Proposition 8

Let $\delta > 0$ be an integer, x an element of $q^\delta$, $\xi$ its class in $\mathrm{gr}_\delta(A) = q^\delta/q^{\delta+1}$ and $\varphi$ the multiplication by $\xi$ in the gr(A)-module gr(M).

a) The dimension of the A-module $M/xM$ is equal to $\dim_A(M)$ or to $\dim_A(M) - 1$. In the second case, one has $e_q(M/xM) \geq \delta e_q(M)$.

b) Suppose that one has $\dim_A(M) \geq 1$ and that the kernel of $\varphi$ is of finite length over $A/q$. Then one has $\dim_A(M/xM) = \dim_A(M) - 1$. Moreover: (i) If $\dim_A(M) > 1$, one has $e_q(M/xM) = \delta e_q(M)$. (ii) If $\dim_A(M) = 1$, one has for every integer $n \geq 0$

where $\varphi^n$ is the n-th iterate of the endomorphism $\varphi$, and

Put $M'' = M/xM$; consider the Hilbert-Samuel series $H_M = H_{M,q}$ and $H_{M''} = H_{M'',q}$, together with the Poincaré series $P(T) = \sum_{n \geq 0} \mathrm{long}_{A/q}(\mathrm{Ker}\ \varphi_n) \cdot T^n$. By § 4, No. 3, th. 2 and No. 4, remark 2, one has

with $d = \dim_A(M)$, $d'' = \dim_A(M'')$, $R_M$ and $R_{M''}$ in $\mathbf{Z}[T]$, $R_M(1) = e_q(M)$, $R_{M''}(1) = e_q(M'')$. By lemma 6 of § 4, No. 5, one has in $\mathbf{Z}((T))$ the inequalities

Putting $R(T) = (1 - T^\delta)/(1 - T) = 1 + T + \cdots + T^{\delta-1}$, this may also be written

(1)

By Lemma 2 of § 4, No. 1, the first inequality (1) implies either $d'' \geq d$, or $d'' = d - 1$ and $R(1) R_M(1) \leq R_{M''}(1)$, that is to say $\delta e_q(M) \leq e_q(M'')$. This proves a), since $d'' \leq d$.

Under the assumption of b), one has $P(T) \in \mathbf{Z}[T]$ and $P(1) = \operatorname{long}_A(\operatorname{Ker} \varphi)$. The second inequality (1) is written

Suppose that one has $d > 1$; then Lemma 2 of § 4, No. 1 yields $d'' + 1 \leq d$, whence $d'' = d - 1$ by part a) of the proof; one then has

(loc. cit.), whence (i).

Now suppose $d = 1$. By loc. cit., one has $d'' = 0$ and

Consequently, $M''$ is of finite length equal to $e_q(M'') = R_{M''}(1)$, and one obtains

(2)

Let $n \geq 1$ be an integer. Replace $x$ by $x^n$ in (2); one therefore has

(3)

It is immediate that the submodules $\operatorname{Ker} \varphi^n$ of the noetherian gr(A)-module gr(M) form an increasing sequence and hence are stationary, and that each of them is of finite length over $A/q$. Dividing by $n \geq 1$ in inequality (3) and making $n$ tend to $+ \infty$, one finds $e_{xA}(M) = \delta e_q(M)$ by definition of $e_{xA}(M)$.