Lemma 27.11.5. With hypotheses and notation as in Lemma 27.11.1 above. Assume $A_ d \to B_ d$ is surjective for $d \gg 0$ and that $A$ is generated by $A_1$ over $A_0$. Then
$U(\psi ) = Y$,
$r_\psi : Y \to X$ is a closed immersion, and
the maps $\theta : r_\psi ^*\mathcal{O}_ X(n) \to \mathcal{O}_ Y(n)$ are isomorphisms.
Proof. By Lemmas 27.11.4 and 27.11.2 we may replace $B$ by the image of $A \to B$ without changing $X$ or the sheaves $\mathcal{O}_ X(n)$. Thus we may assume that $A \to B$ is surjective. By Lemma 27.11.3 we get (1) and (2) and surjectivity in (3). By Lemma 27.10.3 we see that both $\mathcal{O}_ X(n)$ and $\mathcal{O}_ Y(n)$ are invertible. Hence $\theta $ is an isomorphism. $\square$
Your email address will not be published.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: