Lemma 15.51.2. Let $R \to \Lambda$ be a homomorphism of Noetherian rings. Assume $P$ has property (B). The following are equivalent

1. the fibres of $R \to \Lambda$ have $P$,

2. the fibres of $R_\mathfrak p \to \Lambda _\mathfrak q$ have $P$ for all $\mathfrak q \subset \Lambda$ lying over $\mathfrak p \subset R$, and

3. the fibres of $R_\mathfrak m \to \Lambda _{\mathfrak m'}$ have $P$ for all maximal ideals $\mathfrak m' \subset \Lambda$ lying over $\mathfrak m$ in $R$.

Proof. Let $\mathfrak p \subset R$ be a prime. Then the fibre over $\mathfrak p$ is the ring $\Lambda \otimes _ R \kappa (\mathfrak p)$ whose spectrum maps bijectively onto the subset of $\mathop{\mathrm{Spec}}(\Lambda )$ consisting of primes $\mathfrak q$ lying over $\mathfrak p$, see Algebra, Remark 10.17.8. For such a prime $\mathfrak q$ choose a maximal ideal $\mathfrak q \subset \mathfrak m'$ and set $\mathfrak m = R \cap \mathfrak m'$. Then $\mathfrak p \subset \mathfrak m$ and we have

$(\Lambda \otimes _ R \kappa (\mathfrak p))_\mathfrak q \cong (\Lambda _{\mathfrak m'} \otimes _{R_\mathfrak m} \kappa (\mathfrak p))_\mathfrak q$

as $\kappa (\mathfrak q)$-algebras. Thus (1), (2), and (3) are equivalent because by (B) we can check property $P$ on local rings. $\square$

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