Bourbaki

Proposition 19

Tag 04AJ, permanent. Cite it as [Bourbaki, Tag 04AJ].

Commutative Algebra, chapter II, § 4. Spectra of rings and supports of modules

Let A,BA, B be two rings, ϕ:A→B\phi : A \to B a homomorphism and

aϕ:Spec⁡(B)→Spec⁡(A) ^{a}\phi : \operatorname{Spec}(B) \to \operatorname{Spec}(A)
the continuous mapping associated with ϕ\phi (Proposition 13). For every AA-module MM,
Supp⁡(M(B))⊂aϕ(Supp⁡(M));if also M is finitely generated, then \operatorname{Supp}(M_{(B)}) \subset ^{a}\phi(\operatorname{Supp}(M)) ; \text{if also } M \text{ is finitely generated, then}
Supp⁡(M(B))=aϕ−1(Supp⁡(M)). \operatorname{Supp}(M_{(B)}) = ^{a}\phi^{-1}(\operatorname{Supp}(M)).
Let qq be a prime ideal of BB and p=ϕ−1(q)p = \phi^{-1}(q). Suppose that qq belongs to Supp⁡(M(B))\operatorname{Supp}(M_{(B)}); then M(B)⊗BBq=(M⊗AB)⊗BBq=M⊗ABq=(M⊗AAp)⊗ABqM_{(B)} \otimes_B B_q = (M \otimes_A B) \otimes_B B_q = M \otimes_A B_q = (M \otimes_A A_p) \otimes_A B_q, since the homomorphism A→B→BqA \to B \to B_q factors into A→Ap→BqA \to A_p \to B_q (§ 2, no. 1, Proposition 2); the hypothesis M(B)⊗BBq≠0M_{(B)} \otimes_B B_q \neq 0 implies therefore M⊗AAq≠0M \otimes_A A_q \neq 0, whence the first assertion. As the homomorphism ϕq:Ap→Bq\phi_q : A_p \to B_q is local, the second assertion follows from 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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