Lemma 30.3.4. Let $f : X \to Y$ be a quasi-compact morphism with $X$ and $Y$ quasi-separated. If $R^1f_*\mathcal{I} = 0$ for every quasi-coherent sheaf of ideals $\mathcal{I}$ on $X$, then $f$ is affine.
Proof. Let $V \subset Y$ be an affine open subscheme. We have to show that $U = f^{-1}(V)$ is affine. The inclusion morphism $V \to Y$ is quasi-compact by Schemes, Lemma 26.21.14. Hence the base change $U \to X$ is quasi-compact, see Schemes, Lemma 26.19.3. Thus any quasi-coherent sheaf of ideals $\mathcal{I}$ on $U$ extends to a quasi-coherent sheaf of ideals on $X$, see Properties, Lemma 28.22.1. Since the formation of $R^1f_*$ is local on $Y$ (Cohomology, Section 20.7) we conclude that $R^1(U \to V)_*\mathcal{I} = 0$ by the assumption in the lemma. Hence by the Leray Spectral sequence (Cohomology, Lemma 20.13.4) we conclude that $H^1(U, \mathcal{I}) = H^1(V, (U \to V)_*\mathcal{I})$. Since $(U \to V)_*\mathcal{I}$ is quasi-coherent by Schemes, Lemma 26.24.1, we have $H^1(V, (U \to V)_*\mathcal{I}) = 0$ by Lemma 30.2.2. Thus we find that $U$ is affine by Lemma 30.3.1. $\square$
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Your email address will not be published.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: