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Journal of Algebra 264 (2003) 582–612

www.elsevier.com/locate/jalgebr

The Euler class group of a polynomial algebra

Mrinal Kanti Das

Harish-Chandra Research Institute, Chhatnag Road, Jhusi, Allahabad 211 019, India

Received 28 May 2002

Communicated by Craig Huneke

1. Introduction

Let A be a smooth affine domain of dimensionn over a fieldk andI be a prime ideaof A[T ] of heightr such thatA[T ]/I is smooth and 2r � n+ 3. Letf1(T ), . . . , fr (T ) ∈ I

such thatI = (f1(T ), . . . , fr (T ))+(I2T ). Furthermore, assume thatA/(f1(0), . . . , fr (0))is also smooth. In this set up Nori asked the following question (for motivation, seeIntroduction]):

Question. Do there existg1, . . . , gr such thatI = (g1, . . . , gr ) with gi − fi ∈ (I2T )?

This question has been answered affirmatively by Mandal [M2] whenI contains amonic polynomial, even without any smoothness assumptions.

WhenI does not contain a monic polynomial, Nori’s question has been answeredaffirmative in the following cases:

(1) A is a local ring of a smooth affine algebra over an infinite field [M-V, Theorem 4(2) A is a smooth affine algebra over an infinite field andr = n (i.e., dimA[T ]/I = 1)

[B-RS1, Theorem 3.8].

Moreover, an example is given in [B-RS1, Example 6.4] for the case dim(A[T ]/I)= 1,which shows that the question of Nori does not have an affirmative answer in gewithout the smoothness assumption.

So, in view of this example of [B-RS1] one wonders where the obstruction forI to havea set of generators satisfying the required properties lies. In this paper we investigaquestion when dimA[T ]/I = 1. Taking a cue from the above result of Mandal, we prthe following theorem.

E-mail address:[email protected].

0021-8693/03/$ – see front matter 2003 Elsevier Science (USA). All rights reserved.doi:10.1016/S0021-8693(03)00240-0

M.K. Das / Journal of Algebra 264 (2003) 582–612 583

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Theorem. LetA be a Noetherian ring of dimensionn� 3, containing the field of rationalsLetI ⊂A[T ] be an ideal of heightn. Suppose thatI = (f1, . . . , fn)+(I2T ) and there exisF1, . . . ,Fn ∈ IA(T ) such thatIA(T ) = (F1, . . . ,Fn) and Fi = fi mod I2A(T ). Then,there existg1, . . . , gn such thatI = (g1, . . . , gn) andgi = fi mod(I2T ).

Let A be a Noetherian ring of dimensionn � 2 and letJ ⊂ A be an ideal of heighr such thatJ/J 2 is generated byr elements. It is of interest to know when a sof r generators ofJ/J 2 can be lifted to a set ofr generators ofJ . This question wasinvestigated in [B-RS3] for ideals of heightn, where, an abelian groupEn(A), calledthe Euler class group ofA is defined and corresponding to a set of generators ofJ/J 2

an element of this group is attached and it is shown that this set of generators ofJ/J 2

can be lifted to a set of generators ofJ if the corresponding element ofEn(A) iszero.

Now let R = A[T ] whereA is a Noetherian ring of dimensionn. Since every ideaI ⊂ A[T ] of heightn+ 1 contains a monic polynomial, it can be shown, using a resuMandal [M1], thatEn+1(R)= 0.

As one of the interesting consequences of our theorem, we can define a notionnth Euler class group ofA[T ] (denoted byEn(A[T ])), whereA is a Noetherian ring odimensionn. Further, to any set ofn generators ofI/I2, whereI ⊂ A[T ] is an idealof heightn, we attach an element of this group and show that if this element isthen the set of generators ofI/I2 can be lifted to a set of generators ofI (Theorem 4.7).Moreover, there is a canonical injective homomorphism fromEn(A) to En(A[T ]) whichis an isomorphism whenA is a smooth affine domain over an infinite field. (This isalgebraic analogue of a well-known result in algebraic topology as ifA is a smooth affinedomain over reals, then the setX of real points of SpecA is a manifold of dimensionn andthe groupsEn(A) andEn(A[T ]) are algebraic analogues of thenth cohomology groupHn(X) andHn(X × I).)

The layout of this paper is as follows: In Section 3, we prove our main theo(Theorem 3.10). In Section 4, as an application of our main theorem, we definnotion of thenth Euler class groupof A[T ], as mentioned above. We also definenth Euler classof a pair (P,χ), where,P is a projectiveA[T ]-module of rankn (withtrivial determinant) andχ is a generator of

∧n(P ). In this section we also prove ou

main theorem in a more general form (Theorem 4.8) and derive several analogresults of [B-RS3, Section 4] as consequences (see, for example, Corollaries 4.10In Section 5, a “Quillen–Suslin theory” for Euler class groups is developed. We proa local–global principlefor Euler class groups (Theorem 5.4), which is an analoguthe Quillen localization theorem. An analogue of local Horrocks theorem is also prIn Section 6, we define the notion of thenth weak Euler class groupEn

0(A[T ]) whichis a certain quotient ofEn(A[T ]). Section 7 deals with the case when dimensionthe base ring is two. In this section we prove a weaker version of the main theand apply it to obtain results similar to those in Sections 4–6. In Section 2, we dsome of the terms used in the paper and quote some results which are usedsections.

584 M.K. Das / Journal of Algebra 264 (2003) 582–612

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2. Some preliminaries

In this section we define some of the terms used in the paper and state somefor later use. All rings considered in this paper are commutative and Noetherian amodules considered are assumed to be finitely generated. For a moduleM over a ring,µ(M) will denote the minimal number of generators ofM.

Definition 2.1. Let A be a ring. A row(a1, a2, . . . , an) ∈ An is said to beunimodularifthere existb1, b2, . . . , bn in A such thata1b1 + · · · + anbn = 1.

Definition 2.2. Let A be a Noetherian ring. LetP be a projectiveA-module. An elemenp ∈ P is said to beunimodularif there exists a linear mapφ :P →A such thatφ(p)= 1.

We now state a theorem of Serre [Se].

Theorem 2.3. LetA be a Noetherian ring withdimA= d . Then any projectiveA-modulehaving rank> d has a unimodular element.

As an immediate consequence we have the following corollary.

Corollary 2.4. LetA be a Noetherian ring withdimA= 1. Then any projectiveA-modulehaving trivial determinant is free.

The following lemma has been proved in [Bh].

Lemma 2.5. LetA be a ring andJ ⊂ A be an ideal of heightr. LetP ,Q be projectiveA/J -modules of rankr and let α :P � J/J 2 and β :Q � J/J 2 be surjections. Leψ :P →Q be a homomorphism such thatβψ = α. Thenψ is an isomorphism.

The following lemma is easy to prove and hence we omit the proof.

Lemma 2.6. LetA be a Noetherian ring andP a finitely generated projectiveA-module.Let P [T ] denote projectiveA[T ]-moduleP ⊗ A[T ]. Let α(T ) :P [T ] � A[T ] andβ(T ) :P [T ] � A[T ] be two surjections such thatα(0) = β(0). Suppose further that thprojectiveA[T ]-moduleskerα(T ) andkerβ(T ) are extended fromA. Then there exists aautomorphismσ(T ) of P [T ] with σ(0)= id such thatβ(T )σ(T )= α(T ).

The next lemma follows from the well-known Quillen Splitting Lemma [Qu, Lemmand its proof is essentially contained in [Qu, Theorem 1].

Lemma 2.7. LetA be a Noetherian ring andP be a finitely generated projectiveA-module.Let s, t ∈ A be such thatAs + At = A. Let σ(T ) be anAst [T ]-automorphism ofPst [T ]such thatσ(0) = id. Then,σ(T ) = α(T )sβ(T )t , whereα(T ) is anAt [T ]-automorphismof Pt [T ] such thatα(T )= id modulo the ideal(sT ) andβ(T ) is anAs[T ]-automorphismof Ps [T ] such thatβ(T )= id modulo the ideal(tT ).


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Now we state two useful lemmas. The proofs of these can be found in [B-RS1].

Lemma 2.8. Let A be a Noetherian ring and letI be an ideal ofA. Let J,K be idealsof A contained inI such thatK ⊂ I2 andJ +K = I . Then there existsc ∈ K such thatI = (J, c).

Lemma 2.9. LetA be a Noetherian ring containing an infinite fieldk and letI ⊂A[T ] bean ideal of heightn. Then there existsλ ∈ k such that eitherI (λ) = A or I (λ) is an idealof heightn in A, whereI (λ)= {f (λ): f (T ) ∈ I }.

Now we quote a theorem of Eisenbud–Evans [E-E], as stated in [Pl].

Theorem 2.10. Let A be a ring andM be a finitely generatedA-module. LetS bea subset ofSpecA and d :S → N be a generalized dimension function. AssumeµQ(M)� 1+ d(Q) for all Q ∈ S. Let(m,a) ∈M ⊕A be basic at all prime idealsQ ∈ S.Then there exists an elementm′ ∈M such thatm+ am′ is basic at all primesQ ∈ S.

As a consequence of Theorem 2.10, we have the following corollary. For a proocan look at [B-RS3, 2.13].

Corollary 2.11. Let A be a ring andP be a projectiveA-module of rankn. Let(α, a) ∈ (P ∗ ⊕ A). Then there exists an elementβ ∈ P ∗ such thatht(Ia) � n, whereI = (α+aβ)(P ). In particular, if the ideal(α(P ), a) has height� n thenhtI � n. Further,if (α(P ), a) is an ideal of height� n andI is a proper ideal ofA, thenhtI = n.

The following lemma is an application of Lemma 2.8 and Corollary 2.11. Its proessentially contained in [B-RS3, 2.14].

Lemma 2.12. LetA be a Noetherian ring of dimensionn � 2 and letP be a projectiveA[T ]-module of rankn. Let I ⊂ A[T ] be an ideal of heightn and letα :P/IP � I/I2

be a surjection. Then there exists an idealI ′ ⊂A[T ] and a surjectionβ :P � I ∩ I ′ suchthat:

(i) I + I ′ =A[T ].(ii) β ⊗A[T ]/I = α.(iii) ht(I ′)� n.(iv) Furthermore, given finitely many idealsI1, I2, . . . , Ir of height� 2, I ′ can be chosen

with the additional property thatI ′ is comaximal with each of them.

Definition 2.13. Let A be a commutative Noetherian ring,P a projectiveA[T ]-module.Let J (A,P ) ⊂ A consist of all thosea ∈ A such thatPa is extended fromAa . It followsfrom [Qu, Theorem 1] thatJ (A,P ) is an ideal andJ (A,P ) = √

J (A,P ). This is calledtheQuillen idealof P in A.

Remark 2.14. It is easy to deduce htJ (A,P )� 1 from Quillen–Suslin theorem [Qu,Su1If determinant ofP is extended fromA, then htJ (A,P )� 2 by [B-R, 3.1].


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The following result is due to Lindel [Li, Theorem 2.6].

Theorem 2.15. Let A be a commutative Noetherian ring withdimA = d and letR =A[T1, . . . , Tn]. LetP be a projectiveR-module of rank� max(2, d + 1). ThenE(P ⊕R)

acts transitively on the set of unimodular elements ofP ⊕R.

3. Main theorem

In this section we prove the main theorem. This section is divided into two parts.

3.1. The semilocal case

In this part we prove the main theorem in semilocal situation. We need the follolemmas and propositions.

Lemma 3.1. LetB be a Noetherian ring withdimB = n andJ ⊂ B be an ideal containedin the Jacobson radical ofB. LetI ⊂ B[T ] be an ideal such thatI +JB[T ] = B[T ]. Thenany maximal ideal ofB[T ] containingI has height� n.

Proof. SupposeM ⊂ B[T ] is a maximal ideal of heightn+ 1. ThenM ∩B is a maximalideal ofB. HenceM ∩ B containsJ . SinceI + JB[T ] = B[T ], it follows that I is notcontained inM. This proves the lemma.✷

The following proposition is implicit in [Na]. We give a proof for the sake of compleness.

Proposition 3.2. LetA be a semilocal ring andI ⊂ A be an ideal,I = (a1, . . . , an)+ L,whereL is an ideal contained inI2 andn� 1. Then,I = (b1, . . . , bn) with ai = bi modL.

