On Bezout Inequalities for non-homogeneous Polynomial Ideals
aa r X i v : . [ c s . S C ] J a n On B´ezout Inequalities for non-homogeneous PolynomialIdeals ∗ Amir Hashemi ( a ) , Joos Heintz ( b ) , Luis Miguel Pardo ( c ) , Pablo Solern´o ( d ) ( a ) Department of Mathematical Sciences, Isfahan University of Technology,Isfahan, 84156-83111, IranSchool of Mathematics, Institute for Research in Fundamental Sciences (IPM),Tehran, 19395-5746, Iran( b ) Departamento de Computaci´on and ICC, UBA-CONICET,Facultad de Ciencias Exactas y Naturales, Universidad de Buenos Aires,Ciudad Universitaria, 1428, Buenos Aires, Argentina( c ) Depto. de Matem´aticas, Estad´ıstica y Computaci´on Facultad de Ciencias,Universidad de Cantabria, Avda. Los Castros s/n E-39071 Santander, Spain( d ) Departamento de Matem´atica and IMAS, UBA-CONICET,Facultad de Ciencias Exactas y Naturales, Universidad de Buenos Aires,Ciudad Universitaria, 1428, Buenos Aires, ArgentinaE-mail addresses:[email protected], [email protected], [email protected], [email protected] October 18, 2018
Abstract
We introduce a “workable” notion of degree for non-homogeneous polynomial ide-als and formulate and prove ideal theoretic B´ezout Inequalities for the sum of twoideals in terms of this notion of degree and the degree of generators. We computeprobabilistically the degree of an equidimensional ideal.
Keywords.
Non-homogeneous polynomial ideal, ideal degree, idealistic B´ezoutInequality.MSC 13F20, 14A10, 13P10
Introduction
Motivated by the aim to formulate and prove an idealistic version of B´ezout’s Theorem(see [18]) and by applications to transcendence theory (see [15, 4]), the notion of degree of ∗ Partially supported by the following Iranian, Argentinean and Spanish grants: IPM No.95550420(A.H), UBACyT 20020130100433BA and PICT-2014-3260 (J.H), MTM2014-55262-P (L.M.P). omogeneous polynomial ideals became intensively studied. In general this work relied onthe notion of degree of homogeneous polynomial ideals based on the Hilbert polynomial(see [14] for a different view).In this paper we propose an alternative and self-contained approach for non-homogeneouspolynomial ideals leading to specific results which are not simple consequences of their ho-mogeneous counterparts.We introduce a “workable” notion of degree for non-homogeneous polynomial idealsand formulate and prove ideal theoretic B´ezout Inequalities for the sum of two ideals interms of this notion of degree. However it turns out, that, due to the presence of embeddedprimes, a B´ezout Inequality in completely intrinsic terms (depending only on the degreesof the two given ideals) is unfeasable. Hence in some place the degrees of generators ofat least one of the ideals comes into play and our main B´ezout Inequality will be of thismixed type.We finish the paper with a probabilistic algorithm which computes the degree of anequidimensional ideal given by generators. Organization of the paper
The first three sections are devoted to the development of the tools we are going touse in the sequel. The technical highlight is Proposition 19 in the third section whichanticipates in some sense the “correctness” of our ideal theoretic notion of degree (seeTheorem 21 and Definition 22) and the main result of the paper, namely the mixed typeB´ezout Inequality, Theorem 29, in the fourth section.These results become combined in the fifth section with techniques going back toMasser and W¨ustholz [15] in order to estimate the degree of a polynomial ideal in termsof the degrees of generators (Theorem 32).Finally the sixth section contains a probabilistic complexity result concerning the com-putation of the degree of an equidimensional ideal.
Let K be an algebraically closed field, ~X = ( X , . . . , X n ) with X , . . . , X n indeterminatesover K and K [ ~X ] the ring of n − variate polynomials with coefficients in K . The affinespace K n with the Zariski topology is denoted by A n . For any ideal a of K [ ~X ], we denoteby V ( a ) the set of its common zeros in A n and by A := K [ ~X ] / a the associated factor ring.We shall use freely standard notions and notations from Commutative Algebra andaffine Algebraic Geometry. These can be found for example in [1, 16, 17].Less standard is the notion of degree of closed (affine) subvarieties of A n we are goingto use.For an irreducible closed subvariety V of A n we define the degree deg V of V as themaximum number of points that can arise when we intersect V with an affine linearsubspace E of A n such that V ∩ E is finite (observe that it is a nontrivial fact thatdeg V < ∞ holds). The degree deg V of an arbitrary closed subvariety V of A n is the sumof the degrees of the irreducible components of V .It is a remarkable fact that for this notion of degree that for closed subvarieties V and2 an intrinsic B´ezout Inequality holds: deg( V ∩ W ) ≤ deg( V ) . deg( W ) (with intrinsicwe mean that the degree of V ∩ W is estimated in terms of the degrees of V and W only).For more details we refer to [10, 6, 18]. Let a ⊂ K [ ~X ] be an arbitrary ideal of dimension m (i.e. the Krull dimension of the ring A equals m ) such that all its isolated primes have dimension m (or, equivalently, the variety V ( a ) is equidimensional of dimension m ). Under this condition we say that the ideal a is equidimensional of dimension m . Definition 1
Let ~f := f , . . . , f m ′ , m ′ ≤ m , be a sequence of polynomials of K [ ~X ] . Then ~f is called a secant sequence for a of length m ′ if for any index j , ≤ j ≤ m ′ , thedimension of the ideal a + ( f , . . . , f j ) is m − j . Moreover, ~f is called a regular sequenceof length m ′ with respect to a if a + ( f , . . . , f m ′ ) is a proper ideal and if for any index j , ≤ j ≤ m ′ the polynomial f j is not a zero-divisor modulo the ideal a + ( f , . . . , f j − ) . Ifthis is the case the residue classes f + a , . . . , f m ′ + a are said to form a regular sequencein A . Remark 2
In the sequel we shall only consider secant sequences of maximal length m . Observe that for a secant sequence ~f := f , . . . , f m for a , all the isolated componentsof the ideal a + ( f ) have dimension m − f , . . . , f m is a secant sequence for a + ( f ).More generally, for 1 ≤ j < m the polynomials f j +1 , . . . , f m form a secant sequence for a + ( f , . . . , f j ).Any regular sequence of maximal length m with respect to a constitutes a secantsequence for a , because any member of the regular sequence drops the Krull dimension byone at each step, up to reach dimension 0. Proposition 3
Let ~f := f , . . . , f m be a secant sequence for the equidimensional ideal a of dimension m . Let us consider the regular morphism: ~f : V ( a ) −→ A m ,x ~f ( x ) := ( f ( x ) , . . . , f m ( x )) . (1) Then ~f ( V ( a )) is Zariski dense in A m . Said otherwise, ~f : V ( a ) → A m is dominant. Proof.
The proof is an easy consequence of the Theorem of the Dimension of the Fibers(see for instance [17, § ~f is secant, the fiber ~f − (0) isa zero-dimensional algebraic set and, therefore, there exists an irreducible component of V ( a ) which intersects this fiber. If p is the corresponding minimal associated prime of a , theTheorem of the Dimension of the Fibers applied to the restriction ~f | V ( p ) : V ( p ) −→ A m implies that ~f | V ( p ) is dominant. Thus ~f ( V ( p )) is Zariski dense in A m and hence also3 f ( V ( a )).From Proposition 3 we deduce immediately the following statement: Corollary 4
With the same notations as in Proposition 3, the following is a monomor-phism of K − algebras: ~f ∗ : K [ Y , . . . , Y m ] −→ A = K [ ~X ] / a ,Y i f i + a , ≤ i ≤ m. In particular, the residual classes { f + a , . . . , f m + a } are algebraically independent over K . Corollary 5
Let ~f be a secant sequence for the equidimensional ideal a of dimension m . Then there exists a non-empty Zariski open subset U of A m , such that for all ~a :=( α , . . . , α m ) ∈ U ∩ K m the sequence ~f − ~a := f − α , . . . , f m − α m is a secant sequencefor a . Proof.
Fix an index j, ≤ j ≤ m and let ~f j : V ( a ) → A j be the polynomial map( f ( x ) , . . . , f j ( x )). As in the proof of Proposition 3, we deduce that there exists at leastone irreducible component of V ( a ) such that the restriction of ~f j is dominant and itstypical fiber has dimension m − j . On the other hand, if p is an isolated associated primeof a such that the typical fiber of the restriction of ~f j to V ( p ) has not dimension m − j ,then the Zariski closure ~f j ( V ( p )) is properly contained in A j . Therefore, the set of points ~y ∈ A j such that the fiber ~f − j ( ~y ) is m − j dimensional contains a nonempty Zariskiopen set U j ⊂ A j . One verifies immediately that U := m \ j =1 ( U j × A m − j ) ⊂ A m satisfies thestatement of the corollary. Remark 6
There exists in ( A n ) m a nonempty Zariski open set O of linear forms of K [ ~X ] such that each element ~f = f , . . . , f m of O constitutes a secant family for a . Proof.
The statement is a consequence of Noether’s Normalization Lemma as in [10,Lemma 1] applied to the equidimensional variety V ( a ).We need in the sequel the following technical lemma concerning Zariski dense subsets: Lemma 7
Let be given positive integers n , . . . , n s and a subset U of A n × · · · × A n s .For each i = 1 , . . . , s denote by π i the canonical projection of the product space onto A n i .Assume that the set U satisfies the following conditions:i) π ( U ) is Zariski dense in A n .ii) For each i = 1 , . . . , s − and each a := ( a , . . . , a i ) ∈ (cid:0) π × . . . × π i (cid:1) ( U ) the set π i +1 (cid:16) ( { a } × A n i +1 × · · · × A n s ) ∩ U (cid:17) ⊂ A n i +1 is Zariski dense in A n i +1 . hen, the set U is Zariski dense in A n × · · · × A n s . Proof.
By induction on s . For s = 1 there is nothing to prove because condition i )implies already the conclusion (the second condition is vacuous). Suppose now s > i = 1 , . . . , s denote by T i the n i − tuple of coordinates of A n i and suppose that apolynomial F ( T , . . . , T s ) vanishes on U . Consider F as polynomial in the variables T s : F = X α ∈ N ns f α ( T , . . . , T s − ) T αs . For an arbitrary point a := ( a , . . . , a s − ) ∈ (cid:0) π × . . . × π s − (cid:1) ( U ) the n s − variate polynomial F ( a, T s ) vanishes at any point b ∈ A n s with ( a, b ) ∈ U . In other words, the polynomial F ( a, T s ) vanishes on the set π s (cid:0) ( { a }× A n s ) ∩ U (cid:1) , which is Zariski dense in A n s by condition ii ) for i = s −
1. Hence the coefficients f α ( a ) are zero for all subindexes α and for all a ∈ (cid:0) π × . . . × π s − (cid:1) ( U ).We consider now the set U ′ ⊂ A n × · · · × A n s − defined as U ′ := (cid:0) π × . . . × π s − (cid:1) ( U ) . It is easy to see that U ′ satisfies conditions i ) and ii ). Hence, by induction hypothesis, theset U ′ is Zariski dense and therefore each polynomial f α must be identically zero. Hence F = 0, which implies that U is Zariski dense in A n × · · · × A n s . Corollary 8
There exists in ( A n +1 ) m a nonempty Zariski open set O of polynomials ofdegree one such that each element ~f = f , . . . , f m of O constitutes a secant sequence for a . Proof.
The condition to form a secant sequence is expressible in first order logic and henceconstructible. From Corollary 5, Remark 6 and Lemma 7 we deduce that the sequences ~f = f , . . . , f m of degree one polynomials which are secant sequences for a constitutes aZariski dense subset of ( A n +1 ) m . Since this set is also constructible the corollary follows.We are now going to analyze the ubiquity of regular sequences with respect to anequidimensional polynomial ideal a ⊂ K [ ~X ] of positive dimension. We start with thefollowing simple observation. Remark 9
Let p be a prime ideal of the polynomial ring K [ ~X ] of positive dimension.Then the set T := { ( a , . . . , a n ) ∈ K n | there exists λ ∈ K such that a X + · · · + a n X n + λ ∈ p } is a proper linear subspace of K n . Proof.
The set T is clearly a linear subspace because p is an ideal. If T = K n theelements of the canonical basis of K n belong to T . Hence, there exist scalars λ , . . . , λ n such that X + λ , . . . , X n + λ n belong to p . Thus the ideal p is maximal. This contradictsthe assumption that p is of positive dimension.From Remark 9 and Lemma 7 we deduce now the following result about the densityof degree one regular sequences with respect to an equidimensional polynomial ideal.5 roposition 10 Let a ⊂ K [ ~X ] be an equidimensional ideal of dimension m > . Thenthere exists a Zariski dense subset U of ( A n +1 ) m such that for all ( ~a , . . . , ~a m ) ∈ U with ~a i := ( a i , . . . , a ni , a ( n +1) i ) , ≤ i ≤ m , the polynomials ℓ , . . . , ℓ m of degree defined by ℓ i := a i X + · · · + a ni X n + a ( n +1) i form a regular sequence with respect to the ideal a . Proof.
We start by the construction of a suitable regular generic polynomial of degree1. Let p , . . . , p t be the associated primes of a of positive dimension and let p t +1 , . . . , p r those associated primes which are maximal ideals. Observe 1 ≤ t since the ideal a isof positive dimension. For each j = 1 , . . . , t let T j be the proper linear subspace of A n associated to p j following Remark 9. Thus U := A n \ S tj =1 T j is a constructible Zariskidense subset of A n . Observe that for 1 ≤ j ≤ t and any (homogeneous) linear form ℓ whose coefficients belong to U , the constructible set ℓ ( V ( p j )) is Zariski dense in A .Thus, the intersection U ℓ := T sj =1 − ℓ ( V ( p j )) is constructible and Zariski dense too. Hence U := { ( ℓ, u ) ; ℓ ∈ U , u ∈ U ℓ } is a constructible subset of A n +1 which is Zariski densefollowing Lemma 7.Now, for each maximal ideal p j , t < j ≤ r , associated to a we consider W j ⊂ A n +1 , the n − dimensional linear subspace of the polynomials of degree one contained in p j . Since U is Zariski dense and constructible in A n +1 , we conclude that U := U \ r [ j = t +1 W j is constructible and Zariski dense in A n +1 .Now we consider an arbitrary polynomial ℓ := a X + · · · a n X n + a n +1 with ( a , . . . , a n +1 ) ∈U . Then, ℓ / ∈ p j for t < j ≤ r since the vector ( a , . . . , a n +1 ) does not belong to ∪ rj = t +1 W j . On the other hand, we have ℓ / ∈ p j for all 1 ≤ j ≤ t because the homogeneouspart of ℓ is in U .In other words, for ℓ ∈ U we infer that ℓ does not belong to any associated primeof a . Thus ℓ is not a zero divisor mod a .In order to prove that ℓ is a regular element it suffices to show that ℓ is not a unitymodulo a . Otherwise ℓ must be also a unity mod p , i.e. ℓ ( p ) = 0 for all p ∈ V ( p ) andthen − a n +1 is not in the image ℓ ( V ( p )) which contradicts the fact that a n +1 ∈ U ℓ (recallthat 1 ≤ t holds).Summarizing, we have shown that there exists a (constructible) Zariski dense subset U of A n +1 such that any polynomial of degree one with coefficients in U is regular mod a . Now the corollary follows by an inductive argument based on Lemma 7: take any ℓ with coefficients in U . Then by Krull’s Principal Ideal Theorem, the ideal a := a + ( ℓ )is equidimensional of dimension m −
1. If m = 1 there is nothing to prove.Assume m >
1. We build a constructible Zariski dense subset U ,ℓ of A n +1 (dependingon ℓ ) such that for any ℓ with coefficients in this set, the pair ℓ , ℓ is a regular sequenceof length 2 for a . By repeating this argument for each ℓ , Lemma 7 ensures the existence ofa Zariski dense subset U of A n +1 × A n +1 representing the coefficients of regular sequences6f length 2 for a , formed by polynomials of degree 1. The corollary follows now inductivelyafter m steps. Throughout this section let notations and assumptions be the same as in the previoussection. Let us now consider a minimal primary decomposition of the equidimensionalideal a , of dimension m , namely a = q ∩ · · · ∩ q r , where each ideal q i is p i − primary. Assume ~f is a secant sequence with respect to the ideal a . Let us denote by K [ ~f ] the image of K [ Y , . . . , Y m ] in A = K [ X , . . . , X n ] / a .Let us recall the polynomial mapping ~f : V ( a ) −→ A m and let ~f i be its restriction toeach of the components V ( p i ): ~f i := ~f | V ( p i ) : V ( p i ) −→ A m , where 1 ≤ i ≤ r. Taking into account Proposition 3 we may assume without loss of generality that thereexists an index 1 ≤ s ≤ r such that ~f i is dominant if and only if 1 ≤ i ≤ s . Lemma 11
With these notations the following statements hold:i) For any i , s + 1 ≤ i ≤ r , there exists a non-zero polynomial Q i ∈ K [ Y , . . . , Y m ] suchthat Q i ( f , . . . , f m ) belongs to q i .ii) For any i , ≤ i ≤ s , and for any j , ≤ j ≤ n , there is a polynomial Q ij ∈ K [ Y , . . . , Y m ][ T ] , of positive degree in the variable T such that Q ij ( f , . . . , f m )( X j ) belongs to q i . Proof.
For the proof of statement i ), observe that for any i , s + 1 ≤ i ≤ r , the image ~f ( V ( p i )) is contained in a proper hypersurface of A m . Taking G i ∈ K [ Y , . . . , Y m ] theminimal equation of this hypersurface, we conclude that G i ( f , . . . , f m ) belongs to p i .Since the radical of q i is p i , there exists some power Q i := G k i i such that Q i ( f , . . . , f m )belongs to q i , as wanted.For the proof of statement ii ), let us write by p := p i where 1 ≤ i ≤ s . As ~f i ( V ( p ))is Zariski dense in A m , and a is m − equidimensional we have a monomorphism of finitelygenerated K − algebras of the same Krull dimension: ~f ∗ i : K [ Y , . . . , Y m ] ֒ → K [ X , . . . , X n ] / p , with ~f ∗ i ( Y k ) = f k + p , 1 ≤ k ≤ m . Thus we may consider K [ ~f ] as a subalgebra of K [ X , . . . , X n ] / p . Let S := K [ ~f ] \ { } , Then K ( ~f ) := S − K [ ~f ] is the quotient field of K [ ~f ] and we have an extension of domains of the same Krull dimension: K ( ~f ) ֒ → S − ( K [ X , . . . , X n ] / p ) , which is therefore a finite field extension. Then, for any 1 ≤ j ≤ n the residue class X j + p is algebraic over K ( ~f ). In particular, there is a polynomial H j ∈ K [ Y , . . . , Y m ][ T ]7depending on p ), of positive degree in T such that H j ( f , . . . , f m )( X j ) belongs to p . As p = p i , let us write G ij := H j . Since p i is the radical of q i , there is some power Q ij := G k ij ij such that Q ij ( f , . . . , f m )( X j ) belongs to q i as wanted.For the next statement observe that Corollary 4 implies that for any nonzero poly-nomial q ∈ K [ Y , . . . , Y m ] we have q ( ~f ) = q ( f , . . . , f m ) = 0. Thus K [ ~f ] \ { } formsa multiplicative closed set of A which we denote in the sequel by S ( ~f ). Observe that K ( ~f ) := S ( ~f ) − K [ ~f ] is the quotient field of K [ ~f ]. Proposition 12
There is a nonzero polynomial q ∈ K [ Y , . . . , Y m ] such that for q ( ~f ) = q ( f , . . . , f m ) ∈ S ( ~f ) the localizations of K [ ~f ] and A by q ( ~f ) define an integral ring exten-sion: K [ ~f ] q ( ~f ) ֒ → A q ( ~f ) . In particular A q ( ~f ) is a finite K [ ~f ] q ( ~f ) − module. Proof.
We use the notations of Lemma 11. Let us consider the polynomial q ∈ K [ Y , . . . , Y m ]given by q := s Y i =1 n Y j =1 a ij ( Y , . . . , Y m ) × r Y i = s +1 Q i ( Y , . . . , Y m ) ! ∈ K [ Y , . . . , Y m ] , where for 1 ≤ i ≤ s and 1 ≤ j ≤ n the polynomial a ij is the non-zero leading coefficient ofthe polynomial Q ij ∈ K [ Y , . . . , Y m ][ T ] which satisfies the condition a ij ( f , . . . , f m ) p i (see the proof of Lemma 11). As q is a non-zero polynomial we have q ( ~f ) = q ( f , . . . , f m ) a . Thus we obtain a K − algebra extension K [ ~f ] q ( ~f ) ֒ → A q ( ~f ) . As q ( ~f ) ∈ q i for s + 1 ≤ i ≤ r , the minimal primary decomposition of the ideal (0) in A q ( ~f ) is given by: (0) = q e ∩ · · · ∩ q es , where q ei is the extension of q i to A q ( ~f ) . In particular, the ring extension above is integralsince for every 1 ≤ j ≤ n the residue class X j + a satisfies the algebraic dependenceequation given by Q si =1 Q ij described in Lemma 11. In particular, as A q ( ~f ) is a finitelygenerated K [ ~f ] q ( ~f ) − algebra and as every residue class X j + a , 1 ≤ j ≤ n is integral over K [ ~f ] q ( ~f ) , the algebra A q ( ~f ) is a finite K [ ~f ] q ( ~f ) − module.We deduce from Proposition 12 the following statement. Corollary 13
The localized ring S ( ~f ) − A is a K ( ~f ) − algebra of finite dimension. roposition 14 There is a nonzero polynomial p ∈ K [ Y , . . . , Y m ] such that p ( f ) := p ( f , . . . , f m ) ∈ S ( ~f ) has the following property. The localizations by p ( f ) define an integralring extension: K [ ~f ] p ( ~f ) ֒ → A p ( ~f ) , and A p ( ~f ) is a free K [ ~f ] p ( ~f ) − module of finite rank. Moreover, its rank satisfies the condi-tion rank K [ ~f ] p ( ~f ) ( A p ( ~f ) ) = dim K ( ~f ) S ( ~f ) − A. Proof.
Let q ∈ K [ Y , . . . , Y m ] be the non-zero polynomial of Proposition 12. Then, wehave • a finite subset β of A , such that β is a basis of S ( ~f ) − A as K ( ~f ) − vector space. • a finite subset M of A , such that M is a system of generators of the K [ ~f ] q ( ~f ) − module A q ( ~f ) .As β is a basis of S ( ~f ) − A as K ( ~f ) − vector space, there is some nonzero polynomial h ∈ K [ Y , . . . , Y m ] such that h ( ~f ) := h ( f , . . . , f m ) = 0 in A and such that all elementsin M are linear combinations of the elements in the basis β with coefficients in K [ ~f ] h ( ~f ) .Then, p := qh ∈ K [ Y , . . . , Y m ] is a non-zero polynomial such that K [ ~f ] p ( ~f ) ֒ → A p ( ~f ) is an integral ring extension and such that A p ( ~f ) is a free K [ ~f ] p ( ~f ) − module of finite rank.As rank is stable under localizations, we concluderank K [ ~f ] p ( ~f ) ( A p ( ~f ) ) = dim K ( ~f ) S − A. Lemma 15
Suppose that ~f is a regular sequence. Then the inequality dim K A/ ( ~f ) ≤ dim K ( ~f ) S ( ~f ) − A holds. Proof.
By induction on the Krull dimension m of A .If m = 0 there is nothing to prove. Suppose m ≥ B := A/ ( f ).Since ~f is a secant sequence, the ring B has Krull dimension m − ~f ′ := f , . . . , f m is a regular sequence for the ideal a + ( f ). Therefore, if T denotes the multiplicativesubset K [ ~f ′ ] \ { } of B and L ′ := K ( ~f ′ ) we have by induction hypothesis the inequalitydim K B/ ( ~f ′ ) ≤ dim L ′ T − B .Since B/ ( ~f ′ ) = A/ ( ~f ) it suffices to show the inequality dim L ′ T − B ≤ dim K ( ~f ) S − A .Let g , . . . , g s be polynomials in K [ ~X ] whose classes form a basis of T − B over L ′ . Weconsider the images of g , . . . , g s in the ring S ( ~f ) − A . We show that these elements arelinearly independent over K ( ~f ). Suppose on the contrary that they are K ( ~f ) − linearlydependent. Cleaning denominators we may suppose that there exists a non trivial lin-ear combination P si =1 p i g i which belongs to the ideal a where the p i ’s are polynomials in9 [ ~f ]. Since the polynomials f , . . . , f m are algebraically independent modulo a , f ∈ K [ ~f ]may be viewed as an irreducible element in a factorial domain. Moreover, since f is notzero-divisor modulo a we may suppose without loss of generality that f is not a commonfactor of all polynomials p , . . . , p s . Taking the class of each polynomial p i modulo theideal ( f ) ⊂ K [ ~f ] we conclude that the polynomials g i are not linearly independent in T − B , contradiction.Lemma 15 fails for secant sequences: Example 16
Consider the one dimensional ideal a = ( X , X X ) ⊂ K [ X , X ] whoseprimary decomposition is a = ( X ) ∩ ( X , X ) . The polynomial X is a secant sequencefor a and A/ ( X ) = K [ X , X ] / ( X , X ) ≃ K [ X ] / ( X ). Therefore dim K A/ ( X ) = 3.On the other hand, if S := K [ X ] \ { } and L := K ( X ), we have S − A ≃ S − K [ X , X ] / ( X , X ) ≃ K ( X )[ X ] / ( X )(since X is invertible in S − A ). But then dim L S − A = 2.Observe that X is a secant family for a but it is a zero divisor in A since X X ∈ a and X / ∈ a .Typically we have equality in Lemma 15. Proposition 17
Let ~f be a secant sequence for a , as before. Then there exists a non-empty Zariski open subset O of A m such that for any ~a ∈ O the equality dim K A/ ( ~f − ~a ) =dim K ( ~f ) S ( ~f ) − A holds. Proof.
By Proposition 14, there exists a non-zero polynomial p ∈ K [ Y , . . . , Y m ] suchthat the localizations by p ( ~f ) define an integral ring extension K [ ~f ] p ( ~f ) ֒ → A p ( ~f ) and A p ( ~f ) is a free K [ ~f ] p ( ~f ) − module of finite rank N := rank K [ ~f ] p ( ~f ) ( A p ( ~f ) ) = dim K ( ~f ) S ( ~f ) − A. Let H := { p = 0 } ⊂ A m and let U be the nonempty Zariski open subset of A m ofCorollary 5. Then, O := H ∩ U is also a non-empty Zariski open subset of A m . Let ~a ∈ O be a point in this open set and let us denote by m ~a the ideal in K [ ~f ] generated by thesequence f − a , . . . , f m − a m in K [ ~f ]. As K [ ~f ] is a polynomial ring, m ~a is a maximalideal in K [ ~f ]. Let us denote by m ~a A its extension to A . Moreover, as p ( ~a ) = 0, we have p ( ~f ) m ~a . In particular, the extension m (1) ~a := m ~a K [ ~f ] p ( ~f ) is also a maximal ideal in K [ ~f ] p ( ~f ) . Let us consider the submodule m (1) ~a A p ( ~f ) of A p ( ~f ) as K [ ~f ] p ( ~f ) − module. Notethat ( A/ m ~a A ) p ( ~f ) is isomorphic as K [ ~f ] p ( ~f ) − module to (cid:16) K [ ~f ] p ( ~f ) (cid:17) N / m (1) ~a (cid:16) K [ ~f ] p ( ~f ) (cid:17) N . K [ ~f ] p ( ~f ) − module, ( A/ m ~a A ) p ( ~f ) is isomorphic to (cid:16) K [ ~f ] p ( ~f ) / m (1) ~a (cid:17) N ∼ = (cid:16) K [ ~f ] / m ~a (cid:17) N ∼ = K N . Hence dim K A/ ( ~f − ~a ) = N = dim K ( ~f ) S ( ~f ) − A for all ~a ∈ O . Corollary 18
There exists a Zariski dense subset U of ( A n +1 ) m such that for any sequence ~f = f , . . . , f m of degree one polynomials of U the sequence ~f is secant and the equality dim K ( ~f ) S ( ~f ) − A = dim K A/ ( ~f ) holds. Proof.
Combine Corollary 8 with Proposition 17 and Lemma 7.Let 1 ≤ q ≤ r be the number of isolated primes of the ideal a . Without loss ofgenerality we may assume that these are p , . . . , p q . Then for each 1 ≤ j ≤ q the ring A p j is local and Artinian with maximal ideal ( p j / a ) p j . In the sequel we denote by ℓ j thelength of A p j .As in Lemma 11 we may assume without loss of generality that there exists an index1 ≤ s ≤ q such that ~f i = ~f | V ( p i ) : V ( p i ) → A m is dominant if and only if 1 ≤ i ≤ s .We say that the ring extension K ( ~f ) ⊆ S ( ~f ) − A is separable if for every 1 ≤ i ≤ s thefield extension K ( ~f ) ⊆ S ( ~f ) − A/ ( p i / a ) is separable (compare Corollary 13).With these notions and notations we may formulate the following result. Proposition 19 i) Assume that the ring extension K ( ~f ) ⊆ S ( ~f ) − A is separable. Then we have dim K ( ~f ) S ( ~f ) − A ≤ s X i =1 deg V ( p i ) ℓ i ! m Y k =1 deg( f k ) ≤≤ q X j =1 deg V ( p j ) ℓ j m Y k =1 deg( f k ) . ii) There exists a nonempty Zariski open subset O of ( A n +1 ) m such that for any sequence ~f = f , . . . , f m of degree one polynomials of O , the sequence ~f is secant for a andsuch that dim K ( ~f ) S ( ~f ) − A = q X j =1 deg V ( p j ) ℓ j holds. Proof.
We are going to show statement i ). Let us abbreviate L := K ( ~f ), S := S ( ~f ) and R := S ( ~f ) − A = S − A .By virtue of Lemma 11 i ) the localizations m i := S − ( p i / a ), 1 ≤ i ≤ s , are exactlythe maximal ideals of R . By Corollary 13 the L − algebra R is finite dimensional and11herefore Artinian. From the Chinese Remainder Theorem we deduce now R = s M i =1 R m i .Let 1 ≤ i ≤ s and observe that the rings R m i and A p i are isomorphic.Therefore ℓ i is also the length of the Artinian local ring R m i . This implies dim L R m i =[ R/ m i : L ] ℓ i .Putting all this together we obtaindim L S − A = dim L R = s X i =1 [ R/ m i : L ] ℓ i . (2)Fix again 1 ≤ i ≤ s . We are now going to analyze in geometric terms the quantity[ R/ m i : L ].Observe that the dominating morphism ~f i = ~f | V ( p i ) : V ( p i ) → A m induces canonicalfield isomorphisms R/ m i ∼ = S − ( A/ ( p i / a )) ∼ = K ( V ( p i )) , where K ( V ( p i )) denotes the fraction field of V ( p i ). Since by assumption the field extension K ( ~f ) ⊂ S − ( A/ ( p i / a )) is separable, we see that ~f i is generically unramified. In particular,we are in conditions to apply [17, Theorem 4, Chapter II, § W of A m such that for any point ~a ∈ W the fiber ~f − i ( ~a ) isunramified and of cardinality [ R/ m i : L ].Choose ~a ∈ W . From the B´ezout Inequality [10] we deduce now[ R/ m i : L ] = ♯ ~f − i ( ~a ) ≤ deg V ( p i ) m Y k =1 deg( f k ) . From (2) we infer finally the statement i ) of the proposition, namelydim K ( ~f ) S ( ~f ) − A = dim L S − A = s X i =1 ♯ ~f − i ( ~a ) ℓ i ≤≤ s X i =1 deg V ( p i ) ℓ i ! m Y k =1 deg( f k ) ≤ q X j =1 deg V ( p j ) ℓ j m Y k =1 deg( f k ) . We are now going to prove statement ii ) of the proposition.Following Corollary 8 and [10, Lemma 1] we may choose a nonempty Zariski opensubset of ( A n +1 ) m such that for any sequence ~f = f , . . . , f m of degree one polynomials of O the following conditions are satisfied: • ~f is a secant sequence for a ; • for 1 ≤ j ≤ q the morphisms ~f j := ~f | V ( p j ) : V ( p j ) → A m are dominant, genericallyunramified and of degree deg V ( p j ).Thus, in particular, any sequence ~f = f , . . . , f m of O fulfills the assumption of thestatement i ) of the proposition. 12rom the proof of statement i ) we deduce:dim K ( ~f ) S ( ~f ) − A = q X j =1 ♯ ~f − j ( ~a ) ℓ j for a suitable, generically chosen point ~a ∈ A m . Since f j is generically unramified of degreedeg V ( p j ) we may choose ~a such that ♯ ~f − j ( ~a ) = deg V ( p j ) holds for 1 ≤ j ≤ q . This impliesthe statement ii ) of the proposition, namely dim K ( ~f ) S ( ~f ) − A = P qj =1 deg V ( p j ) ℓ j . As at the end of the last section let be given an equidimensional ideal a of K [ ~X ] ofdimension m with isolated primes p , . . . , p q . Let A := K [ ~X ] / a , recall that for 1 ≤ j ≤ q the ring A p j is local and Artinian and let ℓ j be the length of A p j .We shall need the following technical result. Lemma 20
There exists a nonempty Zariski open subset U of ( A n +1 ) m such that for anysequence ~f = f , . . . , f m of degree one polynomials of U , the sequence ~f is secant for a and dim K A/ ( ~f ) is constant, independently from ~f . Proof.
