Bourbaki

Exercise 10, § 4

Commutative Algebra, VII, 573

Bourbaki marks this one as harder than the rest.

Let AA be a commutative ring. For every submodule RR of a finitely generated free AA-module L=AnL = A^n, let c1(R)c_1(R) denote the ideal generated by the (x,x∗)(x, x^*), where xx runs through RR and x∗x^* runs through the dual L∗L^*; write ck(R)=c1(Im(⋀kR))c_k(R) = c_1(\mathrm{Im}(\bigwedge^k R)), Im(⋀kR)\mathrm{Im}(\bigwedge^k R) being the canonical image of the kk-th exterior power of RR in ⋀kL\bigwedge^k L. If R1,R2⊂RR_1, R_2 \subset R, are two submodules of LL, then ck(R1)⊂ck(R2)c_k(R_1) \subset c_k(R_2) for all kk.

(a) Let MM be a finitely generated AA-module; MM is isomorphic to a quotient module L/RL/R, where L=AnL = A^n for a suitable nn. Show that the sequence of ideals (ak)k≥0(a_k)_{k \geq 0} such that ak=cn−k(R)a_k = c_{n-k}(R) for k≤nk \leq n and ak=Aa_k = A for k>nk > n is independent of the expression of MM in the form L/RL/R. (Consider first the case where L/RL/R and L/R′L/R' are isomorphic; then note that An/RA^n/R is isomorphic to An+h/(R×Ah)A^{n+h}/(R \times A^h) for all h>0h > 0.) We write dk(M)=akd_k(M) = a_k for all k≥0k \geq 0 and say that these are the determinantal ideals associated with MM. Then dk(M)⊂dk+1(M)d_k(M) \subset d_{k+1}(M) for k≥0k \geq 0.

(b) If AA is a principal ideal domain and M=L/RM = L/R where LL is free and finitely generated, show that the ideals ck+1(R)(ck(R))−1c_{k+1}(R)(c_k(R))^{-1} are the invariant factors of RR in LL.

(c) Let r=γ(M)r = \gamma(M) be the least of the cardinals of the systems of generators of MM and let r0r_0 be the least integer hh such that dh(M)=Ad_h(M) = A. Show that r0≤rr_0 \leq r and give an example where r0<rr_0 < r (take AA to be a Dedekind domain).

(d) If aa is the annihilator of MM, show that d0(M)⊂ad_0(M) \subset a and ar−k⊂dk(M)a^{r-k} \subset d_k(M) for k≤r=γ(M)k \leq r = \gamma(M).

(e) Suppose that MM is a direct sum of submodules MiM_i (1≤i≤h1 \leq i \leq h). Show that

dk(M)=∑dk1(M1)…dkn(Mn) d_k(M) = \sum d_{k_1}(M_1) \ldots d_{k_n}(M_n)
where the sum is over the finite sequences (ki)1≤i≤h(k_i)_{1 \leq i \leq h} such that ∑ki=k\sum k_i = k.

(f) If NN is a finitely generated submodule of MM, then dk(M/N)i=1⊂dk(M)d_k(M/N)^{i=1} \subset d_k(M) for all k≥0k \geq 0. Show that:

∑j=0kdj(N)dk−j(M/N)⊂dk(M). \sum_{j=0}^k d_j(N) d_{k-j}(M/N) \subset d_k(M).

(g) Let KK be a field, AA the polynomial ring K[X,Y,Z]K[X, Y, Z], mm the maximal ideal AX+AY+AZAX + AY + AZ and qq the ideal generated by X,Y2,YZX, Y^2, YZ and Z2Z^2. Consider the AA-module M=A/qM = A/q and its submodule N=m/qN = m/q. Show that d0(M)=qd_0(M) = q, d1(M)=Ad_1(M) = A, d0(N)=m2d_0(N) = m^2 and d1(N)=md_1(N) = m.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.