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There is a step in 3.5.3 of Hartshorne that I am stuck at.

The setup is this: Let $\mathcal{F}$ be a coherent sheaf on a scheme $X$ which is proper over $\text{Spec}(A)$ for a noetherian ring $A$. Let $\mathcal{L}$ be an invertible sheaf on $X$. Let $P$ be a closed point of $X$ and let $\mathcal{I}_P$ be the ideal sheaf of the closed subset $\{P\}$. Let $\mathcal{F'}= \mathcal{F} \otimes \mathcal{L}^n$, $k(P)= \mathcal{O}_X/\mathcal{I}_P$.

Suppose that $H^0(\mathcal{F'})\to H^0(\mathcal{F'}\otimes k(P))$ is surjective. The claim is that there exists $s_i\in H^0(\mathcal{F'})$ such that the images of the $s_i$ in $\mathcal{F'_P}$ generate $\mathcal{F'_P}$ as an $\mathcal{O}_{X,P}$-module.

Hartshorne says this is by Nakayama's lemma over the local ring $\mathcal{O}_{X,P}$.

What I have tried:

I know that $k(P)$ has the property that it's stalk at $P$ is $\mathcal{O}_{X,P}/\mathfrak{m}$, where $\mathfrak{m}$ is the max ideal of $\mathcal{O}_{X,P}$. So the stalk of $\mathcal{F'}\otimes k(P)$ at $P$ is of the form $M/\mathfrak{m}M$ with $M$ being the stalk of $\mathcal{F'}$ at $P$. Now it looks like Nakayama would be useful, but I do not see how to use it here. Furthermore, I cannot see where surjectivity of global sections comes into play.

I have also tried to prove it in the case where $X=\text{Spec}(R)$ is affine, since then I can write $\mathcal{F'}$ as $\widetilde{M}$ for some finitely generated $R$-module $M$. Also, this affine case originally appealed to me because I have explicitly calculated what $\mathcal{I}_P$ is in this case.

I know that the surjectivity of global sections gives a surjection between germs of global sections.

Finally, I wanted to add that for my question I do not think $X$ proper over $A$ matters, nor do I think the presence of $\mathcal{L}$ matters. This is why I have used the notation $\mathcal{F'}$.

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    $\begingroup$ Let $(A, \mathfrak{m})$ be a local ring and let $M$ be a f.g. $A$-module. Let $x_1, x_2, \cdots ,x_n $ be elements of $M$ whose images in $M/\mathfrak{m}M$ forms a basis of this vector space (over $A/\mathfrak{m}$). Then $x_1, x_2, \cdots ,x_n$ generate $M$. $\endgroup$
    – Krish
    Commented Nov 26, 2014 at 22:58
  • $\begingroup$ @Krish, yes this is what I would like to use, however the problem is how to get the $x_i$. $\endgroup$
    – john w.
    Commented Nov 26, 2014 at 23:28

2 Answers 2

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I agree that $A$ and $\mathscr{L}$ don't play any role in this step.

I think the version of Nakayama you want to use is the following: $(A, \mathfrak{m})$ is a local ring and $M$ is a finitely generated $A$-module. If the images of $x_1, \dots, x_r \in M$ in $M/\mathfrak{m}M$ form a set of generators over $A/\mathfrak{m}$ then $x_1, \dots, x_r$ generate $M$ over $A$. This is version 4 on Wikipedia.

In this case, $A = \mathscr{O}_{X, P}$, $M = \mathscr{F}'_P$, and $x_i = (s_i)_P$. Here the $s_i$ are global sections of $\mathscr{F}'$ whose images in $\mathscr{F}' \otimes k(P)$ generate.

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  • $\begingroup$ I guess my issue is how to get these $s_i$. Also, when you say that '$s_i$ are global sections whose images in $\mathcal{F'}\otimes k(P)$ generate', do you mean generate as in the germs they determine generate the stalk of $\mathcal{F'}\otimes K(P)$ at $P$? $\endgroup$
    – john w.
    Commented Nov 26, 2014 at 23:26
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    $\begingroup$ @johnw. You assumed that $H^0(\mathcal{F}') \to H^0(\mathcal{F}' \otimes k(P))$ was surjective and the target is a finite dimensional vector space over $k(P)$ so pick a finite generating set. Then lift to global sections. $\mathcal{F}' \otimes k(P)$ lives at the closed point $P$ so I just think of it as a vector space -- there's no extra data -- and I mean that the images of the $s_i$ generate the vector space. $\endgroup$
    – Hoot
    Commented Nov 26, 2014 at 23:57
  • $\begingroup$ Thank you for your time. There is still something I am confused about. As in your answer, I need to find $s_i\in H^0(\mathcal{F'})$ such that $\overline{(s_i)_P}\in \mathcal{F'_P} /\mathfrak{m}\mathcal{F'_P}$ generate $\mathcal{F'_P}/\mathfrak{m}\mathcal{F'_P}$. Maybe I am just being stupid, but it is not clear to me how to find such $s_i$. $\endgroup$
    – john w.
    Commented Nov 27, 2014 at 1:43
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    $\begingroup$ @johnw. I'm worried that I don't know what you're confused about. You say that $H^0(\mathcal{F'})\to \mathcal{F}'_P/\mathfrak{m}\mathcal{F}'_P$ is surjective, so just lift any basis. Is the trouble in transitioning from the sheaf $\mathcal{F}' \otimes k(p)$ to this vector space? $\endgroup$
    – Hoot
    Commented Nov 27, 2014 at 2:10
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    $\begingroup$ Happy to look over what you've done. r $\endgroup$
    – Hoot
    Commented Nov 27, 2014 at 4:18
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The surjection $H^0(\mathcal{F'})\to H^0(\mathcal{F'}\otimes k(P))$ says that $H^0(\mathcal{F'}\otimes k(P)) = \mathcal{F'_P}/\mathfrak{m}_P\mathcal{F'_P}$ (that you have already proved) is generated by global sections of $\mathcal{F}',$ but it does not says directly that $\mathcal{F'_P}$ is generated by global sections. To prove that $\mathcal{F'_P}$ is generated by global sections of $\mathcal{F}'$ we need Nakayama Lemma. Choose $s_i \in H^0(\mathcal{F}')$ such that $\overline{(s_i)_P}$ generates $\mathcal{F'_P}/\mathfrak{m}_P\mathcal{F'_P}.$ Then $(s_i)_P$ generates $\mathcal{F'_P}.$

I hope that it will clear your doubt.

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