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I've been trying to prove the following result:

Let $X$ be an affine variety over an algebraically closed field $k$, $A=\Gamma(X,\mathcal{O}_X)$, $M$ an $A$-module of finite type, $\mathcal{F}=\widetilde{M}$ the associated sheaf on $X$, and $x\in X$.

Let $\mathfrak{m}_x=\{f\in A\mid f(x)=0\}$, and $k(x)=A/\mathfrak{m}_x\cong k$. We define $\mathcal{F}(x)=M\otimes_Ak(x)$.

Then $\mathcal{F}(x)$ is a finite dimensional $k$-vector space, and $$U_n=\{x\in X\mid\text{rank}_k(\mathcal{F}(x))\leq n\}$$ is an open subset of $X$ for any $n\geq1$.

I think I have a proof, but am unsure about a few steps and would appreciate some verification.

For the first claim, since $M$ is of finite type we have $M=Am_1+\cdots+Am_s$ for some $s\geq1$ and $m_j\in M$. Then for any $x\in X$, we have that $m_1\otimes_A1,\ldots,m_s\otimes_A1$ generate $\mathcal{F}(x)$ over $k(x)\cong k$, so $\text{rank}_k(\mathcal{F}(x))\leq s$ and $\mathcal{F}(x)$ is a finite dimensional $k$-vector space.

For the second claim, choose an arbitrary $n\geq 1$. My aim is to prove that for each $x\in U_n$ we can find some $f_x\in A$ such that $x\in D(f_x)\subseteq U_n$. Then we would have that $U_n=\cup_{x\in U_n}D(f_x)$ is open and the claim would follow.

Take any $x\in U_n$. We have $$\mathcal{F}(x)=M\otimes_AA/\mathfrak{m}_x=M/\mathfrak{m}_xM$$ and we know that $\mathcal{F}(x)$ is generated by some $\overline{x_1},\ldots,\overline{x_n}$ over $A/\mathfrak{m}_x$. Then by Nakayama's Lemma we have that $x_1,\ldots,x_n$ generate $M_f$ as an $A_f$-module for some $f\in A\setminus\mathfrak{m}_x$, where $x_j+\mathfrak{m}_x=\overline{x_j}$.

We know $x\in D(f)$ since $f\notin\mathfrak{m}_x$, now take any $y\in D(f)$. The aim is to show that $y\in U_n$ to conclude that $D(f)\subseteq U_n$. We have that $\mathcal{F}(y)$ is generated by some $\overline{y_1},\ldots,\overline{y_t}$ over $A/\mathfrak{m}_y$ since we proved any fibre was finite dimensional over $k\cong k(x)$, and we can find $y_i\in M$ with $y_i+\mathfrak{m}_y=\overline{y_i}$. Then for each $i$ we have $$y_i=\frac{a_1x_1+\cdots+a_nx_n}{f^m}$$ for some $a_j\in A$, $m\geq0$, since the $x_j$ generate $M_f$ as an $A_f$ module (we can assume a common denominator by adding more $f$s into the $a_j$ if necessary).

Then $$f^l(f^my_i-(a_1x_1+\cdots+a_nx_n))=0$$ for some $l\geq0$ from the definition of equality in the localisation. Since $A/\mathfrak{m}_y$ is a field, there exists some $g\in A$ such that $(gf^{l+m}-1)\in\mathfrak{m}_y$. Then setting $b_j=gf^l$ we get $$y_i-(b_1x_1+\cdots+b_nx_n)\in\mathfrak{m}_y\tag{*}$$ by multiplying through by $g$, so $$y_i+\mathfrak{m}_y=b_1x_1+\cdots+b_nx_n+\mathfrak{m}_y$$ Then the $(x_j+\mathfrak{m}_y)$ generate $\mathcal{F}(y)$ as an $A/\mathfrak{m}_y$-module, so $\text{rank}_k(\mathcal{F}(y))\leq n$ and we are done.

It's the step $\text{(*)}$ that I'm unsure about, are the $A_f$ and $A/\mathfrak{m}_y$ actions compatible in the way I've used them? I think since they're both ultimately derived from the action of $A$ on $M$ it should be alright, but I just wanted to make sure.

If I've made any other mistakes I'd also appreciate having them pointed out, this seems an important exercise and so I want to make sure my proof is correct! Many thanks in advance.

Note: I've updated the first step of my proof to try to clarify some issues raised in the comments, and have also made a few steps more explicit.

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Dave
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  • How do you know $M=Am_1+\cdots Am_n$ for the same $n$ that you started with? It may need many more generators than $n$. – Mohan Jun 05 '19 at 21:59
  • @Mohan I'll edit to clarify, in the statement of the problem I should have said that $U_n$ is open for any $n$, not meaning to fix a specific one. Then I meant that $M$ has $n$ generators for some $n$ – Dave Jun 05 '19 at 22:05
  • It is still unclear. Write $M=Am_1+\cdots+Am_p$. Then, why is $U_n$ open with your argument? – Mohan Jun 05 '19 at 22:07

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The standard argument for this question is as follows. Since $A$ is Noetherian and $M$ is finitely generated, it is finitely presented, so we have an exact sequence $A^p\to A^q\to M\to 0$ for some $p,q\in\mathbb{N}$. The map $A^p\to A^q$ is given by a $p\times q$ matrix over $A$. Then $U_n$ is given by the open set of points where at least one of the $q-n\times q-n$ minor does not vanish, by Nakayama.

Mohan
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  • Many thanks for your answer, when you say the points where at least one $q-n\times q-n$ minor does not vanish, are they the points $x\in X$ for which $(a_{ij}(x))$ has such a minor which does not vanish (where $(a_{ij})$ is our $p\times q$ matrix)? If so, would you mind explaining why the set of such points is open, and why Nakayama's Lemma implies that this set is $U_n$, as I can't seem to see where to apply the forms of the lemma I'm familiar with. Finally, do we add generators when $n>q$ to avoid issues with the size of the minors? Apologies for all the basic questions, and thanks again – Dave Jun 06 '19 at 01:01
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    @Dave Is it clear to you that one $q-n\times q-n$ minor does not vanish at a point, then $M$ is generated by at most $n$ elements? Conversely, if at a point it is generated by $n$ elements, Nakayama says that $n$ of the given $q$ generates $M$ and thus the remaining $q-n$ elements are not really necessary and then you should be able see that the corresponding $q-n$ minor does not vanish. Finally, a minor gives a an element $f\in A$ and $f\neq 0$ is an open set. So, $U_n$ is (finite) union of these pen sets and thus open. If $n>q$, then clearly $U_n=X$. – Mohan Jun 06 '19 at 01:14
  • Please ignore my previous comments, I've spent some more time looking at your proof, and I think I follow the first half of the argument now. We have $A^p\rightarrow A^q\rightarrow M\rightarrow0$, and tensoring with $k(x)$ for a point $x$ gives us $k(x)^p\rightarrow k(x)^q\rightarrow\mathcal{F}(x)\rightarrow0$. Then if we have a $q-n\times q-n$ minor, by the Rank-Nullity Theorem and exactness we have $\text{dim}(\mathcal{F}(x))\leq n$, so $x\in U_n$. I'm struggling to use Nakayama to show the converse though. The version I'm familiar would say that... – Dave Jun 07 '19 at 22:56
  • ...if $x_1,\ldots,x_n$ generate $M/\mathfrak{m}_xM=\mathcal{F}(x)$, then they generate $M_f$ as an $A_f$-module for some $f\notin\mathfrak{m}_x$. I can see that if we found a $q-n\times q-n$ minor in the matrix $k(x)^p\rightarrow k(x)^q$ then we would have $f\notin\mathfrak{m}_x$, but I can't quite see how Nakayama gives us the desired minor. Assuming it did though, then we have showed that $x\in U_n$ is equivalent to us finding a $q-n\times q-n$ minor, and that $x\in D(f)$, but then how does it follow that $D(f)\subseteq U_n$, which I think we need to conclude the proof? – Dave Jun 07 '19 at 22:57