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(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.

2 added 897 characters in body

(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 $(1-a)n=0$.

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.

1

(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 $(1-a)n=0$.

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 $p^{-1}(t)$.