Bourbaki

Proposition 20

Tag 0549, permanent. Cite it as [Bourbaki, Tag 0549].

Commutative Algebra, chapter VII, § 4. Modules over integrally closed Noetherian domains

(i) For a finitely generated A-module M to be pseudo-zero, it is necessary and sufficient that M⊗ABM \otimes_A B be a pseudo-zero B-module. (ii) If M is a finitely generated torsion A-module, M⊗ABM \otimes_A B is a finitely generated B-module and: (22)

χB(M⊗AB)=i(χA(M)). \chi_B(M \otimes_A B) = i(\chi_A(M)).
(iii) If M is a finitely generated A-module, M⊗ABM \otimes_A B is a finitely generated B-module and: (23)
cB(M⊗AB)=i(cA(M)) c_B(M \otimes_A B) = i(c_A(M))
(24)
rB(M⊗AB)=rA(M). r_B(M \otimes_A B) = r_A(M).

(i) Let p\mathfrak{p} be a prime ideal of BB, p=P∩A\mathfrak{p} = \mathfrak{P} \cap A; then (M⊗AB)p=M⊗ABp(M \otimes_A B)_{\mathfrak{p}} = M \otimes_A B_{\mathfrak{p}} (Chapter 11, § 2, no. 7, Proposition 18) and on the other hand

M⊗ABp=(M⊗AAp)⊗ApBp=Mp⊗ApBp; M \otimes_A B_{\mathfrak{p}} = (M \otimes_A A_p) \otimes_{A_p} B_{\mathfrak{p}} = M_p \otimes_{A_p} B_{\mathfrak{p}};
the relation Mp=0M_p = 0 is therefore equivalent to (M⊗AB)p=0(M \otimes_A B)_{\mathfrak{p}} = 0 (Chapter II, § 4, no. 4, Lemma 4). It suffices to apply this remark to the ideal P=(0)\mathfrak{P} = (0) and the ideals p∈P(B)\mathfrak{p} \in P(B) to prove (i), taking account of no. 4, Definition 2.

To prove (ii), we shall use the following lemma:

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

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Cites

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