vrrA Name : Merle Eugene Manis Place of Birth : St . I gnat i us , Montana Dat e of Birth : August 20, 1934 Schools Atte.n ded: University of Montana University of Washington University of Oregon Degrees Awarded: Bachelor of Arts , J une 1960, University of Montana .. Master of Arts , August 1961 , University of Montana Doctor of Philosophy, June 1966, University of Oregon Area of Special Interest: Mathematics (Algebra ) Professional Experience: Teaching Assistant, University of Washington , Sept. 1961 Instructor, University of Montana, January 196_2-July 1963 Teaching Assistant , University of Oregon, 1964 - 1966 • VALUATIONS ON A COMMUTATIVE RING by MERLE EUGENE MANIS A THESIS Presented to the Department of Mathematics and the Graduate School of the University of Oregon in partial fulfillment of the requirement s for the degree of Doctor of Philosophy June 1966 " APPROVAL David K. Harrison, Advisor TABLE OF CONTENTS Introduction . 1 1. Valuations and Valuation Pairs •• 4 2. Independence and Extensions ._ • • . •. 21 3. The Inverse Property, Approximation Theorems • . • • • • • . ~ • • • • • • 36 4. Galois Extensions. · .57 Bibliography• •• . 75 ACKNOWLEDGMENT This thesis was motivated by an as yet unpublished work on a 11 theory of primes" now being developed by Professor D. K. Harrison, whom I wish to thank for making his concepts and results available to me and for being a continuous source of inspiration and encouragement during my years in Graduate School. Special thanks are due to Hoyt D. Warner and Gerson Levin for laboring through this thesis while it was still in its handwritten form and making many valuable suggestionso I am also deeply indebted to the National Science Foundation for its generous financial support during four years of Graduate School. 1 I NTRODUCTION The purpose of this paper is to extend some results from the theory of valuations on a field to an arbitrary commutative ring with identity. The results obtained are well known when interpreted in the context of a field and comprise only a bare introduction -to the theory for fields, however, the modified proofs give some added insight even in this case. In Section I of Chapter 1, the concept of valuation on a field is extended to an arbitrary commutative ring and a natural correspondence is obtained between valuations and what we call valuation pairs . Section II shows that these valuation pairs are the same as those in [1], where they are the subject of an exercise. Sections III and IV of Chapter 1 relate valuation pairs to the generalized primes of D. K. Harrison. The results presented here predate the rest of this paper, · being developed t? investigate primes [3]. Proposition 1.14 evolved during the course of a seminar given by Professor Harrison during the fall of 1965 f while 1. 12 and 1. 13 appeared in the unrevised form of [3]. The outline followed for Chapters 2, 3 and 4 is essent ially that used in [6] in developing the theory for fields. Many of the arguments used are almost verbatum 2 those used in this source . Standard results for fini t e rank valuations are not given since their proofs (and statement s) are not significantly altered in the more general context . Section I of Chapter 2 deals with the concept of independence of valuations and Section II with the concept of extension~ Section III combines these to obtain some results essential to Chapter 3. Sections I and II of Chapter 3 are used to develop the machinery and setting for the "approximation theorems" of Section III. The approximation theorem is applied in / Section TV to obtain the classical inequality "~e. f. < n 11 • ]. ]. - The paper ends with the proof of the classical equation efg1rd = !GI in the context of a commutative ring R which is Galois over a ring K with group G. The generalized Galois theory necessary for this result is outlined in Section I of Chapter 4. The rest of the chapter is devoted to the definitions and relations (which are interesting in their ovm right) necessary for its statement and proof . General ring theory comparable to that found in [5] is assumed, but beyond that the treatment is largely self contained. A notable exception is Sec tion I of Chapter 4 where several results are quoted from [2] without proof . In order to cut down on verbiage, much notation is assumed as standard once it is introduced . Thus R is always a commutative ring with identity, K is always a 3 subring of R, Vis always a valuation on R, et c. 11 Ring 11 will always mean "commutative ring with identity". 11K a subring of R11 means the same thing as 11R is an extension of 11 ; in both c ases meaning that K is a ring, Kc Rand the identity of K is the identity of R. Ring Homomorphisms will always take identity to identity. Prime ideals are al •ays proper . The word 11 iff 11 is a contraction of "if and only if 11 , and is sometines denoted by ¢::;; . 11A ⇒ B" means •" A implies B", "3 11 means "there exists II and 11 'd II means 11 f or all 11 • If A and Bare sets, A'B = (x I x EA, x t B} and should not be confused with A/B, which denotes a quotient of rings, groups , etc. 4 1. VALUATIONS AND VALUATION PAIRS Section I By an ordered group, we mean an abelian group r* (written multiplicatively) which is linearly ordered by a relation 11 < 11 · satisfying a < 13 ⇒ a:y < ~ for all a, /3 , -y Er*. We will always denote the identity of an ordered group bye, and we admit the group (e} as an ordered group. DEFINITION 1.1. A valuation semigroup r is the di sjoint union of an ordered group r* and an element 0, where the .order and multiplication of r* is extended to I" by: i. ) O < a for all a € f' ii.) O ·a= a· O = O for all a € I" DEFINITION 1.2. A valuation Von a commutative ring Risa map V from R to a valuation semigroup r satisfying i.) V(xy) = V(x)V(y) for all x,y € R ii.) V(x+y) ~ max(V(x),V(y)} for all x,y in R iii.) Vis onto r. We shall sometimes like to think of r as embedded in a larger valuation semi - group, in which case we relax iii.) to "V(R)~V(O) is a g roup". 5 One can check that V(l) = e and V(O) = O for all valuations. If Risa field and i.) holds, then V( R"-.(O)) is always a group so iii.) can be replaced by V(l) f V(O) or one can work with ordered groups rather than semi- groups . Condition ii.) is the non-Archimedian condition in a field. PROPOSITION 1.3. Let V be a valuation on a ring R, set Av = {x € R V(x) < e) pv = {x E R V(x) < e) (JV = (x € R V(x) = O} Then Av is a subring of R, Pv is a prime ideal of Av and crv is a prime ideal of R. Further, if a is an ideal of R, CJ C Av,~ f R, then CJ C crv. PROOF. Note that v(-x) = v(-l)V(x) and v(-1) = v(-1)- 1 , thus that v(-1) = ·e and v(x) = V(-x) for all x € R. Thus we have A = -A p v' = -P and V V V (JV = - (1 • By condition ii.) of Definition 1.2 we have V A + ~c ~, p + p C p and V V V V (JV + (JV C (JV. By i.) we have AP c p and Rcrv c crv, thus A,v is a subring of R, V V V Pv is an ideal of Av and crv is an ideal of R. If a· b E Pv, then e > V( ab ) = V(a)V(b) so either e > V(a) ore> V(b). Thus Pv is a prime ideal of Av (V(l) = e so 1 f Pv). If ab€ (JV' then O = v(ab) = V(a)V(b), 6 so V(a) = O or V(b) = 0, so crv is a prime ideal of R. Finally, suppose Av f Rand cr is an ideal of R. If cr ¢ crv, then V(a) f O for s ome a E cr. But then 1 - V(b) = V(a)- for some b ER and V(c) > e for some c ER (since~ f R by hypothesis). But then abc E cr while V(abc) = V(a)V(b)V(c) = eV(c) = V(c~ > e so cr ¢A. V PROPOSITION 1.4. If Vis a valuation on a ring R, X E R"-..A , then there is a y E P with xy E A~ P . V V V V PROOF . If XE R"-..Av, then V(x) > e and for some y E R, V(y) = V(x)- 1 . e = V(x)- 1v(x) > V(x)- 1 e = v(x)-1 ~ - 1 so y E p • Now V( xy ) = v(x)V(y) = V(x)V(x)- = e so V xy E A~P • V V DEFINITION 1.5 . By a valuation pair of a ring R, we mean a pair (A,P), where A is a subring of Rand Pis a prime ideal of A, such that x E R"-..A ⇒ xy E A......_____P for some y E P. Note that (A ,P) is a valuation pair of R for any V V valuation V or R. We have the converse; PROPOSITION 1.6. If (A,P) is a valuation pair of R, then there is a valuation V of R with A= Av and P = Pv. Furthermore if v1 is another valuation of R with either P = P , or with A= A and A+ R, then there is an order vl -. vl preserving isomorphism ¢ :r v -r with ¢ 0 v1 = V. 1 V 7 PROOF. Let (A,P) be a valuation pair of R. For - x, y E R define x "-- y if ( z E R I xz E P} = ( z E R I yz E P} . """"" is clearly an equivalence relation on R. Let V(x) = (y I Y""--'X} and r = (V(x) J x E R}. V CLAIM 1. V(xy) = V(x 1y 1 ) for all x 1 E V(x), y 1 E V(y) . . - Thus defining V(x)V(y) = V(xy) malrns r v into a semi group. - Furthermore, r ""-tV(O)} is a group withe= V(l) = A"- P. V SUBPROOF. Suppose x 1 E V(x), y 1 E V(y). Then (xy)z E P iff x(yz) E P iff x'(yz) E P 1ff y(x'z) E P - iff y'(x'z) E P iff (x'y')z E P, so V(iy) = v(x'y'). The - - operation V(x)V(y) is thus well defined; iu is associative and commutative since multiplication in R is. V(l) is clearly an identity and V(l) f v(o) since 1 • O e P but 1 · 1 f P. If x f A, then xy E A' °'p for some y E P. Since 1 · y E P, xQC-1 . Thus V(x) f V(l) and V(l) CA • . If z E P, then z • 1 E P, 1 · 1 f P so V(l) f V(z) and V(l) c A""---.P. (Note that we have also shown that V(x) n A = if x f A.) Suppose x E A"-P and xy E P. If y f P (i.e., if x f V(l)); then y f A, since Pis a prime ideal of A. But then yz € A""---.P for some z e P, while x(yz) = (xy)z E P, contradicting Pa prime ideal of A. Thus A""---. P c V(l), which gives V(l) = A""'- P. 8 Finally , if x f v(o), then we have xy f P for some y ER. If xy E A""-.P, we have V(xy ) = V(x)V(y) = V(l); otherwise xy f A and xyz E A~ P for some z E P and V( xyz ) = V(x)V(yz) = V( l) . Thus r ~(v(o)} is a group . V CLAIM 2 . Define v(x) < V(y) if 3 z ER with xz E P , - - yz E A""-.P. Then 11< 11 is a linear order on f' v' r ~£v(o)} is an ordered group and r is a valuation semi-group. V - SUBPROOF. Note that V(x) < V(y) iff V(xz ) C P while - - - V( yz) = V(l) for some z E P iff V(y) f O and · - 1 V( x)V(y )- C P . Thus 11< 11 is well defined . - If V(x) f V(y), then for some z ER, either xz € P - and yz f P, or xz f P and yz E P. Suppose xz € P and yz f P. If yz E A'-..