Euler class group of a polynomial algebra II

MRINALKANTIDAS

Department of Mathematics

University of Kansas, Lawrence, Kansas 66045 e-mail : mdas@math.ku.edu

1. INTRODUCTION

In this paper we continue our study on the theory of the Euler class group of a polynomial algebraA[T], whereA is a commutative Noetherian ring (containing Q) of dimensionn. For such a ringA, in [Da], we defined the notion of the nth Euler class groupEⁿ(A[T]) ofA[T]. For simplicity let us call itE(A[T]). In [Da]

we also studied the relations betweenE(A[T]) andE(A), whereE(A) is thenth Euler class group ofA. For example, there is canonical mapΦ :E(A)−→E(A[T]) which is an injective group homomorphism and it is an isomorphism whenAis a smooth affine domain [Da, Proposition 5.7]. In general, these two groups are not isomorphic (see discussion preceding [Da, Proposition 5.7]). In this context, the following question is natural.

Question. Does there exist a group homomorphism, say,Ψ : E(A[T]) −→ E(A) such that the compositionΨΦis the identity map onE(A)?

In this paper we give an affirmative answer to the above question.

A few words about the proof are in order. LetR denote A or A[T]. We may recall thatE(R) is a free abelian group modulo some relation (see the section on Preliminaries for definition) and elements ofE(R)are classes of pairs(I, ωI)where I is an ideal ofRof heightnandω_I : (R/I)ⁿ I/I² is a surjection. One attempt to define a map fromE(A[T])toE(A)could be by restriction at T = 0, meaning, given (I, ωI) ∈ E(A[T]) we may try to associate it to something like(I(0), ω_I(0)) inE(A) whereI(0) = {f(0)|f(T) ∈ I}andω_I(0) : (A/I(0))ⁿ I(0)/I(0)² is the surjection induced byωI. ButI(0)may not have heightnand therefore(I(0), ω_I(0)) may not be a legitimate element ofE(A). On the other hand, since AcontainsQ, there existsλ∈Qsuch thatI(λ)is an ideal of height≥n. In this case(I(λ), ω_I(λ)) is an element of E(A) but since the element λ depends on I and may vary for

1

different ideals, we may not find a singleλfor the whole groupE(A[T])so that we can apply the restrictionT =λ.

To tackle this problem we note that, however, the idealI(0)/I(0)² is generated by n elements and applying some “moving lemma” (which is an application of Eisenbud-Evans theorem) we can find an ideal K of Aof height≥ n, residual to I(0), and a surjectionω_K : (A/K)ⁿ K/K². We defineΨ(I, ω_I) =−(K, ω_K)and prove the following

Theorem 1.1. The mapΨ : E(A[T]) → E(A), described above, is a homomorphism of groups such that if(I, ω_I)∈E(A[T])has the property thatI(0)is an ideal ofAof height n, thenΨ((I, ω_I)) = (I(0), ω_I(0)).

To prove thatΨis well defined and is a group homomorphism we require the so called “addition” and “subtraction” principles in a little more generality which we prove to fit our needs.

IndeedΨis surjective and the compositionΨΦ :E(A)→E(A[T])is the identity map. Furthermore,Ψis an isomorphism whenAis a smooth affine domain.

Let us recall one important result from [Da] which is very much relevant to this context. In [Da], given an ideal I ⊂ A[T]of height n and a surjectionω_I : (A[T]/I)ⁿ I/I², we associated an element (I, ω_I) ∈ E(A[T]). One of the prime objectives was to show that this element(I, ωI) ∈ E(A[T])is the precise obstruc- tion for the surjectionω_I to lift to a surjectionθ : A[T]ⁿ I ([Da, Theorem 4.7]).

For proving this theorem we first showed that we can assume htI(0) = n and then argued “since (I, ωI) = 0 in E(A[T]), we have (I(0), ωI(0)) = 0 in E(A)”.

This argument essentially assumes the existence of the group homomorphismΨ : E(A[T]) → E(A) with the property mentioned in Theorem 1.1 above. However, the question of existence of such a group homomorphism has not been addressed there. In this sense, we now get a complete proof of [Da, Theorem 4.7].

As discussed above, while working on the group homomorphismΨ, our prime concern was the fact that given an ideal I of A[T]of heightn, I(0)may not have height n. But the form of Theorem 1.1 led us to believe that that while working onE(A[T]), we may restrict ourselves to the “nice” idealsIofA[T]for whichI(0) has heightn. In this context, we define a “restricted” Euler class groupE⁰(A[T])of A[T]which concerns only those “nice” ideals and prove thatE(A[T])is isomorphic toE⁰(A[T])(Proposition 3.7).

LetAbe an affine algebra of dimensionnover an algebraically closed fieldkof characteristic zero. ThenE(A)is isomorphic toE₀(A)(can be easily deduced from [B-RS 2, Lemma 3.4]). In Section 4 we investigate the Euler class group E(A[T]) and the weak Euler class groupE0(A[T])whenAis an affine algebra over an alge- braically closed field and prove thatE(A[T])andE₀(A[T])are canonically isomor- phic (Corollary 4.4).

In Section 4 we also address the following question.

Question. LetAbe a Noetherian ring (containingQ) of dimensionn≥3. Let(I, ω_I) ∈ E(A[T])be an arbitrary element. Does there exist a projective moduleP of rankn(with trivial determinant) together with an isomorphismχ:A[T]' ∧ⁿ(P)such thate(P, χ) = (I, ω_I)?

In general, the answer to this question is negative as one can takeA to be the coordinate ring of an even dimensional real sphere, any real maximal idealJ of A and setI = J[T]. We show, using Corollary 4.4 and the following theorem of Bhatwadekar-Raja Sridharan that the above question has an affirmative answer if Ais an affine domain over an algebraically closed field of characteristic zero.

Theorem 1.2. [B-RS 5, Theorem 2.7] Let A be an affine domain of dimension n ≥ 3 over an algebraically closed fieldkof characteristic zero. LetI ⊂A[T]be a local complete intersection ideal of heightnsuch thatI/I²is generated bynelements. Then there exists a projectiveA[T]-moduleP of ranknwith trivial determinant and a surjectionΦ :P I.

We also give an alternative proof of Theorem 1.2 using Euler class computations.

Our proof appears simpler with the use of Euler class techniques. We may note that when Bhatwadekar-Raja Sridharan proved this result, the Euler class group of a polynomial algebra was not defined.

Let dimA = nandP be a projectiveA[T]-module of ranknwith trivial deter- minant together with an isomorphism χ : A[T] → ∧^{∼ n}(P). To the pair(P, χ), in [Da], we associated an elemente(P, χ)inE(A[T]), called the Euler class of(P, χ).

We proved thatP has a unimodular element if and only ife(P, χ) = 0inE(A[T]).

In the proof we crucially used a theorem of Bhatwadekar-Raja Sridharan [B-RS 4, Theorem 3.4]. Here we give an independent proof of [Da, 4.11].

Acknowledgement

I sincerely thank S. M. Bhatwadekar for his criticism and comments on [Da], from which various questions addressed in this paper came out, and for generously

sharing his ideas with me. Thanks are due to Raja Sridharan for many helpful discussions and to Satya Mandal for his constant encouragement.

2. PRELIMINARIES

In this section we define some of the terms used in the paper and record some results which are used in later sections.

All rings considered in this paper are commutative and Noetherian and all mod- ules considered are assumed to be finitely generated. The projective modules are assumed to have constant rank.