Proof. SinceI = (a1, . . . , an)+L andL⊂ I2, from Lemma 2.8 we getI = (a1, . . . , an, e)

wheree ∈ L is such thate(1 − e) ∈ (a1, . . . , an). So, if L is contained in the Jacobsoradical ofA, we haveI = (a1, . . . , an). Suppose thatL is not contained in the Jacobsoradical ofA and say,M1, . . . ,Mr are those maximal ideals which do not containL. Afterrearranging theMi ’s we may assume thata1 belongs toM1, . . . ,Mt and does not belontoMt+1, . . . ,Mr . By ‘Prime avoidance’ we can chooseb ∈L∩Mt+1 ∩ · · · ∩Mr such thatb /∈M1 ∪ · · · ∪Mt . Thena1 + b, a2, . . . , an generateI as they do so locally. ✷

As a consequence we have the following corollary.

Corollary 3.3. LetR be a semilocal ring andK ⊂ R[T ] be an ideal containing a monipolynomial. Suppose thatK = (g1, . . . , gn)+ (K2T ) wheren� 2. ThenK = (k1, . . . , kn)

such that,k1 is monic andki = gi mod(K2T ).


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Proof. Let f ∈ K be monic. Addingf pT (p � 2) to g1 for suitably largep we canassume thatg1 is monic. LetA= R[T ]/(g1) and bar denote reduction modulo(g1). It isclear thatA is a semilocal ring. Now,K = (g2, . . . , gn) + (K2T ). From the proof of theabove proposition it follows thatK = (g2 + h, . . . , gn) whereh ∈ (K2T ). So,h= k+ g1l,wherek ∈ (K2T ) andl ∈R[T ]. Now it is clear thatK = (g1, g2 + k, g3, . . . , gn). ✷

The following lemma is an analogue of Mandal’s theorem ([M2, 2.1] or [M-RS, 2.2

Lemma 3.4. LetC be a ring andM ⊂ C[Y ] be an ideal which contains a monic polynomand is such that the ringC1+L is semilocal, whereL=M ∩C. Suppose thatM/(M2Y ) isgenerated byn elements(n� 2). Then, any set ofn generators ofM/(M2Y ) can be liftedto a set ofn generators ofM.

Proof. Suppose thatM = (F1, . . . ,Fn) + (M2Y ). We can clearly assume thatF1 ismonic. Going toC1+L[Y ] and applying the corollary above we get thatM1+L =(F1,G,F3, . . . ,Fn) whereG − F2 ∈ (M2Y )1+L. Therefore we can finds ∈ L such that,M1+s = (F1,G,F3, . . . ,Fn) andG − F2 ∈ (M2Y )1+s . Now we can adapt the proof o[M2, 2.1] to get the result. ✷

With the above lemma in hand one can prove the following analogue of a theorMandal–Raja Sridharan [M-RS, Theorem 2.3].

Lemma 3.5. LetC be a ring. LetM,N ⊂ C[Y ] be ideals such that

(1) M contains a monic polynomial;(2) C1+L is semilocal whereL=M ∩C;(3) N =N(0)[Y ] is an extended ideal;(4) M +N = C[Y ].

LetJ =M ∩N . SupposeJ (0)= (a1, . . . , an) andM = (F1(Y ), . . . ,Fn(Y ))+M2 suchthatFi(0)= ai modM(0)2. ThenJ = (G1(Y ), . . . ,Gn(Y )) with Gi(0)= ai .

Proof. Same as in [M-RS]. ✷Now we turn to the main problem. We first state a lemma whose proof is the sa

that of [B-RS3, 5.3].

Lemma 3.6. Let R be a semilocal ring of dimensionn � 3, I ⊂ R[T ] be an idealof heightn such thatI + JR[T ] = R[T ] whereJ is the Jacobson radical ofR. LetωI : (R[T ]/I)n � I/I2 be a surjection. Suppose thatωI can be lifted to a surjectionα :R[T ]n � I . Let f ∈ R[T ] be a unit moduloI and θ ∈ GLn(R[T ]/I) be such thatdet(θ) = f 2. Then, the surjectionωI θ : (R[T ]/I)n � I/I2 can be lifted to a surjectionβ :R[T ]n � I .

Now we prove the semilocal version of the main question in the following form.


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Theorem 3.7. Let R be a semilocal ring withdimR = n � 3 and I be an ideal ofR[T ]of heightn such thatI +JR[T ] =R[T ], whereJ is the Jacobson radical ofR. Supposethat I = (f1, . . . , fn)+ (I2T ). Also suppose that,IR(T )= (u1, . . . , un) such thatui = fimod I2R(T ). Then, there existh1, . . . , hn ∈ I such thatI = (h1, . . . , hn) and hi = fimod(I2T ).

Proof. SinceI + JR[T ] = R[T ], it follows that I is not contained in any ideal whiccontains a monic polynomial and hence, any monic polynomial ofR[T ] is unit moduloI .In particular,I (0)= R.

We also note that, to prove the theorem, it is enough to prove thatI = (h1, . . . , hn)

with hi = fi mod I2. Let us briefly explain why this is so. Consider the unimodurows (f1(0), . . . , fn(0)) and (h1(0), . . . , hn(0)) over the semilocal ringR. SinceEn(R)

acts transitively on the set of unimodular rows of lengthn over a semilocal ringR,there is aθ ∈ En(R) such that(h1(0), . . . , hn(0))θ = (f1(0), . . . , fn(0)). Suppose thaθ = ∏

Eij (rij ), whererij ∈R. SinceI (0)=R, we can findfij ∈ I such thatfij (0)= rij .We consider the elementary matrixθ = ∏

Eij (fij ) ∈En(R[T ]). Then, it is easy to see th(h1, . . . , hn)θ is a desired set of generators ofI .

We give the proof in steps.

Step 1. We have,I = (f1, . . . , fn) + I2. From the given condition we see that there imonic polynomialf ∈R[T ] such thatIf = (u1, . . . , un), whereui = fi modI2

f . Letf k beso chosen such thatf 2kui ∈ I, 1 � i � n. Write f 2kui = gi . Then,I = (g1, . . . , gn)+ I2

andgi − f 2kfi ∈ I2.Now I = (g1, . . . , gn) + I2 implies that(g1, . . . , gn) = I ∩ K whereK is an ideal of

R[T ] such thatK + I = R[T ]. SinceIf = (g1, . . . , gn)f , we see thatKf = R[T ]f , i.e.,K contains a monic polynomial. Also note thatK = (g1, . . . , gn)+K2.

Sincef is a unit moduloI , by Lemma 3.6, it is enough to getI = (m1, . . . ,mn) wheremi − f 2kfi ∈ I2. Therefore it is enough to getI = (m1, . . . ,mn) with mi − gi ∈ I2.

Step 2. Using Corollary 3.3 we find thatK = (k1, . . . , kn) such thatki = gi modK2.Note thatk1 is a monic polynomial. The row(k1, . . . , kn) is unimodular modI2. Sincek1 is monic, it is unit modI2. Therefore we can elementarily transform the above rowthat kn = 1 modI2 (note that this transformation does not affectk1). Since elementartransformations can be lifted via surjection of rings, we can findσ ∈ En(R[T ]), which isa lift of the above elementary transformation. Let

(g1, . . . , gn)σ = (h1, . . . , hn).

Therefore, we haveI ∩K = (h1, . . . , hn), K = (k1, . . . , kn), andhi = ki modK2. (We arestill calling the new set of generators ofK as(k1, . . . , kn).)

WriteC =R[T ] and consider the following ideals in the polynomial extensionC[Y ]:

M = (k1, . . . , kn−1, Y + kn), N = IC[Y ], J =M ∩N.


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ClearlyM +N = C[Y ]. SupposeL= M ∩ C. Sincek1 ∈ L is a monic polynomial inT ,we see thatC/L = R[T ]/L is integral overR/(L ∩ R). ThereforeC/L is a semilocalring asR/(L ∩ R) is so. Consequently,C1+L is also semilocal. So conditions (1)–(4)Lemma 3.5 are satisfied.

We haveJ (0) = I ∩K = (h1, . . . , hn) andM = (k1, . . . , kn−1, Y + kn) whereki = himodM(0)2. Therefore, applying Lemma 3.5, we getJ = (G1(Y ), . . . ,Gn(Y )) such thatGi(0) = hi . PuttingY = 1 − kn we getI = J (1 − kn) = (G1(1 − kn), . . . ,Gn(1 − kn)).Now kn = 1 mod I2 implies thatGi(1 − kn) = hi mod I2. Writing li for Gi(1 − kn)

we haveI = (l1, . . . , ln) with li = hi modI2. Let (l1, . . . , ln)σ−1 = (m1, . . . ,mn). Then,clearlymi = gi modI2 whereI = (m1, . . . ,mn). This completes the proof.✷3.2. The general case

Now we proceed to prove the main theorem. We begin with a lemma.

Lemma 3.8. Let A be a Noetherian ring containing the field of rationals withdimA =n� 2, I ⊂A[T ] an ideal of heightn. LetP be a projectiveA[T ]-module of rankn. WriteJ = I ∩ J (A,P ) whereJ (A,P ) is the Quillen ideal ofP in A andB = A1+J . Supposethat there is a surjection

φ :P � I/(I2T

).

Assume further that there exists a surjection

θ :P1+J � I1+J

such thatθ is a lift of φ ⊗B. Then there exists a surjectionΦ :P � I such thatΦ is a liftof φ.

Proof. We choose an elements ∈ J such thatθ :P1+s � I1+s is surjective. Note thasinces ∈ J (A,P ), the projectiveAs[T ]-modulePs is extended. Letφs(0) denote the map(P/T P)s � I (0)s induced fromφs by settingT = 0 andγ = φs(0) ⊗ As[T ] :Ps � Is(=As[T ]). Then the elementsθ ⊗As(1+sA)[T ] andγ ⊗As(1+sA)[T ] are unimodular elements ofP ∗

s(1+sA) and they are equal modulo(T ). Since dimAs(1+sA) � n−1, rankP = n,andA contains the field of rationals, by [Ra, Corollary 2.5], the kernels of the surjecθ ⊗ As(1+sA)[T ] andγ ⊗ As(1+sA)[T ] are locally free projective modules and henceQuillen’s local–global principle [Qu, Theorem 1], these kernels are projective mowhich are extended fromAs(1+sA). Hence, by Lemma 2.6, there exists an automorphσ of Ps(1+sA) such thatσ = id modulo(T ) and(θ ⊗As(1+sA)[T ])σ = γ ⊗As(1+sA)[T ].Therefore, there exists an elementt ∈A of the form 1+ rs such thatt is a multiple of 1+ s

andσ is an automorphism ofPst with σ = id modulo(T ) and(θ⊗Ast [T ])σ = γ ⊗Ast [T ].We note that, sinces ∈ J (A,P ), it follows thatPst is extended fromAst .

SincePst is extended, we can adjoin a new variableW and consider theAst [T ,W ]-automorphism ofPst [W ] given byτ (W)= σ(TW). Sinceτ (0) is identity, by Lemma 2.7it follows thatτ = αsβt , whereα is anAt [T ,W ]-automorphism ofPt [W ] such thatα = id


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modulo the ideal(sTW) andβ is anAs[T ,W ]-automorphism ofPs [W ] such thatβ = idmodulo the ideal(tT W). PuttingW = 1 and using a standard patching argument, wethat the surjections(θ ⊗ At [T ]).α(1) :Pt � It and (γ ⊗ As[T ]).β(1)−1 :Ps � Is patchto yield a surjectionΦ :P � I . It is easy to see thatΦ is a lift of φ. This proves thelemma. ✷

The following is a restatement of [B-RS1, Lemma 3.6] in our set up. We give the pfor the sake of completeness.

Lemma 3.9. LetA be a Noetherian ring of dimensionn� 3, I ⊂A[T ] an ideal of heightnand J be any ideal contained inI ∩ A such thathtJ � 2. Let P be a projectiveA[T ]-module of rankn. Supposeψ :P � I/(I2T ) is a surjection. Then we can find a lφ ∈ HomA[T ](P, I) ofψ , such that the idealφ(P )= I ′′ satisfies the following properties:

(i) I ′′ + (J 2T )= I .(ii) I ′′ = I ∩ I ′, whereht(I ′)� n.(iii) I ′ + (J 2T )=A[T ].

