There is a following result in Mumford's red book of schemes (Chapter II Section 8). Here $R$ is a valuation ring with algebraically closed fraction field $k$.
Let $Z \subset \mathbb{P}^n_k$ be an irreducible closed subset of dimension $r$ and let $\mathcal{Z} = \overline{ \{ i(Z)\} }$, where $i: \mathbb{P}^n_k \to \mathbb{P}^n_R$. Thee exist $(r+1)$ linear forms defined by $\ell_0 = ... = \ell_r = 0$, which we denote by $\mathcal{L}$, is disjoint from $\mathcal{Z}$. Let $\tau$ be the projection $\mathbb{P}^n_R \backslash \mathcal{L} \to \mathbb{P}^r_R$. Then giving any structure of closed subscheme, $\tau: \mathcal{Z} \to \mathbb{P}^r_R$ is a finite surjective morphism.
Then later on he applies the Going-Down theorem (Question about the statement of the going-down theorem of Cohen-Seidenberg in Mumford) to this morphism $\tau: \mathcal{Z} \to \mathbb{P}^r_R$. One of the conditions is that no non-zero element of $O_{\mathbb{P}_R^r, \tau(x)}$ is a $0$-divisor in $O_{\mathcal{Z},x}$. I have been trying to prove that this condition is satisfied, and I would greatly appreciate any explanation of this. Thank you.
Below is my attempt:
By restricting to affine open $D(\ell_j)$, where $\ell_j \neq 0$, without loss of generality it suffices to consider
$$
\operatorname{Spec}{R[\frac{X_0}{\ell_0}, \ldots, \frac{X_n}{\ell_0}]/I} \to
\operatorname{Spec}{ R[\frac{X_0}{\ell_0}, \ldots, \frac{X_n}{\ell_0}]} \to
\operatorname{Spec}{R[\frac{Y_1}{Y_0}, \ldots, \frac{Y_r}{Y_0}]},
$$
where $D(\ell_0) \cap \mathcal{Z} = \operatorname{Spec}{ R[\frac{X_0}{\ell_0}, \ldots, \frac{X_n}{\ell_0}]/I}$.
Thus it suffices to consider the ring homomoprhisms
$$
\tau^{\#}: R[\frac{Y_1}{Y_0}, \ldots, \frac{Y_r}{Y_0}] \to R[\frac{X_0}{\ell_0}, \ldots, \frac{X_n}{\ell_0}] \to R[\frac{X_0}{\ell_0}, \ldots, \frac{X_n}{\ell_0}]/I,
$$
where the first map is defined by
$$
\frac{Y_j}{Y_0} \mapsto \frac{\ell_j}{\ell_0}.
$$
Let $Q$ be a prime ideal of $R[\frac{X_0}{\ell_0}, \ldots, \frac{X_n}{\ell_0}]/I$ and $Q' = (\tau^{\#})^{-1}(Q)$. Then we consider the map of stalks by looking at the following commutative diagram $\require{AMScd}$ \begin{CD} R[\frac{Y_1}{Y_0}, \ldots, \frac{Y_r}{Y_0}] @>>> R[\frac{X_0}{\ell_0}, \ldots, \frac{X_n}{\ell_0}]/I \\ @VVV @VVV\\ R[\frac{Y_1}{Y_0}, \ldots, \frac{Y_r}{Y_0}]_{Q'} @>>> (R[\frac{X_0}{\ell_0}, \ldots, \frac{X_n}{\ell_0}]/I)_{Q}. \end{CD} But then it seems possible that for some $j$, $Y_j/Y_0 \mapsto \ell_j/\ell_0$ and this $\ell_j/\ell_0$ may be in $I$. And in this case we don't have that no non-zero element of $O_{\mathbb{P}_R^r, \tau(x)}$ is a $0$-divisor in $O_{\mathcal{Z},x}$...
