Lemma 37.4.2. Let $X \subset X'$ be a thickening. Let $n$ be an integer invertible in $\mathcal{O}_ X$. Then the map $\mathop{\mathrm{Pic}}\nolimits (X')[n] \to \mathop{\mathrm{Pic}}\nolimits (X)[n]$ is bijective.
Proof for a finite order thickening. By the general principle explained following Definition 37.2.1 this reduces to the case of a first order thickening. Then may use Lemma 37.4.1 to see that it suffices to show that $H^1(X, \mathcal{I})[n]$, $H^1(X, \mathcal{I})/n$, and $H^2(X, \mathcal{I})[n]$ are zero. This follows as multiplication by $n$ on $\mathcal{I}$ is an isomorphism as it is an $\mathcal{O}_ X$-module. $\square$
Proof in general. Let $\mathcal{I} \subset \mathcal{O}_{X'}$ be the quasi-coherent ideal sheaf cutting out $X$. Then we have a short exact sequence of abelian groups
\[ 0 \to (1 + \mathcal{I})^* \to \mathcal{O}_{X'}^* \to \mathcal{O}_ X^* \to 0 \]
We obtain a long exact cohomology sequence as in the statement of Lemma 37.4.1 with $H^ i(X, \mathcal{I})$ replaced by $H^ i(X, (1 + \mathcal{I})^*)$. Thus it suffices to show that raising to the $n$th power is an isomorphism $(1 + \mathcal{I})^* \to (1 + \mathcal{I})^*$. Taking sections over affine opens this follows from Algebra, Lemma 10.32.8. $\square$
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Your email address will not be published.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: