The Stacks project

Proof. Fully faithfulness. Suppose that $(\mathcal{E}, n)$ and $(\mathcal{F}, m)$ are objects of $\mathop{\mathrm{colim}}\nolimits _ F \textit{Vect}(X)$. Let $(a, b) : G(\mathcal{E}, n) \to G(\mathcal{F}, m)$ be a morphism in the RHS. We may choose $N \gg 0$ and think of $a$ as a map $a : F^{N - n, *}\mathcal{E}|_ Z \to F^{N - m, *}\mathcal{F}|_ Z$ and $b$ as a map $b : F^{N - n, *}\mathcal{E}|_{X'} \to F^{N - m, *}\mathcal{F}|_{X'}$ agreeing over $E$. Choose a finite affine open covering $X = X_1 \cup \ldots \cup X_ n$ such that $\mathcal{E}|_{X_ i}$ and $\mathcal{F}|_{X_ i}$ are finite free $\mathcal{O}_{X_ i}$-modules. For each $i$ the base change

is another almost blow up square as in Example 38.37.11. For these squares we know that

by Lemma 38.38.2 (see proof of the lemma). Hence after increasing $N$ we may assume the maps $a|_{Z_ i}$ and $b|_{X'_ i}$ come from maps $c_ i : F^{N - n, *}\mathcal{E}|_{X_ i} \to F^{N - m, *}\mathcal{F}|_{X_ i}$. After possibly increasing $N$ we may assume $c_ i$ and $c_ j$ agree on $X_ i \cap X_ j$. Thus these maps glue to give the desired morphism $(\mathcal{E}, n) \to (\mathcal{F}, m)$ in the LHS.

Essential surjectivity. Let $(\mathcal{F}, \mathcal{G}, \varphi )$ be a triple consisting of a finite locally free $\mathcal{O}_ Z$-module $\mathcal{F}$, a finite locally free $\mathcal{O}_{X'}$-module $\mathcal{G}$, and an isomorphism $\varphi : \mathcal{F}|_ E \to \mathcal{G}|_ E$. We have to show that after replacing this triple by a Frobenius power pullback, it comes from a finite locally free $\mathcal{O}_ X$-module.

Noetherian reduction; we urge the reader to skip this paragraph. Recall that $X = \mathop{\mathrm{Spec}}(A)$ and $Z = \mathop{\mathrm{Spec}}(A/(f_1, f_2))$, $X' = \text{Proj}(A[T_0, T_1]/(f_2T_0 - f_1T_1))$, and $E = \mathbf{P}^1_ Z$. By Limits, Lemma 32.10.3 we can find a finitely generated $\mathbf{F}_ p$-subalgebra $A_0 \subset A$ containing $f_1$ and $f_2$ such that the triple $(\mathcal{F}, \mathcal{G}, \varphi )$ descends to $X_0 = \mathop{\mathrm{Spec}}(A_0)$ and $Z_0 = \mathop{\mathrm{Spec}}(A_0/(f_1, f_2))$, $X_0' = \text{Proj}(A_0[T_0, T_1]/(f_2T_0 - f_1T_1))$, and $E_0 = \mathbf{P}^1_{Z_0}$. Thus we may assume our schemes are Noetherian.

Assume $X$ is Noetherian. We may choose a finite affine open covering $X = X_1 \cup \ldots \cup X_ n$ such that $\mathcal{F}|_{Z \cap X_ i}$ is free. Since we can glue objects of $\mathop{\mathrm{colim}}\nolimits _ F \textit{Vect}(X)$ in the Zariski topology (Lemma 38.39.1), and since we already know fully faithfulness over $X_ i$ and $X_ i \cap X_ j$ (see first paragraph of the proof), it suffices to prove the existence over each $X_ i$. This reduces us to the case discussed in the next paragraph.

Assume $X$ is Noetherian and $\mathcal{F} = \mathcal{O}_ Z^{\oplus r}$. Using $\varphi $ we get an isomorphism $\mathcal{O}_ E^{\oplus r} \to \mathcal{G}|_ E$. Let $I = (f_1, f_2) \subset A$. Let $\mathcal{I} \subset \mathcal{O}_{X'}$ be the ideal sheaf of $E$; it is globally generated by $f_1$ and $f_2$. For any $n$ there is a surjection

Hence the first cohomology group of this module is zero. Here we use that $E = \mathbf{P}^1_ Z$ and hence its structure sheaf and in fact any globally generated quasi-coherent module has vanishing $H^1$. Compare with More on Morphisms, Lemma 37.72.3. Then using the short exact sequences

and induction, we see that

is surjective. By the theorem on formal functions (Cohomology of Schemes, Theorem 30.20.5) this implies that

is surjective. Thus we can choose a map $\alpha : \mathcal{O}_{X'}^{\oplus r} \to \mathcal{G}$ which is compatible with the given trivialization of $\mathcal{G}|_ E$. Thus $\alpha $ is an isomorphism over an open neighbourhood of $E$ in $X'$. Thus every point of $Z$ has an affine open neighbourhood where we can solve the problem. Since $X' \setminus E \to X \setminus Z$ is an isomorphism, the same holds for points of $X$ not in $Z$. Thus another Zariski glueing argument finishes the proof. $\square$