ON A THEOREM OF MOHAN KUMAR AND NORI

Transcription

1 ON A THEOREM OF MOHAN KUMAR AND NORI RICHARD G. SWAN Abstract. A celebrated theorem of Suslin asserts that a unimodular row x m 1 1,..., xmn n is comletable if m 1 m n is divisible by (n 1!. Examles due to Suslin and to Mohan Kumar and Nori show that this result is the best ossible in all characteristics. We give a new version of the roof of Mohan Kumar and Nori which avoids the need to use Grothendieck s Riemann Roch theorem or other dee results of algebraic geometry. We also adat the roof to give examles of stably free modules which are not self dual in all characteristics. 1. Introduction Let R be a commutative ring and let (a 1,..., a n be a unimodular row over R. We write P (a 1,..., a n for the kernel of R n a1,...,an R. This is dual to the definition used in [18] and the early sections of [16] but the resent usage agrees with that of [8] and of [16, 17]. If necessary we secify R by writing P R (a 1,..., a n. Suslin [12] has shown that if m ν > 0 for all ν then P (a m1 1,..., amn n is free if m 1... m n 0 mod (n 1!. while in [15] it was shown that P (x m1 1,..., xmn n over A n = C[x 1,..., x n, y 1,..., y n ]/( x i y i 1 is not free if m 1... m n 0 mod (n 1!. Examles due to Suslin himself [13] [14] and, indeendently, to Mohan Kumar and Nori [16, 17], have extended this result to all characteristics as follows. Theorem 1.1. Let R be any non zero commutative ring and let A = R[x 1,..., x n, y 1,..., y n ]/( x i y i 1. Let m ν > 0 for all ν = 1,..., n. If P A (x m1 1,..., xmn n is free over A then m 1... m n 0 mod (n 1!. If P is free over A then P R k will be free over k[x 1,..., x n, y 1,..., y n ]/( x i y i 1 where k = R/m is a field. Therefore it is sufficient to consider the case where R is a field. The roofs given by Suslin and by Mohan Kumar and Nori are quite different. Suslin s roof makes use of his theorem that SK 1 (A = Z is generated by a matrix with first row x 1, x 2, x 2 3,..., x n while the roof given by Mohan Kumar and Nori makes use of Chern classes and uses a consequence of Grothendieck s Riemann Roch theorem. 1 The main urose of the resent aer is to give a more elementary roof avoiding the use of this difficult result. Hoefully this will 1 Unfortunately a sign ( 1 i 1 was omitted in both equations in the statement of this result in [16, Theorem 13.2]. 1

2 2 RICHARD G. SWAN make this roof accessible to a wider audience. We will, however, use the relations satisfied by the Adams oerations. This requires the slitting rincile so some basic results on schemes are still needed. I will also show how to adat the roof of the theorem to extend the main result of [18] to all characteristics. This can also be roved using Suslin s methods as Ravi Rao has shown in [11]. Theorem 1.2. Let R and A be as in the revious theorem Let n be odd and let m ν > 0 for all ν. Let P = P A (x m1 1,..., xmn n. If P P = Hom(P, A then 2m 1... m n 0 mod (n 1!. In [18] this was roved for R = C. We refer to [18] for further discussion of this result. As above it will suffice to rove the theorem for the case where R is a field. 2. λ, γ, and ψ Oerations I will assume familiarity with classical algebraic K theory but will begin by recalling some standard results on Grothendieck s λ and γ oerations and Adams ψ oerations. Standard references for this material are [4], [6], and [10] as well as [1] which considers the toological case. Let R be a commutative ring and let P be a finitely generated rojective R module. Let Λ n (P be its n th exterior ower and let λ n (P = [Λ n (P ] in K 0 (R. We set Λ 0 (P = R as usual. Since Λ(P Q = Λ(P Λ(Q as graded rings we have Λ n (P Q = Λ (P Λ q (Q. +q=n and therefore λ n (P Q = +q=n λ (P λ q (Q Let λ t (P be the formal ower series n=0 λn (P t n in K 0 (R[[t]]. Then λ t (P Q = λ t (P λ t (Q so λ t is an additive function with values in the abelian grou 1 + tk 0 (R[[t]] and hence factors through K 0 (R giving a homomorhism λ t : K 0 (R 1 + tk 0 (R[[t]]. We write λ t (x = n=0 λn (xt n. It follows that λ n (x + y = +q=n λ (xλ q (y for all x and y in K 0 (R. In addition to the λ n, Grothendieck also defines new oerations on K 0 by letting γ t (x = λ t 1 t (x and writing γ t(x = n=0 γn (xt n. In articular, γ 0 (x = 1, γ 1 (x = λ 1 (x = x, and γ n (x + y = +q=n γ (xγ q (y for all x and y in K 0 (R. It is not hard to exress the relation between the λ n and the γ n in the form of finite sums. Lemma 2.1. For n 1, (1 γ n (x = (2 λ n (x = q=0 q m=1 λ +1 (x ( 1 q γ q+1 (x. Proof. We have γ t (x = λ t (x = 1 + λ m t m (x 1 t (1 t m = 1 + ( m λ m (xt m k Now, for m 1, ( m k m=1 k=0 ( ( m + k 1 m + k 1 = ( 1 k = ( 1 k k m 1 ( 1 k t k.

