The definition of a smooth morphism is too abstract. Can we make it simpler in a special case?

Question

The definition of a smooth morphism of schemes $f:X \rightarrow S \space$ given here on stacks project is so abstract that it is intractable to me.

$f$ is called smooth if for every point $x$ of $X$, there exists an affine open neighborhood $U$ of $x$, and an affine open set $V$ of $S$, with $f(U) \subset V$, such that the induced ring homomorphism $\mathcal O_S(V) \rightarrow \mathcal O_X(U)$ is a smooth ring homomorphism. A smooth ring homomorphism is defined in terms of the "naive cotangent complex."

I just have no idea how to think about or work with this definition. So let me consider a more special case: let $R$ be a discrete valuation ring with quotient field $K$ and residue field $k$. Let $X = \operatorname{Spec} A$ be a scheme of finite type over $R$, say $X \subset \mathbb A_R^n$.

What does it mean to say that $X$ is smooth over $R$? Can we give a simpler criterion than the one given on stacks project? I'm sure a necessary condition is that $A$ is a flat $R$-algebra. Can we talk about the smoothness of $X$ over $R$ in terms of the smoothness of the generic and special fibres? That is, in terms of $X_K = \operatorname{Spec} A \otimes_R K$ and $X_k = \operatorname{Spec} A \otimes_R k$ being smooth varieties over $K$ and $k$?

Answer 1

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 $\overline{\kappa(y)}$.