Bourbaki marks this one as harder than the rest.
Let A be a commutative ring. For every submodule R of a finitely generated free A-module L=An, let c1(R) denote the ideal generated by the (x,x∗), where x runs through R and x∗ runs through the dual L∗; write ck(R)=c1(Im(⋀kR)), Im(⋀kR) being the canonical image of the k-th exterior power of R in ⋀kL. If R1,R2⊂R, are two submodules of L, then ck(R1)⊂ck(R2) for all k.
(a) Let M be a finitely generated A-module; M is isomorphic to a quotient module L/R, where L=An for a suitable n. Show that the sequence of ideals (ak)k≥0 such that ak=cn−k(R) for k≤n and ak=A for k>n is independent of the expression of M in the form L/R. (Consider first the case where L/R and L/R′ are isomorphic; then note that An/R is isomorphic to An+h/(R×Ah) for all h>0.) We write dk(M)=ak for all k≥0 and say that these are the determinantal ideals associated with M. Then dk(M)⊂dk+1(M) for k≥0.
(b) If A is a principal ideal domain and M=L/R where L is free and finitely generated, show that the ideals ck+1(R)(ck(R))−1 are the invariant factors of R in L.
(c) Let r=γ(M) be the least of the cardinals of the systems of generators of M and let r0 be the least integer h such that dh(M)=A. Show that r0≤r and give an example where r0<r (take A to be a Dedekind domain).
(d) If a is the annihilator of M, show that d0(M)⊂a and ar−k⊂dk(M) for k≤r=γ(M).
(e) Suppose that M is a direct sum of submodules Mi (1≤i≤h). Show that
dk(M)=∑dk1(M1)…dkn(Mn) where the sum is over the finite sequences
(ki)1≤i≤h such that
∑ki=k.
(f) If N is a finitely generated submodule of M, then dk(M/N)i=1⊂dk(M) for all k≥0. Show that:
j=0∑kdj(N)dk−j(M/N)⊂dk(M).
(g) Let K be a field, A the polynomial ring K[X,Y,Z], m the maximal ideal AX+AY+AZ and q the ideal generated by X,Y2,YZ and Z2. Consider the A-module M=A/q and its submodule N=m/q. Show that d0(M)=q, d1(M)=A, d0(N)=m2 and d1(N)=m.