Let T ij , 1 ≤ i ≤ m , 0 ≤ j ≤ n be new indeterminates, ~T = ( T ij ) ≤ i ≤ m ≤ j ≤ n and let F i := P mj =1 T ij X j + T i , 1 ≤ i ≤ m . Fix any monomial order of ~X and let g , . . . g s ∈ K [ ~X ]be a set of generators of a . Observe that the ideal ( g , . . . g s , F , . . . , F m ) of K ( ~T )[ ~X ] iszero-dimensional.Consider an arbitrary Gr¨obner basis computation β of this ideal.The leading coefficients occurring in β form a finite set of nonzero rational functionsof K ( ~T ). Hence, there exists a nonempty Zariski open subset U of A ( n +1) m where none ofthe numerators and denominators of these rational functions vanishes.Let ~f = f , . . . , f m be a sequence of degree one polynomials of U . Then β may bespecialized to a Gr¨obner basis computation of ( g , . . . g s , f , . . . , f m ) in K [ ~X ] which yieldsthe stair of β .In view of Corollary 8 we may assume without loss of generality that for every sequence ~f = f , . . . , f m of degree one polynomials of U the sequence ~f is secant for a .Therefore everything is well defined and dim K A/ ( ~f ) is finite and constant on U . Theorem 21
There exists a nonempty Zariski open subset O of ( A n +1 ) m such that forany sequence ~f = f , . . . , f m of degree one polynomials of O the sequence ~f is secant for a and dim K A/ ( ~f ) = P qj =1 deg V ( p j ) ℓ j holds. Proof.
Combining Proposition 19 ii ) with Lemma 20 we find a nonempty Zariski opensubset O of ( A n +1 ) m such that for any sequence ~f = f , . . . , f m of degree one polynomials of O the sequence ~f is secant for a such that dim K A/ ( ~f ) is constant and dim K ( ~f ) S − ( ~f ) A = P qj =1 deg V ( p j ) ℓ j holds. 13rom Corollary 18 we conclude that there exists a sequence ~f = f , . . . , f m of degreeone polynomials of O satisfying the equalitydim K ( ~f ) S − ( ~f ) A = dim K A/ ( ~f ) . This implies dim K A/ ( ~f ) = q X j =1 deg V ( p j ) ℓ j . Hence, for an arbitrary sequence ~f = f , . . . , f m of degree one polynomials belonging to O the sequence ~f is secant for a and it holdsdim K A/ ( ~f ) = dim K A/ ( ~f ) = q X j =1 deg V ( p j ) ℓ j . We may simplify the somewhat complicated formulation of Theorem 21 saying that ageneric sequence ~f = f , . . . , f m of degree one polynomials is secant for a and dim K A/ ( ~f ) = P qj =1 deg V ( p j ) ℓ j holds. In this sense the word generic refers always to the existence of anonempty Zariski open set which not always is made explicit.Using this terminology we may define the degree of the equidimensional ideal a in twodifferent ways as follows. Definition 22
The degree deg ( a ) of the equidimensional ideal a may be equivalently de-fined as i ) deg( a ) := dim K A/ ( ~f ) for a generic sequence ~f = f , . . . , f m of degree one polyno-mials of K [ ~X ] or ii ) deg( a ) := P qj =1 deg V ( p j ) ℓ j , where p , . . . , p q are the isolated primes of a and ℓ , . . . , ℓ q are the lengths of the Artinian local rings A p , . . . , A p q .The formulation ii ) for the degree of a was introduced in [2] for homogeneous ideals whereasthe formulation i ) seems new for non-homogeneous ideals and represents our “workable”notion of degree (see Section 6). The next statement is a straightforward consequence of this definition:
Proposition 23
Let a and b two equidimensional ideals in the ring K [ ~X ] of the samedimension m . If a ⊆ b then deg( b ) ≤ deg( a ) . Proof.
Since the ideals have the same dimension we can take the same generic degree onepolynomials ~f for both ideals. Thus a + ( ~f ) ⊆ b + ( ~f ) and the proposition follows.If the ideal a is generated by a single polynomial its degree agrees with the total degreeof the polynomial which generates it. 14 roposition 24 Let a be the ideal generated by a non constant polynomial g . The deg( a ) = deg( g ) . Proof.
It suffices to observe that after a generic linear change of coordinates, the degreeof the polynomial g does not change if n − g .In the case of general polynomial ideals, as customary, we extend our notion of degreeas follows. Definition 25
Let I ⊂ K [ ~X ] be an arbitrary proper polynomial ideal with isolated primarycomponents q , . . . , q t . We define: deg( I ) := t X h =1 deg( q h ) . Remark 26
For radical polynomial ideals Definitions 22 and 25 coincide with the usualnotions of geometric degree of (equidimensional or arbitrary) algebraic closed subvarietiesof affine spaces following [10].
In view of the B´ezout Inequality [10, 6, 18] for affine varieties one might expect that forarbitrary ideals
I, J ⊂ K [ ~X ] the following estimation holds:deg( I + J ) ≤ deg( I ) . deg( J ) . That this may become wrong shows the following example.
Example 27
Consider the one-dimensional ideal I = ( X , X X ) ⊂ K [ X , X ] whoseprimary decomposition is I = ( X ) ∩ ( X , X ). The primary isolated component is ( X )while ( X , X ) is the embedded one. By Definition 22 we have deg( I ) = deg(( X )) = 2,since for any generic linear polynomial f := aX + bX + c the ring K [ ~X ] / ( X , f ) hasthe K -basis { , X } .Take the ideal J := ( X ) ⊂ K [ ~X ], which has degree one, and consider the degree of thesum I + J . We have I + J = ( X , X X , X ) = ( X , X ) , which is a 0-dimensional ( X , X )-primary ideal. Clearly deg( I + J ) = 3.But on the other hand we have deg( I ) . deg( J ) = 2 . < Example 28
Let k ∈ N and I = ( X k , X X ) ⊂ K [ X , X ]. It is easy to see that theprimary decomposition of I is: I = ( X ) ∩ ( X k , X ) and then deg( I ) = deg(( X )) = 1.By adding the ideal ( X ) we havedeg( I + ( X )) = deg( X k , X X , X ) = deg( X k , X ) =15 length ( K [ X , X ] / ( X k , X )) = k. Observe that the degree of the sum of the ideals depends on the degree of the generatorsof I but not on the degree of I .Nevertheless, Proposition 19 implies the following B´ezout-type Inequality for equidi-mensional ideals. Theorem 29
Let K be of characteristic zero and let a ⊂ K [ ~X ] be an equidimensionalideal of dimension m > . Let f , . . . , f k be a regular sequence, not necessarily maximal,for a . Then the inequality deg( a + ( f , . . . , f k )) ≤ deg( a ) k Y i =1 deg( f i ) holds. Proof.
Since a is assumed equidimensional and f , . . . , f k is a regular sequence, Krull’sPrincipal Ideal Theorem implies that ideal b := a + ( f , . . . , f k ) is also equidimensional.Combining Proposition 10 with Theorem 21 we see that there exists a regular sequence f k +1 , . . . , f m of degree one polynomials such that deg( b ) = dim K K [ ~X ] / b + ( f k +1 , . . . , f m )holds.Let S := K [ f , . . . , f k , f k +1 , . . . , f m ] \ { } and L := K ( f , . . . , f k , f k +1 , . . . , f m ), thenLemma 15 states the inequalitydeg( b ) = dim K A/ ( f , . . . , f k , f k +1 , . . . , f m ) ≤ dim L S − ( A ) . On the other hand, taking into account that the characteristic of K is zero, from Propo-sition 19 i ) we deducedim L S − ( A ) ≤ q X j =1 deg( V ( p j )) ℓ j m Y i =1 deg( f i ) = deg( a ) k Y i =1 deg( f i ) . This implies the theorem.
The constructions in this section are inspired by [15].Let a be an arbitrary non-zero proper ideal of the polynomial ring R := K [ ~X ] where K is an algebraically closed field of characteristic zero and ~X := ( X , . . . , X n ) is a set ofvariables. Denote by r := ht( a ) the height of the ideal a .From a primary decomposition of a we obtain a decomposition of a as follows: a = Q r ∩ Q r +1 ∩ · · · ∩ Q n ∩ I , j = r, . . . , n the ideal Q j is the intersection of all isolated primary compo-nents of a having height j , or the whole ring R otherwise. The ideal I is the intersectionof the embedded primary components.For any j = r, . . . , n such that Q j = R , let Q j = T s j i =1 q ji be its primary decomposition.Observe that Q j is unmixed of height j . a In this section we introduce suitable simple multiplicative sets such that the respectivelocalizations detect each equidimensional component of the ideal a (see Proposition 30below).With the previous notations, for each pair ( k, ℓ ) such that Q k = R and 1 ≤ ℓ ≤ s k , thereexists a point z kℓ ∈ A n lying in the variety V ( q kℓ ) but outside the union of the remainingirreducible components V ( q ji ), with j = k or i = ℓ if j = k , and the immerse variety V ( I ). In purely idealistic terms, there exists a maximal ideal m kℓ such that q kℓ ⊆ m kℓ but q ji * m kℓ for all the other pairs ( j, i ) and I * m kℓ .For each index j = r, . . . , n such that a has isolated components of height j we introducethe multiplicative set S j := R \ s j [ i =1 m ji . If there is no isolated component of a with height j we define S j := R \ { } .From q kℓ ⊆ m kℓ we infer q kℓ ∩ S k = ∅ . On the other hand, for j = k , we have q kℓ ∩ S j = ∅ . If S j = R \ { } this is obvious. If S j = R \ { } the assumption q kℓ ∩ S j = ∅ implies the inclusion q kℓ ⊂ S s j i =1 m ji . By [1, Proposition 1.11], there exists then an index i with q kℓ ⊂ m ji , in contradiction with the choice of the maximal ideals and j = k . Asimilar argument shows I ∩ S j = ∅ for all j = r, . . . n . Namely, if I is disjoint from S j then I must be included in suitable maximal ideal m ji , which again contradicts the choice ofthe maximal ideals.With these considerations we have Proposition 30
For any index k = r, . . . , n the equality S − k ( a ) = S − k ( Q k ) holds in the fraction ring S − k R .In particular, if Q k = R , the ideal S − k ( a ) is unmixed of height k (or equivalentlyunmixed of dimension n − k ). Proof.
By the previous arguments we have S − k ( a ) = S − k ( Q r ) ∩ S − k ( Q r +1 ) ∩ · · · ∩ S − k ( Q n ) ∩ S − k ( I ) == \ ji S − k ( q ji ) ∩ S − k ( I ) = \ ki S − k ( q ki ) = S − k ( Q k ) . Thus, in case Q k = R , the ideal S − k ( a ) is unmixed of height k because Q k has thisproperty. 17 .2 A suitable local regular sequence contained in a Let ~g := g , . . . , g s be a system of generators of a with degrees D ≥ D ≥ · · · ≥ D s ,respectively. Since a is assumed generated by s many polynomials, Krull’s Principal IdealTheorem (see [1, Corollary 11.16]) implies that in the primary decomposition of a onlyunmixed components Q k with k ≤ s may appear. Lemma 31
Fix an index k = r, . . . , n with Q k = R . Then, there exist polynomials p , . . . , p k ∈ a such that for all j , ≤ j ≤ k , the following conditions are satisfied:i) The polynomial p j is a generic linear combination of the polynomials g j , . . . , g s (inparticular, deg( p j ) = deg( g j ) = D j ).ii) p , . . . , p j is a regular sequence in the localized ring S − k ( R ) .iii) If a j := ( p , . . . , p j ) and a ∗ j := S − k ( a j ) ∩ R , the ideal a ∗ j is an unmixed ideal of height j in R .iv) The inequality deg( a ∗ j ) ≤ deg( a ∗ j − ) D j holds. Proof.
