The Stacks project

A flat degeneration of a disconnected scheme is either disconnected or nonreduced.

Lemma 37.26.7. Let $f : X \to S$ be a morphism of schemes. Assume that

  1. $S$ is the spectrum of a discrete valuation ring,

  2. $f$ is flat,

  3. $X$ is connected,

  4. the closed fibre $X_ s$ is reduced.

Then the generic fibre $X_\eta $ is connected.

Proof. Write $S = \mathop{\mathrm{Spec}}(R)$ and let $\pi \in R$ be a uniformizer. To get a contradiction assume that $X_\eta $ is disconnected. This means there exists a nontrivial idempotent $e \in \Gamma (X_\eta , \mathcal{O}_{X_\eta })$. Let $U = \mathop{\mathrm{Spec}}(A)$ be any affine open in $X$. Note that $\pi $ is a nonzerodivisor on $A$ as $A$ is flat over $R$, see More on Algebra, Lemma 15.22.9 for example. Then $e|_{U_\eta }$ corresponds to an element $e \in A[1/\pi ]$. Let $z \in A$ be an element such that $e = z/\pi ^ n$ with $n \geq 0$ minimal. Note that $z^2 = \pi ^ nz$. This means that $z \bmod \pi A$ is nilpotent if $n > 0$. By assumption $A/\pi A$ is reduced, and hence minimality of $n$ implies $n = 0$. Thus we conclude that $e \in A$! In other words $e \in \Gamma (X, \mathcal{O}_ X)$. As $X$ is connected it follows that $e$ is a trivial idempotent which is a contradiction. $\square$


Comments (1)

Comment #1116 by Simon Pepin Lehalleur on

Suggested slogan: A flat degeneration of a disconnected scheme is either disconnected or non-reduced.

There are also:

  • 2 comment(s) on Section 37.26: Connected components of fibres

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Your email address will not be published.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 055J. Beware of the difference between the letter 'O' and the digit '0'.