The Stacks project

Lemma 52.14.3. Let $A$ be a Noetherian domain which has a dualizing complex and which is complete with respect to a nonzero $f \in A$. Let $f \in \mathfrak a \subset A$ be an ideal. Assume every irreducible component of $Z = V(\mathfrak a)$ has codimension $> 2$ in $X = \mathop{\mathrm{Spec}}(A)$, i.e., assume every irreducible component of $Z$ has codimension $> 1$ in $Y = V(f)$. Then $Y \setminus Z$ is connected.

Proof. This is a special case of Lemma 52.14.2 (whose proof relies on Proposition 52.12.3). Below we prove it using the easier Proposition 52.12.6.

Set $U = X \setminus Z$. By Proposition 52.12.6 we have an isomorphism

\[ \mathop{\mathrm{colim}}\nolimits \Gamma (V, \mathcal{O}_ V) \to \mathop{\mathrm{lim}}\nolimits _ n \Gamma (U, \mathcal{O}_ U/f^ n \mathcal{O}_ U) \]

where the colimit is over open $V \subset U$ containing $U \cap Y$. Hence if $U \cap Y$ is disconnected, then for some $V$ there exists a nontrivial idempotent in $\Gamma (V, \mathcal{O}_ V)$. This is impossible as $V$ is an integral scheme as $X$ is the spectrum of a domain. $\square$

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Your email address will not be published.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 0EG3. Beware of the difference between the letter 'O' and the digit '0'.