TITLE: Verifying claims in the proof of the Rigidity Lemma (Mumford, GIT)
QUESTION [6 upvotes]: 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 $s \in S$.

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.
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!

REPLY [2 votes]: 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.