Exercise 8, § 2

Let $A$ be a noetherian local ring, $M$ a finitely generated $A$-module. We propose to prove the equivalence of the following two conditions:

(i) $M$ is a Buchsbaum module ; (ii) there exists a positive integer $i(M)$ such that one has $\lg(M/qM) - e_q(M) = i(M)$ for every ideal $q \subset m_A$ such that $M/qM$ is of finite length.

(a) Under assumption (i), let $x$ be a weak sequence for $M$, and let $x$ be the ideal generated by it. With the notations of Exercise 6, d), prove the equality $\lg(M/xM) - ex(M) = \lg(K_n)$. Let $x'$ be another weak sequence for $M$, and let $K'_n$ be the corresponding module ; prove the equality $\lg(K_n) = \lg(K'_n)$ (reason by induction on $n$, considering an element $t$ of $m_A$ such that the sequences $(x_1, \ldots, x_{n-1}, t)$ and $(x'_1, \ldots, x'_{n-1}, t)$ are weak sequences for $M$).

(b) Assume that assumption (ii) is satisfied. Let $x = (x_1, \ldots, x_n)$ be a maximal weak sequence for $M$, and let $M_n = M/(x_1M + \ldots + x_{n-1}M)$ ; applying the formula of Exercise 6, d) to the sequences $x$ and $(x_1, \ldots, x_{n-1}, x_n^2)$ (taking into account Exercise 6, e)), prove that one has $\ker(x_n)_{M_n} = \ker(x_n^2)_{M_n}$. Conclude with Exercise 7, d)).

(c) For $M$ to be a Buchsbaum module, it is necessary and sufficient that this be so for $\widehat{M}$ ; in this case $i(\widehat{M}) = i(M)$.

(d) Let $M$ be a Buchsbaum module, and let $(x_1, \ldots, x_p)$ be a sequence of elements of $m_A$ which is a weak sequence for $M$, with $p < \dim(M)$ ; prove that one has $i(M/(x_1M + \ldots + x_pM)) = i(M)$ (use the expression of $i(M)$ given in (a)).