__P, we have V(x) < V(y).' Otherwise - yz f A so yzw E A"-.... P for some we P; · then (xz )w E P and again V(x) < V(y). Thus 11<" is a linear order on r v· - - . 1 Let V(x) < V(y) and V(z) f o. Let V(w) = V(z)-. Now xt E P and yt e A'-..__P for some t ER, so we have (xt)(zw) = xz(tw) € P and (yt)(zw) = (yz)(tw) E A""-.P so - -- - - - that V(x)V(z) < V(y)V(z). Thus f'v"-._(V(O)} is an ordered group. Clearly v(o ) ~ v(x) and v(o)v(x) = v(o) for all x ER, so f'v is a valuation semi-group . Thus Vis a valuation on R. By construction , A=~ and P = Pv . Now suppose v1 is a valuation on R with A= A f R vl 9 or P = P . If P = P , then A = [x €RI xP c P } vl vl vl 1 1 = [x €RI xP c P} = A. If A= A f R, then vl P = [x € A zy € A for some y f A} = [x EA xy EA for some y f A } = P . Thus ~f vl vl vl vl A= A f R or P = P , then (A,P) = (A ,P ) . vl vl - vl vl Claim for for x ER that v 11- ([ v 1 (x)}) = V(x) . SUBPROOF. This is clear if V(x) = v(o). 1 - - Leto t V( x ), V(z) = V(x)-. Then y € V(x) iff yz E A'--...P iff v1 (zy) -= e ,iff v1 (z)v1 (y) = e iff v1 (y) = v1 (z) - l iff vl(y) = vl(x). Thus V(x) = ·vl-1 ([Vl(x)}) . - - - Now v1- 1 ([v1 (xy)}) = v 1 1- ([v1 (x )v1 (y)}) = v(xy) - - - - .. = V(x)V(y), so :rv rv 1 is an isomorphism. Also v1 (x) < v1 (y) iff v1 (x)V1 (y)-l < e iff vl-1([vl(x)vl(y)-1}) = V(x)V(y)- 1 c P iff V(x)V(y)-1 < e - - iff V(x) < V(y), so order is preserved . - 1 Thus v1- ([ }) is the order preserving isomorphism c laimed in the proposition. Henceforth, we will speak of the valuation determined by (A,P) and will refer to the coset representation of r V derived above as the normal representation and wherever desired assume this is the representation under consideration. 10 COROLLARY 1.7 . I f (A,P) is a valuation pair of R, then i . ) R'--. A is closed under multiplication ii.) R""-P is closed under multiplication iii.) xy E A ⇒ X E A or y E p iv. ) xn E A ⇒ x E A v_. ) xn E A-.......__P ⇒ x € A"--P vi.) A= ( x ER I xP c-P} vii.) A= R or P = (x EA I xy EA for some y f A). PROOF. Let V be the valuation associated with (A,P) in 1. 8. Translating, we have i.) V(x)V(y) > e if V(x) > e and V(y) > e; - - - ii.) V(x)V(y) >: e if V(x) ::. e and V(y) > e; - - - - iii.) V(x)V(y) < e ⇒ V(x) ~ e or V(y) < e; iv . ) V(x)n -~ e ⇒ V(x) < e; v. ) V(x)n = e ⇒ V(x) = e; - - vi.) v(x) ~ e ¢=> V(x)V(y) < e for all V(y) < e; - - - vii.) If V(z) > e for some z then V ( x) < e ¢:I V( x) V( t ) < e for some V( t ) > e. Section II DEFINITION 1.8. For Ra commutative ring, let T = T(R) = ((A,5) I A is a subring of R, 5 is a prime ideal - of A). For (A,5), (B,cr) ET write (A,5) < (B,a) if Ac B 11 and 5 = An 11 11 (5. < is clearly an inductive partial order on T, so by Zorns l emma, T has maximal elements . We (temporarily) call maximal elements of T maximal pairs . - Note that if (A,6) ET, then there is a maximal pair (B,cr) with (B,-cr) ~ (A,6). PROPOSITION 1 . 9 . If (A,6) is a maximal pair of R, then A is integrally closed in R. PROOF. Let A be the integral closure of A in R. Then there is a prime ideal cr of A with cr n A = 5 ( see [5],'p. 257). That is (A,cr) > (A,5) so A= A. PROPOSITION 1.10. (A,5) is a maximal pair of R, iff - (A,5) is a valuation pair of R. PROOF. It is clear that valuation pairs are maximal pairs, so it is the converse that is of interest . Let (A,5) be a maximal pair of R, x + A, B = A[x] and cr = B5 . cr is an ideal of B with 5 C cr n A. If · 5 == A f\ cr, then A. ......__,_ 5 is a multiplicative subset of B with (A"-...5)n cr = ~. Then by Krulls lemma, there is a prime ideal cr' of B with cr c cr' and (A -......_,_5) (\ cr' = ~. - That is 5 = cr' n A and (B,cr') ~ (A,5). But since A f B, this is a contradiction, hence cr -('\ A f 5. n i Thus there are p. E 5, a E A"-. 5 with(* ) ~ x p. = a. l i=O l 12 We can assume n is minimal for an expression of this form. 1 n-1 . . 1 We have apnn - = ( xpn )n + .~ ( xpn ) 1 Pnn - -1p i, an i=O -. integral expression for xpn , thus xpn EA by 1.9. is an expre~sion of form(*) with lower degree, contra- dicting the choice of n. Thus xpn E A'--..5 and (A,5) is a valuation pair. We now dro-p the terminology 11maximal pair" in favor of "valuation pair" . Section III DEFINITION 1.11. We call a valuation pair (A,P) of - Ran· H (Harrison) pair (Pis what is called a finite prime - in [ 3]) if A/Pis a locally finite field. That is if every finite s ubset of A/Pis cont ained in a finite sub- field of A/P. PROPOSITION 1 . 12 . (See [3]): (A,P) ET is an H pair of Riff i. ) Q is closed under • and , and P c Q ⇒ P = Q or 1 E Q, ii.) A= ( x I xP c P}. 13 PROOF . Let (A,P) be an H pair . Then (A,P) is a - valuation pair so ii .) is c lear . Suppose Q is c losed under multiplication and P ~ Q. Let x E Q'-...P. If x f A, t hen for some y E P, xy E A"-.P,since (A,P) is a valuation pair and (xy)n = 1 + z for some integer n > O, z € P, - sinc e A / P is a locall y finite field . But then xy E Q, (xy)n E Q, z E Q, so (xy)n z = 1 E Q. If x EA then x n = 1 + z for s.ome n > 0, z E P a- nd x n - z = 1 E Q. Conversely, suppose (A,P) satisfies i.) and ii .). If (B,cr) ET and (B,cr) > (A,P), then a= P by i.) and then B = A by ii.),so (A,P) is a valuation pair. Assume (A,P) satisfies i .) and ii .) and let p:A ~A/P be the natural map. Then if cr is a non-zero subset of A/P (e.g., an ideal of A/ P) c losed under· and-, then 1 1E p~ (cr) and 1 E cr. Hence A/Pis a field. Als o 1 e Z · p(l) • p for all prime integers p with p(l) · pf 0, so Z · p(l) = Zp = z/1p) for some prime integer p. Also, if x e A/P, x f 0, then 1 E xzp[x] so ~ is algebraic over Zp , hence is in the finite field Zp [x] · of A/ P. This gives case n = 1 of the inductive hypothesis: 11If Eis a finite subset of A/ P and IE!= n, then Eis contained in a finite subfield of A/P. 11 Assume the hypothesis true for n and let IE!= n + 1 , a e E. Then jE "-,_ (a}I = n so there is a finite subfield F of A/P with E"-.(a} c F. If a= O we are done, otherwis e 1 E aF[a] so a is algeb raic over F, 14 hence F[a] is a finite subfield of A/P containing E. Thus A/ Pis locally finite. COROLLARY 1 . 13. If Sis a subset of R closed under - and·, and 1 f S, then there is a H pair (A,P) of R with Sc P. If B = (x ER I XS c S} and B/S is a locally finite field, then (A,P) ~ (B,S). PROOF . (cr c R I Sc cr, cr - a c cr, cr · cr c cr, 1 f cr} is inductively partially ordered by c, thus by Zorns lemma c ontains a maximal element P. Then if A= (x I xP c P}, (A, P) satisfies i.) and ii.) of 1 . 12 . By the maximality of P, Pis a maximal ideal of A, hence a prime ideal so (A,P) € T. Thus (A,P) is an H pair of R. If B/S is a locally finite field, x € B"'-S, then x n = 1 + S for some integer n > 0, some s e S, thus xn E A""-.P. But then by 1. 7, x € A""-.P. Thus B"'-S c A"-.P so B C A and S = B f\ P. PROPOSITION 1.14. Let Ebe a finite subset of R, _cr a subset of R with cr - cr c cr, cHJ c cr. If cr and the multiplicative subset generated by E have void intersection, and Ecr c cr, then there is an H pair (A,P) of R with cr C P and E C A- ....____p. PROOF. Consider the finitely generated subring S = Z • l[ E] of R. µ = cr I\ S is an ideal of S which has void intersec tion with the multiplicative subset generated 15 by the finite subset E of S. Thus by the integer version of the Nullstellensatz (see [l], pp . 67,68) there is a maximal ideal 6 of S with E (\ 6 = and µ c 6. Also by the Nullstellensatz , s/6 is a locally finite field. Now 6 + cr is closed under· and - (since Scr c cr), and 1 f 6 + cr (for 1 = p + a gives a = 1 - p € µ = cr n S and 1 = p + a E 6). Thus by 1 .3, there is an H pair - (A,P) of R with 6 + cr C P. Then 5 c P so again by 1.3, S '---,_6 c A......__,__P since S/6 is a locally finite field. But then E C 3--....__5 C A......____P. COROLLARY 1. 15 . If N1 = fl (P I (A, P) is a valuation pair of R}, N2 = ncp I (A,P) is an H pair of R}, N = (x €RI xn = 0 for some integer n}, then N = N1 = N2 • PROOF. N c N1 by 1 .7 and N1 c N2 since the set being intersected to obtain N1 contains that being intersected to obtain N2 • If x f N, then with E = (x} and cr = (0) in 1.14, we have x E A......