Satyagopal Mandal
Department of Mathematics
University of Kansas
Office: 624 Snow Hall  Phone: 785-864-5180
  • e-mail: mandal@math.ukans.edu
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    Some Results in Complete Intersections
    964th AMS Meeting of AMS, March 30-31, 2001
    University of Kansas, Lawrence, Kansas 66045



    Boratynski (J. algebra, 1978) proved the following theorem.

    Theorem(Boratynski). Let A be a commutative ring and I be an ideal with

    I=(f1, ¼, fk) + I2.

    Let
    J=(f1, ¼, fk-1) + I(k-1)!.

    Then J is image of a projective module Q of rank k:
    Q ® J


    This theorem of Boratynski had a far reaching impact in the study of complete intersections in affine varieties. Mohan Kumar used this theorem of Boratynski to prove the following theorem.

    Theorem(Mohan Kumar). Let A be a reduced affine ring with dim(A)= n over an algebraically closed field k, and let Q be a projective A-module of rank n. Suppose that Q maps onto a complete intersection ideal J = (f1, ¼, fn) of height n. Then Q = Q0 ÅA.


    This theorem of Mohan Kumar also had a far reaching consequence. Eventually, using Monhan Kumar's theorem, Murthy (Annals of Mathematics, 1994) proved :

    Theorem(Murthy). Let A be a reduced affine ring with dim(A)= n over an algebraically closed field k, and let Q be a projective A-module of rank n.
    Then Q = Q0 ÅA if and only if the top Chern Class Cn(Q) = 0 in the Chow Group of zero cycles.


    The theorem of the day that I want present is as follows:

    Theorem. Let A be a noetherian commutative ring. Let r1, ¼, rk be nonnegative integers and k £ dim(A). Let f1, ¼, fk be a regular sequence in A. Suppose Q is a projective A-module of rank k. Suppose we have a surjective map
    Q ® (f1r1, ¼, fkrk)

    If (k-1)! divides r1r2¼ rk, then [Q] = [Q0Å A] in K0(A), for some projecive A-module Q0.


    In fact, we give a more general theorem on such decomposition of projective modules as follows.

    Theorem. Let A be a noetherian commutative ring. Let r1, ¼, rk be positive integers and k £ dim(A).
    Let
    J0 =(f1, ¼, fk) + J02.

    be a locally complete intersection ideal of height k. Define
    J1 =(f2, ¼, fk) + J0r1,

    J2 =(f1r1, f3, ¼, fk) + J1r2,

    ¼     ¼     ¼    

    J = Jk =(f1r1, f2r2, ¼, fk-1rk-1) + Jk-1rk .
    ( So, J = Jk =(f1r1, f2r2, ¼, fk-1rk-1 , fkrk) + J2).

    Suppose that Q is a projective A-module of rank k and there is a surjective map :
    f : Q ® J

    Assume that (k-1)! divides r1r2¼ rk. Then

    [Q] = [Q0Å A] - (r1r2¼ rk)¤(k-1)!)[A/J0]


    for some projective A-module Q0 of rank k-1.


    A key ingrediant in the proof involves the completability of unimodular rows to invertible matrices. For a long time, I knew that the determinant
    ÷
    x y
    z w
    ÷   =   xw-yz


    So, if (x,y) is a unimodular row then (x,y) is the first row of an invertible matrix. Swan-Towber wrote down that the determinant (Krusemeyer's version)

    ÷ 2    y   z   ÷
    ÷   -y + 2cx    c 2    - a - bc  ÷
    ÷   - z - 2bx    a - bc   b 2 ÷
      =   (ax + by +cz) 2

    So, if (x, y, z) is a unimodular row then (x 2, y, z) is the first row of an invertible matrix.


    Suslin extended the above:

    Theorem. Let A be a commutative ring. Let r0, r1, ¼, rn be nonnegative integers. Suppose
    ( x 0, x 1, x 2, ¼ , x n )

    is a unimodular row in A.

    Assume that n! divides r0r1¼ rn.

    Then there is an invertible matirx a with it's forst row
    ( x 0r0, x 1r1, x 2r2, ¼ , x nrn )



    Boratynski used this theorem of Suslin to prove his theorem. Another ingrediant is the Murthy's extension of Boratynski's theorem.

    Theorem. Let A be a commutative ring and I be a locally complete intersection ideal of height k with

    I=(f1, ¼, fk) + I2.

    Let
    J=(f1, ¼, fk-1) + I(k-1)!.

    Then J is image of a projective module P of rank k with [P] - k = -[A/I] in K0(A).


    The proof of this theorem is done by looking at the "universal ring"
    Ak = Z[X1, X2, ¼ ,Xk,Y1,Y2, ¼ ,Yk, Z] ¤( åXi Yi - Z(1+Z))

    and it's Chow Ring.
    I used the flowing "universal ring"
    Bk = Z[X1, X2, ¼ ,Xk,Y1,Y2, ¼ ,Yk, S, T, U, V] ¤(SU - TV -1, åXi Yi - ST))

    and it's Chow Ring.