Gluing finitely presented quasi coherent sheaves

Question

Let $X$ be a quasi-compact, separated scheme, and $\{\text{Spec}(A_i)\subset X\}_{i=1,\ldots, n}$ a finite affine open cover.

Suppose a quasi-coherent $\mathcal{O}_X$-module $\mathcal{F}$ is such that $\mathcal{F}(\text{Spec}(A_i))$ has a finitely presented $A_i$-submodule $M_i\subset \mathcal{F}(\text{Spec}(A_i))$ for each $i$.

Does there exist a finitely presented $\mathcal{O}_X$-submodule $\mathcal{F}_0\subset\mathcal{F}$ such that $\mathcal{F}_0(\text{Spec}(A_i))$ contains $M_i$ as an $A_i$-submodule?

Attempt Since $X$ is separated, calling $j_k : \text{Spec}(A_k)\to X$ the open immersion, $j_k$ is quasi-compact, so $(j_k)_*(\widetilde{M}_k)$ is quasi-coherent. Take the sum and call $\mathcal{F}_0$ the image of the map $\bigoplus_{k=1}^n (j_k)_*(\widetilde{M}_k)\to\mathcal{F}$. This attempt doesn't work, since for quasi-coherent sheaves the map $\text{id}\to j_*j^*$ is almost never an isomorphism, so there usually is no useful map $\bigoplus_{k=1}^n (j_k)_*(\widetilde{M}_k)\to\mathcal{F}$. The question remains, however.

Slogan I'd like to know if, for a quasi-coherent sheaf that on each member of a finite affine cover has a finitely presented sub-sheaf, has a finitely presented subsheaf whose restrictions on each member of that cover contain each of them.

Answer 1

This is not always possible. Note that finitely presented modules are always quasi-coherent (Tag 01BO); this helps a little bit to streamline the argument.

Example. Let $R = k[x,y,\{a_i\}_{i \in I},\{b_i\}_{i \in I}]/(\{xa_i-yb_i\}_{i\in I})$ for some infinite set $I$, and let $X = \operatorname{Spec} R$. Then $M = (x,y)$ is a finitely generated $R$-module that is not finitely presented, because $x$ and $y$ satisfy the relations $xa_i = yb_i$ for all $i \in I$. Let $\mathscr F = \tilde{M}$.

Let $U_1 = D(x-1) = \operatorname{Spec}(A_1)$ and $U_2 = D(x+1) = \operatorname{Spec}(A_2)$ be opens that cover $X$ and both contain $V(x,y)$ (say $\operatorname{char} k \neq 2$, so that $U_1 \neq U_2$). Write $U_{12} = U_1 \cap U_2 = \operatorname{Spec} A_{12}$. Let \begin{align*} M_1 = (x) \subseteq \mathscr F(U_1) = (x,y) \subseteq A_1,\\ M_2 = (y) \subseteq \mathscr F(U_2) = (x,y) \subseteq A_2. \end{align*} Note that both $x$ and $y$ are nonzerodivisors on $X$, so $M_1$ and $M_2$ are both free of rank $1$; in particular finitely presented. On the other hand, $M_1|_{U_{12}} + M_2|_{U_{12}} = (x,y) = \mathscr F(U_{12})$, which is not finitely presented.

Now suppose that there exists a finitely presented $\mathcal O_X$-submodule $\mathscr F_0 \subseteq \mathscr F$ such that $\mathscr F_0(U_i)$ contains $M_i$. Then $\mathscr F_0(U_{12})$ contains both $M_1|_{U_{12}}$ and $M_2|_{U_{12}}$, hence equals $\mathscr F(U_{12})$. But this contradicts finite presentation of $\mathscr F_0$. $\square$

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Remark. The problem is basically that sums of finitely presented submodules need not be finitely presented. This problem disappears for example for Noetherian rings, in which case the answer is actually positive. Indeed, [Hart, Exc. II.5.15(d)] shows that there exists a coherent subsheaf $\mathscr F_i \subseteq \mathscr F$ such that $\mathscr F_i|_{U_i} = \tilde M_i$. Taking $\mathscr F_0 = \sum_{i=1}^n \mathscr F_i$ gives the result. $\square$

Note that the positive result in the Noetherian case does not use separatedness, whereas the negative result above already happens in the affine case!

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References.

[Hart] Hartshorne, Robin, Algebraic geometry. Graduate Texts in Mathematics 52. Springer-Verlag, New York-Heidelberg-Berlin (1977). ZBL0367.14001.