We start with an easy lemma.

Lemma 2.1. LetB be a Noetherian ring of dimensionnandJ ⊂ B be an ideal which is contained in the Jacobson radical ofB. Suppose thatK ⊂ B[T]is an ideal such that K+J B[T] =B[T]. Then any maximal ideal ofB[T]containingK has height≤nand hence dim(B[T]/K) = 0.

Now we state a useful lemma. The proof of this lemma can be found in ([B-RS 1], 3.3).

Lemma 2.2. LetAbe a Noetherian ring containing an infinite fieldkand letI ⊂A[T]be an ideal of heightn. Then there existsλ∈ksuch that eitherI(λ) =AorI(λ)⊂Ais an ideal of heightn, whereI(λ) ={f(λ) :f(T)∈I}.

Definition 2.3. Let A be a commutative Noetherian ring and P be a projective A-module of rank n ≤dimA. By a generic surjection of P we mean a surjection α : P J where J is an ideal of A of height n. It follows from a theorem of Eisenbud-Evans ([E-E],[Pl]) that generic surjections exist.

Definition 2.4. LetAbe a commutative Noetherian ring,Pa projectiveA[T]-module.

LetJ(A, P)⊂Aconsist of all thosea∈Asuch thatP_ais extended fromA_a. It fol- lows from ([Qu], Theorem 1), that J(A, P) is an ideal and J(A, P) = p

J(A, P).

This is called theQuillen idealofP inA.

Remark 2.5. Let A, P, J(A, P) be as in the above definition. Then it is easy to deduce from Quillen-Suslin theorem ([Qu], [Su1]) that height ofJ(A, P)is at least one. If determinant ofP is extended fromA, then by [Pl, Corollary 2], htJ(A, P)≥ 2.

Definition 2.6. Let A be a ring A[T]be its polynomial extension. We denote by A(T), the ring obtained fromA[T]by inverting all the monic polynomials inA[T].It can be proved easily that dimension ofA(T)is same as dimension ofA.

The proof of the following lemma can be found in [B-RS 1, Remark 3.9].

Lemma 2.7. Let A be a ring, I ⊂ A[T]be an ideal such that I = (f1,· · · , fn) +I². Assume further that eitherI(0) = AorI(0) = (a₁,· · ·, a_n)such thatf_i(0) = a_i mod I(0)². Then we can find g1,· · ·, gn ∈ I such thatI = (g1,· · ·, gn) + (I²T) with the properties: (1)gi =fimodI², (2)gi(0) =ai.

We now quote a theorem of Mandal. The following version is implicit in [M 1, Theorem 1.2].

Theorem 2.8. LetAbe a Noetherian ring andI ⊂ A[T]be an ideal containing a monic polynomial. Suppose thatI = (f₁,· · · , f_r) +I², wherer≥dim(A[T]/I) + 2. Then, there existg₁,· · · , g_r∈Isuch thatI = (g₁,· · ·, g_r)andf_i =g_imodI².

The following theorem is also due to Mandal [M 2, Theorem 2.1].

Theorem 2.9. LetAbe a Noetherian ring andI ⊂ A[T]be an ideal containing a monic polynomial. Suppose thatI = (f₁,· · ·, f_r) + (I²T), wherer ≥dim(A[T]/I) + 2. Then, there existg1,· · ·, gr ∈I such thatI = (g1,· · · , gr)andfi =gimod(I²T).

The following result is a special case of [M-RS, Theorem 2.3].

Theorem 2.10. LetAbe a Noetherian ring. SupposeK₃ =K₁∩K₂be the intersection of two comaximal idealsK₁,K₂ ofA[T]such that :

(1) K₁contains a monic polynomial inT. (2) K2is an extended ideal.

(3) K1= (f1(T),· · ·, fn(T))withn≥dimA[T]/K1+ 2.

(4) K₃(0) = (c₁· · ·, c_n)withc_i−f_i(0)∈K₁(0)². ThenK₃= (h₁(T),· · · , h_n(T))withh_i(0) =c_i.

We will refer to the following lemma as “moving lemma”. This lemma can easily be proved adapting the proof of [B-RS 3, Corollary 2.14].

Lemma 2.11. LetAbe a Noetherian ring of dimensionn≥ 2. LetJ be an ideal ofAof height≥1such thatJ = (a₁,· · · , a_n) +J². LetKbe any ideal ofAof height at least one.

Then there exists an idealJ⁰⊂Asuch that

(1) J⁰ is comaximal withJ∩Kand htJ⁰≥n.

(2) J∩J⁰ = (c₁,· · · , c_n)wherec_i=a_imodJ².

In the rest of this section we briefly sketch the definitions of theEuler class groups E(A[T])and theweak Euler class groupsE0(A[T])(whereAis a commutative Noe- therian ring containing Q of dimension n ≥ 2) and quote some results that are relevant to this paper. The notions ofE(A[T])andE0(A[T])have been defined and studied in [Da]. We refer the reader to [Da] for a detailed account of these topics.

Definitions ofE(A[T])andE₀(A[T]):

LetAbe a Noetherian ring of dimensionn≥2containingQ. LetI ⊂A[T]be an ideal of heightnsuch thatI/I²is generated bynelements. Two surjectionsαand βfrom(A[T]/I)ⁿI/I²are said to berelatedif there existsσ ∈SLn(A[T]/I)such thatασ=β. This is an equivalence relation on the set of surjections from(A[T]/I)ⁿ toI/I². Let[α]denote the equivalence class ofα. We call such an equivalence class [α]alocal orientation ofI.

It was shown in ([Da], Proposition 4.4) that ifα: (A[T]/I)ⁿI/I² can be lifted to a surjection θ : A[T]ⁿ I then so can anyβ equivalent toα. We call a local orientation[α]ofIaglobal orientation ofIif the surjectionα: (A[T]/I)ⁿI/I²can be lifted to a surjectionθ:A[T]ⁿI.

Let G be the free abelian group on the set of pairs (I, ωI) where I ⊂ A[T]is an ideal of heightn such that Spec(A[T]/I) is connected,I/I² is generated byn elements andωI: (A[T]/I)ⁿI/I²is a local orientation ofI.

LetI ⊂ A[T]be an ideal of heightnandω_I a local orientation ofI. NowI can be decomposed uniquely asI = I₁∩ · · · ∩I_r, where theI_k’s are ideals ofA[T]of heightn, pairwise comaximal and Spec(A[T]/Ik)is connected for eachk. Clearly ω_Iinduces local orientationsω_IkofI_kfor1≤k≤r. By(I, ω_I)we mean the element Σ(Ik, ωIk)ofG.

LetH be the subgroup ofGgenerated by set of pairs(I, ω_I), whereI is an ideal ofA[T]of heightngenerated bynelements andω_Iis a global orientation ofIgiven by the set of generators ofI. We define the Euler class group ofA[T], denoted by E(A[T]), to beG/H.

The weak Euler class groupE0(A[T])is defined in a similar way, just dropping the orientations, as follows:

Let F be the free abelian group on the set of ideals I where htI = n, I/I² is generated by nelements and Spec(A[T]/I) is connected. For an ideal I of A[T]

of heightnsuch thatI/I² is generated bynelements, we take its decomposition into connected components (as above), say,I =I₁∩ · · · ∩ I_r, and associate toIthe element(I) := ΣI_kofF. LetKbe the subgroup ofF generated by elements of the type(I), whereI ⊂ A[T]is an ideal of heightnandI is generated bynelements.

We defineE₀(A[T])to beF/K.

LetP be a projectiveA[T]-module of ranknwith trivial determinant. Fix a triv- ialization χ : A[T] ' ∧ⁿ(P). Letα : P I be a generic surjection (i.e., I is an ideal of heightn). Note that, sinceP has trivial determinant and dimA[T]/I ≤ 1, P/IP is a freeA[T]/I-module. Composingα⊗A[T]/I with an isomorphismγ : (A[T]/I)ⁿ'P/IP with the property∧ⁿ(γ) =χ⊗A[T]/Iwe get a local orientation, sayω_I, ofI. Lete(P, χ)be the image inE(A[T])of the element(I, ω_I)ofG. (We say that(I, ω_I)is obtained from the pair(α, χ)). It can be proved that the assignment sending the pair(P, χ)toe(P, χ)is well defined (see [Da]). We define theEuler class ofP to bee(P, χ).

3. MAIN RESULTS

We begin this section with the following addition and subtraction principles.

Here we have only relaxed the condition on height of the ideals concerned. The methods of proof are similar to the usual addition and subtraction principles (one can look at [B-RS 6, Propositions 3.1, 3.2]). However we include the proofs for the sake of completeness.

Proposition 3.1(Addition Principle). LetAbe a Noetherian ring of dimensionn ≥ 3 and I, J be two comaximal ideals of A, each of height ≥ n−1. Assume further that I = (a1,· · · , an) andJ = (b1,· · ·, bn). Then,I ∩J = (c1,· · · , cn)such thatci = ai

modI²andc_i=b_imodJ².

Proof. Note that we can always perform elementary transformations on(a₁,· · · , a_n) and(b₁,· · · , b_n)and no generality is lost doing so. To see this, let us assume that (a1,· · · , an)is elementarily transformed to(ae1,· · ·,afn)and(b1,· · · , bn)is elemen- tarily transformed to(be₁,· · ·,be_n). Suppose we can find a set of generatorsce₁,· · ·,ce_n ofI ∩J satisfyingcei = aei modI² andcei = bei mod J². Then we can use the sur- jectivity of the canonical mapEn(A/I∩J) −→ En(A/I)×En(A/J)to transform (ce₁,· · ·,ce_n) to (c₁,· · ·, c_n), so thatI ∩J = (c₁,· · · , c_n) with c_i = a_i mod I² and ci =bimodJ².

Let B = A/(b1,· · · , bn) and bar denote reduction mod (b1,· · · , bn). Since I + J = A, (a₁,· · ·, a_n) ∈ U m_n(B). Now dimB ≤ 1 andn ≥ 3. Therefore we can elementarily transform(a₁,· · · , a_n)to(1,· · · ,0). Applying [RS1, Lemma 2] we can apply an elementarily transformation and assume that ht(a1,· · ·, an−1) = n−1.

Note that this transformation preserves the fact thata₁ = 1moduloJ. Therefore, (a1,· · · , an−1) +J =A.

Now let C = A/(a₁,· · ·, an−1) and bar denote reduction mod (a₁,· · · , an−1).

Consider the unimodular row(b₁,· · · , b_n)∈U m_n(C). Using similar arguments as in the above paragraph we finally obtain :

(1) (a1,· · · , an−1) + (b1,· · · , bn−1) =A.

(2) ht(a₁,· · ·, an−1) =ht(b₁,· · ·, bn−1) =n−1.

InA[T]we consider the ideals

I1 = (a1,· · ·, an−1, T +an), I2= (b1,· · · , bn−1, T +bn)

and let K = I1 ∩I2. Note that I1 +I2 = A[T]. Therefore, using the Chinese remainder theorem we can chooseg1(T),· · · , gn(T)∈Ksuch that

K = (g1(T),· · · , gn(T)) +K²

satisfyingg_i(T) =a_imodI₁²,g_i(T) =b_imodI₂²,1≤i≤n−1;g_n(T) =T+a_nmod I₁²,gn(T) =T +bnmodI₂².

Now ht(a₁,· · · , an−1) =ht(b₁,· · ·, bn−1) =n−1. Also note that dimA[T]/I₁= dimA/(a₁,· · ·, an−1) and dimA[T]/I₂ = dimA/(b₁,· · ·, bn−1). Therefore, it fol- lows that dimA[T]/K ≤ 1. Since n ≥ 3, the conditions of Theorem 2.8 are sat- isfied for K. Applying Theorem 2.8 we obtainK = (h₁(T),· · · , h_n(T))such that hi(T) =gi(T)modK². Lethi(0) =ci. ThenI ∩J = (c1,· · ·, cn)withci =ai mod

I² andci=bi modJ². 2

Proposition 3.2(Subtraction Principle). LetAbe a Noetherian ring of dimensionn≥3 andI, J be two comaximal ideals ofA, each of height≥ n−1. Assume further thatI = (a1,· · · , an)andI∩J = (c1,· · · , cn)such thatci =aimodI². ThenJ = (b1,· · · , bn) such thatc_i =b_imodJ².

Proof. First note that we can perform elementary transformations on(a1,· · ·, an) because we can apply the same elementary transformations on(c1,· · · , cn) to re- tain the relation that c_i = a_i mod I². Let B = A/J² and bar denote reduction moduloJ². Since ht(J) = n−1, dimB ≤ 1. Therefore, performing elementary

transformations as in the proof of the above proposition we may assume that: (1) ht(a₁,· · · , an−1) =n−1, (2)a_n= 1modJ².

Consider the following ideals inA[T]:

I1= (a1,· · ·, an−1, T +an), I2=J A[T], K =I1∩I2.

Applying Theorem 2.10 we obtain a set of generators(h₁(T),· · ·, h_n(T))ofKsuch thathi(0) =ci. Letbi = hi(1−an). ThenJ = (b1,· · · , bn). Sincean = 1modJ², b_i−c_i =h_i(1−a_n)−h_i(0) = 0modJ². This proves the proposition. 2 We are now ready to show that there is a group homomorphism fromE(A[T]) to E(A) such that if(I, ω_I) ∈ E(A[T])has the property thatI(0)is an ideal of A of heightn, then this group homomorphism takes(I, ω_I) to(I(0), ω_I(0))in E(A), where ω_I(0) is the local orientation of I(0) induced by ωI. This is done in Theo- rem 3.3 below. To prove this theorem we mainly need addition and subtraction principles proved above and Lemma 2.11.

Theorem 3.3. LetA be a Noetherian ring containingQof dimension n ≥ 3. There is a group homomorphismΨ : E(A[T]) −→ E(A)such that if(I, ωI) ∈ E(A[T])has the property thatI(0)is an ideal ofA of heightn, thenΨ((I, ω_I)) = (I(0), ω_I(0))inE(A), whereω_I(0)is the local orientation ofI(0)induced byωI. IfI(0) =A,Ψ((I, ωI)) = 0.

Proof. We give the proof in steps.

Step 1. Recall thatE(A[T])is defined asG/H, where Gis the free abelian group on the set of pairs (I, ωI), where I ⊂ A[T] is an ideal of height n having the property that Spec(A[T]/I) is connected and I/I² is generated by n elements, and ω_I : (A[T]/I)ⁿ I/I² is a local orientation of I. Let us pick one such el- ement (I, ωI). Let the local orientation ωI be given by I = (f1,· · · , fn) +I². Now I(0) is an ideal of A (not necessarily proper) with ht(I(0)) ≥ n −1 and I(0) = (f₁(0),· · ·, f_n(0)) +I(0)². LetJ =I∩A.

Now applying Lemma 2.11 we can find an ideal K ⊂ A of height ≥ n such that K is comaximal with J andK ∩I(0) = (a₁,· · · , a_n), where a_i = f_i(0) mod I(0)². First assume that bothI(0)andKare proper ideals ofA. Let us call the local orientation ofK, induced bya₁,· · ·, a_n, to be ω_K. To the element (I, ω_I)of Gwe associate the element −(K, ω_K) of E(A). In the case whenI(0) = A or K = A, (I, ωI)is associated to the zero element ofE(A).

We need to show that this association does not depend on the choice ofK. To prove this, letK⁰be another ideal ofAof heightnsuch thatK⁰ is comaximal with

J andK⁰ ∩I(0) = (b1,· · · , bn), wherebi = fi(0)mod I(0)². LetωK⁰ be the local orientation ofK⁰induced byb₁,· · · , b_n. We claim that(K, ω_K) = (K⁰, ω_K⁰)inE(A).

In the next paragraph we prove this claim.

First note that, by repeated use of addition and subtraction principles and mov- ing lemma (Lemma 2.11), we may assume thatK⁰ is comaximal withK. Now we can find an idealL ⊂A of heightnand a local orientationωLofLsuch thatLis comaximal with each ofJ,KandK⁰and(L, ω_L) + (K, ω_K) = 0inE(A). Therefore it is enough to prove that(L, ωL) + (K⁰, ωK⁰) = 0inE(A). In order to do so, we first apply the addition principle to the two comaximal idealsL∩KandK⁰∩I(0)to see thatL∩K∩K⁰∩I(0)is generated bynelements (with appropriate set of genaer- ators). Next we apply the subtraction principle to the comaximal idealsK∩I(0) andL∩K⁰ to conclude thatL∩K⁰ is generated bynelements (with appropriate set of generators), i.e.,(L, ω_L) + (K⁰, ω_K⁰) = 0inE(A). Thus the claim is proved.

Step 2. Extending to whole ofG, we get a group homomorphismψ:G−→E(A).

Note that in the above definition we nowhere used the fact that Spec(A[T]/I) is connected. So, given any(I, ωI)∈E(A[T])(where Spec(A[T]/I)is not necessarily connected), following the above procedure, we can also associate an element, say

−(K, ω_K)ofE(A). We claim that the image of(I, ω_I)underψis actually−(K, ω_K).

Proof of the claim : Consider a decomposition ofI into its connected componenets, say, I = I₁ ∩ · · · ∩I_r. Now since I_i’s are pairwise comaximal, ω_I induces local orientationωIiofIi,i= 1,· · · , rand we have,(I, ωI) =Pr

i=1(Ii, ωIi). Suppose that ψ((I_i, ω_Ii)) =−(K_i, ω_Ki)(∈E(A)). In view of Lemma 2.11, we can clearly assume thatKi’s are pairwise comaximal. For simplicity we work out the case whenr= 2.

By definition ofψ, we haveψ((I, ω_I)) =−(K₁, ω_K1)−(K₂, ω_K2)inE(A). Since K₁ andK₂ are comaximal, we can writeψ((I, ω_I)) = −(K₁∩K₂, ω_K1∩K₂), where ωK1∩K2 is induced byωK1 andωK2.

Now by the definition ofψwe haveK₁∩I₁(0)isn-generated with appropriate set of generators. Same is true forK2∩I2(0). Since the idealsK1∩I1(0)andK2∩I2(0) are comaximal and each has height≥n−1, applying addition principle we have K1 ∩I1(0)∩K2 ∩I2(0)n-generated with appropriate set of generators. In other words,I(0)∩(K1∩K2)isn-generated. Now keeping track of the generators and proceeding as in last paragraph of Step 1, we can easily conclude that(K, ω_K) = (K1∩K2, ωK1∩K2). This proves the claim.

Step 3. Recall that E(A[T]) = G/H, where H is the subgroup ofGgenerated by pairs(I, ω_I)∈Gsuch thatω_Iis a global orientation. We now show thatHis in the kernel ofψ.

First let(L, ω_L)∈Gbe such thatω_Lis a global orientation. This means that there exist f₁,· · ·, f_n ∈ L such that L = (f₁,· · · , f_n) andω_L is induced by this set of generators ofL. But thenL(0) = (f1(0),· · · , fn(0))and hence from the definition ofψit follows thatψ((L, ω_L)) = 0inE(A). Now an element ofHis of the form

(I, ωI) =

r

X

i=1

(Ii, ωIi)−

s

X

j=r+1

(Ii, ωIi),

whereωIi, ωIj are global orientations. It is now clear thatψ((I, ωI)) = 0inE(A)as each of the elements on the right hand side is mapped to zero.

Therefore, we have a group homomorphismΨ :E(A[T])−→E(A).

Step 4. Let(I, ω_I) ∈ E(A[T])be such that ht (I(0)) = n. In this caseω_I induces a local orientationω_I(0) of I(0) and(I(0), ω_I(0)) ∈ E(A). The way we picked up K andωK in Step 1 actually means in this case that(I(0), ω_I(0)) + (K, ωK) = 0in E(A). Therefore,Ψ((I, ω_I)) = −(K, ω_K) = (I(0), ω_I(0)). This completes the proof

of the theorem. 2

We now use the above theorem to give a complete proof of the following theorem ([Da, Theorem 4.7]).

Theorem 3.4. LetAbe a ring of dimensionn≥3,I ⊂A[T]be an ideal of heightnsuch thatI/I²is generated bynelements and letω_I : (A[T]/I)ⁿI/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,Iis generated bynelements andω_Ican be lifted to a surjectionθ:A[T]ⁿI. Proof. LetΨ : E(A[T]) −→ E(A)be the group homomorphism, as defined in the theorem above. Supposeω_I is given byI = (f₁,· · · , f_n) +I². We first assume that I(0) is a proper ideal ofA. We have, I(0) = (f1(0),· · ·, fn(0)) +I(0)². Suppose that Ψ((I, ω_I)) = −(K, ω_K), where K ⊂ A is an ideal of height ≥ n such that K∩I(0) = (c₁,· · ·, c_n)wherec_i =f_i(0)modI(0)²andω_Kis induced byc₁,· · ·, c_n. Since(I, ωI) = 0in E(A[T]), Ψ((I, ωI)) = 0 inE(A) and therefore, (K, ωK) = 0 in E(A). This implies, by [B-RS 3, Theorem 4.2], thatK = (a₁,· · · , a_n) such that ai =cimodK². Now applying subtraction principle (Proposition 3.2) we see that I(0) = (b1,· · · , bn) such that bi = ci mod I(0)². Consequently, bi = fi(0) mod I(0)². Therefore, using Lemma 2.7, it follows that ω_I can be lifted to a surjection θ:A[T]ⁿI/(I²T).

On the other hand, ifI(0) =A, then again, applying Lemma 2.7 we can liftωIto a surjectionθ:A[T]ⁿI/(I²T).