3 ON A THEOREM OF MOHAN KUMAR AND NORI 3 so γ t (x = 1 + and therefore, for n 1, n γ n (x = m=1 m=1 k=0 ( m + k 1 m 1 ( n 1 λ m (x = m 1 λ m (xt m+k ( n 1 λ +1 (x. A similar argument alies to λ t (x = γ t 1+t (x Corollary 2.2. Let a 1,..., a N and b 1,..., b N be finite sequences of integers. Either of the equivalent conditions (1 N 1 a kt(1 + t k 1 = N 1 b kt k (2 N 1 a kt k = N 1 b kt(t 1 k 1 imlies N 1 a kγ k = N 1 b kλ k. Proof. The first condition imlies the second by substituting t 1 for t while the second imlies the first by substituting t + 1 for t. If the first condition holds then N N k 1 ( k 1 a k t(1 + t k 1 = a k t +1 so b m = N k=m a ( k 1 k m 1 and therefore N N k a k γ k = a k m=1 k=1 ( k 1 λ m = m 1 N b m λ m. Remark 2.3. This result can be reinterreted as follows: The coefficient of λ k in Lemma 2.1(1 is the coefficient of t k in t(1 + t and the coefficient of γ k in (2 is the coefficient of t k in t(t 1. We can write this symbolically as γ n = λ(1+λ and λ n = γ(γ 1 for n 1 with the convention that a k th ower of the symbol λ on the right hand side should be relaced by the oerator λ k and similarly for γ. Therefore a linear combination of the λ k can be converted to a linear combination of the γ k by relacing each λ n by t(1 + t, collecting owers of t, and relacing each t n by γ n. For simlicity, I will assume in the remainder of this section that the ring R is connected so that each finitely generated rojective module P has a well defined rank rk P. This induces a ma ɛ : K 0 (R Z with kernel K 0 (R. If rk P = n then rk Λ k (P = ( n k. In articular Λ k (P = 0 for k > n and Λ n (P is invertible, i.e. a rank 1 rojective, so λ n (P is a unit in K 0 (R. We define ɛλ t (x = n=0 ɛλn (xt n. Lemma 2.4. ɛλ t (x = (1 + t ɛ(x. Proof. This is clear if x = [P ]. The general case follows since both sides are homomorhisms from K 0 (R to 1 + tz[[t]] Corollary 2.5. ɛγ t (x = (1 t ɛ(x. m=1 This follows by substituting t/(1 t for t in Lemma 2.4.

4 4 RICHARD G. SWAN Corollary 2.6. If x lies in K 0 (R then λ n (x and γ n (x also lie in K 0 (R for n > 0. Lemma 2.7. If x = [P ] where rk P = 1, then λ t (x = 1 + tx, and γ t (x = (1 + t(x 1/(1 t. This is clear from the definitions. Corollary 2.8. If x = [P ] where rk P = 1, then (1 λ 0 (x = 1, λ 1 (x = x, and λ n (x = 0 if n > 1 (2 γ 0 (x = 1, and γ n (x = x if n 1. Lemma 2.9. Let x = [P ] [R n ] K 0 (R. Then γ i (x = 0 for i > n and γ n (x = ( 1 n λ 1 (P = n i=0 ( 1n i λ i (P. Proof. γ t (x = λ t (P /λ t (Rn = 1 t 1 t Corollary K0 (R is a nil ideal of K 0 (R. n λ i t i (P (1 t i / 1 n (1 t n = λ i (P t i (1 t n i 0 Proof. If x K 0 (R then γ t (x and γ t ( x are olynomials in t whose roduct is γ t (0 = 1 but in a reduced ring a olynomial which divides 1 must be a constant. Since γ 1 (x = λ 1 (x = x, x must be nilotent. More details may be found in [1, Cor ]. The Adams oerations are defined by ψ 0 (x = ɛ(x and ψ t (x = ψ n (xt n = ψ 0 (x tλ t (x 1 d dt λ t(x. n=0 The right hand side is usually written as ψ 0 (x t d dt log λ t(x with the observation that although log introduces denominators, the oeration d dt log does not. These oerations are additive since λ t (x + y = λ t (xλ t (y and therefore differentiating and dividing by λ t (x + y gives ψ t (x + y = ψ t (x + ψ t (y. Lemma ɛψ t (x = ɛ(x(1 t 1 and therefore ɛψ n (x = ɛ(x for all n. Proof. ɛψ t (x = ɛ(ψ n (xt n = ψ 0 (x tɛ(λ t (x 1 d dt ɛ(λ t(x. 0 The result now follows immediately from Lemma 2.4 and the fact that ψ 0 = ɛ. Alying the definition to the rank one case gives us the following result. Lemma If x = [P ] where rk P = 1, then ψ t (x = (1 tx 1 so ψ 0 (x = 1, and ψ n (x = x n if n 1. The Adams oerations behave articularly well with resect to roducts and comosition. The analogous formulas for the λ and γ oerations are given by universal olynomials with coefficients in Z but these seem to be very comlicated and aarently have never been written down exlicitly. The following theorem of Adams is the only dee result which we will need. In fact, only (2 is required. 0

5 ON A THEOREM OF MOHAN KUMAR AND NORI 5 Theorem 2.13 (Adams. (1 ψ n (xy = ψ n (xψ n (y. (2 ψ (ψ q (x = ψ q (x. Proof. According to the slitting rincile [4], [5], [6] [10] we can embed K 0 (R in K 0 (X for a scheme X reserving all oerations and such that any finite number of secified elements become sums of rank one elements and negatives of such elements. Therefore it is enough to check the relations for elements x and y of this form. This is immediate by Lemma A very secial case In this section we rove some facts alying to the rather secial case where K 0 (R 2 = 0 and where K 0 (R is torsion free. As above we assume R is connected so ɛ : K 0 (R Z is defined. Lemma 3.1. If K0 (R 2 = 0 then λ n (x + y = λ n (x + λ n (y and γ n (x + y = γ n (x + γ n (y for x, y K 0 (R and n > 0. Proof. Write λ t (x = 1 + f t (x and similarly for y. By Lemma 2.6, f t (x and f t (y have all coefficients in K 0 (R so f t (xf t (y = 0. Therefore 1 + f t (x + y = (1 + f t (x(1 + f t (y = 1 + f t (x + f t (y. A similar argument alies to γ. Lemma 3.2. If K 0 (R 2 = 0 then ψ n (x = ( 1 n+1 nλ n (x for x K 0 (R. Proof. Let f t (x = λ t (x 1 = 1 λn (xt n as above. Since f t (x has coefficients in K 0 (R we have λ t (x 1 = (1 + f t (x 1 = 1 f t (x. Therefore ψ t (x = 0 t(1 f t (x d dt (1 + f t(x = t d dt f t(x = t d ( 1 n λ n (xt n dt 1 = ( 1 n λ n (xnt n Corollary 3.3. Suose K 0 (R 2 = 0 and K 0 (R is torsion free. If x K 0 (R then λ (λ q (x = ( 1 (+1(q+1 λ q (x for, q > 0. Proof. Since ψ (ψ q (x = ψ q (x by Theorem 2.13 we have ( 1 +1 λ (( 1 q+1 qλ q (x = ( 1 q+1 qλ q (x. 1 Since, q > 0 and K 0 (R is torsion free we can divide this by q. Proosition 3.4. Suose K 0 (R 2 = 0 and K 0 (R is torsion free. If x K 0 (R then γ n (γ n (x = ( 1 (n 1!γ n (x + a k γ k (x for some integers a k. k=n+1

6 6 RICHARD G. SWAN Proof. By Lemma 2.1, Corollary 2.6, Lemma 3.1, and Corollary 3.3 we have γ n (γ n ( (x = λ +1 (γ n (x = = q=0 q λ +1 (λ q+1 (x q=0 To exress this in terms of the γ k we use Corollary 2.2. Let S = q=0 q ( 1 q t(t 1 (+1(q+1 1 = = q=0 q=0 q ( 1 +q t(1 t q++q q ( 1 +q t(1 t (+1q (1 t Now 1 (1 t +1 = ( + 1t + O(t 2 so q ( 1 q λ (+1(q+1 ( = ( 1 t(1 t {1 (1 t +1 } ( S = ( 1 ( + 1 t n + O(t n+1. The required result now follows from Corollary 2.2 and the following lemma. Lemma 3.5. ( 1 ( + 1 = ( 1 (n 1!. Proof. Let ( f(t = ( 1 e (+1t = e t (1 e t = ( 1 t + O(t n. The lemma follows by differentiating n 1 times and setting t = 0. Theorem 3.6. Let R be a connected commutative ring with K 0 (R = Z Z. Let P be a finitely generated rojective R module with rk P = n. Then λ 1 (P 0 mod (n 1!. Proof. Let K 0 (R = Zξ. Then ξ 2 = rξ for some r Z and therefore ξ m = r m 1 ξ for m > 0. By Corollary 2.10, ξ is nilotent so r = 0 and ξ 2 = 0. Therefore K 0 (R 2 = 0 and K 0 (R is clearly torsion free. Let x = [P ] [R n ]. By Lemma 2.9, γ n (x = ( 1 n λ 1 (P so we have to show that γ n (x 0 mod (n 1!. By Corollary 2.6, γ n (x lies in K 0 (R and therefore γ n (x = kξ for some k Z. If k = 0 we are done. Assume k 0. By Lemma 2.9, γ i (x = 0 for i > n so by Proosition 3.4, γ n (γ n (x = ( 1 (n 1!γ n (x. By Lemma 3.1, γ n (γ n (x = γ n (kξ = kγ n (ξ. Therefore kγ n (ξ = ( 1 (n 1!γ n (x = ( 1 (n 1!kξ. Since we are assuming k 0 it follows that γ n (ξ = ( 1 (n 1!ξ. If x = mξ, then γ n (x = mγ n (ξ = ( 1 (n 1!mξ 0 mod (n 1!

7 ON A THEOREM OF MOHAN KUMAR AND NORI 7 Next we recall some standard facts about Cohen Macaulay rings. If R is a noetherian local ring and x m R then dim R 1 dim R/(x dim R and dim R/(x = dim R 1 if x is regular. It follows by induction on n that if x 1,..., x n m R then dim R n dim R/(x 1,..., x n dim R and also that dim R/(x 1,..., x n = dim R n if x 1,..., x n is a regular sequence. Lemma 3.7. If R is a Cohen Macaulay local ring and x 1,..., x n m R then dim R/(x 1,..., x n = dim R n if and only if x 1,..., x n is a regular sequence. If so, R/(x 1,..., x n is again Cohen Macaulay. This is roved for n = 1 in [17, Lemma 8.6] and the general case follows by induction. A more general version is given in [3, Th (c]. Corollary 3.8. Let R be a Cohen Macaulay local ring and let I = (x 1,..., x n be an ideal with ht I n. Then x 1,..., x n is a regular sequence. Proof. dim R/(x 1,..., x n = dim R/I dim R ht I dim R n. Since dim R/(x 1,..., x n dim R n, we have dim R/(x 1,..., x n = dim R n and the lemma shows that x 1,..., x n is a regular sequence. Let R be a commutative ring, let P be an R module and let f : P R. The Koszul comlex Kosz(f of f is defined to be the exterior algebra Λ(P with its usual grading and with d : Λ k+1 (P Λ k (P by d( 0 k = k m=0 ( 1m f( m 0 m k. This has the augmentation ɛ : Λ 0 (P = R R/ im f. Lemma 3.9. Let R be a Cohen Macaulay ring. Let I be an ideal of R with ht I n and let P be a finitely generated rojective R module of rank n. If there is an eimorhism f : P I then Kosz(f is a rojective resolution of R/I. Proof. It is sufficient to check this locally. After locallizing at a rime ideal, P will become free with base e 1,..., e r maing to a set of generators a 1,..., a r of I and the Koszul comlex localizes to the usual Koszul comlex K(a 1,..., a r. If I localizes to R, a 1,..., a r will be a unimodular row so K(a 1,..., a r will be exact and therefore a resolution of R/I = 0 (locally. Otherwise the localization of I will be a roer ideal of height at least n so r n but since rk P n, r n and therefore r = n. By Corollary 3.8, a 1,..., a n will be a regular sequence and therefore the Koszul comlex will be a resolution of R/I. The following is the main result of this section. Corollary Let R be a Cohen Macaulay ring with K 0 (R = Z Z. Let I be an ideal of R with ht I n and let P be a finitely generated rojective R module of rank n. If there is an eimorhism f : P I then R/I has finite rojective dimension so [R/I] is defined in K 0 (R and [R/I] 0 mod (n 1! in K 0 (R. Proof. The revious lemma shows that R/I has finite rojective dimension and that λ 1 (P = [R/I] in K 0 (R. The final statement follows from Theorem A Patching Lemma The remainder of the roof is essentially the same as the original roof of Mohan Kumar and Nori. I will give here a slightly more general version of one of their lemmas which will also be useful in roving Theorem 1.2. We use the notation R s for the localization R[s 1 ].

8 8 RICHARD G. SWAN Lemma 4.1. Let R be a commutative ring and let M be a finitely generated R module. Let R = Ra + Rb and let P and Q be finitely generated rojective of rank n over R a and R b resectively. Suose we have resolutions and over R a and R b. If and 0 L P M a 0 0 N Q M b 0 0 L b P b M ab 0 0 N a Q a M ab 0 slit (e.g. if M ab is rojective over R ab and if L b N a over R ab, then there is a finitely generated rojective R module S of rank n with an eimorhism S M. Proof. Since P b L b M ab and Q a N a M ab we can use the isomorhism L b N a to get P b Q a and a commutative diagram From this we get 0 L b P b M ab 0 0 N a Q a M ab 0 Q Q a P b P M b M ab M ab M a The ullback S of the uer line mas to the ullback M of the bottom line. By localizing we see that S a P and S b Q showing that S is rojective of rank n. The ma S M is onto since it localizes to S a = P M a and S b = Q M b. 5. A Useful Ring The roof of Mohan Kumar and Nori makes use of the following auxiliary ring. Let B = B n = k[x 1,..., x n, y 1,..., y n, z]/( x i y i z(1 z where k is a field. We have B 0 = k k and B n is a domain for n > 0 since the olynomial f = xi y i z(1 z is irreducible. Also x i and y i are non zero in B n since f does not divide x i or y i. As above, we use the notation R s for the localization R[s 1 ]. Lemma 5.1. Let B = B n. (1 B z = k[x 1,..., x n, u 1,..., u n ] 1 P xiu i where u i = y i /z. (2 B 1 z = k[x 1,..., x n, v 1,..., v n ] 1 P xiv i where v i = y i /(1 z. Proof. Letting u i = y i /z we can write B z = k[x 1,..., x n, u 1,..., u n, z, z 1 ]/( x i u i (1 z. Therefore z = 1 x i u i can be eliminated rovided we ensure its invertibility. A similar argument alies to (2 where z = x i v i. Corollary 5.2. B n is a regular domain for n 1.

9 ON A THEOREM OF MOHAN KUMAR AND NORI 9 Proof. It is sufficient to rove the regularity locally so it is enough to show that B z and B 1 z are regular. This is clear from the lemma. Remark 5.3. The fact that B n is a domain for n 1 can also be deduced from the lemma using the following elementary result. Lemma 5.4. Let R be a commutative ring and let R = Rs i. If all R si are domains and if s 1 mas to a non zero element of R si for all i, then R is a domain. Proof. We claim that R R s1 is injective. It is enough to rove this locally by considering the mas R si R s1s i. These are injective since we are localizing a domain R si at a non zero element s 1. The following calculation is due to J. P. Jouanolou [7] who also considered the case of higher K theory. The roof here uses only classical K theory. Proosition 5.5. K0 (B n = Z generated by B n /I n where I n = (x 1,..., x n, z. For the roof we will use the following localization sequence, a secial case of [2, Ch. IX, Th. 6.5]. Lemma 5.6. Let R be a commutative regular ring and let s be a non zero divisor in R. Then the sequence K 1 (R K 1 (R s G 0 (R/(s K 0 (R K 0 (R s 0 is exact and the ma is given by sending α M n (R s to [ckr(s m α] [R n /s m R n ] for any m such that s m α lies in M n (R. Proof. The standard localization sequence [2, Ch. IX, Th. 6.3] has the form K 1 (R K 1 (R s K 0 (H K 0 (R K 0 (R s where the ma is as in the lemma and where H is the category of finitely generated R modules M such that M s = 0 and d R (M <. Since R is regular the second condition is always satisfied so K 0 (H is K 0 of the category of finitely generated R modules M with M s = 0. By devissage [2, Ch. VIII, Th. 3.3] this is equal to G 0 (R/(s. We can add a zero on the right since K 0 (R = G 0 (R and similarly for R s. If R/(s is also regular we can relace G 0 (R/(s by K 0 (R/(s. Proof of Proosition 5.5. The result clearly holds for B 0 = k k. We use induction on n. We aly Lemma 5.6 with R = B n and s = x n getting K 1 (B n K 1 ((B n xn K 0 (B n /(x n K 0 (B n K 0 ((B n xn 0 Note that B n /(x n = B [y n ] which is also regular. Now (B n xn = k[x 1,..., x n, x 1 n, y 1,..., y, z] so by standard K-theoretic calculations [2, Ch. XII] we have K 0 ((B n xn = Z and K 1 ((B n xn = k Z where the Z is generated by x n U((B n xn. It follows that ker[k 0 (B n K 0 ((B n xn ] is just K 0 (B n and, since k is in the image of K 1 (B n, the image of K 1 ((B n xn K 0 (B n /(x n is generated by the image of x n which is [B n /(x n ] and therefore the cokernel of K 1 ((B n xn K 0 (B n /(x n is K 0 (B n /(x n. It follows that K 0 (B n /(x n K 0 (B n. Now since B n /(x n =

10 10 RICHARD G. SWAN B [y n ] and B is regular, we have K 0 (B K 0 (B n /(x n. This takes the generator [B /I ] of K 0 (B to [B /I B B n /(x n ] = [B n /(x n, I ] = [B n /(x 1,..., x, z, x n ] This is just [B n /I n ] so this element generates K 0 (B n. Next we investigate some useful ideals of B n. The following simle observation will be useful. Lemma 5.7. If I is an ideal of a commutative ring R and if b R is a unit modulo I then ab I imlies a I. The roof is immediate: If bc 1 mod I then a abc 0 mod I. Lemma 5.8. If N m i then z N (1 z N lies in the ideals (x m1 1,..., xmn n and (y m1 1,..., ymn n. Proof. If N m i, then in z N (1 z N = ( x i y i N each term in the exansion of ( x i y i N will contain a ower at least m i of x i for some i. Therefore z N (1 z N lies in (x m1 1, xm2 2,..., xmn n and similarly it lies in (y m1 1,..., ymn n Lemma 5.9. The ideal (x m1 1, xm2 2,..., xmn n, z N of B n is indeendent of N for N m i. Also (x 1, x 2,..., x n, z N = (x 1, x 2,..., x n, z for all N 1. Proof. Suose M m i. Then, by Lemma 5.8, z M (1 z M lies in the ideal (x m1 1, xm2 2,..., xmn n, z N and therefore, by Lemma 5.7 z M lies in this ideal. The same alies with M and N interchanged. For the last statement, the same argument alies with M, N 1 since z(1 z = x i y i so z(1 z lies in (x 1, x 2,..., x n, z N. Define J m1,m 2,...,m n to be the ideal considered in Lemma 5.9. In articular, J 1,1,...,1 = I, the ideal considered in Proosition 5.5. Lemma J m1,m 2,...,m n /J m1+1,m 2,...,m n B n /J 1,m2,...,m n and similarly for each m i. Corollary [B n /J m1,m 2,...,m n ] = m 1 m n [B n /I] in K 0 (B n. This follows by induction on the m i since J 1,...,1 = I. For the roof of the lemma we use the following. Lemma Let I be an ideal of a commutative ring R such that R/I A[X]/(X m+1 f where A[X] is a olynomial ring in one variable and f = f(x A[X]. Let x R ma to X modulo I. Then (x m, I/(x m+1, I R/(x, I. Proof. (x m, I/(x m+1, I is generated by x m and (x, Ix m (x m+1, I. We must show that rx m (x m+1, I imlies r (x, I. Now rx m = ax m+1 + i imlies tx m I where t = r ax and we need to show that t lies in (x, I. Let t = t mod I. Then tx m = 0 in A[X]/(X m+1 f but in A[X]/(X m+1 f the annihilator of X m lies in (X since hx m (X m+1 f imlies h (Xf. Therefore t lies in (X in R/I and hence t lies in (x, I as required.

11 ON A THEOREM OF MOHAN KUMAR AND NORI 11 Proof of Lemma Since J m1,m 2,...,m n = (x m1 1, xm2 2,..., xmn n, z N is indeendent of N for large N we can choose N > m i. We aly the lemma with R = B n, x = x 1, m = m 1, and I = (x m2 2,..., xmn n, z N. In B n /(z N, 1 z is a unit so by Lemma 5.1 R/I = R 1 z /I 1 z = k[x 1,..., x n, v 1,..., v n ]/(x m2 2,..., xmn n, ( x i v i N where we have substituted x i v i for z. Let A = k[x 2,..., x n, v 1,..., v n ]/(x m2 2,..., xmn n. Then R/I = A[x 1 ]/(( x i v i N. Since N > m i, if we multily out ( x i v i N, each term will be divisible by some x mi+1 i. Excet for i = 1 all such terms will be 0 in A[x 1 ]. Therefore, in A[x 1 ], ( x i v i N will have the form x m1+1 1 f(x 1. Therefore Lemma 5.12 alies and finishes the roof. 6. Proof of Theorem 1.1 Let B = B n and let J = J m1,m 2,...,m n. Since J = (x m1 1, xm2 2,..., xmn n, z N we have J z = B z. Since z N (1 z N lies in (x m1 1, xm2 2,..., xmn n by Lemma 5.8, Lemma 5.7 shows that J 1 z = (x m1 1, xm2 2,..., xmn n B 1 z. Ma B1 z n with base e i to J 1 z by sending e i to x mi i and ma B z to J z = B z by sending e 1 to 1 and e i to 0 for i > 1. We get short exact sequences and 0 L B n 1 z J 1 z 0 0 N B n z J z 0. These slit when localized to B z(1 z since J z(1 z = B z(1 z. Since J z(1 z = (x m1 1, xm2 2,..., xmn n B z(1 z = B z(1 z we see that L z = P (x m1 1, xm2 2,..., xmn n over B z(1 z. This is induced from the module P = P (x m1 1, xm2 2,..., xmn n over the ring A = R[x 0,..., x n, y 0,..., y n ]/( x i y i 1 via the ma A B z(1 z sending x i to x i and y i to y i /z(1 z. Since N 1 z is free it follows that if P is free then L z N 1 z. Therefore Lemma 4.1 alies and shows that there is an eimorhism Q J where Q is finitely generated rojective of rank n. By Corollary 3.10 we see that [B/J] 0 mod (n 1! in K 0 (B. Since [B/J] = m 1 m n [B n /I] by Corollary 5.11 and K 0 (B = Z generated by [B/I], it follows that m 1 m n 0 mod (n 1! if P is free. 7. A Duality Theorem The next two sections contain reliminary results needed in the roof of Theorem 1.2. We first recall a standard result. Lemma 7.1. Let R be a commutative ring and let n 1 a ib i = 1 in R. Then P (a 1,..., a n P (b 1,..., b n. Proof. We abbreviate a 1,..., a n to a and let a b = a i b i. We have R n = P (a Rb = P (b Ra because z R n has the form z = w + rb with a w = 0 if and only if r = z a. The bilinear form (x, y x y induces a airing P (a P (b R and therefore gives a ma P (b P (a. This is injective since if y mas to 0

12 12 RICHARD G. SWAN then y z = 0 for all z in P (a. Since y b = 0 it follows that y R n = 0 so y = 0. If f : P (a R, extend f to R n by f(b = 0. Then f(z = c z for some c in R n. Since f(b = c b = 0, c lies in P (b and mas to f showing that our ma is onto. Proosition 7.2. Let R be a commutative ring and let n 1 x iy i = 1 in R. Let m 1,..., m n be ositive integers. Then P (x m1 1,..., xmn n P (y m1 1,..., ymn n. Proof. By a theorem of Suslin [12], if m = m 1... m n then P (x m1 1,..., xmn n P (x m 1, x 2,..., x n so it will suffice to treat the case where m 1 = m and m i = 1 for i > 1. A new roof of this result of Suslin was given by Mohan Kumar. An account of this roof is given by Mandal in [9, Lemma 5.3.1]. The following roof is based on this idea. We can assume that n > 2 since otherwise P (x m 1, x 2 is free, being stably free of rank 1. Let Then so by Lemma 7.1. Let x m 1 y m 1 z = 1 + x 1 y (x 1 y 1 m 1. + z n x i y i = x m 1 y1 m + z(1 x 1 y 1 = 1 2 P (x m 1, x 2,..., x n P (y m 1, zy 2,..., zy n z t = 1 + tx 1 y t m 1 (x 1 y 1 m 1. Then P t = P (y m 1, z t y 2,..., z t y n over R[t] is extended. To see this we can assume that R is local by Quillen s atching theorem. Therefore one of the y i is a unit. If y 1 is a unit then P t is free. Suose y 2 is a unit. Then by an elementary transformation we can relace z t y 3 by z t y 3 y 1 2 y 3(z t y 2 = 0 and therefore P t is free. It follows that P 1 = P (y1 m, zy 2,..., zy n P (x m 1, x 2,..., x n is isomorhic to P 0 = P (y1 m, y 2,..., y n as required. 8. Automorhisms We determine some automorhisms of K 0 (B n using the following simle fact. Lemma 8.1. Let f be an automorhism of a commutative noetherian ring R. Then f : G 0 (R G 0 (R sends [R/I] to [R/f(I]. Proof. More generally let f : R R where R is finitely generated and flat as an R module. Then f : G 0 (R G 0 (R sends [R/I] to [R R R/I] = [R /R f(i]. Let n 1 and let B = B n = k[x 1,..., x n, y 1,..., y n, z]/( x i y i z(1 z as above. Let α i be the automorhism of B which interchanges x i and y i and fixes the remaining x j and y j as well as z. Let β be the automorhism of B which sends z to 1 z and fixes the x i and y i. Lemma 8.2. The α i and β induce the automorhism x x on K 0 (B n.

13 ON A THEOREM OF MOHAN KUMAR AND NORI 13 Proof. Let C = B/(x 2,..., x n = (k[x 1, y 1, z]/(x 1 y 1 z(1 z[y 2,..., y n ]. Then C = B 1 [y 2,..., y n ] so C is a domain. Now C/(x 1 = k[z]/(z(1 z[y 1,..., y n ]. Since k[z]/(z(1 z = k k we have C/(x 1 = B/(x 1,..., x n, z B/(x 1,..., x n, 1 z. The exact sequence 0 C x1 C C/(x 1 0 shows that [C/(x 1 ] = 0 in K 0 (B and therefore [B/(x 1,..., x n, z]+[b/(x 1,..., x n, 1 z] = 0. Since [B/I] = [B/(x 1,..., x n, z] generates K 0 (B, this shows that β acts as 1. It will suffice to treat the case of α 1. I will write x for x 1 and y for y 1 here. Let a = (x, z and b = (y, z in C. Then a b = (z. It is sufficient to check this in C/(z = (k[x, y]/(xy[y 2,..., y n ] where it is clear that (x (y = (xy = 0. We have a cartesian diagram which leads to an exact sequence C/(z C/a C/b C/(a + b 0 C/(z C/a C/b C/(a + b 0. Now [C/(z] = 0 and C/(a + b = B/(I, y also has [B/(I, y] = 0 since B/I = k[y 1,..., y n ] and y = y 1 is regular on it so 0 B/I B/I B/(I, y 0 is exact. It follows that [C/a]+[C/b] = 0 and C/a = B/I while C/b = B/α 1 (I. Corollary 8.3. Let θ = α 1... α n β. Then θ induces ( 1 on K 0 (B n. In this roof we use the ideal 9. Proof of Theorem 1.2 J = ((1 z N x m1 1,..., (1 zn x mn n, z N y m1 1,..., zn yn mn of B n where N is some integer such that N m i. Lemma 9.1. If M m i, then z M (1 z M J. Proof. By Lemma 5.8, z M (1 z M lies in (x m1 1,..., xmn n and also in (y m1 1,..., ymn n. Therefore z M+N (1 z M and z M (1 z M+N lie in J. Since (z M+N (1 z M, z M (1 z M+N = z M (1 z M (z N, (1 z N = (z M (1 z M, the result follows. Lemma 9.2. J z = (y m1 1,..., ymn n B z and J 1 z = (x m1 1,..., xmn n B 1 z. Proof. Since z N (1 z N lies in J, (1 z N lies in J z and the first statement follows immediately. The second statement is roved similarly.

14 14 RICHARD G. SWAN and We therefore get short exact sequences 0 L B n 1 z J 1 z 0 0 N B n z J z 0. Since J z(1 z = B z(1 z the sequences slit after localizing to B z(1 z and we see that L z = P (x m1 1,..., xmn n over B z(1 z and N 1 z = P (y m1 1,..., ymn n over B z(1 z. So L z and N 1 z are induced from P (x m1 1,..., xmn n and P (y m1 1,..., ymn n over A = k[x 1,..., x n, y 1,..., y n ]/( x i y i 1 using the ma A B z(1 z sending x i, y i to x i /z, y i /(1 z. If P (x m1 1,..., xmn n is self dual then P (x m1 1,..., xmn n P (y m1 1,..., ymn n by Proosition 7.2 and therefore L z N 1 z. Therefore Lemma 4.1 gives us an eimorhism Q J where Q is finitely generated rojective of rank n. By Corollary 3.10 we see that [B/J] 0 mod (n 1! in K 0 (B. By Lemma 9.1, B/J = B/(J, z N B/(J, (1 z N Now (J, z N = ((1 z N x m1 1,..., (1 zn x mn n, z N. Since (z N, (1 z N = B, it follows that (J, z N = (x m1 1,..., xmn n, z N = J m1,...,m n. Similarly (J, (1 z N = (y m1 1,..., ymn n, (1 z N = θj m1,...,m n where θ is as in Corollary 8.3. By Corollary 5.11 [B/J m1,...,m n ] = m 1 m n [B/I] in K 0 (B. By Corollary 8.3, [B/θJ m1,...,m n ] is ( 1 times this so and therefore [B/J] = m 1 m n (1 + ( 1 [B/I] m 1 m n (1 + ( 1 [B/I] 0 mod (n 1!. This is vacuous for n even while for n odd it is equivalent to 2m 1 m n 0 mod (n 1!. References 1. M. F. Atiyah, K-Theory, W. A. Benjamin, New York H. Bass, Algebraic K-Theory, W. A. Benjamin, New York W. Bruns and J. Herzog, Cohen Macaulay Rings, Cambridge Studies in Adv. Math. 39, Cambridge P. Berthelot, A. Grothendieck, L. Illusie, et al., Théorie des Intersections et Théorème de Riemann Roch, SGA6, 1966/1967, Lecture notes in Math. 225, Sringer Verlag, Berlin J. Burroughs, The slitting rincile for Grothendieck rings of schemes, Toology 9(1970, W. Fulton and S. Lang, Riemann Roch Algebra, Sringer Verlag, New York J. P. Jouanolou, Quelque calculs en K théorie des schémas, in Algebraic K Theory I, Lect. Notes in Math. 341, Sringer Verlag, Berlin T. Y. Lam, Serre s Problem on Projective Modules, Monograhs in Mathematics, Sringer Verlag, Berlin Heidelberg New York, S. Mandal, Projective modules and comlete intersections, Lecture notes in Math. 1672, Sringer Verlag, Berlin Ju. I. Manin, Lectures on the K functor in algebraic geometry, Russ. Math. Survey 24 No. 5(1969, 1 89.

15 ON A THEOREM OF MOHAN KUMAR AND NORI Ravi A. Rao, A study of Suslin matrices; their roerties and uses, Lecture at International Centre for Theoretical Physics, Trieste, Italy June 7, A. Suslin, Stably free modules, Mat. Sb. 102(1977, A. Suslin, Mennicke symbols and their alication to the K theory of fields, in Algebraic K Theory, Part I, Lecture Notes in Math. 966, Sringer Verlag, Berlin Heidelberg New York 1982, 14. A. Suslin, K theory and K cohomology of certain grou varieties, in Algebraic K Theory, Adv. Soviet Math. 4, Amer Math. Soc R. G. Swan and J. Towber, A class of rojective modules which are nearly free, J. Algebra 36(1975, R. G. Swan, Vector bundles, rojective modules, and the K-theory of sheres, Proc. of the John Moore Conference, Algebraic Toology and Algebraic K-Theory, ed. W. Browder, Ann. of Math. Study 113 (1987, R. G. Swan, Neron-Poescu desingularization (exository aer, in Algebra and Geometry, Proc. Internat. Conf. of Algebra and Geometry, ed. M-c. Kang, International Press R. G. Swan, Some stably free modules which are not self dual, htt:// swan/stablyfree.df Deartment of Mathematics, The University of Chicago, Chicago, IL address: swan@math.uchicago.edu