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!

  • $\begingroup$ 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


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 $p^{-1}V\subseteq U$.

  • 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) 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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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