InE(A(T))also, the element(IA(T), ω_IA(T))is zero, which, by [B-RS 3, Theorem 4.2], implies thatω_IA(T)(and henceθ⊗A(T)) can be lifted to a set of generators of IA(T). Applying [Da, Theorem 3.10], we conclude thatθcan be lifted to a surjec- tionα:A[T]ⁿI. Clearlyαliftsω_I. Soω_Iis a global orientation. 2 Remark 3.5. Let us review the group homomorphismΨ :E(A[T])−→ E(A). Re- call that we already have a canonical group homomorphismΦ :E(A) → E(A[T]) and it follows from Theorem 3.4 thatΦis injective. Further, it is easy to see that the compositionΨΦis the identity onE(A). ClearlyΨis surjective. Of particular inter- est is the kernel. Let(I, ω_I)∈E(A[T])be an element of Ker(Ψ). Then, as shown in the proof of Theorem 3.4 above,ωIcan be lifted to a surjectionθ:A[T]ⁿI/(I²T).

Ker(Ψ) precisely consists of these elements (i.e., roughly, the ideals I of A[T] of heightnsuch that I/(I²T) is generated by nelements). We may recall that from [B-RS 1, Theorem 3.8] it follows that ifAis a smooth affine domain over an infinite perfect field, Ker(Ψ) is trivial and henceΨbecomes an isomorphism. IfAis not smooth, there is an example [B-RS 1, 6.4] of a normal affine domainA for which Ker(Ψ)is not trivial. We expect Ker(Ψ) = 0whenAis a regular ring containingQ.

The “local-global principle” for Euler class groups [Da, Theorem 5.4] suggests that it is enough to prove Ker(Ψ) = 0whenAis a regular local ring containingQ.

The main point of Theorem 3.4 is that for an idealI ⊂A[T]of heightn,I(0)may not have heightnand therefore given(I, ωI)∈E(A[T]), something like(I(0), ω_I(0)) may not make sense. This makes sense only whenI(0)has heightnor I(0) = A.

Then we started wondering what happens if we work only with those idealsI for whichI(0)has heightnor I(0) = A. This is reflected in the following definition and the proposition after that.

Definition of a group :

We define a group E⁰(A[T]) which may be regarded as the “restricted” Euler class group ofA[T]. The definition ofE⁰(A[T])is similar to that ofE(A[T]).

Let G⁰ be the free abelian group on the set of pairs (I, ω_I), whereI ⊂ A[T]is an ideal of heightnhaving the properties : (i) I(0) ⊂ A is an ideal of heightnor I(0) = A(we point out here that this is the “restriction” and only at this point the definition differs from that ofE(A[T])) (ii) Spec(A[T]/I)is connected (iii)I/I² is generated bynelements; andωIis a local orientation ofI.

LetI ⊂A[T]be any ideal of heightnsuch thatI/I²is generated bynelements.

Let I = I₁ ∩ · · · ∩ I_k be the decomposition ofI into its connected components.

Letω_I be a local orientation of I. Thenω_I induces local orientationsω_Ii of I_i for i= 1,· · ·, k. By(I, ωI)we mean the elementΣ(Ii, ωi)ofG⁰.

LetH⁰ be the subgroup ofG⁰ generated by the set of pairs(I, ωI)whereωI is a global orientation ofI.

We defineE⁰(A[T])to be the groupG⁰/H⁰.

Remark 3.6. Clearly the obvious map ∆ : E⁰(A[T]) −→ E(A[T]) which sends (I, ω_I)∈E⁰(A[T])to(I, ω_I)∈E(A[T]), is a group homomorphism.

Proposition 3.7. The map∆ :E⁰(A[T])−→ E(A[T]), as described above, is an isomor- phism of groups.

Proof. By the very definition ofE⁰(A[T])and by Theorem 3.4, it follows that ∆is injective. To prove the surjectivity, let(I, ωI)∈E(A[T])be an arbitrary element. So I(0)may not necessarily be of heightn. To prove that∆is surjective it is enough to find some(I⁰, ω_I⁰) ∈ E(A[T])such that ht(I⁰(0)) ≥ nand(I, ω_I) = (I⁰, ω_I⁰) in E(A[T]).

Suppose thatωIis given byI = (f1,· · ·, fn)+I². Then,I(0) = (f1(0),· · · , fn(0))+

I(0)². Let J = I ∩A. Since htJ ≥ n−1 ≥ 2, we can find an elements ∈ J² such that ht(s) = 1. Let bar denote reduction modulos. Since dimA ≤ n−1, it follows by a result of Mohan Kumar [Mo, Corollary 3], that I(0) = (a1,· · ·, an), where a_i = f_i(0)modulo I(0)². By adding suitable multiples of sto a₁,· · · , a_n, we may assume by the Eisenbud-Evans theorem (see [B-RS 3, Corollary 2.13]) that (a₁,· · · , a_n) = I(0)∩K, whereK ⊂ Ais an ideal of heightnandK + (s) = A.

Note thatK = (a₁,· · · , a_n) +K². Let us call the local orientation corresponding to this set of generators ofK/K²byωK.

LetI1=I∩K[T]. ThenI1is an ideal ofA[T]of heightn. SinceIandK[T]are co- maximal ideals, the local orientationsω_Iandω_K⊗A[T], ofIandK[T]respectively, induce a local orientationωI1 ofI1, say, given byI1= (g1,· · · , gn) +I₁². Now

I1(0) =I(0)∩K= (a1,· · ·, an)

and we have g_i(0) = a_i mod I₁(0)². Therefore we can lift g₁,· · ·, g_n to a set of ngenerators ofI1/(I₁²T), which also corresponds toωI1. In E(A[T])we have the equation :

(I1, ωI1) = (I, ωI) + (K[T], ωK⊗A[T]).

SinceωI1 is given by a set of generators ofI1/(I₁²T), we can apply [Da, Lemma 3.9] to find an idealI₂ofA[T]of heightnand a local orientationω_I2 ofI₂such that (i)I₂(0) =Aand (ii)(I₁, ω_I1) + (I₂, ω_I2) = 0inE(A[T]).

Therefore we have the equation

(I, ω_I) + (K[T], ω_K⊗A[T]) + (I2, ω_I2) = 0

inE(A[T]). Since(K[T], ω_K⊗A[T])and(I₂, ω_I2)both belong toE⁰(A[T]), the result

follows. 2

LetAbe a ring (containingQ) of dimensionn ≥ 3. LetP be a projectiveA[T]- module of ranknhaving trivial determinant andχ be a trivialization of∧ⁿP. To the pair(P, χ)we can associate an elemente(P, χ)inE(A[T]), called the Euler class of(P, χ)(see Preliminaries for the definition). In [Da] we proved thate(P, χ) = 0in E(A[T])if and only ifPhas a unimodular element. In that proof we crucially used a theorem of Bhatwadekar-Raja Sridharan [B-RS 4, Theorem 3.4]. Here we give an independent proof of this result. Further, we derive a version of [B-RS 4, Theorem 3.4] using our theorem.

Theorem 3.8. Let A be as above. LetP be a projective A[T]-module of rank n having trivial determinant andχbe a trivialization of∧ⁿP. Then,e(P, χ) = 0inE(A[T])if and only ifP has a unimodular element.

Proof. Letα:P I1be a surjection whereI1is an ideal inA[T]of heightnandωI1

be the local orientation ofI1induced by(α, χ). Then,e(P, χ) = (I1, ω_I1)inE(A[T]).

Suppose thatP has a unimodular element. We show, under this condition, that (I₁, ω_I1) = 0inE(A[T]). In view of the isomorphism∆in Proposition 3.7, we can assume that eitherI1(0) =Aor htI1(0) =n. SinceP has a unimodular element, it follows that the projectiveA-module P/T P and the projectiveA(T)-moduleP ⊗ A(T) both have unimodular elements. Consequently, by [B-RS 3, Corollary 4.4], we have(I1(0), ω_I1(0)) = 0inE(A)and(I1A(T), ω_I1⊗A(T)) = 0inE(A(T)). Now following the arguments as in Theorem 3.4, it is easy to see that(I₁, ω_I1) = 0 in E(A[T]).

Let us now assume that e(P, χ) = 0 in E(A[T]). We prove that then P has a unimodular element. We give the proof in steps.

Step 1.In this step we show that the projectiveA-moduleP/T P has a unimodular element.

Recall that we have α : P I1, a generic surjection of P andωI1 is the local orientation of I₁ induced by(α, χ). Therefore,e(P, χ) = (I₁, ω_I1) in E(A[T]). As usual, we may assume that eitherI₁(0) = A or I₁(0) is a proper ideal of height n. IfI1(0) =A, then clearly theA-moduleP/T P has a unimodular element. Now suppose thatI₁(0)is a proper ideal of heightn. Then, following the definition of the Euler class of a projective module we have,e(P/T P, χ⊗A[T]/(T)) = (I1(0), ω_I1(0)) inE(A). Since(I1, ω_I1) = 0, inE(A[T]), it follows that(I1(0), ω_I1(0)) = 0inE(A).

Therefore,e(P/T P, χ⊗A[T]/(T)) = 0inE(A)and hence by [B-RS 3, Corollary 4.4], P/T P has a unimodular element.

So in any caseP/T P has a unimodular element.

Step 2. LetJ(A, P)denote the Quillen ideal ofP inA. WriteJ = J(A, P). In this step we prove, using a theorem of Mandal, thatP_1+Jhas a unimodular element.

Since determinant ofPis extended (actually free), by Remark 2.5, htJ(A, P)≥2.

Since dimA/J ≤n−2, it follows that the projective(A/J)[T]-moduleP/J[T]Phas a unimodular element i.e., there is a surjectionP/J[T]P (A/J)[T]. Using this fact and the Eisenbud-Evans theorem ([E-E],[Pl]) it is easy to see that (since P is projective), there is a generic surjectionβ : P I such thatI is comaximal with J[T].

Letω_I be the local orientation of I induced by(β, χ). Thene(P, χ) = (I, ω_I) in E(A[T]).

Consider the ringB =A_1+J. We want to prove that the projectiveB[T]-module P_1+Jhas a unimodular element. IfIB[T] =B[T], we are done. Therefore, suppose thatIB[T]is a proper ideal ofB[T]of heightnand note that it is comaximal with J B[T]andJ Bis contained in the Jacobson radical ofB.

Let us elaborate howωIis obtained from(β, χ). SinceP has trivial determinant, P/IP is a freeA[T]/I-module. We choose an isomorphismλ: (A[T]/I)ⁿ' P/IP such that∧ⁿλ=χ⊗A[T]/I. ω_I is the surjection(β⊗A[T]/I)λfrom(A[T]/I)ⁿto I/I², say, given byI = (f1,· · ·, fn) +I².

Sincee(P, χ) = 0, we have(I, ω_I) = 0in E(A[T]) and hence by Theorem 3.4, I = (g1,· · · , gn)such thatgi =fimoduloI². So we haveIB[T] = (g1,· · ·, gn)and IB[T] +J B[T] =B[T]. Therefore(g1,· · ·, gn)is a unimodular row over(B/J B)[T]

and since dim(B/J B) ≤n−2, it is elementarily completable. Using this and the fact that elementary matrices can be lifted via surjection of rings, it is easy to see

that we can alter the above set of generators of IB[T]by an elementary matrix σ∈E_n(B[T])and assume that

(1) ht(g₁,· · ·, gn−1) =n−1,

(2) (g1,· · · , gn−1) +J B[T] =B[T]and hence, (3) dimB[T]/(g₁,· · · , gn−1)≤1.

We setC =B[T],R=C[Y],K = (g1,· · ·, gn−1, Y +gn). Let us denoteP1+J by P⁰. Note that

C[Y]/K 'B[T]/(g1,· · ·, gn−1),

and so we have dimC[Y]/K ≤1. Therefore, it follows that the projectiveC[Y]/K- moduleP⁰[Y]/KP⁰[Y]is a free module of rankn. We choose an isomorphism

τ(Y) : (C[Y]/K)ⁿ→^∼ P⁰[Y]/KP⁰[Y]

such that ∧ⁿτ(Y) = χ⊗C[Y]/K. Since ∧ⁿλ = χ ⊗B[T]/IB[T], it follows that τ(0) and λ differ by an element of SL_n(B[T]/IB[T]). Since IB[T] + J B[T] = B[T]and J B is contained in the Jacobson radical of B, by Lemma 2.1, we have dim(B[T]/IB[T]) = 0. Therefore,SLn(B[T]/IB[T]) =En(B[T]/IB[T]). Since ele- mentary transformations can be lifted via surjection of rings, we may alterτ(Y)by an element ofSLn(C[Y]/K)and assume thatτ(0) = λ. Letγ(Y) : (C[Y]/K)ⁿ K/K² denote the surjection induced by the set of generators(g₁,· · · , gn−1, Y +g_n) ofK.

Thus, we obtain a surjection

δ(Y) =γ(Y)τ(Y)⁻¹ :P⁰[Y]/KP⁰[Y]K/K².

Since τ(0) = λ, β ⊗B[T]/IB[T] = ω_Iλ⁻¹ and γ(0) = ω_I, we haveδ(0) = β ⊗ B[T]/IB[T].

Therefore, applying Mandal’s theorem [M 2, Theorem 2.1], we obtain a surjec- tionη(Y) : P⁰[Y]K. Specializing atY = 1−gn, we obtain a surjection fromP⁰ toB[T].

Step 3. So far we have proved thatP/T P has a unimodular element (Step 1) and P_1+J has a unimodular element (Step 2), whereJ is the Quillen ideal ofP inA. In this step we combine these two facts and appeal to a patching argument of Plum- stead to conclude thatP has a unimodular element

NowP1+J has a unimodular element. Let us call itp1. We have already seen that P/T P has a unimodular element, sayp. We claim that there is an elementary auto- morphismσofP_1+J such thatσ¯p¯₁ = ¯p, where “bar” denotes reduction moduloT. To see this, let us consider the ringD=B/J(B)whereJ(B)denotes the Jacobson radical ofB. Since dimD ≤n−2it follows that there is an elementary automor- phismτ ofP1+J ⊗Dsuch thatτ p1 =poverD. Since elementary automorphisms can be lifted via a surjection of rings, we have, by repeated use of this argument, a σ∈E(P_1+J)such thatσ¯p¯₁= ¯p. Letqdenote the unimodular elementσp₁ofP_1+J.

SinceP_1+J has a unimodular element, we can finds∈ J such thatP_1+sAhas a unimodular element. We still call itq. SincePs is extended fromAs, it has a uni- modular element, namelyp. Sincepandqare equal moduloT, i.e. overA_s(1+sA), it follows using a patching argument of Plumstead [Pl] thatP has a unimodular

element. 2

We can now derive the following version of a theorem of Bhatwadekar-Raja Srid- haran [B-RS 4, Theorem 3.4].

Theorem 3.9. Let A be a Noetherian ring containing Q of dimension n ≥ 3. Let P be a projectiveA[T]-module of rankn with trivial determinant. Suppose that Pf has a unimodular element for some monic polynomial f ∈ A[T]. Then P has a unimodular element.

Proof. SincePfhas a unimodular element andfis monic, by [B-RS 4, Lemma 3.1] it follows that there is an idealIofA[T]of height at leastnand a surjectionα:P I such thatI contains a monic polynomial. If htI > n, it follows thatI =A[T]and there is nothing to prove. So we assume htI =n. Fix an isomorphismχ :A[T]→^∼

∧ⁿ(P). Now(α, χ)induces a local orientationω_I ofI and hencee(P, χ) = (I, ω_I) in E(A[T]). Since I contains a monic polynomial, it follows from a theorem of Mandal (Theorem 2.8) thatω_Iis a global orientation, i.e.,(I, ω_I) = 0. Consequently, e(P, χ) = 0inE(A[T]). By the above theorem,P has a unimodular element. This

proves the theorem. 2

We end this section with another nice application of Proposition 3.7, giving an alternative proof of the main theorem of [D-RS].

Theorem 3.10. LetAbe a commutative noetherian ring containing the field of rationals withdimA = n(neven) and letP be a projectiveA[T]-module of rank nsuch that its determinant is free. Suppose there is a surjectionα:P I whereI is an ideal ofA[T]of

heightnwhich is generated bynelements. Assume further thatP/T P has a unimodular element. ThenP has a unimodular element.

Proof. Fix a trivializationχ : A[T] ' ∧ⁿP. Then (α, χ) inducese(P, χ) = (I, ω_I) in E(A[T]), where ω_I is a local orientation of I (induced byα and χ). NowI is generated bynelements, say,f1· · ·, fn. Therefore, applying [Da, Proposition 6.7]

we see that there exists a stably free A[T]-moduleQ⁰ of rankn, a generatorχ₁ of

∧ⁿ(Q⁰)such thate(Q⁰, χ1) = (I, ωI)inE(A[T]). SinceQ⁰is stably free of ranknand AcontainsQ, by a result of Ravi Rao [Ra],Q⁰is extended. Therefore,Q⁰ =Q[T]for some stably freeA-moduleQ. So we havee(Q[T], χ₁) = (I, ω_I)inE(A[T]).

Therefore, in order to prove that P has a unimodular element it is enough to prove thatQ[T]has a unimodular element. In what follows we prove that theA- moduleQhas a unimodular element.

Note that, in view of Proposition 3.7, we may assume that I(0) is an ideal of heightn orI(0) = A. SinceQ[T]maps ontoI, it follows thatQ maps onto I(0).

Therefore, ifI(0) =A, then Qhas a unimodular element and we are done in this case. So assume that htI(0) =n.

We have a surjectionα ⊗A[T]/(T) : P/T P I(0). Then (α⊗A[T]/(T), χ⊗ A[T]/(T))induces the Euler class ofP/T P as

e(P/T P, χ⊗A[T]/(T)) = (I(0), ω_I(0)),

whereω_I(0)is also the local orientation induced byω_I.

On the other hand we have,e(Q[T], χ1) = (I, ωI)inE(A[T]). Restricting atT = 0 we obtain

e(Q, χ1⊗A[T]/(T)) = (I(0), ω_I(0)) =e(P/T P, χ⊗A[T]/(T)).

ButP/T P has a unimodular element and it implies that(I(0), ω_I(0)) = 0. Conse- quently,e(Q, χ₁⊗A[T]/(T)) = 0and thereforeQhas a unimodular element. This

proves the theorem. 2

4. POLYNOMIAL EXTENSION OF AN AFFINE ALGEBRA OVER AN ALGEBRAICALLY CLOSED FIELD AND A THEOREM OFBHATWADEKAR-RAJA SRIDHARAN

Let Abe an affine algebra of dimension nover an algebraically closed field of characteristic zero. It is known that in this case, the canonical map from E(A) to E₀(A)is an isomorphism of groups (can be easily deduced from [B-RS 2, Lemma 3.4]). In this section we investigateE(A[T])andE0(A[T])for a ringAas above and

prove thatE(A[T])andE0(A[T])are canonically isomorphic. To prove this we first show that ifB is an affine algebra of dimensionnover aC₁-field of characteristic zero thenE(B) →^∼ E₀(B). Then we use the injectivity of the canonical map from E(A[T])toE(A(T)), proved in [Da, Proposition 5.8].

Proposition 4.1. Let R be an affine algebra of dimension n ≥ 3 over a C1 field k of characteristic zero. Let J ⊂ R be an ideal of height n such that J is generated by n elements. Then any set ofngenerators ofJ/J²can be lifted to a set ofngenerators ofJ. Proof. SupposeJ = (a1,· · · , an). Let us take an arbitrary set of generators ofJ/J²:

J = (b1,· · · , bn) +J².

We want to show that there existsc₁,· · ·, c_n ∈ J such thatJ = (c₁,· · · , c_n) and bi =cimodJ².

Clearly we may assume that ht(a3,· · · , an) = n −2. Since any two surjec- tions from (R/J)ⁿ to J/J² differ by an element of GL_n(R/J), there exists a ma- trixδ ∈ GL_n(R/J)such that(a₁,· · · , a_n)δ = (b₁,· · ·, b_n). Let u ∈ R be such that u=det(δ)⁻¹. Then(u, a1,· · · , an)∈U mn+1(R). LetB =R/(a3,· · · , an). ThenBis an affine algebra overkof dimension≤2. Therefore, all stably free modules over B are free by [Su2, Theorem 2.4]. So the unimodular row(u, a1, a2) ∈ U m3(B)is completable. Applying [RS2, Lemma 2.4] we have a set of generators ofJ, sayJ = (d₁,· · · , d_n), and a matrix δ⁰ ∈ SL_n(R/J) such that (d₁,· · · , d_n)δ⁰ = (b₁,· · · , b_n).

Since dim(R/J) = 0, we haveSLn(R/J) =En(R/J)and therefore we can liftδ⁰to a matrix∆∈E_n(R). Suppose(d₁,· · · , d_n)∆ = (c₁,· · ·, c_n). ThenJ = (c₁,· · · , c_n)

is the desired set of generators. 2

Corollary 4.2. LetRbe an affine algebra of dimensionn≥3over aC₁fieldkof charac- teristic zero. ThenE(A)'E0(A).

Proof. We know that the canonical map fromE(A)toE0(A)is surjective. To prove injectivity, let(J, ω_J)be in the kernel. Then, by [B-RS 2, Lemma 3.3] we have

(J, ω_J) +

r

X

i=1

(J_i, ω_i) =

s

X

k=r+1

(J_k, ω_k),

where Ji, J_k are ideals of height n such that each of them is generated by n el- ements. By the above proposition, each of the local orientations ω₁,· · ·, ω_s is a global one. Consequently(J, ωJ) = 0inE(A). This proves the corollary. 2

Proposition 4.3. Let A be an affine algebra of dimensionn ≥ 3 over an algebraically closed fieldkof characteristic zero. LetI ⊂ A[T]be an ideal of heightn. Assume thatI is generated bynelements. Then any set ofngenerators ofI/I²can be lifted to a set ofn generators ofI. In other words,(I, ωI) = 0inE(A[T])for any local orientationωI ofI. Proof. We will use the injectivity of the canonical map fromE(A[T]) toE(A(T)), where A(T) is the ring obtained from A[T] by inverting all monic polynomials.

This has been proved in [Da, Proposition 5.8]

Note thatA(T)is an affine algebra over aC1 field. Therefore, if we consider the image(IA(T), ω_I ⊗A(T))of(I, ω_I) inE(A(T)), it follows by Proposition 4.1 that (IA(T), ωI⊗A(T)) = 0asIA(T)is generated bynelements. Therefore,(I, ωI) = 0

inE(A[T]). 2

The following corollary is now immediate.

Corollary 4.4. LetAbe an affine algebra of dimensionn≥3over an algebraically closed fieldkof characteristic zero. ThenE(A[T])'E₀(A[T]).

4.1. A theorem of Bhatwadekar-Raja Sridharan. LetAbe any commutative Noe- therian ring of dimensionncontainingQ. Now we may ask the following ques- tions.

Question 1. Let(J, ωJ)be any element ofE(A). Does there exist a projectiveA-module of ranknwith trivial determinant together with an isomorphismχ:A→ ∧^{∼ n}(P)such that e(P, χ) = (J, ω_J)?

Question 2. Let(I, ω_I)be any element ofE(A[T]). Does there exist a projectiveA[T]- module of rank n with trivial determinant together with an isomorphismχ : A[T] →^∼

∧ⁿ(P)such thate(P, χ) = (I, ω_I)?

These questions do not have affirmative answers in general. One can takeAto be the coordinate ring of the real three sphere and J be any real maximal ideal.

Then it is known thatJ is not surjective image of a projectiveA-module of rankn.

IfAis an affine algebra over an algebraically closed field, it follows from a theo- rem of Murthy [Mu, Theorem 3.3] that Question 1 has an affirmative answer.

In this note we discuss a theorem of Bhatwadekar-Raja Sridharan [B-RS 5, Theo- rem 2.7] which essentially says that Question 2 has an affirmative answer whenA is an affine algebra over an algebraically closed field of characteristic zero. We may note that when Bhatwadekar-Raja Sridharan proved this theorem, the Euler class

group of a polynomial algebra was not defined. Below we give a proof of their theorem using Euler class computations.

The following lemma is an improvement of [B-RS 4, Lemma 4.1] and is crucial for later discussions.

Lemma 4.5. LetB be a ring of dimensionn≥3such that height of the Jacobson radical J(B)is at least one. LetI ⊂ B[T]be an ideal of heightnsuch that : (1)I +J(B)[T] = B[T](so I is zero dimensional) (2)I = (a₁,· · ·, an−1, f(T))wherea₁,· · · , an−1 ∈ B and ht(a1,· · ·, an−2) =n−2. Then any set ofngenerators ofI/I²can be lifted to a set ofngenerators ofI(i.e., any local orientation ofI is a global one).

Proof. Letω_Ibe a local orientation ofI, corresponding to a set of generators ofI/I². We show that(I, ωI) = 0inE(B[T]). We do this using [B-RS 4, Lemma 4.1] and the local-global principle for Euler class groups [Da, Theorem 5.4].

Letmbe any maximal ideal ofBof heightn. ConsinderBm[T]. Now by [B-RS 4, Lemma 4.1],(I, ω_I) = 0inE(B_m[T]). Since it happens for every maximal idealm ofBof heightn, we have, by the local-global principle for Euler class groups [Da, Theorem 5.4] that(I, ωI)comes fromE(B). But since htJ(B) ≥1, it follows from [Mo, Corollary 3] thatE(B) = 0. Therefore,(I, ω_I) = 0inE(B[T]). 2

We now quote the following two propositions from [B-RS 5].

Proposition 4.6. LetAbe a Noetherian ring with dimR = d≥ 1. LetI ⊂ R[T]be an ideal with ht(I)≥2. Suppose thatI/I²is generated bynelements wheren≥d+ 1. Then I is generated bynelements.

Proposition 4.7. LetA be an affine domain of dimensionnover an algebraically closed field of characteristic zero. LetI ⊂A[T]be an ideal and letb∈I∩Abe a nonzero element such thatAb is regular. Suppose there exists a projectiveA1+bA[T]-moduleP⁰ of rankn with trivial determinant and a surjectionβ : P⁰ I_1+bA. Assume further thatP_b⁰ is free. Then there exists a projectiveA[T]-moduleP of ranknwith trivial determinant and a surjection fromP toI.

We are now ready to prove the following theorem of Bhatwadekar-Raja Sridha- ran [B-RS 5, Theorem 2.7].

Theorem 4.8. LetAbe an affine domain of dimensionn≥3over an algebraically closed fieldkof characteristic zero. LetI ⊂A[T]be a local complete intersection ideal of heightn such thatI/I² is generated bynelements. Then there exists a projectiveA[T]-moduleP of ranknwith trivial determinant and a surjectionΦ :P I.

Proof. We will replace Step 1 and Step 2 of the proof of [B-RS 5, Theorem 2.7] by some Euler class computations. We briefly outline the part preceding these steps from their proof.

Letω_I be a local orientation ofI given byI = (g₁,· · · , g_n) +I². LetJ =I ∩A.

Letbbe a nonzero element ofJ²which also belongs to the singular locus ofA. Let R = A/(b). Then dimR ≤ n−1. Therefore applying Proposition 4.6, we have I = (f₁,· · · , f_n, b) where f_i = g_i modulo I². Applying Swan’s Bertini theorem [B-RS 2, Theorem 2.11], and adding suitable multiples ofbtof1,· · ·, fnthey obtain an element(I⁰, ω_I⁰)∈E(A[T])such that :

(1) (I, ω_I) + (I⁰, ω_I⁰) = 0inE(A[T]).

(2) I⁰+ (b) =A[T]and henceI⁰+I =A[T].

(3) I⁰is a prime ideal of heightn.

Let B = A_1+bA. IfI⁰B[T] = B[T], IB[T] is image of a free module and the theorem is proved in this case using Proposition 4.7. Therefore assume thatI⁰B[T]

is proper. Since it is prime and is comaximal with the Jacobson radical ofB, it is a maximal ideal of heightn. To be consistent with their notation, letI⁰B[T] =M.

Using techniques from [Bh] it follows that there is an idealL1 ⊂B[T]of height nsuch that

(1) M∩L₁= (b₁,· · · , bn−1, f(T))whereb_i ∈Bandf(T)∈B[T].

(2) L1+M =B[T]andL1+bB[T] =B[T]

Using a theorem of Murthy they show that there is a projectiveB[T]-moduleP⁰ with trivial determinant and a surjectionα :P⁰ L₁. Some additional arguments imply thatP_b⁰ is free.

We now use Euler class computations to show that thisP⁰maps ontoIB[T].

We fix an isomorphismχ:B[T]' ∧ⁿ(P⁰). Then(α, χ)induces a local orientation ω_L1 of L₁ and we have e(P⁰, χ) = (L₁, ω_L1) in E(B[T]). We also have(I, ω_I) + (M, ω_M) = 0 in E(B[T]). SinceM andL₁ are comaximal, ω_M andω_L1 together induce a local orientation of M ∩L1, say ωM∩L1. Note that by Lemma 4.5, the ideal M ∩L₁ = (b₁,· · ·, bn−1, f(T)) has the property that any local orientation of M ∩L1 is a global one. Therefore, ωM∩L₁ is a global orientation and hence, (M, ω_M) + (L1, ω_L1) = 0inE(B[T]). Consequently,e(P⁰, χ) = (I, ω_I)inE(B[T]).

Now using [Da, Corollary 4.10] it follows that there is a surjectionβ :P⁰ IB[T].

Applying Proposition 4.7, the theorem follows. 2