Lemma 33.43.11. Let $X$ be a separated scheme of finite type over $k$. If $\dim (X) \leq 1$ and no irreducible component of $X$ is proper of dimension $1$, then $X$ is affine.
Proof. Let $X = \bigcup X_ i$ be the decomposition of $X$ into irreducible components. We think of $X_ i$ as an integral scheme (using the reduced induced scheme structure, see Schemes, Definition 26.12.5). In particular $X_ i$ is a singleton (hence affine) or a curve hence affine by Lemma 33.43.10. Then $\coprod X_ i \to X$ is finite surjective and $\coprod X_ i$ is affine. Thus we see that $X$ is affine by Cohomology of Schemes, Lemma 30.13.3. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: