONS

complete and of ranks n - 1 and 1 respectively. By the induction hy­ pothesis, H = S2 (x) ... (x) Sn with = 1 - Let B be a rational basis of S/Hand for b in B fixb* in Sin the residue class b. Let B* = {b*}. Then B* isrationallyindependent. Let S1be the set of all elements of S rationally dependent on B*. Then S^ is a rationally complete subgroup of S and hence under the canonical order isomorphism of S onto S//H, S1 gets mapped onto S//H. Therefore S = S 1 ® H = S, (x) ... ® Sn . EXERCISE. In Proposition 2.10, is the strongerconclusion true without the restriction of rational completeness? LEMMA 2.11 . Let S be an ordered abelian group. following two statements are equivalent: (1 ) (2)

r(S) = r < oo. p(S) = n < 00 and if 0 = SQ < S^ < ... < Sn = S the sequence of isolated subgroups of S, then r(Si/Si_i ) = ri < 00 fop

When

Then the

(1 )and hence (2) holds, then

1

= 1 9 •*

is

n*

r = r1 + ... + rR .

PROOF. By 2.10.

discrete

LEMMA 2.12. Let S be an ordered abelian group. Then < = > S is an integral direct sum and p(S) =r(S) = = > PROOF.

S is S = Zr .

Straightforward, in view of what has been proved up till

now. PROPOSITION 2.13. Let S be an ordered abelian group and a subgroup of S of finite index. Then (1) (2)S (3)

P(S) = p(T) and r(S) = r(T); is an integral direct sum < = > T direct sum; and S is discrete < = > T is discrete.

PROOF.

T

is anintegral

Straightforward.

9. Valuations. Let K be a field and let Kx denote the multiplicative group of the nonzero elements of K. A valuation v of K is a homomorphism of Kx into an ordered abelian group S such that v(a + b) ^ min (v(a), v(b)) for all a, b in Kx . v(a) is called the value (or rather the v-value) of a. Since v(Fx ) is a subgroup of S, by replacing S with v(Fx ), we may assume that v(Fx ) = S; thenS will be called the value group of v and will be denoted by Sy . The ad­ jectives of Sy may then be applied to v, e.g., p(v) = p(Sy ), r(v) = r(S ), etc.Let k be a subfield of K. If v(a) = 0for all a e k, we say that v is over k or v is a valuation of K/k. Given K we

II.

8

VALUATION THEORY

can set v(a) = 0 for all a g K* and get the trivial valuation. Unless otherwise stated, we shall exclude this from the discussion; we may thus add in the definition that: there exists a g F* with v(a) / 0 or equivalently, that Sy = v(F*) is nontrivial. Observe that: v is a homomorphism of F* into S = = > the unit of F* goes into the unit of S , i.e., v(l ) = o and hence v(- 1 ) = 0 [since (- 1 ) = 1] so that v(- a) = v(a), and v(a - b) ^ min (v(a), v(b)) for all a, b e F*. Hence if v(a) < v(b) then v(a + b) = v(a). We set v(o) = oo with the convention that: for all s € s , s - i - o o = oo + s = oo + oo = oo and s oo j then we need not always say "if a / o." We define Ry = (the valuation ring My = {a

g

of v) = (a

g

K | v(a) ^

0}

,

F | v(a) > 0}

Then Ry is a ring and xe P, x 4 Ry = > v(x) < o = > x 4 0 and v(l/x) > 0 = > 1/x e R y . Now let a g Ry . Then a 4 v(a) = 0 < = > a 4 0 and v(l/a) = o < = > 1/a e Ry . ThereforeMy is the set of nonunits in Ry and hence by 1 .5 , Mv is the unique maximal ideal in Ry . The field iscalled the residue field of v and is denoted by Dv;. if v is over k we may assume that k C Dy . The image of a g Ry under the canonical homomorphism of Ry onto Dy is called the v-residue of a; if a 4 Ry we set oo = the v-residue of a with the convention that for all d e D , d + oo = oo + d =oo. Let A be a subdomain of K and P a prime ideal in A; we shall say that v has center P in A if (R , M ) has center P in A, i.e., if Ry 3 A and My fl A = P. If K 1 -is a subfield of K and v 1 is the restriction of the valuation v of K to K ’, then v ! is a valuation (possibly trivial) of K 1 and v is an extension of v f to K. PROPOSITION 2 .1 4 . Let R be a subdomain of a field from K. Then R is the valuation ring of a valuation of K < = = > [a g K, a £ R ===> 1/a g R] .

K

different

PROOF. (==> ) Has been proved above. ( < = ) [Observe that if a valuation v of K with R = Rv does exist, then Ry - lV^ is the kernel of- v so that Sy is canonically (group) isomorphic to Kx/(Ry - My) and My/(Rv - My) --> S^; to prove the existence of v,we invert this procedure.] Let I be the set of units in R. Then I is amultiplicative subgroup of Fx . LetS = Fx/I, and let us write the operation in S additively. For a g Fx let a' denote the residue class of a modulo I. Then a 1 = b f < = > b = au with u g I = > [a g R < = > b g R] and [a g I < = = > b g I]. Hence if we let P = (a* g S | a g R, a / I) and N = [af g S | a / R ) then P and N are well defined disjoint subsets of S such that 0 / P, / N and (o) U P U N = S and for a 1 g S,

9.

VALUATIONS

^9

a* € P < = = > - a ! € N. Also a 1, b'e P = > a, b € R, / I = > ab € R, / I = > a f + b f = (ab)f € P. Hence if we set P = S+ (i.e., a 1 > b ’ < = > a 1 - b 1 e P, i.e., a ! > b 1 < = > a/b e R, / I, i.e., a 1 ^ b ’ < = > a/b € R) then S becomes an ordered abelian group. So it is enough to show that a, b e Fx = > (a + b ) 1 > rain (a!, b ’)* Say a* < b 1. Then we have to show that (a + b ) ! - a* ^ o, i.e.,

(- » ! ) ’ > »

•

But a' < b' = >

| € R —

>

(1 + | ) e R = >

(1 +

^ 0

.

DEFINITION 2 .1 5 . Let v be a valuation of a field K and let R = Ry.. The valuation constructed in 1.1 k will be called the canonical valuation given by v or by R. DEFINITION 2.16. Let v and w be two valuations of a field K. Then v and w will be said to be equivalent if there exists an order isomorphism f of Sw onto Sy such that for all a e Kx we have: v(a) = f(w(a)). This is clearly an equivalence relation. LEMMA 2.17. Let v be a valuation of a fieldK and let u be canonical valuation given by v. Then there exists a unique order iso­ morphism f of Su onto Sy such that v(a) = f(u(a)) for all a e Kx . PROOF. We shall use the notation of the proof of 2.1^. (Ex­ istence)Let f (aT) = v(a). Then a ! = b 1 = > a/b e I = Ry - My = > v(a/b) = 0 = > v(a) = v(b). Therefore f is well defined. Also f(a! +b f) = f((ab)!) = v(ab) =v(a) + v(b) = f(a’) + f(b!), and a ! ^ b 1 = > a/b e R = > f(a1) - f(bf) = f(a’/b!)= v(a/b) 1 0 = > f(a!) ^ f(b*)‘ (Uniqueness) v(a) = f(u(a)) = f(a?)» LEMMA 2.18. Let v and w be two valuations of a field K. Then v and w are equivalent < = > Rv = R^ = = > there exists a unique order isomorphism g of S¥ onto Sy such that v(a) = g(w(a)) for all a e Kx .

of

Sw

PROOF, onto Sy

v and w such that

are equivalent v(a) = g(w(a))

= = > an order isomorphism g for all a e Kx = = > [a e Ry

< = = > v(a) > 0 < = > g(w(a)) > 0 (since g is an order isomorphism) w(a) > 0 < = = > a € Rw_] = = > Ry = R^.. The restof the assertions follow from 2.17. DEFINITION 2.18.

From now on, unless otherwise stated, we shall

50

II.

VALUATION THEORY

not distinguish between equivalent valuations. DEFINITION 2.19. Let R be asubdomain of a field K and let p be a homomorphism of R into a field D suchthat p(R) 4 0. We shall say that p is a place of K if p cannot be extended, i.e., if [p* is a homomorphism of a subdomain S of K such that R C S and p* | R = p] = > [S = R] . Now assume that p is a place. Let M = p~1(o). Then M is the unique maximal ideal in R (for otherwise p could be extended to R^) and hence p(R) is a subfield of D so that replacing D by p(R) we may assume that D = p(R). We define Rp = (the valuation ring of Mp = M

p) = R

;

;

and Dp = (the residue field of

p) = D

Since any isomorphism of a subdomain of K can be extended to K and since a nonzero homomorphism of a field is an isomorphism we conclude that Rp = K < = > p is an isomorphism. An isomorphism of K is called a trivial place. Unless otherwise stated, by a place we shall mean a non­ trivial place. Two places p and q of K are said to be isomorphic if there exists an isomorphism f of Dponto D^ such that fp = q. It is clear that p and q areisomorphic if and only if R = R and Sr when that is so, the isomorphism f is unique. The canonical homomorphism of Rp onto Rp/Mp will be called the canonical place isomorphic to p. PROPOSITION 2.20. Let R be a subdomain of a field K differ­ ent from K. Then there exists a place p of K with R = Rp < = > [x € K, i R = > 1/x g R]. PROOF. ( < = )By 2.1I4- there exists a valuation v of K with Ry = R. Let p be the canonical homomorphism of Ry onto Dy = Rv/Mv Let S be a subdomain of K with R C Sand let p* be a homomorphism of Sinto an overfield D* of Dy such that p* | R = p. Then [x e S, i R] = > [1/x g R = Ry so that 1/x g My ]= > [p*(l/x) = p(l/x) = 0] = > [p*(l ) = p*(x)p*(l/x) = 0] which is a contradiction. Therefore S = R. Hence p is a place. ( = > ) Let D = Dp and M = Mp. We assert that given u g K, there exist c1,•.., cn g M such that 1 + c1u + ... +cnun = 0 0 Let A = (f(X) g R[X] | f(u) = 0}. Then A is a prime ideal in R[X] . Let p 1 : R[X] -- > D[X] be the natural extension of p and let A 1 = p.j (A). Suppose if possible that A 1 4 D[X] . Then B* = p ^ ^ A ^ / R[X] and hence there exists a maximal ideal B in R[X] with B D B*. Since B* D A and B* D M, we have B 5 A and B 5 M. Let q : R[X] -- > R[u] be the R-homomorphism with q(X) = u. Then kernel q = A and hence i R

9.