Proof. We choose any liftφ ∈ HomA[T ](P, I) of ψ . Since φ(P ) + (I2T ) = I , byLemma 2.8 we can chooseb ∈ (I2T ) such that(φ(P ), b) = I . LetC = A[T ]/(J 2T ) andbar denote reduction modulo(J 2T ). Now applying Corollary 2.11 to the element(φ,b)of P ∗ ⊕A[T ]/(J 2T ), we see that there existsβ ∈ P ∗ such that ifN = (φ + bβ)(P ) thenht(Nb)� n.

Sinceb ∈ (I2T ), the elementφ + bβ is also a lift ofψ . Therefore, replacingφ byφ + bβ , we may assume thatN = φ(P ).

Now as(N,b) = I andb ∈ (I2T ), it follows thatN = I ∩ K, (K,b) = A[T ]. Sinceb ∈ I , Nb =Kb. Therefore we have

(1) N = I ∩K with ht(K)= ht(Kb)= ht(Nb)� n.(2) (b)+K = C.

We claim thatK = C. Assume, to the contrary, thatK is a proper ideal ofC. Sinceb ∈ (I2T ), in view of (1) and (2), we have

n � ht(K

) = ht(KT

)� dim(CT )= dim

(A/J 2)[T ,T −1] = dim(A/J )+ 1

� (n− 2)+ 1 = n− 1.

This is a contradiction. ThusK = C andφ(P )+ (J 2T )= I . This proves (i).We choose, by Lemma 2.8, an elementc ∈ (J 2T ) such that(φ(P ), c) = I . As before,

using Corollary 2.11, we can add a suitable multiple ofc to φ and assume that the ideφ(P )= I ′′ satisfies (ii) and (iii). This proves the lemma.✷

Now we prove the main theorem. The proof of this theorem is motivated by [B-Theorem 3.8].


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Theorem 3.10. Let A be a Noetherian ring of dimensionn � 3, containing the field orationals. LetI ⊂A[T ] be an ideal of heightn. Suppose thatI = (f1, . . . , fn)+(I2T ) andthere existF1, . . . ,Fn ∈ IA(T ) such thatIA(T )= (F1, . . . ,Fn) andFi = fi modI2A(T ).Then, there existg1, . . . , gn such thatI = (g1, . . . , gn) andgi = fi mod(I2T ).

Proof. We give the proof of the theorem in steps.

Step 1. Let J = I ∩A. Applying Lemma 3.9 we getk1, . . . , kn ∈ I and an idealI ′ ⊂A[T ]of heightn such that

(i) (k1, . . . , kn)+ (J 2T )= I whereki = fi mod(I2T );(ii) (k1, . . . , kn)= I ∩ I ′;(iii) I ′ + (J 2T )=A[T ].

Let J ′ = I ′ ∩A. We claim that dim(A/(J + J ′))= 0.

Proof of the claim. Since ht(I) = n and ht(I ′) � n, we have dimA/J � 1 anddimA/J ′ � 1. Without loss of generality we may assume that dimA/J = dimA/J ′ = 1.It suffices to show that there is no prime idealQ of A containingJ andJ ′ and having theproperty that dimA/Q= 1. Suppose, to the contrary, that such a prime ideal exists.

Let I ′ = K1 ∩ K2 ∩ · · · ∩ Kr be a primary decomposition ofI ′, with√Kl = Pl .

Then J ′ = (K1 ∩ A) ∩ · · · ∩ (Kr ∩ A). Since dimA/J ′ = dimA/Q = 1, it followsthat Q is minimal overJ ′. ThereforeQ = Pl ∩ A for somel. We have

√Kl = Pl ⊃

I ′ +QA[T ] ⊃ I ′ + JA[T ]. But by property (iii) of Lemma 3.9,I ′ + (J 2T )=A[T ]. Thisyields a contradiction and proves the claim.

Step 2. WriteB =A1+J . We have, from Lemma 3.9,I ∩ I ′ = (k1, . . . , kn) and the idealsIandI ′ are comaximal. Going to the ringB[T ], we getIB[T ]∩I ′B[T ] = (k1, . . . , kn)B[T ].Therefore,I ′B[T ] = (k1, . . . , kn)B[T ] + I ′2B[T ]. Now asI ′ + (J 2T ) = A[T ], we haveI ′(0)=A. Note thatJB is contained in the Jacobson radical ofB.

We claim that we can lift the above set of generators ofI ′B[T ]/I ′2B[T ] to a set ofgenerators ofI ′B[T ], i.e., there existl1, . . . , ln ∈ I ′B[T ] such thatI ′B[T ] = (l1, . . . , ln)

andli = ki modI ′2B[T ].

Proof of the claim. SinceI ′B[T ] = (k1, . . . , kn)+ I ′2B[T ] andI ′(0) = A, it is easy tosee, using the Chinese remainder theorem, that there existα1, . . . , αn such thatI ′B[T ] =(α1, . . . , αn)+ (I ′2T )B[T ], whereαi = ki modI ′2B[T ].

Write R = B1+J ′B . In order to prove the claim, in view of Lemma 3.8, it is enoughshow thatI ′R[T ] = (β1, . . . , βn) such thatβi = αi modulo(I ′2T )R[T ].

SinceR = B1+J ′B = A1+J+J ′ , it follows from Step 1 thatR is semilocal. Now weconsider the ringR(T ). Using the subtraction principle [B-RS3, 3.3], we getI ′R(T ) =(v1, . . . , vn) with the property thatvi = αi moduloI ′2R(T ). Therefore, by Theorem 3.7we obtain a set ofn generators ofI ′R[T ] with the desired property. Thus the claimproved.


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Step 3. Recall that we have:

(i) (k1, . . . , kn)+ (J 2T )= I .(ii) (k1, . . . , kn)= I ∩ I ′, ht(I ′)= n.(iii) I ′ + (J 2T )=A[T ].(iv) (l1, . . . , ln)= I ′B[T ], such thatli = ki modI ′2B[T ].

SinceI ′B[T ] + (J 2T )B[T ] = B[T ], it follows that the row(l1, . . . , ln) is unimodularmodulo (J 2T )B[T ]. Let D = B[T ]/(J 2T )B[T ] and bar denote reduction modu(J 2T )B[T ]. We want to show that(l1, . . . , ln) ∈ Umn(D) can be elementarily transformeto (1,0, . . . ,0).

SinceJB is contained in the Jacobson radical ofB, it is easy to see thatJD is containedin the Jacobson radical ofD. Therefore, it suffices to show that the row(l1, . . . , ln) can beelementarily completed over the ringD/JD. But this follows from Theorem 2.15 becauD/JD � (A/J )[T ], dimA/J � 1, andn� 3.

Since elementary transformations can be lifted via surjection of rings, it followsthere is an elementary automorphismτ ∈En(B[T ]) such that(l1, . . . , ln)τ = (m1, . . . ,mn)

where(m1, . . . ,mn)= (1,0, . . . ,0) modulo(J 2T )B[T ]. Applying Corollary 2.11, we canfind d1, . . . , dn−1 ∈ B[T ] such that ht((m1 + d1mn, . . . ,mn−1 + dn−1mn)mn) � n − 1.We write hi = mi + dimn, i = 1,2, . . . , n − 1. SinceI ′B[T ] = (h1, . . . , hn−1,mn) andht(I ′) � n, it is easy to verify that ht(h1, . . . , hn−1) = n − 1. We have(h1, . . . , hn−1) +(J 2T )B[T ] = B[T ] and we note that(J 2T )B[T ] is contained in the Jacobson radicof B. Therefore, it follows from Lemma 3.1 that dimB[T ]/(h1, . . . , hn−1) � 1. We takehn = h1 +mn. Thenhn = 1 modulo(J 2T )B[T ].

Note that the automorphism ofB[T ] which transforms(l1, . . . , ln) to (h1, . . . , hn) iselementary. Let us call itσ .

Step 4. Recall that we haveIB[T ]∩I ′B[T ] = (k1, . . . , kn)B[T ] andI ′B[T ] = (l1, . . . , ln)

such thatli = ki modI ′2B[T ] for i = 1, . . . , n. From Step 3 we haveσ ∈En(B[T ]) suchthat(l1, . . . , ln)σ = (h1, . . . , hn). Therefore,I ′B[T ] = (h1, . . . , hn).

Let (k1, . . . , kn)σ = (u1, . . . , un). Then, sinceσ is elementary,IB[T ] ∩ I ′B[T ] =(u1, . . . , un). Also note thathi = ui moduloI ′2B[T ] for i = 1, . . . , n.

Let C = B[T ], R = C[Y ]. Let K1 be the ideal(h1, . . . , hn−1, Y + hn) of R,K2 =IC[Y ], andK3 =K1 ∩K2.

Since from Step 3 we have dimB[T ]/(h1, . . . , hn−1)� 1, we see that all the conditionof [M-RS, Theorem 2.3] are satisfied. Applying that theorem, it follows easilyK3 = (H1(T ,Y ), . . . ,Hn(T ,Y )) such thatHi(T ,0)= ui . PuttingY = 1− hn, we see thaIB[T ] = (H1(T ,1−hn), . . . ,Hn(T ,1−hn)). Sincehn = 1 modulo(I2T )B[T ], it followsthatHi(T ,1− hn)=Hi(T ,0) (= ui) modulo(I2T )B[T ].

Let (H1(T ,1 − hn), . . . ,Hn(T ,1 − hn))σ−1 = (w1, . . . ,wn). Then we haveIB[T ] =

(w1, . . . ,wn) whereaswi = ki modulo(I2T )B[T ] and hencewi = fi modulo(I2T )B[T ].Now we can apply Lemma 3.8 to obtain the desired set of generators ofI . This proves

the theorem. ✷


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It is not hard to see that, adapting the same proof, we can prove the main theoremfollowing form.

Theorem 3.11. LetA,I be as above andP be a projectiveA-module of rankn with trivialdeterminant. Suppose there exists a surjection

φ :P [T ] � I/(I2T

).

Suppose thatφ ⊗ A(T ) can be lifted to a surjectionφ′ :P [T ] ⊗ A(T ) � IA(T ). Then,there is a surjectionψ :P [T ] � I which liftsφ.

Another similar result will be proved in the next section (Theorem 4.8).

4. The Euler class group of A[T ]

For the rest of the paper,A will denote a commutative Noetherian ring containingfield of rationals.

Remark 4.1. LetA be a commutative Noetherian ring containing the field of rationalsdimA = n � 3. Let us describe the general method of approach to the problems wbe considering. In the following two sections, we will frequently use our main theoproved in the last section. In most cases, we will try to find a suitable idealI ⊂ A[T ]of heightn and a surjectionωI : (A[T ]/I)n � I/I2 in such a manner that the questireduces to finding a set ofn generators ofI which lifts ωI . Now, sinceA containsQ, itfollows from Lemma 2.9, that there is aλ ∈ Q such that eitherI (λ) = A or I (λ) is ofheightn. Therefore, if necessary, we can replaceT by T − λ and assume thatI (0)=A orI (0) is of heightn.

To lift ωI to a surjection fromA[T ]n to I , we will consider the induced surjectionωI(0) (= ωI ⊗ A[T ]/(T )) : (A/I (0))n � I (0)/I (0)2 over the ringA andωI ⊗ A(T ) :(A(T )/IA(T ))n � IA(T )/I2A(T ) overA(T ). Since dim(A)= dim(A(T ))= n and thereis a well-studied description of Euler class group of an arbitrary Noetherian ring wdeals with top height ideals only, using results on them (mostly from [B-RS3]), weensure thatωI(0) andωI ⊗ A(T ) can be lifted. Then we appeal to the main theoremconclude thatωI is liftable. An explicit description of this method is given in the followiproposition.

Proposition 4.2 (Addition principle).Let dimA = n � 3, and I, J be two comaximaideals inA[T ], each of heightn. Suppose thatI = (f1, . . . , fn) and J = (g1, . . . , gn).ThenI ∩ J = (h1, . . . , hn) wherehi = fi modI2 andhi = gi modJ 2.

Proof. WriteK = I ∩J . SinceI andJ are comaximal, the generators ofI andJ induce aset of generators ofK/K2. Say,K = (k1, . . . , kn)+K2 whereki = fi modI2 andki = gimodJ 2.


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re

SinceA contains the field of rationals, by Lemma 2.9 we get someλ ∈ Q such thatK(λ) = A or K(λ) has heightn. Therefore, if necessary, we can replaceT by T − λ andassume thatK(0)=A or ht(K(0))= n.

If K(0) = A, by [B-RS1, Remark 3.9], we getK = (l1, . . . , ln)+ (K2T ) with li = kimoduloK2. Now assume that ht(K(0))= n. SinceI andJ are comaximal ideals inA[T ],it is easy to see thatK(0) = I (0) ∩ J (0). Therefore, ht(I (0)) � n and ht(J (0)) � n.Both of them cannot equalA, asK(0) is proper. If one of them do, sayI (0) = A, thenK(0)= J (0)= (g1(0), . . . , gn(0)) whereasgi(0)= ki(0) moduloJ (0)2. SinceI (0) = A,it follows thatgi(0)= ki(0) moduloK(0)2. Therefore, again by [B-RS1, 3.9] we can gK = (l1, . . . , ln)+ (K2T ) with li = ki moduloK2.

Now assume that bothI (0) andJ (0) are proper ideals. In this case, by the additprinciple [B-RS3, Theorem 3.2], we getK(0) = (a1, . . . , an) such thatai = fi(0)modulo I (0)2 and ai = gi(0) modulo J (0)2. Therefore,ai = ki(0) modulo I (0)2 andai = ki(0) moduloJ (0)2, implying thatai = ki(0) moduloK(0)2. Then, as before, b[B-RS1, 3.9] we getK = (l1, . . . , ln)+ (K2T ) with li = ki moduloK2.

So, in any case, we can lift the given set of generators ofK/K2 to a set of generators oK/(K2T ). Now we go to the ringA(T ). Note that dimA(T ) = n. Applying the additionprinciple [B-RS3, 3.2], we getKA(T ) = (k̃1, . . . , k̃n) such thatk̃i = fi moduloI2A(T )

and k̃i = gi modulo J 2A(T ). Therefore,k̃i = ki modulo I2A(T ) and k̃i = ki moduloJ 2A(T ) implying thatk̃i = ki moduloK2A(T ).

Now we can appeal to the main theorem (Theorem 3.10) and obtain the desiredgenerators forK. ✷Proposition 4.3 (Subtraction principle).Let dimA = n � 3 and I, J be two comaximaideals inA[T ], each of heightn. Suppose thatI = (f1, . . . , fn) andI ∩ J = (h1, . . . , hn)

such thathi = fi modI2. Then,J = (g1, . . . , gn) wherehi = gi modJ 2.

Proof. The method of proof is the same as that of Proposition 4.2 and therefore wjust outline the proof here. LetK = I ∩ J . As above, we can assume thatK(0) = A orht(K(0)) = n. First note thatJ = (h1, . . . , hn) + J 2. If J (0) = A or I (0) = A then, asbefore, we can lift the above set of generators ofJ/J 2 to a set of generators ofJ/(J 2T ).So assume that ht(K(0)) = n and bothI (0) andJ (0) are proper. Then we can apply tsubtraction principle [B-RS3, Theorem 3.3] and conclude thatJ (0)= (a1, . . . , an) whereai = hi(0) modJ (0)2. Therefore, by [B-RS1, 3.9] again, we getJ = (l1, . . . , ln)+ (J 2T ).

Next we go to the ringA(T ) and apply the subtraction principle [B-RS3, 3.3] theto conclude thatJA(T ) = (h̃1, . . . , h̃n) with h̃i = hi moduloJ 2A(T ). Therefore, usingTheorem 3.10 we get the desired set of generators forJ . ✷

Now we proceed to define thenth Euler class groupof A[T ] whereA is a commutativeNoetherian ring with dim(A)= n� 3 and which contains the field of rationals.

Let I ⊂ A[T ] be an ideal of heightn such thatI/I2 is generated byn elements. Letαandβ be two surjections from(A[T ]/I)n to I/I2. We say thatα andβ arerelatedif thereexists an automorphismσ of (A[T ]/I)n of determinant 1 such thatασ = β . It followseasily that this is an equivalence relation on the set of surjections from(A[T ]/I)n to I/I2.


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Let [α] denote the equivalence class ofα. We call such an equivalence class[α] a localorientation ofI .

We note that following Remark 4.1 it is not hard to derive that if a surjectionα from(A[T ]/I)n to I/I2 can be lifted to a surjectionθ :A[T ]n � I then so can anyβ equivalentto α (however, we give a proof below for the convenience of the reader). Thereforenow on we shall identify a surjectionα with the equivalence class[α] to which it belongs.

Proposition 4.4. Let α andβ be two surjections from(A[T ]/I)n to I/I2 such that thereexistsσ ∈ SLn(A[T ]/I) with the property thatασ = β . Suppose thatα can be lifted to asurjectionθ :A[T ]n � I . Thenβ can also be lifted to a surjectionφ :A[T ]n � I .

Proof. SinceA containsQ, by Lemma 2.9 we can find someλ ∈ Q such thatI (λ) = A

or I (λ) is an ideal of heightn. If necessary, we can replaceT by T − λ and assume thaeitherI (0)=A or htI (0)= n.

If I (0) = A, by [B-RS1, Remark 3.9], we can liftβ to a surjectionγ :A[T ]n �I/(I2T ). On the other hand, if htI (0) = n, then we consider the surjectionsα(0), β(0) :(A/I (0))n � I (0)/I (0)2 and note thatα(0)σ (0) = β(0). Since dim(A/I (0)) = 0, wehave SLn(A/I (0)) = En(A/I (0)) and, since elementary matrices can be lifted visurjection of rings, it follows thatσ(0) can be lifted to an elementτ ∈En(A). Composingτ with θ(0) (which is a lift ofα(0)), we get a lift ofβ(0) to a surjection fromAn to I (0).So, again by [B-RS1, Remark 3.9], we can liftβ to a surjectionγ :A[T ]n � I/(I2T ).

Now we move to the ringA(T ) and consider the induced surjectionsα ⊗ A(T )

and γ ⊗ A(T ). Since dim(A(T )/IA(T )) = 0, it follows that SLn(A(T )/IA(T )) =En(A(T )/IA(T )). Following the same method as in the above paragraph, we getof γ to a surjection fromA(T )n to IA(T ). Now we can apply Theorem 3.10 and concluthatβ can be lifted to a surjectionφ :A[T ]n � I . ✷

We call a local orientation[α] of I a global orientation of I if the surjectionα : (A[T ]/I)n � I/I2 can be lifted to a surjectionθ :A[T ]n � I .

Let G be the free abelian group on the set of pairs(I,ωI ), whereI ⊂ A[T ] is an idealof heightn having the property that Spec(A[T ]/I) is connected andI/I2 is generated byn elements, andωI : (A[T ]/I)n � I/I2 is a local orientation ofI .

Let I ⊂ A[T ] be an ideal of heightn. ThenI can be decomposed asI = I1 ∩ · · · ∩ Ir ,where theIk ’s are ideals ofA[T ] of heightn, pairwise comaximal and Spec(A[T ]/Ik) isconnected for eachk. The following lemma shows that such a decomposition is uniWe shall say thatIk are the connected components ofI .

Lemma 4.5. The decomposition ofI into its connected components, as described abis unique.

Proof. Suppose that we have another decompositionI = I ′1 ∩ · · · ∩ I ′

s , where I ′i are

pairwise comaximal and Spec(A[T ]/I ′i ) is connected. By set topological arguments

follows thatr = s and, after suitably renumbering, that Spec(A[T ]/Ii) = Spec(A[T ]/I ′i )

for everyi. Hence√Ii = √

I ′i . SinceIi are pairwise comaximal, it follows that there exi

f ∈ I1 andg ∈ I2 ∩ · · · ∩ Ir such thatf + g = 1. Let S be the multiplicative closed se


s.

n

idealsn

,

{1, g, g2, . . .}. It follows easily using the fact thatf + g = 1, thatI1 = S−1I ∩A[T ]. Now,using the facts that

√Ii = √

I ′i andI = I ′

1∩· · ·∩I ′r , it follows easily thatI ′

1 ⊂ S−1I∩A[T ].Hence,I ′

1 ⊂ I1. Similarly, I1 ⊂ I ′1. Therefore,I ′

1 = I1. Similarly, I ′i = Ii for everyi. This

proves the lemma. ✷Now assume thatI ⊂ A[T ] be an ideal of heightn such thatI/I2 is generated byn

elements. LetI = I1 ∩ · · · ∩ Ir be the decomposition ofI into its connected componentThen, ht(Ik) = n andIk/I2

k is generated byn elements. LetωI : (A[T ]/I)n � I/I2 be asurjection. ThenωI induces surjectionsωIk : (A[T ]/Ik)n � Ik/I

2k . By (I,ωI ) we mean the

elementΣ(Ik,ωIk ) of G.Let H be the subgroup ofG generated by the set of pairs(I,ωI ), whereI is an ideal

of A[T ] of heightn generated byn elements andωI : (A[T ]/I)n � I/I2 has the propertythatωI can be lifted to a surjectionθ :A[T ]n � I (in other words, a global orientatioof I ). We define thenth Euler class groupof A[T ], denoted byEn(A[T ]), to beG/H .

By a slight abuse of notation, we will writeE(A[T ]) for En(A[T ]) throughout thepaper.

LetP be a projectiveA[T ]-module of rankn having trivial determinant. Letχ :A[T ] �∧nP be an isomorphism. To the pair(P,χ), we associate an elemente(P,χ) of E(A[T ])

as follows:Let λ :P � I0 be a surjection, whereI0 is an ideal ofA[T ] of height n. Let bar

denote reduction moduloI0. We obtain an induced surjectionλ :P/I0P � I0/I20 . Note

that, sinceP has trivial determinant and dim(A[T ]/I0) � 1, by Corollary 2.4,P/I0P isa freeA[T ]/I0-module of rankn. We choose an isomorphismγ : (A[T ]/I0)

n � P/I0P ,such that

∧n(γ ) = χ . Let ωI0 be the surjectionλγ : (A[T ]/I0)

n � I0/I20 . Let e(P,χ) be

the image inE(A[T ]) of the element(I0,ωI0). We say that(I0,ωI0) is obtainedfrom thepair (λ,χ).

Lemma 4.6. The assignment sending the pair(P,χ) to the elemente(P,χ), as describedabove, is well defined.

Proof. Let µ :P � I1 be another surjection whereI1 ⊂ A[T ] is an ideal of heightn. Let(I1,ωI1) be obtained from(µ,χ).

Applying Lemma 2.12, we can find an idealK ⊂ A[T ] of height n such thatKis comaximal withI0, I1 and that there is a surjectionν :A[T ]n � I0 ∩ K such thatν ⊗ A[T ]/I0 = ωI0. SinceK and I0 are comaximal,ν induces a local orientationωKof K. Clearly,(I0,ωI0)+ (K,ωK)= 0 inE(A[T ]).

Let L = K ∩ I1. Note thatωK andωI1 together induce local orientationωL of L. Wewish to show that(L,ωL) = 0 in E(A[T ]), which proves the lemma because(L,ωL) =(K,ωK)+ (I1,ωI1) in E(A[T ]).

It is easy to see that applying Lemma 2.9 we can assume that each of theK(0), I0(0), I1(0) in A is either of heightn or equal toA. We take the case wheht(K(0))= ht(I0(0))= ht(I1(0))= n (other cases can be handled similarly).

Since e(P/T P,χ ⊗ A[T ]/(T )) is well defined inE(A) (see [B-RS3, Section 4])it follows that (L(0),ωL(0)) = 0 in E(A). Therefore, we can liftωL to a surjectionA[T ]n � L/(L2T ). On the other hand, using the fact thate(P ⊗A(T ),χ ⊗A(T )) is well


s

],

f

2]

t

defined inE(A(T )), it follows thatωL⊗A(T ) is a global orientation ofLA(T ). Therefore,in view of Remark 4.1, using Theorem 3.10 we see thatωL is a global orientation. Thiproves the lemma. ✷

We define theEuler classof (P,χ) to bee(P,χ).

Theorem 4.7. Let A be of dimensionn � 3, I ⊂ A[T ] be an ideal of heightn such thatI/I2 is generated byn elements and letωI : (A[T ]/I)n � I/I2 be a local orientation ofI .Suppose that the image of(I,ωI ) is zero in the Euler class groupE(A[T ]) ofA[T ]. Then,I is generated byn elements andωI can be lifted to a surjectionθ :A[T ]n � I .

Proof. Without loss of generality we can assume that eitherI (0) = A or ht(I (0)) = n.SupposeI (0) �=A. Now (I,ωI ) gives an element(I (0),ωI (0)) of E(A). Since(I,ωI )= 0in E(A[T ]), we have(I (0),ωI (0)) = 0 in E(A). Therefore, by [B-RS3, Theorem 4.2ωI(0) can be lifted to a set of generators ofI (0), which in turn implies thatωI can be liftedto a set of generators ofI/(I2T ). If I (0)=A, we can also liftωI to a set of generators oI/(I2T ).

InE(A(T )) also, the element(IA(T ),ωIA(T )) is zero, which, by [B-RS3, Theorem 4.implies thatωIA(T ) can be lifted to a set of generators ofIA(T ).

Using Theorem 3.10, the theorem follows.✷Now we prove our main theorem in a more general form.

Theorem 4.8. LetA be a Noetherian ring containingQ with dimA= n� 3 andI ⊂A[T ]be an ideal of heightn. LetP be a projectiveA[T ]-module of rankn whose determinanis trivial. Assume that we are given a surjection

ψ :P � I/(I2T

).

Assume further thatψ ⊗A(T ) can be lifted to a surjection

ψ ′ :P ⊗A(T )� IA(T ).

Then, there exists a surjectionΨ :P � I such thatΨ is a lift ofψ .

Proof. We fix an isomorphismχ :A[T ] � ∧n P .Let J (A,P ) denote the Quillen ideal ofP in A. Let J = J (A,P ) ∩ I . Since the

determinant ofP is trivial, we have, htJ (A,P )� 2. So it follows that, htJ � 2. Thereforewe can apply Lemma 3.9 and obtain a liftφ ∈ HomA[T ](P, I) of ψ and an idealI ′ ⊂A[T ]of heightn such that

(1) I ′ + (J 2T )=A[T ],(2) φ :P � I ∩ I ′ is a surjection, and(3) φ(P )+ (J 2T )= I .


a

and

m

m

al

ngs,

It follows thate(P,χ) = (I ∩ I ′,ωI∩I ′) in E(A[T ]) where the local orientationωI∩I ′is obtained by composingφ ⊗ A[T ]/(I ∩ I ′) with a suitable isomorphismλ : (A[T ]/I ∩I ′)n � P/(I ∩ I ′)P , as described above in the definition of an Euler class.

Therefore,e(P,χ) = (I,ωI )+ (I ′,ωI ′). We note that sinceI ′(0)= A, we can liftωI ′to a surjection fromA[T ]n � I ′/(I ′2T ). Moreover, considering the equation

e(P ⊗A(T ),χ ⊗A(T )

) = (IA(T ),ωI ⊗A(T )

) + (I ′A(T ),ωI ′ ⊗A(T )

)

in E(A(T )) and using the condition of the theorem, it is easy to deduce that(I ′A(T ),ωI ′ ⊗ A(T )) = 0 in E(A(T )). (Actually, the condition of the theorem tells thate(P ⊗A(T ),χ ⊗A(T ))= (IA(T ),ωI ⊗A(T )).) As a result, by Theorem 3.10,(I ′,ωI ′)= 0 inE(A[T ]). Therefore, by Theorem 4.7,I ′ = (l1, . . . , ln) such that this set of generators islift of ωI ′ .

Let us writeB =A1+J . Note that we haveI ′ + (J 2T )=A[T ] and thatJB is containedin the Jacobson radical ofB. Therefore, proceeding as in Step 3 of Theorem 3.10using Theorem 2.15 we can alter the above set of generators ofI ′B[T ] by an elementarytransformationσ ∈En(B[T ]) and can assume that

(i) ht(l1, . . . , ln−1)= n− 1,(ii) dimB[T ]/(l1, . . . , ln−1)� 1, and(iii) ln = 1 modulo(J 2T )B[T ].

We setC = B[T ],R = C[Y ],K1 = (l1, . . . , ln−1, Y + ln),K2 = IC[Y ],K3 =K1 ∩K2.Let us denoteP1+J by P ′.

We claim that there exists a surjectionη(Y ) :P ′[Y ] �K3 such thatη(0)= φ ⊗B[T ].We first show that the theorem follows from the claim. Specializingη atY = 1− ln, we

obtain a surjectionθ :P ′ � IB[T ].Sinceln = 1 modulo(J 2T )B[T ], the following equalities hold modulo(J 2T ):

θ = η(1− ln)= η(0)= φ.

Thereforeθ lifts ψ ⊗B[T ]. Now using Lemma 3.8, the theorem follows.

Proof of the claim. Recall that we have chosen an isomorphismλ : (A[T ]/I ∩ I ′)n �P/(I ∩ I ′)P such that

∧nλ = χ ⊗ A[T ]/(I ∩ I ′). This induces an isomorphis

µ : (A[T ]/I ′)n � P/I ′P such that∧n µ = χ ⊗ A[T ]/I ′. Also note thatφ ⊗ A[T ]/I ′ =

ωI ′µ−1.SinceC[Y ]/K1 � B[T ]/(l1, . . . , ln−1), we have dimC[Y ]/K1 � 1. Therefore, the

projectiveC[Y ]/K1-moduleP ′[Y ]/K1P′[Y ] is free of rankn. We choose an isomorphis

τ (Y ) : (C[Y ]/K1)n � P ′[Y ]/K1P

′[Y ] such that∧n

τ (Y )= χ ⊗C[Y ]/K1. Since∧n

µ=χ ⊗B[T ]/I ′B[T ], it follows thatτ (0) andµ differ by an element ofSLn(B[T ]/I ′B[T ]).Since I ′B[T ] + (J 2T )B[T ] = B[T ] and JB is contained in the Jacobson radicof B, we have, by Lemma 3.1, dim(B[T ]/I ′B[T ]) = 0. Hence,SLn(B[T ]/I ′B[T ]) =En(B[T ]/I ′B[T ]). Since elementary transformations can be lifted via surjection of riwe see that we may alterτ (Y ) by an element ofSLn(C[Y ]/K1) and assume thatτ (0)= µ.


tors

ma.

t

s

Let α(Y ) : (C[Y ]/K1)n � K1/K

21 denote the surjection induced by the set of genera

(l1, . . . , ln−1, Y + ln) of K1. Thus, we obtain a surjection

β(Y )= α(Y )τ(Y )−1 :P ′[Y ]/K1P′[Y ] �K1/K

21 .

Sinceτ (0) = µ, φ ⊗ B[T ]/I ′B[T ] = ωI ′µ−1, andα(0) = ωI ′ , we haveβ(0) = φ ⊗B[T ]/I ′B[T ]. Therefore, applying [M-RS, Theorem 2.3], we obtainη(Y ) :P ′[Y ] � K3such thatη(0)= φ ⊗B[T ].

Thus, the claim is proved and hence the theorem.✷To derive some corollaries of the above two theorems, we need the following lem

Lemma 4.9. LetA be a ring,I ⊂ A[T ] be an ideal, andP be a projectiveA[T ]-module.Suppose that we are given surjectionsα :P � I/I2 andβ :P � I (0) = I/I ∩ (T ) suchthat α ⊗A[T ]/I A/I (0) = β ⊗A A/I (0). Then there is a surjectionθ :P � I/(I2T ) suchthat θ lifts α andβ .

Proof. We choose liftsψ1,ψ2 ∈ HomA[T ](P, I) of α andβ , respectively. We note tha(ψ1 −ψ2)(P )⊂ I2 + (T ).

Since

I2 + (T )

I2 ∩ (T )= I2

I2 ∩ (T )⊕ (T )

I2 ∩ (T ),

considering(ψ1 − ψ2) as a map fromP to I2 + (T )/I2 ∩ (T ), we can decompose it a(ψ1 −ψ2)= (η1, η2).

Write θ1 = ψ1 − η1, θ2 = ψ2 − η2. Note thatθ1 andθ2 lift α andβ , respectively. It isalso easy to see thatθ1 = θ2. We call itθ .

So we have a mapθ :P → I/I2 ∩ (T ) which lifts bothα andβ . Write θ(P ) = K andconsider the idealK + (I2T ). SinceK + I ∩ (T ) = I andK + I2 = I , it follows thata maximal idealM of A[T ] containsI if and only if it containsK + (I2T ). Note thatsinceK + I2 = I , for every maximal idealM of A[T ] containingI we haveKM = IM .Therefore, it follows thatK + (I2T ) = I . In other words,θ is a surjection. Sinceθ liftsbothα andβ , this proves the lemma.✷Corollary 4.10. LetA be of dimensionn� 3. LetP be a projectiveA[T ]-module of ranknhaving trivial determinant andχ be a trivialization of

∧nP . Let I ⊂ A[T ] be an ideal of

heightn such thatI/I2 is generated byn elements. LetωI be a local orientation ofI .Suppose thate(P,χ)= (I,ωI ) in E(A[T ]). Then, there exists a surjectionα :P � I suchthat (I,ωI ) is obtained from(α,χ).

Proof. Since determinant ofP is trivial, P/IP is a freeA[T ]/I -module of rankn. Wecan choose an isomorphismλ :P/IP � (A[T ]/I)n such that

∧nλ = (χ ⊗ A[T ]/I)−1.

Therefore, we get a surjectionωIλ :P/IP � I/I2.


a

by

it is

ing

Let,te

We can assume that eitherI (0) = A or ht(I (0)) = n. First suppose thatI (0) = A.Then it follows using Lemma 4.9 that there is a surjection fromP to I/(I2T ) which liftsωIλ :P/IP � I/I2.

If ht(I (0))= n, then sincee(P/T P,χ⊗A[T ]/(T ))= (I (0),ωI (0)) inE(A), it followsfrom [B-RS3, 4.3], that there is a surjectionα :P/T P � I (0) such that(I (0),ωI (0)) isobtained from(α,χ ⊗ A[T ]/(T )). Therefore, it follows from Lemma 4.9 that we havesurjectionP � I/(I2T ) which is a lift ofωIλ.

So, in any case, we have a surjectionγ :P � I/(I2T ) which lifts ωIλ.Sincee(P ⊗ A(T ),χ ⊗ A(T )) = (IA(T ),ωI ⊗ A(T )) in E(A(T )), it follows from

[B-RS3, 4.3] that there is a surjectionΓ :P ⊗ A(T ) � IA(T ) such that(IA(T ),ωI ⊗A(T )) is obtained from(Γ,χ ⊗A(T )). This actually means thatΓ is a lift of γ ⊗A(T ).

Therefore, by Theorem 4.8, the result follows.✷Corollary 4.11. LetA be as above. LetP be a projectiveA[T ]-module of rankn havingtrivial determinant andχ be a trivialization of

∧nP . Then,e(P,χ) = 0 if and only ifP

has a unimodular element. In particular, ifP has a unimodular element thenP maps ontoany ideal ofA[T ] of heightn generated byn elements.

Proof. Let α :P � I be a surjection whereI is an ideal inA[T ] of height n. Lete(P,χ)= (I,ωI ) in E(A[T ]), where(I,ωI ) is obtained from the pair(α,χ).

First assume thate(P,χ) = 0. Then e(P ⊗ A(T ),χ ⊗ A(T )) = 0 in E(A(T )).Therefore, by [B-RS3, 4.4],P ⊗ A(T ) has a unimodular element. Consequently,[B-RS4, Theorem 3.4], it follows thatP has a unimodular element.

Now we assume thatP has a unimodular element. But then, following Remark 4.1,easy to see that(I,ωI )= 0 in E(A[T ]).

The last assertion of the corollary follows from Corollary 4.10.✷Corollary 4.12. Let dimA = n � 2 and I ⊂ A[T ] be an ideal of heightn. Let P be aprojectiveA[T ]-module of rankn with trivial determinant andα :P � I be a surjection.Suppose thatP has a unimodular element. ThenI is generated byn elements.

Proof. If n = 2, then by Corollary 2.4P is a free module and henceI is generated bynelements. Therefore, in what follows, we assumen� 3.

Let us fix an isomorphismχ :A[T ] � ∧nP . Suppose that(I,ωI ) ∈ E(A[T ]) is

obtained from the pair(α,χ). Then we havee(P,χ) = (I,ωI ) in E(A[T ]). SinceP hasa unimodular element, it follows from Corollary 4.11 thate(P,χ)= 0. Now the corollaryfollows from Theorem 4.7. ✷

Now we prove the “subtraction principle” in a more general form in the followcorollary.

