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I can't quite follow Chapter III, Example 12.7.2 in Hartshorne (I've written it in the case of affine varieties only here since I haven't worked with schemes in general yet and this is all I need).

In it, he gives an alternative proof of a result I asked about in this question.

Let $X$ be an affine variety over an algebraically closed field $k$, $A=\Gamma(X,\mathcal{O}_X)$, and $\mathcal{F}$ a coherent sheaf on $X$.

Let $x\in X$, $\mathfrak{m}_x=\{f\in A\mid f(x)=0\}$, and $k(x)=A/\mathfrak{m}_x$. Then the function $$\varphi(x)=\dim_{k(x)}(\mathcal{F}\otimes_A k(x))$$ is upper semi-continuous.

I'll go through his reasoning and my thoughts so far. Since $\mathcal{F}$ is coherent, I have taken an $A$-module $M$ with $\mathcal{F}=\widetilde{M}$.

He starts by saying that by Nakayama's Lemma we have that $\varphi(x)$ is the minimum number of generators of the stalk $\mathcal{F}_x$ as an $\mathcal{O}_x$-module.

Here is my attempt to justify this:

Suppose $x_1,\ldots,x_n$ generate $\mathcal{F}_x=M_{\mathfrak{m}_x}$ as an $\mathcal{O}_x=A_{\mathfrak{m}_x}$ module, with $n$ minimal. Then for any $m\in\mathcal{F}_x$ we can find $a_i\in A$, $f\in A\setminus\mathfrak{m}_x$ such that $$m=\frac{a_1}{f}x_1+\cdots+\frac{a_n}{f}x_n$$ (we can assume a common denominator by taking terms into the $a_i$). Since $k(x)$ is a field we can find $g\in A$ with $fg-1\in\mathfrak{m}_x$. Then for $b_i=a_ig$ we have $m=b_1x_1+\cdots+b_nx_n$ in $\mathcal{F}\otimes_A k(x)=M/\mathfrak{m}_xM$ and so $\varphi(x)\leq n$.

Conversely, suppose that $\varphi(x)=l$ and $x_1,\ldots,x_l$ generate $\mathcal{F}\otimes_A k(x)$ over $k(x)$. Now, $\mathcal{F}\otimes_A k(x)=\mathcal{F}_x/\mathfrak{m}_x\mathcal{F}_x$, so by Nakayama's Lemma there exists some $f\in A\setminus\mathfrak{m}_x$ such that the $x_i$ generate $(M_{\mathfrak{m}_x})_f$ as an $A_f$-module. But $(M_{\mathfrak{m}_x})_f=M_{\mathfrak{m}_x}$ since $f\notin\mathfrak{m}_x$, and $A_f\subseteq A_{\mathfrak{m}_x}$, so they certainly generate $\mathcal{F}_x$ as an $\mathcal{O}_x$-module. Then $n\leq\varphi(x)$, so $\varphi(x)=n$ as desired.

He goes on to say that if $x_1,\ldots,x_n$ form a minimal set of generators, then they generate $\mathcal{F}$ in some neighbourhood.

From my justification of the first statement, it appears that this neighbourhood would be $D(f)$. We know they generate $M_{\mathfrak{m}_x}$ as an $A_f$-module, but I'm not sure how this would show they generate $M_f$. I also can't see why we would need to use the minimum number of generators for this step in particular to be true.

He concludes that if we take any $y$ in this neighbourhood, then $\varphi(y)\leq n$.

This is where I am particularly confused. As I understand it (and I may well have already misunderstood something), we would (hopefully) have that $x_1,\ldots,x_n$ generate $M_f$ as an $A_f$-module, with $y\in D(f)$. Then why does this imply that we can generate $\mathcal{F}_y=M_{\mathfrak{m}_y}$ as an $\mathcal{O}_y=A_{\mathfrak{m}_y}$ module using $n$ elements?

Perhaps I'm not using Nakayama's Lemma in the way he intended, my justification seems very laboured in comparison to how it is presented, so I feel I'm really missing something here. Any help would be much appreciated.

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Dave
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    For your last question, the tensor product if right exact. – Ahr Jun 20 '19 at 20:29
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    For the generation, notice that the generators can be written $a_i/s_i$ with $s_i(x)\neq 0$, and look at what happens in $U=D(s_i...s_n)$, you should easily see that $\mathcal F (U)$ is generated by the same generators. – Ahr Jun 20 '19 at 20:34
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    The first point is simply Nakayama as pointed by Hartshorne, take $K$ the cokernel of $O_x^n\to F_x$ mapping $(a_1,...,a_n)$ to $a_1x_1+...+a_nx_n$ with $(x_1,...,x_n)$ a set of generators of the fiber mod $\mathfrak m_x $ of length $\varphi(x)$, it is a finitely generated $O_x$ module, you have $K\otimes k(x)=0$ by definition, in other words $K=\mathfrak m_x K$ and Nakayma tells you that it implies $K=0$. – Ahr Jun 20 '19 at 20:49
  • Many thanks, this has made things a lot clearer! Just one question about your first comment, would we be using the right exactness on the sequence giving a presentation of $\mathcal{F}(U)$, and tensor with $\mathcal{O}_y$, since I think $\mathcal{F}(U)\otimes\mathcal{O}_y=\mathcal{F}_y$? – Dave Jun 20 '19 at 21:03
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    Yes, saying that $a_i$ is a set of generators of an $A$-module $M$ is saying that $A^r\to M $ mapping $(t_1,...,t_r)$ to $t_1a_1+...+t_ra_r$ is onto. But as tensoring is right exact, you deduce from the surjectivity of $A^r\to M $ the surjectivity of $A_m^r\to M_m$, thus the fact that if you have "global" generators, their germs generate each stalks. – Ahr Jun 20 '19 at 21:08
  • Thanks again, I can follow his argument now. If you wanted to copy any of this into an answer I'd very happily accept it, no worries if not! – Dave Jun 20 '19 at 21:23

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