, ‘ 3 1‘. ;_,4 "2*: . 3m;- ": c- ;s ‘ 4:1: “u "11"; £“‘ a. .3 . u,,._.§}n. ‘ ‘: . -.« “gha- ‘- _ 1%. E‘s?“ ,4 x V 332 L uJ'.’ . 5.73.. C a ’3 “5“ 3 1; r’ém" 12 6.1” - ‘ '1 ”I 115;“; w:- ' J5 1' 703$- 3 u ‘ ‘12?“ {5' ~- “fir £5675! , T V . m .. . ‘ , , . .— _. .4 w. rt. 3‘ . ~wm ”103' ' “Via“ ¢ ‘ . *3 .y‘" . "F . '. ‘5 .» ‘ErL’.’}‘-';~’"w 'a’fiSQ-Iiw , . , ..: , *3“? v ~ 4r ~ w - 1.. ' , ‘7 ~34 . n . “a , 1 may; a. 4.. .92“ ‘ '. a ’~" v . . .. . ‘ , N), ,A A .. . ~ , L . 3'. ‘ ‘ *Ek": ,-.;;“~£:s,",- 3;. ”$- .‘£!'.‘:\ 3. '4. “$9". :-1_ . - ‘ :9 This is to certify that the dissertation entitled Low Dimensional Formal Fibers in Characteristic p 0 presented by Peter Donald Shelburne has been accepted towards fulfillment of the requirements for dggggna] degreein mathematics. (jaw was Major professor Date 6/22/94 MS U is an Affirmative Action/Equal Opportunity Institution 0-12771 PLACE II RETURN BOXto romavothb chockomm your record. TO AVOID FINES return on or before date duo. DATE DUE DATE DUE DATE DUE usu In An Affirmative ActtotvKu-I Oppottuntty Institution WNDJ Low Dimensional Formal Fibers in Characteristic p > 0 By: Peter Donald Shelbume A DISSERTATION Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Department of Mathematics 1994 ABSTRACT Low Dimensional Formal Fibers in Characteristic p>0 By: Peter Donald Shelbur'ne The goal of this dissertation is to answer a question originally asked by Matsumma in [3] : is it possible for a local ring to have formal fibers with dimension 1 through the dimension of the ring minus 3? Rotthaus (in [6]) answered this question positively for rings of characteristic 0; in the following I again answer it in the affirmative for rings of characteristic p > 0. I also prove that the examples are excellent rings. The proofs rely on Weierstrass Preparation, differential algebra in characteristic p > 0, and the existence of hypertranscendental elements. To prove that the examples have A low dimensional formal fibers the crucial result was to show that trdegRsRs = no . For this, I discovered a new proof quite different from that of Rotthaus which takes advantage of certain Weierstrass automorphisms, under which R8 is particularly well behaved. To show the examples are excellent, it was enough to show that Q(Rs)—E—)Q(Rs) is a separable extension. I computed explicitly a p—basis of a general power series ring and then used the basis, along with the actual construction of R5, to prove this. Acknowledgments I would especially like to thank Christel Rotthaus for her guidance and advice during the last three years. Thanks are also due to Lewis and Clark College and, in particular, to Bob Owens, for giving me access to their computers, on which I typed this dissertation. TABLE OF CONTENTS Introduction ........................................................................................................... p.1 1. Definition of Rs and Amt ......................................................................... p.5 2. Properties of Power Series Rings and Weierstrass Preparation .................... p.7 3. The Elementary Properties of Rs ............................................................... p. 16 4. The Excellence of Rs and the Separability of Q(Rs) ——‘—+ Q(fis) ............. p.19 5. The Separability of Q(Rs) —£-> Q(Rs) .................................................... p.23 6. Derivations in Characteristic p and an Important Lemma ............................. p.32 7. 'I‘rdegR‘Rs = 09 ......................................................................................... p.50 8. Low Dimensional Formal Fibers .................................................................. p.57 9. Summary .................................................................................................... p.61 List of References ................................................................................................. p.62 Introduction In this paper we obtain the results of Rotthaus' 0n Rings with Low Dimensional Formal Fibers [6] for rings of positive characteristic p>O. We show that there exist suitably nice ("excellent") rings of characteristic p which have formal fibers of dimension 1 through the dimension of the ring minus three. Together with what was known previously, this means that excellent rings of any characteristic may have formal fibers of dimension 0 through the dimension of the ring minus one. Our proof relies heavily on "Weierstrass automorphisms", which motivated Rotthaus' original construction in characteristic 0. Any homomorphism of commutative rings f : R—)S induces a mapping af : spec S-)spec R which takes a prime ideal of S to its preimage under f, which will be a prime ideal of R. Given an ideal p6 spec R, one can examine its preimage under the map 3f i.e. the fiber of pa spec R under a‘f . It can easily be shown that this fiber can be identified with the spectrum of the ring S§k(p), where k(p) = Rp/ pRp (see [2], p. 47). In the special case where the above map f is the natural inclusion i : R -> R , where R is a local ring and R its completion with respect to its maximal ideal, the rings figkfli) are called the formal fibers of R. When R is a domain, the fiber of the prime ideal (O), i.e. figom), is called the generic formal fiber. One of the main goals of commutative algebra has been to understand under what conditions certain properties of rings will pass between the ring and its completion. Grothendieck formulated the definition of an excellent ring with this in mind: his excellent rings preserved "Serre's conditions", which could be used to express many interesting 1 2 properties. Formal fibers play a central role in Grothendieck's definition. In order for a ring to be excellent the ring must have geometrically regular formal fibers. This definition paved the way for formal fibers to become important objects of study for the commutative algebraist. More recently, formal fibers have been used to create a new class of examples of rings via birational extensions (see Rotthaus, Heinzer, and Sally [1]). Here maximal ideals in the generic formal fiber are an intrinsic part of the construction of certain Noetherian rings. The occurrence of formal fibers in their research and in Grothendieck's definition of excellence justify the further study of these rings and their properties. Yet with all that said, it is surprising how little is really known about formal fibers. It was not until 1987 that Matsumura first launched the study of the dimension of formal fibers, surely one of the most fundamental questions that can be asked about a ring. In his paper, 0n the Dimension of Formal Fibres of a Local Ring [3], he shows that the maximum value of the dimension of a ring's formal fibers (which he denotes by a(R) and shows is equal to dim R§Q(R) for a local noetherian domain R) can be either 0, the dimension of the ring minus one, or the dimension of the ring minus two. He then posed the question whether or not 0t(R) could ever be any of the intermediate values, one to the dimension of the ring minus three. In 1990, Christel Rotthaus was able to answer this question in the affirmative in her paper 0n Rings with Low Dimensional Formal Fibers [6]. She produced examples of excellent rings in characteristic 0 which had all the above intermediate values as the dimension of their generic formal fiber, Matsumura's a. She began by defining a ring Rs which is the basic building block of her examples. The construction of R5 was at least partly motivated by the so called Weierstrass automorphisms: because her ring Rs behaves nicely with respect to these automorphisms, she is easily able to prove that R, is Noetherian and even excellent. By adding power series variables to the ring Rs, she is able to obtain the desired values for a. 3 A Perhaps the centerpiece of Ronhaus' argument is the fact that trdegRsRs = co. In order to prove this she uses differential methods: she proves that a hypertranscendental element of k[[T]] (which is contained in Rs) will still be hypertranscendental over Rs. She relies on, of course, the well known fact that hypertranscendentals exist in the formal power series ring k[[T]] when k is a field of characteristic 0. It has now been recently shown by Shildshima-Tsuji and Katsura that such hypertranscendental elements also exist in characteristic p. In their paper Hypertranscendental Elements of a Formal Power Series Ring of Positive Characteristic [7] they use Hasse-Schmidt derivations to find sufficient conditions for a power series to be hypertranscendental and then produce several examples. It was the results of this article that first led us to believe that a generalization of Rotthaus' example to characteristic p might be possible. Two major obstacles stood in the way of proving that our example in characteristic p satisfied the desired properties. The first arose from fundamental differences between the behavior of derivations in characteristic 0 and in characteristic p: even though we knew of the existence of hypertranscendentals in either characteristic, Rotthaus' proof that A trdegRsRs = co no longer went through. Binomial coefficients, which in characteristic 0 were certainly adequate nonzero elements, may suddenly vanish modulo p. In addition, Hasse-Schmidt derivations do not behave exactly like normal derivations: a Hasse- Schmidt derivation is really a family of mappings {dn : n e N} , where du replaces the composition 5“ of an ordinary derivation 5 . These facts combined to make one of Rotthaus' lemmas (which happened to be used no less than three times in her main argument) completely false in characteristic p. The task was then to try to reformulate her lemma in such a way that it was true in characteristic p, and was still strong enough to A show that trdegksRs = co . What we ended up with was a lemma that mimicked the proof of Rotthaus' lemma more than the statement, and a completely new argument for A trdegRsRs = co, which relies heavily on Weierstrass automorphisms. The new proof also has the advantage of working well in either characteristic 0 or p. 4 A Once we knew that trdegRsRs = oo, it followed easily that we could construct rings with low dimensional formal fibers, just as Rotthaus did. What didn't follow was that the rings we had constructed were still excellent. This was the second major obstacle. For this result, it suffices to prove that the extension Q(Rs)-—c—)Q(Rs) is a separable extension. For the proof of this we utilized the notion of a p—basis, and the natural p-basis that exists for a power series ring. Philosophically speaking, there was no reason why rings with low dimensional formal fibers shouldn't exist in characteristic p just as they do in characteristic 0. This dissertation confirms this by showing that the difficulties of generalizing Rotthaus' work to characteristic p can be overcome. We now know that excellent rings exist, in either characteristic, with generic formal fibers of any dimension from 0 to the dimension of the ring minus one. Because the new proof works well in both characteristic p and 0, I hope to have also shown how Weierstrass automorphisms, which are an intrinsic part of the structure of these rings, can be a powerful tool in their study. 1. Definition of Rs and An,t Let k be an infinite field of characteristic p>O such that kp = k, with x1, . . . ,x8 and '1‘ variables over k. Let M g k be an infinite additive subgroup of k. (1.1) Definition: Let s be an integer, s>0. For (n) = (n1, . . . ,ns)e Ms , define : A0,) =k[[x1+n1T, . . . ,xs +nsT]] c; k[[x1, . . . ,xs,T ]] R0 = k[ Am): (n)e Ms ] c; k[[x1, . . . ,xs,T 1] R8 = Q(R0) n k[[x1, . . . ,xs,T]] (Here Qdenotes the quotient field). R, is the fundamental building block of our consu'uction. The definition here differs from that of Rotthaus' only in the set M, which replaces the role of the integers in characteristic 0 (see [6]). We have also added the condition kp = k , which will aid in the proof that R,s is an excellent ring (see section 5). We now introduce the definition of Am, which, though the ultimate goal of the paper, we will hardly encounter until section 8. (1.2) Definition: Let me N be integers with n > 2 and 1 S t S n-3. Lets = n - 2 - t (521) and v1, . . . ,le be variables over Rs. Define: AM = Rs[[v1, . . . ,v,,1]]. The following definition, first formulated by Matsumura in [3], was intended to numerically measure the 'difference between a local Noetherian ring and its completion: (1.3) Definition: For R a local Noetherian ring, define a(R) to be the supremum of the dimension of the formal fibers of R: a R a(R)=sup[dim R®k(p) :pespec(R)) wherek(p)= VR . R P p (1.4) Note: Matsumura shows in [3] that if R is also a domain, then a(R) = dim R§Q( R) , the dimension of the generic formal fiber of R. The goal of this dissertation is to show that Amt is an excellent ring of dimension n, and that “(An,t) = t. 2. Properties of Power Series Rings and Weierstrass Preparation In this section, we review the fundamental properties of power series rings necessary for the main part of the proof. I also include the results of Weierstrass Preparation, which motivated the construction of R8 and play critical roles in most of the proofs that follow. (2.1) Definition: We can write an element P of the power series ring k[[x1, . . . ,xs,T]] as 2P; where the Pi arehomogeneous polynomialsin i=0 x1, . . . ,xs andT ofdegree i. With this in mind, make the following definitions: o(P) = "the order ofP" = min { ie N: Pi it 0} (the order of 0 is defined to be 09 "the initial form of P" = P003) . (2.2) Proposition: The power series ring k[[x1, . . . ,xs,T]] is local with maximal ideal generated by {x1, . . . ,xs,T],and the dimension ofk[[x1, . . . ,xs,T]] =s+ l. (2.3) Proposition: Let P1, . . . ,PS“ be s + 1 power series contained in the maximal ideal (x1, . . . ,xs,T) ofk[[x1, . . . ,xs,T]]. Assume that the degree 1 forms ofthe p1 , . . _ ,pS’rl are linearly independent over k and that the determinant of the coefficients in these degree one forms is not zero. Then the map : k[[x1, . . . ,xs,T]] —)k[[x1, . . . ,xs,T]] xi+--)Pi '= 1 to s 17t—91PS+1 isanautomorphism ofthe ring k[[x1, . . . ,xs,T]]. Proof: See Corollary 2, p. 137: Zariski and Samuel [8]. (2.4) Definition: An element P of the power series ring k[[x1, . . . ,xs,T]] is said to be regular_i.n__’[ifaterchn occursinP,where0¢ ce kandn>0. (2.5) Theorem: Let PE k[[x1, . . . ,xs,T]] be a power series which is regular in T, and not a unit. Let r be the smallest r such that cTr occurs in P, with 0 it ce k. Then there exist power series 86 k[[x1, . . . ,xs,T]], which isaunit,and aie k[[x1, . . . ,xs]] fori=0tor- l suchthat: P=t»:-(Tr + a,_lT"1+... + a0). Proof: See Theorem 5 (Weierstrass Preparation Theorem), p. 139; Zariski and Samuel [8])- (2.6) Proposition: Let Pe k[[x1, . . . ,xs ,T]] be a power series which is regular in T and not a unit. By (2.5) we write: P=e-(Tr + a,,1T"1+... + a0). LetQ=Tr + a,,1r"1 + . .. + aoe k[[x1, . . . ,xs]][T]. Then: kllxt. . . . .xs.T]]/(P) E kllxl. . . . .Xsll [Tl/(Q)- Here (P) and (Q) denote the principal ideals generated in the rings k[[x1, . . . ,xs,T]] and k[[x1, . . . ,xs]] [T] respectively. Also note that the isomorphism is that induced by the natural inclusion k[[x1, . . . ,xs]] ['1'] C k[[x1, . . . ,xs,T]]. Proof: See Corollary 2, p. 146; Zariski and Samuel [8]. The previous results are the basic results of Weierstrass. Now we show how these apply to the rings at hand : Rs in (2.9), (2.10) and (2.11), Amt in (2.12). We begin with the following: (2.7) Definition: For (n) = (n1, . . . ,ns)e Ms set: 00,) : k[[x1, . . . ,xs,T]] -—)k[[x1, . . . ,xs,T]] xii—>xi + niT fori = l to s T|—)T 10 (2.8) Note: (i) It follows from (2.3) that such a map is an automorphism of k[[x1, . . . ,xs,T]]. (ii) °(n)°°(m)=°(n)+0). Then there existsan (n)e M8 such that 00,) (P)e k[[x1, . . . ,xs,T]] is regularinT. 