__,__P for some H pair (A,P). That is, x f N2 . Thus since N c N2 , we have N = N2 • COROLLARY 1.16. If (B,Q) is a valuation pair of R, then Q = n (P I (A, P) is an H pair of R, Q C p}. Further, if E is a finite sub set of R with En Q = ¢,, then there is an H pair (A,P) of R with E nB c A"-..P and (E"-....B)nA = <1>. 16 PROOF . It s uffices to prove the second statement. Let Ebe a finite subset of R with EnQ = . Let E1 = En B, E2 = E'-.__ B. For x € E2 , chose qx € Q with xqx € B""- Q, and let E 12 = (xqx I x € E2 }. Applying 1 . 14 to E1 U E' 2 and Q, there is an H pair (A,P) of R with Q c P and E 11 U E 2 c A'-.....P. But then if X € E2 , xqx €. A'-..._ P, so x f A. That is El c A'-. P and E2 n A= <1>. COROLLARY 1.17. If A= R for all H pairs (A,P) of R~ then A= R for al l valuation pairs (A,P) of R. PROOF . If (A,P) is a valuation pair of R, then by 1.16 and hypothesis, Pis the intersection of (maximal) ideals of R,. hence is an ideal of R. That is, --- A= (x I xP c P} = R. Section N DEFINITION 1.18. Let A be a subring of a ring R. If p:A -:-+Sis a homomorphism we call pa partial homo- morphism on R.If, whenever Bis a subring of R, Ac B, T:B -Ta homomorphism, µ:(image p) -Ta homomorphism and TIA=µ o pone also has B = A, then we call p maximal . One - can show using 1.10 that if A is a subring of R, Pan ideal of A, then (A,P) is a valuation pair of R if and only if, the natural map A -;+A/Pis a ~aximal partial 17 homomorphism of R into a domain . A place on a field is a maximal partial homomorphism into a field . However, if (A,P) is a valua t ion pair of a field F, A"-....._P, then xx- 1 XE E A"-....._P gi. ves -1 , X E A,P by 1 . 17 , so A/ Pis a field . That is, a pair (A,P) of a field Fis a valuation pair if and only if the natural map A-A/P is a place . Thus at f'irst glance one might expect "maximal partial homomorphism into a domain" to generalize "place". This generalization is unsatisfying since such maps do not compose (see 1 . 20) as do places on a field . The generalized places of [3] do compose and satisfy the hypothesis of 1 .20. DEFINITION 1.19 . A valuation pair (A, P) of R is called a prime pair if A/ Pis a field . PROPOSITION 1 . 20. Let p be a partial homomorphism from R to S with dom p +R . If the composite part~al homomorphism rfl~ -A/Pis maximal for all H pairs (A,P) of S, then (d om p, ✓ker p) is a prime pair of R. Conversely , if (dom p, ✓ker p ) is a prime pair of R the composite is maximal for all valuation pairs (A,P) of S. p PROOF . Suppose R -s -A/P is maximal whenever (A, P) is an H pair of S. Let B = dom p, cr = ker p. CLAD'1 1. Every H pai r (A 1 ,P 1 ) of B with ..jcr c P 1 18 is a valuation pair of R. SUBPROOF. Let (A 1 ,P 1 ) be an H pair of B, v1""5 C P'. By 1.13 there is an H pair (A,P) of S with (A,P) ~ (p(A'), p(P 1 )). Then since R .f:.s -+A/Pis maximal, (A',P 1 ) is a maximal pair of R. CLAlli 2. If x f B, then xy EB for some y € B. Also of xy € B, then y E fa. SUBPROOF. Let (At,P 1 ) by any H pair of B with ~ c P 1 • Then since (A 1 ,P 1 ) is a valuation pair of R, there is a y E pt with xy E At " P' c B. If y f✓cr, then ( xy ,y}n ~ = <1>,so by 1. 14 ,the re is an H pair (A 11 ,P 11 ) of B with (xy,y) c A11 "'-.P 11 and v1cr c P 11 • But since x f A11 , this cannot happep by 1 .7. CLAIM 3. ~ is a maximal ideal of B. SUBPROOF . Suppose 6 is a maximal ideal of B with fa c 6 and 6"-fa f ¢, say y E 6'-Jcr. Let x E R"-._B. Then xy f B by Claim 2. Let {A',Pt) be an H pair of B with 6 c P 1 • Then xy f pt so z{xy) E At "'-._ pt for some z E pt. But {zx)y EA'~ B implies zx EB by Claim 2, and then {zx)y E 6 c pt, a contradiction . Thus (BJcr) is a valuation pair of Rand B/fa is a field. Now suppose (BJcr) is a valuation pair of Rand 19 ·B/Jcr is a field. Let (A ,P) be a valuation pair of S. Since p(.,/cr) is il, p(.,/cr) C P. Let x € p(B)'-A. Since . B/.,/cr is a field, there is an x' E p(B), y E p(Ja) with xx' ,... l +ye A', P. Thus x' e Pf\p(B); i.e ., (p(B)('\ A, p(B)n P) is a valua t ion pair of p(B). Thus (p-1 (A), p - 1 (P)) is a valuation pair of B, for if (A',P') ~ (p-1 (A), p - 1 (B)), then (p(A'), p(P')) > (Anp(B), Pnp(B)) . Also Jcr Cp-1 (P) . CLAIM. Every valuation pair (A',P') of B with Jcr c Pis a valuation pair of R. SUBPROOF . Suppose x f A' . If x EB, then :l x' E B"-..../cr with xx 1 E 1 + ./cr; since x f A', x' E P'. If x f B, then 3 y E ./cr with xy E B"-...,fi. xy(xy) 1 = x(y(xy) 1 ) E 1 +./cr, and since x f B, y(xy) 1 E.fac P'. Thus (p- 1 (A), p- 1 (P)) is a valuation pair of R so the composite R f:+ S --+ A/P is maximal. Proposition 1.20 gives some insight into generalized places as defined in [3] and these provide motivation for occas ionally including special results for valuations corresponding to prime pairs. EXAMPLE 1.21. Not all valuation pairs (A,P) are prime pairs, even when one requires A+ R. PROOF . If A= R is allowed, one needs only produce 20 a ring R that has a non maximal prime ideal. For the second case, let R = Q[x] where Q is the rational numbers and xis an indeterminate. Let p be a prime integer, AP={~ (m,n) = 1 = (n,p) , or m = 0), A= ~[x] , P =A· p, a= P + .A.x . One can easily check that (A,f) is a valuation pair of Rand that a is a proper ideal of A with Pc a. Thus (A,P) is not a prime pair of R. 21 2. INDEPENDENCE AND EXTENSIONS Section I Throughout this section, Vis a fixed valuation on a fixed ring R. Let¢ be an order homomorphism of rv into a valuation semi-group r with ¢ ( e ) = e, $(0) = 0. Since ~ is a homomorphism, ¢(r~(O}) i s an ordered group (inherited order ), and since e f o, ¢ o Vis a valuation on R. With this notation we have: PROPOSITION 2.1. ¢- 1 ( e ) is an isolated subgr oup of r and P~ is a prime V-closed ideal of A, where V -v o V V DEFINITION 2.2. A subgroup Hof a valuation semi- group r is said to be isolated if Of Hand whenever a,f3,-y Er with a.::. f3.::. -y and a,-y E H then f3 € H. DEFINITION 2.3. An ideal a of Av is said to be V- x c losed if z Ea, y €Rand V(y).::. V( x ) implies y € a. PROOF. If a,f3,-y Er V' a < f3 < 'Y and $(a) = ¢(-y) = e, then e =¢(a).::. ~(f3).::. ¢(-y) = e since¢ is order preserving. Also e = ¢(e)~(-y) -= ¢(aa~ 1 )¢(-y) = ¢(a)¢(a-1-y) = ¢(a- 1-y) so a -1- y ¢ - 1( e-€ ) and"~' - l (e) is a group, hence an isolated subgroup of r v· 22 Let ~ E f' . If ( f3) < e, then~ < e so P~ c Pv. V 'I' o V - If~.::_ e , then (~) .::_ ¢(e) = e so A c A~ v · Thus V - 'I' o p¢ CA CA and since p¢ is a prime ideal of () V - V - ¢ o V 0 V A¢ v' p 0 V is a prime ideal of A . If X € 0 V p 0 v' y ER and V(y) .::_ V(x) then ¢ o V(y) .::_ ~ V(x) < e so y E P v That is P vis a V-closed ideal of Av. 0 0 The first step towards a converse is: PROPOSITION 2.4 . If His an isolated subgroup of rv, then there is an order homomorphism¢ of f'v onto a valuation semigroup r~ with 1- (e) = H. 'I' o V PROOF. Set (a) = aH for all a Er . Then V (r v'--,,(0)) = r v'----(o)/H is a group and Hf o = o • H. Suppose a<~ and aH f ~H. Then if h1 ,h2 EH, h 1a < h2~ , for otherwise h 1a::... h 2~ gives e - 1 - 1 ~ ~ a~ h 1 h 2 and ~- la EH, since e h 1h EH and His isolated. Thus 1 1 2 the order 11aH .::_ ~H ~ a .::_ ~ 11 is well defined on ¢ (r v). One can easily check now that (r v ) is a valuation semi- group with the usual coset multiplication and that is an order homomorphism onto. Set VH = 0 V and note that P = (x ER I VH(x) < e) VH = ( x E R IV ( x) H < H) = ( x E R I V ( x) < a , 'ti a E H) . If ~ E r V and ~ f H, then ~H < H or ~-lH < H, so ~ € V(P ) VH or ~-l E V(P ). That is H = (a Erv I V{x) < VH 23 -1 min ( a , a } tf x E P } . 1 VH PROPOSITION 2.5. H--+ P and VH O so :J x ' E R with V(x') = V( x )-1• Now V(y).:::. V(x) < e gives y E Pv - - and V ( yx 1 ) < V ( xx 1 ) = e gives yx' EA. Thus xyx' E cr ,i V - but xx ' f cr, soy E cr since cr is a prime ideal of A. V Thus cr is V-closed. Now suppose cr and 5 are V-closed ideals of Av. Suppose x e cr '--.5 and y E 5"--.,cr. Then V( x ) < V(y) g ives ' - x E 5 while V(y) ~ V( x ) gives ·y E cr;~ Thus 6 c cr or cr c 6. In particular the V-closed prime ideals are linearly ordered by c. PROPOSITION 2.7. The set of V-closed ideals of A V and the set of V-c losed prime ideals of Av are order complete with respec t to the order c. 25 . DEFINITION 2.9. If V' is a valuation with Av C Av' and cr c P •1 c P , we say V' dominates V and write V' > V. V - V - V We say V and V" are dependent if V' :::_ V and V' :::_ V" for ooma v' w·th v•(R) + (e,O), and independent. otherwise . PROPOSITION 2.10. If V' > V then P I is a V-closed - V . prime ideal of Av. If V' and V are dependent, then CJ V ' • PROOF. Let V' > v;.J Then P I c A c A I shows that V - V - V Pv' is a prime ideal of~- Then since CJ c P , c P, V - V - V Pv, is V-closed by 2.6. Since (JV is an ideal of R, CJ C p ,, CJ C CJ I by V- V V- V 1.3. crv" c Pv gives crv, c crv also by 1.3, so crv = crv,· Now if V' and V are dependent, say V 11 :::_ V, V" :::_ V', then CJ = a 11 = a , . V V · V PROPOSITION 2. 