VALUATIONS

51

N = q(B) isa prime ideal in R[u] with N D M. Therefore N fl R = M and hence p can he extended to R[u]. This is a contradiction. There­ fore A 1= D[X] and hence 1€ A1, i.e., there exist f , ..., fn in R with (fQ+ f^X + ... +fnXn ) c A such that p(f1 ) = . . . = p(fn ) = 0 and p(f*0 ) c^ =

u i R

= 1 - Since p(f*Q ) = 1 > f*0 a € M and 1 + c-jU + ... + cnun = o.

To complete the proof, assume if possiblethat and 1/u i R. Let n be minimum such that 1 + c.jU + ... + cnun = o,

Let

m

with

u€ K

with

c^ € M

(l )

be minimum such that 1 + d.j(l/u) + ... + dm (l/u)m = 0,

Say

^ hence This proves our assertion.

n ^ m.

with

d^ € M

.

Then um + d lum-1 + ... + djjj = o

,

and hence Cnun + cnd lUn-1 + ... + c ^ u ™

- 0

,

i.e., eQ + e,u + ... + enun = o,

with

eQ, ..., en-1 (2)

in

M and

en = cn

Subtracting equation (2) from equation (1), we obtain (1 -eQ ) + Since

(c1 - e1)u + ... + (cn-1

-en_T )un“1 = 0

.

1 - eQ i M we can divide by itand get 1 + f^u + ... +

which contradicts the minimality of

= 0, n.

with

Therefore

REMARK 2.21. Thus 2.1^ and 2.20 spondence between valuation v of K and p of K (each such class may be replaced such that Rv = Rp. If Rv = R^ we shall

fi e M , u e K, u i R = >

1/u e R.

result in a one-to-one corre­ the classes of isomorphic places by the canonical member in it) say that v is the valuation

II.

52

VALUATION THEORY

associated with p and setv = vp . If v = Vp we shall say that p is a place associated withv and set p = py;sometimes we shall let py uniquely denote the canonical place associated with v. Observe that v is trivial < = = > v is trivial. If A is a subdomain of K and P a ir prime ideal of A, we shall say that p has center P in A vp has center P in A. PROPOSITION 2.22. Let A be a subdomain of a field K other than K and let H be an ideal in A with o / H / A. Then there ex­ ists a place of K such that Rp D A and D H. If H is prime, there exists a place of K having center H in A. PROOF. By Zorn's lemma, there exists a prime ideal P in A with P 0 H. Let f be the canonical homomorphism of A onto A/P = D. Let D* be an algebraic closure of a purely transcendental extension of D whose transcendence degree over D is equal to the cardinal number of K. Then it is clear that for a subdomain S of K containing R (there exists a homomorphism of S into an overfield of D which coincides with f on R) if and only if(there exists a homomorphism of S into D* which coincides with f on R). Let W be the set of homomorphisms gB of domains B with A C B C K into D* such that gB | A = f . Set gB > S*b* B D B* and gB | B* = 8*3*- Then W becomes a partially ordered set whichhas the Zorn property and hence W contains a maximal element p = gB - Now 0 / H / A = = > 0 / P / A =*=> pheve exists 0 / a e P and b € A - P so that p(a) = f(a) = 0 and p(b) = f(b) / 0 = = > p is neither the zero homomorphism nor an isomorphism. Hence the maximality of p implies that p is a place of K with R^ = B D A and Mp n A = P D H. If H is prime, we can take P = H. COROLLARY 2.23. extension of K. Then p

tension of

Let p be place of a field K and can be extended to a place of K*.

K*

an

It is enough to observe that a place p* of K* is an ex­ p < = > vp* has center in Ry = Rp.

Examples of valuations: (1 ) Prime spots in the field Q + % q € Q has a unique expression: q = - irp the prime numbers. Set vp(o.) = Op*

of rational numbers:

Each

where the product is over all

(2) Orders of rational functions of a complex variable (zeros or poles): k(x), k the field of complex numbers. For a e k, let va (f(x)) = the order of the zero (or the negative of the order of the pole) of f(x) at x = a. In both of the above examples, all the valuations are real discrete. These are enough for algebraic number theory or

9.

53

VALUATIONS

algebraic function theory of one variable, i.e., the theory of algebraic curves, i.e., the theory of (closed) Riemann surfaces — a modernistic (meaning thereby, valuation theoretic, due to Dedekind-Weber, 1882), defi­ nition of which is the set of all valuation of the given function field. (3) k(x, y); (k is a field and x, y are transcendental over k), u =a positive irrational number. For f(x, y) = sf^, e k[x, y] (with f ^0), set v(f(x, y)) = min (a + bu). For f, g e k[x, y] with g 4 0, set v(f/g) = v(f) - v(g). This gives a valuation v of k(x, y)/k and the value group of v consists of the real numbers of the form m + nu where m and n are integers. (4 ) In the above notation, set w(f(x, y)) = min {(a, b)} the pairs ofintegers (a, b) are lexicographically ordered.Another of getting w: f(x, y) = xdF(x, y), F(x, y) = yeF*(y) Then

w(f) = (d, e).

ation of

(5) In the k(x, y)/k.

Then

w

,

with

F(o, y) 4

with

F*(o) 4 0

0

where way

, .

is a valuation of k(x, y)/k.

above notation, set

w*(f) = d. Then

w*

is a valu­

(6)

The valuation of k(x, y)/k gotten by substituting a formal series y = za^xbi (where b 1 < b2 < ... are real numbers and a^ e k) and looking at the lowest degree term. LEMMA 2 .2 4 . Let

vbe a valuation of a field

K.

Then

Ry is

normal. PROOF. Let 0 ^ u e K with u /R y integral. Then u11 +a^u11”1 + .. . + an = °with a^ € Ry . Then u i Ry = > v(u) < 0 = > v(a^un“i) > v(un ) for i = 1, ..., n = > u(o) = v(un ) which is a contradiction. PROPOSITION 2.25. Let A be a subdomain of a field K and let B be the Integral closure of A in K. Then B = the intersection of all the valuation rings of valuation of K containing A where we allow the trivial valuation. PROOF. By 2 .2 4 , B is contained in the intersection. Hence it is enough to show that given a € K, a i B there exists a valuation v of K with Ry 3 A and a i Rv *Now a i B ===> a i A[l/a] = > 1/a is a nonunit inA[l/a] = > there exists a nonzero prime id^al P in A[l/a] with 1/a e P. By 2.22, there exists a valuation v of K having center P in A[l/a]. Then Ry 0 Ah/a] 0 A and My 0 Ah/a] = P, so that 1/a e My, i.e., v(l/a) > 0 so that v(a) < 0 and hence a i Ry .

5k

II.

VALUATION THEORY

COROLLARY 2.26. K/A algebraic < = > K = B < = > exist any nontrivial valuations v of K with Ry DA.

there does not

COROLLARY 2 .2 7 * Another proof of 1.16: The second assertion followsfrom the first. To prove the first assertion, in view of 2.25, we may assume that A is the valuation ring of a valuation v of K (if A = K there is nothing to prove). For a(X) = Za^X^ and 0 / h(X) = Ebj.x'k in KtX]set w(a(X)/b(X)) = min {v(at )J - min{v(bt )}. Then w is an extension of v to K(X). h(X) e Ry [X] = > w(h(X)) ^ 0 . Since f(X) and g(X) are monic, w(f(X)) < 0 and w(g(X)) < 0 . Now f(X) i Ry [X] = > w(f (X) ) < 0 = > w(h(X) ) = w(f (X)) + w(g(X)) < 0 which is acontradiction. Hence f(X) g Ry [X]and similarly g(X) e Ry [X]. COROLLARY 2.28. Another proof of 1.28 (1): Let P* be any prime ideal in A* lying above P. By 2.22 there exists a valuation v of K* having center P* in A*. Then v has center P in A and hence v has center M in R. Since Rv is normal, we must have Ry DR*. Since (My. D R*) D R = M, we must have My n R* = M? for some i. Therefore P* = My fl A* = M? n A* = p£. Now let v be a valuation of a field K. Let I be a nonunit ideal in R V . Then a e I, b e K with v(b) ^— v(a) = > b = a ~ and d v(b/a) ^ 0, i.e., b/a € Ry = > b 6 I. Also I C = > for all a € I, v(a) > 0. Thus U(I) = tv(a) | a € I) is an upper segment of Sy . Conversely, given an upper segment U of S , let I(U) = {a g Ry | v(a) g U] . Then a, b and hence

g

I(U) = = > v(a), v(b)

v(a

- b)

g

U'= >

g

a - b

a g I(U), c

U = > I(U).

Also

Ry = >

v(a)

g

g

v(a - b)

g

> min(v(a), v(b))

,

U

and v(ca) = v(c) + c(a) > v(a) = > Alsov(l ) = 0 Now

1i I(U). Therefore

= >

l(U)

ca is

g

I(U)

a nonunitideal

in Ry .

a g I(U(I)) < = > v(a) g U(l) < = > there exists a* g I such that v(a) = v(a*), i.e., such that a = a*d with d = a/a* so that v(d) = 0, i.e., d a unit in Ry < = > Thus

I(U(I)) = I.

a

g

I.

Similarly we can show that I -- > U(I) I(U) S(I) = {s g Sv

|s| < v(a)

for all

a e 1}

between the nonunit ideals in Ry and the segments of Sy . Therefore the set of ideals in Ry is simply ordered by inclusion. Now the upper seg­ ment U is isolated < = > [a, b e S* with a ^ U, b £ U = > a + b i U] < = > [a, b € Ry with a i I(U), b{ I(U) = > ab i I(U)] < = > I(U) is a prime ideal. In other words I Is a prime ideal < = > S(I) is an isolated subgroup. Observe that S(M^) = 0 . If we define the rank p(v) of Ry to be the order type of the set of all nonmaximal prime ideals of Ry, then we can state our conclusions as follows: PROPOSITION 2 .2 9 * There is a one-to-one inclusion preserving correspondence between the nonunit ideals in Ry and the segments of Sy and in this correspondence prime Ideals correspond to isolated subgroups and we have p(v) = p(S )• 1o.

Specialization and composition of valuations.

LEMMA 2.3 0. Let v be a valuation of a field K and S 4K subdomain of K containing Ry . Then S is the valuation ring of a valuation of K. PROOF.

that

a

A direct consequence of 2 .1 4 .

LEMMA 2.31 . Let R^ 0 Ry . Then

v

and

w be

valuations of a field

K

such

(1 ) P = M,r fl Rtt is a nonzero prime ideal in R . Rw = (Rv )p and M^ = P; (2) If w has center My in Ry, then R^ = R , i.e., 0 = >

w = v.

PROOF. Proof of (2). a e K, a i Ry = > v(a) < 0 = > v(l/a) > 1/a € My — > 1/a € = = > w(l/a) > 0 = > w(a) < 0 = > a i Rw Proof of (1).

P 4 0

since

w

is a (nontrivial) valuation and

K isthe quotient field of Ry . By 2.30, (Rv )p = Rq and PRq = where u is a valuation of K. Then w has center M^ in Ru *Hence by (2), R^ = Ru . It remains to be shown that M^. = P, i.e., that M^ = P, i.e., that M Q C P. Now 0 ^ a € M^ = > 1/a ^ R^ = > 1/a i Ry = = > a e Ry = = > a e Mu n Ry = P. with

DEFINITION 2.32. If v and w are valuations ofa field K R^ 0 Ry we shall say that v is a specialization of w, in symbols:

II.

56

VALUATION THEORY

w -- > v; if w -- > v, we shall also write w^ > Vp for corre­ sponding places, and say that vp is a specialization of wp . Observe and that (w -- > v and v -- > w) < = > (w = v) and (w^ -- > Vp -- > Wp) < = > (Wp and vp are isomorphic). More generally, if (R, M) is a local domain and (S, N) is aquotient ring of R with respect to a nonzero prime ideal, then we shall say that (R, M) (orR) is a specialization of (S, N) (or S), in symbols: (S, N) -- > (R, M) (or S -- > R). Again, observe that (S -- > R and R >S) < = = > (S = R). PROPOSITION 2.33* Let v be a valuation of a field K. Let A be the set of prime ideals in Ry . For P e A let S(P) = (Ry )p. Let A* = {S(P) | P € A). If P / o then S(P) is the valuation ring of a valuation of K (if P = o, then S(P) = K) and we have PS(P) = P. Also A* is exactly the set of domains between Ry and K and it is simply ordered by inclusion. PROOF.

A direct consequence of 2.2 9, 2.30 and 2.31 .

Now let wand v be distinct valuations of a field K with w -- > v so that is a prime ideal in Ry and 0 /M^ / My . Let S be the isolated subgroup of Sy corresponding to the prime ideal M^, let fbe the canonical order homomorphism of Sy onto S* = Sy //S and let w* = fv. Since (R^mM^) is the quotient ring of with respect to Mw, we have that Rw - M^. consists exactly of the set of elements of the form a/b with a, b g Rv - M^. Now a, b e Ry = > v(a), v(b) g S = > v(a/b) g S = > a/b g Mv « Thus x g Rw = > x g v” 1 (S) Conversely, y g v “ 1 (S) = > v(y) g S = > [if y g Ry then y g Ry - M^ so that y = y/l g Rw - My; if y / Ry, then 1 /y g and v(l/y) g S so that 1/y g Ry - and hence y = l/(l/y) g R^ - M^J = > y g R^ - M^ Thus we have shown that kernel w = R^ = v’ 1 (S) = kernel w*. Now let x g Kx . . Then w*(x) > 0 < = > v(x) is in the positive complement of S in Sy < = > x g M^. Therefore w* is a valuation of K equivalent to w, i.e., w* = w and we may take S* to be the value group S¥ of w. Furthermore, let pw be the canonical place associated with pw is the canonical homomorphism

i.e.,

Pw : \

= Dw

Let Py-CRy) = ) = My* and for a € that p^1 (l£) = Ry. - ^ = v"1 (S). For a e v(a) = v(a) Then

v

is a transformation of

I? T

g

onto

S S.

w,

•

let let

a = pw (a).

Observe

10. (1 )

(2) (3) (^)

decomposed

SPECIALIATION AND COMPOSITION

57

v Issingle valued: a, b g R^with a = b 4 0 = > a, b i and a = b +d with d e ^ so that v(d) > v(a), v(b) = > v(a) = v(b). v is a homomorphism: Obvious. v is a valuation of D^: Leta, b g D^. Then v(a + 15) = v(a + b) ^ min (v(a), v(b)) = min (v(a), v(b)). R- = Ry and M- = 5^: v(a) \ 0 < = > a g Ry < = > a g Ry . Also v(a) > 0 < = > a g My < = > a g My .

Thus, given valuations w -- > v of K, v into w and v (a valuation of

with w ^ v, We write

we have

w = w o v

of D*. W v of K

Conversely, let w be a valuation of K, and v a valuation Let R =Wp~1(R-)Then R is the valuation ring of a valuation V and v = w 0 v*

PROOF, a g Rw, i R = > Pw (a) i R- = > a 4 Mw so that 1/a g R^ and pw (l/a) e R- = > 1/a g R. Furthermore, a 4 Rw = = > 1/a g = > pw 0 /a) = 0 = > 1/a g R. Therefore, by 2.1 if, R = Ry, where v is a valuation of K. Since R- / D^ we have Ry 3 Rv and R^ 4 Ry, i.e., w --- > v and w 4 v. Therefore v = wo v*, where v* is the valuation of I?Wr with RTVr** = pTWT(R,r) = PTr(R) = R-V so that v* = v V W and v = w o v. PROPOSITION 2.3if. Let A be a subdomain of a field K and P a prime ideal in A. Let R = Ap, M = PR, and S = the integral closure of R in K. Let ¥ be the set of valuations of K having center P in A. Then S 4 K < = > ¥ is nonempty = > S = n V€yRy * PROOF. First observe that ¥ = the set of valuations of K having center M in R. If we Include the trivial valuation in W then it is enough to prove S= ^VGyRy * Let W* be the set of all valuations of K, including the trivial one, whose valuation rings contain R. Then by 2.25, S = n V€y*Rv C ^ v g W ^ v ' Rence It is enough to prove that w g W*= > there exists v g ¥ with w--> v. Let N = R fl M^. Then N C M. If N = M, nothing to prove. So assume N 4 M. Let PW (R) = R and pw (M) = M. Then M is a nonzeroprime Ideal In R, (in fact, M is the unique maximal Ideal in R) and hence by 2.22 there exists a valuation v of having center N in R. Let v = w 0 v. Then v g ¥ and w -- > v. LEMMA 2.35* Let K be a field and K* and overfield. Let v and v* be valuations of K and K* respectively. Then v* is an ex­ tension of v < = = > v* has center M^ in Ry and we may consider Sy

II.

58 to be a subgroup of

VALUATION THEORY

Sy#.

PROOF. (==>) Since v = v* | Kx,Sy = v(Kx ) which is a sub­ group of Sv* since Kxis a subgroup of (K*)x . Also R^ - (o) = v“ 1 (s+ u (o)) = (v*)- 1 (s* u (o)) n kx = (v*)_ 1 (s+# n (o)) n kx = rv* (o) n k. - (o) = v-1 (s+) = (v*)- 1 (s+*) n kx = mv* (o) n k. has

( a g R*. We may assume a / 0. K*/K algebraic = > k0an + ... + kn = 0, Let Let

with k^ € K,

t = min (v(kQ ), ..., v(kn )). h± = k±/kj. Then

Let j

beminimum

hQan + ... + h j ^ a 11- ^ 1 + an‘J + and

v(h^) > 0

an”^

we get

if

i < j

and

£h0aJ* + h ^ * " 1 + ...

+

j _

1j+1 + hj +1 (a) Let

b =the c = the

i a

+

kQ / 0 with

v (kj)

+ .. . + hR =

v(hi ) ^ 0

^

with

if

= t.

0

i > j. Dividingout

by

1

+ ” * + hn (a)

= 0

sum in the first bracket, and sum. in the second bracket.

Let v 1 bean arbitrary extension (2.23) of v to K*. Then [vT(a) ^ 0 ===> v !(b) ^ 0 = > v ’(c) = v !(- ab) > 0] and [v*(a) < 0= > v ‘(c) ^ 0 = = > v T(b) = v !(- c/a) > 0], i.e., v !(b) ^ 0 and v !(c) \ 0. Therefore, by 2.36, b, c e S. Now v*(a) > 0, and hence v*(b) = 0, so that b / P. Hence a = - c/b g Sp = R*.

11 . RAMIFICATION THEORY 11.

59

Ramification theory of valuations.

PROPOSITION 2.38. Let K be a field, v a valuation of K, K* an algebraic extension of K, (R*, M*), (R*, M*), ... the local rings in K* lying above Ry .Then r £ is the valuation ring of a valu­ ation v^ of K* and v*, v*, ... are exactly the extensions of v to K*. PROOF. In 2.37, we have proved that the valuation rings of the extensions of v to K* must be among the rings R^. To prove the con­ verse: By 2.22, there exists a valuation vf of K*having center M? in r £. Let S be the integral closure of Ry in K* and let Pj_ = fl S. Then v£ has center P^ in S and center My in Rv so that, by 2.35, vf is an extension of v and hence by 2.37, R^ = Ry* • DEFINITION 2.39. Let K* be a galois extension of a field K and let v be a valuation of K. Let R*, ...,R£ be the local rings in K* lying above RV . Then by 2.38 Rt ..., vf J = RVj* where v*, I L- are exactly the extensions of v to K*. Hence we can apply the entire con­ siderations of Section 7 » We define: Gs(v?/v) = (the

splitting group of

v? over

v) = GS (R */R ) j Fs (v?/v) = (the splitting field of v? over v) = FS (R */R ) j G 1 (v^/v) = (the inertia group of v? over v) = Gi (R */R ) j F^vt/v) = (the inertia field of v? over v) = F^(R */R ). J J Vj PROPOSITION 2 .1+0 . Let K*/K be agalois extension, v* a valuation ofK* and v the K-restriction of v*. Let Ks = Fs (v*/v) and let vs be the Ks-restriction of v*. Then S = S . vs v REMARK 2.1+1 . We shall not prove this proposition here but we shall only show how it follows from a result of Krull. In [K3] Theorem 21 Krull has proved that there exists an intermediate field Kz of K and K* such that if v^ is the K-restriction of v* to Z* K_, then v* is Zi Zk the only K*-extension of vz and Sy = Sv « It then follows by 1 .1+6 s z that Kz D K and hence Sy C S g C Sy ; therefore S s = Sy . The proof given in [K3] is quite involved and hence we pose the following question: Is it possible to prove that , PROPOSITION 2.1+2. F^(v*/v)

and let

S s = Sy

Letthe notation be

v^* be the K^-restriction of

from our Theorem 1 .4 7 :

as in 2.1+0. Let v*.

Then

K1 =

S . = S . v v

6o

II.

By 1 .1+8 , u

VALUATION THEORY

PROOF. Replacing K by Ks, we may assume that K = Fs (v*/v). D -^/Bv is separable and hence has a primitive element u. Fix

in

R . in the residue class u. Then u/R v _ by 1 .16 the minimal monic polynomial un + a.^un has coefficients in Ry . Then un + a^u11”1 + ... the residue classofa^ mod M^.. Hence n > [D ^

is integral and hence + ... + an = 0 of u/K + aR = o, where a^ is : D ]. By 1.^8,

[K^ : K] = [D . : D ] and hence u is a primitive element of v . Given 0 £ y g K , we may thus write y = x* + x*u + ... + x * ^ u n ~ \

with

K^/K.

x^. £ K

Chooose j such that v(x?) < v(xf) for t = o, 1, ..., n - 1. Let x^ = x^/Xj. Then with x^ e Rv for t = 0, 1, ..., n and Xj £ Since u is of degree n overDv, we must have xQ + x 1u + ... + xn-lun“1 1 o (mod My) and hence v(xQ + x^ u + ... + xn-1un‘1 ) = o . Therefore v*(y) = v*(xj) = v(xp e Sy

.

PROPOSITION 2.1+3. Let the notation be as in 2.1+0 and assume that Dy is of zero characteristic. Then G^(v*/v) is isomorphic to Sv*/Sy (and hence G1 (v*/v) is abelian). REMARK 2.1+1+. This is Satz 3 of Krull [2] (although Krull assumes p(v) =1, the proof works in the general case), the proof is quite in­ volved and will be omitted. COROLLARY 2.1+5. Let K*/K be a finite algebraic extension, v a valuation of K and v* a K*-extension of K. Then ST v^r„/ST v r is finite. If Dy is of characteristiczero and K*/K is separable then the proof follows (by passing to a Galois extension of k* containing K) from 2.1+0, 2.1+2 and 2.1+3. In the general case, we can prove the stronger assertion; : sv ] < [K* : K] directly as follows: Sv-cosets.

Let u 1, ..., u^ be elements in Sy* lying in distinct Fix u^ e K* with u*(u^) = u^. Suppose, if possible, there

exist elements a^ not all zero in K Then for some i 4 j we must have a^ i.e., v*(u^) =v*(aj/ai )v*(Uj), i.e.,

such that

ai'u-| + ••• + ^t^t = 0# and v*(a^Uj_) = v*(ajUj), u^ = vfa./a^Uj, i.e., and £ 0 £ a^

12. Uj Uj.

are in the same Sv-coset which are linearly independent over DEFINITION 2 .45A.

is a contradiction.Therefore u ^ , ... K. Therefore [Sy* : S ] < [K* : K].

In the above notation we define

r(v*/v) = ramification index of =

61

ALGEBRAIC FUNCTION FIELDS

r s v * : Sv ]

v*

over

v

.

12. Valuations of algebraic function fields. Let K field and K* an overfield of K, let v be a valuation of K and v* an extension of v to K* where v and v* are allowed to be trivial. We define (v-dimension of

v*) = (the transcendence degree of

Dy*/Dy)

If K* is a finitely generated extension of K of transcendence degree s, then K* is said to be an algebraic function field of dimension s over the ground field K. PROPOSITION 2 .4 6 . Let K be a field and K* an extension of K offinite transcendence degree s. Let v* be avaluation of K* and let v be the K-restriction of v* where we allow v to be trivial. Let d be the v-dimension of v*. Let r and r* be the rational ranks of v and v* and let p and p* be the ranks of v and v* respectively. Then we have the following: (1) If r* is finite, then r* + d < r + s. (2) If v is an integral direct sum, if K*/K is finitely generated, and if r* + d = r + s, then v* is an integral direct sum and Ry*/Mv* is finitely generated over Ry/My (note that Rv/Mv = K if v is trivial). (1*) If p is finite, then p* + d < p + s. (3) If v is discrete, if K*/K is finitely generated, and if p*+d=p+s, then v* is discrete and Rv*/Mv* is finitely generated over Ry/My. PROOF. First assume that the weaker inequality (A)

r

is finite.

We begin by proving

r* < r + s

Suppose s = 0. Given 0 ^ u € K*, let f(X) = aQXn + a ^ 31”1 + ... + a , a = 1, be the minimal monic polynomial of u over K. Since f(u) = 0, there exist distinct integers i and j such that v*(a^u ’ ) = v*(ajn”J) 4- 00 and hence v*(u) = v(a^/a. )/(i - j), i.e., the value of u depends rationally on the value of (a^/a.) e K. Therefore r* = r = r + s. Now suppose s > 0 andassume that (A) is true for s - 1. Let z.j, z2, ..., z3-1 be part of a transcendence basis of K*/K. Let

be a

62

II.

VALUATION THEORY

K 1 = K(z.j, z2, ..., zg_1 ), let be the restriction of v* to K 1 (v1 may be trivial), and let r1 be the rational rank of v 1. By our induction hypothesis, r1 < r + s - i. If the value of every nonzero element of K* isrationally dependent on the values of elements of K 1, then r * = r ^ < r + s - 1 < r - s, and we are through. Now suppose that there is a nonzero element z in K* such that h = v*(z) does not depend rationally on the values of elements of K 1. Then, by the s = o case, z is transcendental over K-. Let f(X) = f + f-X + ... + fnXn n* and g(X) = gQ + g.jX + ... + gn*X be nonzero elements of K 1 [X] . Let ap = v*(fp) if fp 4 o and bp = v*(gp) if gp 4 o. Since h depends rationally neither on the ap nor on the bp, there exist integers p and q such that « 4 v^f^z^) < v-^fpZ1 ) whenever i 4 P and fp 4 °> and oo 4 v*(g z^) < v*(gizi )whenever i 4 U .and gp 4 °; i.e., v*(f(z)/g(z)) = v*(^p/gq) + (p - q)h. Thus, the value of any nonzero element of K 1 (z) isof the form a + mh where a is in the value group of v 1 and m is an integer,i.e., if r2 is the rational rank of t h e restriction of v* to K ^ z ) (this restriction may be trivial), then r2 = r1 + 1 < r + (s - 1 ) + 1 = r + s Since ¥L*/¥L^(z) is an algebraic extension, by the case s = 0, we have r* = r2 < r + s. Thus the induction is complete and (A) has been proved. Also observe that if v2 is the restriction of v* to K2 = K^z), then the residue fields of v 1 and v2 coincide. For, in the above notation, since v ^ f p Z 1 ) > v2 (f z^) whenever i 4 P and fp 4 °> must have that f(z)/(f zp ) belongs to R and that P "n f ( z ) / ( f p zp )

Similarly,

g(z)/(g^z^)

Now assume that

=

1

+

)

belongs to

f(z)/g(z)

(f i / f p ) z ^

Ry

belongs to

P

and

E

1

(mod

Mv

}

g(z)/(g z01) = 1 (mod My ).

R . We want to

show that we can

2

find to

e

M

in ,

R

1

with

we can take

f(z)/g(z) = e (mod M e = o.

2

Now suppose that

).

If

f(z)/g(z) belongs

f(z)/g(z)

does not belong

2

to

M^. ,

then

o = v2(f (z )/g(z)), v2(f /g

) + (p - q)h

does not depend rationally on v2(fp/gq ), we must have that p = q and v2(f /gp) = 0. Let e = fp/gp* Then

and

p - q = o,

f(z)/g(z) = (fp/gp)(f(z)/fpzp )(gpzp/g(z)) = e (mod ^ since

f (z)/fpzp

and

sinceh

gpzP/g(z) are both congruent to one

)

i.e.,

,

modulo

Mv .

12.

ALGEBRAIC FUNCTION FIELDS

63

This proves our second underlined assertion. To prove (1 ), let us retain our assumption that r is finite, and let y 1, y2, ..., yd be a transcendence basis of Rv*/Mv* over Rv/Mv and fix yd in Ry* belonging to the residue class y^. Let K f = K(y1# y2, ..., yd ) and let v 1 be the restriction of v* to K f. Given 0 / f*(X-,, X2, ..., Xd ) in KtX^ X2, ..., Xd ], choose a co­ efficient q of f having minimum v-value and let F(X.j, X2, ..., Xd ) = (l/q)f(X1, X2, Xd ). Then all the coefficients of F ( X V X2 , ..., Xd ) belong to Ry and at least one of them is equal to 1. Let F(X.j, X2, .. ., Xd ) be the polynomial gotten by reducing the coefficients of F ^ , X2, ..., Xd ) modulo My. Since F(X.j, X2, ..., Xd ) has a coefficient equal to 1 and since y 1, y2, ..., yd are algebraically independent over Rv/Mv, we must have P(y-, > f 2 > " ■ > ^ °’ i,e>> v*(F(y.,, y2, •••> yd )) = 0, i.e., y2, ..., yd )) = v(q) / and hence f (y,, y 2 > • • • > yd ) 4 o. Thus y 1# y2, ..., yd are algebraically independent over K and the valuegroups of vand v l are identical. Since the transcendence degree of K*/K! is s - d, (1) follows by applying (A) to K*/K!. Nowlet g(X^ Xg, ..., Xd ) and h(X^ X2, ..., Xd ) be arbitrary nonzero elements of K[X1, X2, ..., Xd ] and let y d )/g (y1 •i 2 ,

y = f(y-,» j 2 >

•••, y d )

•

Fix coefficients a and b of g and h respectively having minimum v 1-values, and let p = a/b. Let G(X^ X2, ..., Xd ) = (l/a)g(X1, X2, ..., Xd ) and H(X.j, X2, ..., Xd ) = (l/bJhfX^ X2, ..., Xd ). Then, as above, v'CgCy,, y2, •••, yd )/h(y1, y2, ..., yd ) = v 1(a/b)

.

Hence y = g(y.,, y2,•••, yd )/h(y1, y2, ..., yd ) belongs to Ry , if and only if p = a/b belongs to Rv , D K = Rv * Now assume that y does belong to Ryi* Let y and p be the residue classes modulo Myl con­ taining y and p respectively. Let G(X^ X2, ..., Xd ) and H(X.j, X2, ..., Xd ) be the polynomials obtained respectively from G(X.j, X2, ..., Xd ) and K( X^, X2, ..., Xd ) by reducing their coefficients modulo Mv ,. Since H has a coefficient equal to one and since y 1, y2, ..., yd are algebraically independent over Rv/Mv, we have that f

= pG(y-i> y2> •••>

f 2 > •••*

Therefore Rv ,/Mv , = (Ry/MyMy^ y2, ..., yd ), Rv ,/Mvt is finitely generated over Ry/My. Now assume that

v

•

and hence in particular

is an integral direct sum, that

K*/K

is

6k

II.

VALUATION THEORY

finitely generated, and that r* + d= r + s. Let K* and v T he as above. Then v and v ! have thesame value groups, K*/K! is a finitely generated extension of transcendence degree e = s - d = r * - r , and Rv ,/Mv., is finitely generated over Rv/Mv . Fix an integral basis t-j, t2, ..., tp of the value groupof v 1. Let x 1, x2, ..., xe be a transcendence basis of K*/K!. Let = K ! (x1, xg, ..., x^), v| = the restriction of v* to k [, and r^ =the rational rank of v^. Since r* = r + e, we must have, in view of (1), r^ = r^_1 + 1 for i = 1, 2, e. Let v*(x^) = By applying the first of the above under­ lined remarks successively to the extensions K^/KT, K2/K*, .••, , we conclude that for any nonzero element x of we have v*(x) = a + mr+ltr+1 + mr+2tr+2 + ... + mr*tr*

,

where a is the value of an element of K 1 and where mr+2* *’*' mr* are integers; since a = m lt1 + m2t2 + ••• + where m 1, m2, ..., mr are integers, we finally have: v*(x) = m 1t1 + m2t2 + ... + mr*tr*. There­ fore v* is an integral directsum. SinceK*/K* is a finite algebraic i extension, the value group of v Q is a subgroup of the value group of v* of finite index and hence v* is an integral direct sum. Now by the second underlined remark above, the residue field of v^ coincides with the residue field of v !. Since the residue field of v ! is finitely generated over the residue field of v and sinceK*/K@ is a finite algebraic extension, we conclude that Ry */Mv* is finitely generated over Rv/My . This proves (2). The proof of (1*) is entirely similar to that of (1). Finally, assume that v is discrete, p * + d = p + s , and that K*/K is finitely generated. The discreteness of v implies that p = r. Since by (1 ), r* + d < r + s and since r* > p*, it follows that r* = p* and that r* + d = r + s. Hence by(2), v* is an integral direct sum and Rv*/Mv* is finitely generated over Rv/Mv . Since r* = p*, v* is discrete. This proves (3). COROLLARY 2.if7. In the notation of the above lemma, assume that v is trivial, i.e., that v* is avaluation of K*/K. Then:(1 ) p* + d < r* + d < s. Furthermore, if K*/K is finitely generated (i.e., if K*/K is an algebraic function field of dimension s), then we have the follow­ ing: (2) If r* + d = s, then v* Is an integral direct sum and Rv*/My# is finitely generated over K. (3) If p* + d = s, then v* is discrete and Rp*/Mr* is finitely generated over K. (4 ) If d = s - 1, then v* is real discrete and Rv*/My# is finitely generated over K. DEFINITION 2.if8. In the notation of 2.k 7 , If K*/K is finitely generated and if d = s - 1, then we shall say that v is a prime divisor of K*/K.

12.

ALGEBRAIC FUNCTION FIELDS

65

COROLLARY 2 .k 9 . Let K*/K be a two-dimensional algebraic func­ tion field and let v be a valuation of K*/K. By 2 .1+7 , it follows that v is of one of the following four types: (1 ) v is a prime divisor of K*/K, i.e., v is real discrete and Dv/K is a one-dimensional alge­ braic function field; (2) v is discrete of rank two; (3) v is real of rational rank one, but v is not a prime divisor of K*/K; (k) v is real of rational rank two; in this case v is necessarily an integral direct sum. In cases (2) and (4 ), Dy/K Is finite algebraic, and in case (3), Dv/K is algebraic.

CHAPTER III:

NOETHERIAN LOCAL RINGS

For proofs of some of the propositions of Section 13 we shall refer to Chapters III and IV of Northcott's Ideal Theory, [N]. 1 3 * The dimension of noetherian local rings. Let P be a prime ideal in a ring A.The upper bound of integers nfor which there exist prime ideals P1,..., PR in A with P > P 1 > P2 ••• > pn is called the A-rank of P, in symbols: rank ^P. The upper bound of in­ tegers n for which there exist prime ideals P1, ..., Pn in A with P < P1 < ... < Pn is called the A-dimension of P, in symbols: dim ^p. When the reference to A is clear, rank p ? and dim ^P will be re­ placed by rank P and dim P respectively. Observe that rank ^P = rank p PAp and dim p P = dim

PROPOSITION 3 •1 • ideal is finite. PROOF.

In a noetherian ring

A

the rank of any prime

Theorem 7 , page 60 of [N].

PROPOSITION 3-2. Let A be a domain which is of finite trans­ cendence degree n over a subfield k. Then (1) for any prime ideal P of A we have rank P + dim P < n. (2) If A is finitely generated over k, thenthe equality holds. PROOF. (1) Since rank (0) ^ rank P + dim P, it is enough to prove that rank(0) < n. Let ° = PQ < P., < ••• < Pt he a chain of prime ideals in A. We have to show that t then v1, ..., v^. are algebraically independent over k, hencethe inequality. Now assume that the equality holds: let U 19 **• 9 ^n he a transcendence basis of B/k and fix v^ in A with f(Vj_) = u_p then v1, ..., vR is a transcendence basis of A/k. Suppose if possible that for some o / a e A , f ( a ) = o . Multiplying the minimal monic polynomial of a/kCv^ ..., vn ) by the product of the denominators 66

13-

THE DIMENSION 3

67 S— 1

of the nonzero coefficients, we obtain: pga + Ps-1& + ••• + P0 = 0 with p± e k[v1, vn ] and pQ / 0. Applying f we get: f ( p Q) = 0, which contradicts the k-algebraic independence of u 1, 1^.] There­ fore, for any non-zero prime ideal Q in A we have: [tr. deg (A/Q)/k] < n - 1. Therefore: [tr. deg (A/P^)/k] x n of A/k such that A is integral over B = k[x^, ..., xn ],(this is Noether’s normalization theorem, for a proof see [Z1], section 2). Secondly, observe that it follows from 1 .22, 1 . 2 k and 1.2^B that rank^P = rankgP fl B and dirn^P = dim-gP fl B. Now we shall make induction on rank P. If rank P = 0,then P = P fl B = 0 ; and since Q,^ = (x1, • • • , x^)B is a prime ideal in B with 0 < Q1 < ... < we have dimA (o) = dimg(o) > n and hence, by (1 ), rankA (o) + dimA (o) = dimA (o) =n. Now consider the case of rank P = 1. Fix 0 / f(x1, ..., xR ) in Q = P fl B. Since Q, is prime, it contains an irreducible factor g(x^ ..., xn )of f. Since gB is prime and since rankgQ =1, we must have Q = gB. By relabeling the x^ we may assume that xn actually occurs in g. Let x^ be the Q-residue class of x^. Then x^, ..., xn_1 are algebraically independent over k and xR is algebraically dependent over ktx^ ..., xR ]. Hence, tr. deg (B/Q)/k = n - 1 . Since A is integral over B and and since Q = P fl B, A/P is an algebraic extension of B/Q and hence tr. deg (A/P)/k = n - 1. Therefore dim^P = dim^yp(o) = n - 1 so that rank^P + dim^P = n. Now assume that rank P = r > 1 and assume that (2) is true for all smaller values of rank P. Let P1 < ... < Pp < P be a chain of prime ideals in A. Then rank P = r - 1 and hence by the induction hypothesis tr. deg (A/Pp )/k = dim^yp (0) = dimAPp = n. - (r - 1) = n r + 1.

Since

ranklyp P/Pp = 1 we have:

dimAP = dimAyp P/Pp = [tr. deg (A/Pp )/k] - 1 = (n - r + 1) - 1 = n - r.

M

DEFINITION3 -3 - Let (R, M) be a noetherian local ring. is generatedby a finite number of elements. Wedefine dim R

=dimension of R = minimum value of n such that there exist nonzero elements a1, ..., an in R for which (0, a.j, ..., &n )R is primary for M.

Then

68

III.

NOETHER IAN LOCAL RINGS

Let dim R = d. If a1, ..., a^ are nonzero elements in R such that (o, a1, ..., a^)R is primary for M, then (a^ ..., a^) are said to form a system of parameters in R. Observe that M/M2 can be considered as a vector space over the field R/M.We define: emdim R= embedding dimension of R - dimR/MM/M2 = the (R/M)-dimension of the vector space Observe that

M/M .

emdim R ^ dim R.

LEMMA 3 *^* Let be an ideal in R. Then PROOF.

(R, M) be a noetherian local ring and let H n“ = 1 (H + M1 ) = H. In particular = (o).

[N], Corollary 1 , on page 65.

PROPOSITION 3 *5 . Let (R, M) be a noetherian local ring. Then (1 ) dim R = rank^M; (2) Every irredundant basis of M contains ex­ actly (emdim R) elements. PROOF.

[N], Theorem 1 on page 63, and Proposition 6 on page 69.

REMARK 3.6. Let (R, M) be a noetherian local ring of dimension d. Then d = 0 < = = > M is the only prime ideal of R (0) is primary for M < = > M consists entirely of nilpotents;also observethat R is a field

M =(0) (and hence d = 0). From now on, by a noetherian local ring, unless otherwise stated, we shall mean a noetherian local ring of positive dimension. DEFINITION 3 *7 . Let R be a local ring. R is said to be regular if R is noetherian and dim R = emdim R. [We remark that Auslander-Buchsbaum-Serre have proved that the Horndim (Homological dimension of) R is finite < = 3c> R is regular =*=> Horndim R = dim R.] PROPOSITION 3.8. Let (R, M) be an n-dimensional noetherian local ring, and letx ^ ..., xm be a minimal basis ofM. Then R is regular if and only if oneof the following is true: (1 ) If f (X-j, ..., Xm ) is a form of degree s with co­ efficients in R not all in M then f(x.j, ..., xm ) 4 M S+1 . (2) If f (X.j, •.., Xm ) is a form (of arbitrary degree) with coefficients in R not all in M then f(x1' ^ ^ °* Now assume that Ris regular so that n = m. Then R is a normal domain. Let P^ = (x1, ..., x^)R. Then for1 = 0, 1, ..., n - 1 , P is a prime ideal in R,rank P^ = i, dim P^ = n - i, R/P^ is a regular local ring of dimension n - i, andRp is a regularlocal ring of r i_ dimension i .

13PROOF.

69

THE DIMENSION

([N], Section k . 6 ) + (Exercise).

DEFINITION 3 *9 . the quotient field of R. integer such that z e M 8^

Let(R, M ) bea regular local ring and K For o / z e R, let 8(z) be the nonnegative

and z / m 5 ^z ^+1

(8(z)

exists by 3-^).

We call 8(z) the leading degree of z. We set 5 (o) = «. For z^, z2 in R we have 5 (z1z2 ) = 8(z1 ) + S(z2 ) [for proof, see Lemma 3 on page TO of N], hence if we set5 (z1/z2 ) = 5 (z1 ) - S(z2 ), 5 becomes a real discrete valuation of K, andwe callthis the M-adic valuation of K. Now let T be a representative system of R/M in R and fix a minimal basis x 1, ••.,xn of M. Then given o / z g R, wecan write z = f(x^ ..., xn ), where f is a form of degree 8(z) with coefficients in R not all in M. Let F(X^ ..., Xn ) be the form of degree 8(z) in T[X^, ..., XQ] gotten from f after replacing the coefficients by their T-representatives. Then z - F(x1, ..., xn ) € m 6^z ^+1 It follows from 3*8 (2 ) that F(X^ ..., Xn) is the only form of degree 5 (z) in TtX^ ..., Xn ] which satisfies the above inclusion. We define t(z) «= F ( X U ..., Xn ) = the leading form of z with respect to the representative system T of R/M and the minimal basis x 1, ..., xn of M. DEFINITION 3.10. Let (R, M) be a local domain. If R con­ tains a subfield k and a finite number of elements a1, ..., an such that R = Ap where A = k[a^, ..., a^] and P = A flM, then weshall say R is algebraic with ground field k. We set:k-rank of R = tr. deg. (R/M)/k. Observe that by 3*2: k-rank R + dim R = tr. deg. R/k. R will be said to be algebraic if R is algebraic with some ground field. Observe that an algebraic local ring is noetherian. LEMMA 3.11. Let R be a normal local domain with quotient field K and let R* be a local ring lying above R in a finite alge­ braic extension K* of K. Then (1) (2)

If K*/K is separable and R is noetherian, then R* is noetherian with dim R = dim R*, and If R is algebraic with ground field k, then R* is algebraic with ground field k with k-rank R = k-rank R* and dim R = dim R*.

III.

TO

NOETHERIAN LOCAL RINGS

PROOF. If K*/K is separable everything follows from the fact that the integral closure in K* of a normal noetherian domain A quotient K is a finite A-module [W2, page 7 9 ]• If K*/K is not separable, then in the proof of (2), one has to invoke the following: If A is a domain with quotient field K such that A is finitely generated over a subfield k, then the integral closure of A in K* is finitely generated over k [for proof see the last paragraph on page 507 of Z1]. PROPOSITION 3-12. Let (R*, M*) be an algebraic local domain with ground field k and quotient field K* such that k-rank R* = 0. Let K be a subfield of K* such that k C K and K*/K is finite alge­ braic. Then each of the following two is a necessary and sufficient con­ dition for the existence of a local ring R in K lying below R*: (1 ) The ideal (K fl M*)R* is primary for M*. (2) There exist a finite number of elements u 1, ..., in K fl R* such that the ideal (u1, ..., u^ )R* is primary for M*. Furthermore, when

R

exists, it is algebraic with ground field

k.

PROOF. The proof of this is rather involved, and hence we refer to Proposition 1 of [A6]. DEFINITION 3 -1 3 - Let (R, M) and (S, N) be local domains with a common quotient field K. If S has center M in R and if there exist a finite number of elements u 1, .•., u^ in S such that S = Ap where A = R[u^ ..., u^] and P = A fl N, then we shall say that S is a finite transform of R. LEMMA 3.1^. (1) (2) (3)

R

S

be a finite transform of a local domain

If R is noetherian, then S Is noetherian. If R is algebraic with ground field k, then S is algebraic with ground field k. If T Is a finite transform of S, then T is a finite transform of R.

PROOF.

Then

Let

Exercise.

PROPOSITION 3 .1 5 « Let R be a one-dimensional local domain. is regular if and only if it Is normal. PROOF.

See Theorem 8 on page 76 of [N].

PROPOSITION 3*16. (1) (2)

Let

R

be a domain.

Then

R is the valuation ring of a real discrete valuation v < = = > R is a regular one-dimensional local domain. If R is the valuation ring of a valuation v, then R Is noetherian v is real discrete.

R.

13 • THE DIMENSION (3)

71

If R is a local domain, then R is the valuation ring of a valuation R is regular of dimension one.

ideal and in ideal erndim M-adic

PROOF. (1) ( = > ) Fix x in R with v(x) = 1 . For an H in R, let v(H) = min v(h) for h in H. Then H = xv ^ R particular My = xR. Hence R is noetherian (in fact, a principal domain) and M^, is the only prime ideal in R. Hence dim R = 1 = R. ( < = ) Let M he the maximal ideal of R and let v be the valuation. Then R = Ry .

(2) (==>) Let P be a prime ideal in R other than M. Since P is prime, C P would imply M = P which is not the case, hence P. Since the ideals in Ry are simply ordered by inclusion, we must have P C for all i. By 3 ‘^, P C = (°)> i.e., P = (0). Therefore v is real. Now v is nondiscrete implies there exist real numbers a^e Rv with a 1 > a2 > ... > 0, i.e., H 1 < H2 < ..., where H^ = (r e Ry | v(r) = a^} is a strictly ascending infinite sequence of ideals in R which contradicts the noetherian character of r. ( < = ) This has been proved in (1 ). (3)

Follows from (1) and (2).

DEFINITION 3 *1 7 * Recall that a domain dedekind domain if (1 ) (2) (3)

R isnormal R isnoetherian, and Every prime ideal

It follows that domainsuch that

inR

R

is said to be a

is maximal.

(R, M 1, ..., Mn ) is a normal noetherian semilocal dim ■ 1 for i = 1, ..., n < = = > R is a dedekind

domain with a finite number of prime ideals

, ..., Mn .

PROPOSITION 3-18. Let R be a one-dimensional noetherian local domain with quotient field K and let S be the integral closure of R in K. Then S is a principal ideal ring with only a finite number of prime ideals P1, ..., Pn and hence S is the intersection of the real discrete valuation rings Rp , ..., Rp . 1 n PROOF. See [K5 l and Section 39 of [ K k ] . LEMMA 3.18A. Let (R, M) be a two-dimensional algebraic regular local domain with quotient field K, let P be a minimal prime ideal in R, let R = R/P and M = M/P. Let K be the quotient field of R and let T be the integral closure of R in K. Then T is a finite R-module. PROOF.

See page 511 of [Zl]-

PROPOSITION 3.18B.

Let

(R, M)

be a normal algebraic local

III.

72

NOETHERIAN LOCAL RINGS

domain with quotient field K, let K* he a finite separable algebraic extension of K and let (R*, M*) be a local ring in K* lying above R. Assume that R* is regular, R*/M* = R/M and MR* = M*. Then R is regular. PROOF. The proof of this proposition follows from the consider­ ations made in Section 2 of [A2]; since these considerations use the notion of the completion of a local ring, we shall omit the proof. 1k . Quadratic transforms. LEMMA 3.19. Let (R, M) be a regular local domain of dimension s > 1 with quotient field K and residue field k. Let {x^ x2, ..., xs) be a system of parameters in R. Let y 1 = x 1, y^ = x^/x1 for i = 2, 3, ...s; and let S = R[y2>•••> ys]* Then y 1S flR = M, and k = R/M can be canonically identified with a subfield of S/(y.jS). Furthermore, the residues y2, y , ..., ys modulo y.,S of y2, j y ..., ys are algebraically independent over k, and can be canonically identified with the polynomialring k[y2, • • • > ys3 in s - 1 in­ dependent variables. PROOF. Let v betheM-adic valuation of K. Now Ry 0 S and v(y1 ) > 0, and hence 1 /y^S. Therefore, y 1S fl R / R. Since y ^ 3 M and since M is the maximal ideal in R, it follows that y^S fl R = M; and hence wecan canonically identify k = R/M with a subfield of S/(y.jS), inthe usual fashion. With this identification, it is clear thatS/Cy-jS) is generated, as a ring over k, byy2, y^, ..., y g . Note that the center of vin R is the maximal ideal M, while the center of v in S is the minimal prime ideal y.,S. The valuation ring Ry is the quotient ring of S with respect to y^S. The only thing that remains to be shown is that y2, y^, ..., y s are algebraically independent over k. Assume the contrary. Then there exists a polynomial f(y2, i y

ys )= Sf±

i ,...,±72

f. . . in 2, 3, **• > :La f(y2> 1 3 > • • • > la ') - 0 (mod. y^), i.e., f(y2» 1 3 >

1 3 ) =

is ys

S

j

with the coefficients

12 ^ y3

R,

but not all in

1 3>

> l s)

M,

such that

>

where g(y2, l y •••, ys ) is a polynomial in y2, j y •••, y3 with coefficients in R. Multiplying both sides of the above equation by a suit­ able high power x^ of x 1 and setting

1k . QUADRATIC TRANSFORMS

73

t-ip-i ~ - .•• i

Pt (x1, x2,

xs ) =

.-,3-gX!

ip

i„ x3

x2

we get pt (x i» x2, ..., xs ) = x 1t+1g(y2, y 3,

ys )

.

Hence v(Ft (x1, x2, ..., xg )) ^ t + 1, i.e., pt3 x 1' x 2 > •••> x3 ) 13 contained in Mt+1 . This is a contradiction, since F ^ ^ , x2, ..., xs ) is a form of degree t in x^, x2, ..., xg with coefficients in R, not all in M. Hence f 2 > f y • • • , J Q are algebraically independent over k, and this completes the proof. LEMMA 3.2 0. Let (R, M) be a regular local domain of dimension s > 1 and let X y x2, ..., xs be a minimal basis of M. Let v be a valuation of the quotient field K of R having center M in R. Suppose we have arranged the x^ so that v(x^ ) < v (xq) ^OP p = ..., s. Let A = RCXp/x^ • • • > XS/X -|J> P = A fl My, S = Ap and N = PS. Then (S, N) is a regular local domain of dimension t < s, v has center N in S and if by d and d* we denote respectively the R-dimension and the S-dimension of v then we have: s - t = d - d*. PROOF. By 3 *19 , ( A/ x^A) can be canonically identified with a polynomial ring A* = (R/M)(X2, X y .. ., Xg ) in s - 1 variables over R/M. Let P* ~ P/(x-A). Then A*p*is a regular local ring [Z2, Section * * * 4 .1] of dimension h < s - 1. Fix y2, y^, • • • > y^-j A* such that P*A*p* = (y*, y*, Fix

y^

in

A

..., y^+1)A*P*

belonging to the residue class

(x-|, y2, y3> •••> y h+1 )s = N t = dim S

y^.

-

Then it is clear that

and that

= rank N = 1 + rank P* = 1+ h

Therefore S is regular. Also t < 1 + h < (s - 1) + 1 = s. Since N fl R = M, we may canonically assume that R/M C S/N C Ry /Mv . Since the transcendence degree of S/N = A*/P* over R/M is s - l - h = s - t , we conclude that: s - t = (R-dimension of v) - (S-dimension of v) = d. - d*. DEFINITION 3*21. In the notation of the above lemma, S is called "the first (or immediate) quadratic transform of R along v." Let now RQ = R and let R^ be the first quadratic transform of Rj__-| along

v

assuming that

dim

> 1•

^

d;i-m Rj_ > 1>

P°r

III.

7^

NOETHERIAN LOCAL RINGS

i = 1, 2 , n - 1, then Rn will be defined and we shall say that MRn is the n-th quadratic transform of R along v." If S is the n-th quadratic transform of R along v for some n, we shall say that nS is a quadratic transform of R along v." Finally, if S is a quadratic transform of R along some valuation v having center M in R, we shall say that MS is a quadratic transform of R.M

domain

LEMMA 3*22. Let S be a quadratic transform of a regular local R. If R is algebraic with ground field k, then so is S. PROOF.

Follows from 3 -1^-

CHAPTER IV:

TWO-DIMENSIONAL LOCAL DOMAINS

In the next two sections we shall mostly deal with two-dimension­ al regular local domains which are algebraic. However, all (respectively, almost all) the results of Sections 15 and 16 have been extended by us in [A4 ] to absolute (respectively, arbitrary) two-dimensional regular local domains. 15 • Limits of quadratic sequences.

PROPOSITION 4 .1 . Let (R1, M 1 ), (R2, Mg ), ..., be a sequence of normal local domainswith a common quotient field K such that Rj_+1 has center in for i = 1, 2, ... . Assume thatU^=1R^ is not the valuation ring of a valuation of K. Then there exist infinitely many valuations w of K which havecenter in R^ and for which Dw is of positive transcendence degree over for each i. PROOF. Let R = ^ =1Rj[ and M = We may canonically assume that R 1/M1 C R2/M2 C ... . Let D = qR^/M^. Then D is a field and R/M = D, i.e., M is a maximal ideal in the domain R. In fact, by 1.5, (R, M) is a local ring. Also observe that R is normal. Since R is not a valuation ring, by 2 .14 there exists x € K with x 4 R and 1/x 4 R. Let h be the canonical homomorphism of R onto D and X be an indeterminate. For f(x) = wpth f^ e R we set H(f(x)) = zi= 0*1^ i

‘

assert that

H

is a homomorphism of

R[x]

onto

D[X].

[PROOF. It suffices to prove that H is single-valued, I.e., that f(x) = g(x) = = > ^h(pi )Xi = Zh(gi )Xi . Let q± = fi>- g±. Then we have to show that q-j_ e R and z^Qq^x1 = o = > Z^=0h(qi )Xi = 0, i.e., h(q^) = o for all i. Assume the contrary. Then for some i, h(q^) 4 o, i.e., q^ is a unit in R; let t be the maximum value of i for which this is so. Then we can divide the equation iqj[X:d = o by qt and obtain:

Pr/1 + Pn-1xI1"1 + ••• + Pt+1xt+1 + xt + Pt - l xt_1 + ••• + Po = 0

76

IV.

TWO-DIMENSIONAL LOCAL DOMAINS

with

Pj_ = 1^/% e R>

and

h^pt+1 ) = h(pt+2^ = ••• = h(pl

= °"

Let r

- p nx n _ t + pn - 1 x n " t _ 1

+

•••

+ Pt + i x + 1

»

and s = pt_ 1 (l/x) + Pt_2 (l/x) 2 + ... + p0 (l/x)t

,

so that rx^ + x^s =o,

i.e.,

r = - s

Let v be a valuation of K having center M in R, then v(x) > o = = > v(r) ^ o and v(x) < o = > v(r) = v(- s ) > 0 ; hence in either case v(r) ^ o, i.e., r e R . Hence by 2.3^ r e R. Again by 2 . 3 k , x i R = > there exists a valuation v of K having center M in R such that v(x) < 0 . Then v(l/x) > 0 so that v(r) = v(- s) > 0 and hence r e My fl R = M. Thus (1

- r) + Pt+1x + ... +

Pnxn-t

= 0

,

with P-fc+i* •••> Pn ^ ^ an(^ r G M. Now r e M = > 1 - r is a unit in R and hence dividing the above equation by (1 - r)xn“^ we would ob­ tain an equation of integral dependence of 1/x over R. This is a con­ tradiction since 1 /x i R and R is normal.] It Is clear that H is the unique extension of h with H(x) = X. Let p be any one of the infinitely many prime ideals in D[X]. Let P = H~ 1 (p). By 2.22 there exists a valuationw of K having center P in R[x]. Since Dw 0D(x), it is clear that w has the required properties. The infinitely many choices of p give us infinitely many w of the required type. LEMMA ko2 . Let (R, M ) be a two-dimensional local domain with quotient field K. Let P be a minimal prime ideal in R. Then: (1) (2)

Rp Is the valuation ring of a real discrete valuation w of K; There exists at least one and at most a finite number of valuations v of K having center M in R which are composed with w, i.e., for which Ry C R^j

15.

LIMITS OP QUADRATIC SEQUENCES

77

(3) If R is algebraic, then each such valuation v is discrete of rank two and Rv/Mv a finite algebraic extension of R/M (hence in particular v is of R-dimension zero). PROOF.

(1 ) follows from 1 .1 7 and 3 *1 5 * The proof of (2) is as

follows: R/(R fl M^) is a local domain of dimension one with quotient field Rw /Mw . Hence, by 3 -18, the integral closure of R/(R fl M^) in R^/M^ is a Dedekind domain D with a finite number of prime ideals P^, P2, ..., P-^. Let v£ be the real discrete valuation of with

V

= DP *

vi

valuation °f

ls composed of

w

and v£. Then v1, v2, — , vh are exactly the valuations described in (2). Finally, (3) follows at once from 2.49. LEMMA 3. Let (R, M) be a two-dimensional algebraic regular local domain with quotient field K. Then: (1) R is a unique factorization domain, i.e., equiva­ lently, every minimal prime ideal in R is principal; and (2) There existinfinitely many valuations of K having center M in R which have R-dimension zero and which are discrete of rank two. PROOF. For the proof of (1), we refer to Theorem 5 of [Z2]. To prove (2) it is enough to show, in view of 4 .2 , that there exist infinitely many relatively prime irreducible nonunits in R.To show this, let x, y be a minimal basis of M. Let P = xR, R = R/P, M = M/P, K = Rp/(PRp), y = the residue class modulo P containing y, and w = the real discrete valuation of K with R^ = Rp. Then M = yR and hence R is a regular one-dimensional local domain, i.e., R = R- where v is a real discrete valuation of K with v(y) = 1. Let v be the valuation of K which is composed of w and v, and let us write the elements of the value group of v as lexicographically ordered pairs of integers. Let xra = x + y01 where m is a positive integer. Since (xffl, y) is a basis of M, xm is an irreducible nonunit. Since v (x ) = (0, vCy331)) = (0, m), we have that v(xm ) 4 v (xn^ whenever m 4 n. Therefore x 1, x2, x^, ..., are in­ finitely many pairwise relatively prime irreducible nonunits in R. PROPOSITION 4 .4 . Let (R, M) be an n-dimensionalalgebraic regu­ lar local domain with quotient field K withn > 1, and let v be a valuation of K which has center M in R such that Dy is of trans­ cendence degree n - 1 over R/M. Then the quadraticsequence along v starting from R is necessarily finite, i.e., if R =RQ and if R^ is the first quadratic transform of R-j__i along v provided dim R^ ^ > 1

78

IV.

