Constructive Krull Dimension. I: Integral Extensions.

Transcription

1 Constructive Krull Dimension I: Integral Extensions Thierry Coquand ( ) Lionel Ducos ( ) Henri Lombardi ( ), Claude Quitté ( ) May, 2006 Abstract We give a constructive approach to the well known classical theorem saying that an integral extension doesn t change the Krull dimension MSC 2000: 13C15, 03F65, 13A15, 13E05 Introduction In this paper and the following one (Constructive Krull Dimension II: Noetherian Rings) we investigate some classical topics about Krull dimension from a constructive point of view The fact that a constructive theory of Krull dimension, avoiding Choice and Third Excluded Middle Principle is indeed possible was made clear by the work of Joyal and Español ([Joy, Esp, Esp2] 1975, 1982, 1986) Concrete applications appeared when more easily manageable characterizations of Krull dimension appeared ([Lom, CL] 2002, [CLR, CLQ2, CL2] 2005) Notice that a similar elementary definition follows easily from a result of Brenner ([Bre] 2003) Some celebrated theorems of commutative algebra as the Serre s splitting off, the Bass stable range theorem, the Bass cancellation theorem, the Forster-Swan theorem, the Brewer-Costa- Maroscia theorem and the Eisenbud-Evans-Storch theorem, have now a completely algorithmic version (see [Coq, CLQ, Duc, CLS, LQY]) For the Serre s splitting off and the Forster-Swan theorem the constructive approach has eventually lead to stronger versions than the previously existing ones, giving a positive answer to a question of Heitmann in the memorable non-noetherian paper [Hei] In this paper we give a constructive approach to the well known classical theorem saying that an integral extension doesn t change the Krull dimension So we get that for such an extension A B we have an algebraic machinery that transforms the production of identities whose meaning is KdimA n in the production of identities whose meaning is KdimB n, and vice-versa The paper is written in the usual style of constructive algebra, with [MRR] as a basic reference Chalmers, University of Göteborg, Sweden, coquand@cschalmersse Laboratoire de Mathématiques, SP2MI, Boulevard 3, Teleport 2, BP 179, FUTUROSCOPE Cedex, FRANCE, ducos@mathlabouniv-poitiersfr Equipe de Mathématiques, CNRS UMR 6623, UFR des Sciences et Techniques, Université de Franche- Comté, BESANCON cedex, FRANCE, henrilombardi@univ-fcomtefr Laboratoire de Mathématiques, SP2MI, Boulevard 3, Teleport 2, BP 179, FUTUROSCOPE Cedex, FRANCE, quitte@mathlabouniv-poitiersfr 1

2 2 1 CONSTRUCTIVE KRULL DIMENSION 1 Constructive Krull dimension In this section we recall some elementary characterizations of the Krull dimension Proofs may be found eg in [CLR, CLQ2] Let us consider a commutative ring A A filter is a saturated multiplicative monoid in A A prime filter is a filter not equal to A whose corresponding localization gives a local ring Within classical mathematics, prime filters are exactly the complements of prime ideals, a maximal filter is prime and its complement is a minimal prime The radical of an ideal a will be noted as D A (a) We write D A (x 1,,x m ) for D A ( x 1,, x m ) The Zariski (distributive) lattice ZarA is defined as the set ZarA = {D A (x 1,,x m ) m N, x 1,,x m A} ordered by inclusion Within classical mathematics, ZarA is isomorphic to the lattice of compact open subsets of the Zariski spectrum SpecA The isomorphism is given by where D A (a) = {p SpecA a / p } D A (x 1,,x m ) D A (x 1 ) D A (x m ) 11 Krull boundaries Recall that for an ideal a and an element a of A we have the notation (a : a ) A = n N(a : a n ) A = {y A n N, ya n a} Definition 11 Let x 0,,x l be a sequence of elements of a commutative ring A 1 We define inductively an iterated boundary monoid S A (x 0,,x l ) for this sequence by: S A () = {1} and S A (x 0,,x l ) = x N 0 (S A(x 1,,x l ) + Ax 0 ) (1) Eg, S A (x 0, x 1, x 2 ) = x N 0 (x N1 (x N2 (1 + Ax 2 ) + Ax 1 ) + Ax 0 ) 2 We define inductively an iterated boundary ideal N A (x 0,,x l ) for this sequence by: N A () = {0} and N A (x 0,,x l ) = (N A (x 0,, x l 1 ) : x l ) A + Ax l (2) Eg, N A (x 0 ) = (0 : x 0 ) + Ax 0, N A (x 0, x 1 ) = ( ((0 : x 0 ) + Ax 0) : x 1 ) + Ax 1 For any a A, N A (a) meets any maximal ideal and S A (a) meets any maximal filter The inductive definition of S(x 0,,x d ) can be understood with the constructor M a := S a N (S + Aa) where a A and S is an arbitrary multiplicative monoid in A More precisely S(x 0,,x d ) = M x0 M x1 M xd ({1})

