The question comes from my attempt to understand the following question. height of contracted prime ideals in power series rings

$\bullet$ My original question: Let $(R,m)$ be a Noetherian local ring and $\hat{R}$ the completion of $R$ with respect to $m$. Let $Q$ be a prime ideal in $\hat{R}$ and $P = Q \cap R$. Then is height (codimension) of $Q$ equal to height of $P$?

$\bullet$ Remark: The inequality ht $Q \ge$ ht $P$ follows from the going-down property.

In particular, I wanted to know an answer to the question in the following setting.

Setting: Let $R$ be the localization of a polynomial ring over a field at the origin, i.e., $R = k[x_1,\dots,x_d]_{(x_1,\dots,x_d)}$. Let $\hat{R}$ be the completion of $R$ with respect to its maximal ideal $m := (x_1,\dots,x_d)$. After observing that the question has a positive answer when $R = \mathbb{C}[x]_{(x)}$, I was optimistic about the question. However I found that Matsumura (1988) showed that the formal fiber is of dimension $d-1$. I was surprised by the result since I thought polynomial rings over a field is one of the nicest rings.

The result of Matsumura says that the following corollary in Hartshorne's Algebraic Geometry book does not hold true without the assumption that $X$ is of finite type over a field; let $R := \mathbb{C}[x_1,\dots,x_d]_{(x_1,\dots,x_d)}$. Let $X = \hbox{Spec} (\hat{R}), Y = \hbox{Spec} (R)$, and the morphism $f: X \rightarrow Y$ is induced by the natural ring homomorphism from $R$ to $\hat{R}$. Since $\hat{R}$ is a domain, $\hbox{Spec} (\hat{R})$ is reduced and irreducible. Hence it has only one irreducible component which is of dimension $d$. However the fiber $X_y$ has dimension $d-1$ where $y \in Y$ is the point corresponding to the zero ideal of $R$.

$\bullet$ Corollary 9.6 (Chapter 3 of Hartshorne's Algebraic Geometry book)

Let $f: X \rightarrow Y$ be a flat morphism of schemems of finite type over a field $k$, and assume that $Y$ is irreducible. Then the following conditions are equivalent:

(i) every irreducible component of $X$ has dimension equal to $\dim Y + n$;

(ii) for any point $y \in Y$ (closed or not), every irreducible component of the fibre $X_y$ has dimension $n$.

I would like to ask the following two questions.

$\bullet$ Question 1: is there an algebraic or geometric intuition (or explanation) why the dimension of $X_y$ is $d-1$?

$\bullet$ Question 2: how restrictive is the assumption of being of finite type over a field in statements in algebraic geometry? For instance, what are examples of "good" or "big" statements in Hartshorne's Algebraic Geometry book (or from other sources) fail if one removes the condition?