The proof runs by induction on j , where 1 ≤ j ≤ k .For j = 1 let p be a generic linear combination of the generators g , . . . , g s . Clearly p satisfies the conditions i ) , ii ) , iii ) for j = 1 because p is not invertible in S − k ( R ) sinceit belongs to the unmixed ( n − k )-dimensional ideal S − k ( a ) (see Proposition 30). Remarkthat, even if condition iv ) is vacuous for j = 1, the inequality deg( a ∗ ) ≤ deg( p ) holds (seePropositions 23 and 24).Suppose now that the lemma holds for 1 ≤ j < k and let p , . . . , p j be polynomialsverifying conditions i ) − iv ). Let S − k ( a j ) = q \ i =1 h i be a primary decomposition of the ideal S − k ( a j ) in the ring S − k R . Remark that allthe primary components in this decomposition are isolated and ( n − j )-dimensional be-cause S − k R is a Cohen-Macaulay ring and S − k ( a j ) is generated by the regular sequence p , . . . , p j .For each i = 1 , . . . , q consider the linear subspace T i := { µ ∈ K s − j − | X t ≥ j +1 µ t g t ∈ p h i } , where √ h i denotes the radical of h i . If for some i , T i is the whole space K s − j − ,then g j +1 , . . . , g s ∈ √ h i . Then, since p j ∈ h i and p j is a generic linear combination of g j , g j +1 , . . . , g s we infer that g j also belongs to √ h i and by repeating this argument we18onclude that S − k ( a ) ⊂ √ h i . But then, by Proposition 30, we have the inequality ofheights k ≤ j , which contradicts the choice of j .Therefore, any T i is a proper linear subspace of K s − j − , and in particular, a generic vec-tor µ ∈ K s − j − verifies µ / ∈ S i T i and so, the associated polynomial p j +1 := P t ≥ j +1 µ t g t is not a zero divisor modulo the ideal S − k ( a j ). Moreover, p j +1 is not invertible modulo S − k ( a j ) because of the inclusion S − k (( p j +1 ) + a j ) ⊆ S − k ( a ) and Proposition 30. Hence, p , . . . , p j +1 is a regular sequence in S − k R and conditions i ) and ii ) are satisfied for j + 1.Condition iii ) is a consequence of ii ) by Macaulay’s Theorem applied to the Cohen-Macaulay ring S − k R and the well-known fact that the contraction to R of a primary idealin S − k R remains primary of same dimension.We finish the proof showing condition iv ): Since p , . . . , p j is a regular sequence in S − k R , the ideals S − k ( a j ) and a ∗ j are both equidimensional of dimension n − j . On theother hand, as p j +1 is regular with respect to S − k ( a j ), then it is also regular with respectto S − k ( a j ) ∩ R = a ∗ j . In particular, a ∗ j + ( p j +1 ) is equidimensional of dimension n − j − a ∗ j ⊆ a ∗ j +1 and p j +1 ∈ a ∗ j +1 holds, we have the inclusion a ∗ j + ( p j +1 ) ⊆ a ∗ j +1 andboth ideals are equidimensional of dimension n − j −
1. Thus, Proposition 23 impliesdeg( a ∗ j +1 ) ≤ deg( a ∗ j + ( p j +1 )) . Finally, by Theorem 29 applied to the ideal a ∗ j and the polynomial p j +1 , we have theinequalities deg( a ∗ j +1 ) ≤ deg( a ∗ j + ( p j +1 )) ≤ deg( a ∗ j ) deg( p j +1 ) = deg( a ∗ j ) D j +1 and the lemma is proved. Lemma 31 implies the following B´ezout Inequality which appears in [15] in the projectivecase with the usual notion of degree for homogeneous ideals:
Theorem 32
Let K be an algebraically closed field of characteristic zero, ~X := ( X , . . . , X n ) variables over K and a ⊂ R := K [ ~X ] a non-zero and proper polynomial ideal generatedby polynomials g , . . . , g s of total degrees D ≥ · · · ≥ D s , respectively. Fix an index k , ≤ k ≤ n , such that in the primary decomposition of a there exist isolated components ofheight k and denote by Q k the intersection of this components. Then:i) The inequality deg( Q k ) ≤ D . . . D k holds.ii) The inequality deg( a ) ≤ P k ∈C D . . . D k , holds, where C is the set of those k suchthat the ideal a has isolated primary components of height k . Proof.
Following the previous lemma, consider the polynomials p , . . . , p k defining aregular sequence in the localized ring S − k ( R ). Since a k := ( p , . . . , p k ) ⊆ a , we have S − k ( a k ) ⊆ S − k ( a ) = S − k ( Q k ) (Proposition 30). Hence, a ∗ k := S − k ( a k ) ∩ R ⊆ Q k , andboth ideals are unmixed of height k (and in particular, equidimensional of dimension n − k ).From Proposition 23 we conclude deg( Q k ) ≤ deg( a ∗ k ).19y iteration of Condition iv ) in Lemma 31 we obtain deg( a ∗ k ) ≤ D . . . D k which impliesinequality i ).The assertion ii ) is immediate from i ) and the definition of degree of arbitrary ideals(Definition 25). Let a be an equidimensional ideal of the polynomial ring Q [ ~X ] in n variables ~X :=( X , . . . , X n ). Assume that the dimension m and a system of generators g , . . . , g s ∈ Z [ ~X ]of the ideal a are known. Let d and σ be upper bounds for the degrees of g , . . . , g s andthe bit-sizes of their coefficients.The goal of this section is to discuss the complexity character of the problem of com-puting deg( a ).Of course, one could compute (uniformly and deterministically) a primary decompo-sition of the ideal a (see [8]) and determine deg( a ) by means of Definition 22 ii ). Thiswould involve a computational cost which is doubly exponential in n .We present here a probabilistic approach which is more efficient and discuss thenwhether its complexity can be improved.Let T ij , 1 ≤ i ≤ m , 0 ≤ j ≤ n be new indeterminates, ~T = ( T ij ) ≤ i ≤ m ≤ j ≤ n and let for1 ≤ i ≤ m f i := n X j =1 T ij X j + T i . Let ~f := f , . . . , f m and fix a monomial order for ~X .Let b the ideal generated by a . From the proof of Lemma 20 and Theorem 21 wededuce deg( a ) = dim Q ( ~T ) Q ( ~T )[ ~X ] / b . Since b is an ideal of dimension zero in Q ( ~T )[ ~X ] we have just to compute a Gr¨obnerbasis of b from the generators g , . . . , g s and f , . . . , f m .Following [5, Theorem 3.3] this can be done using ( sd n ) O (1) arithmetic operations in Q ( ~T ).Applying [11, Theorem 4.4] we obtain a non-uniform deterministic or uniform proba-bilistic algorithm which computes deg( a ) by means of ( sd n ) O (1) arithmetic operations in Q . The non-uniform deterministic version of the algorithm is based on a hitting sequenceof integers having bit-size ( sd n ) O (1) . This sequence has to be chosen probabilistically inthe uniform complexity model. The whole algorithm requires therefore ( σsd n ) O (1) bitoperations.Putting everything together we obtain the following complexity statement. Theorem 33
There exists a uniform probabilistic algorithm implementable on a TuringMachine with advice which computes deg( a ) in time ( σsd n ) O (1) uniform deterministic complexity ofcomputing deg( a )? This question seems to be out of reach with the actual techniques.The other question asks whether deg( a ) can be computed probabilistically using ( σsd n ) O (1) bit operations.This question can be answered positively if it is possible to guess probabilisticallygeneric degree one polynomials f , . . . , f m of Z [ ~X ] of bit size ( σsd n ) O (1) .In this case the probabilistic algorithm of Lakshman [13], techniques of [9], a suitablearithmetic B´ezout Inequality (see e.g. [3, Th´eor`eme 2] or [12, § Z (see for instance [7, Corollary 16.25]) can becombined to obtain the desired complexity result. We do not go into the (lengthy) detailsof this approach. Conclusion
We introduced a suitable notion of degree for non-homogeneous polynomial ideals andproved extrinsic B´ezout Inequalities for this notion. We argued that an intrinsic B´ezoutInequality for the sum of two ideals is unfeasable. We exhibit a probabilistic algorithm ofsingle exponential complexity which computes the degree of an equidimensional ideal.
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