3 12 Characterizations of Krull dimension 3 Similarly the inductive definition of N(x 0,,x d ) can be understood with the dual constructor I a := a (a : a ) A + Aa where a is an arbitrary ideal of A More precisely We have the equivalences N(x 0,,x d ) = I xd I x1 I x0 ({0}) 0 S A (x 0,,x d ) 1 N A (x 0,,x d ) N A (x 0,,x i 1 ) S A (x i,,x d ) (3) This justifies reversing the order between M x0 M x1 M xd and I xd I x1 I x0 When 0 S A (x 0,,x d ) we will say that the sequence x 0,,x d is pseudo singular Remark In [CLR], where the Krull boundaries are defined for the first time, the boundary ideal uses a slightly different constructor a ( a : Aa) A + Aa Let us denote by V A (a) the closed subset of SpecA defined by a (ie, the complement of D A ) The name boundary of a for b = (D A (0) : a) + a comes form the fact that V A (b) = D A (a) V A (a) is the boundary of V A (a) inside SpecA (in classical mathematics) This gives an intuitive explanation for the fact that the dimension on A/b is stricty lesser than the dimension of A: the boundary of any subvariety in a variety is always strictly lesser than the variety itself Next, Fred Richman defined in [Ric] another boundary with the constructor we use here In fact the boundary ideal ( 0 : Aa) A + Aa of [CLR] contains the Richman boundary ideal N A (a) and they have the same radical So the two quotient rings have isomorphic Zariski lattices, and the difference is not really important Perhaps the most intrinsic definitions would be to consider N A (x 0,,x n ) and the saturation of the monoid S A (x 0,,x n ) 12 Characterizations of Krull dimension First recall that a ring has Krull dimension 1 if and only if it is trivial Theorem 12 Let A be a commutative ring and d N The following are equivalent: 1 (classical definition) Any increasing chain of primes has length d (ie, the number of primes in the chain is d + 1) The maximal length of such a chain belongs to N { } and is denoted by KdimA d 2 (induction using ideal boundary) For any a A, Kdim (A/N A (a)) d 1 3 (induction using monoid boundary) For any a A, Kdim (S A (a) 1 A) d 1 4 (iterated boundaries) Any sequence x 0,,x d in A is pseudo singular 5 (symmetric form) For any x 0,,x d A, there exist a 0,, a d A such that a 0 x 0 D A (0) a 1 x 1 D A (a 0, x 0 ) a 2 x 2 D A (a 1, x 1 ) a d x d D A (a d 1, x d 1 ) 1 D A (a d, x d ) (4) Moreover points 2, 3, 4, 5 lead to constructively equivalent definitions of the Krull dimension

