Bourbaki

Remark 2

Tag 0D6W, permanent. Cite it as [Bourbaki, Tag 0D6W].

Commutative Algebra, chapter X, § 5. Intersections complètes

Let A be a ring, J an ideal of finite type of A and p a prime ideal of A containing J. By cor. 2 of th. 1 of § 1, No. 3, one has

prof⁡Ap(Jp;Ap)≤[κ(p)⊗AJ:κ(p)] \operatorname{prof}_{A_p}(J_p; A_p) \leq [\kappa(p) \otimes_A J : \kappa(p)]

and there is equality if and only if J is completely secant at p.

Suppose that J is distinct from A and generated by a completely secant sequence (x1,…,xr)(x_1, \ldots, x_r). One then has (prop. 4 of § 1, No. 3, and prop. 1 above)

prof⁡A(J;A)=inf⁡p∈V(J)prof⁡Ap(Jp;Ap)=r. \operatorname{prof}_A(J; A) = \inf_{p \in V(J)} \operatorname{prof}_{A_p}(J_p; A_p) = r.

If moreover A is noetherian, one has codim⁡(V(J),Spec⁡(A))≤r\operatorname{codim}(V(J), \operatorname{Spec}(A)) \leq r (VIII, § 3, No. 3, Prop. 4) and prof⁡A(J;A)≤codim⁡(V(J),Spec⁡(A))\operatorname{prof}_A(J; A) \leq \operatorname{codim}(V(J), \operatorname{Spec}(A)) (§1§ 1, No. 7, Prop. 12), whence finally

prof⁡A(J;A)=r=codim⁡(V(J),Spec⁡(A))=ht⁡(J). \operatorname{prof}_A(J; A) = r = \operatorname{codim}(V(J), \operatorname{Spec}(A)) = \operatorname{ht}(J).

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

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