Lemma 27.13.8. Let $R$ be a ring. Let $\mathcal{F}$ be a quasi-coherent sheaf on $\mathbf{P}^ n_ R$. For $d \geq 0$ set
Then $M = \bigoplus _{d \geq 0} M_ d$ is a graded $R[T_0, \ldots , R_ n]$-module and there is a canonical isomorphism $\mathcal{F} = \widetilde{M}$.
Proof. The multiplication maps
come from the natural isomorphisms
see Equation (27.10.1.4). Let us construct the map $c : \widetilde{M} \to \mathcal{F}$. On each of the standard affines $U_ i = D_{+}(T_ i)$ we see that $\Gamma (U_ i, \widetilde{M}) = (M[1/T_ i])_0$ where the subscript ${}_0$ means degree $0$ part. An element of this can be written as $m/T_ i^ d$ with $m \in M_ d$. Since $T_ i$ is a generator of $\mathcal{O}(1)$ over $U_ i$ we can always write $m|_{U_ i} = m_ i \otimes T_ i^ d$ where $m_ i \in \Gamma (U_ i, \mathcal{F})$ is a unique section. Thus a natural guess is $c(m/T_ i^ d) = m_ i$. A small argument, which is omitted here, shows that this gives a well defined map $c : \widetilde{M} \to \mathcal{F}$ if we can show that
in $M[1/T_ iT_ j]$. But this is clear since on the overlap the generators $T_ i$ and $T_ j$ of $\mathcal{O}(1)$ differ by the invertible function $T_ i/T_ j$.
Injectivity of $c$. We may check for injectivity over the affine opens $U_ i$. Let $i \in \{ 0, \ldots , n\} $ and let $s$ be an element $s = m/T_ i^ d \in \Gamma (U_ i, \widetilde{M})$ such that $c(m/T_ i^ d) = 0$. By the description of $c$ above this means that $m_ i = 0$, hence $m|_{U_ i} = 0$. Hence $T_ i^ em = 0$ in $M$ for some $e$. Hence $s = m/T_ i^ d = T_ i^ e/T_ i^{e + d} = 0$ as desired.
Surjectivity of $c$. We may check for surjectivity over the affine opens $U_ i$. By renumbering it suffices to check it over $U_0$. Let $s \in \mathcal{F}(U_0)$. Let us write $\mathcal{F}|_{U_ i} = \widetilde{N_ i}$ for some $R[T_0/T_ i, \ldots , T_0/T_ i]$-module $N_ i$, which is possible because $\mathcal{F}$ is quasi-coherent. So $s$ corresponds to an element $x \in N_0$. Then we have that
(where the subscripts mean “principal localization at”) as modules over the ring
This means that for some large integer $d$ there exist elements $s_ i \in N_ i$, $i = 1, \ldots , n$ such that
on $U_0 \cap U_ i$. Next, we look at the difference
on $U_ i \cap U_ j$, $0 < i < j$. By our choice of $s_ i$ we know that $t_{ij}|_{U_0 \cap U_ i \cap U_ j} = 0$. Hence there exists a large integer $e$ such that $(T_0/T_ i)^ et_{ij} = 0$. Set $s_ i' = (T_0/T_ i)^ es_ i$, and $s_0' = s$. Then we will have
on $U_ a \cap U_ b$ for all $a, b$. This is exactly the condition that the elements $s'_ a$ glue to a global section $m \in \Gamma (\mathbf{P}^ n_ R, \mathcal{F}(e + d))$. And moreover $c(m/T_0^{e + d}) = s$ by construction. Hence $c$ is surjective and we win. $\square$
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)