Bourbaki

Proposition 7

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

Commutative Algebra, chapter VIII, § 6. Dimension des anneaux gradués

Suppose that H is an H0H_0-algebra of finite type. Then the function p↦dim⁡(H⊗H0κ(p))p \mapsto \dim(H \otimes_{H_0} \kappa(p)) is upper semicontinuous on Spec(H0)\mathrm{Spec}(H_0).

Since H is a finitely generated H0H_0-algebra, each HiH_i is a finitely generated H0H_0-module (III, § 1, No. 2, corollary to Prop. 1) and H is generated as an H0H_0-algebra by H0⊕H1⊕⋯⊕HrH_0 \oplus H_1 \oplus \cdots \oplus H_r for a suitable integer r≥0r \geq 0. Let p∈Spec(H0)p \in \mathrm{Spec}(H_0) and put dim⁡(H⊗H0κ(p))=n≥0\dim(H \otimes_{H_0} \kappa(p)) = n \geq 0. By Corollary 1 to Prop. 6, there exist elements a1,...,ana_1, ..., a_n of H, all homogeneous of the same degree d>0d > 0, such that the κ(p)\kappa(p)-homomorphism φ‾:κ(p)[X1,...,Xn]→H⊗H0κ(p)\overline{\varphi} : \kappa(p)[X_1, ..., X_n] \to H \otimes_{H_0} \kappa(p) which maps XiX_i to ai⊗1a_i \otimes 1 for 1≤i≤n1 \leq i \leq n, makes H⊗H0κ(p)H \otimes_{H_0} \kappa(p) into a finite κ(p)[X1,...,Xn]\kappa(p)[X_1, ..., X_n]-algebra. Let φ\varphi denote the H0H_0-homomorphism of H0[X1,...,Xn]=RH_0[X_1, ..., X_n] = R into H which maps XiX_i to aia_i for 1≤i≤n1 \leq i \leq n. If, for every m∈Zm \in \mathbf{Z}, we put Hm′=∑(m−1)d<i≤mdHiH'_m = \sum_{(m-1)d < i \leq md} H_i, we obtain a Z\mathbf{Z}-grading on H, compatible with the structure of R-module given by φ\varphi. Each Hm′H'_m is finitely generated over H0H_0. By Lemma 5, HpH_p is a finitely generated RpR_p-module. By Lemma 6, there therefore exists f∈H0−pf \in H_0 - p such that HfH_f is a finitely generated RfR_f-module. For every q∈Spec(H0)fq \in \mathrm{Spec}(H_0)_f, H⊗H0κ(q)H \otimes_{H_0} \kappa(q) is a finite κ(q)[X1,...,Xn]\kappa(q)[X_1, ..., X_n]-algebra, therefore dim⁡(H⊗H0κ(q))≤n\dim(H \otimes_{H_0} \kappa(q)) \leq n (§ 2, No. 3, Theorem 1), which completes the proof.

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

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