Exercise 10, § 3

We retain the hypotheses of Exercise 9; we suppose that the complex $L$ is exact in degree $\neq 0$.

a) Prove the inclusion $V(\mathfrak{d}(d_{i+1})) \subset V(\mathfrak{d}(d_i))$ for all $i \geq 1$ (observe that according to Exercise 8, a prime ideal $p$ does not contain $\mathfrak{d}(d_i)$ if and only if the $A_p$-module $(\mathrm{Coker}\ d_i)_p$ is free).

b) We now suppose that the annihilator of $H_0(L)$ is not reduced to zero. Prove that we have $\mathrm{rg}(d_1) = \mathrm{rg}_A(L_0)$ (if $f$ is a cancellable element of $\mathfrak{d}(d_1)$, the $A_f$-module $H_0(L)_f$ is projective of constant rank according to Exercise 8, hence necessarily zero); deduce that the support of $H_0(L)$ is $V(\mathfrak{d}(d_1))$.

c) Put $p = \mathrm{prof}_A(\mathfrak{d}(d_1); A)$. Prove that one has $V(\mathfrak{d}(d_i)) = \mathrm{Supp}(H_0(L))$ for $1 \leq i \leq p$ (reason as in the proof of a), observing that if a prime ideal $p$ of the support of $H_0(L)$ did not contain $\mathfrak{d}(d_k)$, one would have $\mathrm{dp}_{A_p} H_0(L)_p < k \leq \mathrm{grade}_{A_p} H_0(L)_p$.