The Stacks project

Lemma 33.43.2. Let $X$ be a separated, irreducible scheme of dimension $> 0$ over a field $k$. Let $x \in X$ be a closed point. The open subscheme $X \setminus \{ x\} $ is not proper over $k$.

Proof. Since $X$ is irreducible, $U = X \setminus \{ x\} $ is not closed in $X$. In particular, the immersion $U \to X$ is not proper. By Morphisms, Lemma 29.41.7 (here we use $X$ is separated), $U \to \mathop{\mathrm{Spec}}(k)$ is not proper either. $\square$


Comments (4)

Comment #7492 by WhatJiaranEatsTonight on

I think we need to be separated here. Otherwise \the projecitve line with double original point is a counterexample.

Comment #7514 by David Holmes on

Dear WhatJiaranEatsTonight,

I quite agree with your counterexample. Moreover, Lemma 26.22.1 refered to in the proof uses separatedness. Though happily this tag is only applied in once place, and in that place it is applied to a separated scheme (over a field).

Best wishes, David

Comment #7516 by Laurent Moret-Bailly on

Here is a more elementary (and, I think, more natural) proof: since is irreducible, is not closed in . In particular, the immersion is not proper. By Lemma 29.41.7 (and because is separated), is not proper either.

There are also:

  • 4 comment(s) on Section 33.43: Curves

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Your email address will not be published.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 0A24. Beware of the difference between the letter 'O' and the digit '0'.