11. If V 1 > V, then there is an order homomorphism :r -r I with V' = o V. Also, there is V V a valuation (V 1 ,V) on A ,/P I such that if ~:A 1 -A / P , V V V V 1 V is the natural homomorphism , then the following diagram commutes. V (Av~Pv,)--- r v l~ (v 1 ,v) A I / p I V1/'V 26 Further ~A( V', V), p (VI, V)) = (Av/PV (, p V/ PV I); cr ( v , , v ). = 11 ( Pv ., ) ; and ( V< V) ( Av 1/Pv 1 ) = - l ( e ) U (O } • (V',V) is called the induced valuation. PROOF .. Using 2.10 and preceeding results,V' = V Pv' V(A "-....P) =· <1>-l(e). V V The remainder of the proposition is clear once i t is shovm that x E A ~ P , implies V(x) = V(y) for all V V y € x + P 1 • But if x € A i"- P 1 , p € P , then V V V V 1 V(x) > V(p),so V(x) = V( x + p - p) ~ max(V(x + p), V(- p )} ... - -- - < max(V(x ), V(p)} = V(x). PROPOSITION 2.12. Let V1 and V2 be distinct dependent valuations on R. Then there is a valuation Von R with V:::. v1 and V:::. v2 such that (v,v1 ) and (v,v2 ) are independent valuations on ~/Pv. PROOF. S~nce v1 and v2 are dependent, A = (Pv, I V' a valuation on R, V' :::_ v1 , V' :::. v2 ) is non- empty. Thus P = inf A is a Vi. closed ideal of A and V . Vi V > V., i = 1, 2. - J. Now suppose Vis a valuation on ~/Pv with V > (v,v1 ) and V:::. (v,v2 ). Let P = (x €~Ix+ Pv € ~). Since _· F;j: is a prime ideal of A /P, i = 1, 2, and Vi V 27 P- c(P / P ) n (P / p ), it follows that v _ vl v _ v2 v a = ·a · C p C p C p . n p and pis a prime ideal of v 1 v 2 - v ~ - v 1 v 2 A , and thus V. closed b y 2.6, i = 1, 2. Thus P = Pv' V. l l for s ome valuation V' on R with V 1 > V., i = 1, 2, by 2.6 - J. and 2.5 . But then V' € .ti so V' > V and P C P. v- Thus p = p and P- = V V P/Pv = Pv/Pv is an ideal (zero) - - of Av/Pv so v(A /P ) = (e,o}. That is (v,v1 ) and (v,v2 ) V V are independent . Section II Throughout this section, let v be a fixed valuation 0 on a ring Kand let R be an extension of K. We will consider the problem of "extending" v to R. 0 PROPOSITION 2.13. There is a valuation pair (A,P) of R with -A- vo = An K and P = P (l K. Further, if . vo (¾ ,Pv) is a prime (H) pair of K, then (A,P) 1 can be - 0 0 chosen as a prime (H) pair of R. PROOF. (A ,P ) e T(R) so there is a valuation - vo vo - pair (A,P) € T(R) with (A,P) ::_ (¾ ,Pv ). Then since - - - - 0 0 (A , P ) , the first statement follows. - vo vo 28 If (A ,P ) is an H pair of K, the existence of an - vo vo H pair (A,P) with (A,P) > (A ,P ) is given by 1.13. - - - vo vo Now suppose (A ,P ) is a prime pair of K. Let . - vo vo S = ((A,P)j(A,P) > (Av ,Pv) and (A,P) a valuation pair _ - - 0 0 of R}. For. (A1,P1), (A2,P2) € s, define a partial order - - ~ on s by (A1 ,P1 ) ~ (A2,P2) if A2 C A1 and P2 C P1• If A. is a chain in s, then A = u (A I (A,P) e A for some P} 'J& is . a ring, PA aa: U (P I (A,P) e ./& for some A} is a prime ideal of A~, and (A~ ,P&) ::_ (A , P ). Now there is a Jlli - J/."l:.. Jlli - vo vo valuation pair (A,P) of R with (A,P) ::_ (AA,P.A). Then - - - (A,P) is in S and (A,P) is • an upper bound for A. .That is, ~ is an inductive partial order on S, hence S has maximal elements by Zorns lemma. Let (A,P) be maximal ins. Let 6 be a maximal ideal of A with P c 6. Then P c 6 nA , and since (A ,P ) vo - . vo - vo vo is a prime pair, P = 5 (\ A • Thus if (A 1 , P') ·is ?1- v O v0 valuation pair of R with (A 1 ,P') ~ (A,5), then (A 1 ,P') e S - - and (A' ,P') ~ (A,P). Since (A,P) is maximal in S, A= A', pt= Pc 5 c P'. That is, Pis a maximal ideal of A. shows that a maximal partial homomorphism of K into a domain (field, locally finite field) can always be 29 extended to a maximal partial homomorphism of R into a domain (field, locally finite field) . This is a classical result on extension of places. Due to the trivial ideal structure of a field, thi s also is an extension theorem for valuations on fields, as will be seen from the following interesting, but misleading result . PROPOSrrroN 2.14 . If v 1 is _a valuation on R with (A ,P ) > (A ,P ), then there is an order isomorphism - vl vl - - Vo Vo cl> of (f' ""'- ( O}) into F' such that o v (x) = v (x) for - vo vl 0 1 all XE K with Vo( x ) + o. PROOF. Let z EK, v (x ) f O. Using the standard 0 representation of l[" and r , it will suffice to show that vo vl V0 (x) = V1 (x)nK, for then c!>(V0 (x)) = V1 (x) is as advertised. - - 1 - - Let x I E v (x )- , y E v (x). Then x 1y E A "'-._P 0 0 vo vo = Vo(l) c Av "-.Pv = vl(l), so Vl(x')-1 = vl(y) . . That is - 1 1 - - - v 1 (y) = v 1 (v0 (x)) = v 1 (x) :3 v 0 (x). If z e v 1 (x)() K, then zx 1 € Vl ( 1) n K = - ( A "-. P - ) n K = A -.......___ P - _--vl vl Vo Vo = v 0 (1),so v 0 (z) = v (x 1 0 )-l = v 0 (x). Thus V1 (x)n K = v 0 (x). The above result is misleading since in general there are many x E cr with v1 (x) f O. VO 30 DEFINITION 2. 15 . Let R be an extension of K, V0 a valuation of K. A valuation v1 on R is called an extension of v to R if there is an order is omorphism 0 ¢ of y into r s u c h ·that¢ o v0 (x) = v ( x ) for all Vo vl 1 X € K. By the proof of 2 . 14 , an immediate result is . iii. ) ⇒ i. ) of the following . PROPOSITION 2. 16 . Let R be an extension of K, V0 a valuation on K, v1 a valuation on R. Then the following are equivalent . i.) v is an extension of v0 to R. 1 ii.) (A , P ) > (A , P ) and V 1 IK is a - v l v l - - vo vo valuation on K. ~ii.) PROOF . If v1 is an extension of vO to R, then v1(K) = ¢ o v0;K) is a valuation semi- group contained in rv , so v1 IK is a valuation on K. If x € K, then 1 ,.. v1 (x) ~ ·e iff ¢ a v0 (x) ~ e iff v0 (x) < e so A n K = A . --y 1 V 0 Also V l ( x ) < e iff ¢ o v0 (x) < e iff v0 (x) < e so Pv n - -K = P • That is (A ,P ) > (A ,P ), which gives 1 VO ~ vl vl - - Vo Vo i.) ⇒ ii. ). 31 is a valuation on K, then (A I , vl K = av by 1.6 . But cr = 0 vl K ii.)~· iii.). THEOREM 2.17 . (Extension Theorem ) v0 has extensions to R iff K n Rcr = ; • Further, if v0 has extensions vo vo to Rand (~ ,Pv) is a prime (H) pair of K, then v - 0 0 0 has an extension v1 such that (Av ,Pv) is a prime (H) 1 1 pair of R. PROOF . If v0 has an extension v1 to R, then cr C cr by 2.16 so Kn Rcr C Kn Rcr = K n C5 = a • vo - vl vo - vl vl vo Conversely, suppose Kn Rcr = cr • Then vo vo 5 = P + Ra is an ideal of B = A + Rcr with A = B II K vo vo vo vo . vo One can check that A /P I"- B /5 , so 5 vo vo - is a prime ideal of Band (B,5) > (A ,P ). Now if - - - V 0 V 0 (A ,P ) is any valuation pair of R with (A ,P ) > (B,5), - vl vl - vl vl , - Rcr c cr by 1. 3. That is a c cr , so v1 is an extension Vo - vl Vo - vl 32 of v0 by 2.16. Finally, if (A ,P ) is a prime (H) pair of K, then - vo vo (B,5) is a prime (H) pair of B, and since R is an extension - bi' B, (above) may be chosen as a prime (H) pair of R by 2.11. PROPOSITION 2.18. Let R be an extension of K, and suppose R is integral over K. Then every valuation on K has extensions to R. In particular , if v0 is a valuation on K, v1 a valuation on R with (A ,P ) > (A ,P ), - --y 1 V 1 - - VO V 0 t hen v1 extends v • 0 PROOF. It suffices to prove the last statement, and for this proof we are indebted to D. K. Harrison. By 2.16 and 2.17 we need only s~ow implies Rcr c A • Let a E vo - vl X € R. Since R is integral over K, there are ai € ~, n n-1 . n> with Z a.x J.. 0 X + = 0. Then a n . 0 = i=O 1. for i = o, 1, , n-1, that is, ax is integral over Since A is integrally closed , ax E ~. vl 1 33 Section III In this section, we assume R is an extension of K. The results obtained will be needed in Chapter 3. PROPOSITION 2. 19 . Let v 10 , v0 be valuations on K wi:th V 1 O ::::_ VO. Then 1~) v has extensions to Riff v0 r has 0 extensions to R, ii .) If v1 is an extension of v0 to R, then the set of extensions V' of v 1 0 to R with V' ::_ v1 is non- empty, linearly ordered and has a smallest element. If ·· (r / ll:" )"'- ( O} is torsion , then there is a unique - vl vo extension V' of v 1 0 to R with V'::. v1 • PROOF. Rcr n K = crv . Thus i.) follows by 2.15. VO 0 Let v , = o v , an order homomorphism off' 0 0 · V 0 onto r . Then 1<1>- (e) is an isolated subgroup of r Vor VO and H = (- y E r f 3. a, t3 E - l ( e ) with a ::. -y ::. t3} is an vl isolated subgroup of ll:" • If e is the natural map vl f' v -► ll:" v / H, then V1 ' = e o v1 1 1 is a valuation on R with Let V' = 1 0 o v1 be an extension of v0 r . Then . e ' - l ( e) n r = -l ( e ) so H C e '-l ( e) . That is , - VO V' ~ v 11 • The linear order property now follows from 2.6 and 2. 9. Now suppose (r /r )"-.( O} is torsion and e '(a ) = e . - vl Vo Then there is an integer n > 0 with a n € r , so VO a n E - l( e). If a~ e , th en a n ~a~ e so a€ H, while i f e > a, then (an)- l ~ a-1 > . e so a-l € H. Thus e , - 1 (e) = H and. V 1 = V 1 '. PROPOSITION 2.20. Let v , v 1 1 be valuations on R 1 with v1 r ~ v1 . If v1 I K is a valuation on K, then so is V 1 t I K and V 1 , I K ::. V 1 I K. PROOF. Let V 1 ' = o V 1 , ,;,-,rhere is an order homo- morphism of r v onto F' , • Then V1 (K) a valuation semi-1 vl group gives v1 '(K) = o v1 (K) a valuation semi-group. Since V1 t I K = o V1 K' V1 ' I K ::_ V1 I K. PROPOSITION 2 . 21 . Let v , v 1 0 0 be valuations on K with v0 ' ~ v 1 0 and v1 , v1 be corresponding extensions to R with v 1 1 ::_ v1 • Then the induced valuation (v 11 ,v1 ) is an - extension of t he induced valuat ion (v0 r,v0 ). PROOF. A.r / P = A.r, n K/ P 'n K l'v O' VO' 1 v 1 35 meaningful . Using the fact that v1 r extends v0 r and 2.11 one sees that the solid part of the diagram -i- ( (r v /Hp )"-.(Hp }) _ o v v 0 0 I I I -!t I ((.A vl,/Pvl ,)''-,-JPvl,})} -i- .(.( r vl/Hp )"-.(Hp }} vl' vl, commutes. This induces the dotted part. The proposition is that the zeros of the inside parts can be included and t he diagram will remain commutative. This is clear. PROPOSITION 2.22. Let Vo be a valuation on K, vl, v2 dependent extensions of v O to R. Then there is a valuation V' of R with V' ~ v 1 and V' ~ v2 such that the induced valuations (v,v1 ) and (V',V2 ) are independent extensions of (V' I K'VO). PROOF . There is a V' ~ Vi, i = 1, 2 with (V' ,V1 ) and (V 1 ,v2 ) independent by 2.12. V' I K is a val uation on K and V' I K~ v O by 2 . 20 and (V' ,Vi) extends (V' I K,vO), i = 1, 2 by 2.21. 3. T"tlE INVERSE PROPERTY, APPROXIMATION THEOREMS Section I A key fact about fields that is indispensible in proving theorems about valuations is that the set of all valuations on a field satisfy: DEFINITION 3. 1. We say that a set A of valuations on a ring R has the inverse property if for every x in R there is an x ' in R such that V(xxt) = e whenever Vis in A and V(x) f O. A is said to have the strong inverse property if for every x in R there is an x ' in R with V(xx 1 - 1) < e whenever Vis in A and V(x) f O. Note that (V} has the strong inverse property iff (Ay,Pv) is a prime pair of R. PROPOSITION 3.2. Let A be a set of valuations on R which has the inverse property , A t a set of valuations on R such that for every vr in A' there is a Vin A with V' > V. Then ff;. U A' has the inverse property. In particular, A 1 has the inverse property. PROOF. Let x , xt ER with V(xx') = e whenever VE A with V(x) + O. Let V' E 1!i t and suppose V' -> V , V E A .. and V'(x) f O. Then V(x) +Oby 2.10 so V(xx') = e . - Then xx t E A ~ P c A 0 P I so V 1 ( xx t) = e. V V V V 37 PROPOSITION 3. 3. Let A be a set of valuations on R with the i nverse property and V' a va luation on R s uch that V' > V for all Vin A . Then JA* = ((V',V) IV e ~ } has the inverse property. PROOF. Let p be the natural map A , - A ,/P , . V V V For x E Av '' let V(xx') = e whenever V E .f!l , V( x ) + 0 • . Since (V',Vi(p(x) ) = v(x) if XE A ,"-.P ,; - V V (V 1 ,V)(p(x)) = 0 if x e P , ; we have (V 1 ,V)(p( xx ')) = e V if (V' ,V)(p(x)) f 0 . Thus it remains OLlly to show that x e Av , if x E Av 0 Pv , . Since xx r e A ~ P c A ~ P , , V V V V this follows by 1. 7. In general the set of all valuations on a r ing does not satisfy the inverse condition . In orde r t o discover some sets which do , some preliminary result s are needed . PROPOSITION 3.4. Let V be a val uation on a ring R, a,b ER with V(a ) f V(b) . Then V( a + b) = max(V(a),V(b )}. PROOF. Without loss of generality , we may assume V(a) > V(b). Then V(a) = V(a' + b - b ):: max(V(a+b ), V(b)} - . :: max(V(a),V(b)} = V(a), so max(V (a+b ),V(b) } = V( a+b ) = V(a ). COROLLARY 3.5 . Let V be a valuat i on on a r ing R, n a. ER, i = 1, 2, m. If V( Z a .)< max V( a .), then l i=l l l n PROOF. Then since V( Z a i ) i=l n n = v( z a. + a.) -.. max[V( Z a.),V(a.)}, i=l i J - J·.--1 l J ifj if j n n v( z a.)= V(a.) by 3.4. But V( Z a.)< max V(a.), so - i= l l - J i=l J. - ifj - J. if j ifj ~!~ V(ai) ~ v(aj), that is V(ak) = max V(a.) . .l . l. l TJ - - lTJ - some k + j. COROLLARY 3.6. Let V be a valuation on a ring R, a. E R, i = 1, 2, ···, n, n+l, ···, k, with V(ai) = O l k n - for n < i < k. Then v( _z ai) = V( _z ai). - i=l - i=l k n k PROOF. V( Z a.) = V( Z a. + Z a.) < - i=l l i=l l i=n+l 1 n k n max [V ( Z a . ) , V ( Z a. ) } = V(Za.). The last equality - l. = 1 l. - J. .=n+ 1 l i=l 1 k holds since V( Z a.)= O. Equality now follows from 3.2. - i=n+l l Section II For the remainder of this chapter, R is assumed to be an extension of a ring Kand v0 a valuation on K which has extensions to R. If Va is any extension of v0 to R, we will consider I" as a sub-semi- group of f' v • vo a 39 PROPOSITION 3.7 . Let A be a set of valuations on R extending Vo , 6 an ideal of R contained in n{crv I V € A} , such that 6 n K = a • If x € R has x + 6 algebraic over . VO K/a , then there is an x ' ER with V(xx') = e for a ll VO V E 11 with V ( x) f O. Tf (A ,P ) is a prime pair of K, - vo vo then x' may·be chosen so that V(x~ 1 - 1) < e. PROOF. Note that V(t) = 0 f or all t ~ 6, VE A. If x + 6 is algebraic over K/crY ,o , then there are a. EK, I J.. f n i t E 6 with ar 6 and ~ a.x = t , (v(an) f o). Let J.. = 0 J. s = min(i I V(ai) f o}. Then for V EA, 0 = V(t) n . Thus if V(x) f O, then V( ~ a.x1 -s) = o • J. - 1.=S n . n . = V( ~ a .xi- s +as)< max(V( ~ a.xi-s),V(a )), so by - i=s+l J. i=s+l J... . s · n . n • 3.4, V( ~ a.x1 -s) = V(as) = V( x ) V ( ~ a.x1 -s-l) • - i=s+l 1 - i=s+l 1 I Choose a' EK with Vo(a'as) = e, (v0(a'as + 1) < e if (A , P ) is a prime pair of K). Then with - vo vo n i - s-1 . x' = a' · ~ _a.x V(.xx') = e whenever V e A with i=s+l J... 40 V(x) f O. If (A ,P ) is a prime pair, V(x) f O, VE A, then - vo vo n . V(xx ' + a 1as) = V• (a 1 )V( . ~ ai. x 1 ) = 0 so by 3.6, V(xx ' - 1) - i=s COROLLARY 3.8. Let A be a set of valuations on R extending Vo~ 5 = n (av I V €A}, and suppose R/6 is algebraic over K /Kn 5. Then A has the inverse property; A has the strong inverse property if (A ,P ) is a prime - vo vo pair of K. PROOF. This is clear by 3.7. Note that K n 5 = cr • VO If V extends v , then there is a natural homomorphism 0 P :I"V - ~ ~lr' v"- ( 0}) /~lr VO"-._ ( 0 })»LJ ( 0}, namely p(a) = a(lrv"-._(O}). Rather than carry the zeros, we denote . - p(r ) by rv/r and p(x) by J o,ana that _ vo vo o . I" v/r v is torsion if every element is torsion. Note that 0 aF" is torsion iff an E r for some integer n > O. V VO With this notation, we have a companion proposition to 3.8. PROPOSITION 3.9. Let A be a set of valuations on R 41 extending vo,6 = n(crv I V € A}, and suppose R/6 is algebraic over K /K r'\ 6. Then r /r is torsion for all / J V VO PROOF . This is immediate from 3.10.· PROPOSITION 3. 10. Let V extend v0 , 6 be an ideal of R with CJ c 6 c cr . Let x € R wi th x + 6 algebraic VO · V -.J\f) .(' -1,, over K/ K n 6. Then .xr is torsion . VO PROOF. If V(x) = 0, there is nothing to show, so s uppose V( x ) =J: O. Then ~ a i € K, t € 6, ar f 6 with r i ~ a.x = t. Since V(arxr ) =J: o, we have O = v ·(t) 1.. == 0 l. = max(V(a.xi)} = V(a.xj) f O for some if j . l. J Assume i > j ~d let V( x ') = V(x)-1 , V(a') = V(ai)-1 • Then V(xi-j) = V(aixi)V(x' )jv(a') = ·v{ajxj)V(x' )jv{_a') = v( aJ. )v( a') € I." • VO I PROPOSITION 3 . 11 . Let V be an extension of v 0 , V 1 > V and v 0 ' = V' !K. If r / r is torsion, then so is I" ,/r , V Vo V Vo and f' ( V r, V )/r (VO t , VO)• PROOF. Let :r v __,, r v I be the homomorphism such that V' = 0 <1> v. Then v , = <1> 0 0 v 0 , r(v',v) = {ker <1>)U(o) 42 and f'(v f V ) = ~ker )/)f' V } U(o} = r(v' v)nr·v • 0 1 0 - 0 , 0 If (a) Er 1 , then an E r for some n > O so V VO (an) = (a)n € r so r /r is torsion. If - vo, VO' VO' a € f'(v' v) then an € ]["' V nr(v' v) = f'(v V ) for some , 0 ' 0'' 0 n > o so r ( , )/r (v ) is torsion~ V ,v 0''V0 A trivial but useful remark is REMARK 3.12. If Vis an extension of vO and rv/rv is torsion, t hen V(R) = (e,o} iff v0(K) = (e,o}. 0 REMARK 3.13. If R is integral over K, 6 any ideal of R, then R/6 is integral (hence algebraic) over K/K (J 6 . Section III In this section we assume R is an extension of K, v0 is a valuation on Kand A is a set of extensions · of .V0 to K with the inverse property and such that r ft is V VO tors.ion for each V € A. In some of the results we also require Pv g; Pv' if V, V' EA and V +V '. The following proposition indicates the effect of this last restriction. PROPOSITION 3.14. Let V1 and V2 be distinct elements of A with P c P • Then P is an ideal of Kand R is vl - v2 vo not integral over K. 43 PROOF • . lf P is an ideal of K then P and P VO Vl v2 are ideals of R by 3.12. Then A = A = R, and if R were vl v2 integral over K we would also have P = P (see [5], vl V2 - page 259), contradict'ing P and P distinct. vl v2 It remains only to show that if P is not an ideal VO If P Vo is not an ideal of K, then P and Pv are vl 2 not ideals of R, so by 1.6, A f A • vl v2 CASE 1. Av'-._ Av f ~. Let y E Av"' A • Then 1 2 1 --v2 ' . v,1 (y) ~ e < v 2 (y). Since r /r is torsion, there is V. VO ' J an integer n > o, and a e K with v 2 (yn) = v 0 (a). Then v (y) = v (yn+la 1 ) > e while v (yn+la') = V (yn+l)v· (a')< e, 2 2 1 1 1 . since v {a~) < e. Thus yn+la' ~ P '-.P • 0 vl v2 CASE 2. A "'-._A +$ . By Case 1, there is y E R --y2 VI with V l ( y) > e > V 2 ( y) • Then V l ( 1 + y) = V l ( y) > e while v;(l + y) = v;(1) = e so V~ ( (1 +y ) ') ~ e while V 2 ( ( 1 + y) ') = e. Thus ( 1 + y) ' € p "--- p • vl v2 44 PROPOSITION 3.15. Let vl, v2, ... Vn be distinct elements of A. with Pv . ~ Pv if if 1. Then there is an ]. 1 I x e R with v1 (x)?:. e and v1 (x) < e for if 1. Further, if P is not an ideal of Kone can require v (x) > e. VO 1 PROOF. Case 1: P an ideal of K. Then P is a vo vi prime ideal of R, i = 1, 2, .... n. n i = 2, 3, • • • , n and let x = 1r x .