4 4 1 CONSTRUCTIVE KRULL DIMENSION Two sequences a 0,,a d and x 0,,x d satisfying the point 5 will be called complementary sequences In the lattice notation this gives D A (a 0 ) D A (x 0 ) 0 ZarA D A (a 1 ) D A (x 1 ) D A (a 0 ) D A (x 0 ) D A (a 2 ) D A (x 2 ) D A (a 1 ) D A (x 1 ) D A (a d ) D A (x d ) D A (a d 1 ) D A (x d 1 ) 1 ZarA D A (a d ) D A (x d ) Here are light variations on the formulations for the point 4 1 For any x 0,, x d there exist a 0,,a d A and m 0,,m d N such that x m 0 0 ( (xm d d (1 + a dx d ) + ) + a 0 x 0 ) = S(x 0,,x d ) = x N 0 (x N 1 ( (x N d (1 + Ax d) + Ax 1 ) + Ax 0 ) 3 1 N(x 0,, x d ) = (( (0 : x 0 ) + Ax 0 ) : x d ) + Ax d 4 For any x 0,, x d A, there exist n N such that: (x 0 x d ) n A(x 0 x d 1 ) n x n+1 d + A(x 0 x d 2 ) n x n+1 d Axn+1 0 Remarks In constructive mathematics the sentence KdimA l is well defined but KdimA is not in general a well defined element of N { } It is remarkable that most classical theorems using Krull dimension may be put under the form If KdimA l then The basic fact that KdimK[X 1,,X l ] = l when K is a discrete field has a very simple proof, see [CL2] As a consequence, the usual geometrical rings have a well defined Krull dimension in constructive mathematics This means eg, that the construction of complementary sequences is given by an effective procedure in such rings See also [Lom] for an explicit generalization of the Nullstellensatz 13 Some basic facts Following simple facts show a kind of strong duality between addition and multiplication, ideals and filters, and the two kinds of Krull boundaries Fact 13 (Krull boundaries, localizations and quotients) Let x 0,,x l A, S a monoid and a an ideal One has: 1 S A/a (x 0,,x l ) = S A (x 0,,x l ) mod a 2 N S 1 A(x 0,,x l ) = S 1 N A (x 0,,x l ) ) 3 (a) S S 1 A(x 0,,x l ) = S (x 1 N 0 (xn 1 ( (xn d (S + Ax d) + Ax 1 ) + Ax 0 ) (b) If S = S(x l ) then S S 1 A(x 0,,x l 1 ) = S 1 S A (x 0,,x l ) ( ) 4 (a) N A/a (x 0,,x l ) = (( (a : x 0 ) + Ax 0 ) : x d ) + Ax d /a

5 5 (b) If a = N(x 0 ) then N A/a (x 1,,x l ) = N A (x 0,, x l )/a The fact that Krull dimension cannot increase by localization or quotient is direct and constructive from the constructive definition Some converse implications are given in the following lemmas Lemma 14 (Krull dimension and quotients) Let a, b be ideals of A Then KdimA/(a b) = KdimA/ab = max {KdimA/a, KdimA/b} This remains true for finite intersections and products of ideals Proof The first equality comes from D A (a b) = D A (ab) If x 0,,x n A, 0 S A/a (x 0,,x n ) means that S A (x 0,,x n ) meets a So KdimA/a n and KdimA/b n means that S A (x 0,,x n ) meets a and b for any x 0,,x n A, which is equivalent to S A (x 0,,x n ) meets the product a b Lemma 15 (Krull dimension and localizations) Let S 1,, S l be comaximal monoids in A, ie, any ideal meeting all the S i equals A Then KdimA = max { KdimS 1 i A i = 1,, l } Proof Straightforward and constructive 2 Integral extensions Let A B be an integral extension Within classical mathematics the Lying Over is equivalent to the following inclusion for any ideal a: A ab D A (a) The following classical lemma gives a slightly more precise result, without using prime ideals It is easily proven with a determinant trick Lemma 21 Let A B be an integral extension and an ideal a A Then any b ab is integral over the ideal a, ie, As a consequence we get a 1 a, a 2 a 2,,a n a n, b n + a 1 b n a n 1 b + a n = 0 We give now a slight generalization A ab D A (a), 1 + ab (1 + a) sat B Lemma 22 Let A B be an integral extension with an ideal a A, an ideal b B, and a monoid S A Then one has A (b + ab) D A ((A b) + a), S + ab (S + a) sat B