11 Proof: Let P, be the initial form of P (r = o(P) ). Because P, is homogeneous, P, (x1, . . . ,xs,1 )isanonzero polynomialinthe variables x1, . . . ,xs . SinceMis infinite, there existsan(n)=(n1, . . . ,ns)e M8 such that P, (n1, . . . ,ns,l ) at 0. Now 60,) (P,) = P, (X, + an , . . . ,xs +n,T,T) which has the term P,(n1, . . . ,n,, 1)T1'. Since 0i P,(n1, . . . ,ns, l)ek, wehave 6(a) (P,), and hence 0(a) (P), is regular in T. (2.10) Corollary: Let 0 it PS Rs , which is not a unit. Then we can write: P: e' (Tr + b,,1T’°l +. . .+ b0) where e is a unit in Rs and b0, . . . ,b,-1 e A(m) for some suitable (m)e Ms . Proof: We apply a 02;) to the results of (2.5) as follows: By (2.9), choose (n) = (n1, . . . ,ns)e M8 such that o(n) (P) is regularin T. By (2.6) we can write: C“) 60,) (P) = e-(Tr + a,_1T”1 + . . . + a0) where 86 k[[x1, . . . ,xs,T]] isaunit and aie k[[x1, . . . ,xs ]] fori=0tor-1. Since kflxrt . , , ,xs ]]=A(0, ....0) ; R0 5 Rs , e is actuallyaunitin Rs. We apply 02,1!) to (*) : P= o;},,(e>(T’ + o;}.,(a.-1>T'-‘+...+ 02;)(a1)T+62,1,)(ao)) 12 Now let 02,1.)(8)=e,whichwillstillbeaunitinkflxl, . . . ,xs,T]] . Also let bi = 62;)(ai) for i = 0 to r-l. Now the bie k[[x1 - an, . . . ,xs - nsT]] = A_(n). Hence: P = e' (Tr + b,_1T"l +. . .+ b0) as desired. Note that since the his A_(,,) CR0 t; Rs , eis actuallyaunitin Rs. (2.11) Proposition: Let 0 ¢ Pe k[[x1, . . . ,xs,T]] which is not a unit. By (2.10) we write : P=e' (Tr + b,_1T"l +. . .+ b0) whereeisaunitinkflxl, . . . ,xs,T]] and b0, . . . ,b,-leA(n). We let Q: T’ + b,_1T"1 +. . .+boe A(,,)[T]. Then: A(n)[TJ/(Q) a k[[x1.. . . .x,.'rn/ k[[x1, . . . ,xs,T]] Now since 02,1,“ (0(a) (P)) ) = (P) and 02,11“ (o(n)(Q)) ) = (Q), it is clear that : A-(n)[T]/(Q) E k[[xlr ' ° ° 9xsrT]]/(P) and that this map is induced by the natural inclusion of A-(,,) ['1'] into k([x1, . . . ,xs,T]]. Now we apply Weierstrass Preparation to the larger ring Ann- (2.12) Proposition: Let0 it P e An,,\(v1, . . . ,vm). Then we can expressPas: P=e' (T’+b,,lTr'1+ . . . +b0) 14 where eisaunitin A“, =k[[x1, . . . ,xs,T, v1, . . . ,v,+1]]. andthe bieA(m)[[vl, . . . ,v,+1]] for some (m)e Ms. Proof : This fact is essentially the same as (2.9) and (2. 10) , but we must make adjustments for the extra variables v1, . . . ,v,+1. We extend each map 6“,), as follows: 08,): k[[x1, . . . ,xs,T, v1, . . . ,v,+1]] —-)k[[x1, . . . ,xs,T, v1, . . . ,v,+1]], xii—9x, + niT fori= 1 to s, THT, v,|—> Vi fori = l to t+1. Hence of“) trim ..... r..'rn = 0.(fl) ' Note that of“) isan automorphism of k[[x1, . . . ,x,,r, v1, . . . ,v,+1]] by (2.3), with inverse of“). First we show that cfn)(P) is regular for some (n)e Ms. We f'u'st write P as 2P, where the P, are homogeneous polynomials in x1, . . . ,xs ,T, i=0 v1, . . . ,vm of degree i. Chooser such that P, containsamonomial of the form Tr°x{‘-. . .-x;'. The existence of such anris guaranteed by Pt! (v1, . . . ,v,+1). Now P,(x1, . . . ,xs,T,0, . . . ,0) isanonzero polynomial in X], . . . ,xs,T, whichis also homogeneous. Consequently, P,(x1, . . . ,xs,1 ,0, . . . ,0) is a nonzero polynomial in x1, . . . ,xs. SinceMis infinite, there existsan(n)=(n1, . . . ,ns)e M8 such that P,(n1, . . . ,ns,1,0, . . . ,0) $0. Now ofn)(P,) = P,(x1+n1T, . . . ,xs+nsT,T, v1, . . . ,v,+1) which has the term P,(n1, . . . ,ns,l,0, . . . ,0)Tr. Since P,(nl, . . . ,ns,1,0, . . . ,0) $0, ofn)(P,) and hence ofn)(P) is regularinT. Since ofn)(P) is regular in T, by (2.5) we can write ofn)(P) as: cfn)(P) = e-(Tr + a,_1Tr'1 + . . . + a0) 15 where e isaunitink[[x1, . . . ,xs,T, v1, . . . ,v,+1]] and the aie kHXl, . . . ,XS,V1, . . . ,Vt+1]]. -1 As in (2.10), we apply (001)) to obtain: P=e' (Tr + b,,1T"l +. . .+ b0) whereeisaunitinkflxl, . . . ,xs,T, v1, . . . ,v,+1]] and the bie A_(,,)[[v1, . . . ,v,+1]], as desired. 3. The Elementary Properties of Rs In this section, elementary facts are presented which will enable us to prove that R, is excellent in sections 4 and 5. All the results are immediate consequences of the basic facts about power series rings and Weierstrass Preparation. (3.1) Proposition: R, is a local ring of dimension s + 1 , with maximal ideal (x1, . . . ,x,,T), and completion 13., =k[[x1, . . . ,x,,T]]. Proof: An element of the power series ring k[[_x1, . . . ,x,,T]] is invertible if and only if it has anonzerotermin degree zero. Since R, = Q(R0)flk[[x1, . . . ,x,,T]] . the same is true of R, . Hence R, is a local ring with maximal ideal generated by x,, . . . ,x, ,and T. Clearly R, =k[[x1, . . . ,x,,T]] which implies that dim R, =s+ 1. (3.2) Proposition: R, is a Noetherian ring. Proof: Let 1; Rs be a proper ideal. It must be shown that I is finitely generated. Pick 0 at P6 I . By (2.10), express P as: P=e' (Tr + b,.1T"1 +. . .+ b0) whereeisaunitinkflxl, . . . ,x,,T]] and b0, . . . ,b,-1eA(,,) for some suitable (n)EMs . 16 17 Let Q = Tr + b,.1T"l +. . .+boe A(,,)[T]. Consider the following commutative diagram of maps: R,/(P) (-—=—) R,/(P) aT bl A(n)[T]/(Q) —°=——“ai—> kllxt. . . . ,x,,T]]/(P) Claim: The map b is an isomorphism. Proof (of claim): (i) Let Fe R, / (P) , where re R, , such that b(i"') = 0. Then r= I Pwhere E E k[[x1, . . . ,x,,T]] . Then I =r/Pe Q(R,) . Since 11.: Q(Ro) . Q(R,); Q(Ro) or Q(R,)nkllxt. . . . stern; Q(R0)nk[[x1, . . . ,x,,T]] = R, . Hence 8 e R, , which shows that f = 0 in R, /(P), and that b is injective. (ii) By (2.11), the map c is an isomorphism. Now by the commutativity of the diagram, boa is surjective, hence b is surjective. But k[[x1, . . . ,x,,T]] /(P) is a Noetherian ring, so R, /(P) must also be Noetherian. Hence the ideal It; R, /(P) is finitely generated by say Pl, . . . ,P, where H, . . . ,P,e R, . Now clearly I is generatedin R, by P, Pl, . . . ,P, ,which is all that needed to be shown. (3.3) Proposition: For any proper ideal I; R, , the quotient ring R, /I is complete. Proof: From the proof of (3.2), we have that for some 0 at P6 I : l8 .RS/(P) E k[[x1, . . . ,x,,T]] / (P). Hence: Rs/I E Rs/(I’yi E kllxl, . . . ’XS’TJ]/(Pyi E kHXl, . . . ,XS,T]]/(I) and the latter is a complete ring. (3.4) Proposition: a(R,) = 0. Proof: It is enough to show that for any prime ideal (0)¢qe spec ( k[[x1, . . . ,x,,T]]), an, it (0). Let 0 it x6 q . Since xe k[[x1, . . . ,x,,T]], by (2.10) express x =eQ where e is a unit in k[[x1, . . . ,x,,T]] and Q6 A(,,)[T] for some suitable (n)e Ms . Since eQe q, a prime ideal, and e is a unit, Q must be in q. Hence 0 #3 Q6 qfl R, , which is all that needed to be shown. 4. The Excellence of Rs and the Separability of Q(R,)—‘HQ(R,) In this section, we reduce the problem of excellence for R, to the separability of the extension Q(R,)—L9Q(R,) , using the elementary properties from section 3. Let R be a commutative ring. (4.1) Proposition: R is reduced if and only if Rq is reduced for all qe spec R. Proof: "=>": Let r/seRq such that (r/s)n = 0 in RC1 . Then trn = 0 = t"rn = (tr)n for some is R\q. R reduced implies that tr = 0 and hence r/s = 06 Rq. "4:": Suppose X'1 = 0, x6 R\(O). Then ann (x) is properly contained in R, so there exists an me Max (R) such that ann (x): m. By hypothesis R,n is reduced. Since (x/ 1)n = 0 in Rm, x/l = 0 or there exists a te R\m such that tx = 0. But since ann (x); in this is a contradiction. (4.2) Proposition: Assume R is Noetherian, dim R = 0. Then R is regular if and only if R is reduced. Proof: R is regular if and only if Rq is regular for all qe spec R. 19 20 Dim R = 0 implies that dim Rq = 0, so that Rq regular means that qRq is generated by zero elements, which can happen if and only if RC, is reduced. Applying (4.1), the result follows. (4.3) Proposition: (R, m) a local Noetherian ring, a(R) = 0. Then: R has geometrically regular formal fibers if and only if R has geometrically reduced formal fibers. Proof: Let qe spec R , L finite over k(q) = Rq/qRq . Since a(R) = o , dim it? k(p) = 0. Since the extension R g k(p)—L) fig L is finite and flat, dim fr g L = 0 also. Now by (4.2), a? L is regular ifand only if it? L is reduced. (4.4) Proposition: (R, m) local Noetherian domain, a(R) = 0, and for all 0 it qe spec R the ring R/q is complete. IfL is a finite extension field of k(q), where 0 #5 qe spec R then R? L = L . In particular, R? L is reduced. Proof: R®L=R® k(q) e L=R® R/q®Rq e L R R k(q) R R k(q) 21 = R/ “an ®L=(R/®R )®L ( Q) R qk(q) qR ‘1 k(q) = k(q) a L = L. k(q) (4.5) Proposition: (R, m) local Noetherian domain with R" also a domain. Assume also that Q(R)-—E—>Q(R) is separable. If L is a finite extension field of Q(R), then R g L is reduced. Proof: R a domain implies that R—C—)Q(R). Q(R) a flat R-module implies that fig Q(R)—C)Q(R)§ Q(R) = Q(R). Since Q(R) is a field, L is flat as a Q(R) module : rigour.) a L -—C-> Q(R) e L Q(R) Q(R) = i = 1 R®L —‘—> Q(R)®L R R But out)? L = Q(R)Qt(81>{)L which, by the definition of separable, is reduced. By the above inclusion then R? L is reduced. (4.6) Corollary: (R, m) local Noetherian domain, a(R) = 0 and R also a domain. Assume that R is universally catenary, that R/q is complete for all 0 ¢ qESpec R, and finally that Q(R)—c—-)Q(R) is separable. Then : R is excellent. 22 Proof: It is enough to show that the formal fibers of R are geometrically regular (see Matsumrrra [2] for this result, and also the definition of excellence). By (4.3), it is enough to show that the formal fibers of R are geometrically reduced. This is shown for q it 0 by (4.4), and for q = 0 by (4.5). (4.7) Corollary: If the extension Q(R,)—C—-)Q(R,) is separable, then R, is an excellent ring. Proof: In view of (4.6), the result follows from (3.1), (3.2), (3.3), and (3.4). Note that R, is universally catenary because R, is regular (which also follows from (3.1) and (3.2)). s. The Separability of Q(R,)—L+Q(R,) In section 4, we saw that R, is excellent if Q(R,)¢)Q( R,) is a separable extension of fields. In characteristic 0 this meant that R, is clearly excellent. In characteristic p, it meant that a considerable amount of work still remained. In this section, I include not only the proof of separability in characteristic p, but also a result that we obtained en route to the proof. The proposition didn't prove useful in the end, but is interesting in its own right. Our initial attack on the problem of separability was based on the idea that Q( R,) = Q(Ro ), and that we could obtain Q( R,) as the direct limit of a family of finite products of fields (namely the composita of finitely many Q(A(,,))). It was known that Q(A(,,))—:—>Q(R,) is separable, and that the direct limit of fields which form separable extensions when extended to Q( R,) would also form a separable extension when extended to Q(R,). This meant that all we needed to show was that a finite product of subfields, each of which extend to a larger field in a separable fashion, will also extend to the larger field in a separable fashion. In what follows I present our results in this direction, before preceding in (5.6) to the actual argument that Q(R,)—E—9Q( R,) is a separable extension. Suppose K—c—éE and L—c—eE are both separable extensions of characteristic p. We are interested in what conditions are sufficient to conclude that KL—c—>E is also a 23 separable extension. This is always the case if, say, K—5—)E is algebraic, but fails in general , as the following example shows. (5.1) Example: Let K be a field of characteristic p, x and y variables over K. Let E = K(x , y , (xy)1/p ) . The extensions K(x)—‘——)E and K(y)—‘—>E are both separable extensions while K(x)K(y) = K(x, y)-—‘->E is not a separable extension. Proof: E = K(x , y , (xy)1/P) = K(x , (xy)1/P ) , and the element (xy)1/P is transcendental over K( x ). Since purely transcendental extensions are separable, K(x)—‘->E (and by symmetry K(y)—‘—>E ) is a separable extension. 0n the other hand, the irreducible polynomial of (xy)1/p is: Vp-xy which is not separable. Hence the extension K( x , y )—‘:->E is not separable. The following Proposition gives sufficient conditions on the extensions K—c—eE and L——‘—-)E for KL—c—)E to be separable. It was the strongest result we found in this direction, though it did not help us to show that Q( R,)—c-—)Q( R,) is separable. First we make the following definitions: (5.2) Definition: Let KQL be an extension of fields of characteristic p. Let B = {b1 , . . . , b,};; L. 25 i) Define I‘B ={b?‘-...-b:" :OSaiSp-l, i=0,...,n] ii) B is WCL if I}; is linearly independent over L’ (K) . iii) B is a M if B is p-independent for KcL and L" (K,B) = L. (5.3) Note: B is a p-basis of L over K 4: B is a differential basis of L over K ( i.e. {db : be B} is a basis of the L-vector space QL/K: where flux is the module of differentials of L over K, and d the universal derivation). See [2], p.202. (5.4) Proposition: Let K, LgE be fields of characteristic p. Let F; KP" be a subset such that : (IF-(F3)p (KOL) = Ep(K). If K-—:—)E and L—‘—-)E(F) are separable extensions, then KL—‘—-)E is separable. Proof: K—ie E separable implies that K—5——)KL is separable, or that we have the following sequence: OaflngL-aQKL—iQKL/KAO Tensoring this sequence over KL with E, and comparing it to the sequence above with E replacing KL, we obtain the following commutative diagram: 26 K KL KL =l fi gl 0—)9K®E——) 95 ——-> (rpm—>0 K Now KL-—9—>E is separable if and only if f is injective. By the above diagram, it is enough to show that g is injective. For this, it must be shown that there exists a p-basis of KL over K which remains p-independent for KQE. Now QKL/K is generated as a KL-vector space by [db : be KL}, where d: KL Ann”, is the universal derivation. Writing b = 2‘. xiyi where xie K and yie L, we have: d = d(2xm) = thdm). Thus QKL/K is also generated as a KL-vector space by the smaller set {dy : yeL}. Now choose a p-basis BgL of KL over K. This implies that I]; is linearly independent over (KL)p (K) = IP(K) . Thus TB is also linearly independent over LP (K 0L) . Since BQL, B is p-independent for KnL ; L. Because L—c-eECF) is separable, the map : QL/KnL €503) —-’ QE(F)/KnL is injective (see [2], Theorem 26.6, p.202-3). This means that any set which is p- independent for K nL ; L is also p-independent for K nL g: E(F). Hence B is also p- independent for K nL ; E(F) . But this means that [‘3 is linearly independent over (13p (F))(K nL) = Ep (K), by hypothesis. Thus B is p-independent for K; E, which is all that had to be shown. 27 (5.5) Corollary: K, LQE fields of characteristic p. If K—E—)E is separable and -l L-—c——>E( Kp ) is separable, then KL—c—-)E is also separable. Proof: Let F=Kp .1 in (5.4) . This result also follows directly from MacLane's criterion for separability (Theorem 26.4, p.201 [2]). Useful as the above Corollary appears, neither it nor (5.4) were sufficient to show that a compositum of finitely many Q(A(,,)) extended separably to k((xl , . . . ,x, ,T)). Instead, we computed a differential (p—) basis of a power series ring, and relied on the specifics of our example to prove the rest. The result is surprisingly straightforward, and exactly what we needed. (5.6) Proposition: Let K be a field of characteristic p, K = Kp , y1,... ,y, variables over K. Then the set [ y1,.. . , y, l is a p-basis of the extension it (the prime field) ;K((yi....,yr))- Proof: Let B = {y1,...,y, ]. All that needs to be shown is: i) [K((ylsmsyr))]p(B) = K((ylsu-syr». ii) I]; is linearly independent over [K((y1,... , yr ))]p. For i), let ot=2amy§1y§ eK[[y1,...y,]] where (i)=(i1,...,i,)e N’. (i) For each (i), express i, = jt + m,p where 0 S j, S p-l for all t = 1 to r. 28 Thus: a(i)y;‘-...-y;’ = yi"...-y:' -a(,)yfm1-...-yfm‘. Let Mm = yii‘ -...-y;" eI‘B , so that now: a = ZMmAw where A,» e K([yf,...yf]]. M(D€r3 Since K=Kp, this shows that: Kllyi.....y,]l = (Kllyi.....y.ll)"lyl....,y,] and clearly i) follows. For ii), it is enough to show linear independence over K[[yf ,.. . y? l]. or that the above representation of or is unique. Thus we show that: if 0 = 2M(j)A(j) then A0) = o forall (j). M(j)€r3 But since the coefficient of yf‘ “mp-...y} +m'p in szA0) is just the coefficient of yfml'W'YEm' in AGu-ul) , itisclearthat ifany AG) #0 , then also 2M(J')A(i) #0, which is what needs to be shown. (5.7) Note: (5.6) implies that the dimension of QK((yt.-«.y,))/1t as a K((y1,...,y, ))-vector space is r. Compare this to characteristic 0: If L is a field of characteristic 0, then the dimension of the vector space QL((y1.....y,))/1t is equal to infinity (in characteristic 0, the 29 notions of transcendence basis and differential basis coincide). In this sense, the characteristic p case is surprisingly "simpler" than characteristic 0. Now we apply (5.6) in successive corollaries to show R, is an excellent ring: (5.8) Corollary: [x1 + an,..., x, + n,T} is a differential basis of the field extension 75; Q( A(n))- Proof: Immediate from: Q(A(,,))=k((x1 +n1T,...,x, + n,T))Ek((y1,...,y,)), where yi = Xi + niT. (5.9) Corollary: If L = k(A(,,) : (n)e Q) where Q C M8 and |Ql 2 2 , then {xl ,...,x,,T} is a differential basis of ngL. Proof: By (5.8), Q(A(,,)) = Q(A(n))p(x1 + n1T,... ,x, +n,T), and it is also true that LP = k(Q(A(,,))p i (106 Q). Now: Lp(x1+ antuosxs +nST : (n) E Q) = = k(Q(A(,,))p : (n)eQ)(x1+n1T,...,x,+n,T : (n)eQ), = k(Q(A(,,)) 1 (MEQ) = L- 30 Since IQ] 2 2, there exist (11) = (n1,...,n,)e Q and (m) = (m1....,m,)e Q with some m, it 11,. (_xi + miT) - (Xi + II'LT) m, - n, Now: T = e k(xi + miT,xi + niT). Hence TeLp(x1+n1T,...,x,+n,T : (n)eQ). But then: Lp(x1+n1T,...,x,+n,T : (n)eQ)=Lp(x1,...,x,,T) and thus Lp(x1,...,x,,T)=L from above. By (5.6), r{x1.....X. ,T} is linearly independent over k((xf ,...,xg ,T)) :3 Lp . Thus FIX: ”0",. ,T] is linearly independent over Lp , which, combined with the above, shows that {x1,... ,x, ,T} is the desired differential basis. (5.10) Corollary: IfL is as in (5.9), then L-—C—)k((x1,...,x,,T)) is separable. Proof: A differential basis of ngL, {x1,... ,x,,T}, is also a differential basis of it; k((x1,. .. ,x, ,T)), which means that the map: 91.. %k((x1,. .. ,Xs ,T» -) aka,“ ,...,x, ,T)) is injective. This shows that the extension is separable. 31 (5.11) Note: It is also true that Q(A(,,))——C—-)k((x1,...,x,,T)) isaseparable extension. (5.12) Corollary: Q(R,)—Csoa‘ig is separable. Proof: Q(R,) satisfies the conditions for L in (5.9) and (5.10), and Q(R,) = k((x1,...,x,,T)). (5.13) Corollary: R, is an excellent ring. Proof: This follows from (4.7). 6. Derivations in Characteristic p and an Important Lemma A To show trdegRsR, = no , we rely heavily on the theory of derivations. In this section the basic information we will need is presented. In particular, we define Hasse- Schmidt derivations, which are a generalization of the iteration of ordinary derivations for rings of characteristic p. After reviewing the elementary properties of these derivations, we move on to prove the major stepping stone in the proof that trdegRsR, = co, a lemma generalized from the work of Rotthaus and the most technical result of this paper. In section 7, we will finish off the proof that udegRsR, = co. Let R be a ring of characteristic p. (6.1) Definition: {dn : ne N} , a sequence ofmaps dn:R —-) R , is adefiyatipnofR if: (D1). do =idR (D2): d..(x+y> = dn(X)+dn(Y) (03): dn(X)') = 2di(X)dn-i()’) i=0 m +11 (D4): d“ odm(X) =[ m )dm+n(X) for all x,yeR and m,neN. Note: (11 is an ordinary derivation; the higher "nth derivative" differ from the composition d? by a constant, arising from the binomial coefficient which occurs in (D4). 32 33 (6.2) Example: Let R[[T]] be the formal power series ring in one variable T. Set: dogglfli) = i(;},Ti'n for n e N. i=n Then {dn : n e N] is a derivation of R[[T]]. Note that the binomial coefficients appearing in this definition and in (6.1) above are understood as integers modulo p. The fact that many of these vanish (=0) is what makes the characteristic p case so much more complicated than the characteristic 0 case. The following proposition proved exceedingly useful in determining when these binomial coefficients do not vanish. (6.3) Proposition: Let m,neN. Let: c . e , m= 2mip‘ and n = znip‘ i=0 i=0 be the p-adic expressions ofm and n (0 Smi,ni Sp-l forall i). (:l'ittl-(t‘l~tiflW m Thus (“)i0modpc m,2ni fori = 0toe. Then: Proof: See Shikishima-Tsuji and Katsura [7], Lemma 1, p.94. For (6.4) and (6.5), assume that [d,, : n e N] is a derivation on a ring R of characteristic p. Let xe R. (6.4) Proposition: d,,(xi)=i-xi'l-dnx + a, where aieZ/pZ[d0x,...,d,,,1x]. (6.5) Proposition: i) dnxP‘ =0iin¢o mod p°. ii) dnp.xP° = (d,x)P°. Proof: For proofs of (6.4) and (6.5) see Okugawa, Proposition 2 and its corollary, p.12. In Rotthaus, the following useful lemma appears: (6.6) Lemma: Suppose Kg. is an extension of fields of characteristic 0. Let 5 be a derivation of L with 8 (K) gK. Then for any element aeL, the following conditions are equivalent: i) the set {8“ (or) : n e N} is algebraically independent over K. ii) tr'degKK(8n (a) : n e N) is infinite. In particular, if (1) fails then K(Bn (or) : n e N)=K(ct,8a,...,8ma) for some meN. 35 Proof: See Rotthaus' paper for the details. The idea is to suppose that i) fails, then there exists an algebraic relation among the first, say, m + 1 derivatives of or: or,8a,...,6ma. We apply 5 to this equation to obtain a new equation, which now involves 5""! (or) also. Continuing in this manner one produces enough relations to conclude that tr'degKK(5n (a) : n e N)<°°. A The above lemma is used in Rotthaus' proof that trdegRsR, = on no less than three times - in seemingly critical places. It was clear immediately that we needed a similar statement for char p, and yet here the lemma is patently false, as the following example shows. (6.7) Example: Let KQL be fields of characteristic p. Let {(1,1 : n e N} be a derivation of L, under which K is closed (dn (K) c K for all n e N). Let me L be such that {dna : n e N }is algebraically independent over K (these do exist in certain cases, see discussion (7.3)). Let a = dla. Then : i) {dna : n e N} is algebraically dependent over K and ii) trdegKK(d,,a : n e N) = 0°. n+1 Proof: dna = dn(d1a) =[ n )dna by (D4). in m Ifm>0, then [p Jjalmodpand [p 41:1]50modpby(6.3) sothat P P ' dpm a = dprn +1(It and dprn _la = 0 for all m>0. These facts imply ii) and i) respectively. 36 In view of (6.7), the question is how to restate Rotthaus' lemma (6.6) so that it will be true for char p, and still be strong enough to show that trdegRsR, = co. We follow the exact same method of proof as Rotthaus - applying successive derivations to an algebraic relation - but vanishing coefficients will make the restatement of the lemma and the argument quite technical, and at least slightly tricky. The following notion, interesting in its own right, proves to be useful to our understanding of the revised lemma. (6.8) Definition: Let K; L be fields of characteristic p (resp. 0), [tin : n e N} (resp. 8) a derivation on L under which K is closed. Let are L. Set : vK(a)= lim —:1-(trdegKK(doa,...d,ct)) {-900 1' (resp. = lim -1—1(trdegK(a,5a,. . . ,a'ot)). r l’—)°° (6.9) Note: (i) It follows from (6.6) that, for the analogous definition in characteristic 0, vK (or) = 0 or vK(a) = 1. (ii) Shikishima—Tsuji and Katsura, in Theorem 3 of [7] , show that for the extension of fields Kg K((X)) in characteristic p, for every real number OSrSI, there exists an aeK[[X]] such that vK (a) = r. Here we see how much more complicated the characteristic p case is in an explicit manner. (iii) unfortunately, we do not know that the limit actually exists for any element (in characteristic p), and allowances for this must be made in the following proposition. 