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

\[ M_ d = \Gamma (\mathbf{P}^ n_ R, \mathcal{F} \otimes _{\mathcal{O}_{\mathbf{P}^ n_ R}} \mathcal{O}_{\mathbf{P}^ n_ R}(d)) = \Gamma (\mathbf{P}^ n_ R, \mathcal{F}(d)) \]

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

\[ R[T_0, \ldots , R_ n]_ e \times M_ d \longrightarrow M_{d + e} \]

come from the natural isomorphisms

\[ \mathcal{O}_{\mathbf{P}^ n_ R}(e) \otimes _{\mathcal{O}_{\mathbf{P}^ n_ R}} \mathcal{F}(d) \longrightarrow \mathcal{F}(e + d) \]

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

\[ (T_ i/T_ j)^ d m_ i|_{U_ i \cap U_ j} = m_ j|_{U_ i \cap U_ j} \]

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

\[ (N_ i)_{T_ j/T_ i} \cong (N_ j)_{T_ i/T_ j} \]

(where the subscripts mean “principal localization at”) as modules over the ring

\[ R\left[ \frac{T_0}{T_ i}, \ldots , \frac{T_ n}{T_ i}, \frac{T_0}{T_ j}, \ldots , \frac{T_ n}{T_ j} \right]. \]

This means that for some large integer $d$ there exist elements $s_ i \in N_ i$, $i = 1, \ldots , n$ such that

\[ s = (T_ i/T_0)^ d s_ i \]

on $U_0 \cap U_ i$. Next, we look at the difference

\[ t_{ij} = s_ i - (T_ j/T_ i)^ d s_ j \]

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

\[ s_ a' = (T_ b/T_ a)^{e + d} s_ b' \]

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.
$\square$

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