6 6 3 ALGEBRAIC EXTENSIONS Proof Let a A (b + ab) Using Lemma 21 in the integral extension B/b of A/(A b), we deduce that a N meets a + b But a A and a A, so a N meets a + (b A) Let s S + ab We use Lemma 21 in the integral extension S 1 B of S 1 A Since s belongs to (1 + as 1 B) sat B, it belongs also to (1 + as 1 A) sat B, and this implies in B that s belongs to (S + a) sat B Proposition 23 Let A B be an integral extension and a 0,,a d A Then A N B (a 0,,a d ) D A (N A (a 0,,a d )), S B (a 0,, a d ) S A (a 0,,a d ) sat B Proof by induction on d First, A N B (a 0,,a d ) = A ((N B (a 0,,a d 1 ) : a d ) B + Ba d) (definition) D A (A (N B (a 0,,a d 1 ) : a d ) B + Aa d) (Lemma 22) = D A (((A N B (a 0,,a d 1 )) : a d ) A + Aa d) D A ((D A (N A (a 0,,a d 1 )) : a d ) A + Aa d) (induction) = D A ((N A (a 0,,a d 1 ) : a d ) A + Aa d) = D A (N A (a 0,,a d )) (definition) Second, S B (a 0,, a d ) = a N 0 (S B (a 1,,a d ) + Ba 0 ) (definition) a N 0 (S A (a 1,,a d ) sat B + Ba 0 ) (induction) a N 0 (S A (a 1,,a d ) + Ba 0 ) sat B a N 0 (S A (a 1,,a d ) + Aa 0 ) sat B (Lemma 22) ( a N 0 (S A (a 1,,a d ) + Aa 0 ) ) sat B = S A (a 0,,a d ) sat B (definition) Corollary 24 If A B is an integral extension then Kdim A Kdim B The reverse inequality will be shown in a more general setting in Section 3 Proof The constructive meaning of Kdim A Kdim B is the implication KdimB d = KdimA d (for any d 1) The case d = 1 is clear Assume KdimB d with d 0 Let a 0,, a d A, we have 0 S B (a 0,,a d ), so 0 S A (a 0,,a d ) sat B and 0 SA (a 0,, a d ) 3 Algebraic extensions Recall that elements in a ring are comaximal when they generate the ideal 1 Equivalently the monoids generated by these elements are comaximal Definition 31 Let A B be an extension of commutative rings We say that x B is algebraic over A if there exist comaximal elements a 0,,a k A such that i a ix i = 0 We say that B is algebraic over A when any element of B is algebraic over A

7 7 Remark When A is a Bezout domain and B contains the fraction field of A, we find the usual notion of algebraic elements in field extensions But in general it seems that the subset of B made of elements that are algebraic over A is not necessarily a subring of B Lemma 32 If B is algebraic over A then any quotient B/b is algebraic over A/(b A) Lemma 33 Let A be a reduced ring One has N A (x) = Ax + Ann A (x) = (Ax i+1 : x i ) i 1 Let a 0,,a k, x A such that i a ix i = 0 Then N A (a 0 ) N A (a k ) N A (x) + k i=0 Ann A(a i ) N A (x) + Ann A ( a 0,, a k ) In particular if the a i s are comaximal in A and x B A (this means that x is algebraic over A) we get N B (a 0 ) N B (a k ) N B (x) Remark The last point means that the boundary of V B (x) is contained in the union or the boundaries of the V B (a i ) s Proof We write x for Ann A (x) The first point is straightforward In light notation this gives in particular N A (x) = x + x Let us see the second point Let a = N A (x) + i Ann(a i) We write the proof for k = 3 (the general case is similar), ie, a 3 x 3 + a 2 x 2 + a 1 x + a 0 = 0 ( ) We have (1) a 0 Ax N A (x) a from ( ) (2) a 1 a 0 (Ax 2 : x 1 ) a multiplying ( ) by a 0 (3) a 2 a 1 a 0 (Ax 3 : x 2 ) a multiplying ( ) by a 1 a 0 (4) a 3 a 2 a 1 a 0 (Ax 4 : x 3 ) a multiplying ( ) by a 2 a 1 a 0 (5) a 3 a 2 a 1 a 0 a Thus a 0 + a 1 a a 3 a 2 a 1 a 0 a, whence the conclusion since i ( a i + a i ) a 0 + a 1 a a 3 a 2 a 1 a 0 (remark that the right hand side is a sum of 5 = terms and the left hand side a sum of 16 = 2 4 terms) For the last point, since a 0,,a k = 1, Ann B ( a 0,,a k ) = 0 Remark If A is not reduced a similar proof gives N A (a 0 ) N A (a k ) D A (N A (x))+ i (0 : a i ) Theorem 34 If B is algebraic over A then Kdim B Kdim A Corollary 35 Let A B be an integral extension, then Kdim B = Kdim A Proof Follows from proposition 23 and Theorem 34

