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The main theorem

Dans le document On Stably Free Modules over Affine Algebras (Page 15-21)

Using again that cd(k(xp)) ≤3−p for any xp ∈X(p) ([31, §4.2, Proposition 11]), we get HpZar(X,Hq(m))=0 for any q≥4. Since X is affine, this implies H2(X,H3(m))= H5et(X, μlnm)=0 by [21, Chapter VI, Theorem 7.2].

Now there is a commutative diagram ([9, Theorem 2.3])

x1X(1)

K2(k(x1))/ln

x2X(2)

K1(k(x2))/ln

x3X(3)

K0(k(x3))/ln

x1X(1)

H2(k(x1), μln2)

x2X(2)

H1(k(x2), μln)

x3X(3)

H0(k(x3),Z/ln)

whose vertical maps are isomorphisms ([20]). This gives H2

X,K3/ln

=H2

X,H3(3)

=0.

The result follows.

7. The main theorem

In this section, we prove that any stably free module P of rank (d −1) over a d-dimensional normal affine algebra R over an algebraically closed field k, for which gcd((d−1)!,char(k))=1, is indeed free.

Recall first that a unimodular row of length n2 over R is a row a=(a1, . . . ,an) with aiR such that there exist b1, . . . ,bn∈R with

aibi =1. We denote by Umn(R) the set of unimodular rows of length n and we consider it as a pointed set with base point e1:=(1,0, . . . ,0). If M∈GLn(R)and aUmn(R), then aM is also unimodular, and in this way GLn(R)acts on Umn(R). If a is any unimodular row, the exact sequence

0 P(a) Rn a R 0

yields a projective module P(a)which is free if and only if there exists M∈GLn(R)such that aM=e1. It is clear that any projective module P such that P⊕RRncomes from a unimodular row, and it suffices therefore to understand the (pointed) set Umn(R)/GLn(R) to understand stably free modules of rank n−1 satisfying P⊕RRn.

Of course, any subgroup of GLn(R)acts on Umn(R), and one is classically inter-ested in the (pointed) orbit set Umn(R)/En(R) which classifies unimodular rows up to elementary homotopies. An orbit in Umn(R)/En(R)is usually called an elementary orbit.

We now recall a lemma of L. N. Vaserstein ([39, Corollary 2]).

Lemma 7.1. — Let(a1,a2,a3)Um3(R), uR, and Ra1+Ru=R. Then the rows (a1,a2,a3)and(a1,ua2,ua3)are in the same elementary orbit.

Consequently, by Whitehead’s lemma the rows(a1,ua2,ua3)and(a1,a2,u2a3)can be seen to be in the same elementary orbit. In particular, if −1 is a square in R, then (a1,a2,a3)and(a1,a2,a3)are in the same elementary orbit.

Given v=(v0, v1, v2), w=(w0, w1, w2)Um3(R), withv·wt=1, we shall de-note by V(v, w)the class in WE(R)of the 4×4 skew-symmetric matrix

⎜⎜

0 v0 v1 v2

v0 0 w2w1

v1w2 0 w0

v2 w1w0 0

⎟⎟

of Pfaffian 1. For simplicity, we may just write V(v), as the class of V(v, w) does not depend on the choice of w such that v·wt =1. Associating V(v) to v yields a well defined map V:Um3(R)/E3(R)→WE(R)called the Vaserstein symbol ([40, §5]).

Theorem7.2. — Let R be a smooth affine algebra over a field k of cohomological dimension≤1 which is perfect if the characteristic is 2 or 3. Then the Vaserstein symbol

V:Um3(R)/E3(R)→WE(R) is a bijection.

Proof. — See [27, Corollary 3.5].

This bijection gives rise to an abelian group structure on the set Um3(R)/E3(R), when R is as above. We recall Vaserstein’s rule in WE(R)(see [40, Theorem 5.2 (iii)]). If (a,b,c)and(a,x,y)are in Um3(R)/E3(R), then

V(a,b,c)⊥V(a,x,y)=V(a, (b,c)

x y

y x

)

for any x,ysuch that xx+yy=1 modulo(a).

Lemma7.3. — Let R be a ring. The following formulas hold in WE(R):

(i) V(a,b,c)1=V(−a,b,c), where a is an inverse of a modulo(b,c).

(ii) V(ab2,c,d)=V(a,c,d)⊥V(b2,c,d).

Proof. — Since Vaserstein’s rule holds in WE(R), it suffices to follow the proof of

[38, Lemma 3.5(iii), Lemma 3.5(v)].

We now recall the Antipodal Lemma of Rao in [25, Lemma 1.3.1].

Lemma 7.4. — Let R be a ring and (a,b,c)Um3(R). Suppose that (a,b,c) and (a,b,c)are in the same elementary orbit. Then for any nN, we have

nV(a,b,c)=V(an,b,c) in WE(R).

Proof. — The proof of [25, Lemma 1.3.1] applies mutatis mutandis using the

for-mulas of Lemma7.3.

We now state and prove our main theorem:

Theorem7.5. — Let R be a d-dimensional normal affine algebra over an algebraically closed field k such that gcd((d −1)!,char(k))=1. If d =3, suppose moreover that R is smooth. Then every stably free R-module P of rank d1 is free.

Proof. — Let P be a stably free module of rank d−1. Since the result is clear when d2, we assume that d ≥3. Using Suslin’s cancellation theorem [34, Theorem 1], we can suppose that there is an isomorphism P⊕RRd, and therefore that P is given by a unimodular row(a1, . . . ,ad). In view of [34, Theorem 2], to prove that P is free it suffices to show that there exists a unimodular row(b1, . . . ,bd)such that (a1, . . . ,ad)= (b(d11)!, . . . ,bd)in Umd(R)/Ed(R).

Suppose that d≥4. Let J be the ideal of the singular locus of R. Since R is normal, J has height at least 2 and dim(R/J)d −2. It follows from [6, Theorem 3.5] that Umd(R/J)=e1Ed(R/J)and we can therefore assume, performing elementary operations if necessary, that ad≡1 (mod J)and a1, . . . ,ad1∈J. Using now Swan’s Bertini theorem [37, Theorem 1.5], we can perform elementary operations on(a1, . . . ,ad)such that B:=

R/(a4, . . . ,ad)is either empty, either a non-singular threefold outside the singular locus of R. In the first case, the row(a4, . . . ,ad)is unimodular, and therefore the row(a1, . . . ,ad) is completable in an elementary matrix. Thus we can restrict to the second case. In this situation, we see that B is actually smooth since ad≡1 (mod J).

Given a unimodular row(a,b,c)on B, we can choose lifts a,b,c∈R and consider the unimodular row(a,b,c,a4, . . . ,ad)on R. It is straightforward to check that this gives a well-defined map

Um3(B)/E3(B)Umd(R)/Ed(R),

showing that(a1, . . . ,ad)comes from the unimodular row(a1,a2,a3)on B. We are thus reduced to the case where R is the affine algebra of a smooth threefold. By Theo-rem 7.2, the set Um3(R)/E3(R)is in bijection with WE(R) and is thus endowed with the structure of an abelian group. Since −1 is a square in k, Lemma 7.4 shows that n·(a1,a2,a3)=(a1n,a2,a3)in Um3(R)/E3(R)for any nN. Now Propositions 5.1and 6.1show that Um3(R)/E3(R)is a divisible group prime to the characteristic of k. Since gcd((d−1)!,char(k))=1, there exists a unimodular row(b1,b2,b3)Um3(R)such that

(a1,a2,a3)=(d−1)! ·(b1,b2,b3)=(b(d−1)1 !,b2,b3)

in Um3(R)/E3(R). The result follows.

Remark 7.6. — The proof of the theorem actually gives a bit more. Let R be an affine algebra of dimension d4 over an algebraically closed field k such that gcd((d− 1)!,char(k))=1. Let J be the ideal of the singular locus of R, and let(a1, . . . ,ad)be a unimodular row of rank d which is in the elementary orbit of e1 in Umd(R/J)/Ed(R/J).

Then the projective module associated to(a1, . . . ,ad)is free. To see this, observe that we can use [28, Theorem 2] to reduce to the case of a reduced ring and then apply the proof of the theorem.

To conclude, we derive from the above theorem some stability results for the group K1(R)where R is a smooth affine algebra over an algebraically closed field.

Corollary7.7. — Let R be a smooth algebra of dimension d3 over an algebraically closed field k. Assume that d!k=k. Then SLd(R)∩Ed+1(R)=Ed(R).

Proof. — Letσ ∈SLd(R)∩Ed+1(R). In view of T. Vorst’s theorem in [42], it suffices to show that there exists a matrix N(X)∈SLd(R[X])with the property that N(0)=Id and N(1)=σ.

Since 1⊥σ ∈Ed+1(R), there exists α(X)∈GLd+1(R[X]) with α(0)=Id and α(1)=1⊥σ. It follows that the unimodular row v(X):= e1α(X)Umd+1(R[X]) is congruent to e1 modulo f , where f :=X2−X. We now prove that there exists a ma-trix M(X)∈GLd+1(R[X])congruent to Id modulo f satisfying the equality e1M(X)= e1α(X). This will prove the corollary since these properties imply that

α(X)M(X)1=

1 0 N(X)

with N(X)∈GLd(R[X])such that N(0)=Id and N(1)=σ.

To find M(X), we argue as in [27, Proposition 3.3]. Set B=R[X,T]/(T2Tf) and write e1α(X)=e1+(X2−X)w(X), withw(X)∈R[X]d+1. Observe that u(T):=e1+ Tw(X)is a unimodular row of length d+1 over B with u(f)=v(X)and u(0)=e1. Since R is smooth, it follows that the ideal of the singular locus of B is precisely(T,f). Now

u(T)is congruent to e1modulo T and we can use Remark7.6to findε(T)∈Ed+1(B)such that u(T)ε(T)is of the form(bd1!, . . . ,bd+1). By [27, Proposition 3.3] or [26, Proposition 5.4], there is aδ(T)∈SLd+1(B)congruent to Id modulo f such that u(T)ε(T)=e1δ(T).

Finally, we set M(X):=β(0)β(f)1 where β(T)=ε(T)δ(T)1. It is straightforward to check that M(X)satisfies the conditions stated above, and therefore the result is proved.

Corollary 7.8. — Let R be a smooth algebra over the algebraic closure k of a finite field of dimension d3. Assume that d!k=k. Then the natural map SLd(R)/Ed(R)→SK1(R) is an isomorphism.

Proof. — Using [40, Corollary 17.3], we see that Umn(R)=e1En(R)for nd. It follows that SLd(R)/Ed(R)→SK1(R)is onto. We conclude using Corollary7.7.

Corollary7.9. — Let R be a smooth algebra of dimension d4 over an algebraically closed field k. Assume that d!k=k. Then the Vaserstein symbol

V:Um3(R)/E3(R)→WE(R) is an isomorphism.

Proof. — L.N. Vaserstein in [40, Theorem 5.2(c)] has shown that V is onto. Apply [40, Lemma 5.1] to conclude that V is injective as SL4(R)∩E(R)=E4(R) in view of

Corollary7.7.

Acknowledgements

The first author wishes to thank the Swiss National Science Foundation, grant PP00P2_129089/1, for support.

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J. Fasel

Mathematisches Institut der Universität München, Theresienstrasse 39,

80333 München, Germany jean.fasel@gmail.com R.A. Rao

Tata Institute of Fundamental Research, 1, Dr. Homi Bhabha Road, Navy Nagar, Mumbai 400 005, India

ravi@math.tifr.res.in R.G. Swan

University of Chicago, Chicago, IL, 60637, USA swan@math.uchicago.edu

Manuscrit reçu le 13 novembre 2011 Manuscrit accepté le 7 juin 2012 publié en ligne le 30 juin 2012.

Dans le document On Stably Free Modules over Affine Algebras (Page 15-21)

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