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This is Vakil 13.7 E, self-study.

We are to show that if $X$ is a scheme and $\mathcal F$ is a finite type quasicoherent sheaf on $X$, then if $p \in U \subset X$ is an open neighborhood of $p$ and $a_1, ... , a_n \in \mathcal F(U)$ have images generating the fiber $\mathcal F_p \otimes \kappa(p)$, then there must be an affine open neighborhood $p \in \operatorname{Spec} A \subset U$ such that the $a_i$ each restricted to $\operatorname{Spec} A$ generate $\mathcal F(\operatorname{Spec}A)$ as an $A$-module, and for each $q \in \operatorname{Spec} A$, the (images of) the $a_i$ generate $\mathcal F_q$ as an $\mathcal O_{X, q}$-module.

Here is my attempt, but something feels off about it:

If we assume $U$ is already an affine open $\operatorname{Spec}A$, then we know $\mathcal F$ is locally a finite type $A$-module $M$ on $U$. Then the fiber at $p$ is isomorphic to $M_p/pM_p$. Since being a finite type $A$-module is a local property, $M_p$ is a finite type $A_p$-module. Since $p$ is a prime ideal in $A$, by version 8 of Nakayama's Lemma from the Stacks Project's tag 07RC, $M_p$ is generated by the images of the $a_i$. Since $p$ was arbitrary, again by the localness of being finite type, $M$ is finitely generated by the $a_i$.

Using localness once more, $M_q$ is generated by the images of the $a_i$ for any $q \in \operatorname{Spec}A$.

Something about assuming $U$ was affine feels off, almost like I did not quite show what was asked. Also, we did not show that finite type was local in the notes thus far, only that $M$ is finite type over $A$ if and only if $M_{f_i}$ is finite type over $A_{f_i}$, where the $f_i$ generate $A$. I am not sure this allows me to conclude the same about localizing at a prime. It also feels like I used localness "too much."

Johnny Apple
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  • This is fine on a conceptual level. If you're worried about assuming $U$ is affine, perhaps you should write down the reduction in detail: since the affine opens form a basis of the topology, there exists some affine open $U'$ containing $p$ contained in $U$, and then verify that all the hypotheses still hold over $U'$. You may have to use the trick where the intersection of two affine opens is covered by simultaneously distinguished affine opens. – KReiser Nov 06 '20 at 00:37
  • You say in your argument that $p$ was arbitrary to conclude $M$ is finitely generated over $A$. But in the statement $p$ is actually fixed. So I think a bit more needs to be done – Alex Mathers Nov 06 '20 at 00:43
  • Thank you both. I did indeed wrongly assume p was arbitrary, but am unsure how to rectify this. – Johnny Apple Nov 06 '20 at 00:56
  • Sorry, I missed the issue with $p$ and $q$ - I agree with Alex Mathers that there's more to do there. A starting point would be to write down what the generators of $\mathcal{F}_p$ that you're guaranteed by the assumptions look like. – KReiser Nov 06 '20 at 01:05

1 Answers1

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You have already reduced to the affine case (and there is nothing wrong with your reduction) so I will write things purely in terms of rings and modules.

We are in a situation where we have a ring $A$, a finitely-generated $A$-module $M$, and elements $a_1,\dots,a_n\in M$ whose images generate $M_p$ as an $A_p$-module. Our goal is to find some $g\in A\smallsetminus p$ for which the images of $a_i$ generate $M_g$ as an $A_g$-module, because then $\operatorname{Spec}(A_g)$ is the affine open neighborhood of $p$ you are looking for.

Now you should use the fact that $M$ is finitely generated over $A$ and write down a generating set, say $x_1,\dots,x_m$, and notice that the $x_i$ will generate $M_g$ over $A_g$ for any $g$ we choose. Now we know that the $a_i$ generate $M_p$ over $A_p$, so for each $i$ we can write $$x_i=\sum_j \big(\frac{b_{ij}}{s_{ij}}\big)a_j\:\:\:\:\:\:\:\:\text{in $M_p$}$$ for some elements $b_{ij}/s_{ij}\in A_p$, so $s_{ij}\in A_p\smallsetminus p$. I claim we should take $g:=\prod_{i,j}s_{ij}$; since the $x_i$ already generate $M_g$ over $A_g$, you just need to verify that each of these can be generated with the $a_i$ in $M_g$, and this is easy to see by our choice of $g$.

Edit: the comments below are correct, one needs a tweak to the argument above that I'm too lazy to type out right now.

Alex Mathers
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  • Does this choice of $g$ really work? Equality in $M_p$ and $M_g$ are not the same thing after all - that equality in $M_p$ is an equality in $M$ up to multiplication by some $\varepsilon \in A \setminus p$. Maybe I‘m missing something here. But at least a slight variation of this proof would work for sure (one adds these factors appearing in the equalities in $M_p$ to $g$). – Qi Zhu May 20 '21 at 14:26
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    Starting from $x_i = \sum_j (b_{ij} / s_{ij}) a_j$, multiplying through by $g$, we get an expression of the form $x_i = (\sum_j (c_{ij} a_j)) / g $ in $M_p$. This implies there is some $t_i \notin p$ such that $t_i (x_i g - (\sum_j (c_{ij} a_j))) = 0$ in $M$. So yes, we do need to localize at $g * \prod_i t_i$ – David Lui Dec 04 '22 at 10:40
  • Hi @Alex Mathers ,may I ask a silly question why in the second paragraph of the answer , that image of $a_i$ generate the fiber $\mathcal{F}|_p = M_p/(pM_p)$ will implies that $a_1,...,a_n$ will generated $M_p$ as $A_p$ module? – yi li Feb 22 '23 at 03:43
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    @yili This is a form of Nakayama's lemma: if $(A,m)$ Is a local ring and $M$ an $A$-module such that the images of $m_1,\dots,m_n$ generate $M/mM$ as an $A/m$-module then $m_1,\dots,m_n$ generate $M$ as an $A$-module – Alex Mathers Feb 22 '23 at 06:20
  • I realize it thank you Alex – yi li Feb 22 '23 at 06:35