6
$\begingroup$

In Chapter 6 of Mumford's Geometric invariant theory, during the proof of the rigidity lemma, there are two statements I'm not sure how to verify. The general setup is:

$p : X \rightarrow S$ is flat, $S$ connected, and $H^0(X_s, o_{X_s}) \cong k(s)$ for all points $s \in S$.

  1. In the first part, we're assuming $\epsilon : S \rightarrow X$ is a section, and that $S$ consists of one point. Mumford says: "One checks that $p_*(o_X) \cong o_S$." I found a proof when $p$ is projective (and even proper, I think), which works because this is going to be used on projective abelian schemes, but the general case is still bothering me.

  2. In the second part, $X$ still has the section $\epsilon$, but $S$ is now general (i.e. not just a point), and $p$ is a closed map. During the proof, $Z$ is a closed subscheme of $X$. Mumford claims the statement:

If $p^{-1}(t) \subset Z$ (set-theoretically), for any $t \in S$, then for all artin subschemes $T \subset S$ concentrated at $t$, $Z$ contains $p^{-1}(T)$ as a subscheme.

implies that $Z$ contains an open neighborhood of $p^{-1}(t)$. Intuitively, I think of the artin subscheme as a thickening of the point, and so if I contain an entire fiber then I get "a little bit extra", making $Z$ contain an open neighborhood. I'm wondering how I should do this more formally.

Thanks for any help, it is much appreciated!

$\endgroup$
2
  • $\begingroup$ I wonder what $k(s)$ means here. It cannot be the residue field. $\endgroup$
    – Y. Li
    Sep 28, 2016 at 10:32
  • $\begingroup$ @Y.Li: why not? $\endgroup$ Sep 28, 2016 at 12:24

3 Answers 3

4
$\begingroup$

1) This is really simple. If $S=\{s\}$, then $X=X_s$ and hence $p_*\mathscr O_X=H^0(X,\mathscr O_X)=k(s)=\mathscr O_S$.

2) This may be a little trickier, but still not too hard.
-- Since the statement is local on $S$, we may assume that $S=\mathrm{Spec}A$ is affine.
-- We may also assume that $X=\mathrm{Spec}B$ is also affine. Indeed, we may cover $p^{-1}(t)$ with open affines, so we have an open set on each affine that is contained in $Z$. Their union is open, contained in $Z$, and contains $p^{-1}(t)$. Let $I\subseteq B$ denote the ideal of $Z$ in $B$.
-- Therefore $p$ comes from a morphism $\phi:A\to B$. Let $\mathfrak q=I(t)\subseteq A$ be the ideal of the point $t\in S$ and let $Q=\phi(\mathfrak q)B\subseteq B$ be the ideal generated by it in $B$. This is the ideal of $p^{-1}(t)$.
-- Next, define $J=\cap_{r\in\mathbb N_+}Q^r$ and observe that (by definition) $Q\cdot J=J$ and hence by Nakayama's lemma there exists an $f\in Q$ such that $(1-f)J=0$. The condition in the gray area implies that $I\subseteq J$, and hence $(1-f)I=0$.
-- Finally observe that now we have that $$ p^{-1}(t)\subseteq U=\mathrm{Spec}B_{1-f}\subseteq Z $$ where $U\subseteq X$ is open.

-- In fact a little more is true: $Z$ contains the pre-image of an open set on $S$, but this is a simple consequence of the assumption that $p$ is closed. Let $U\subseteq Z$ be the open set that contains $p^{-1}(t)$ and let $W=X\setminus U$. Since $p$ is closed, $p(W)\subseteq S$ is closed and hence $V=S\setminus p(W)\subseteq S$ is open. By construction $t\in V$ and $p^{-1}V\subseteq U$.

$\endgroup$
1
  • 1
    $\begingroup$ Sorry to unbury this. About 1), I think it's only correct if S is the spectrum of a field, but the interesting case is when S is local Artinian (that's what Mumford uses later in the proof of his lemma). How to prove this case? $\endgroup$
    – user159150
    Jun 7, 2020 at 16:55
2
$\begingroup$

(2) is basically a rephrasing of Krull's intersection theorem, which for us will be the following statement:

Let $A$ be a Noetherian ring, let $I\subset A$ be an ideal, and let $M$ be an $A$-module. Consider the intersection $N=\bigcap_{n\geq 1}I^nM$: for every $n\in N$, there exists $a\in I$ such that $(1-a)n=0$. (This might not be true. See Sandor's comment below)

Let us apply this when $M=B$ is a Noetherian $A$-algebra. Then $N\subset B$ is an ideal and is finitely generated over $B$. Therefore, we can find $a\in I$ such that $(1-a)N=0$. In particular, if there is an ideal $J\subset B$ such that $J\subset N$, then $(1-a)J=0$. Moreover, $1-a$ is a unit in $B/IB$ (in fact it maps to $1$).