8 8 REFERENCES Proof of Theorem 34 We prove by induction on n that KdimA n implies KdimB n The case n = 1 is clear Without loss of generality we assume A and B are reduced rings Assume that KdimA n with n 0 For any x B, we have to prove KdimB/N B (x) n 1 As x is algebraic over A, there exist comaximal elements a i of A such that l i=0 a ix i = 0 Lemma 33 gives i N B(a i ) N B (x) Thus Lemma 14 gives ( / ) KdimB/N B (x) Kdim B N B(a i ) = max KdimB/N B (a i ) i i Induction hypothesis applies to B i = B/N B (a i ) and A i = A/(N B (a i ) A) Moreover A i is a quotient of A i = A/N A(a i ), so KdimA i KdimA i and KdimB i KdimA i KdimA i n 1 Remark Let K be the fraction field of a principal ideal domain A Then K is algebraic over A and 0 = KdimK < KdimA = 1 if A K References [Bre] Brenner H Lifting chains of prime ideals J Pure Appl Algebra, 179 (2003), 1 5 [Coq] [CL] [CL2] [CLQ] [CLQ2] [CLR] [CLS] Coquand T Sur un théorème de Kronecker concernant les variétés algébriques C R Acad Sci Paris, Ser I 338 (2004), Coquand T, Lombardi H Hidden constructions in abstract algebra (3) Krull dimension of distributive lattices and commutative rings, in: Commutative ring theory and applications Eds: Fontana M, Kabbaj S-E, Wiegand S Lecture notes in pure and applied mathematics vol 231 M Dekker (2002), Coquand T, Lombardi H A short proof for the Krull dimension of a polynomial ring American Math Monthly 112 (2005), no 9, Coquand T, Lombardi H, Quitté C Generating non-noetherian modules constructively Manuscripta mathematica 115 (2004), Coquand T, Lombardi H, Quitté C Dimension de Heitmann des treillis distributifs et des anneaux commutatifs Preprint 2005 Coquand T, Lombardi H, Roy M-F An elementary characterisation of Krull dimension in: From Sets and Types to Analysis and Topology: Towards Practicable Foundations for Constructive Mathematics (L Crosilla, P Schuster, eds) Oxford University Press (2005) Coquand T, Lombardi H, Schuster P A nilregular element property Archiv der Mathematik 85 (2005), [Duc] Ducos L Vecteurs unimodulaires et systèmes générateurs Journal of Algebra 297 (2006), [Esp] Español L Constructive Krull dimension of lattices Rev Acad Cienc Zaragoza (2) 37 (1982), 5 9 [Esp2] Español L Dimension of boolean valued lattices and rings J Pure Appl Algebra 42 3 (1986), [Joy] Joyal A Le théorème de Chevalley-Tarski Cahiers de Topologie et Géometrie Differentielle, (1975)

9 CONTENTS 9 [Hei] Heitmann, R Generating non-nœtherian modules efficiently Michigan Math 31 2 (1984) [Lom] Lombardi H Dimension de Krull, Nullstellensätze et Évaluation dynamique Math Zeitschrift 242 (2002), [LQY] [MRR] Lombardi H, Quitté C, Yengui I Hidden constructions in abstract algebra (6) The theorem of Maroscia, Brewer and Costa Preprint 2005 Mines R, Richman F, Ruitenburg W A Course in Constructive Algebra Springer-Verlag Universitext (1988) [Ric] Richman F A colon approach to Krull dimension Manuscript December 2004 Contents Introduction 1 1 Constructive Krull dimension 2 11 Krull boundaries 2 12 Characterizations of Krull dimension 3 13 Some basic facts 4 2 Integral extensions 5 3 Algebraic extensions 6 References 8