• l.. = 2 l. Case 2: P not an ideal of K. Proof by induction VO on n. n = 2, Choose y € P , P • . Then v (y) ~ e > v (y). V2 Vl 1 2 Since rv /rv is torsion and :r f ( 0, e}, there is an 2 0 • VO n > O and a € K'\.. '¾- with e > v2 (a) > v2 (yn). Then with 0 x = a'yn we have v1 (x ) ~ v1{a') > e while v2(aa') = e > v2 (x). Now assume 3.15 holds for r = n - 1, n > 2. For i = 2, 3, choose yi € R with v1(yi) > e arid Vj(yi) < e if j f 1 and j f i. If Vi(yi) ~ e, let xi= yi, otherwise let xi= (1 + y 11 ) yi. SUBPROOF. This is aut omatic i f xi= yi. Otherwise 45 - - V. ( 1 + y. ) = V. ( y. ) > e and V. ( ( 1 + y. ( 'y . ) = e if j f 1, i 1 l 1 - 1 l - 1 - l - - V . ( 1 + y. ) = V . ( 1) = e and V . (( 1 + y. ) 'y. ) = V . ( y. ) < e. J - 1 J - J 1 l J - l Thus we have v1 (x2x3) ~ e and Vi{x2x3) < e if if 1. - - Let z = x2x3• Again since r /r is torsion and Vi VO . r f (o,e), there is an n> o and an a e K,a with vo vo - . e > V. (a) > V. (zn) f'or all i =f= 1 and x = a 1 zn has l 1 - - v1 (x) > · e, Vi(x) < e for all if 1. PROPOSITION 3.16. Assume P is not an ideal of K VO and v1 , v2 , ···, Vn EA are pairwise independent. Then if a 1 E r "'- ( 0) , i = 2, 3, · · · n, there is an x E R vi 3 , •• • , n . PROOF. Since r /r is torsion for i = 2, • .. , n, Vi VO n. there are n. > O with a. i Er "-.(O). Let l , l VO n. a= min({e) U (ai i I i = 2, ···, n)}. It suffices to show there is an x E .R with v1{x) > e and Vi{x) < a, i = 2, 3, ···, n. Let H ={a€ I" jx € R with V {x) ~ e, Vi{x) VO 1 < min{a,a- 1) if 1). Then e EH by 3.15, and it is easily ( checked that His an isolated subgroup of rv. The 0 proposition will be established if H = r "--.(0), or VO 46 equivalently, that if v 1 0 is the valuation determined by H, then v 10 (K) = Te,o} = rv /H. 0 Since v r > v0 and r /r is torsion for each i, 0 - vi vo by 2.19 there is a unique V. ' > V which extends V 11 1. - i O ' I i = 1, 2, ···, n. Since the Vi are independent, either Vi'(R) = {e!O} for some i, in which case v 10 (K) = {e,o} by 3.12 and the proposition is es~ablished; or the v1 r are distinct. Assume the Vl .' are distin. ct. By 3.2 and 3.11, 3.15 applies to v1 •, v2 •, ···, Vn'· Thus there is an x e R with v 11 (x) > e and Vi'(x) < e, i = 2, 3, , n. There is an integer r > 0 and b in K with Vi' (xr) > Vi' (b) = v O' (b) < e for i ;i; 2, 3, • • • , n. That is Vi(xr) < v;(b) e. This is a contra- diction since then v0 (b) e H, v 1 0 (b) = e. Thus Vo'(K) = {e,O}. COROLLARY 3.17. (Approximation Theorem) Suppose Pv is not an ideal of K and v1 , v2 , • • ·, Vn e A are 0 pairwise indepe.n dent • Then if ai e r v ~ { o} , i = 1, 2, • • • , n, J. - then there is an x e R with Vi(x) = ai, i = 1, 2, • •• , n. PROOF. For each i, choose zi e R with v1 (zi) = a .• l.. Choose x1 e R with Vi(xi) > e, and for jf i, with 47 Vj(xi) < min(ajVj(zi)-1 ,e}, if Vj(zi) f O, with VJ.(x1 ) < e if v.(z.) = o. Lett.= x.(1 + x .)'. Then J l. l. 1 1 Vi(t1 ) = e and Vj(ti) = Vj(xi) if if j. Now Vi(tizi) = vi(zi) = ai, and if if j, Vj(t1 z1 ) o if vj(zi) = o, = v .(ti)v .(zi) = J J • vj(xi)Vj(zi) < aj, if vj(zi) f o. That is V.(t.z.) = max V.(tkzk) only if i = j, so J J.. J.. k J n vj(.~ t.z.) = v.(t .z.) = a., j = 1, 2, . . . n by 3.2 • i=l J.. J.. J J J J COROLLARY 3.18. (Strong Approximation Theorem) Suppose Ii. has the strong inverse property and v1 ,v2 ,···,Vn e A . are pairwise independent. If ai e R have Vi(a1 ) f 0, i = 1, 2, ··•, n, then there is an x e R with Vi(x) = vi(ai) > vi(x - ai), i = 1, 2, ···, n. PROOF. Case 1: P an ideal of K. Then the Pv. VO ' l. are maximal ideals of R so P 1 P if if j,and 3.15 vi vj applies. For each i,choose x. e R with v.(x.) = e, J.. l. l. Choose x 1 • e A "-. P with i vi vi while Vi(x1x 1iai - a1 ) = Vi(a1t 1 ) = O < v1 (a1 ) = Vi(x1x• 1ai) = e. · 48 n Let x = ~ x.x'.a., then V.(x - ai) i=l 1 1 1 1 Case 2: P not an ideal of K. Choose a'i so that VO V.(a.a'.) = e whenever VJ.(a .) f O. For each i, choose J J. J_ • 1 x . e R with V-i(x-i) > e; v.(x.) <- min(V.(a.)v.(a'.),e} 1 ~ ~ J J. J J J J. if V . (a. ) f O, V . ( x. ) < e if V . (a. ) = O. Chaos e y. e R J J. J J. J J_ J. with V .(y.) = V .(1 + x. )-l if V .(1 .t- x.) f O and so that J J. J J. J J. vi(yi(l + xi) -1) < e. Then y.(1 + x.) = 1 + t. where Vi(t.) < e; J. J. J. J. (xiyi -1)(1 +xi)= xiyi(l + xi) - 1 - x1 = •x1 t 1 - l; vi ( x. Yi -1 )v. ( 1 + x. ) < max V. ( x. y. ) , V. ( 1) < V. ( x. ) J. J. - J. - J. J. J. J. - J. J. = V. ( l + x ~ ) ; so V. ( x. y. -1) < e and vi ( x. yiai - a,f ) J. J. J. J. J. J. . ~ < v.(a.). J. J. -Also if i f j , V . ( y. ) = V . ( 1 + x. ) - l = V . ( 1) - l = e , . J J. J J. J so v.(x.y.ai) = v.(x.)v.(a.) < v.(a.). J J. J. J J. J J. J J n Now if x = ~ x.y.a. we have v.(x - a.) j=l J J J i _ i = V.((x .y .a . - a .) + ~ x.y.a.) < max{{v.(x 1 1 1 1 1 jfi J J J - J. 1 y.a. - ai)) J. J. U{vt(xjyjaj) I i + j}} < v 1(ai). 49 Section r:v DEFINITION 3.19. Let D be a domain, Ir its field of quotients and San extension of D. Then the ring of quotients SD'-..(O) (see [5]) is a vector space over Ir. Set [S;D] = dimr?,SD""-.(o)• [S;D] is called the rank of S over D. One can show using "common denominator" arguments, that if r ~ [S;D], there are a1 , a2 , •·•, ar ~ S such that n ~ d.a = O, die D implies di= . o, i = 1, 2, ···, r. If i=l 1 1 s > [S;D] and a1 , a2 , ···, as e S, there are die D, not s all zero, with ~ d.a. = O. In the first case we call 1. .= 1 1. 1. the a. "independent", and in the second, "dependent". 1. DEFINITION 3.20. Let R be an extension of K, v0 a valuation on K with extensions to R. Let A 0 =(VI V extends v0 to R), and for Ac J/10 let crA = n(cr I V e 11). Set nil\ = [R /crJI\ ; K/cr ] , and note 'that V ~ il VO .A c A I gives n.A ~ n11 ,. For V e 11 0 , set f v r [ A_ /P-0 v ; AV o /PV o ] . f is called vn A the relative degree of V (with respect to vO). Set ev = (rv : rv) (the index of the group rv '-. (O} in rv'--(O}). - 0 - 0 ev is called reduced ramification index of V (with respect 50 to v0) • Note that if nA < co then for each x e R/aA the set x, x2 , • • • , x h.11. is dependent over K/av • Thus R /a.A 0 is algebraic over K/cr ,so that A has the inverse property VO by 3. 8 and r /r ·i s torsion for each V e A by 3. 9. , . V VO PROPOSITION 3.21 . Let R be an extension of K with nA < co, where A = (Vl, v2, . . . , vn) c A • Suppose the 0 Vi are pairwise independent and if p is an ideal of K, VO n also assume p v. _fPv . for i + j. Then :Z e f < nA. l. . J i =l vi vi - PROOF. First s uppose P is not an ideal of K. For VO each i, i =l, 2, •··, n, choose y ,y , . .. 12 21 y l in R ni such that the cosets V.(yk.)r are non zero and distinct. l. l. VO Note that n. < e • Since r /r is torsion for each i, l. - Vi Vi VO there is t3 er vo with o < f:3 < Vt(Ykt) for all t, k, and an ai er v with a.VJ.(_y ri. ) < (:3 for all j +i and all r. · By 0 i By 3.16 there e R with v.(a.) = e, < a. l. l. l. if i +j • . V. (bk.) = V. (yk. ), so the cosets l. J. l. l. 51 Also if k f i, V J. _(b k. ) = V J. (_a . ) V J. (_ Y k . ) < a . V . ( Yk. ) - < f3 1 1 1 _ 1 J 1 < Vt(Yst) for all t, s. That is, since Vt(y8 t) = Vt(b t) 6 we have: (a) Vj~bki) < Vt~bst) for all s,t if it j. Let x1i·, x2i, ···, ~.i be ill Av with the xki + Pv i i i linearly independent over Av /Pv • Note that m. < f v . 0 0 1 - 1 As in the above argument, there is an a 1 € r v with 0 ai + O and aiVj~xr, ) < e if if j. Choose bi€ R with V. ( b . ) = e and V J. (_ b . ) < a. if i f j. 1 1 1 1 Then Vi(aki) = e so the aki + P are linearly independent over A /P • vo vo vo Also vj~aki) = vj~b1 )vj(xki) so (b) Vj(aki) ~ e iff if j, for all k. If P is an ideal of K, using 3.15 (note n. = 1 VO 1 for all i),one can chose aki' bki with the properties described above, including (a) and (b). The arguments are similar but simpler. The proof of the proposition will be complete if we 52 n can show that the Z n.m. elements ak.b J.1. + o& are linearly i. = l i J. i a independent over K/o . To show this, it suffices to show VO that if ak . . € K has Vt ( . Z ak .. ak. b .. ) = o, t = 1, 2, ••• , n, J.J k , J. , i Ji i J1 then v0 ~akji) = o for all k, i, j. Withou~ loss of generality, we can assume v0 (a111 ) = maxv0 (a .. k). We have v 1 (~(~kcrkjl¾:l )bjl + z.akjiakibji) = O, ijk J.J . J k, J il Consider the second te rm of (c). For if 1, . v 1 ~akji) ~ v 1 ~a111) by assumption; v1 ~aki) < e by (b); and V1 (bji) < V1 (bjl) for all j, by {a). In particular . - - then, unless v 1 (a111 ) = v 0 (a111 ) = o, since v 1 (b11 ) f o, . one has v 1 ~akjiakibji) < v1 ~a111 )v1 ~b11 ) for all k, j 1 when i f 1, so that Vl (kz .akjia kib ji) < Vl (a111 )V 1_(b11 ) · , J . i>l Thus if we show v 1 (z(zak.lak1 )b. 1 ) ~ V1 (alll)V (b- j -k J J 1 - 11 ) , it will follow that v 1 (a111 ) = 0 and the proposition will be established.' 53 If Claim Dis true, since max v1 (ak.1 ) e rv and k - J 0 the v 1(bJ.1 ) determine distinct cosets v1 (b. 1 )rv, we - J 0 have vl(~ak'lakl)b ·1) f vl(~aksl~l)vl(bsl) ifs f j. Then - k J J k - Vl~~~~akjlakl)bjl) = m~x Vl((~akjlakl)bjl) .::_ Vl(alll)Vl~bll) by 3.5 and Claim D. Thus all that remains is to establish D. Let ' 1 so assume v1 ~aljl) f o. Lett e K with v 1(t) = v1 ~aljl)- • Then v 1 ~taljl) = e and v 1(takjl) < e if k f 1, so v1 (Ztakjlak1 ) ~ e, since v 1 (ak1 ) = e for all k. k - .. Let p be the natural map A ~A /P • If vl vl vl V1 ~~takjlak1 ) < e then ~p~takjl)p~ak1 ) = O, but takjl € AvO for all k, j, ,p(taljl) f O and the p(ak1 ) were chosen linearly independent over p(~) which gives a - 1 - contradiction. Thus Vl (~takjla kl) = e = Vl ( t )(~akjla k1.), so - k - - k vl~~akjlakl) = vl~t)-1 = vl~aljl) = m~x vl~akjl). Let V be an extension of vO to Rand V' > V. By 2.20 V'IK is a valuation on K with V'IK :::_ vO• Let ev' and fv' be the reduced ramification index and the relative degree for (V', V). PROPOSITION 3.22. With the above notation, we have ev,e{v',v) = ev and f(v•,v) = fv. PRO OF. r v /r (v , , v) ~ r v , , r v / r (v , IK , v ) ~ r v , IK 0 O by 2.11, so .