37 The next result is the strongest reformulation of Rotthaus' lemma that we could find; then we deduce a weaker corollary which turns out to be sufficient for the result of section 7, D'dchsRs = 0°. Let L, g by g E be fields of characteristic p, [du : n e N] a derivation of E under which both L, and L; are closed. Let me E. (6.10) Lemma: If {duct : n e N } is algebraically independent over L1 and algebraically dependent over IQ, then there exists a it e by such that either le (it) does not exist, or v11(3.)>0. Proof: Since A: [duo : n e N} is algebraically dependent over 1.4, there exist wo, . . . , w,e14 such thatAis algebraically dependent over 14(w0, . . . , w,). LetI‘,= {dnwizneNl fori=0,...,r. Clearly A is algebraically dependent over 14(1‘0, . . . , 1“,). By reducing r if necessary, it can be assumed that: i) A is algebraically independent over 14(F0 , . . . , I‘ ,-1), ii) A is algebraically dependent over 1.1(1‘0, . . . , I‘,). WritingL=L1(I‘o, . . . ,1",_1) , w= w, and I‘=I‘,, these become: 38 i) A is algebraically independent over L, ii) A is algebraically dependent over L[F], whereLis closed under {dnz ne N}. Thus there exists an meN such that [d001, . . . , dmct} is algebraically dependent over L[F]. Choose 0¢f(vo, . . . ,vm)e L[I‘][vo, . . . ,vm] (vi arevariables) ofminimal total degree in the vi's such thatO = f(doct, . . . , dmct). As in the proof of Rotthaus' lemma, we would like to apply a derivation to the equation 0 = f(d0a, . . . , dma) and obtain a new relation, but in the characteristic p case this is much more difficult. How the above equation will behave under different (1,, is determined by numerical aspects of the polynomial f, defined below. These will help us decide which (1,1 we should apply, and what to expect afterwards. Define: e= max{ e :f(v0, . . . ,vm)EL[l"][v3 , . . . , v31” 0 e 0+1 9+1 j= min[j :f(v0, . . . ,vm)eL{r][vg ,... ,v? ,v§?+,, ,x,};I ]} c = [Egg—J" = the greatest integer less than or equal to 2:21;. mP-J'ZmP-m = m. This implies that CZm. p-l p-l Note that jSm so 39 Letll besuch that f(v0, . . . ,vm)eL[d0w, . . . , duw][vo, . . . ,vm]. The significance of e, j, audit is that now f(vo, . . . ,v,,,)canbeexpressedas: 71 ip‘ f(v0, . . . ,vm)= 2%)F,(vo, . . . ,vm )v‘i where 1: F pe Po Pul 0+1 ,(vo, . . . ,vm)eL[dow, . . . ,d'_,w][v0 , . . . ’VJ-l ,vJ+1 , . . . ,vm ] and there existsaniiEOmodp such that F,(v0, . . . , vm)¢0. One should think of vj as corresponding to the "highest derivative" which occurs in f, even though derivatives which are higher (up to m) certainly may exist. This is because the higher derivatives all occur raised to a multiple of a strictly greater power of p (at least e+1). When we apply higher and higher derivatives to the equation 0 = f(v0 , . . . ,vm) , according to (6.5) the terms derived from d Jan eventually will be the highest derivatives in our new relation. For this reason, tip is isolated and kept track of specifically, while the other derivatives are of only secondary importance. Certain derivatives of a will never be able to be expressed as derivatives of tip ( n the duo for which [j] a 0 mod p, see (D4)). so we essentially add these to the base field L in the claim that follows. We also add the first c derivatives, in view of the word eventually in the previous paragraph. The point of the following claim is, more or less, to express how many derivatives of w have to be added to obtain a new derivative of a. By comparing w and or in this way , we'll subsequently be able to argue that certain 40 derivatives of w are algebraically independent over L, and finally that v1,I (w)>0 (or doesn't exist), which is the goal of the lemma. Let B={dnot: (3)50modp} C={door, . . . , d,ct} Wi={d,,w : nSi}. Claim: For all teN, d,0t is algebraically dependent over L[B, C, Wu”). (H) ]. Proof (of claim): Suppose the claim is false, then there exists a minimal t such that d,a is algebraically independent over the above ring. Clearly d,a e B and d,or e C, so we can t assume that t>c and (j) #0 mod p. We want to show that d,a is algebraic over the above ring. To do this we apply dlp'(t-D to the equation: it . , o = f(doot, . . . , dma) = 2t,.(d,o)‘P where i=0 f, = 1:,(d0o, . . . ,dma)eL[w,][(d0a)P°, . . . (dj-,ot)P',(d,-,,a)P°”, . . . , (dmot)P°”] For greater easeinnotation, let DP. = {(doa)p°, . . . ,(dj_1a)p°} opal = {(d,,1a)P°”, . . . ,(dma)P"‘ }. Now f,eL[W,,, Dp., opal 1. 41 ‘1 ip‘ Thus: 0: d”, gamma) . p°(t-D 1) i , By (D3): 0= [2%) gtdpmflygn-(dttdja)P ). = 1: Ifi>0, r :50 mod p° then by (6.5) d((d,ct)ip°= o, and ifi = o, e > 0, then still d3 (dja)ip° = 0. Using these facts we obtain a new equation: H n . , 0 = (20 20(dp‘(t-j-l)fi) ' (dip: “Pom )° = 1: Now apply the formula from the second part of (6.5) to obtain: t-l' 1) . , O = (2;) 20(dP°(t-j-e)fi)'(d3(dl“)l)p ° = 1: We separate the terms for which 8 = t- j : t-i-l 1! . . 11 . . <*> 0= 2 2~sewer)" + Err-(dander)? - (=0'=0 i=0 We examine each of the factors in the above equation in turn: i) What is dp°(t-j-£)fi ? Or even better, what ring is it in ? We know fieL[W,,, DP° , Dpe+l ]. We compute what happens to W“, DP° , and Dpen intum: 42 a) d,,_.(,,-_p,)(L[W]) elestdtw) 5-.- 0top° k= own] ;L[W ]for e=otot-j-1. n+1) °(-t D b) Similarly, dp.(,_j_,)(L[Dp.]) ;L[d5(dkot)P° : k = 0toj-1, 5 = Oto p°(t-j)] by (6.5) ; L[(dka)p° : k = o to j - 1 + t -j] ; L[door, . . . , d,_1a]. 0) Finally, )(LlemD Quartile)?” =k=j+1.-~.m.5=0.~-.P°(t-J')] dp °-(tl'l p"+1 - Pe(t'j) by (6.5) ; L[(dkor) I k=j+1, . . . , 111+ T ]. P mp J - Note that t>cimplies thatt>— , or that tp - t > mp - J and -0 -0 e -0 t>m+22m+ t——J . Hence m+ £(t—fl- S t-l. p I) 1,344 This, with the above, shows that : dp 0.“ -j- ‘POOJD e+1])CL[doadl_la] . 43 Combining a), b) and c), we conclude that : d p_e(, -j- -()fi 6 ”Wu 11+? .t_(t D][doa,. , d,_1a] . This ends i). ii) As for (d¢(d,-a)i)P° , the second factorin (*), we first note that e s t- j-l. By (6.4) and (D4) , - - £+° d((dtx)l =i:(da)l—l- J -dg+-a + a, where aieZ/pZ[dka :k=j, . . . , Z +j - 1] J J 3 J But i + j .<_t- 1, soclearly: (d,(dja)i)P°eL[doa, . . . ,d,-1a]. . . t iii)r=inally,d.-(d-ot)l = i-(do)"1- , ~da+ai where a-eL[dct, . . . ,dHa] tJ J J J t 1 J by (6.4). Thus we rewrite (*) as: t- -r'-1 n '1 . . e 0= 2 23%.“. J _-,Jf. (d,(d o) i")P +2f a.P° +[2ii.i.(;).(d,a)("1)l’ ]-(d,ct)P (=0 0i=0 i=0 i=1 t-j-l n Letting co- — J20 dem J._,Ji (d[(dJ-a) )P + 2013° .g’a ~01=01" we 10 w) we get that 0 = co + cl-(dtot)pe where by i), ii), and iii), co, CIEL[Wu+p¢(,_D][doa, o . o , dt_la]o We now Show that in fact c1¢0 : First note that the polynomialin v0, . . . ,vm: 1] 213010,... ,vm)i[J]vJ(‘l)p e L[dow,. .. ,duw][v0, . . . ,vm] i=1 is not equal to 0 because: i) [J )920 mod p and ii) there existsanii-‘PO modp such thatF,(v0, . . . , vm)¢0. If F,(v0, . . . ,vm)¢0 andi>0, their: Tot.Dego(F.(vo. . . . .vm)viJ-P')2p°+TotDeg.[P,(vo, . . . ,vm) i (J) ngnP' ] It follows that : 45 n . . Tot.Deg.[2F,(vo, . . . ,vm)va ]> i=0 Tot De 11 ° t (i-1)P° . g. 2F,(v0, . . . ,vm)-r- J N, . i=1 So that, by choice of f: n t (H) ° 0 ¢ c1= ZF,(doot, . . . ,dma)-i-(J)-(dja) P . i=1 Hence 0: c0 4» c1.(d,or)P° where c0, cleL[WJJ+JJ.(J_JJ][doa, . . . , duct] and c1¢0. Thus d,a is algebraically dependent over L[WJJ +p° (J_D][d0a, . . . , duct]. Let u < t be an integer. By our choice of t, we can assume that duo is algebraically dependent over L[B, C, WJJ )]' Hence duet is also algebraically ]. This shows that L[B, C, +P‘(u-j dependent over the larger rlng L[B, C, Wu +p° (l-D Wu+p°(t-j)][doa’ . . . , duct] rs an algebrarc extensron of L[B, C, wu+p‘(t-j)]‘ Now we have shown d,0t is algebraically dependent over L[Wu+p°([-j)][d0a’ . . . , d,,1a] and hence over L[B, C, WJJ+pe(,_D][doa, . . . , duct]. Hence d,a is algebraically dependent over L[B, C, WJ1 +p° (,_ ))l’ which is a conuadiction to the choice of t, and ends the proof of the claim Because A = {duct : n e N] is algebraically independent over L, A\(Bu C) = n {dna : [j] #0 mod p , n>c} is algebraically independent over L[B, C]. We would like 46 to describe how 'big' this algebraically independent set A\(B U C) is, or more precisely, since |A\(B U C)| = co , how frequently can we expect to find a member of A\(BU C) . First choose e' such that p°'> j. By the computational result of Shikishima-Tsuji and Katsura (6.3), we know that: e' - [np_+1)almodp foralln. J This shows that once past c, at least one out of every p° derivatives of a will be in A\(BUC). On the other hand, by the claim and the above Comment for u < t, for all c < u .<_ t duet is algebraically dependent over L[B, C, Wu+p‘ (t-))]’ or: tl'dcguB’ C, wJJJJJJJJJJJJ)L(B, C, wll'i'PcU'J) )(doa, . . . ,d,or) = 0. We now have: ( MB’ C’ wn+p°(t-D)(d°a’ ° ' ° ,d,a) trdeg=0 t-C Udegz [ij L(Bs C9 wll+P°(l'j)) K L(B, C) t- C So that we must also have tr'degL(BJ C)L(B, C, Wu +p° (t- 1)) 2 l—J. pd 47 Since ngL ;L(B, C), this implies that: t- C "dcgl‘llell+p°(t-j)] >;eT—1 . trde L ) We let ‘Dg = 812+ :(wt . the term which appears in the limit of definition (6.8). The above has shown that: 33—1 e 1Slal‘tPVt-j) > 1)e o = e+e' e+e'- e' e' n+1) (t-l)+1 p H) l+p n+1) t-c-pe t+ (a constant) p°+°t + (a constant) For large t, this latter quotient approaches l/(pe+e'), so that : . 1 J11?” fiu+p°(t-j) 2 F > 0, if the limit actually exists. But this implies that the limit vLJ (w)¢0, if it exists. Since we 112 , setting w = l proves the lemma. Were we to attempt to reformulate this lemma (6.10) for characteristic 0, what would we end up with? We first make a definition, and then prove a similar lemma for characteristic 0. The fact that the proof will rely exclusively on Rotthaus' lemma (6.6) shows that (6.10) is an appropriate analogy to (6.6). 48 (6.11) Definition : Let R; S be rings of characteristic p (resp. char 0). Let {do : n e N} (resp. 5) be a derivation of S under which R is closed. An element aeS ishmmnsmdenntmemif {dna tn 6N) (resp- {5“a:neN}) isalgcbmically independent over R. Let L, t; by t; E be fields of characteristic 0, 5 : E-—)E be a derivation of B such that L] and IQ are closed under 5. Let ae E . (6.12) Lemma : If or is hypertranscendental over L1 and not hypertranscendental over Ly , then some element of L2 is hypertranscendental over L1. Proof : As in characteristic p (see beginning of proof of (6. 10)) , the situation can be reduced to the following: LJQL is a field, closed under 5 , and web; such that : i) or is hypertranscendental over L, ii) a is not hypertranscendental over L(8“w : n e N). By Rotthaus' lemma (6.6), trdeguaa L(5“w, find: it e N) is finite. Since w: neN) trdegLL(8"w, 5nd: n e N) = 0°, it must be true that trdegLL(5“w : n e N) = 0°. But 49 , by lemma (6.6) once more, this implies that we L2 is hypertranscendental over L, and hence over Ll. Thus we obtain the following corollary: (6.13) Corollary : Let L1 c; 14 g; E be fields of characteristic p (or 0), {(1n : n e N} ( or 5) a derivation of E under which both L1 and L, are closed. If 0: era is hypertranscendental over L1 and not hypertranscendental over 14, then trdegLJI/z = co. Proof : In characteristic 0 this follows immediately from (6.12). In characteristic p, (6.10) guarantees a it. ell; such that vLJ (3.) either doesn't exist or is greater than zero. Either case implies trdegL1 L1 (duh : n e N) = co , which in turn implies that trdegLJLZ = co. 7. TrdegRs R, = oo A In this section we present the proof that udegR,R, = co, which will be essential in the next section when we prove that the ring AM, constructed from R,, has low dimensional formal fibers. The proof here depends on the corollary (6.13) from the previous section and the Weiersuass automorphisms introduced earlier (2.7). The proof is quite different than that of Rotthaus for characteristic 0. The advantage is that the new proof works in any characteristic, as we shall see in (7.8). First we state a result of Shikishima-Tsuji and Katsura (Theorem 1, p.9 , [7]): (7.1) Notation: Let [cln : n e N} represent the Hasse-Schmidt derivation with respect to the variable T on the ring k[[x1, . . . ,x, ,T]] (or on k[[T]] ) introduced as example (6.2). (7.2) Theorem: There existsan ore k[[x1, . . . ,x, ,T]] which is hypertranscendental over k(T). Note: This is a crucial result. Shikishima—Tsuji and Katsura's proof does not require that kp = k, or that k be infinite. Before stating and proving the next theorem, we prove the following lemma, which shows that our rings A(,,) = k[[x1 + an, . . . ,x, + n,T}] are closed under the derivation {dn : n e N} . We will need this to show that certain fields are closed under {dn : n e N} , which will enable us to apply the corollary (6.13). 50 51 (7.3) Lemma: Let (n)=(n1, . . . ,n,)e MS, r>0. Then d,(A(,,)) c; A0,). Proof: First note that d,: k[[x1, . . . ,x, ,T]] —> k[[x1, . . . ,x, ,T]] takes terms n+’) g; nrn (wherem= of total degree n to terms of total degree n - r. Hence d, (m (x1, . . . ,x, ,T), themaximalideal ofk[[x1, . . . ,x, ,T]]). Thus d, is continuous with respect to the topology on k[[x1, . . . ,x, ,T]] induced by m. Defineamap o :k[[y1, . . . ,y,]J ——> k[[x1, . . . ,x, ,T]] yi H Xi +1'liT fOfi=1IOS. Then by (2.3) we have that : k[[y1, . . . ,y,]] 5 Am = k[[x1 + n,T, . . . ,x, + n,T]] g; k[[x1, . . . ,x, ,T]]. Let a 6 Am. We write a as a = 2a, , where each a, is a homogeneous polynomial of i=0 degreei (in X], . . . ,x, ,T). Claim: d, (a,) e A(,,) is a homogeneous polynomial of degree i - r ( if i < r then d, (3,) =0). Proof (of claim): Let i, + . . . + i, =i. We compute d, onamonomial: d,((x1 + an)i‘° . - - -(x, + HST)i‘) = 2er (x1 + an)i1- . . . ~d,J(x, + n,T')i' rl+ o o o +r'=r by the product rule. Similarly : 52 er(Xj + 11jT)ij = 2 ng(Xj + an)' . . . °leJ(Xj + an) (1+ . . . +£,J=rj 0 if (m >1 and dlmo‘j + DJ,“ = I‘lj if [m =1 x, + an e A(,,) ime = 0 by the definition (6.2) of our derivation. Hence: d ij _ 11 ’1‘ . .Tii'rl' A ri("i+“in ’ r '“i'(xl+nl) E (n)- N°w= «mm + nmil- . . . «x. + nsT)i') = i . i J 2 [r:)-n{1.(xl 4" 111131141 - , , . .[rs].n;3 .(xs 4,, “313"“, r, + . . . +r,=r s which is homogeneous of degree i - r. Note that for a term in the above sum to be nonzero, i ,- must be greater than or equal to r,- for all j. Hence d, (a,) = 0 if i < r. This proves the claim. Since d, (a,) e A(n)’ we can consider the elements o'1(d, (a,)) , which will also be homogeneous polynomials (in the y, 's) of degree i — r. Since k[[y1, . . . ,y,]] is n on complete, lint 2¢—1(d,(ai )) = 2¢‘l(d,(ai)) is in k[[y1, . . . ,y,]]. n 00 i=r i=r Now ¢[i ¢-1(dr (30)] = i ¢(¢-l(dr (39)) i=r i=r = 2d,(a,) = d,[2ai]= d,(a) i=r i=0 since (1, is continuous. Hence d, (a) e Im¢ = A(,,) which proves the lemma. 53 Now for our theorem: (7.4) Theorem : Let aek[[T]] which is hypertranscendental over k(T). Then or is also hypertranscendental over R,. In particular, trdegRsR, = no. Proof: Let A = [d,a: n e N ]. Assume 0t is not hypertranscendental over R, , i.e. A is algebraically dependent over R, . Since R, ; Q(Ro), this means that A is also algebraically dependent over Q(Ro) = k( A(,,) : (n) e M3 ). Since only finitely many A(,,) can be involved in an algebraic relation, A must be algebraically dependent over: k(A(,,J), . . . , A(,,r)) for some (n1), . . . ,(n,)e M3. By reducing r if necessary, it can be assumed that : i) A is algebraically independent over k(A(,,J), . . . ,A(,,JJJ)), ii) Ais algebraically dependent over k(A(,,J), . . . , A(,,J)). Again, an algebraic relation can only involve finitely many elements I] , . . . ,t, e A(nJ). LettingL= k(A(,,J), . . . ,A(,,J_J)), we get: i) A is algebraically independent over L, ii) A is algebraically dependent over L(1:1, . . . ,t,). We would like to argue that this is a contradiction to (6.13), but the field 14(11, . . . ,t,) is not necessarily closed under our derivation. To arrange for this, we could add to the field all the derivatives of t1, . . . ,x,, but then we could potentially have infinite transcendence degree between the two fields, so no contradiction. Instead, we apply a Weierstrass automorphism to adjust the situation. Apply the automorphism o_(,,J) to i) and ii) above: Gingham) = 0-(n.)