Vector bundles over affine surfaces - u-bourgogne.frVector Bundles over Affine Surfaces 127 After this paper was written, we received an expanded version of [3]. The authors show that

Inventiones math. 36, 125 - 165 (1976) l_~l vefl t iofles matbematicae �9 by Springer-Verlag 1976

Vector Bundles over Affine Surfaces

M. Pavaman Murthy (Chicago) and Richard G. Swan (Chicago)

To Jean-Pierre Serre

In algebraic terms, our basic result reads as follows.

Theorem 1. Let k be an algebraically closed field and let A be a finitely generated commutative k-algebra of Krull dimension

126 M.P. Murthy and R.G. Swan

The infinite transcendency of k is used only for the last statement and for showing that (d) implies (e), a result due to Roitman [42]. Without this hypothesis our proof gives the following result.

Theorem 3. Let V=SpecA be a non-singular irreducible affine surface over an algebraically closed field k. Then the statements (a), (b), and (c) of Theorem 2 are equivalent, imply (d), and are implied by (e). Furthermore, SKo(A ) is divisible and there is an epimorphism SAo(X)--, SKo(A ) where X is any non-singular projective surface birationally equivalent to V.

In particular, since SAo(X)=O if X is a ruled surface, this includes the main results of [38]. Artin and Lieberman have shown that A o (X)= 0 if X is an Enriques surface [3]. In w we will show that for char k=0 , SAo(X)=O for the surfaces which have pg(X)=0 and q>0 . It is conjectured that SAo(X)=O whenever pg(X)=0 [7]. We will also show that for any X, SAo(X ) is the direct sum of a finite group and a divisible group. We have not been able to decide whether the finite group can be non-trivial.

If char k---0, k is uncountable and p~(X)=gO, Mumford [36] has shown that Ao(X ) is not finite-dimensional. This extends to the case where k is countable but of infinite transcendency over the prime field (see w Using this result, Mumford has shown how to construct examples of indecomposable bundles of rank 2 on affine surface. No examples are known in the case char k 4: 0. We will give Mumford's examples in w We also will show that invertible ideals in A may require more than two generators.

We will derive Theorem 1 from the following result which extends Theorem 2.1 of [61] to the non-free case

Theorem 4. Let A be a commutative ring with unit such that m - S p e c A is a noetherian space of dimension < 2. Let P be a finitely generated projective A-module of rank 2 and let (a2, 7t)6A G P be unimodular. Then (A O P)/A(a 2, n)~, P.

Surprisingly, it is not possible to omit the restriction on dimension. A 7- dimensional counterexample is given in w Here is the best result we have been able to obtain in the general case.

Theorem 5. Let A be a commutative ring with unit and let P be a finitely generated projective A-module of rank 2. Let (a z, ~)cA G P be unimodular. I f P/aP ~ (A/aA) 2, then (A G P)/A( c~2, 7t) .~ P.

The hypothesis on P/aP is equivalent to the seemingly weaker condition that Q / a Q ~ A / a A where Q=A2P. This and other equivalent conditions are given in Lemma 1.1 below. We will give two proofs of Theorem 5 corresponding to the two proofs of Theorem 2.1 in [61]. We will then derive Theorems 4 and 1 from it. In w 5, we give the above applications to surfaces. We then conclude with a discussion of the topological analogues of Theorems 4 and 5.

We would like to thank M. Artin, S. Block, and D. Lieberman for showing us the work referred to in w A. Geramita for showing us the examples referred to in w D. Mumford for the example given in Theorem 6.1 and for much useful information concerning his paper [36]; and R. Narasimhan for many helpful conversations and remarks.

Vector Bundles over Affine Surfaces 127

After this paper was written, we received an expanded version of [3]. The authors show that SAo(X)=O for all surfaces over ~ which have pg=0 and which are not of general type (i.e., have ~c < 2 [45]). They also show that SAo(X ) is divisible for surfaces over ~. We also received a letter from A. A. Roitman in which similar results are stated. Roitman also asserts that the torsion subgroup of Ao(X) maps isomorphically onto that of Alb (X) but he does not state explicitly what hypotheses are needed for this. If it is true in general, it implies (by our Theorem 5.9) that SAo(X) is a torsion free divisible group and, in particular, is zero for surfaces over the algebraic closure of a finite field.

1. First Proof of Theorem 5. This proof will reduce Theorem 5 to the main theorem of [61].

Lemma 1.1. Let A be a commutative ring with unit and let P be a finitely generated projective A-module of rank 2. Let (a, ~)~A GP be unimodular. Then the following conditions are equivalent.

(a) P/aP.~(A/aA) 2. (b) Q/aQ~ A/aA where Q= A2 p. (c) There is a unimodular element (a, ~)~A OQ. (d) There exist (~P and q~: P - * A such that q~(Tz+a()=0 and Aa+q~(P)=A. Furthermore, if (d) holds and q~(q) is prime to a, then P/aP is free with base ~, ~l,

the images of ~z, tl in P/aP.

Proof We write P=P/aP, A=A/aA, Q=Q/aQ. The image (0, ~) of (a, ~) in P is unimodular. Therefore P = A ~ Q M . Now M=AZP=A-ZP=O. so P ~ A ~ | This shows that (b) implies (a). The reverse implication is trivial. If (a, ~)~A ~)Q is unimodular, so is its image (0, ~)~AOQ so (~Q is unimodular. Since (~ has rank 1, Q = A 4. Thus (c) implies (b). In the reverse direction, if (b) holds, lift the inverse isomorphisms A ~ Q ~ A to maps A s , Q_~g A. Then g f(1) = 1 rood a and if follows that (a, ()~A @Q is unimodular if ~ =f(1). Suppose now that (b) holds. Lift the projection P -- A ~ �9 Q--~ Q ~ A to q~: P - * A. Since P--~ .4 is onto, it follows that A a + I = A where l=~o(P). Also, ~00z)-0 m o d a so tp(zr)~InAa=la. Let ~p(rc) = ia with i~1 and write i = (p((). Then q~(~z- a 4) = 0. Finally suppose (d) holds. Choose r/e P so that A a + A (p(rl)= A. Then ~/~ P is unimodular and t~:,4 ~/~ 5,. Now

e ker q3 and ,4 ~ is a direct summand of P so we can write ker q~ =,4 ~ (~ N. There- fore P = A g ~ A ~ I ~ N . Since r k P = 2 , it follows that N = 0 . This proves (a) and the final statement of the lemma.

We now turn to the proof of Theorem 5. Let (~, aZ)~PO)A be unimodular and suppose P/aP~(A/~tA) z. Then Lemma 1.1(c) holds for a = ~ but (a,~) is unimodular if and only if (a2, 4) is. Therefore, Lemma 1.1 (d) holds for a = az. The transvection (y, x) ~ (y + x (, x) sends (n, az) to (~', a2) where ~p(z~') = 0. Since POA/A(~, a2),.~P~A/A(r(, ~ ) by the transvection we can replace ~ by re' and assume that q~ (n) = 0. The last statement of Lemma 1.1 shows that P-- A n + A r/+ a2 p where ~o(t/) is prime to ~.

We make P~)A a module over the polynomial ring A[t] by defining t(y,x)=(O,~o(y)). Note that t2(p~A)=O. Since t (~,a)=0, we see that M = P~)A/A(n, ~) is a module over A[t]. Our first objective is to find a projective


resolution for M. Since (n, a) is unimodular , M is projective over A and hence Pdatt]M


This construction commutes with localization and gives the usual detf, t r f if P is free. Therefore detf, t r f can be characterized as the unique elements of R whose images in Rp are detf~, trf~ for all p. The following properties are easily checked by localizing and checking the case where P is free by direct calculation.

(1) de t ( fg ) = ( de t f ) . (det g), t r ( f + g ) = t r f + t r g for f, g: P---, P. (2) If f : P--*Q, g: Q--~P, then det( fg)=det(gf) , t r ( fg )= t r (g f ) . (3) f : P ~ P is an isomorphism if and only if d e t f is a unit in A.

(4) If r k P = 2 , f : P ~ P , and a~A, then t r ( a + f ) = 2 a + t r f and d e t ( a + f ) = a 2 + a t r f + d e t f

(a 0) Any endomorphism ~0 of A G P can be represented by a matrix f where ae A, n e P, f : P-~ A, O: P ~ P, and q)(x, y)=(a x + f (y), xn +0(y)). Suppose (a ,~)~A~P is unimodular. Then A(~P/A(a, rc)~P if and only if there is an automorphism q~: A ( ~ P ~ A O P with q0(1,0)=(a, r0, or, equivalently, by (3), if

and only if there is a matrix ( f 0 ) a s above whose determinant is a unit in A.

(a 0) Lemma 2.1. Let rk P = 2 and let ~o = f as above be an endomorphism of

AOP. Then

det ~o = a det 0 - f ( ~ ) tr 0 +f(O(~)).

Proof. As above we reduce to the case where P is free by localizing. In this case the result is easily checked by direct calculation.

Remark.. The following results came up in the course of this work. They will not be needed here but may perhaps be of some interest. Suppose rk P = 2 and f: P- - ,P by f ( x ) = ~ gi(x)v i where vieP and gi: P---~A. Then t r f = ~ gi(vi) and det f = ~ [gi(vi) gj(vi)- gi(vj) gj(vi)].

i< j We next recall a standard result about localization.

Lemma 2.2. Let A be a ring with unit. Let M and N be A-modules with M finitely presented. Let S be a central multiplicatively closed set of A.

(a) I f f: M---~N and fs: Ms---~Ns is zero, then there is some s~S with sf=O.

(b) I f g: Ms---~N s, we can find f: M---~N and sES so that fs=sg.

This is immediate from [58, Prop. 5.8, Cor. 5.9]. The following result is classical, and almost trivial, for torsion free modules

over an integral domain.

Lemma 2.3. Let A be a commutative ring with unit. Let P and Q be finitely generated projective A-modules such that rk Pp = rk Qp for each prime ideal p of A. Let f: P-*Q and suppose f induces an isomorphism f : P/IP,~Q/IQ for some ideal I of A. Then we can find g: Q--~P and eeA with e = 1 m o d I such that f g = e and g f = e.

Proof Let C = c k r f Since C is finitely generated and C/IC=O, we can find c - = l m o d I with cC=O [26, Th. 76]. Let S={1, c,c 2 . . . . }. Then Cs=O so fs: Ps~Qs is an epimorphism. Since P and Q are projective, we see that kerfs is


projective of rank = rk P - rk Q = 0. Therefore kerfs = 0 and fs is an isomorphism. By Lemma 1.2(b), we can find h: Q-*P and c"eS such that hs=c"fff 1. Therefore, hs f s = c" and fshs = c". By Lemma 1.2(a), there is some c"e S such that cm(hf - c n) =0 and c'( fh-cn)=O. Set g=c 'h and e=c m+".

3. Second Proof of Theorem 5. To begin with, we will consider any unimodular

(a 0) element (a,~)eA(~P and attempt to construct an endomorphism q~= f with det q~ a unit in A. Let A = A/aA and t5 = P/aP. The image (0, ~) of (a, r 0 in A (~P is still unimodular so ~ is unimodular in/5 and we can write P=A ~@Q. If Q=A2P, then A215=Q/aQ which shows that Q,~Q/aQ. Since Q is projective we can lift the map Q / a Q - ~ Q c P to a map fl: Q--~P. Let g: A O Q ~ P by h(a, q)=art+fl(q). Clearly ~: . 4 ~ Q ~ P . By Lemma 1.3, we can find h: P-- ,AGQ and eeA such that e = l m o d a , gh=e and hg=e. Define f : P---,A to be the composition P h,A(~Q, c"5~A where ceA, 6: Q- ,A. Define 0': P---~P to be the composition

p h ~ ~p , A @ Q A O Q ~

where u, yeA, It: Q--*A and veQ. This matrix represents the most general automorphism of A G Q since Homa(Q, Q)=A. Define 0: P---,P to be O(x)=px+O'(x) where peA.

We now compute A = det f using Lemma 2.1. By (2),

t r 0 = t r [ g ( ; v ~ ) h ] = t r [ h g ( ; ~)]=e(u+v)

since h g = 1 . Similarly,

d e t 0 ' = t r [ g ( ; ~ ) h ] = d e t [ h g ( ~ ~)]=e2[uv-#(v)] .

The last equality is again verified by first localizing to make Q free of rank 1. By (4), we get

det 0 = det (p + 0') = p2 + p e(u + v) + e 2 [ u v - #(v)],

tr 0 =2p+e(u+v).

Now f(n)=(c, 6)h(rc) but re=g(1,0) so h(n)=(e,O) and f(n)=ce. Also O'(n)=

(; ~) h(n)=g(eu, ev)so f(O'(n))=(c,g))hg(eu, ev)=e2[cu+f(v)] and f(O(n))-- g

f (p n + 0'(~)) = p c e + e 2 [c u + 6(v)]. Using Lemma 2.1 we have

A = a det O- f (n ) t r O+f(O(n))

=a[_pZ + pe(u + v) + e2(uv-#(v))] + e2 6 (v ) -pce -eZ cv


after simplification. Now write q=p+ev, s = a ( u - v ) - c . These can be made arbitrary elements of A by an appropriate choice of p and c. Similarly, ~ = 6 - a # can be made to be any element of Horn(Q, A) by appropriate choice of 6. After a bit of computation, we get

A =aq2+seq+eZe(v), (,)

where s, qeA, veQ, eeHom(Q, A) are arbitrary, and we know that A a + A e = A . Suppose now that we have a unimodular element (a, tl)eA•Q, i.e. that (,)

is satisfied. Then there is a map (r, ~): A@Q---~A with ra+~( t / )= 1. In (*), choose v = t / a n d e=t~ where teA. Let b=~(t/) so that A a + A b = A . Then

A =aq 2+seq+te 2 b

where Aa+ Aeb= A. We want to find q,s and t in A so that A is a unit. This can be done i f a = a 2, aeA. If so, we have A a + A e b = A . Write ~ q + e b x = l . Squaring gives a q2+2eqabx+x2e2b2=l and we can make A = I choosing s=2abx and t = b x 2.

Remark. We can obtain an expression identical with (.) by choosing/~ = u = v = 0 which gives

A =ap 2-pce+e26(v).

Therefore, it would have been sufficient to consider a 0 having the form 0(x)= p x + l(x)~ where ~ e P, 1: P--*A, but this only leads to very minor simplifications and also obscures the fact that the more general method used here gives no better results. There is one case, however, in which this approach gives a simpler proof, namely, the case where A 2 p ~ A . In this case P has a symplectic structure ( , ). We choose ~ so that ( n , ~ ) = e is prime to a and set l ( x ) = ( x , ~ ) , f ( x ) = (x, c~+dn). If O(x)=px+l(x) ~, an application of Lemma 2.1 gives

a d e t ( f o ) = a p 2 - p c e - e 2 d

and we conclude the argument as above if a = a2. If P is free, this argument is equivalent to the one in [61].

3. Proof of Theorems 1 and 4. Suppose A and P are as in Theorem 4. Let Q =A2p and X = m - spec A. Let T c X be a finite set meeting each irreducible component

of X. Since Q , ~ A m for all meX, we can find maps A - ~ Q g~A inducing isomorphisms Aim ~', Q/mQ ~ Aim for all meT(cfi [59], proof of Theorem 3.1). Let q=f(1) . If I = aq(_t/) then mz~I for m e T and it follows that m-specA / I has dimension


Theorem 5 now applies to (a 'z,n ') but A ~ P / A ( a z, n ) ~ A ~ P / A ( a 'z, ~') by the transvections constructed above.

We now turn to the proof of Theorem 1. Suppose A ~ P ~ A ~ Q . As in [61] we can reduce to the case where P and Q have constant rank and A is reduced. If this is > 3, use Bass' cancellation theorem [5, Ch. IV, Cor. 3.5]. If it is 1 use exterior powers [54], i.e., apply det: Ko(A)---} Pic A. Thus we can assume rk P = r k Q = 2 . Let (1 ,0)~AOQ map to (a ,n )~A~P under A G Q ~ A G P . Then A ~ P / A ( a , n ) ~ Q . By Bertini's theorem [60] we can find h: A ~ P so that ~ ' = ~ + h ( a ) is such that l=op(~ ' ) defines a reduced subscheme of SpecA of dimension 0. This is proved by the same argument as that of [60, w 1]. The smooth- ness assumed there is not essential to the argument. Therefore, by a transveetion, we can assume that A/oQ(~)=k I x .-. x k, where k i = k since k is algebraically closed. Since k 2 = k , we can find an a~A with an a ~ 2 rood oQ(n). By another transvection, we replace a by ~2 and apply Theorem 4.

Remark. Suppose k is infinite but not algebraically closed. If we assume that A is geometrically reduced, the above argument allows us to reduce to the case A/oj,(~)=k a x ... x k, where the ki are finite separable extensions of k. If all such extensions are closed under square roots, we can complete the argument as above. This will certainly be the case if k is perfect of characteristic 2. Also, if k is perfect, then A/nil A is geometrically reduced. Therefore, Theorem 1 holds if k is an infinite perfect field of characteristic 2.

Recently, Geramita, Krusemeyer, and Ojanguren have applied Theorem 2.1 of [61] to the unimodular row (x, y, z) over the ring IR [x, y, z]/(1 + x" + y" + z"), n even, generalizing a result of Towber for n = 2. This example suggests that Theorem 1 should hold over R if Spec A has no points rational over IR. We will prove a more general statement.

Theorem 3.1. Let k be a real closed field and let A be a finitely generated commutative k-algebra of Krull dimension 2. Suppose that all points of Spec A which are rational over k tie on a closed subscheme of dimension < 1. Then cancellation holds for finitely generated projective A-modules (as in Theorem 1).

Proof Let J be the intersection of all maximal ideals m of A with Aim = k. The hypothesis says that A/J has dimension < 1. It is clearly reduced.

Suppose A G P ~ A ~ Q . As above, we reduce to the case where A is reduced and P has constant rank 2. Let ( 1 , 0 ) ~ A ~ Q correspond to (a,n)eA@P. By [60, Th. 1.4], there is a finite set S c P so that if T=S is finite and y is a general linear combination of elements of T, then n' = n + ay is such that I = op(n') defines a reduced 0-dimensional subscheme of Spec A. By the same theorem applied to ( ~ , ~ ) e A G P with A=A/J , P=P/JP we can find a finite set S l o p so that if T 1 = S 1 is finite and z is a general linear combination of elements of TI then ~ + fi z is unimodular. Choose T ~ S such that the image of T in/5 contains S~ and let y be a sufficiently general linear combination of elements of T. Then the image of ~' = zc + ay in/5 will be unimodular. Since P is projective, the image of I = op(W) in .4 wil be op(ff')=.4. Therefore I + J = A and the subscheme of Spec A defined by I does not contain any maximal ideal m with A / m = k . Therefore A/ l= k~ x -.. x k, where the k, are non-trivial algebraic extensions of k. Since k is real


closed, the k~ will be algebraically closed and we can complete the argument as before.

In the most interesting examples where Theorem 3.1 applies, the set of points rational over k is actually empty. The reason for this was pointed out by Nara- simhan: the set will be empty whenever Spec A is non-singular.

Theorem 3.2 (Narasimhan). Let X be an irreducible algebraic scheme over a real closed field k such that the points of X rational over k lie on a proper closed subscheme of X. Then all points of X rational over k are singular on X.

Proof Suppose X is smooth at a point rational over k. By 1-20, Exp. II, Def. 1.1; Exp. I, Th. 7.6] a neighborhood of this point will be isomorphic to a neighborhood U of 0 in a hypersurface V c A "+1 defined by a polynomial f ( x I . . . . . x , , y)

with f (0 ,0 )=0 , Of (0,0)#0. Given c < 0 < d in k, we can find elements a i, blek

with a i < 0 < b i so that f ( x , y ) = 0 has a solution yEk with c < y < d whenever x~ek with a~

134 M. P. Murthy and R.G. Swan

is proper, C is a non-singular irreducible curve, and f~'k(C). We choose this definition of B(X) and say z l, z z ~Z(X) are rationally equivalent if za- z 2 ~B(X). A similar approach has been used by Fulton [15] to define the Chow ring in all dimensions.

Suppose now that C is any irreducible curve, possible singular. IffE(gp-{0}, define ve(f) = dim Cp/(gef and extend v e to all fEk(C)* using vp(fg) = ve(f) + vp(g). Define ( f ) = ~ re(f) P.

Lemma 4.1. If q): C'--*C is a finite morphism of curves and f sk (C ' )* then q~,((f)) = (Nk(c,)/k(C)f).

Proof We first look at the case where C' is the normalization of C in k(C). If P c C, the integral closure of (9 e is R = (-] Cp, where P' runs over the points of C' over P. Since R is a principal ideal domain, we see easily that ~ Vp,(f)= dim R/ fR for f eR . Let fe(gp and consider the map of the exact sequence O~(_ge--*R---~(ge/R-*O into itself given by multiplication by f Applying the snake lemma gives

O-*M---~(gp/f(Pe---,R/fR---~N---,O where O--,M--~R/(geJL~ R/(gv-*N--,O.

Since dim k R/Cp< oo, it follows that Vp(f)=dim (ge/.f(Pe=dim R / fR=~ , Vp,(f) and our result follows easily. In the general case, let C', C be the normalizations of C' and C in their function fields. Using the result just proved, we are reduced to looking at C ' ~ C. If P c 0', (gp is a discrete valuation ring. Its integral closure in k(C') is R=(~(9 e, over p,~q~-l(p). In this case it is a standard result that Vp(Nk(c,)/k(C)f)=-E Ve,(f) , [52, Ch. I, w 5].

This lemma shows that in our definition of B(X) it is not necessary to restrict ourselves to non-singular curves C since if ~o: C--,X is proper, we can replace C by its normalization C without changing ~0,((f)). Once this is done, we can then restrict ourselves to closed immersions qo: C-~X in defining B(X). Suppose ~: C ' ~ X is proper. If ~k(C') is a point, then C' is complete so d e g ( f ) = 0 and ~k,((f)) = 0. Otherwise, C'--~ C = ~k(C') is finite and ~k,((f)) = (N f ) by the lemma.

The classical definition of rational equivalence is somewhat different in appearance. We consider 1-cycles D on X • IP' and let B'(X) consist of all 0-cycles of the form Prx[D. (X • 0 ) ] - P r x [ D . (X • oo)]. It is, however, quite easy to see that B'(X)=B(X). To get generators for B(X) as a subgroup, it is sufficient to look at the case where D is an irreducible curve. In this case, q~=Prx: D ~ X is proper and f=pr~,, : D ~ I P ' is a function on D. An easy calculation shows that the above expression is equal to q~,((f)). Conversely, given ~k: C--~X proper and g: C---~IP' we get (~k, g): C---,X • IP'. This map may be obtained as the composition F--~ C x IP' ~' • 1 X • IP' where r is the graph of g. Since F is closed and ~, x 1 is proper, we see that D =(~k, g)(C) is closed. Except for the trivial case where D is a point, C--*D will be finite and Lemma 4.1 shows that ~, , ( ( f ) )= ~p,((Nktc)/ktO ~ f)) where tp, f are defined as above. Therefore B'(X)= B(X) and our definition of rational equivalence agrees with the classical one.

The definition used by Mumford [36] and Roitman [41, 42] is again somewhat different in appearance, but is shown in [41, w 1] to be equivalent to the classical definition.


In [19] and [21], Grothendieck shows that if X is a quasi-projective surface, then the graded ring associated with the natural (topological) filtration of K~ is isomorphic to the Chow ring of X. The proof makes use of his theory of Chern classes. We will give here a very elementary proof of this fact using a presentation for K ~ discovered by Eagon [13]. We will treat only the affine case, leaving the more general quasi-projective case to the reader since it will not be used here.

If V= SpecA is a non-singular irreducible affine surface, then A is regular so Ko(A)= Go(A) ([5]). Let FiKo(A)=FiGo(A ) be the subgroup of Go(A) generated by all [M] where supp M has codimension > i. Thus F o K o = K o, F 1K o is generated by all [A/p] where p is a non-zero prime ideal, F 2 K o is generated by all [A/m] with m maximal, and F~ K o = 0 for i > 3.

Theorem 4.2 (Grothendieck). Let V = S p e c A be a non-singular irreducible affine surface over an algebraically closed field. Then

(a) K~ is a filtered ring, i.e., FiFjcFi+j,

(b) rk: Ko(A)/F 1K o (A) ~ Z so F 1Ko(A) = [(o(A),

(c) det: F 1Ko(A)/F 2 Ko(A ) ~ Pic A so F 2 Ko(A ) = SKo(A), (d) S K o ( A ) = F z K o ( A ) ~ A ( V ) with [A/mx] corresponding to the zero cycle x

in A(V), (e) the pairing Pic(A)• Pic(A)--~A(V), induced by multiplication in Ko(A),

is given by taking intersections on V.

Proof The multiplication in Ko(A ) induces one in Go(A ) given by the well- known formula [51]

[M] [N] = ~ ( - 1) i [Tor~ a (M, N)]. (1)

It is clear from this that F~ K 0 is an ideal of K o.

Lemma 4.3. Let I be an invertible ideal of A and let m~ . . . . . m, be a finite number of maximal ideals of A. Then there is an invertible ideal I ' ~ I such that I 'egmifor i=1 . . . . . n.

Proof Let Ira, =A,,,a~ with a ~ I . By the Chinese remainder theorem, find a6I with a - a i ~ m i I . Then I , , = A m a. If S = A , - [ . ) m i we then have I s = A s a so there is s~S with s l c A a . Let I' =a-i ' s I . Then I,,, =Am,.

We note that all prime ideals of height 1 in A are invertible since all A., are regular local and hence factorial. To show F 1F 2 = 0, we must compute [A/p] [A/m]. By Lemma 4.3 find I ~ p with I e m. Then [A/p] = [A/I] and all terms of (1) for [A/I] [A/m] are annihilated by m + I - - A.

Now consider [A/p] [A/q] where p and q are primes of height 1. Let m 2 q be maximal and let I~ .p with Iegm. Then [A/p]=[A/I] and all terms of (1) for [A/I] [A/q] are annihilated by I + q. This is not contained in any prime of height 1. The only possibility would be q but q c m and I e m . Therefore all terms in (1) have finite length which shows that F 1F 1 ~ F 2 . This proves (a).

For (b), note that F1Ko(A ) clearly lies in the kernel of rk: K o ( A ) - * Z but Ko.(A)/F 1 has only one generator [A]. Since de t (A)=0 in P i c a we also see that det: F 1 K ( A ) ~ Pic A is onto. To see that F 2 lies in its kernel we proceed as follows. Let m be a maximal ideal of A. Then Ext 1 (m, A) = Ext 2 (A/m, A) is annihilated by


m so we can compute it by localizing first to A,,. This is regular local of dimension 2 so we get Ext 1 (m, A) ~ A/m. Let

0-~ A ---, P--~ m---~ 0 (2)

be an extension corresponding to a generator of Ext I (m, A). By [55] P is projective of rank 2.

Lemma 4.4. Let A be a regular noetherian ring of dimension 2. Let 0 ~ Q --, P ~-~ m --~ 0 be exact with m a maximal ideal of A and P finitely generated projective of rank 2. 7hen Q ~ A 2 P.

Proof The map ~0: P ~ A gives us a Koszul complex by extending q~ to a derivation of A*P. In our case, the result is

O --~ A z P ~ P ~-~ m --~ O (3)

where O(x ^ y)=q~(x)y-q)(y)x. This is exact since it localizes to the ordinary Koszul complex

~ r n ----~ 0

and a, b is an R-sequence (trivially if p + m and by regularity if p = m). Comparing (3) with the given exact sequence gives Q,~ A 2 P.

This lemma shows immediately that detF2=0 since [A/m] = [ A ] - [ m ] = [A z] - [ P ] in Ko(A).

Define p: Pic A --+ Ko(A ) by p(l)= [I] - [ A ] . This is not a homomorphism but det o p = id. Now p(I) p(J)= p ( I J ) - p ( l ) - p(J) lies in F 2 Ko(A) since p(l~ p(J)~FIKo(A ). It follows that p: Pic A --} F 1 Ko(A)/F 2 Ko(A) is a homomorphism and is an inverse for det: FI/F 2 ~ PicA since podet is the identity on the generators [ I ] - [A].

To prove (d), we use Eagon's presentation of Ko(A) [13]. His generators are symbols (0), (p) , (m) corresponding to the prime ideals of A. Let C be the free abelian group on these. Since dim A = 2, there are only 2 types of relations

(a) For each prime p of height 1 and fCp, choose a composition series for A/(p, f ) . Let nm be the number of times Aim occurs. The relation is then ~.. nm(m ).

(b) For each x ~=0 in A choose a prime filtration ~ for A/Ax , let np(~) be the number of times Alp occurs as a quotient of ~, and similarly for n,~(~). The relation is then

r(x) = + Y

Note that for htp = 1, n~(o~) is independent of the choice of ~ We see this by passing to Ap. This is a discrete valuation ring and we get n~(~)= ordp x.

If S c C is generated by relations of type (a) and (b), Eagon's theorem asserts that C/S,~ Go(A) by the map taking (p ) to [A/p]. Therefore/(o(A)= C/S where C c C is generated by the ( p ) and (m) , omitting (0).

Let Tc C be the subgroup generated by relations of type (a) only. Then the image of r(x) in C/T is independent of the choice of ~ This is clear from Eagon's arguments. Furthermore, just as in Eagon's paper, the sequence 0 ~ A / A x ~ A / A x y ~ A / A y ~ 0 shows that r(x y) = r(x) + r(y) mod T so we can extend r to a homomorphism r: K * ~ ~/T.

Vector Bundles over Afline Surfaces 137

Let or be the group of invertible ideals of A and define C - - , J by (p)~--~t0 and ( m ) ~--,A. This leads to a commutative diagram

0 -~ A* , K* ~ , C /T ., Ko(A) ,0

(4)

0 ~A* ~K* ~ J ~ PicA ~0.

To see that ker r = A* note that A* c ke r r while the kernel of K* ~ J is precisely A*.

Remark. Using Eagon's arguments it is easy to see that C/T.~ Ko(~- ) where ~- is the category of torsion A-modules. The diagram (4) is then

KI(A ) , KI (K ) ~ ~ K o ( Y ) --, Ko(A ) ~ Ko(K)

1 l U(A) , U(K) 0 -*Pic(A,S) ~ PicA , PicK.

Since the vertical maps in (4) are onto, their kernels form an exact sequence which reduces to an isomorphism C' /T~F2Ko(A)=Ker(de t ) where C ' c C is generated by the ( m ) with m maximal. We identify C' with Z(A) by Jetting (rex) correspond to the 0-cycle x. The subgroup T c Z ( A ) is now obtained as follows. We look at height one primes p in A, i.e., at irreducible curves C c V. We then look at non-zero elements f of A/p, i.e., regular functions on C. The relation of type (a) is then seen to be exactly the 0-cycle (f). This differs from our description of rational equivalence only m that f is here required to be regular and not merely rational. But, since C is affine, any rational function is a quotient of two regular ones. Since ( f g ) = ( f ) + ( g ) this shows that T = B ( V ) and so F2Ko(A)=A(V).

We will leave the proof of (e) to the reader. It is immediate from (1) and Serre's formula for intersection multiplicity [51].

Remark. By using some of the easier parts of Theorem 4.2 for the complete case we can get a simple proof of the Riemann-Roch theorem for surfaces. Define p: Pic X --~ K~ as above. Since (F 1 K ~ 3 = 0, we have p(I1) p(I2) P(I3) = 0. From this is easily follows that fl(I, J) = p(I) p(J) = p (1J ) - p (1 ) - p(J) is bi-additive. (This only requires A to be noetherian with dim m - spec A < 2 by [5, Ch. IX, Prop. 4.4].) Let X: K~ --~ 7l be the Euler characteristic. By an easy calculation, we see that X(fl(I, J ) )= (I. J) and it follows that g(p(1))-�89 = g(I) is additive in I. Therefore g(1)=�89 Now Z ( I ) = z ( K - I ) by Serre's duality. Putting this in our formula and using the additivity of g we conclude that g ( I ) = - � 8 9 1) so X(1)= �89 the Riemann-Roch theorem. A similar and easier argument works for curves.

We conclude this section by proving the equivalence of (a), (b), and (c) in Theorem 2.

Theorem 4.5. Let A be a regular affine domain of dimension 2 over an algebraically closed field. Then the following conditions are equivalent.

(a) SKo(A)=O.


(b) Every finitely generated projective A-module is the direct sum of a free module and an invertible ideal.

(c) Every maximal ideal of A has 2 generators.

Proof If (a) holds, then det:/s Pic(A). Let P be finitely generated projective of rank r, let I = det P = ArP and let P' = A r- 1 • 1. Then ~ = [P] - [P ' ] ~/(o(A) and det ~ = 0 so that ~ = 0. Therefore P and P' are stably isomorphic so P ~ P' by Theorem 1.

Now suppose (b) holds. Let m be a maximal ideal and consider the sequence (2) above. By (b), P ~ A @ I but Lemma 4.4 shows that I = det P ~ A. Thus P = A �9 A and m has 2 generators.

Finally, assume (c). If m is any maximal ideal we have an exact sequence 0 -+ Q --~ A2-+ m - * 0 where Q isprojectivesincepdm


to the sequence 7 / ~ 7 / ~ 0 ---, 0. The snake l emma then gives

A o ( X - V) -+ Ao(X) -~ A(V) -+ O. (2)

If X = X 1 u X 2 where X 1 and X 2 are closed, we get

A(X 1 n X2) ~ A(X1)• A(X2) --~ A(X) ~ O. (3)

This follows from the fact that any irreducible curve in X lies in X~, or in X 2 . If X is complete and X 1 n X z 4: ~J, we can m a p (3) to 7 / ~ 7 / � 9 77 ~ 7 / ~ 0 and the snake l e m m a gives

Ao(X ~ n X2) -+ Ao(X1) �9 Ao(X2) --* Ao(X) -~ O. (4)

If X = X 1 u . . . u X , where the X i are closed, then G A(Xi)--~ A(X) is onto. This is trivial and also a consequence of (3). The similar result for A 0 requires a bit more work.

L e m m a 5.1. Let X = X 1 u ... u X , , n < ~ where the X i are closed in X. I f X is complete and connected, then O Ao(Xi)--* Ao(X ) is onto.

Proof. This is clear for n = 1. Assume it for n - 1 . Let F be the graph with vertices 1 . . . . . n where i and j are jo ined by an edge i fX i n Xj ~ . Then F is connected since X is. Let T be a maximal tree of F. By renumber ing if necessary, we can assume the vertex 1 is an end point of Tso T - {1} is connected. Let Y= X 2 u . . . u X . Then Y is connected and Y n X 1 ~ . By (4), Ao(X1)OAo(Y )--~ Ao(X ) is onto and our result follows by induction.

We will now finish the p roof of T h e o r e m 2. We first observe that it is sufficient to consider the case where V is an open set of X. Embed V in A N and let [7 be its closure in lP N. By the resolut ion of singularities [1] there is a proper m a p ~p: X'--~ 17 with X' nonsingular and project ive and q ~ - l ( V ) ~ V. Since Theo rem 2 involves birat ional invar iants of X, we can replace X by X'.

Remark. We can avoid the use of the resolut ion of singularities as follows. Let Ybe the normal iza t ion of 14. By [46, p. 45] we can blow up points of X to get X" with a birat ional morph i sm f : X"--~ Y. By [46, p. 55], f - ~ ( V ) is obta ined f rom V by blowing up points. By Cas te lnuovo ' s theorem [46, p. 102] we can blow down the blown up points of X" lying over V. This gives the required X'.

The fact that (a) implies (d) now follows f rom

Theorem 5.2. Let X be a non-singular projective surface and let V e X be a non-empty open set. I f A(V) is finitely generated, then Ao(X) is finite-dimensional in the sense of Mumford.

Remark. Our hypothesis is somewhat artificial. R o i t m a n [42] has shown that Ao(X) is divisible. Therefore, by (2), so is A(V) if V , X . Hence A ( V ) = 0 if it is finitely generated (unless V = X when A(V) = 7/, Ao(X) = 0). However , T h e o r e m 5.2 is complete ly e lementary and uses no proper t ies of abel ian varieties. The p roof is the same as for the case A ( V ) = 0 .

Proof Choose a finite set S c V which generates A(V). By (1) we see that A (X- V ) G A ( S ) ~ A (X) is on to and hence so is A (Y')--~ A (X) where Y'= ( X - V)u S.

140 M.P. Murthy and R. G. Swan

Enlarge Y' as follows. For each isolated point of Y' add a curve through it, getting a larger set Y" which is a union of curves. Let H be a hyperplane section of X and let Y= Y"u {H}. Since H meets every curve on X, Y is connected. Let Y= C 1 u- . . u C, where the C i are irreducible curves. By Lemma 5.1, O)Ao(C~)~Ao(Y) is onto. If W = X - Y, then A(W)=0 by (1) and so Ao(Y)---,Ao(X) is onto by (2). Let C~ be the normalizationof C~. Then Ao(C~) ~ AQ(Ci) is onto. Therefore • Ao(C~) ~ Ao(X ) is onto. But Ao(Ci) is the Jacobian of C~ and hence finite-dimensional. It then follows that Ao(X ) is also finite-dimensional.

To show that (d) implies (a) in Theorem 2 we use the following result.

Theorem 5.3. Let X be a non-singular projective surface. Let V be a non-empty affine open set of X. Then the composition Ao(X-V)---~Ao(X)--~Alb X is onto.

Proof. Let ~o: X---~ Alb X be the canonical map. Then the image of Ao(X- V) in AlbX is the subgroup generated by all ~0(yl)-(p(y2) with y l , y a e Y = X - V . Our first task is to show that this subgroup is closed.

Lemma 5.4. Let G be an algebraic group and let (p : Y---~ G where Y is connected. Then the subgroup H of G generated by all elements of the form tp(yl) ~o(y2) -1 for Yl, Y2 e Y is closed and connected.

Proof. Let Y1, ..., I1, be the irreducible components of Y. Let ~'i: Y/x Y/I> G by ~q(Yl, Y2)=~o(Y0 q~(Y2) -I. By 1-9, Ch. I. Prop. 2.2] the subgroup K generated by the images of the ~,~ is closed and connected. We claim that H = K. This is clear for n = 1. We assume it for n - 1. As in the proof of Lemma 5.1, we can renumber so that Y= YIuZ where Z = Y2w.. .u Y, is connected and YlnZ:~fJ. Let xe Ylc~Z. If yl,y2EY1, clearly tp(yl)~p(y2)-lEK. If z1 ,7 .2r , then q)(zl)tp(z2)-lEK by the induction hypothesis applied to ~olZ. Suppose now yE Y~ and zeZ. Then ~0(y) q~(z) -~ =(q~(y) q~(x)-l)(~o(x) ~o(z)-l)eKK c K and similarly for ~o(z) ~o(y) -~ Therefore H c K, while K c H is clear.

Now, returning to the proof of Theorem 5.3, we know that Y = X - V is connected by [17] [23, Ch. II, Cor. 6.2]. Therefore, by Lemma 5.4, the elements r with y l , y z ~ Y generate an abelian subvariety B c A l b X . Let A = A l b X / B and let if: X---~A be the composition X ~~ Then ~(Y) is a point. We will show that ~O(X) is a point. Since ~k(X) generates A, it then follows that A = 0 being a divisible group.

Let Z=ff (X) and p=(p(Y). Suppose first that d imZ--1 . If z~Z, zCp, then ~,-l(z) has dimension > 1 and is closed in X and hence complete. But ~k-l(z) is disjoint from Yc r (p) and so lies in the affine set V which is absurd.

Now suppose dim Z = 2 . By a theorem of Mumford [37] [46, p. 96], the intersection form is negative definite on the group of divisors of X supported by q~-l(p). But by [17] [46, Ch. II, Th. 4.2] Y~tp-~(p) supports an ample divisor D and such divisors satisfy (D 2) > 0.

It follows that dim Z must be 0 which proves the theorem.

Corollary 5.5. Let V be an affine open set of a non-singular projective surface X. Then SAo(X ) ~ A(V) is onto.

This is immediate from (2), Theorem 5.3 and the definition of SAo(X) as the kernel of Ao(X) ~ Alb X.


If k is uncountable, (d) implies (a) by a result of Roitman [42], who has shown that SAo(X ) is finite if and only if Ao(X) is finite-dimensional. He assumes that char k = 0 and shows that SAo(X)=O in this case. Without this assumption, his proof shows that SAo(X) is a finite p-group. He obtains a bijection A 1 --, Ao(X ) with an abelian variety A1, and, by taking the pullback along the canonical map z: X ~ Ao(X), gets a bijective map f : W o ~ X. Using this he produces a map Alb X -* A1 of degree equal to that off. If char k = 0, this degree is 1 but in characteristic p it may be a power of p.

We will show below that Roi tman's result extends to the case where k is countable but of infinite transcendency over the prime field.

Remark. We have observed above that A(V) is a divisible group. Therefore it will follow that A(V)= 0 if SAo(X) is finite.

Now, assuming k uncountable, we will prove the last statement of Theorem 2. Suppose SKo(A)=A(V)4:0. Let aEA(V), a+O and let beAo(X ) map onto a. By [42, Lemma 13] there is a "regular" map q~: A ~ A o (X) of an abelian variety A into Ao(X) whose image contains b. Let Y = X - V. By [17], [23, Ch. II, Prop. 3.1, Cor. 6.2], Y is connected and pure of dimension 1 so that Y= C a u . . . w C, where the C i are irreducible curves. Let F~ be the normalization of Ci and let Ji be the Jacobian of F i. By Lemma 5.1 we see that J=I-IJ i=l- IAo(Fi) -~Ao(Y) is onto, so by (2), J ~'~ Ao(X)-~A(V)---~O is exact. Now ~: J ~ A o ( X ) is a regular map. In fact, for each i, we have a surjection S ~' F~ • S ~' F i ~ Ji so we get

l](sg, I) ,s"x•

I~ Ji --~ ao(X)

where n = ~ gi" Therefore ~: J ~ Ao(X) is regular [42, Def. 1]. By [42, Lemma 3], the set W = A • J of all (x, y) with q~(x)=~(y) is c-closed. Therefore so is its projection B on A but B is the kernel of the composition A ~ Ao(X)-~ Ao(V). Let B o be the connected component of B. It is a closed subgroup of A so A' = A/B o is an abelian variety. Also BIB o is countable. This gives us a map A'- - ,A(V) with countable kernel B' = BIB o. Since a + 0 lies in the image, A' 4:0 and so dim A' > 0. The torsion subgroup T of A' is countable so A' /T is a torsion free divisible group of r ank= lk l since clearly A has Ikl points (cf. Theorem 7.6). Let A " = A ' / T as a vector space over Q and let B" be the Q subspace generated by the image of B'. Then B" has countable rank so the rank of A"/B" is still I k I. Since A"/B" is a quotient of a subgroup of SKo(A) we see that rkSKo(A ) > Ikl. On the other hand it is obvious that Ao(X) has at most Ikl elements.

We will now show how to use the Lefschetz principle to prove Theorem 2 when k is countable. The idea underlying this principle, as formulated by Weil [63, p. 306], can be formulated in a very elementary way. As in [63] we will not at tempt to give a general version since the special cases we shall use will make the underlying idea clear. We will give a detailed treatment since it is not immediately clear that the usual rather vague statement of the Lefschetz principle applies here. The fundamental importance of functors preserving filtered direct limits has also


been pointed out in connection with formalized versions of the Lefschetz principle [14].

Let kobe a field and let cg be the category of algebraically closed fields containing k o, the maps being ko-algebra homomorphisms. The next two lemmas contain the special cases of the Lefschetz principle which we will need.

Lemma 5.6. Let 4: cs Ab be a functor preserving filtered direct limits. Let K6Cg have infinite transcendency over k o and suppose that q~(K) = G has one of the following properties

(1) G = 0 ,

(2) G is finite,

(3) G is of infinite rank.

Then ~(L) has the same property for all L6Cg of infinite transcendency over k o .

Lemma 5.7. Let alp, 7J : cg ~ Sets be functors preserving filtered direct limits and let ~: ~--~ 7 t be a natural transformation. I f q(K): q~(K)--~ 7J(K) is onto jbr some K6Cg of infinite transcendency over ko, then q(L) is onto for all LEc~ of infinite transcendency over k o.

Proof The idea of the proof is that any KeCs is the filtered union of subfields E6Cg which are of finite transcendency over k o and so ~ ( K ) = l i m ~(E). If K is of infinite transcendency over k o, any such E can be embedded in K and this embedding extends to any F ~ E of finite transcendency over E. Using this we will show that each of the properties in question can be reformulated in terms of fields of finite transcendency over k o and hence is independent of the choice of K. We will simply give the required reformulation, the proof being trivial. In these statements, E and F will always refer to algebraically closed fields of finite transcendency over k o. For Lemma 5.6 the required statements are:

(1) If x ~ ( E ) , there is an F ~ E such that x goes to 0 in ~(F).

(2) The fact that ~ ( K ) < n is characterized by: If x o . . . . . x,~O(E), there is an F ~ E such that for some i4:j, x i and x i have the same image in ~(F).

(3) The fact that r k ~ ( K ) > n is characterized by: There are elements x~ . . . . . x , ~ ( E ) for some E such that for each F ~ E , the images of x~ . . . . . x, in �9 (F) generate a free subgroup of rank n.

For Lemma 5.7 the required statement is: For each y e 7~(E), there is an F = E and xeCb(F) such that r/(F)x =y.

To apply these lemmas we observe that X can be defined over an algebraically closed field k o of finite transcendency over the prime field. Thus there is a non- singular surface X o over k o such that X = K | o. For each K ~ , write X r = K | o. Similarly, the algebra A can be defined over some algebraically closed field k o of finite transcendency of the prime field so that A = k | where SpecA o is smooth over k o. For each KeCs let A t = K| o.

Lemma 5.8. The functors on cg taking K to Ao(XK) , Alb(XK), SAo(XK), Ko(AK), /(o(AK), Pic(Ar) , SKo(A~), Sn XK x S" X K preserve filtered direct limits.

Proof For Ao(XK) we observe that each 0-cycle on X K can be defined over some subfield E6Cg having finite transcendency over k o. But if K =liimm K , is a


filtered limit in cg, each subfield of K with finite transcendency over k o hes in some K,. The same applies to each curve and function needed to give the relations in Ao(XK). Similar arguments apply to all the other functors. The only cases for which there may be some doubt are Alb X K and SAo(XK). It will suffice to consider SAo(XK) since Alb X K =Ao(X~:)/SAo(X~). Let z be a 0-cycle on X K defined over some EeCg of finite transcendency over k o . Suppose z represents a point of SAo(XK). This means that z maps to 0 in Alb X K and hence, by the universal property of Alb X/~, that any map f : X K -~ A, where A is an abelian variety over K, will send z to 0. Let B be any abelian variety over E and let g: X E --~ B. Then K | g: XK = K|174 sends z to 0. Since abelian varieties are projective [35], it is clear that B ~ K | is injective. Therefore z goes to 0 in B. This shows that z represents an element of SAo(Xe).

Now, to prove Theorem 2 for k countable, choose k 0 ~ k as above and let L ~ k be uncountable and algebraically closed. By Lemma 5.6 we see that statement (e) of Theorem 2 is equivalent to the same statement for L | X. The same applies to (d) of Theorem 2 since the finite-dimensionality of Ao(X) means that S"X • S"X ~ Ao(X ) is onto for some n and we can apply Lemma 5.7. Therefore the equivalence of our statements for L Q k X implies their equivalence for X. For the last statement, if SKo(A)+ O, then SKo(A L)4= 0 by Lemma 5.6(1). Therefore SKo(AL) has infinite rank and, by Lemma 5.6(3), so does SKo(A ). Since k is countable, this is all we need.

Remark. By the same argument we can extend Mumford's theorem [36] to the case where char k = 0 and k is of infinite transcendency over @. If pg(X)+O, then pg(XL)=pg(X)#:O so Ao(X L) is not finite-dimensional and so, as above, neither is Ao(X). The fact that pg(XL) =pg(X) is well-known but can also be deduced by a Lefschetz principle argument. The appropriate functor sends K to f2(XK)= holomorphic 2-forms on X K. This is a vector space over K. The fact that dimKf2(XK) =pg can be reformulated as follows: If x o . . . . . xp~f2(Xe) there is an F ~ E such that the images of x o . . . . . Xp in O(Xv) are linearly dependent. Also, for some E, there are elements x 1 . . . . . Xp~f2(Xe) whose images in f2(XF) are never linearly dependent. Of course, f2 commutes with extensions of the ground field and this gives the same result with no transcendency restriction.

We conclude this section with a result on the structure of SA o.

Theorem 5.9. Let X be a non-singular projective surface over an algebraically closed field. Then SAo(X ) is the direct sum of a finite group and a divisible group.

Proof Let V be a non-empty affine open set of X and let Y = X - V. Then Y is connected and pure of dimension 1 [17], [23, Ch. II, Prop. 3.1, Cor. 6.2]. As in the proof of Theorem 5.2, let Y= C 1 w. . . w C,, let C~ be the normalization of C~ and let Ji=J(Ci)=Ao(Ci). By Lemma 5.1, J=IJJi---~Ao(Y) is onto. Therefore by Theorem 5.3, the composit ion J - - ~ A o ( X ) ~ A l b X is onto. Furthermore, this composition is a map of abelian varieties since the map Jg -+ Alb X is the map Alb C i - + A l b X induced by the proper map (~i-~X. Let J ' be the connected component of the kernel J" of J ~ Alb X. Then J ' is an abelian subvariety of J. Then J ' is an abelian subvariety of J. By Poincare's complete reducibility theorem [30, 35] there is an abelian subvariety B of J such that B + J ' = J and B n J ' is


finite. Since J"/J' is finite, it follows that B n J " is finite. Therefore B---~Ao(X ) Alb X is onto with finite kernel. Let D be the image of B in Ao(X ). Then D is

divisible so we can write A o ( X ) = D @ E and E is also divisible because Ao(X ) is. Applying the snake lemma to

0--~ D ,Ao(X ) ~ E--~ 0

O ~ A l b X , A l b X , 0 ~ 0

gives 0--* F ~ SAo(X)--~ E --+ 0 where F is the kernel of D ~ Alb X which is finite. Now apply the following elementary, and presumably standard, lemma.

Lemma 5.10. Let G be an abelian group with a finite subgroup F such that E = G/F is divisible. Then G is the direct sum of a finite group and a divisible group.

Proof For any integer n, 0 ~ F ~ G ~ E ~ 0 gives F/nF -~ G/nG ~ E/nE ~ O. since E/nE = O, [G/nG[ < J Ft. Choose n so [G/nGJis maximal. For any m, G/m n G G /nG is onto so I G/m nG [ = [ G /nG [ and so mnG = nG. This shows that nG is divisible so G = nG~)H. Clearly H,~ G/nG is finite.

6. Examples. We begin with Mumford 's construction of indecomposable bundles.

Theorem 6.1. Let k be an algebraically closed field o f characteristic 0 and of infinite transcendency over q). Let X be a non-singular projective surface which is regular (q =0) and has pg(X)+-O. Let V be a non-empty affine open set of X. Then there is a vector bundle ~ on X of rank 2 such that ~ l V is indecomposable.

Proof Since X is regular, its Picard variety is trivial and hence Pic X is finitely generated [29, Ch. V, w 6, Th. 7]. Let D x . . . . . D, generate Pic X and let U c V be a non-empty affine open set which is disjoint from U suppDi. Then Pic U = 0 so all line bundles on U are trivial. Now by Theorem 2 and [36], we have SKo(A ) ~ 0 where U = Spec A. Therefore A has a non-trivial projective module P which we can take to be of rank 2 by Serre's theorem [5]. This P must be indecomposable since all projectives of rank 1 over A are free. All that remains is to extend P to a bundle over X.

Lemma 6.2. Let X be a non-singular surface, V an open set of V and q a vector bundle over V. Then there is a vector bundle ~ over X with ~1V ~ tl.

Proof Let ~- be a coherent sheaf on X with ~ [ V ~ t / . Let g = ~ * * where ~ * = H o m ~ , , ( ~ , (gx). Then g I V g t / but is locally free since reflexive modules over a 2-dimensional regular local ring are free [26, Lemma 6].

We now look at the problem of deciding when invertible ideals have 2 generators. Let A be a commutative noetherian ring with dim m - Spec A < 2, p: Pic A ~ K o by p(I) = [ I ] - [ a ] , and define fl(I, J) = p(I) p(J) = p ( I J ) - p ( I ) - p(J). In w 4 we saw that fl is a biadditive map fl: Pic A x P icA- - ,SKo(A ). Note that f l (I ,J)= [1J] - [I] - [J ] + [A] so, by Theorem 1, fl(1, J ) = 0 if and only i f l G J , ~ A G I J .


L e m m a 6.3. Let A be a commutative noetherian ring with dim m - S p e c A < 2. Let I be an invertible ideal of A. Then the following are equivalent

(a) I has 2 generators,

(b) 101-1 ~ A 2,

(c) fl(I, I ) = 0 .

Proof By the r emark preceding the l e m m a it is clear that (b) holds if and only if fl(I, I - 1) = _ fl(1, 1) = 0. Also it is trivial that (a) holds if and only if 1~) Q ,~ A z for some Q so (b) implies (a). I f I@QmA 2 then Q is projective of rank 1 and applying det: Ko(A)---~ PicA shows that IQ,~A so Q~1-1.

Corollary 6.4. I f A in Lemma 6.3 has SKo(A)=O then every invertible ideal in A has 2 generators.

This is clear since fl(I, J)~SKo(A ). The converse is false because the construction used to prove Theorem 6.1 gives an example where SKo(A)#:O but Pic A = 0 so that every invertible ideal is principal. We will give two part ial converses which will enable us to construct examples of invertible ideals requiring m o r e than 2 generators.

Theorem 6.5. Let A be a commutative noetherian ring with dim m - spec A < 2. Let L c Ko(A ) be the subgroup generated by elements of the form [ I ] - [A] where I is an invertible ideal. Let L o be the kernel of det: L ~ Pic A. I f every invertible ideal of A has 2 generators then 2L o = 0.

Proof Let L I be the subgroup of e lements beL o with 2 b = 0 . By L e m m a 6.3 we have/~(I , I ) = 0 for all 1. Since fl is symmetr ic and biadditive, it follows that 2fl(I, J ) = 0 for all I, J. Therefore ~(I, J ) s L 1 . This implies that

p ' : Pic A & L--~ L/L 1

is a h o m o m o r p h i s m . Since L I c L o, det induces det ' : L / L I ~ P i c A . Clearly, det ' o p ' = id. We claim that p' o det ' = id. Since we are dealing with h o m o m o r p h i s m s , it is sufficient to check this for the genera tors Eli - [A] of L/L~ where it is trivial. Therefore det ' : LIE ~ Pic A so L o = L x .

Corollary 6.6. Let V = Spec A be a non-singular affine surface over an algebraically closed field. I f every invertible ideal in A has 2 generators and Ko(A) is generated by the classes of invertible ideals, then SKo(A)= O.

In this case L = Ko(A ) and L o = SKo(A). Since SKo(A ) is divisible, by Theorem 3, it mus t be 0.

The hypothesis that Ko(A ) is genera ted by invertible ideals will be satisfied if every point on V is the intersection of 2 curves (with multiplicity 1). In this case, if m is any maximal ideal, we can write m =p~ "+P2 where p~, P2 are pr imes of height 1 and hence invertible. This gives 0---~01 n P 2 - - * P l O P 2 ~ m - - - ' 0 . Since A is regular and hence locally factorial, Pl r iP2 = P l P2 is invertible and [A /m] = [A] - [ m ] = [A] + [pl P2] - [P l ] - [ P / ] = fi(Pl, P2). It follows that the p roduc t of 2 curves F~ x F 2 has the desired property. If these curves are not ra t ional we will have p~(F x x F2)>0 so we can apply [36] and Theorem 2 to get


Corollary 6.7. Let k be an uncountable algebraically closed field of characteristic O. Let F1, F 2 be non-singular complete curves over k with generators gl, g2 >0. Let V= Spec A be a non-empty affine open set of F 1 x F 2 . Then A has an invertible ideal I which requires more than 2 generators.

We now give another partial converse to Corollary 6.4.

Theorem 6.8. Let V= Spec A be a non-singular affine surface over an algebraically closed field such that S Ko(A ) ~ O. Then there is a double covering U = Spec B --~ Spec A by a non-singular surface U such that some invertible ideal of B requires more than 2 generators.

Proof. Suppose the conclusion is false. We will show that 2SKo(A)=O. By Theorem 4.2(d), SKo(A ) is generated by the elements ~ = [A/m] with m maximal. Choose a resolution 0-~ A-~ P--~ m-~ 0 as in w 4(2). Then P has rank 2 and

---= [ P ' ] - [ A 2 ] . By a theorem of Schwarzenberger [47], we can find a nonsingular U which is a double covering of V such that the pullback to U of the vector bundle on V defined by P has a non-trivial subbundle. Since U is finite over V it is affine and U = S p e c B where A c B has rank 2 as an A-module. Let Q=B| Then Q has the form Q,,~IOJ where I. and J are invertible ideals of B. By Lemma 4.4, A 2 p ~ A and so A2Q,~B which shows that J ~ I -~. Since we have assumed the conclusion of Theorem 6.8 is false, it follows from Lemma 6.3 that Q ~ B E. There- fore the map i ,:Ko(A)--, Ko(B) defined by the inclusion A c B sends ~ to zero. Since SpecA and SpecB are non-singular, Ko(A)=Go(A ) and Ko(B)=Go(B), and, since B is finite over A we can define a transfer i*: Go(B)--> Go(A) by i*[M] = [M]. The composition i* i , : Ko(A )--, Ko(A) is given by multiplication by [B]. Therefore [ B ] ~ = 0 . But, since ~eSKo(A) and [B]-[AE]eKo(A) we have ~ ( [ B ] - [A2])=0 so [B] ~ = [A 2] ~ = 2~. This shows that 2SKo(A)=O and hence SKo(A)=O by divisibility.

One conceivable application of this theorem is this: The results of w 7 suggest the conjecture that SKo(A ) = 0 for all surfaces if k is the algebraic closure of a finite field. Additional evidence is given by the theorem of Vaser~tein that the stable range of A is < 2 in this case [6]. If one could show that the projective stable range is __


construction could be extended to a complete model X of Spec A we could apply results on the classification of etale coverings [53] to show that P is decomposable in certain cases. Unfortunately this does not seem to work. The basic difficulty is that S(P*) is not the dual of S(P) in a canonical way for char k = 2.

7. Calculations. In this section we will show that SAo(X)=O for surfaces of the type considered in [45, Ch. IV and VII I with pg=O. This work was inspired by an example of Bloch and Lieberman [24] and the work of Artin and Lieberman on Enriques surfaces [3]. We begin by showing how to compute A(X) by descent.

Theorem 7.1. Let to: X --~ Y be a surjective proper morphism of algebraic sets. Then the sequence

A(X • pl,_p2 , A(X)-~:~ A(Y)-~O

is exact, where Pl, P2 are the projections of X • y X on X.

A more naive (and probably more useful) way of expressing the result is this: to, is onto and its kernel is generated by the classes of O-cycles of the form x 1 - x 2 where to(x1) = to(x2).

For the proof we will need a well-known elementary lemma.

Lemma 7.2. Let qo: X - ~ Y be a dominating morphism of algebraic sets. Then there is a closed subset X' ~ X which dominates Y and has dim X' = dim Y.

If q0 is proper, it follows that to IX': X'--. Y is onto. Note that if Y is irreducible we can choose X' irreducible since one of its components will dominate Y

Proof It will suffice to look at to-I(Y~)~ Y~ for each component of Y Therefore we assume that Y is irreducible. Let V be a non-empty affine open set of Y Choose a locally closed affine set U ~ X which dominates V and which has the smallest possible dimension. Let A and B be the coordinate rings of V and U. Then A ~ B and for every non-zero prime ideal P of B we have P c~ A 4= 0; otherwise Spec B/P would dominate V and have smaller dimensionl If S = A - {0}, it follows that B[S -~] must be a field since it has no non-zero prime ideals. Let F be the quotient field of A. Then clearly F B = B [ S - 1 ] = K , the quotient field of B. Since K is of finite type as an F-algebra, the Nullstellensatz shows that K is a finite extension of F. Therefore dim U = dim V. We choose X' to be the closure of U in X.

Proof of Theorem7.1. It is clear that t o , ( p l , - P 2 , ) = 0 . Therefore, if C = c k r ( p l , - p 2 , ) we have a map ~ , : C--~A(Y). Define a map r Z ( Y ) ~ C by sending a 0-cycle z on Y to the image in C of the class of any 0-cycle z' on X with to(z')=z. If to(z')=to(z"), then z ' - z " = ~ ( x , - y , ) where q)(x,)=to(y,). Therefore, using the naive interpretation of the theorem, it is clear that ~ is well-defined. We claim that ~ factors through a map r/: A ( Y ) ~ C. This will prove the theorem because t~, q is clearly the identity and so is q qS, since to compute ~ to(z) we can choose z itself as the lifting of t0(z).

Suppose then that ~O: F - * Y is a proper map of a non-singular curve F into Y and let f be a rational function on F. We must show that r Apply Lemma7.2 to the proper map X xyF--~,F getting a closed curve F ' c X x y F


which maps onto F. This gives us a diagram

F' ~' >X

F 0 )y.

Let F=k(F) and K=k(F ' ) be the function fields o f F and F'. Then K is a finite extension of F. By Tsen's theorem [62] or [28,Th. 6, 13], NK/e: K * - + F * is onto. Choose g e K with NK/e(g)=(f) . By Lemma 4.1 we have 0((g))=(f). Therefore we can choose $'((g)) as the lifting of (f) which represents r But $'((g))~0 in A(X) .

We now consider a special case where Y is a quotient of X by a finite group G. Note that X / G will be an algebraic set if X is quasi-projective [35, Ch. III, w 12, Th. 1]. We write Gab=G/[G, G] =Hx(G, 7/,,).

Corollary 7.3. Let G be a f ini te group acting on an algebraic set X. Assume that Y = X / G is again an algebraic set. Then

(i) A(X) /G ~" , A(Y) .

(2) I f X is complete, there is an exact sequence

Gab -* Ao(X) /G -~ Ao( Y ) -* O.

(3) I f X and Y are complete, irreducible and non-singular, we have exact sequences

Gab -+ h l b ( X ) / G ~ hlb(Y) - . 0

and

H -+ SAo(X) /G -* SAo(Y) --* 0

where H c G.

Proof. The naive interpretation of Theorem 1 gives us (1) and (2) follows by taking the G-homology sequence of 0 -+ Ao(X ) --~ A(X) -~ 7] - . 0 getting

111 (G, Z) --+ A o (X) /G

0 , A o ( Y )

, A(X)/G , z , o

, A ( Y ) , z , o .

The first sequence in (3) is obtained from the diagram

ul (G, z) > Ao(X) /G , A o ( Y ) , 0

A lb (X ) /G , Alb(Y).

(1)

Note that Alb (X) /G is an abelian variety since it is the quotient of Alb(X) by the subgroup generated by the images of the maps a - 1 : AIbX ~ A l b X for a e G .


These sets are all irreducible and contain 0 [9, Ch. I, Prop. 2.2]. Let C be the cokernel of the map H 1 (G, 7z)~ AIb(X)/G obtained from (i). Choose a basepoint xoeX and let z: X ~ Ao(X ) by z(x)=[X-Xo]. Then X ~ > Ao(X)---, A lbX is a canonical map c~ of X to AIbX and induces ~: X ~ C. If eEG, then z ( e x ) - z ( x ) = [ t ~ x - x ] maps to zero in Ao(Y ). Therefore its image in Ao(X)/G lies in the image of H 1 (G, 7/) and it follows that ~(ex)= ~(x). Since X " , Alb(X)--, C are morphisms, so is ~ and therefore ~ induces fl: Y---, C. By the universal property of the canonical map 7: Y---,Alb(Y) with 7(yo)--0, yo=image of x o, we get a map of abelian varieties 0: AIb(Y)---, C with [3=07. The diagram (1) shows that we have a map (: C ~ A l b Y with ([3=7. Since fl(Y) generates C and 7(Y) generates AlbY, it follows that (0 = id, 0 ( = id.

To obtain the second exact sequence in (3), let H be the kernel of H 1 (G, 7/) Alb(X)/G, complete the diagram (1) by putting HI(G, 7/)/H in the lower left corner, and apply the snake lemma.

Remark. The kernels of Ao(X)/G ~ Ao(Y) and Alb(X)/G ---, Alb Y will be non- trivial in general. For example if X is an elliptic curve and G acts by translations, both of these maps are X ~ X/G with kernel G. We can obtain an analogue of Theorem 7.1 for A o by applying the snake lemma to

0 , A o ( X x r X ) , A ( X x r X ) ,7/ ,0 |

Pl * -- P2 * ~ 0

4,

0 > Ao(X ) , A(X) ,7/ ,0

but we cannot deduce the analogue of (1) by the usual categorical argument because A o does not preserve coproducts.

We now come to the main result of this section.

Theorem 7.4. Let G be a finite group acting on two complete non-singular curves C and F such that X=(C x F)/G is a non-singular surface. Suppose that Ao(C)~Ao(F)=O. Then SAo(X ) is finite and Alb(X) is a finite covering of J(C/G) • J(F/G). I f in addition, F/G is rational, then SAo(X ) = 0 and Alb(X) = J( C/G).

We will show that one of F/G or C/G is always rational unless k is the algebraic closure of a finite field. We will also see that, in fact, SAo(X) and the covering group of AIb(X)-*J(C/G)x J(F/G) are subquotients of Gab X Gab. The theorem implies, of course, that Ao(X) is finite-dimensional.

Before giving the proof we will briefly discuss the hypothesis. If D is a divisible group and T is a torsion group then D | T = 0 (where | = ( ~ . Therefore we have Ao( C) | Ao(F) = QAo(C) | QAo(F) where we define QAo(F) = Ao(F)/torsion. This is a vector space over Q. If V and W are vector spaces over Q, then V Q W = V| W. By reducing mod G we see that our hypothesis is equivalent to

QAo(C) Q~QAo(F)=O.

This involves only rational representations of G.

Remark. Since QAo(C)| maps onto QAo(C/G)| ), the hypothesis of the theorem implies that one of Ao(C/G) and Ao(F/G ) is torsion.


Now if k is not the algebraic closure of a finite field, Corollary 7.7 below shows that QAo(C/G)~:O unless C is rational. Therefore, with this restriction on k, one of C/G and F/G is rational and the stronger conclusion holds. We will discuss the case where k is the algebraic closure of a finite field after proving the theorem.

Proof of Theorem 7.4. For any X and Y we clearly have Z(X • Y) = Z(X) | Z( Y). By factoring out relations corresponding to vertical and horizontal curves, we

get A(X)@A(Y) so there is a natural epimorphism A(X)| ~ , A(X • Y). In the situation of Theorem 7.4 we write

0 --~ Ao(C)QAo(F) --~ A(C)@ A(F) ~ P --~ 0

where P=A(C)xzA(F) and the map of A(C)| to P has the projections 1 | deg and deg @ 1. Collapsing this under the action of G and using the hypothesis we get A ( C ) ( ~ A ( F ) ~ - ~ P/G. Now # gives fi: A(C)(~GA(F)--~A(X ) so we get an epimorphism P/G---~A(X). By collapsing O-~Ao(C)(~Ao(F)--~P--.7I-~O under the action of G, we get

Ao( C)/G �9 Ao(F)/G --~ P/G ~ 7l --~ O.

Using our epimorphis m P/G ~ A(X), this gives us an epimorphism

~: Ao(C)/G D Ao(F)/G ~ Ao(X).

Choose any basepoints co,C, fo~F and define i: C--~X by sending c to the image of (C, fo) and j: F--~ X by sending f to the image of (f, Co). Then the diagram

A~176

Ao(C)/GOAo(F)/G- i -~ Ao(X )

commutes. This follows easily from the commutativity of

P

Ao(C)•Ao(F) ~ A(C)|

, P/G

, A(C)(~6A(F)

where P[Ao(C ) sends z to z | and p lA o(F) sends z to c o | z. Now the projection

diagram oi:, \ Ao(X)

AIbX

,A(X)

X ~ C / G induces AIbX-*J(C/G). The commutative

, Ao(C/G )

, J ( c / G )

Vector Bundles over Affine Surfaces

and a similar one for F show that the composition in the bot tom line of

Ao(C) OAo(F)

151

Ao( C)/G 0 Ao(F)/G --~ Ao(X ) --~ Alb(X) --~ J( C/G) • J(F/G)

= Ao(C/G) (~ A o (F/G) (2)

is just the direct sum of the canonical maps

Ao(C)/G -~ Ao(C/G), Ao(F)/G -~ Ao(F/G ).

By Corollary 7.3(2), these are onto and their kernels are quotients of Gab. Since the two maps on the left in the bot tom row of (2) are onto, the conclusions of the theorem follow.

Suppose now that F/G is rational. Since Ao(F)/G is divisible and Ao(F)/G--~ Ao(F/G) = 0 is onto with finite kernel, we see that Ao(F)/G = 0. We saw above that the map j , : Ao(F ) --~ Ao(X ) factors through Ao(F)/G. Therefore j , =0. But j(F) is exactly the fiber p-l(c) of the projection p: X---~ C/G. Applying Theorem 7.1 we see that p , : A(X)~A(C/G) and so p , : Ao(X)~Ao(C/G ). This factors into Ao(X) --~ A l b X -~J(C/G) = Ao(C/G) and the conclusions follow immediately.

Remark. Suppose k is the algebraic closure of a finite field. Then every abelian variety A over k is torsion since A = U A~Fp,). Thus QAo(F)=O for all curves F so Theorem 7.4 applies to all non-singular surfaces X of the form (C x F)/G and, in particular, to C x F itself.

Suppose C = F = E, an elliptic curve, with G acting on E by translations. Then X = E x E/G is abelian so A l b X = X and AlbX--~J (E/G)• J(E/G) has a non- trivial kernel ~ G.

However we have not been able to obtain an example with SAo(X ) finite and non-zero in this way. Suppose, in fact, that Ao(F ) is torsion. Then Ao(C)| Ao(F) = 0 so Theorem7.4 applies with G = I . Since Gab=l we get SAo(CXF)=O. By Corollary 7.3(4), it follows that SAo(X)=0 for X = ( C • F)/G.

We now give three cases in which Theorem 7.4 applies.

(I) (Ruled surfaces) G = 1 and F = IP 1.

(II) C is an elliptic curve, G acts on it by translation, and F/G =IP ~. (III) F is an elliptic curve and if A = k e r [ G --~ Aut J(F)] then H = G/A + 1, C/A

is elliptic, and H acts on C/A by translations. (Note that H4:1 if and only if F/G = I P 1 . )

We must now verify the hypotheses of Theorem 7.4. In each case, G acts freely so X is non-singular. Also F/G=IP 1. It remains to compute Ao(C)~GAo(F). In case (I) this is trivial because A0(F)=0. In case (II), G acts trivially on A0(C ) because Ao(C)~ C with the 0-cycle x - y corresponding to the actual difference x - y computed in C. This is clearly invariant under translations. It follows that QAo(C)|174 but QAo(F)/G=QAo(F/G)=O by Corollary 7.3(2). In case (III) we first compute

Q Ao( C) | Q Ao(F) = Q Ao( C /A ) @r Q Ao(F)


as in case (II). Since H acts on C/A by translations, the same argument now applies to the action of H and we get QAo(C)| QAo(C/A)| but QAo(F)/H = QAo(F)/G = QAo(F/G ) = O.

Corollary7.5. I f c h a r k = 0 and X is a non-singular projective surface with pg = O, q 4= O, then SAo(X ) = O.

In fact, X has one of the above three types by [45, Ch. IV]. It is of course well known and trivial that SAo(X)=O for ruled surfaces. Of the remaining surfaces with p~ = 0, Artin and Lieberman [3] have shown that the Enriques surfaces have Ao(X)=O. Bloch has pointed out that the Godeaux surfaces l-8] would form a good starting point for the investigation of the surfaces of general type, i.e., with ~(X)=2.

We conclude this section by proving the result mentioned in the remark preceding the proof of Theorem 7.4. The proof was suggested by the work of Lutz [32].

Theorem 7.6. Let k be an algebraically closed field which is not the algebraic closure of a finite field. Let A 4= 0 be an abelian variety over k. Then A, considered as an abelian group, has rank equal to the cardinality of k.

We are, of course, identifying A with the set of points of A rational over k.

Proof. Suppose first that k is uncountable. Then the result is trivial since the cardinality of A is equal to that of k (e.g. by the representation of A used below) while the torsion subgroup of A is countable. Therefore we can assume that k is countable. In this case there is a real non-trivial non-archimedian valuation on k whose residue field has characteristic p 4=0. A simple trick for seeing this was given by Lech [31]: Let F c k be Q or IFp(x) and let f2 be the algebraic closure of the completion of F at a non-archimedian valuation. Since ~ is uncountable it has infinite transcendency over the prime field so we can extend the inclusion F e l2 to an embedding of k in f2. We use the restriction to k of the canonical extension to f2 of our chosen valuation on F.

Let R = (x e k l lxl< 1} be the valuation ring and E = R/m the residue field with q~: R --, E the canonical map. We will need the standard fact that E is algebraically closed. To see this, let g be any monic polynomial over E. Lift g to a monic heR[x]. If ~ek is a root of h then I~l


By making U smaller if necessary, we can assume that the qi are never zero on U. Clearly the addition on V corresponds to that on A whenever x, y, x + y e U . Since x + 0 = x and 0 + y = y, we have

pi(x, O) = x i qi(x, O) + ki(x) f (x ),

p,(0, y) = y, qi(O, y) + l,(y) f (y).

Replace pi(x,y) by p i ( x , y ) - k i ( x ) f ( x ) - l i ( y ) f ( y ) . Then, since f ( 0 ) = 0 , we have pi(x, O)=xiqi(x, O) and pi(O, y)=yi qi(O, y). Therefore, pi(x, y ) = x i + yi+higher terms, so we can write

r~(x, y) (x + Y)i= xi + yl-b qi(x, y )

where ri(x, y) has no terms of degree < 1. Similarly, the map x ~ - x on A defines a rational map, also denoted x ~ - x,

on V. Near 0, it will be given by ( - x)i = si(x)/ti(x) where t/(0)= 1 and we can again take U so small that the ti are never zero on U. Therefore the map x ~ - x on V will correspond to the one on A whenever x, - x ~ U. Since x + ( - x) = 0 we see that

x, s,(x)~ [ So(X) [ So(X) + t ~ ) qi kX' to(X ) . . . . ) +ri tX, to(X ) . . . . )

is zero on V. Clearing denominators gives a polynomial which is zero on V and hence a relation of the form

T(x)(ti(x ) x i + si(x)) + Ri(x ) = mi(x ) f (x)

where T (0 )= 1 and R i has no terms of degree < 1. Since ti(0)= 1, this reduces to

x i + si(x) + higher terms = mi(x ) f (x).

Replace si(x ) by si(x ) - m i ( x ) f ( x ). Then si(x ) = - x i + higher terms, so we have

( - x)~ = x~ -~ ui(x) t,(x)

where ui has no terms of degree ___< 1. Define Lxl=maxlx~l and let G~={xeV]lxl 0 . By making U

O < i < n

smaller if necessary we can assume that U is defined by g(x)#O where g(O)= 1. If 6 is small enough, we will have Ig (x ) l= l for I x l < 6 and hence G~c U. If 6 is sufficiently small, we will also have

]qi(x, y)[ = 1, Iti(x)[ = 1, ]ri(x, y)[


claim that 0 is onto if 6 is so small that [fo(Xl . . . . . x,)l < 6 and If1 (xl . . . . . x,)l = 1 for [xll


Therefore we can assume G = U is unipotent and hence we can put G in triangular form. The result is clearly true in this case. The same argument shows that if case (a) applies but cases (b) and (c) do not then G has elements of every order prime to chark since this is true for k* and abelian varieties.

8. Topological Lemmas. In this section we will give some topological results which will be needed in w167 9 and 10. This section is intended mainly as a convenience for the reader and no claim is made for the originality of the results, all of which will presumably be regarded as well-known or obvious by experts in the field.

Let Z be a pathwise connected space and D a subspace of Z. We say (Z, D) is d-connected if D+~J and 7~(Z, D ) = 0 for 1


Corollary 8.3. Let X be a CW complex with subcomplex A. Let Z be a d-connected space. Then IX~A, Z] ~ [X, Z] is surjective if dim A 1 since A ~ ~J. Therefore both vertical arrows are bijections and hence so is the bottom arrow.

Corollary 8.4. Let Y and Z be pathwise connected spaces. Let qg: Y-~ Z such that ni(q~): ni( Y ) --~ ni(Z ) is an isomorphism for 1


where p, and i, are induced by p and the inclusion i: G-~ E. Furthermore, [X, G] is a group and acts on IX, El from the right and, for ct, fl~ IX, El, we have p, (~t)= p,(fl) if and only if f l = e a for some ae[X , G].

Proof" If s, t: X--~G and f : X--+E we define st by x~--*s(x)t(x) and f s by x ~-~f(x)s(x). This clearly gives the required group action. If fl= eta, then p,(fl)= p,(~). Conversely, suppose p,(fl)=p,(cO and let f, g: X ~ E represent c~ and ft. Lift the homotopy pg~-p f to a homotopy g-~ h. Then h represents fl and ph = p f so we can write h(x )= f (x )a (x ) where tr: X ~ G. Since a bundle is locally trivial by definition, it is easy to check the continuity of a. This proves the last statement of the lemma. The exactness at [X, El is an obvious consequence. To define 0, let f : X ~ B. Then f*(E) is a principal bundle over X with group G depending only on the homotopy class c~ of f Let g: X---, BG be a classifying map for this bundle and let 8(e) be the homotopy class ofg. I fB is sufficiently nice, e.g. a C W complex, we can find a classifying map c: B ~ BG for E which will obviously induce 8. Recall that f * ( E ) = {(x, e)~X x E l f ( x ) =p(e)}. I f f = p h where h: X --~ E then f*(E) has a section x~-~(x,h(x)) and so is trivial. Therefore 8p, is trivial. Conversely, suppose 0e is trivial so that f*(E) is trivial. Let s be a section of f *(E) and write s(x)=(x, h(x)) where h: X ~ E. Then f = p h which proves the exactness at IX, B].

If B is sufficiently nice, we can regard this exact sequence as part of the homotopy sequence of the fibration E P , B c , BG.

Corollary 8.6. Let X be a C W complex. Then [X, BSU(n)] -~ IX, BU(n)] is surjective if HE(x, 7Z).~O and injective if Hi(X, 7/)=0.

Proof Since U(n)/SU(n)=S 1, this follows immediately from Lemma8.5 applied to the principal bundle S 1-- - ,BSU(n)~BU(n) once we observe that S 1 =K(Z , 1) and BS 1 = K(;~, 2) so IX, S 1] = H~(X) and IX, BS ~] = Hz(x).

Our final lemma is of a somewhat different nature.

LemmaS.7. Let X be a (n-1)-connected space, n>2, with H,(X,Z)--Z, H,+ 1 (X, 7/)--0. Then there is an exact sequence

7~n+2(3 n) ~ 7~n+2(/) h ~, Hn + 2(X ' Z) -----r ~n+l(S n) ~ 7~n+l (X)-- ) '0

Proof By the Hurewicz theorem, g,(X) = 7~. Let f : S" -* X represent a generator of n,(X). By replacing S" by a space of the same homotopy type, which I will still denote by S", we can replace f by a fibration F ~ S" ~ X. The homotopy sequence shows that F is (n-1)-connected, and an inspection of the spectral sequence shows immediately that H , (F )=0 and z : Hn+2(X) "~ ~H,+I(F) where z is the transgression [48]. By the Hurewicz theorem we have

7t.(F)=O and n,§ ~" ,H.+I(F) .


The required result now follows from the diagram

n.+2(S ) rc.+2(x) ,n.+dF)-,z.+,(S")-,n.+,(X)-,O

H n + 2 ( X ) @ H n + I ( F ) �9

The square commutes by [48, p. 452].

Corollary 8.8. I f n > 3 in Lemma 8.7, the cokernel of h: n, + 2 ( X ) ~ H, + 2(X) is 0 or ~E/2Z. Therefore, if H , + 2 ( X ) = Z , there is a map S "+2 --~ X of degree 2.

S" = 7 / / 2 1 for n > 3. Note that this result in turn follows This follows from nn+ 1 from Lemma 8.7 with X = K(Z, n).

9. An Example. We will now give an example to show that Theorem 5 is false if the hypothesis P/~P ~ (A/c~A) 2 is deleted. Let A be the C algebra with generators Z 1 . . . . , z4 , t l , . . . , t4, p, q, r, s and relations ~ z i t i = 1, p + s = 1, ps=qr, rz3=pz4,

sza=qz4, st3=Pt4, s t3=rG. The relations imply that (~ qs) i s i d e m p o t e n t o f

rank 1 since it has determinant 0 and trace 1. Let p: A2-+ A z be the corresponding endomorphism, i.e., p(x, y) = (p x + q y, r x + s y). Then Q = imp and Q' = ker p are projective of rank 1 and Q G Q ' = A 2. Let v =(z 3, z4)eA 2. Then pv =v so vEQ. Let P = A @ Q and n = ( z 2, v)~P. Let ~ = z 1. Then (~z, n ) ~ A O P is unimodular since in A 4 ~ A @P we have (~2, n) = (z~, z 2, z 3, z4) which is unimodular.

Theorem 9.1. (A • P)/A(cd, n) ~ P.

Remark. The last two relations in the definition of A have not yet been used. Their purpose is to reduce the dimension of A. With these relation A is a regular 7-dimensional domain and SpecA is locally isomorphic to the 7-dimensional affine space A v. To see this, let

B = tr.[p, q, r, s]/(p + s - 1, p s - qr)= lE [p, q, r]/(p(1 - p ) - q r).

We first show that SpecB is a variety locally isomorphic to A 2. Let U be the open set of SpecB defined by q 4:0 and V the open set defined by r 4: 0. Then

U = Spec B[q-1] = Spec C[p, q, q -1]

is locally isomorphic to A 2 and similarly for F.. The complement Y of U u 1/is the closed set defined by q = r = 0. If u, v, w are related to p, q, r, s by (6) below, then B = C[u, v, w]/(u 2 --]- 1)2.4_ w 2 _ 1) and Y is defined by u = v = 0. Since u = v = w = 0 defines the empty set, SpecB is covered by the open sets aU, aV where a runs over the automorphisms of B obtained by permuting u, v, and w. Since these sets

all contain the point u = v = w = 1/]/~, we see also that SpecB is irreducible. We claim now that SpecA looks locally like SpecB x A 4. Since SpecB is covered by the two open sets {p 4: 0} and {s 4: 0}, whose intersection is non-empty, it is sufficient to look at the inverse images of these sets in SpecA which are SpecA[p -1] and


S p e c A [ s - a ] . In A[p -1] we have z 4 = p -1 r23, t 4 = p -1 q t 3. Also

z 3 t 3 + z 4 t4=(1 + p - 2 r q ) z 3 t 3 =(1 + p - 1 s) z a t 3 = p - 1 z3 t3 so

A [ p - 1 ] = B [,p-1] [,Z1 ' Z2 ' Z3 ' tl ' t2 ' t a ] / (Z 1 tl + Z2 t2 + p -1 Z3 t a _ 1).

Locally, one of t 1, t2, t 3 is invertible but

A[ ,p-1] [-q-l] = B[-p-1] [z2, z3 ' tl ' t2 ' ta ' t ( 1 ] ,

etc. The sets { t i+0 } have non-empty intersections so SpecA[,p -1] is connected. A similar a rgument applies to Spec A [s -1]. Since these two meet, it follows that A is a domain locally i somorphic to B[,x I . . . . . x4].

We now t u m to the p roof of Theorem 5.1. If (A (~P) /A(cd , ~ ) ~ P , then

A4/A(z21, z2, z3, z4 )= A4/A(o~ z, ~ ) = (A �9 P (~ Q') /A(~ 2, ~)

, , ~ P O Q ' = A O Q G Q ' = A 3.

Therefore it will suffice to prove the following s t ronger result.

L e m m a 9.2. A4/A(z21, z2, z3, z4) is not f ree .

A similar result was p roved in [-61 ] for the ring IE [,z 1 . . . . . z 4, t 1 . . . . . t4]/( ~ z i t i - 1). As in that proof, we prove L e m m a 9.2 by passing to the topological case [-57]. Consider the 2-sphere S 2 defined by u 2 + v e + w 2 = 1. In C(S 2) let

p=~(1 +w), s = 2x-(1 - w ) , q = l ( u + i v ) , r = � 8 9 (6)

Then p + s = l and p s = q r . Define an a u t o m o r p h i s m p of the trivial bundle

o2 = S / x ~2 by thema t r ix ( ; sq). As above, p is idempoten t o f r ank 1 so o2 = ~ G ~ '

where r ~ ' = k e r p . The s tandard hermit ian metric I lz l tZ=~ [zi[ z on Ir 4 induces a hermit ian metr ic on o 4 = S z x ~;4 and hence, by restriction, gives such a metric on o G o �9 ~ c o 4. Let X be the unit sphere bundle of o @ o �9 r with respect to this metric. Then X is a bundle over S z with fiber S 5 and with a cross-section s defined by taking the element (1, 0, 0) in each fiber. Consider X as a subset of o 4 = S z x q2 4. It is trivial to verify that X is the subset of S z x IU 4 defined by the equat ions ~ z~ 5~ = 1, r z 3 = p .74, and s z 3 ---~ q z4, where z i are the coord ina te functions on ~4. Since ~ = p and ?/= r in C($2), the last two equat ions give, by conjugation, q z3 = P z4 and s z3 = r z4" Therefore there is a IE-algebra h o m o m o r p h i s m A ---, C(X) defined by sending p, q, r, z I . . . . . z 4 to the functions on X denoted by the same symbols, and sending t~ to ~. Under this map, A 4 / A ( z 2, z 2, z 3, z4) is sent to the

module of sections of the bundle ~ which is the cokernel of the m a p 0 ~f' z~. z~. ~,), o4 so it will suffice to show that ~ is non-trivial. The project ion of S 2 • C 4 on IE 4 gives us a m a p f : X ~ S 7. If t / o n S v is the cokernel of the m a p o ~ o 4 by (z 2, z 2, z 3, z4) then clearly ~ =f*(r /) .

N o w let us recall the p roof that q is non-tr ivial given in [,61]. Let e: S 7 ~ S v by e(z) = (d -1 z 2, d -1 z2, d -1 z3, d -1 z4) where d = I-Ix 114 + [z 212 + [Z 3 [2 ..[_ 1Z412]�89 Then e has degree 2 and r /= e*(~) where ~ is the cokernel of o ~ o 4 by (z~, z2, z3, z4). Since ~ z i ~ i = 1, ~ can be identified with the set of (y~ . . . . . Y4) in o 4 such that ~ y g = 0 .


Thus the total space of ~ is the set of (x ,y)eSTx ~4 such that y_kCx. Let x 0 = (0 ,0 ,0 , I ) e S 4 and identify t123 with ~ 3 x 0 ~ C 4. There is an isomorphism U(4) x U(3)~a ___}( by (g, y ) ~ ( g x 0 , gy) which shows that ~ is the vector bundle associated with the canonical principal bundle U(3) --* U(4) ~ S 7. Let c: S 7 ---} BU(3) be the classifying map for this bundle, and hence for (. The exact h o m o t o p y ladder for

u ( 3 ) -

u ( 3 )

gives '~7 S7

' U ( 4 ) , S 7

, E u ( 3 ) , B U ( 3 )

U(3) U(4)=O ~' ~6 ) 7['6

c . =

~7 BU(3) , n 6 U(3)

SO C , : 7~ 7 S 7 ----~ n 7 BU(3) is onto. Since deg e = 2, it follows that

(ce)," n7 $7 e ,=2~nTS 7 c, ,nTBU(3)=Z/67/ [12] (7)

has image 71/37l so ce~O and r/ is non-trivial. To show that ~ is non-trivial we must show that its classifying map ce f is non-trivial. The first step is to show that f has degree 1. Since X and S 7 are closed oriented 7-manifolds, this follows immediately f rom

Lemma9.3 . Let U be the open set of S 7 defined by Iz31Z+lz4[2 ,0. Then f -1 (U) --} U is a bijection.

Proof If p, q, r, s and u, v, w are related as above, the condit ion u 2 + v 2 + w 2 = 1 is equivalent to ps=qr if p + s = 1 is given. The u, v, w will be real if and only if

= p, ~ = s, ?/= r. Given z �9 U, we must find p, q, r, s satisfying these conditions and also r z3 = P z4, s z3 = q z4. If z3 = 0 then z 4 4= 0 and the unique solution is p = q = r = 0, s = 1. If z 4 =0, the unique solution is q = r = s = 0, p = 1. Suppose z 3, z 4 4=0 and let ~=z3z4 ' . Then p=~r, q=~s so p = l - s , q=~s, r = ~ - t ( 1 - s ) . These satisfy p + s = l andps=qr. Thecondi t ionYt=r,~=sgives~s=~-l (1-s)or( l +l~12)s= l. This gives a unique value of s and all condit ions are satisfied.

Corollary 9.4. f , : H 7 (X) --+ H 7 ($7).

Proof Let E be a closed 7 cell lying in f - ' (U) . Then f: E ~ f ( E ) is a bijection and hence is a homeomorph i sm. By invariance of domain, f ( I n t E ) is open in S 7. If x �9 Int E, we have

H7(X)~ H 7 ( X , X - x ) ~ ~ HT(E,E-x)

H7($7) ~ ~ HT(S 7 , S T - f ( x ) ) ~ H 7 ( f ( E ) , f ( E - x ) ) .


Unfortunately the same is not true for n v. In fact, since nT(S 5) and 7~7(S 2) are finite, so is nT(X) by the homotopy sequence of S S ~ X - * S 2. Therefore nv(X ) --~ n7(S 7) =7 / i s zero. We will get around this difficulty by replacing X by a different space.

Since H2($7)=0, Corollary 4.6 shows that c: S 7 ~ B U ( 3 ) is homotopic to a map into BSU(3). Since c is only defined up to homotopy we assume it has this property. Since H 1 (X)= 0, Corollary 8.6 shows that c e f t -0 in B U(3) if and only if c ef t- 0 in BS U(3). Now the fibration S 3 = S U(2) ~ S U(3) --, S 5 shows that S U(3) is 2-connected. Therefore BSU(3) is 3-connected. Embed S 2 in X by the section s of X - + S 2. This $2 = f - 1 ( 1 , 0, 0, 0) so that f factors through Y=X/S 2, say f : X k , y g , S 7 where k is the canonical map. Since BSU(3) is 3-connected, Corollary 8.3 shows that k induces a bijection [Y,, BSU(3)] --~ IX, BSU(3)]. There- fore it will suffice to show that cegq~O in BSU(3) and even in BU(3).

The spectral sequence for $ 5 ~ X ~ S 2 shows that H i ( X ) = 7 / f o r i=0 , 2, 5, 7 and otherwise is zero. Therefore Hi(u ) =71 for i=0 , 5, 7 and is otherwise zero. Also HI(X) ~ ,Hi(Y) for i + 2 and, in particular, we see that g," Hv(Y),~H7(ST). Now Corollary 8.8 applies to Y and gives us a map h: S v ~ Y of degree 2. It will clearly suffice to show that eegh~O. But gh: S 7 ~ S 7 has degree 2 so egh has degree 4. Therefore (eg h), = 4 :~7 $7 ~ n7 $7 while c , : ~v $7 ~ n~ BU(3) = 7//6 7/ is onto. Therefore the image of(cegh), is 7//37/and so cegh ~0.

10. Further Topological Results. In view of Theorem 4, it is natural to ask whether there is a 3-dimensional counterexample to Theorem 5 if P/eP~ (A/o~A) 2 is not assumed. We do not know the answer to this but will show here why a 7-dimensional example was needed to apply topological methods. We first dispose of the real case.

Theorem 10.1. Let A be the ring of real continuous functions on a topological space X. Let M be any A-module and let (a, n ) e A O M be unimodular. I f a>O at every point of X, then (AOM)/A(a, n )~M.

Proof There is an A-homomorphism f: A G M ~ A with f ( a ,n )= l . There- fore we have ba+g(n)=l where beA, g = f l M . Let r be the transvection of A O M defined by q~(x,y)=(x+g(~)g(y),y). Then q~(a,n)=(a',n) where a ' = a + g(7~) 2. Clearly, a ' > 0 at every point of X so a' is a unit of A. By another transvection we send (a', n) to (a', 0) and the required result follows.

It is clear that the argument applies more generally to differentiable or analytic functions, etc., and it would be easy to give an axiomatic version.

This theorem shows why we had to work over �9 in w 9. The next theorem shows why dim A > 7 was needed.

Theorem 10,2. Let X be a CW complex with dim X < 6 . Let A be the ring of continuous complex valued functions on X. Let P be a projective A-module of rank 2 and let (~2, n)~AGP be unimodular. Then (AOP)/A(a2, n)~ P.

We will prove this by considering a universal example. Suppose to begin with that we have a vector bundle 4 over a space B. Let p: E ~ B be the bundle over B obtained by omitting the zero section of 4. Let q = p*(4) = {(e, x) e E < q ]p(e) = p(x)}. This has a canonical section s given by s(e)=(e, e), which is never zero [19, w 1].

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