Problem sheet 4 (some questions are only preliminary sketches) Ex. 1 If x1,..xn is an M-regular sequence and sum_n xi*mi = 0 with mi in M, (*) prove that m_r in (x1,..x_{r-1})M for r in 1..n [Hint: xn is a nonzerodivisor for M/(x1,.. x{n-1})M, and the given equality says xn*mn is 0 in that quotient. Therefore mn is a linear combination mn = sum_{n-1} xi*ni for some ni in M. Now if we subtract of the mn term from (*) we get the same kind of linear dependence relation involving x1.. x_{n-1}, so the required expression holds for all mi.] Ex. 2 If x1,..xn is an M-regular sequence, prove that also x1^s,x2,..xn is also an M-regular sequence. [Hint: use induction on s, with linear combinations based on Ex. 1 as the inductive step.] [Ma], Theorem 16.1. Ex. 3 Check that x1,..x3 regular (respectively x1,..x4 regular) implies the Koszul complex written out as (1.23) and (1.25) of the notes are exact. Ex. 4 Check that the skew tensor product expression for the Koszul complex K.(x1,..xn, M) is given by K.(x1,..xn, M) = K.(x1,..x_{n-1}, M) tensor K.(x, A) Ex. 5 Double complexes [Ma] Appendix B, p.275 A double complex is a double indexed array of objects K_{ij} with two sets of differentials d'_i, d"_j where d' lowers i by 1 and d" lowers j by 1 (that is d'_i: K_{ij} -> K_{i-1,j}). Assume that each horizontal row and vertical column is a complex (that is, d'_{i-1}.d'_i = 0 and sim. for d"), and that the squares commute. This is what you get if you take tensor product of two complexes, or a general complex and make a resolution of it. Check that the single complex described in my notes (Appendix, pp. 14-15) as Ksum := sum_{i+j=k} K_{ij} and putting one minus sign in each square is a complex. [Ma], p. 277 leaves the exercise of proving: if all the rows and columns are exact except at zero, then the homology groups of the bottom row K_{i0} equals those of the associated single complex Ksum equals those of the first column K_{0i} This is just an elaborate diagram chase. Ex. 6 For a module M, any two projective resolutions M <- P. and M <- Q. can be compared by a commutative diagram of maps fi: Qi -> Pi that give isomorphisms on the homology groups. Moreover, any two such set of maps are homotopy equivalent. (Left as exercise in [Ma], p.278.) Ex. 7 Review the material on Supp in [UCA], Chap. 7 or my lecture notes. In particular, check the following: 1. For a cyclic module, Supp(A/I) = V(I). 2. For a finite module Supp N = V(I) is equivalent to I = rad(Ann(N)). 3. Supp(N) contained in V(I) is equivalent to I contained in rad(Ann(N)). Ex. 8 The converse (3) => (1) in the proof of [Ma], Th. 16.6. Assume that depth M > 0, and let x1 be a regular element. N is a module with Supp(N) contained in V(I). Write the standard s.e.s. M -x1-> M -> Mbar. Assume by induction that Ext^i(N,Mbar) = 0 for i = 0..n-2. Use the long exact sequence of Exts to deduce that x1: Ext^i(N,M) -> Ext^i(N,M) is injective for i = 0,..n-1. Now Supp(N) contained in V(I) gives x1^n in Ann(N), so also Ext^i(N,M) = 0. Ex. 9 Criteria for homogeneous ideal in a graded ring. Let A = sum Ai be an NN-graded ring, and write f = f0 + f1 + .. fn for the homogenous pieces of f. By definition, an ideal I in A is homogeneous if I = sum Ii where Ii = I intersect Ai. Equivalently, every f in I is the sum of homogeneous pieces fi that are still elements of I. For a unit u in A0^x, define the automorphism tau_u of A that acts on Ai by multiplication by u^i. That is, tau_u takes f0 + f1 +.. fn |-> f0 + u*f1 +.. u^n*f^n. Prove that tau_u takes a homogeneous ideal I to itself. Under the assumption that A0 contains an infinite field k, prove the converse: an ideal I taken to itself by tau_u for u in k^x is homogeneous. [Solution: if sum_0^n fi in I with deg f_i = i. Then the (n+1)x(n+1) Vandermonde determinant |u^i| for i = 0..n, and u taking n+1 distinct values is nonzero, so that the f_i are linear combinations of the n+1 different expressions tau_u(f) = sum u^i*fi in I. Therefore all the fi in I, as required.]