A NOTE ON RELATIVE UNIMODULAR ROWS

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1 A NOTE ON RELATIVE UNIMODULAR ROWS KUNTAL CHAKRABORTY AND RAVI A. RAO 1. Witt Group and Relative Witt Group R will denote a cmmutative ring with 1 0, unless stated otherwise. I will denote an ideal of R. ( We denote by GL n (R, I) by the kernel of the canonical map GL n (R) GL R ) n I. Let SLn (R, I) denotes the subgroup of GL n (R, I) of elements of determinant 1. Definition 1.1. The Relative Elementary group E n (R, I) : Let R be a ring and I R be an ideal. The relative elementary group is the subgroup of SL n (R, I) generated by the matrices of the form αe i,j (a)α 1, where α E n (R), i j and a I. We identify GL n (R) with a subgroup of GL n+1 (R) by associating the matrix the element α GL n (R). We set ( α ) GL n+1 (R), with GL(R) = lim GL n (R), GL(R, I) = lim GL n (R, I) SL(R) = lim SL n (R), SL(R, I) = lim SL n (R, I) E(R) = lim E n (R), E(R, I) = lim E n (R, I) Let Alt 2n(R) be the set of all skew-symmetric matrices in GL 2n (R). For any r N, define χ r inductively by ( ) 0 1 χ 1 = 1 0 and χ r+1 = χ r χ 1. Clearly χ r Alt 2r(R). For any m n, Alt 2m(R) can be embedded into Alt 2n(R) by the inclusion map i m,n : Alt 2m(R) Alt 2n(R) given by α α χ n m. Define Alt (R) = lim Alt 2n(R). For α Alt 2m(R) and β Alt 2n(R) define α E β if and only if there exists some t N and ε E 2(m+n+t) (R) such that α χ n+t = ε T (β χ m+t )ε. It can be shown that this relation is reflexive, symmetric and transitive i.e. E is an equivalence relation on Alt (R). It can be shown that Alt (R)/ E is an abelian group with respect to the operation [7, Section 3]. Denote this group by W E (R). Similarly one can consider the subset Alt 2n (R) Alt 2n(R), the set of all skew-symmetric matrices in GL 2n (R) with Pfaffian 1. Using the same embedding, as defined earlier, one can construct the set Alt(R) := lim Alt 2n (R). Also E is an equivalence relation on Alt(R) and Alt(R)/ E is an abelian group with respect to the operation. Denote this group by W E (R). This group is called the elementary symplectic Witt group. Let I be an ideal of R. We denote the set {α Alt 2n(R) : α χ ±1 n r( mod I) for some r 0} by Alt 2n(R, I). For m n, Alt 2m(R, I) can be embedded into Alt 2n(R, I) by the inclusion map i m,n : Alt 2m(R, I) Alt 2n(R, I) given by α α χ n m. We put Alt (R, I) = lim Alt 2n(R, I). We can define an equivalence relation I on Alt (R, I) as follows: 1

2 2 KUNTAL CHAKRABORTY AND RAVI A. RAO Let α Alt 2m(R, I) and β Alt 2n(R, I). Then α I β if and only if there exists t N and ɛ E 2(m+n+t) (R, I) such that : α χ n+t = ɛ T (β χ m+t )ɛ We denote the set Alt (R, I)/ I by W E (R, I). We can also define the set Alt 2n (R, I) as the subset of Alt 2n(R, I) consisting skew-symmetric matrices of Pfaffian 1. We put Alt(R, I) := lim Alt 2n (R, I). As above the relation I defines an equivalence relation on Alt(R, I). We denote the set Alt(R, I)/ I by W E (R, I). Definition 1.2. Let f 1 : A C be a ring homomorphism between two commutative rings A and C. Also let f 2 : B C be another ring homomorphism between B and C. A ring D is said to be fibre product of A and B with respect to C if the following two conditions are satisfied: 1) there exist two ring homomorphisms,namely p 1 : D A, p 2 : D B such that the following diagram is commutative: p 1 D A p 2 f 1 f 2 B C 2) If there exists some ring E with two ring homomorphisms q 1 : E A, q 2 : E B such that the following diagram is commutative : E B q 1 A q 2 f 1 then there exists a unique map h : E D such that the following diagram is commutative: E f 2 q 1 C q 2!h D p 1 A p 2 f 1 We denote the fibre product of A and B with respect to C by A C B. Consider the fibre product D R p 2 B p 1 π Here the ring D is called the double ring of the ring R with respect to the ideal I. D can be identified with the set { (a, b) R R : a b I} 1.5 Lemma 1.3. Let α M n (D). Then α can be associated canonically to an element (α 1, α 2 ) M n (R) M n (R) such that α 1 α 2 (mod I). In other words M n (D) is isomorphic to the double ring of M n (R) with itself with respect to the ideal M n (I). f 2 R π R/I C

3 A NOTE ON RELATIVE UNIMODULAR ROWS 3 Proof : The maps p 1 and p 2 induces the ring homomorphisms p 1 : M n (D) M n (R) and p 2 : M n (D) M n (R). We have a commutative diagram M n (D) (p 1) M n (R) (p 2) π M n (R) π M n (R/I) Since the fibre product is unique upto isomorphism it is enough to show that M n (D) = M n (R) Mn(R/I) M n (R). By the universal property of the fibre product there exists a unique map h : M n (D) M n (R) Mn(R/I) M n (R),which is defined by h(a) = (A 1, A 2 ) where A = (a ij ) with a ij = (a 1 ij, a2 ij ) such that a1 ij a2 ij I, A 1 = (a 1 ij ) and A 2 = (a 2 ij ). Now define g : M n(r) Mn(R/I) M n (R) M n (D) as following: Let (B 1, B 2 ) M n (R) Mn(R/I) M n (R). Then B 1 = B 2 in M n (R/I). If B 1 = (b 1 ij ) and B 2 = (b 2 ij ) then we have b 1 ij b2 ij I. Therefore the matrix B M n(d) is well-defined. We define g(b 1, B 2 ) = B. It can be shown that g and h both are ring homomorphisms. Now we have g h = 1 Mn(D) and h g = 1 Mn(R) Mn(R/I) M n(r). Hence M n (D) = M n (R) Mn(R/I) M n (R). Using the above canonical isomorphism we always represent the element of M n (D) as an element of M n (R) Mn(R/I) M n (R), i.e. as a pair of two matrices satisfying certain properties. Lemma 1.4. Let M, N M n (D) be such that M = (M 1, M 2 ) and N = (N 1, N 2 ). Then M N = (M 1 N 1, M 2 N 2 ) in M 2n (D). Proof : This is true in M 2n (R) M 2n (R) hence the relation is true in M 2n (R) M2n(R/I)M 2n (R). Finally the relation is true in M 2n (D) by Lemma 1.3. Definition 1.5. Let R be a ring and I R be an ideal. The excision ring of R with respect to the ideal I is denoted by R I and is defined by the set {(r, i) : r R, i I} with addition is componentwise and multiplication is defined by (r, i)(s, j) = (rs, rj + si + ij). 1.6 Lemma 1.6. [2, Lemma 4.3] Let (R, m) be a local ring with maximal ideal m. Then the excision ring R I with respect to a proper ideal I in R is also a local ring with maximal ideal m I. Definition 1.7. We shall say a ring homomorphism φ : B D is a retract if there exists a ring homomorphism γ : D B so that φ γ is identity on D. We shall also say that D is a retract of B. 3.2 Lemma 1.8. ([2, Lemma 3.3]) Let B, D be rings and let D be a retract of B and let π : B D. If J = ker(π), then E n (B, J) = E n (B) SL n (B, J), n Proposition 1.9. ([4, Proposition 3.1]) Let R be a commutative ring and I R be an ideal. Then the excision ring R I is a fibre product of R and R with respect to R/I. Let φ : D R I be defined by φ(a, b) = (b, a b). It is proved in Proposition 1.9, that φ is an isomorphism. For all n N, let us define a map i 2n : Alt 2n(R, I) W E (D) by i 2n(α) = (χ ±1 n r, α)

4 4 KUNTAL CHAKRABORTY AND RAVI A. RAO where α χ ±1 n r mod I for some r 0. It can be shown that the maps i 2n will induce a map i : Alt (R, I) W E (D). It can be shown that i induce a well defined map between W E (R, I) and W E (D), which we still call i. Theorem Let R be a commutative ring and I R be an ideal. Then The set W E (R, I) has an abelian group structure with the operation [α].[β] = [α β]. Proof : It is enough to check that each element [α] of W E (R, I) has an inverse. Let α Alt 2n(R, I). Consider the element α defined by φ i 2n (α). We have α χ ±1 n r( mod 0 I) for some r 0. In the group W (R I), we have [ α α 1 ] = [χ 1 ]. Hence there exists E E 4n+2t (R I) such that E T ( α α 1 χ t )E = χ 2n+t. Going modulo 0 I we have ĒT (χ (χ ) 1 χ t )Ē = χ 2n+t where χ = χ ±1 n r. Since [χ ±1 n r] = [χ n ] in W (R, I), then there exist F E 4n+2t (R, I) such that F T (χ (χ ) 1 χ t )F = χ 2n+t. Also we have F T (χ (χ ) 1 χ t ) F = χ 2n+t where F E 4n+2t (R I, 0 I). Now replacing E by EĒ 1 F, we may assume that E E4n+2t (R I, 0 I) and E T ( α α 1 χ t )E = χ 2n+t. Now projecting R I onto R, we have E1 T (α α 1 χ t )E 1 = χ 2n+t where E 1 E 4n+2t (R, I). Hence we have [α α 1 ] = [χ 1 ], i.e., [α].[α 1 ] = [χ 1 ]. Similarly we can show that [α 1 ].[α] = [χ 1 ]. Hence W (R, I) is a group with the operation [α].[β] = [α β]. To show this group structure is commutative: Let α Alt 2n (R, I) and β Alt 2m (R, I). Hence we have α χ ±1 r χ ±1 n r( mod I) and β χ ±1 s χ ±1 m s( mod I) for some r 0, s 0. Now consider the element α, β in Alt 2n (R I) and Alt 2m (R I) respectively. In W (R I), we have [ α β α 1 β 1 ] = [χ 1 ]. Hence there exist t N and E E 4(m+n)+2t (R I) such that E T ( α β α 1 β 1 χ t )E = χ 2(m+n)+t. Now going modulo 0 I, we have Ē(χ χ (χ ) 1 (χ ) 1 χ t )Ē = χ 2(m+n)+t where χ = χ ±1 n r and χ = χ ±1 s χ ±1 m s. Now there exist F E 4(m+n)+2t (R I, 0 I) such that F T (χ χ ) (χ ) 1 (χ ) 1 χ t )F = χ 2(m+n)+t. Hence replacing E by EĒ 1 F we may assume that E E 4(m+n)+2t (R I, 0 I) and E T ( α β α 1 β 1 χ t )E = χ 2(m+n)+t. Now projecting R I onto R, we have E1 T (α β α 1 β 1 χ t )E 1 = χ 2(m+n)+t where E 1 E 4(m+n)+2t (R, I). Hence [α β α 1 β 1 ] = [χ 1 ] in W (R, I). Hence [α β] = [β α]. Hence W (R, I) is an abelian group with the operation Lemma The following sequence is exact 0 W E (R, I) i W E (D) (p 1) W E (R) 0 Where i([α]) = [(χ ±1 n r, α)] and p 1 : D R is the projection onto first factor. Proof : Let i([α]) = [χ 1 ] in W E (D). Consider the isomorphism between W E (D) and W E (R I) induced from the isomorphism φ : D R I given by φ(a, b) = (b, a b). Then we have φ i([α]) = [χ 1 ]. Hence we have [ α] = [χ 1 ] where α = φ((χ ±1 n r, α)). We have α χ ±1 n r( mod 0 I). Replacing α by F T αf for some F E 2n (R I, 0 I), we may assume that α χ n ( mod 0 I). Since [ α] = [χ 1 ] in W E (R I), there exists a natural number t and an elementary matrix E E 2(n+t)(R I) such that E T ( α χ t )E = χ n+t.