Corollary 4.13. LetA be a Noetherian ring withdimA= n� 3. LetP andQ be projectiveA[T ]-modules of rankn andn− 1, respectively, such that their determinants are free.χ :

∧n(P ) � ∧n

(Q⊕A[T ]) be an isomorphism. LetI1, I2 ⊂ A[T ] be comaximal idealseach of heightn. Letα :P � I1 ∩ I2 andβ :Q⊕A[T ] � I2 be surjections. Let bar deno


d

e

,

e Euleres

,

reduction moduloI2 and α :P � I2/I22 , β :Q⊕A[T ] � I2/I

22 be surjections induce

from α andβ , respectively. Suppose that there exists an isomorphismδ :P � Q⊕A[T ]such that(i) βδ = α, (ii)

∧n(δ) = χ . Then, there exists a surjectionθ :P � I1 such that

θ ⊗A[T ]/I1 = α ⊗A[T ]/I1.

Proof. Let us fix an isomorphismσ :A[T ] � ∧n(P ). Let (I1 ∩ I2,ωI1∩I2) be obtained

from (α,σ ). Thene(P,σ )= (I1 ∩ I2,ωI1∩I2)= (I1,ωI1)+ (I2,ωI2) in E(A[T ]).On the other hand, let(I2, ω̃I2) be obtained from(β,χσ). It is easy to see, from th

conditions stated in the proposition, that(I2,ωI2) = (I2, ω̃I2) in E(A[T ]). ButQ⊕A[T ]has a unimodular element. Therefore, it follows that(I2,ωI2) = 0. Consequentlye(P,σ )= (I1,ωI1). Therefore, the result follows from Corollary 4.10.✷

5. A “Quillen–Suslin theory” for the Euler class groups

In this section we investigate some questions concerning the relations among thclass groupsE(A), E(A[T ]), andE(A(T )). The motivation for these questions comfrom the Quillen–Suslin theory for projective modules.

Remark 5.1. Let A be a Noetherian ring with dimA = n � 3. Let I ⊂ A[T ] be an idealof heightn andωI : (A[T ]/I)n � I/I2 be a local orientation ofI . Let f ∈ A[T ]/I be aunit. ComposingωI with an automorphism of(A[T ]/I)n with determinantf , we obtainanother local orientation ofI which we denote byfωI . On the other hand, letωI , ω̃I betwo local orientations ofI . Then, it is easy to see from Lemma 2.5, that̃ωI = fωI forsome unitf ∈A[T ]/I .

The following is an improvement of Lemma 3.6.

Lemma 5.2. Let A be a Noetherian ring withdimA = n � 3, I ⊂ A[T ] an ideal ofheight n and ωI : (A[T ]/I)n � I/I2 a surjection. Suppose thatωI can be lifted to asurjectionα :A[T ]n � I . Let f ∈ A[T ] be a unit moduloI . Let θ ∈ GLn(A[T ]/I) withdeterminantf 2. Then, the surjectionωI θ : (A[T ]/I)n � I/I2 can be lifted to a surjectionβ :A[T ]n � I .

Proof. Without loss of generality we can assume that eitherI (0)= A or I (0) has heightn.Suppose that ht(I (0)) = n. Now ωI induces a surjection, sayωI(0) : (A/I (0))n �

I (0)/I (0)2, which can be lifted toα(0) :An � I (0). Note thatf (0) ∈ A is a unit moduloI (0) and θ(0) (∈ GLn(A/I (0))) has determinantf (0)2. Therefore, by [B-RS3, 5.3]ωI(0)θ(0) can be lifted to a surjectionβ :An � I (0). Consequently, we can liftωI θ toa surjectionψ :A[T ]n � I/(I2T ).

On the other hand, ifI (0) = A, we can always liftωI θ to a surjectionψ :A[T ]n �I/(I2T ).

Similarly, we can apply [B-RS3, 5.3] toA(T ) and find thatωI θ ⊗ A(T )/IA(T ) :(A(T )/IA(T ))n � IA(T )/I2A(T ) can be lifted to a surjectionφ :A(T )n � IA(T ).

Now the lemma is a consequence of Theorem 3.10.✷


e as

re,eal

an

nce

l

early,

Applying Lemma 5.2, we obtain the following lemma. (The method of proof is samin [B-RS3, 5.4].)

Lemma 5.3. LetA be a Noetherian ring withdimA= n� 3, I ⊂ A[T ] an ideal of heightnandωI be a local orientation ofI . Letf be a unit moduloI . Then(I,ωI )= (I, f 2ωI ) inE(A[T ]).

Proof. If (I,ωI ) = 0 in E(A[T ]), then the result follows from Lemma 5.2. Therefolet us assume that(I,ωI ) �= 0 in E(A[T ]). Then, by Lemma 2.12, there exists an idI1 ⊂A[T ] of heightn which is comaximal withI and a surjectionα :A[T ]n � I ∩ I1 suchthatα⊗A[T ]/I = ωI . Letα⊗A[T ]/I1 = ωI1. By the Chinese remainder theorem, we cchooseg ∈A[T ] such thatg = f 2 moduloI andg = 1 moduloI1. Applying Lemma 5.2,we see that there exists a surjectionγ :A[T ]n � I ∩ I1 such thatγ ⊗ A[T ]/I = f 2ωIandγ ⊗A[T ]/I1 = ωI1. From the surjectionα we get(I,ωI )+ (I1,ωI1)= 0 inE(A[T ]).From the surjectionγ we get(I, f 2ωI )+ (I1,ωI1)= 0 in E(A[T ]). Therefore,(I,ωI )=(I, f 2ωI ) in E(A[T ]). This completes the proof.✷

A consequence of Quillen’s ‘local–global principle’ [Qu] is that the following sequeof groups is exact.

0 → K̃0(A)→ K̃0(A[T ]) →

∏m

K̃0(Am[T ]),

where the direct product runs over all maximal idealsm of A. The following theoremshows that a ‘local–global principle’ holds for Euler class groups also.

Theorem 5.4. Let A be a Noetherian ring containingQ with dimA = n � 3. Then thefollowing sequence of groups is exact.

0 →E(A)→E(A[T ]) →

∏m

E(Am[T ]),

where the direct product runs over all maximal idealsm ofA such thatht(m)= n.

To prove this theorem, we need the following lemma.

Lemma 5.5. Let R be a semilocal ring(containingQ) of dimensionn � 3, I1 ⊂ R[T ]be an ideal of heightn such thatI1 + JR[T ] = R[T ] whereJ is the Jacobson radicaof R. Suppose thatωI1 is a local orientation ofI1 given byI1 = (f1, . . . , fn) + I2

1 andωI1 ⊗ Rm[T ] can be lifted to a set of generators ofI1 ⊗ Rm[T ] for all maximal idealsmofR of heightn (hence for all maximal ideals ofR). Then there is a set of generators ofI1which liftsωI1 .

Proof. We proceed by induction on the number of maximal ideals in the base ring. Clwhen the base ring is local, we have nothing to prove.


nl

,

eal

by

nd

e

Suppose that max(R)= {m1, . . . ,mk}. LetI1 = I ′1∩· · ·∩I ′

r be a primary decompositioof I1. LetPi = √

I ′i ∩R. SinceI1 +JR[T ] =R[T ], it follows that

√I ′i is a maximal idea

of R[T ] of heightn whereasPi is a prime ideal ofR of heightn− 1.Suppose that, from the family of prime ideals{P1, . . . ,Pr } (after renumbering)

{P1, . . . ,Ps} is the collection such thatPi is not contained inm1. Note that this collectioncan be empty.

We writeT1 =R− (P1 ∪ · · · ∪Ps) andT =R− (P1 ∪ · · · ∪Ps ∪m1); S1 = T −11 R and

S = T −1R. Note thatT ⊂ T1 and henceS1 is a localization ofS.SinceS1 is a semilocal ring such that all the maximal ideals ofS1 are of heightn− 1,

it follows by a theorem of Mandal [M1], that(I1,ωI1) = 0 in E(S1[T ]). Therefore, thereexists t1 ∈ T1 such that(I1,ωI1) = 0 in E(St1[T ]). Since t1 /∈ P1 ∪ · · · ∪ Ps , we have(t1) + I1S[T ] = S[T ]. Now adapting the proof of [B-RS3, 5.6] we can find an idI2 ⊂ S[T ] of heightn such that

(1) tp1 ∈ I2 for some positive integerp,(2) I1S[T ] ∩ I2 = (g1, . . . , gn), and(3) (I1S[T ],ω′

I1)+ (I2,ωI2) = 0 in E(S[T ]) where the local orientations are induced

g1, . . . , gn.

Note that, by Lemma 5.3,(I1S[T ],ωI1 ⊗ S[T ]) = (I1S[T ],ω′I1) and therefore(I1S[T ],

ωI1 ⊗ S[T ])+ (I2,ωI2)= 0 inE(S[T ]).Let J2 = I2 ∩ S. Note that sincetp1 ∈ J2, J2 is not contained inP1 ∪ · · · ∪Ps and hence

S1+J2 is a local ring with maximal idealm1S1+J2. Therefore, it follows that(I2,ωI2) = 0in E(S1+J2[T ]). We claim that this implies(I2,ωI2)= 0 in E(S[T ]).

Proof of the claim. We note that sinceS is semilocal, we can apply Proposition 3.2 a[B-RS1, 3.9] to adjustωI2 so that it is induced by a set of generators ofI2/(I

22T ), say,

given byI2 = (G1, . . . ,Gn)+ (I22T ) (therefore,Gi = gi modI2

2 ). We have obtained in thabove paragraphI2S1+J2[T ] = (F1, . . . ,Fn) such thatFi = gi modI2

2S1+J2[T ]. Therefore,it follows that I2S1+J2(T ) = (F1, . . . ,Fn) such thatFi = Gi mod I2

2S1+J2(T ). ApplyingTheorem 3.10, we getI2S1+J2[T ] = (H1, . . . ,Hn) such thatHi =Gi mod(I2

2T )S1+J2[T ].Now we can apply Lemma 3.8 and conclude that(I2,ωI2)= 0 inE(S[T ]). This proves theclaim.

So it follows that(I1S[T ],ωI1 ⊗ S[T ])= 0 inE(S[T ]).We can repeat the same arguments as above and findt ∈ T and an idealI3 ⊂ R[T ] of

heightn such that

(1) tq ∈ I3 for some positive integerq ,(2) I1 ∩ I3 = (h1, . . . , hn), and(3) (I1,ωI1) + (I3,ωI3) = 0 in E(R[T ]) where the local orientationωI3 is induced by

h1, . . . , hn.


f

.5]),s), that

f

is

Let J3 = I3 ∩ R. Sincetq ∈ J3 andt /∈ P1 ∪ · · · ∪ Ps ∪m1, we see that max(R1+J3) ⊂{m2, . . . ,mk}. Therefore, by the induction hypothesis it follows thatωI3 can be lifted toa set of generators ofI3R1+J3[T ]. Therefore, as above, this implies that(I3,ωI3) = 0 inE(R[T ]). Consequently,(I1,ωI1)= 0 in E(R[T ]). This proves the lemma.✷

The following is an alternative proof of the above lemma in the case whenR is a domain.We do not assume in this proof thatI1 is comaximal withJR[T ].

Proof. Suppose that max(R)= {m1, . . . ,mk}. SinceωI1 ⊗Rm1[T ] can be lifted to a set ogenerators ofI1Rm1[T ], we can finda ∈R −m1 such thatI1a = (g1, . . . , gn) with gi = fimoduloI2

1a. Let J1 = I1 ∩ R. Let b ∈ J 2

1 andc = ab. Let bar denote reduction moduloc.

Note that dimR � n− 1. Adapting the method of Bhatwadekar–Rao (see [B-RS5, 2we obtain,I1 = (h1, . . . , hn ) such thathi = fi mod I1

2. By adding suitable multipleof c to h1, . . . , hn, we may assume by the Eisenbud–Evans theorem (Corollary 2.11(h1, . . . , hn) = I1 ∩ I2, where ht(I2) = n and I2 + (c) = R[T ]. Note that(h1, . . . , hn)

inducesωI1 and a local orientationωI2 of I2 such that(I1,ωI1)+(I2,ωI2)= 0 inE(R[T ]).Applying the subtraction principle (Proposition 4.3), we see thatI2a = (k1, . . . , kn) with

ki = hi mod I22a

. We note thata is a unit moduloI2. Therefore, adapting the proof o[B-RS3, 5.6] we can find an idealI3 ⊂R[T ] of heightn such that

(1) I3 contains a power ofa,(2) I2 ∩ I3 = (l1, . . . , ln), and(3) (I2,ω

′I2)+ (I3,ωI3)= 0 inE(R[T ]).

Since by Lemma 5.3, we have(I2,ωI2) = (I2,ω′I2) in E(R[T ]), it follows that

(I2,ωI2)+ (I3,ωI3)= 0 inE(R[T ]).Let J3 = I3 ∩ R. Sincea ∈ R −m1, whereasJ3 contains a power ofa, it follows that

m1 does not belong to max(R1+J3). Therefore, by the induction hypothesis,(I3,ωI3) ⊗R1+J3[T ] = 0 in E(R1+J3[T ]). As in the first proof (see the claim and its proof), thimplies that(I3,ωI3)= 0 inE(R[T ]). Therefore, it follows that(I1,ωI1)= 0 inE(R[T ]).This proves the lemma.✷Proof of Theorem 5.4. Let I ⊂A[T ] be an ideal of heightn andωI be a local orientationof I such thatωI ⊗ Am[T ] is a global orientation for all maximal idealsm of A ofheightn. We show that there exists an idealJ ⊂ A of heightn and a local orientationωJ : (A/J )n � J/J 2 such that(J [T ],ωJ ⊗A[T ])= (I,ωI ) in E(A[T ]).

SinceA containsQ, we can assume that eitherI (0) is an ideal of heightn or I (0)= A.

Case 1. I (0) is proper.

Applying Lemma 2.12 we can find an idealK ⊂A of heightn which is comaximal withI ∩A and a local orientationωK of K such that(I (0),ωI (0))+ (K,ωK)= 0 inE(A).


f

fators of

a

inivas

n

al

gular.

l

Let L = I ∩ K[T ]. Since the idealsI andK[T ] are comaximal,ωI andωK induceωL : (A[T ]/L)n �L/L2 and we have the following equation inE(A[T ]):

(L,ωL)= (I,ωI )+ (K[T ],ωK ⊗A[T ]).

Since(L(0),ωL(0)) = (I (0),ωI (0)) + (K,ωK) = 0, it follows that we can liftωL toa set of generators ofL/(L2T ). We proceed to prove that(L,ωL) = 0. Note that sinceK[T ] is extended andL= I ∩K[T ], it follows thatωL ⊗Am[T ] is a global orientation oL⊗Am[T ] for all maximal idealsm of A of heightn.

SinceωL is actually induced by a set of generators ofL/(L2T ), adapting the proof oTheorem 3.10 and applying the above lemma, it is easy to see that this set of generL/(L2T ) can be lifted to a set of generators ofL.

Thus, (L,ωL) = 0 in E(A[T ]) and this, in turn, implies that(I,ωI ) = (I (0)[T ],ωI (0) ⊗A[T ]) in E(A[T ]).

Case 2. I (0)=A.

Then we can liftωI to a set of generators ofI/(I2T ). Proceeding as we did forL inCase 1, we conclude that(I,ωI )= 0 inE(A[T ]).

Therefore, the proof of the theorem is complete.✷As a consequence, we get the following interesting corollary.

Corollary 5.6. Let A be a Noetherian ring containingQ with dimA = n � 3 andP bea projectiveA[T ]-module of rankn with trivial determinant. Suppose thatP/T P has aunimodular element. Assume further that the projectiveAm[T ]-moduleP ⊗Am[T ] has aunimodular element for every maximal idealm ofA of heightn. Then,P has a unimodularelement.(Taking P = Q[T ], we see that the condition thatP/T P has a unimodularelement, is necessary.)

Let A be a Noetherian ring containingQ with dimA = n � 3. Note that we havecanonical mapΦ :E(A)→E(A[T ]). It is easy to see thatΦ is injective.

However, there is an example due to Bhatwadekar, Mohan Kumar, and Sr[B-RS1, Example 6.4], where they have constructed a normal affine domainA over Cof dimension 3, an idealI ⊂ A[T ] of height 3 such thatI (0) = A and a surjectionφ̃ :A[T ]3 � I/(I2T ), and it has been shown thatφ̃ cannot be lifted to a surjectiofrom A[T ]3 to I (in fact, I is not a surjective image of any projectiveA[T ]-module ofrank 3 which isextended fromA). It also follows from their example that for any locorientationωI of I , the element(I,ωI ) in E(A[T ]) does not come fromE(A). Therefore,the canonical mapΦ :E(A)→E(A[T ]) is not surjective in general.

We note that in their example the affine domain in question is normal, but not reTherefore, one can ask the following natural question.

Question 1. Let A be a regular ring containingQ with dimA = n � 3. Is the canonicamapΦ :E(A)→E(A[T ]) an isomorphism?


ith

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The following proposition gives a partial answer to the above question.

Proposition 5.7. Let A be a smooth affine domain containing the field of rationals wdimA= n� 3. Then the canonical mapΦ :E(A)→E(A[T ]) is an isomorphism.

Proof. Let (I,ωI ) ∈ E(A[T ]) whereI is an ideal ofA[T ] of heightn andωI is a localorientation ofI . By Lemma 2.9, without loss of generality we may assume that eI (0)=A or ht(I (0))= n.

If I (0) = A, then we can liftωI to a set of generators ofI/(I2T ). Then, by [B-RS1,Theorem 3.8], it follows thatωI is a global orientation.

Now suppose that htI (0)= n. We consider the element(I (0),ωI (0)) ∈E(A), inducedby (I,ωI ). Applying Lemma 2.12, we can find an idealK ⊂ A of height n whichis comaximal withI ∩ A and a local orientationωK of K such that(I (0),ωI (0)) +(K,ωK)= 0 inE(A).

Let L = I ∩ K[T ]. Since the idealsI andK[T ] are comaximal,ωI andωK induceωL : (A[T ]/L)n �L/L2 and we have the following equation inE(A[T ]):

(L,ωL)= (I,ωI )+ (K[T ],ωK ⊗A[T ]).

Since(L(0),ωL(0))= (I (0),ωI (0))+ (K,ωK)= 0 inE(A), it follows that we can liftωLto a set of generators ofL/(L2T ). Now, by [B-RS1, 3.8],ωL is a global orientation. Thus(L,ωL)= 0 and hence it follows that(I,ωI )= (I (0)[T ],ωI (0) ⊗A[T ]) in E(A[T ]).

Therefore,Φ is a surjection. ✷LetA be a Noetherian ring containingQ with dimA= n� 3. Since the ring extensio

A[T ] → A(T ) is flat, we see that there is a canonical mapΓ :E(A[T ]) → E(A(T )). Weend this section discussing the following interesting question, which is, in fact, an anaof the Affine Horrocks theorem.

Question 2. Is the canonical mapΓ :E(A[T ])→E(A(T )) injective?

The following proposition gives a partial answer to the above question.

Proposition 5.8. LetA be a Noetherian ring of dimensionn� 3 containingQ. Then thecanonical mapΓ :E(A[T ])→E(A(T )) is injective in the following cases:

(1) htJ � 1, whereJ denotes the Jacobson radical ofA.(2) A is an affine domain over an algebraically closed fieldk of characteristic zero.

Proof. (1) Let (I,ωI ) ∈ E(A[T ]) be such that,(IA(T ),ωI ⊗ A(T )) = 0 in E(A(T )).Now since ht(J )� 1, it follows that ht(J , T )� 2. Therefore, using Lemma 2.12, we cfind an idealK ⊂ A[T ] of heightn and a local orientationωK such thatK is comaximalwith I ∩ (J , T ) and(I,ωI )+ (K,ωK) = 0 in E(A[T ]). SinceK + (J , T )= A[T ], it iseasy to see thatK(0)= A. Therefore, we can liftωK to a set of generators ofK/(K2T ).


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Now, since(KA(T ),ωK ⊗ A(T )) = 0 in E(A(T )), it follows from Theorem 3.10 tha(K,ωK)= 0 inE(A[T ]). Therefore,(I,ωI )= 0 in E(A[T ]). This proves (1).

(2) From (1) it follows thatΓ is injective whenA is local. Therefore, in view of Theorem 5.4, it is easy to see that to prove the injectivity ofΓ (for any Noetherian ringA con-tainingQ with dimA� 3), it is enough to prove the injectivity of the canonical map frE(A) toE(A(T )). We note that this is exactly how the Quillen–Suslin theorem is pro

Let A be an affine domain over an algebraically closed fieldk of characteristic 0. Le(I,ωI ) ∈E(A) be such that(I ⊗A(T ),ωI ⊗A(T ))= 0 inE(A(T )).

Suppose thatI = (a1, . . . , an) + I2 and this set of generators ofI/I2 correspondsto ωI . By Lemma 2.8, we see that there existsa ∈ I such thatI1−a = (a1, . . . , an). Writeb = 1 − a. Note that ifb is a unit inA, we are done. Therefore assume thatb is not aunit inA. Then, sincek is algebraically closed,b is transcendental overk. We consider themultiplicatively closed setS = {1+ cb | c ∈ k[b]}. Note thata ∈ S.

We consider the surjectionα :Anb � Ib which sendsei to ai and the surjectionβ :

AnS � IS which sendse1 to 1 andei to 0 for i � 2. We note thatAbS is an affine domain

over theC1-field k(b) of dimensionn − 1. Therefore, by a result of Suslin [Su2], tunimodular row(a1, . . . , an) overAbS is completable to a matrixσ ∈ SLn(AbS) and henceby patching we obtain a surjectionγ :P � I whereP is a projectiveA-module of ranknwith trivial determinant.

We fix an isomorphismχ :A � ∧n(P ). Then (γ,χ) induces an element(I, ω̃I )in E(A). It follows from Remark 5.1, thatωI = cω̃I for some unitc ∈ A/I . Sincek isan algebraically closed field of characteristic zero and dim(A/I) = 0, we can find a unid ∈ A/I such thatdn−1 = c. Now by [B-RS3, 5.1], there exists a projectiveA-moduleP1 of rankn with trivial determinant, an isomorphismχ1 :A � ∧n

(P1) and a surjectionδ :P1 � I such thate(P1, χ1)= (I, dn−1 ω̃I ) in E(A). Thus,e(P1, χ1)= (I,ωI ) in E(A).Now, since(IA(T ),ωI ⊗ A(T )) = 0 in E(A(T )), it follows from [B-RS3, 4.4], thatP1 ⊗ A(T ) has a unimodular element. Therefore, by [B-RS4, 3.4],P1 has a unimodulaelement. Hence(I,ωI )= 0 inE(A). This completes the proof.✷

6. The weak Euler class group of A[T ]

Let A be a Noetherian ring of dimensionn � 3 containing the field of rationals. Wdefine thenth weak Euler class groupEn

0(A[T ]) of A[T ], in the following way:

Let S be the set of idealsI ⊂A[T ] with the properties:(i) ht(I )= n,(ii) I/I2 is generated byn elements, and(iii) Spec(A[T ]/I ) is connected.LetG be the free abelian group onS.

Let I ⊂ A[T ] be an ideal of heightn such thatI/I2 is generated byn elements.Now I can be decomposed asI = ⋂k

i=1Ii whereIi ’s are pairwise comaximal anSpec(A[T ]/Ii ) is connected for eachi. We associate toI , the element

∑Ii of G. By

abuse of notation we denote this element by(I).


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LetH be the subgroup ofG generated by elements of the type(I), whereI ⊂A[T ]is an ideal of heightn such thatI is generated byn elements.

We defineEn0(A[T ])=G/H .

By a slight abuse of notation, we will writeE0(A[T ]) for En0(A[T ]) in what follows.

We note that there is a canonical surjective group homomorphism fromE(A[T ]) toE0(A[T ]) obtained by forgetting the orientations.

We first prove some general results onE(A[T ]) in the form of the following lemmasWe will need them to prove results onE0(A[T ]).

The proof of the following lemma is contained in [B-RS3, 2.7, 2.8, and 5.1] and hwe omit the proof.

Lemma 6.1. LetA be a Noetherian ring withdimA= n� 3. LetP be a projectiveA[T ]-module of rankn having trivial determinant andχ :A[T ] � ∧n

P be a trivialization. Letα :P � I be a surjection, whereI ⊂ A[T ] is an ideal of heightn and let (I,ωI ) beobtained from(α,χ). Let f ∈ A[T ] be a unit moduloI . Then, there exists a projectivA[T ]-moduleP1 of rankn having trivial determinant, a trivializationχ1 of

∧nP1, and a

surjectionβ :P1 � I such that:

(i) P is stably isomorphic toP1,(ii) (I, f n−1ωI ) is obtained from(β,χ1).

The following lemma is essentially a translation of Lemma 2.12 in the language ofclass groups.

Lemma 6.2. Let A be a Noetherian ring of dimensionn � 3, I ⊂ A[T ] an ideal ofheightn and ωI a local orientation ofI . Suppose that(I,ωI ) �= 0 in E(A[T ]). Then,there exists an idealI1 of A[T ] of heightn and a local orientationωI1 of I1 such that(I,ωI ) + (I1,ωI1) = 0 in E(A[T ]). Further, given any finite set of idealsK1, . . . ,Kt ofA[T ] with ht(Ki) � 2, I1 can be chosen with the additional property that it is comaximwith eachKi .

Adapting the proof of [B-RS2, 3.7] and using the Eisenbud–Evans theorem (Clary 2.11) in place of “Swan’s Bertini” theorem, the following lemma can be easilyduced.

Lemma 6.3. Let A be a Noetherian ring of even dimensionn � 4. Let P be a stablyfree A[T ]-module of rankn and χ :A[T ] � ∧n

(P ) be an isomorphism. Suppose the(P,χ) = (I,ωI ) in E(A[T ]), whereI ⊂ A[T ] is an ideal of heightn andωI is a localorientation ofI . Then there is an idealI1 ⊂ A[T ] generated byn elements and a locaorientationωI1 of I1 such that(I,ωI )= (I1,ωI1) in E(A[T ]). Moreover,I1 can be chosento be comaximal with any given ideal ofA[T ] of heightn.

The following three propositions can be proved by using Lemmas 5.3, 6.1–6.3 opaper and adapting the proofs of [B-RS2, 3.8, 3.9, 3.10].


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Proposition 6.4. LetA be a Noetherian ring of even dimensionn � 4. Let I1, I2 be twocomaximal ideals ofA[T ], each of heightn. Let I3 = I1 ∩ I2. If any two ofI1, I2, andI3are surjective images of stably free projectiveA[T ]-modules of rankn, then so is the third

Proposition 6.5. LetA be a Noetherian ring of even dimensionn� 4. LetI ⊂A[T ] be anideal of heightn such thatI/I2 is generated byn elements. Then(I) = 0 in E0(A[T ]) ifand only ifI is the surjective image of a stably free projectiveA[T ]-module of rankn.

Proposition 6.6. LetA be a Noetherian ring of even dimensionn� 4. LetP be a projectiveA[T ]-module of rankn with trivial determinant. Suppose thatP maps onto an ideaI ⊂ A[T ] of heightn. Then(I) = 0 in E0(A[T ]) if and only if [P ] = [Q ⊕ A[T ]] inK0(A[T ]) for some projectiveA[T ]-moduleQ of rankn− 1.

Proposition 6.7. LetA be a Noetherian ring of even dimensionn� 4, I ⊂ A[T ] an idealof heightn such thatI/I2 is generated byn elements. Let̃ωI : (A[T ]/I)n � I/I2 bea surjection. Suppose that the element(I, ω̃I ) of E(A[T ]) belongs to the kernel of thcanonical homomorphismE(A[T ]) � E0(A[T ]). Then, there exists a stably freeA[T ]-moduleP1 of rank n and a trivializationχ1 of

∧nP such thate(P1, χ1) = (I, ω̃I ) in

E(A[T ]).

Proof. (We follow the same proof as in [B-RS3, 6.5].) Since(I) = 0 in E0(A[T ]), byProposition 6.5, there exists a stably freeA[T ]-moduleP of rank n and a surjectionα :P � I . Let χ :A[T ] � ∧n

(P ) be an isomorphism. Suppose that(I,ωI ) is obtainedfrom (α,χ). By Remark 5.1, there existsf ∈ A[T ] such thatf ∈ A[T ]/I is a unit andω̃I = fωI . By Lemma 6.1, there exists a projectiveA[T ]-moduleP1 such thatP1 isstably isomorphic toP and an isomorphismχ1 :A[T ] � ∧n

(P1), such thate(P1, χ1) =(I, f n−1ωI ) in E(A[T ]). Sincen is even, by Lemma 5.3 we have(I, f n−1ωI )= (I,f ωI )

in E(A[T ]). Hence,e(P1, χ1)= (I, ω̃I ) in E(A[T ]). ✷

7. The case of dimension two

In this section we briefly outline results similar to those in the previous sections icase when dimension of the base ring is two.

We first note that there is an example [B-RS1, Example 3.15] which shows that thetheorem (Theorem 3.10) is not true if dimA= 2. However, we have the following theorem

Theorem 7.1. LetA be a Noetherian ring of dimension2 (containingQ) and I ⊂ A[T ]be an ideal of height2 such thatI = (f1, f2)+ (I2T ). Suppose that there existF1,F2 ∈IA(T ) such thatIA(T )= (F1,F2) andFi = fi modI2A(T ) for i = 1,2. Then, there exish1, h2, andθ ∈ SL2(A[T ]/I) such that

(i) I = (h1, h2),(ii) (f1, f2 )θ = (h1,h2 ) (bar denoting moduloI2), and(iii) fi(0)= hi(0) for i = 1,2.


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Proof. Since a unimodular row of length two (over any ring) is always completaba matrix with determinant 1, it follows easily using a standard patching argumenthere is a projectiveA[T ]-moduleP of rank 2 with trivial determinant mapping ontoI .Letα :P � I be the surjection. Fix an isomorphismχ :A[T ] � ∧2

P . SinceP/IP is free,α andχ induce a set of generators ofI/I2, sayI = (g1, g2)+ I2.

It follows from Lemma 2.5 that there is a matrixσ ∈ GL2(A[T ]/I) with determinant(say)f such that(f1,f2 )= (g1,g2 )σ . Now following [B-RS3, 2.7, 2.8], we see that, theexists a projectiveA[T ]-moduleP1 of rank 2 having trivial determinant, a trivializationχ1of

∧2P1, and a surjectionβ :P1 � I such that if the set of generators ofI/I2 induced byβ

andχ1 is h1,h2, then(h1, h2 ) = (g1,g2 )δ, whereδ ∈ GL2(A[T ]/I) has determinantf .Therefore, it follows that the two sets of generators,(f1,f2 ) and (h1, h2 ) of I/I2 areconnected by a matrix inSL2(A[T ]/I).

The above discussion makes it clear thate(P1 ⊗ A(T ),χ1 ⊗ A(T )) = (IA(T ),

ωI ⊗A(T )) in E(A(T )), whereωI : (A[T ]/I)2 � I/I2 is the surjection correspondingthe generators(f1, f2 ). Therefore, from the given condition of the theorem it follows tP1 ⊗A(T ) has a unimodular element and hence is free. Therefore, by the Affine Hortheorem,P1 is a freeA[T ]-module. This proves (i) and (ii).

To prove (iii), note thatI (0) = (f1(0), f2(0)) = (h1(0), h2(0)) and there is somγ ∈ SL2(A/I (0)) such that(f̃1(0), f̃2(0)) = (h̃1(0), h̃2(0))γ , where tilde denotes reduction moduloI (0)2. Applying [B-RS3, Lemma 2.3], we getΓ ∈ SL2(A) such that(f1(0), f2(0)) = (h1(0), h2(0))Γ . Changing(h1, h2) by thisΓ , we get the desired set ogenerators ofI . ✷

As applications of the above theorem, we can prove the following additionsubtraction principles. The method of proof is the same as that used in Section 4 andomitted.

Corollary 7.2. LetA be a Noetherian ring containingQ with dimA= 2 andI1, I2 be twocomaximal ideals inA[T ], each of height2. Suppose thatI1 = (f1, f2) andI2 = (g1, g2).Then there existh1, h2 ∈ I1 ∩ I2 and σi ∈ SL2(A[T ]/Ii), i = 1,2, such thatI1 ∩ I2 =(h1, h2), ((h1, h2) ⊗ A[T ]/I1)σ1 = (f1, f2) ⊗ A[T ]/I1, and ((h1, h2) ⊗ A[T ]/I2)σ2 =(g1, g2)⊗A[T ]/I2.

Corollary 7.3. Let A be a Noetherian ring containingQ with dimA = 2 and I1, I2 betwo comaximal ideals inA[T ], each of height2. Suppose thatI1 = (f1, f2) andI1 ∩ I2 =(h1, h2) such thathi = fi modI2

1 . Then there existg1, g2 ∈ I2 andσ ∈ SL2(A[T ]/I2) suchthat I2 = (g1, g2) and((h1, h2)⊗A[T ]/I2)σ = (g1, g2)⊗A[T ]/I2.

Remark 7.4. For a two-dimensional ringA containingQ, we can define the notions othe Euler class groupand theweak Euler class groupof A[T ] in exactly the same waas we did in previous sections. The only difference is that, for an idealI of A[T ] ofheight 2, a local orientation[α] will be called a global orientation if there is a surjectiθ :A[T ]2 � I and someσ ∈ SL2(A[T ]/I) such thatασ = θ ⊗ A[T ]/I . For a projectiveA[T ]-moduleP of rank 2 having trivial determinant, theEuler classof P is defined as inSection 4.


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Theorem 7.5. LetA be a Noetherian ring containingQ with dimA= 2 andI ⊂ A[T ] bean ideal of height2 such thatI/I2 is generated by2 elements. LetωI : (A[T ]/I)2 � I/I2

be a local orientation ofI . Suppose that the image of(I,ωI ) is zero inE(A[T ]). Then,ωIis a global orientation ofI .

Proof. Same as of Theorem 4.7.✷Theorem 7.6. Let A be a Noetherian ring containingQ with dimA = 2 and I ⊂ A[T ]be an ideal of height2 such thatI/I2 is generated by2 elements. LetωI : (A[T ]/I)2 �I/I2 be a local orientation ofI . Let P be a projectiveA[T ]-module of rank2 havingtrivial determinant andχ be a trivialization of

∧2P . Suppose thate(P,χ) = (I,ωI )

in E(A[T ]). Then, there exists a surjectionα :P � I such that(I,ωI ) is obtained from(α,χ).

Proof. It follows adapting the method of Murthy [Mu, Theorem 1.3] that there is ajectiveA[T ]-moduleQ of rank 2, stably isomorphic toP , together with an isomorphismχ1 :A[T ] � ∧2

Q and a surjectionβ :Q� I such that(β,χ1) induces(I,ωI ). Now onecan follow the proof of [B-RS3, Theorem 7.2], and the “Symplectic” cancellation theoof Bhatwadekar [Bh, Theorem 4.8] to prove the theorem.✷Remark 7.7. Let A be as above. LetI ⊂ A[T ] be an ideal of height 2 such thatI/I2 isgenerated by 2 elements andωI be a local orientation ofI . It is clear from Theorem 7.1that there exists a projectiveA[T ]-moduleP of rank 2 together with an isomorphisχ :A[T ] � ∧2

P and a surjectionα :P � I such that(I,ωI ) is obtained from(α,χ).As an immediate consequence of this observation, we see that the ‘local–global pri(Theorem 5.4), holds when dimA = 2 (actually it reduces to the Quillen localizatiotheorem). Since projectiveA[T ]-modules are extended whenA is regular (containingQ),it follows that Question 1 in Section 4 has an affirmative answer in the two-dimenscase. As for Question 2 of Section 4, we see that it reduces to the Affine Horrocks the

The theory of the weak Euler class group described in Section 6 also follows inmanner in the two-dimensional case.

Acknowledgments

I sincerely thank Professor S.M. Bhatwadekar for suggesting the problems tackleand generously sharing his ideas with me. I am grateful to him for giving me a chto work with him. I sincerely thank Dr. Raja Sridharan for many stimulating discusscriticism, corrections, and above all, for training me in this subject, thus giving mnecessary confidence to pursue research. I thank the School of Mathematics, Tataof Fundamental Research, for allowing me to visit in several spells which made this ppossible.


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the euler class group of a polynomial algebra

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