The Stacks project

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 10.18.5. 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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BK8. Beware of the difference between the letter 'O' and the digit '0'.