(rv, : rv'IK) = ev' = -~rv/r(v',v) : rv/r(v'I ,v )) - K 0 f • V PROPOSITION 3.23. Suppose R is an extension of K, A CA0 , nA <~and Pv f Pv' whenever V and V' are independent elements of A. Then if v1 , v2 , ·•·, Vn are distinct elements n of A, one has ~ e v. f v. -< n&. In particular A is a finite . l J. J. .v. i= set. PROOF. (Note that b y 3.14 the restriction Pv f Pv' applies only when~ = Pv .) By induction on n. Proposition VO 0 3.21 gives n = 1, so assume the proposition holds for n > 1. We distinguish three cases, the first which is also covered by 3. 21. 55 CASE 1. • • • Vn+l are independent. CASE 2. v1 and v2 are independent. CASE 3. Vi and Vj are dependent for all i, j. In case 2, assume v1 and Vn+l are dependent. By 2.12, there is a valuation v1 r on R with v 1 , ~ v1 and V1 '::.. Vn+l and (v1 •,v1 ), (v1 •,vn+l ) independent. 'By 2.19, . for i = 2, 3, • • • , r, there are unique V. ' > V. which l. - J. extend v1 •1K• Let Vi l ' vi2 , ···, Vis' be the distinct valuations thus obtained, v 1 , = Vi1• Now (is)::_ n, and by 3.2 and 3.11 the inductive hypothesis applies to (Vil' vi2 , •··,Vis}= A', so s .~ e f < nA , • j=l vij vij - \ Let sij = ( k I vi j ::_ vk} and let Aij = ((vij'vk) I k € sij}. Since v1 and v2 are independent each Aij has nor fewer elements and by 3.3 and 3.11, the inductive hypothesis applies to give Now crA .. J.J ~ fv _; crlt., = cr(V V • • • V } :> er/A so n/A, < n • Now iJ a l' 2' ' n A a - A S, using 3.12 and the above, n 4 ::_ n/A, > ~ e 4 = j=l vij s ·~- 1 ~ e e( J- . ke S ij v1. J. _v 1.. )f( J , vk _v 1. J. , vk ) n+l = ~ e f • . l v. v. i= 1 1 This completes case 2. For case 3, chose V' ::_ v1 and V'::. v2 such that (V', v1 ) and_ (v•,v2 ) are independent (2.12). Continue as in case 2, noting that Ail may haven+ 1 elements, but that two distinct ones are independent, so case 2 allows us to get the equation(*) and complete the argument. 57 4. GALOIS EXTENSIONS Section I DEFINITION 4.1. Let R be a ring, Ga finite group of automorphisms on Rand RG = (x ER I cr(x) = x for all cr E G} = K. We say R is Galois over K with group G if I either of the following conditions hold: n (1) There are x .,yJ.. ER such that ~ x.cr(y.) = 1 i=l 1 - J. 6crl' where 6crl = 1 if cr ~ 1 (the identity of G) and 6crl = O if cr E G, cr f 1. (2) For every ideal 6 of Rand cr e G, with - 6 f R, cr f 1, there is an x e R with x - cr(x) f 6. For the equivalence of the above two conditions, and for the equivalence of either to the "usual" definition of 11R Galois over K with group G11 , the interested reader is referred to [2], page 18. For the main results of this chapter, we will . need an assortment of specialized results~ [2] will be quoted freely as a source of proofs. \ LEMMA 4.2. If R is Galois over K with group G, then there is an a E R· with = 1. PROOF. See [2], page 21. 58 PROPOSITION 4.3. If R is Galois over K with group G, and 6 is a prime ideal of K, then R/R6 is Galois over K /6 ,A.. with group G~ G. PROOF. R is integral over K (see [2] or 4.12) so R6 - is an ideal of R with R6 (\ K = 6 ([5], page 257), thus we can identify·K/6 with a subring of R/R6 . For cr e G, cr(R6) = cr(R)cr(6) = R6, so setting cr(x + Ro) = cr(x) + R6, for all x E R, gives an automorphism of R /R6. The map G-+ (a fa E G) = G is clearly a group homomorphism, and by (2) of 4.1, if cr e G, cr +1 , there is an x ER - with cr(x) - x t R6, so cr + 1 and the map is one-one. Let p:R-+ R/R6 be the natural map. If x1 ,y1 e R n n satisfy (1) of 4.1, then ~ p(x.)cr(p(y.)) = p( ~ x.cr(y.)) - i=l - i - - i - i=l i - i /'- = 601 , so· R/R6 is Galois over (R/R6)G with group 'G. Now suppose x ER and cr(p(~)) = p(x) for all cr e"G. - Then for each cr E G there are t E R6 with x = cr(x) ·+ t . 0 0 Let a ER have 1 = ~ cr(a) as in 4.2. Then cr(a)x = creG . cr(ax) + cr(a)t ; x = ~ cr(a)x = ~ a(ax) + ~ cr(a)t ; 0 aEG - creG - crEG - 0 p(x) = p( ~ a(ax)). Since T( ~ cr(ax)) = ~ cr(ax) for all - crEG - - crEG - crEG 't" E G, ~ a(ax) E K and p(x) . e p(K) = K/6. That is crEG - 59 PROPOSITION 4.4. Let R be a ring, Ka subring which is a domain,~ be the ring of quotients of R with respect to the multiplicative set K"-(O}. Then if cr is an auto- morphism of R with cr(x) = x for all x € K, there is a unique extension. of a to an automorphism cr on¾· Further cr(x) = X for all X € ¾• PROOF. Clear. See [5] for definition and existence PROPOSITION 4 ·.5. If R is a Galois over K with group G and K is a domain, then¾ is Galois over¾ with group G = rcr I (J € G} '::::: G. PROOF. Clear using (1) of 4.1,and 4.4. 1 LEMMA 4.6. If R is Galois over K with group G and K is a field, then di~R = la·!. (Isl= number of elements in S.) PROOF. See [2], page 27. COROLLARY 4.7. If R is Galois over K with group G and 6 is a prime ideal of K, then [R/R6 ; K/6] = !GI. PROOF. Clear by 4.3, 4.5 and 4.6. LEMMA 4.8. If R is Galois over K with group G and R I is a domain, then G is the set of all automorphisms of R such that cr(x) = x for all x e K. 60 PROOF. See [2]. PROPOSITION 4. 9. If Risa domain, Ga finite group of automorphisms on R with K = RG, then (1) ~= RR /'- (2) ~ is Galois over¾ with group G ""- G. - (3) [R;K] = !GI (4) Every automorphism_ cr of R with cr{x) = X for all x E K,is an element of G. PROOF. Let G = { a I a E G}, where cr is as in 4.4. A. Then ~G = Kx, so~ is an integral extension of a field, and is a domain, thus~ is a field and RK = RR. (2) of 4.1. is then satisfied, so~ is Galois over KK with group G. Since ,a,= IGI, (3) follows from 4.6 and the definition of [ R;K]. If cr is an automorphism of R satisfying (4), then the extension a (as in 4.4) has cr(x) = X for all XE ¾ 1so a E G by 4.8. But then cr = crlK E G. 'PROPOSITION 4.10. If R is Galois over K with group G and His a subgroup of G, then \ (1) R is Galois over RH with group H. (2) If His normal in G, then RH is Galois over K with group G/H, where (crH)(x) = a(x) for ,all cr E G, X € RH. PROOF. See (2], page 22. 61 PROPOSITION 4.11. Suppose R is Galois over K with group G f 1, 6 is a prime ideal of R, be Rand {bx Z cr{x)) e 6 for all x e R. Then be 6. aeG - PROOF. There is an x e R, TE G with x - T(x) t 6 by (1) of 4.1. But bx - Z a( x ) ea; creG Z cr{-r{x)) ' = bT(x) Z a(~) € 6 gives O€G - - <1€G - b{x - -r(x)) e 6, so b e 6. This completes the preliminaries. Section II For the remainder of this chapter we will assume that G is a finite group of automorphisms on a ring R,with RG = {x e RI cr{x) = x for all a e G} = K. We let !GI= n. Let v0 be a fixed valuation on K. PROPOSITION 4.12. R is integral over K. PROOF. 1r (x - cr(a)) aeG - n-1 . = ~ + Z a(i)x1 • One computes that a(i) = Z 1r cr{a), i=O - S €A. creS - J. where A1 is the set of al l subset s of G containing n - i elements, and tqat a{i) € K for i = O, 1, 2, ·•·, n-1. Since fa(a) = O, a is integral over K. I 62 Thus v0 has extensions to R by 2.18. PROPOSITION 4.13. Let V be a fixed extension of v0 to R1and for CJ e G, x e R define Vcr(x) = V(CJ(x)). Then - - . V cr is a valuation on R extending v0 and (Vcr I cr E G} = (V' V1 is a valuation on R, (Ay,,Pv,) ::.(Av ,Pv )}. 0 0 Furthermore A== (x ER I V(cr(x)) ~ e,b'CJ e G} = n ~ 'lis the integral closure of A in R~ creG CJ vo (x ER I V(CJ(x)) < e !:;Jcr E G) = (\p =~; ' O'EG V CJ V 0 and (x e RI v(a(x)) = o,t:Ja e G} = t\a = ✓Ra " - - creG V a V o PROOF. V =V O a is a multiplicative homomorphism O' of R onto rv so it is a valuation. V (x) = V(cr(x)) = V(x) 0 = v0 (x) for x e K, so v extends v0 • 0 Since A is integrally closed, the integral closure V (J of~ in R is contained in A. However from the form of 0 fa in 4.12, if a e A, a(i) e A , i = o, 1, •••, n..:l, so VO that a is integral over A • VO It is clear that P C I\ P so that~ C (\ P • Vo CJEG Vo Vo aeG VCJ Conversely, b y the form of f in 4 .12, · if a € (l P , a · CJEG v CJ then a(i) E P , i = O, 1, ···, n-1, so that - V CJ n-1 an = - ~ a(i)ai E AP , and a€~. The argument that i=O - vo vo 63 na =~ is similar. creG vcr vo Now let (Av 1 ,Pv,) be a valuation pair of R with (A ,,P 1 ) > (A ,P ). By 2.18, V' extends v • - V V - - Vo Vo 0 Now by 3.15, if Pv' i Pv 1 Pv' for al l cr e G, there 0 is an x e R with V'(x) = e, Vcr(x) < e for all cr e G, contradicting v'l'-:·-::-7( = ("\ P C - P , • Thus vo creG vcr ~ P , c P (or P c P ,) for some cr e G. If P is not V - Vl'T V - V V . (J VO an ideal, this gives V' = Vcr by 3.14. If Pv is an ideal, 0 then so are Pv' and P , and since R is integral over K, vcr P, = P (see [5], page 259), so V' = V. V V(J _ (J COROLLARY 4 .11~. VO has a finite number g of extensions and for any two extensions V and V' of vO, ev = e , and V fv = fv,• .PROOF. Since G is finite, the number of extensions is also finite by 4.13. If V and V' are two extensions of vO, V' = Vcr for some cr in G. The map cr: F'v-+rv, given by o'(a) = vcr.(v-1 (a)) is an isomorphism with 'cr'(a) = a for - all a e r v , so ev = (r v : r v ) = ( -a (r v) : 'tt(r v ) ) O 0 - - - 0 = (r ' : r ) = e f • _ V VO V 64 = cr-1 (x) + a-1 (pv) = cr- 1 (x) + Pv,,is an isomorphism with cr(x + Pv) = X +P I for all x € A , so cr(A /P ) V Vo - Vo Vo = [cr(A /P ) = f ,. V V V Let e = (r : r ) and f = [ A /P · A /P ] , where - V Vo V V, Vo Vo Vis any extension o~ v • The le~ter e is traditional when 0 used in this way and we rely on the context to distinguish it from V(l). We can "count" the number g of extensioils of v0 • Let V be a fixed extension of v and set 0 G2 d~f{a e GI V = V ) = {cr e GI cr(Pv) = Pv). 0 For the second equality, note that V +V iff V{x) < e 0 and Va(x) ~ e for some x e R. For cr,T e G, v = VT iff 0 p = p iff cr-1 (P) = T-1 (P) iff cr-1T(P) = p iff V 0 VT V V V V -1 cr -r e Gz iff aGz = TG2 - That i~ g = (G: Gz). PROPOSITION 4.15. Let S be any subring of R with Kc Sc R, Van extension of v0 t o R. Then Vis is a valuation on S extending v0 and {V' IV' a valuation on S extending v0) = {V 10 8 I cr e G). PROOF. To sh~~-vj 8 is a valuation on S we need to show that if x e S, V{ x ) + 0 then t here i s aye S with V{y) = v(x)-1• Since r /r i s t or sion;-x ES, V(x) + o, V VO l ' \ I}.., there is an r > o with V( xr) er • If z e K with - VO V{a) = V(xr)-1 , then V(axr-l) = V(x)-l and axr-l e S. Let V' be an extension of v0 to S. Then R is integral overs, so V' has an extension V to R. Vis an extension of v to R, so V = V for some cr e G. But then 0 0 V' =Vis= Vais• NOTATION. For the remainder -of this chapter V will be a fixed extension of v to Rand G = (cr e GI V = V c) 0 2 as above. We will denote subgroups of .G by subscripts such as GB. We l~t ~ = (x €RI a(x) = x for all a€ GB} GB - = R , v B = v IK _, ~ = ~ / Pv , ko = ~ / Pv , • -~ B B O 0 k = 1/PV , eB = ~r V : r VB) and f B :::: [ k ; ~] • PROPOSITION 4.16. Vis the unique extension of v2 to R,and if GA is a subgroup of G such that Vis the unique extension of VA to R, then GA.::. G2 so that K2 a KA. PROOF.· By 4.13 and the d·efinition of G2 , {v• I V' extends v2 to R) = {V I a e G2} = {V}. In the same way, 0 if GA.::_ G, ,(v'IV' extends VA to R} = {V I cr \E GA}, and the 0 latter .set ,is (V} iff V = V for all a EGA iff GA.::_ G 0 2• Section III For the remainder of the chapter, the additional 66 assumption that R is Galois over K with group G will be made. ♦ e vz = f vz = 1. PROOF. Let v , v2, . . . vg be the extensions of v1 0 to K2. By 4. .15, each v1 is of the form (Vz)cr for some a€ G, so ev = ev, fv. = f fo-r each i. Thus i z J. Vz g ~ e f i=l vi vi ~ [Kz/Kzcrv The first inequality follows from 0 3.23 and the second holds since Let r = (G2 j. Since R is Galois over Kz with group Thus there are x1 , x 2' ••• , x r ER with the x. + Rcrv linearly independent i 0 over K2 /cr • ' Let e f g = h. The inequality in the first vz vz vz paragraph gives the existence of y1 , y2 , ··•, yh E K2 , such that they.+ K cr are linearly independent over J. 2 VO K/crv. Thus the hr elements x.y. + Rcr are linearly 0 J. J VO independent over K/cr , and hr < [ R/Rcrv ·, K /o-v ] = n . VO - 0 0 by 4.7, since R is Galois over K. 67 That is e f gr= e f vzn ~ n, thus e = f = 1. vz vz vz vz vz Since cr(~) = Av and a(Pv) = Pv for every a€ Gz, we have a natural map a-+ a of G2 into the group of auto- morphisms of k = .Ay/Pv. We hav_e a = T iff a(x) - x € Pv for all x € ~, so GT d!f(cr € G2 I a= f}= {cr € G2 cr(x) - x € Pv for all x € Av},is a normal sub- group of Gz: Note that v(x - cr( x )) < e whenever V(x) ~ e, - - - gives x - cr(x) € P whenever x €A, and this gives V V cr(x) E Pv whenever 'X E Pv, so GT = {cr € G I V(x - cr(x)) < e for al 1 X € A } • V Let -rr = 1 if the characteristic of k0 is zero, and let -rr be the characteristic of k0 other~ise. For D a domain, let~ be its field of quotients. PROPOSITION 4.18. With the notation above, set A G = {ci O' € Gz} ~ Gz /GT. Then (1) ~ is purely inseparable over ~, - (2) ~ is Galois over ~, · - (3) A nat GA. Autk kT, o· /\. ' (4) G = Autk k, 0 (5) e = 1, • VT (6) e = (r V; f' V ), T (7) f = IG z /GT 1-rr~ for some integer r. 68 PROOF. Let p :~ ~ Av/Pv = k be the natural map. • Note that for a E ~, cr € GT, that a - cr(a) € Pv, so that - p(a) = p(cr(a)). Lett= IGTj. Recall now (replacing G with GT) the -polynomial fa(x) of 4.12, and note that for a€~ that p(a(i)) = (1)p(a)t-i € krr, so that - - t - p(fa(x)) = (x - p(a)) • That is every element p(a) of k is either in. krr or has a purely inseparable minimal polynomial over~- But 0 = l~kby 4.9, so 0 is purely inseparable over~- Since ~ is purely inseparab l e over~, the restriction map Autk*tE ~Aut~ki is an isomorphism, thus the restriction 0 0 . ,. map Autk k ~Autk ~ is also an isomorphism. Let 0 0 ~ ~ G = Calk I cr € GJ. Let S = G kT. Then by 4. 9, ~ is . T Galois over S* with group ~G , so by 4.7 [kT;S] = [K.~T ;S] = I/G\. j. Now kz ~ s, so ~c. s*, so[~;~]= [krr;S~][s~:k2*] A . . = [kT;s][s;k2] = 1Gj[s;'k2] ·= jG2/GT![s;k2] = [kT;k2]. Now R is Ga~ois over Kz with group Gz and GT ( cr )(a) = (a, CJ). PROOF. Recall that V(x) = V{cr(x)) tJo E GT. Thus ~(a)a' e ~ and p(cr(a)a') f o, so {a,CJ) EA*. Also 70 p(x) = p(cr(x)) tix € A • · V Now suppose V(a) = V(b) = a, V(a') = V(b') = a-1 . Then: p(cr(a)a')p(aa')-l = p(cr(a)a')p(ba')p(ba')- 1p(aa')-l = p ( cr (a) a ' ) p ( cr ( b a ' ) ) p ( aa ' ) - l p ( b a r ) - l = p(cr(a)a'cr(b)cr(a'))p(aa')-1p(ba')-l = p(cr(b)a' )p(cr(aa'))p(aa' )- 1p(ba' )- 1 = p(cr(b)a')p(ba') - 1 ../ = p(cr(b)a')p(bb')p(bb')-1p(ba')-l = p(cr(b)a'bb')p(ba')-1p(bb')-l so (a,cr) is well defined. Now let V(a) = a, V(b) = '3. Then (at3,cr) = p(cr(ab)a'b')p(aba'b')-l = p(cr(a)a')p(cr(b)b')(p(aa')p(bb')}-l = p(cr(a)a')p(aa')-1p(cr(b)b')p(bb')-l = (a, cr)(t3, cr) which gives (1). (a,crT) = p(crT(a)a')p(aa')-1 ' = .o ( cr ( T( a) ) a ' ) p ( T( a ) ) a ' ) - l p ( T( a ) a ' ) p ( aa r ) - l = p(cr(b)b')p(bb')-1p(T(a)a')p(aa')-1 = (a,cr)(a,T), where b = -r(a), b' = a' in the third step. This gives (2). It is clear by (1) and (2) that~ and tare h0~omor- phisms: 71 Now cr e ker iff p(er(a)a') = p(aa 1 ) !Jae R'-..erv iff V( er (a) a' - aa' ) < e ti a e R'-.. er V iff V(er(a) - a)V(a ') < e t;Ja e R'-..cr V r PROPOSITION 4.20. If av er* for some integer VO r > O, then a e Ker 1/J. PROOF • . First suppose a er . Then in defining VO (a,cr), we can chose a,a' € K. Then (a,cr) = p(er(a)a 1 )p(aa•)-l = 1, since er(a) = a. Thus r* vr C Ker 1/J. If a e Ker 1/J for some r ::_ O, then 1/J(a) VO - has order vt , for some t. But the only element of Hom(GT,A*) of order a multiple of vis 1, so a e Ker 1/J. 1 The finite abelian group r* /r* may be expressed V VO as the sum of the v group rv and a group rv, with order e0 prime to v. The above proposition shows that a e Ker . cl> if or* er , so there are induced homomorphisms VO V GT /GV ! Hom(:![" v' ,A*) r v' !, Hom(GT/Gv,A*) since (a,er) = (~,T), whenever aKer 1/J = ~Ker 1/J, or erGv = ~Gv· PROPOSITION 4.21. GT/GV is abelian with order prime to v and Gv is av group. 72 PROOF. $ above is one-one, and since Hom(r~,, A*) is abelian and has order prime to~, the same holds for GT/GV. Now let cr € Gv- and suppose a has prime order q. Let H = (cri I i = 1, 2, ···, q}. Since R is Galois over RH with group H, by Proposition 4.11, either V(q) = O (and q =~)or there is an x e R with q V(qx - Z a(x)) +o . Let y = qx - i cri(x) and note that i=l i=l But if V(yy') = e, in the second case we have ,. q i Z y'cr (y) = O, i=l = qp(yy 1 ). Since p(y 1y) f ·O and A /p is a domain, V V p(q) = 0 and q = ~. PROPOSITION 4.22. tis one-one. PROOF. Suppose a € r* v and (a, a) = 1 for all cr .€ GT, i.e., that V(cr(a) - a)< V(a) for all a E GT,whenever V(a) = a. Let V(a) = a, y = TT cr(a). Then y E RGv = Kv crEG - V and V(y) = TT ~u u V(cr(a)) = a , where~ is the order of Gv· crEG V u Since a E Ker t, so is a~u. We wish to show that a~ € r . VO Since Gv <'.I GT, Kv is Galois over KT with group GT/Gv, Kv e'= _z0 and y E gives ZcrGV(y) cri(y) E KT, where e 10 is the i=l - 73 order of GT/Gv and criGV' i = 1 , 2, ···, e 10 , are the distinct elements of GT/GV. Now e'o is prime to v, so V(e 1 0 ) = vu 1. Since a E Ker t, so is a , hence V(cri(y)-y) O. COROLLARY 4.24. If A /p is of characteristic zero, vo vo then efg = I GI, Gv = 1 and :r~/:r! ~ GT. 0 ,. PROOF. efg = jGj by 4.23, since 1r = 1. GV is a 1r group, so Gv = 1. I" 1r is a 1r group so r = 1,and . 11" GT = GT/Gv ~ r 1r' = r;/r~ . 0 \ ,. 75 BIBLIOGRAPHY [l] N. Bourbaki, Algebre Commutative, Chapitre 5, 6, Hermann, Paris, 1964. [2] S. Chase, D. K. Harrison, and A. Rosenberg, Galois Theory and Cohomology of Ccmmutative Rings , Memoirs of Amer. Math. Soc., No . 52, 1965 . [3] D. K. Harrison, Finite and Infinite Primes in Rings and Fields, (in revision). [4] B. L. Van Der Warden, Modern Algebra, Vol.II, Ungar, New York, 1953. [5] O. Zariski and P. Samuel, Commutative Algebra, Vol. I, Van Nostrand, New York , 1960. [ 6] o. Zari.ski and P. Samuel, Commutative Algebra, Vol. II, Van Nostrand, New York, 1960 • • TYPED by Judith Read • ft PM OCT 30 '77 _..., PM •