Proposition 7

Tag 0CSS, permanent. Cite it as [Bourbaki, Tag 0CSS].

Commutative Algebra, chapter VIII, § 7. Multiplicités

Suppose A local. Let s be an integer $\geq 1$ and, for $1 \leq i \leq s$ , let $\delta_i$ be an integer $> 0$ , $x_i$ an element of $m_A^{\delta_i}$ , and $\xi_i$ its class in $m_A^{\delta_i}/m_A^{\delta_i+1}$ . Suppose that $(x_1, ..., x_s)$ is a secant sequence for A. Let x denote the ideal of A generated by $(x_1, ..., x_s)$ . Then one has $e(A/x) \geq \delta_1 ... \delta_s \cdot e(A)$ , with equality if $(\xi_1, ..., \xi_s)$ is a completely secant sequence for gr(A).

Put $B = A/x$ , and consider the formal series

H_A = \sum_{n \geq 0} \operatorname{long}(m_A^n/m_A^{n+1}) \cdot T^n, \quad H_B = \sum_{n \geq 0} \operatorname{long}(m_B^n/m_B^{n+1}) \cdot T^n

and $H_B^{(s)} = (1 - T)^{-s} H_B$ . By prop. 6 of § 4, No. 5, one has in $\mathbf{Z}[[T]]$ the inequality

H_B^{(s)} \geq \left( \prod_{i=1}^s \frac{1 - T^{\delta_i}}{1 - T} \right) H_A,

and there is equality when the sequence $(\xi_1, ..., \xi_s)$ is completely secant. But

R(T) = \prod_{i=1}^s \frac{1 - T^{\delta_i}}{1 - T}

is a polynomial of $\mathbf{Z}[T]$ such that $R(1) = \delta_1 ... \delta_s$ . Put $\dim(A) = d$ ; one has $\dim(B) = d - s$ . By th. 2 of § 4, No. 3, there exist elements $R_A$ and $R_B$ of $\mathbf{Z}[T, T^{-1}]$ such that

H_A = (1 - T)^{-d} R_A(T), \quad R_A(1) = e(A),

H_B = (1 - T)^{-d+s} R_B(T), \quad R_B(1) = e(A/x).

Hence one has

(1 - T)^{-d} R_B(T) = H_B^{(s)} \geq \left( \prod_{i=1}^s \frac{1 - T^{\delta_i}}{1 - T} \right) H_A = (1 - T)^{-d} R(T) \, R_A(T),

and equality holds if the sequence

(\xi_1, ..., \xi_s)

is completely secant. One concludes by Lemma 2 of § 4, No. 1.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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