Okay, now that I am reading a bit more of that page, I see that smoothness can be characterized in more concrete way:
Let $\phi: A \rightarrow B$ be a ring homomorphism making $B$ into a finitely presented $A$-algebra. We say that $\phi$ is standard smooth is $B$ can be realized in the form $B = A[x_1, ... , x_n]/(f_1, ... , f_c)$ for $c \leq n$, such that the image of the polynomial
$$\operatorname{det}\begin{pmatrix} \frac{\partial f_i}{\partial x_j}\end{pmatrix}_{1 \leq i, j \leq c} \in A[x_1, ... , x_n]$$
is a unit in $B$.
Then $f: X \rightarrow S$ is smooth if and only if for every $x \in X$, there exists an affine open neighborhood $U$ of $x$ and an affine open set $V$ of $S$ such that $f(U) \subset V$ and $U \rightarrow V$ corresponds to a standard smooth ring homomorphism. (Lemma 28.32.11).
Also, in the example I asked about, with $R$ a DVR and $X= \operatorname{Spec} A$ a scheme of finite type over $R$, (28.32.3) says that in order to say that $X$ is smooth over $R$, it is sufficient that $X$ be over $R$ and that the generic fibre $X \times_R \operatorname{Spec}K$ and special fibre $X \times_R \operatorname{Spec}k$ be smooth varieties in the usual sense.
Another definition of smoothness (given here): $f: X \rightarrow Y$ is smooth if it is locally of finite presentation, flat, and if for all $y \in Y$, $X \times_Y \operatorname{Spec} \kappa(y)$ is smooth as a scheme over the field $\kappa(y)$. Equivalently, this last condition can be restated as saying $X \times_Y \operatorname{Spec} \overline{\kappa(y)}$ is regular over $\overline{\kappa(y)}$.
Best Answer
I think we have the following (apply it to the case $S = \operatorname{Spec} R$ where $R$ is a local ring and $X$ is a proper $S$-scheme):
Claim: Let $S$ be a topological space containing a point $s \in S$ which is contained in all nonempty closed subsets of $S$. Let $X$ be a topological space admitting a closed continuous map $f : X \to S$. If $f^{-1}(s)$ is connected, then $X$ is connected.
Proof: For any $x \in X$, the closure of $x$ in $X$ contains a point of $f^{-1}(s)$ (i.e. $\overline{\{x\}} \cap f^{-1}(s) \ne \emptyset$) since $f(\overline{\{x\}})$ is a nonempty closed subset of $S$ (hence contains $s$). Let $U_{1},U_{2}$ be two disjoint open subsets of $X$ such that $X = U_{1} \cup U_{2}$. Since $f^{-1}(s)$ is connected, then $f^{-1}(s)$ is contained in (exactly) one of $U_{1},U_{2}$; say $f^{-1}(s) \subseteq U_{1}$; if $U_{2}$ is nonempty, say $x \in U_{2}$, then $\overline{\{x\}} \subseteq U_{2}$ since $U_{2} = X \setminus U_{1}$ is closed, but this contradicts $\overline{\{x\}} \cap f^{-1}(s) \ne \emptyset$.
Remark: The ring $A = \mathbb{Z}_{p}[x]/(x^{p}-1)$ is in fact local: since $A$ is finite over $\mathbb{Z}_{p}$, all maximal ideals lie over $p\mathbb{Z}_{p}$, so the maximal ideals of $A$ correspond to prime ideals of $A \otimes_{\mathbb{Z}_{p}} \mathbb{F}_{p} \simeq \mathbb{F}_{p}[x]/((x-1)^{p})$, which has a unique prime.