5 A NOTE ON RELATIVE UNIMODULAR ROWS 5 Now going modulo 0 I, we have, Ē T χ n+t Ē = χ n+t. Hence replacing E by EĒ 1, we may assume that E T ( α χ t )E = χ n+t and E E 2(n+t) (R I, 0 I) by Lemma 1.8. Hence projecting R I onto R, we have ε T (α χ t )ε = χ n+t, where ε E 2(n+t) (R, I). Hence i is injective. Clearly (p 1 ) is surjective. Now let (p 1 ) ([β]) = [χ 1 ]. By Lemma 1.3, we have β = (β 1, β 2 ) with β 1 β 2 (mod I). Thus [β 1 ] = [χ 1 ] in W E (R). Hence there exists a natural number and F E 2(n+t)(R) such that F T (β 1 χ t )F = χ n+t Now (F, F ) T ((β 1, β 2 ) (χ t, χ t ))(F, F ) = (χ n+t, F T (β 2 χ t )F ). Hence [β] = [(β 1, β 2 )] = [(β 1, β 2 ) (χ t, χ t )] = [(χ n+t, F T (β 2 χ t )F ] = i([f T (β χ t )F ]). Hence the sequence 0 W E (R, I) i W E (D) (p 1) W E (R) 0 is exact Lemma [1] Let C be the kernel of the group homomorphism R (R/I) induced by π, with the convension that C = R if I = R. Then the Pfaffian gives a split exact sequence j 0 W E (R, I) W E (R, I) C 0 P f Proof : Clearly the homomorphism i is injective and the sequence is exact on the middle. For split exact sequence we have to show that there is a map r : C W E (R, I) such that P f r = id C. Define ( ) 0 a r(a) = [ ] a 0 The map r is well-defined since a 1 (mod I). Corollary The set W (R, I) is a group with respect to the operation Lemma The following sequence is exact. i (p 1) 0 W E (R, I) W E (D) W E (R) 0

6 6 KUNTAL CHAKRABORTY AND RAVI A. RAO Proof : Consider the following diagram j 0 W E (R, I) W E (R, I) C 0 i i φ j 0 W E (D) W E (D) D 0 0 W E (R) W E (R) R 0 P f P f (p 1) (p 1) ψ j P f Where the maps φ : C D and ψ : D R are defined by φ(a) = (1, a) and ψ(d 1, d 2 ) = d 1. With respect to the φ and ψ the above diagram is commutative. Each row of the above diagram is an exact sequence by Lemma Observe that all these groups in the above diagram are abelian. The second coloumn of the diagram is an exact sequence by Lemma Also the third coloumn of the above diagram is exact. Hence by diagram chasing the first coloumn is also exact Lemma [7, Lemma 3.1](Karoubi) Let R be a commutative ring with 1. Let α W E (R[X]). Then we have [α] = [l], where l = ψ 0 + ψ 1 X where ψ 0 and ψ 1 are matrices over R Lemma (Karoubi) Let R be a commutative ring and I R be an ideal of R. Let α W E (R[X], I[X]). Then we have [α] = [l], where l = ψ 0 + ψ 1 X where ψ 0 is a matrix over R and ψ 1 is a matrix over I. Proof : Let α Alt 2n (R[X], I[X]). We may assume that α χ n ( mod I[X]). Consider the element α Alt 2n ((R I)[X]). Then we have α χ n ( mod 0 I[X]). Now by Karoubi we have [ α] = [l], where l is a linear matrix in Alt((R I)[X]). It is easy to check that l Alt 2m ((R I)[X], 0 I[X]) for some m. We have [ α l 1 ] = [χ 1 ]. Conjugating α l 1 with some element F E 2(m+n) ((R I)[X], 0 I[X]), we may assume that α l 1 χ m+n ( mod 0 I[X]). Since [ α l 1 ] = [χ 1 ], then there exist t N and E E 2(m+n+t) ((R I)[X]) such that E T ( α l 1 χ t )E = χ m+n+t. Now going modulo 0 I[X], we have ĒT χ m+n+t Ē = χ m+n+t. Hence replacing E by EĒ 1, we have E E 2(m+n+t) ((R I)[X], 0 I[X]) and E T ( α l 1 χ t )E = χ m+n+t. Now projecting (R I)[X] onto R[X], we have E T 1 (α l 1 1 χ t )E 1 = χ m+n+t, where E 1 E 2(m+n+t) (R[X], I[X]) and l 1 Alt 2m (R[X], I[X]) is a linear matrix. Hence we have [α l 1 1 ] = [χ 1] in W E (R[X], I[X]). In other words [α] = [l 1 ] in W E (R[X], I[X]) Proposition [5, Proposition 2.4.1] Let R be a local ring with 1/2k R, and let [V ] W E (R[X]). Then [V ] has a k-th root, i.e. there is a [W ] W E (R[X]) such that [V ] = [W ] k in W E (R[X]). Lemma Let R be a local ring and I R be an ideal of R, 1/2k R and let [α] W E (R[X], I[X]). Then [α] has a k-th root. Proof : Let I = R. By convension we have W E (R[X], R[X]) = W E (R[X]). The lemma is true for I = R by Proposition Hence we may assume that I is a proper ideal of R. Let α Alt 2n (R[X], I[X]).

7 A NOTE ON RELATIVE UNIMODULAR ROWS 7 We may assume that α χ n ( mod I). Consider the element α W E ((R I)[X]). By Lemma 1.6, we have R I is a local ring. And also we have 1/2k R I. Hence by Proppsition 1.17, we have there exists β W E ((R I)[X]) such that [ α] = [β] k. We revisit the proof of Proposition 1.17, to verify that β χ m ( mod 0 I[X]) for some m. We have α Alt 2n ((R I)[X], 0 I[X]). By Lemma 1.16, we have there exist t N and E E 2(n+t) ((R I)[X], 0 I[X]) such that E T ( α χ t )E = χ n+t + rx, where r is a matrix over 0 I. Now let γ = I 2(n+t) χ n+t rx. Clearly γ SL 2(n+t) ((R I)[X], 0 I[X]). Hence χ n+t r is a nilpotent matrix over R I. Since 1/2k R I, we can extract a 2k-th root of γ. Call it δ. It is easy to check that δ SL 2(n+t) ((R I)[X], 0 I[X]). Now M. Karoubi pointed out that E T ( α χ t )E = χ n+t γ = χ n+t δ 2k = (δ k ) T χ n+t δ k. Let β = δ T χ n+t δ. Clearly β χ n+t ( mod 0 I[X]). Aplying Whitehead s lemma one can check that [ α] = [β] k. Let m = n + t. We have [ α β 1..(k times).. β 1 ] = [χ 1 ] (1) in W E ((R I)[X]). Conjugating α β 1 β 1 by some element in E 2(n+mk) ((R I)[X], 0 I[X]) we may assume that α β 1 β 1 χ n+mk ( mod 0 I[X]). By (1) we have there exist p N and F E 2(n+mk+p) ((R I)[X]) such that F T ( α β 1 β 1 χ p )F = χ n+mk+p. Now going modulo 0 I[X], we have F T χ n+mk+p F = χn+mk+p. Hence replacing F by F F 1, we have F E 2(n+mk+p) ((R I)[X], 0 I[X]) and F T ( α β 1 β 1 χ p )F = χ n+mk+p. Now projecting R I onto R we have F1 T (α β1 1 β1 1 χ p )F 1 = χ n+mk+p, where F 1 E 2(n+mk+p) (R[X], I[X]). Hence we have [α β1 1 β1 1 ] = [χ 1] in W E (R[X], I[X]). Hence [α] = [β 1 ] k in W E (R[X], I[X]). Lemma Let R be a local ring of dimension d, d > 1, α R be a non-zero-divisor, v Um d+1 (R[X], (α)). Then v Ed+1 (R[X],(α)) (a 0 (X), αa 1 (X),..., αc d ) such that c d is a non-zero-divisor. Proof : By Excision theorem [3, Theorem 3.21], we may assume that, R is a reduced ring. Let v(x) := (v 0 (X), v 1 (X),..., v d (X)). We assume that deg v 0 (X) 1. Let the leading coefficient of v 0 (X) be a. We may assume that a is a non-zero-divisor of R. Let the overline denote modulo (a) and consider v(x) Um d+1 ( R[X], (α)). By excision and usual stability estimates we have v(x) Ed+1 ( R[X],(α)) e 1. Hence, we may modify v(x) suitably and assume that, v(x) = e 1. As, so crucially, indicated by M. Roitman, this transformation can be perfomed so that every stage the row contains a polynomial which is unitary in R a. And we may ensure that v 0 (X) is unitary in R a and deg v 0 (X) 1. Let v 0 (X) = 1 + v 0(X), v i (X) = αav i (X) for i > 0. One has a li v i (X) = q i(x)v 0 (X) + r i (X), for some l i > 0, r i (X) R[X], with r i (X) = 0 or deg r i (X) < deg v 0 (X). By Excision theorem [3, Theorem 3.21], we can transform v(x) in the relative orbit with respect to (aα) and assume that deg v i (X) < deg v 0 (X), for all i > 0. If deg v 0 (X) = 1, then we are done. Hence we may assume that deg v 0 (X) = d 0 2. Let c 1, c 2,..., c d0(d 1) be the coefficients of 1, X,..., X d0 1 of the polynomials v 2(X),..., v d (X). Since d 0 (d 1) 2. (d+1) 2 > dim R a, we can argue as in the proof of [6, Theorem 5] to conclude that the ideal generated by the polynomials v 0 (X), v 2(X),..., v d (X) contains a polynomial h(x) odf degree (d 0 1) which

8 8 KUNTAL CHAKRABORTY AND RAVI A. RAO is unitary in R a. Let leadiding coefficient of h(x) is ua k where u is a unit in R. Via Excision theorem and the argument in the proof of [6, Theorem 5], we have mse(v 0 (X), αv 1(X),..., αv d(x)) = mse(v 0 (X), a 2k αv 1(X),..., αv d(x)) = mse(v 0 (X), α{a 2k v 1(X) + (1 a k.l(v 1(X)))h}, αv 2(X),..., αv d(x)) = mse(v 0 (X), αv (X), αv 2(X),..., αv d(x)) where l(v 1(X)) is the leading coefficient of v 1(X). Note that v 1 (X) is unitary in R a of degree d 1 < d 0. Rename v 1 (X) as v 1(X). Again we may ensure that deg v i (X) < d 1, for i 2. Repeat the argument above to lower the degree of v 1(X) to 1. Then lower the degree of v i to zero. We then have the desire form of the vector v(x) in the class of MSE d+1 (R[X], (α)). 2. Relative Vaserstein s Theorem Definition 2.1. A row (a 1, a 2,..., a n ) R n is said to be unimodular if there exist a row (b 1, b 2,..., b n ) R n such that n k=1 a kb k = 1. A row (a 1, a 2,..., a n ) R n is said to be relative unimodular row with respect to the ideal I if it is unimodular and (a 1, a 2,..., a n ) (1, 0,..., 0)( mod I), i.e. a 1 1, a 2,..., a n all belong to I. The set of all unimodular rows is denoted by Um n (R) and the set of all relative unimodular rows with respect to the ideal I is denoted by Um n (R, I). The group E n (R) acts on the set Um n (R) by the action E n (R) Um n (R) Um n (R) given by α v = vα. We denote the orbit space Um n (R)/E n (R) by MSE n (R). Similarly the group E n (R, I) acts on the set Um n (R, I) by the same action. We denote the orbit space Um n (R, I)/E n (R, I) by MSE n (R, I). The Relative Vaserstein symbol. Let b = (b 1, b 2, b 3 ) Um 3 (R, I) and a = (a 1, a 2, a 3 ) Um 3 (R, I) be such that a 1 b 1 + a 2 b 2 + a 3 b 3 = 1. Denote θ(a, b) by the matrix 0 b 1 b 2 b 3 b 1 0 a 3 a 2 θ(a, b) = b 2 a 3 0 a 1 b 3 a 2 a 1 0 Define a map s E : Um 3 (R, I) W E (R, I) by s G (b) = [θ(a, b)]. Lemma 2.2. The map s E does not depent on the choice of a, i.e. if c = (c 1, c 2, c 3 ) Um 3 (R, I) be such that c T b = a T b = 1, then [θ(c, b)] = [θ(a, b)] in W E (R, I).

9 A NOTE ON RELATIVE UNIMODULAR ROWS 9 1 d 1 d 2 d Proof : Let ε = where d 1 = c 3 a 2 c 2 a 3, d 2 = c 1 a 3 c 3 a 1, d 3 = c 2 a 1 c 1 a 2. Since d 1, d 2, d 3 I, then ε E 4 (R, I). It is easy to check that θ(c, b) = ε T θ(a, b)ε. Hence [θ(c, b)] = [θ(a, b)] in W E (R, I). Lemma 2.3. Let R be a commutative ring of dimension 2 and I R be an ideal. Then Um 3 (R, I)/E 3 (R, I) is bijectively equivalent to W E (R, I). Lemma 2.4. Let R be a non-singular affine algebra of dimension d over an algebraically closed field k. Let α SL d+1 (R[X], (X 2 X)) E(R[X], (X 2 X)). Then α E d+1 (R[X], (X 2 X )). 3. Rees algebras and extended Rees algebras Definition 3.1. Rees algebra and extended Rees algebras : Let R be a commutative ring of dimension d and I an ideal of R. Then the algebra n R[It] := { a i t i : n N, a i I i } = n 0 I n t n i=0 is called the Rees algebra of R with respect to I. The extended Rees algebra of R with respect to I, denoted by R[It, t 1 ],is defined by n R[It, t 1 ] := { a i t i : n N, a i I i } = n Z I n t n where I n = R for n 0. i= n Clearly the Rees algebra R[It] is a graded ring. For graded ring the local-global principle will be as follows: 2.0 Theorem 3.2. ([?, Theorem ]) Let S = S 0 S 1 S 2... be a commutative graded ring and let M be a finitely presented S-module. Assume that for every maxzmal ideal m of S 0, M m is extended from (S 0 ) m. Then M is extended from S Lemma 3.3. ([?, Lemma 3.1]) Let R be a commutative ring and I, J ideals of R. Then the natural map φ : R[It]/JR[It] R[Īt], where R = R/J, Ī = (I +J)/J, defined by φ(r+a 1t+a 2 t 2 ++a n t n +JR[It]) = r + ā 1 t + + ā n t n, is an isomorphism. 2.2 Theorem 3.4. ([?, Theorem 1.3]) Let R be a Noetherian ring of dimension d and I be an ideal of R. Then dimension of R[It] d + 1. Moreover if I is not contained in any minimal primes of R, then dim(r[it]) = d Theorem 3.5. ([?, Theorem 2.1]) Let R be a Noetherian regular ring. Then R[It] is regular if and only if I = (0) or (1) or generated by single element. 2.4 Theorem 3.6. ([?, Theorem 4.2]) Let R be a ring of dimension d and I R an ideal of R. Then for n max{3, d + 2}, the natural map φ : GL n (R[It])/E n (R[It]) K 1 (R[It]) is an isomorphism.

10 10 KUNTAL CHAKRABORTY AND RAVI A. RAO Here we are giving another examples for dimension 3 for which the Vaserstein symbol may be injective though the algebra is singular. 2.6 Theorem 3.7. Let R be a non-singular affine algebra of dimension 2 and I R be an ideal. Then the Vaserstein symbol V R[It] : MSE 3 (R[It]) W E (R[It]) is bijective. Proof : By Theorem 3.4 we have dimension of R[It] 3. Hence we have Sd(R[It]) 3. By Theorem??, it is enough to show that SL 4 (R[It]) E 5 (R[It]) = E 4 (R[It]). By Theorem 3.6, we have for n 4, the natural map φ : GL n (R[It])/E n (R[It]) K 1 (R[It]) is an isomorphism. So in particular for n = 4, we have SL 4 (R[It]) E 5 (R[It]) = E 4 (R[It]). The above result holds for dimension 3 when R is non-singular over perfect field. Notation 3.8. Let R be a commutative ring and I R be an ideal. Let S R be a subring of R. Then we denote the algebra S[It] by where I 0 = S. S[It] := { n a i t i : n N, a i I i } = n 0 I n t n i=0 S[It] is an algebra since the subring S induces an S module structure on R. Hence I is an S module. 2.8 Lemma 3.9. Let R be a commutative ring and I R be an ideal. Let S R be a subring of R. If there exists some h S for which S R is an analytic isomorphism along h, then S[It] R[It] is an analytic isomorphism along h. Proof : Clearly h is a non-zero divisor of S[It] as well as of R[It]. Since S R is an analytic isomorphism along h, then the natural map i : S R induces an isomorphism between S/hS and R/hR. Therefore the induced map ī : S/hS R/hR induces an isomorphism between S[Īt] and R[It], where S = S/hS, R = R/hR and Ī = (I + hr)/hr. Consider the following diagram S[It]/hS[It] ψ S[Īt] i ī R[It]/hR[It] φ R[ Īt] The diagram is commutative. By Lemma 3.3 we have the map φ is an isomorphism. It can be shown that the map ψ is also an isomorphism. Therefore by commutativity of the above diagram we have ī is an isomorphism. Theorem Let R be a nonsingular affine algebra of dimension 3 over a perfect field k and I R be an ideal. Then the Vaserstein symbol V R[It] : MSE 3 (R[It]) W E (R[It]) is injective. Proof : By Lemma??, it is enough to check that SL 4 (R[It]) E 5 (R[It]) = E 4 (R[It]). By Local global principle (Theorem 3.2) we may assume that, R is a regular local ring. By Theorem??, there exists a subring

11 A NOTE ON RELATIVE UNIMODULAR ROWS 11 L of R and an element h L such that L R is an analytic isomorphism along h. By Lemma 3.9, L[It] R[It] is an analytic isomorphism along h. Thus we have a patching diagram L[It] R[It] L h [It] R h [It] Let α SL 4 (R[It]) E 5 (R[It]). Since dim(r h ) 2, then we have α h SL 4 (R h [It]) E 5 (R h [It]) = E 4 (R h [It]). By Lemma?? we have there exists β SL 4 (L[It]) and γ E 4 (R[It]) such that α = γβ. Therefore it is enough to show that β E 4 (L[It]). By Theorem??, we have L = k[x 1, X 2, X 3 ] (φ(x1),x 2,X 3). Let L = k[x 1, X 2 ] (φ(x1),x 2)[X 3 ]. Now since L L is analytic isomorphism along X 3, then L [It] L[It] is analytic isomorphism along X 3. Hence we have a patching diagram L [It] L[It] L [X ± 3, It] L X 3 [It] Now β SL 4 (L[It]). We have β X3 E 4 (L X3 [It]) by Theorem 3.7. So by Lemma??, we have there exist δ SL 4 (L [It]) and θ E 4 (L[It]) such that β = θδ. Now δ X3 E 4 (L [It] X3 ). Hence by monic inversion theorem ([?]), we have δ E 4 (L [It]). Since E 4 (L [It]) E 4 (L[It]), then β E 4 (L[It]). Hence β E 4 (L[It]). Lemma Let R be a regular local ring of dimension d,d 3, and I R be an ideal. Then The Vaserstein symbol V R[It,t 1 ] W E (R[It, t 1 ]) is injective. Proof : It is enough to check that SL 4 (R[It, t 1 ]) E 5 (R[It, t 1 ]) = E 4 (R[It, t 1 ]). By Theorem??, there exists a Lindel subring L of R, and an element h L such that L R is an analytic isomorphism along h. Therefore L[It, t 1 ] R[It, t 1 ] is analytic isomorphism along h. REFERENCES fr [1] J. Fasel, R.A. Rao; ggr [2] A. Gupta, A. Garge, R.A. Rao; A nice group structure on the orbit space of unimodular rows-ii, J. Algebra 407 (2014), vdk1 [3] W. van der Kallen; A group structure on certain orbit sets of unimodular rows, J. Algebra 82 (2) (1983), keshari [4] M.K. Keshari; Cancellation problem for projective modules over affine algebras. Journal of K-Theory 3 (2009), rao [5] R.A. Rao; On completing unimodular polynomial vectors of length three, Trans. Amer. Math. Soc. 325 (1991),1, roitman [6] M. Roitman; On stably extended projective modules over polynomial rings, Proceedings of the American Mathematical Society Vol. SV 97, No. 4 (Aug., 1986), [7] L.N. Vaserstein, A.A. Suslin; Serre s problem on projective modules over polynomial rings and algebraic K-theory, Izv. Akad. Nauk SSSR Ser. Mat. 40 (1976), (Math of USSR-Izvestia 10 (1976), ).