Krull dimension of generic fiber

Question

Let $p$ be a prime and $A=\mathbb{Z}_p[t_1,\ldots,t_n]/I$ a reduced flat and irreducible $\mathbb{Z}_p$-algebra of finite type and of Krull dimension d. Let $e$ be the Krull dimension of the $\mathbb{Q}_p$-algebra $A[\frac{1}{p}]$ whose spectrum is the generic fiber of $X=\operatorname{Spec}(A)$. Let the special fiber $\operatorname{Spec}(A/(p))$ of $X$ be non-empty, so that $p$ is not a unit in $A$. Is it true that $e<d$?

Answer 1

$\newcommand{\spec}[1]{{\mathrm{spec}#1}}$ $\newcommand{\dim}{\operatorname{dim}}$ $\newcommand{\ZZ}{{\mathbb Z}}$

In Eisenbud, Commutative Algebra with a View.. we have Theorem 13.8:

Let $R$ be a noetherian universally catenary domain and $T \supseteq R$ a domain, finitely generated as an $R$-algebra. Further let $Q$ be a prime ideal of $T$ and $P = Q \cap R$. Then

$$\dim T_Q = \dim R_P + \dim K(R) \otimes_R T$$

if $Q$ is maximal among the $Q' \subseteq T$, prime, with $Q' \cap R = P$.

Now set $T=\ZZ_p[t_1,\ldots,t_n]/I$ and $R=\ZZ_p$. Furthermore let $Q$ be the maximal ideal of the special fiber. Then $d = \dim T \geqslant \dim T_Q$.

On the other hand $\dim R_P = 1$ and $\dim K(R) \otimes_R T = e$. So $e + 1 \leqslant d$ and indeed $e < d$.