°°0. 55 But now a is hypertranscendental over L, , not hypertranscendental over L2, and tr'degLJ L2 5 t < co . This contradicts corollary (6.13), and thus proves the theorem. The proof above works because we isolate a particular A(,,) , and then apply the appropriate 6 to assume that (n) = (0, . . . ,0). Thus A(,,)= k[[x1, . . . ,x,]]. and this ring vanishes under our derivation, which differentiates with respect to T. But these remarks are equally true in characteristic 0. The only changes that have to be made are purely formal, and outlined below: (7.5) Definition: Let k' be an infinite field of characteristic 0, and M; k' be an infinite additive subgroup of k. For (n) = (n1, . . . ,n,)e M" , define: A'(,,) = k'[[x1+n1T, . . . ,x, +n,'1‘]] C; k'[[x1, . . . ,x, ,T ]] R'O = k'[ A'(,,) : (n) e M'‘3 ] (; k'[[x1, . . . ,x, ,T ]] R', = Q(R'o) n k'[[x,, . . . ,xs ,T 1] . (7.6) Note : In Rotthaus' definition, M' = Z, but all the proofs go through with this slight generalization. (7.7) Notation: Let 8 be the derivation on k'[[x,, . . . ,x, ,T ]] which differentiates with respect to T. 56 (7.8) Theorem: Let aek'[[T]] which is hypertranscendental over k'(T). Then a is A also hypertranscendental over R',. In particular, trdegRJJR', = co. Proof : We repeat the proof of (7.4), with the substitution of 5“ in the place of (1,. Note that lenrma (6.12) was proved for any characteristic. Lemma (7.3) will also hold in characteristic 0 : one just has to Show that 8(A'(,,) ) c A'(,,) , which is a quick check. 8. Low Dimensional Formal Fibers In the previous sections we have introduced a ring R,, which has various properties, as we have discussed and proved. But of course, a(R,) = 0 (3.4), so it is not an example of the rings in which we are interested, those with a between 0 and the dimension of the ring minus 2. In this final section of the paper we show that the rings An, constructed from R, (see (1.2)) are excellent and have low dimensional formal fibers. (8.1) Theorem : AM is an excellent Noetherian local domain of dimension n such that (MAM) = t, i.e. the generic formal fiber of An, has dimension t. Proof : It is clear that A“.t is a local domain. It is Noetherian since R, is Noetherian (1.3) and has dimension n because : A“, = k[[x1, . . . ,x,,T,vl, . . . ,v,+1]] and s + t + 2 = n. The ring AM is excellent because R, is excellent (this follows from Rotthaus [5]). It remains to see that dim A“, As Q(AM) = t. "s" : Let Pe A,,\(v,, . . . ,v,+1). By (2.12) we can express P as: 57 58 P: e-(Tm + b,,,_1T"“'l +. . .+b0) where e isaunitin A“, and b, eA(,,)[[v1, . . . v,+1]] foralli=0 to m-l. Thus (P)flA,,, ¢(0), so any prime ideal q t; A", with qmttmt = (0) must be properly containedin(v1, . . . ,vm). Since ht (v1, . . . ,v,+1)=t+l, this shows thatdim AMAO Q(AM) S t. "Z" : Let wl, . . . ,w,e k[[x,, . . ., x,,T]] whicharealgebraically independent over R, (here we rely on the main result of section 7, trdegRsR, = co) . Define a homomorphism: q) : A“, = k[[x1, . . . ,x,,T,v,, . . . ,v,+1]]—>R,[[v,+1]] x,I—>x, fori= l tos T H T vJ-l—) wjvm forj=1, . . .,t vt+l H vt+l Clearly (I) is surjective, and Q = ker (1) must be a prime ideal of height t. It suffices to prove the following claim: Claim: QflAnJ, = (0). Proof (of claim): Let FeAnJ, , F¢0. 59 Then F = 2 f,(v1, . . . ,v,+1) where the f, are homogeneous polynomialsinthe o=0 variables v1, . . . ,vm with coefficientsin R,. Write fu(v1, . . . ’vI-H) = Zbon'Vfl ' . . . 'thll Where bone RS . “1"" ° ”Natl:D NOW 4,03): 2 fo(wlvt+l’ ° ° ° ’wtvwl’ Vt+l) = o=0 _ u ll 1* - 2 VH-l. ZbQJoWll'. . .‘th . u=0 u,+..+tr,=u If Fit 0, then some f,,(v1, . . . ,v,,,,) at 0. But then, since {w,. . . . ,w,] is algebraically independent over R, : Ebonwf‘a . .-th :e o forthat u. ut+~+ltt=v Hence (9 (F) at 0 , or F6 Q, which proves the claim and the theorem. The following corollary, which shows that maximal ideals in a formal fiber may have height strictly less than the dimension of the formal fiber, is proved in Rotthaus [6] for characteristic 0. For the sake of completeness, we also prove it here for characteristic p. (8.2) Corollary: Let w be an additional variable. Then the generic formal fiber of B = An, [w](,,,JJ J J w) contains a maximal ideal of height St+1 < n-l while the generic formal fiber has dimension Zn-l . (Here mAJJJ denotes the maximal ideal of An, .) Proof: B; A,J,[w],mAJJ N); k[[x1, . . . ,x,,T,v1, . . . ,v,,,, w]] = B. 60 By Example 2 of Matsumura, the generic formal fiber of A,J,[w](,,1A J N) has dimension n - 1. Since B = (A,J,[w](,,,JJJu NO) , we must also have: dim B§Q(B) 2 n - 1. Let be AM he a non-invertible element which is algebraically independent over An, . Takeamaximalprimeideal Q; B such thatw-b e Qand QnB =(0). ThenQ=(w- b , q), where q = one.“ is aprime ideal in R,,, with bflAnJ, = (0). By (8.1), ht q s t, thusthSt+l. 9. Summary We have seen that excellent rings of characteristic p can have formal fibers of arbitrary dimension. On the way we have combined facts from Weierstrass Preparation and differential algebra. Hypertranscendental elements played a key role in proving tr'degR,R, = co in both characteristic 0 and p>0. And the intervention of a single Weierstrass automorphism, c-(,,J) , in the midst of Theorem (7 .2) , salvaged a proof which otherwise would have been very technical and messy, if possible at all. We are left with a thorough and complete answer (when taken in conjunction with the work of Rotthaus [6]) to Matsumura's original inquiry into the dimension of formal fibers. Of the many questions that still remain about formal fibers, it might be appropriate to ask about the dimension of the formal fibers of a non-excellent ring. To my knowledge, no examples of low dimensional formal fibers exist for this case. Another major unanswered question involves the definition (6.8). As mentioned, we have been forced to allow for the fact that such a limit may not exist. Surely it should be possible to answer the existence question in a more definitive way. For this I believe that we will have to acquire a better understanding of how derivatives can be applied to algebraic relations, of which Lemma (6.10) is just the beginning. I also believe that the basic method of (7 .2), where we are able to assume that A(n) is indeed A“), . "0) , may be of use in furthering the study of Rotthaus' extraordinary ring R,. 61 List of References [1] [2] [3] [4] [5] [6] [7] [3] List of References W. Heinzer, C. Rotthaus, and J. D. Sally, Formal Fibers and Biratr'onal Extensions , Nagoya Mathematical Journal , 131 (1993), pp.1- 38. H. Matsumura, W , Cambridge University Press, Cambridge, 1986. . 0n the Dimension of Formal Fibres of a Local Ring , Algebraic Geometry and Commutative Algebra in Honor of Masayoshi Nagata, 1987. 139.261 - 266. K. Okugawa, Difi'erential Algebra of Nonzero Characteristic , Lectures in Mathematics, Department of Mathematics Kyoto University, 16 (1987). C. Rotthaus, Komplettierung semilokaler quasiausgezeichneter Ringe , Nagoya Mathematical Journal, 76 (1979) , pp.l73-180. , 0n rings with low dimensional formal fibers , Journal of Pure Algebra, 71 (1991), pp.287-296. K. Shikishima—Tsuji and M. Katsura, Hypertranscendental Elements of a Formal Power Series Ring of Positive Characteristic , Nagoya Mathematical Journal, 125 (1992), pp.93-103. O. Zariski and P. Samuel, W, Volume 11, Van Nostrand, 1958. 62 174 6518 W” AI” Rli mlll L” V" Ilil U U " I'll Tlllll. S" N“ A". mli H" 31293 03