Now set $X=Spec B$, $Y=Spec A$, $t=Spec A/I$ and $Z=Spec B/J$. Take $U\subset X$ to be the basic open $Spec B_{1-a}$. This is an open contained in $Z$ and containing $p^{-1}(t)$.

For the sake of completeness, let me say something about (1). It is a general statement in algebraic geometry that, if you have a proper flat map $p:X\to S$ with geometrically integral fibers (this condition is equivalent to the condition you stated about the cohomology of the fibers) over a connected base $S$, then the natural map $\mathcal{O}_S\rightarrow p_*\mathcal{O}_X$ is an isomorphism. Indeed, $p_*\mathcal{O}_X$ is a finite quasi-coherent algebra over $\mathcal{O}_S$, and so there is a finite $S$-scheme $g:S'\to S$ such that $g_*\mathcal{O}_{S'}=p_*\mathcal{O}_X$. Moreover, $p$ factors as $p'\circ g$, for some map $p':X\to S'$. By construction $\mathcal{O}_{S'}=p'_*\mathcal{O}_X$. You can now use the condition that $p$ has geometrically integral fibers to check that $g$ must actually be an isomorphism, and so we have what we wanted. See EGA III.4.3 for all this.

$\endgroup$
3
  • $\begingroup$ Doesn't the Krull intersection theorem need $M$ to be a finitely generated module? I think your proof still works, but you would have to apply KIT over $B$ and not over $A$. $\endgroup$ Jan 19, 2012 at 5:11
  • $\begingroup$ Well, yes, you're probably right. I was thinking my statement might still be true, and follow from the usual statement for finitely generated modules, but it involves some commuting between inverse and direct limits that I don't see immediately. As you note, we don't really need the non-finitely generated case, anyway. $\endgroup$ Jan 19, 2012 at 5:43
  • $\begingroup$ Regarding (1): The map $p$ need not be proper, $\mathbb P^2 \setminus \{P\} \to \operatorname{Spec} \mathbb C$ is a counterexample. $\endgroup$ Aug 31, 2021 at 17:22
2
+50
$\begingroup$

Regarding 1): Suppose that $A$ is an Artin local ring with maximal ideal $\mathfrak{m}$. By the local criterion for flatness, we have isomorphisms of $\mathcal{O}_X$-modules $\mathfrak{m}^i/\mathfrak{m}^{i+1}\otimes_A \mathcal{O}_X\simeq \mathfrak{m}^i \mathcal{O}_X/\mathfrak{m}^{i+1}\mathcal{O}_X$, in particular the latter is free and $\Gamma(\mathfrak{m}^i \mathcal{O}_X/\mathfrak{m}^{i+1}\mathcal{O}_X)\simeq \mathfrak{m}^i/\mathfrak{m}^{i+1}$ via the natural adjunction isomorphism.

We argue that the natural morphism $\mathfrak{m}^k\to \Gamma(\mathfrak{m}^k\mathcal{O}_X)$ is an isomorphism by descending induction on $k$. For large enough $k$, both sides are zero, so this is true. For the induction step, use the diagram $\require{AMScd}$ \begin{CD} 0@>>>\mathfrak{m}^{k+1} @>>> \mathfrak{m}^{k} @>>>\mathfrak{m}^{k}/\mathfrak{m}^{k+1} @>>>0\\ @. @VVV @VVV @VVV @.\\ 0@>>>\Gamma(\mathfrak{m}^{k+1}\mathcal{O}_X)@>>> \Gamma(\mathfrak{m}^{k}\mathcal{O}_X)@>>> \Gamma(\mathfrak{m}^k\mathcal{O}_X/\mathfrak{m}^{k+1}\mathcal{O}_X)@>>>R^1\Gamma(\mathfrak{m}^{k+1}) \end{CD}

Here, the third vertical arrow is the isomorphism by flatness and the projection formula that I claimed above. It follows that the second vertical arrow is an isomorphism by the five lemma.

$\endgroup$
3
  • $\begingroup$ Thanks for your answer! I have a bit of trouble completing the last step regarding the filtration. In particular I don't know why $\mathfrak m^i \Gamma(\mathcal O_X) / \mathfrak m^{i+1} \Gamma(\mathcal O_X) = \Gamma(\mathfrak m^i \mathcal O_X / \mathfrak m^{i+1} \mathcal O_X)$. Could you help with that? $\endgroup$ Sep 7, 2021 at 7:49
  • $\begingroup$ Good point, the rest of the argument wasn't as clear as I initially thought. I expanded a bit, though it essentially side steps the question you are asking. Let me know whether you think it is clear. $\endgroup$ Sep 7, 2021 at 8:28
  • $\begingroup$ Yes, I think that works. Thanks! $\endgroup$ Sep 7, 2021 at 8:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.