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!



Verifying claims in the proof of the Rigidity Lemma (Mumford, GIT)
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 $p^{-1}V\subseteq U$. 


