Exercise 4, § 7

Let $A$ be a noetherian local ring, with maximal ideal $m$. (a) Suppose that $\mathrm{gr}_m(A)$ is a graded ring which is a complete intersection (cf. exerc. 3). Now prove that the completion $\hat{A}$ of $A$ is a complete intersection, that is to say the quotient of a regular local ring by an ideal generated by a complete intersection sequence. (b) Let $x \in m$, and let $\xi$ be the image of $x$ in $m^v/m^{v+1}$, where $v$ is such that $x \in m^v$ and $x \notin m^{v+1}$. Now prove that one has $e_m(A/x.A) = v.e_m(A)$ if and only if $\xi$ is not a zero divisor in $\mathrm{gr}_m(A)$, and that in this case one has $\mathrm{gr}_m(A/x.A) = \mathrm{gr}_m(A)/\xi.\mathrm{gr}_m(A)$ and $x$ is not a zero divisor in $A$. (Let $I$ be the homogeneous ideal of $\mathrm{gr}_m(A)$ formed by the elements $\eta \in \mathrm{gr}_m(A)$ such that $\xi.\eta = 0$. Consider the ideal $N_n = m^{v+n}:x$ of $A$ and the mapping from $I_n$ into $N_{n+1}/m^{n+1}$ which to $\eta \in I_n$ associates the class mod. $m^{n+1}$ of an element $y \in A$ lifting $\eta$. We shall prove that it is injective, and one will deduce from it the inequality

(i)

moreover, one will prove the equality

(ii)

One will finally remark that $\xi$ is a zero divisor if and only if $I \neq \{0\}$, and using the preceding exercise, one will deduce from (i) that one has $\sum_{n \geq 0} \operatorname{long}_{A/m}(N_{n+1}/m^{n+1}). T^n = \frac{S(T)}{(1 - T)^\nu}$ with $S(1) \neq 0$, and from (ii) the inequality $e_m(A/xA) > e_m(A).$

(c) Let $(x_1, ..., x_s)$ be a sequence of elements of the maximal ideal $m$. Suppose that one has $x_i \in m^{\nu_i}$ and $x_i \notin m^{\nu_i + 1}$, and let $\xi_i$ be the class of $x_i$ in $m^{\nu_i}/m^{\nu_i + 1}$. Now prove that one has $e_m(A/x) \geq e_m(A). \nu_1 ... \nu_s$ and that equality holds if and only if the sequence $(\xi_1, ..., \xi_s)$ is a complete intersection sequence in the ring $\operatorname{gr}_m(A)$. In this case one has an isomorphism of graded rings of $\operatorname{gr}_m(A/x)$ with $(\operatorname{gr}_m(A))/\xi$ with $x = \sum_i x_i A, \quad \xi = \sum_i \xi_i \operatorname{gr}_m(A).$