TWO-DIMENSIONAL LOCAL DOMAINS

then for some integer h we have that is one-dimensional; we also have: = Ry . Furthermore, there exists a field T with R/M C T C Ry/Mv such that T is finitely generated over R/M and Rv/^v as a pure trans­ cendental extension of T of finite positive transcendence degree (we may express this by saying that "Rv/Mv is a ruled extension of R/M"). PROOF. Assume the contrary, i.e., that the quadratic sequence R = RQ < R 1 < R2 < ... along v is infinite. It then follows, by 3.2c, that there exists an integer s such that dim R^ = dim Rg and R^.dimension of v = m - 1 whenever t ^ s, where we have set m = dim R . Let S = up=0Rj_ and N = where M^ is the maximal ideal in R^. Then, as in the proof of 4 .1, N is the unique maximal ideal in S and U?=gR^/M^ = S/N. Since, as in the proof of 3-20, R^ /^t+1 as an alSe" braic extension of whenever t b s, it follows that S/N is an algebraic extension of RfV^t ^or an^ ^ - s • Now v ^as center in R^ for each i and by 2 .47 , v is real discrete. Suppose, if possible, that S is not the valuation ring of any valuation of K. Then by 2 .1 4 there exists x in K such that x / S and 1/x / S. Then x / Rq and 1/x / Rq . Since K is the quotient field of RQ, we can write x = y0/zQ with yQ and z Q in RQ and hence in MQ . Since v has center MQ in RQ, we have: v ( j Q ) > 0 and u (z0 ) > 0# Net be parameters in RQ . Arrange matters so that v(x1 ) < v(x^) for

''*>

xn

i = 1, ..., n. Then x ± / x -\ G R-j Ro*5 i = 1> • • • > n and hence y 1= y0/x1 e R 1 and z1 = z Q/x^ e R 1 . Thus v(yQ )> v(y1), v(zQ ) > v(z1) and x = y0/zQ = y-j/z-j. Proceeding in this manner, we obtain y^, z^ e M^ such that x - J ± / z ± and > v (y-\ ) > > ••• > 0 y (z0 ) > v(z1 ) > v(z2 ) > ... > 0. This contradicts the discreteness of v. There­ fore, S must be the valuation ring of a valuation w of K. Since Rs has a first quadratic transform Rai1, it follows by the definition of o+ I quadratic transforms that m > 1, i.e., that v is of positive R^-dimension whenever t > s. Now Ry D Ry and

( 00

\

00

00

U Ri) n «v = U 0 and b > o. In thiscase, w is of rational rank two and hence v(z-) and v(zp ) are rationally independent. Since a b ai b i z,jZ2 = z = x y d and since Z y z2e R*, wemust havez1= x y d and z2 = x d2

ap

y

d2

where

are units in b3

R*.

a 1, b ^ a2, b2 Let

a2

z^ = z1

are nonnegative integers and d 1 and ”ai b2 "”b 1 z2 and z^ = z^ z2 . Then we

a4

have z^ = y Jd^ and z^ = x d^, where b^ and a^ are integers and d^ and d^ areunits in R*. LetD denote the determinant with first row ( a y b 1 ) and second row (a2, b2 ). Then we have zD^ _ — z2~b 1z^al

and,

D2 _— z^ _b2z^ a2 z

.

17.

UNIFORMIZATION FOR A FUNCTION FIELD

85

Since v(z1) and v(z2 ) are rationally independent, we must have D / 0 and hence in view of the above two equations we conclude that v(z^) and v(z^) are also rationally independent and hence v ( z 4 o 4 v(z^). Therefore b^ 4 0 4 Now we define elements z* and z* as follows: if b^ > o then let z* = z^ and if b^ < o then let z* = z”1; if a^ > o then let z* = z^ and if a^ < othen let z* = z^1. Then * b*-,* z 1 = y d*

, and

* a z2 = x d*

,

where a* and b* are positive integers and d* and d* are units in R*. Since z* and z* belong to M* PI L = Q*, we conclude that P* is primary for M * . 17 - Uniformization for a two-dimensional algebraic function field. Let K/k be a two-dimensional algebraic function field and let v be a valuation of K/k. The theorem of local uniformization asserts that v can be uniformized, i.e., there exists a regular algebraic local domain(R, M) with quotient field K and ground field k such that v has center M in R and k-rank of R = k-dimension of v. This theorem was proved by Zariski for fields of zero characteristic in [Z3 ] and [ Zk] and for perfect fields of nonzero characteristicby the author in [A2 ] and [A6 ]. As an application of ramification theory, we give in this section a simple proof for the case when k is algebraically closed and of characteristic zero. If k-dimension of v is one, we fix a^, ..., an in Ry such that K is the quotient field of A = k[a^, ..., an ], A is normal, and the residue class of a 1 in Dy is transcendental over k, we let P = A D My, R = Ap and M = PR, then (R, M) is regular of dimension one (see 3 .1 5 ) and v has center M in R. Thus it is enough to consider the case when v is zero dimensional (i.e., k-dimension v = o). THEOREM 1+.9. (Theorem of unif ormization for K/k). Let K be a two-dimensional algebraic function field over an algebraically closed ground field k of characteristic zero. Let v be a zero dimensional valuation of K/k. Then v can be uniformized. PROOF. By 2 . k 9 , r(v) = 1 or r(v) = 2 and v is an integral direct sum. If r(v) = 2, choose x 1, x2 In K such that v(x1 ), v(x2 ) is an integral basis for the value group of v; then by 2 . k 6 , (x^ x2 ) is a transcendence basis of K/k. If r(v) = 1, then let (x^, x2 ) be an arbitrary transcendence basis of K/k. Replacing x^ by x^1 if necessary we may assume that v(x^) ^ 0.; since Dy = k, subtracting suitable elements of k from x ^ x2 we .may assume that v(x^) > 0 for i = 1, 2. Let K 1 = k(x.j, xg ), and let (R1, M 1 ) be the quotient ring

TWO-DIMENSIONAL LOCAL DOMAINS

IV.

86

of k[x^ x2 ] with respectto the prime ideal generated by (x^ x2 ). Then v has center M 1 in R 1 anddim R.j = 2 and (x1, x 2 )R1 = M 1 , so that R 1is regular.

v* v.

Let K* be a galois extension of K 1 containing K and let be an extension of v to K* and let v 1 be the K 1-restriction of Observe that if r(v) = 2 , then Sy = Sy . Let v* = v*, v*, v*

be all the K*-extensions of

v 1 . Then

T =

Is the integral

IL = T fl M * for i = 1, ..., n are the i maximal ideals in T. Hence by ) there exists u e T such that u g H.j and u i for i = 2 , ..., n. Let um + alum” 1 + ... + am = 0, with a^ € R^, be the equation of integral dependence of u over Ry . closure of

R__

in

K*,

and

V1

In view of k . 5 , may assume that closure of R 1 P 1 and u i 1 = 2 ,,..., n. Then gP 1 = P 1 By 1 .50 , G 3 (v*/v).

71 (1 .3

by replacing R 1 with a quadratic transform along v 1 we a^ e R 1 for i = 1, ..., m. Let S be the integral in K*, let P 1 = S n My* = 3 .(1 ^ . Then u e S fl H 1 = for i = 2 , ..., n implies that P 1 / S fl for Let g e G(K*/K1) (= the galois group of K* over K 1 ). implies that gH 1 = H 1 . Therefore GS (H1 /MV) D GS (P*/M1^.

Gs (H, /My ) C GS (P*/M1 ). Let

R* = Sp

and

Hence

G 3 (P*/M.,) = GS(H1/MV ) =

M* = P.R*. Then

r 1

(R*, M*)

is the local

1

ring in K* lying above R^ such that v* has center M* in R* and we have GS (R*/R) = Gs (v*/v). Let R = R* n K and M = M* fl K. Let K 3 = PS(R*/R1 ) and K 3 = FS (R*/R). Then by 1 .49 , K 3 is the compositum of K 3 and K. Since K = Dy is algebraically closed, by 1 .48 , we have G^(v*/v) = Gs (v*/V); and by 2 .43 , G^"(v*/v) is abelian. Therefore G(K*/K3 ) = G 1 (v*/v) and this is abelian. Hence K 3 is galois over K 3 and G(KS/K3 ) is abelian. Hence we can find a sequence of fields K 3 = Lq C L 1 C ... C Lt = K 3 such that L^ is a cyclic extension of L ^ -1 of prime degree for i = 1, ..., t. Let w^ be the L^-restriction of v*. Since v* is the only K*-extension of w , w^ is the only ex­ tension of w i-1 to L^ for i = 1, •• •j t .Let S^ = R* fl L^ and N± = M* flL± . By 1. h i , M 1Sq = Nq and hence the regularity of R 1 implies the regularity of SQ . Thus

17. wQ

UNIFORMIZATION FOR A FUNCTION FIELD

87

has been uniforraized. Now we anticipate the following

PROPOSITION 4 .1 0 . Let L be a two-dimensional algebraic func­ tion field over an algebraically closed ground field k of characteristic zero. Let L* be a cyclic extension of L of prime degree p (p > 1). Let w be a zero dimensional rational valuation of L/k having a unique extension w* to L*. Assumethat w can be uniformized. Thenw* can be uniformized. If r(v) = i, then applying this proposition successively to the extensions L^/Lq, L2/L1, ..., L^_/L^