Smooth morphism of rings

Remarks

If $A$ is smooth over $R$, then $A$ is flat over $R$, and the module of Kähler differentials $\Omega_{A/R}$ is a finitely generated projective $A$-module.

Example: the polynomial ring $A = R[x_1, \ldots, x_n]$ is smooth over $R$. It is finitely presented, with no relations, and it is formally smooth, because a lift is obtained by lifting the images of the variables.

Non-example: let $k$ be a field and $A = k[x_1, x_2, x_3, \ldots]$ in infinitely many variables. It is formally smooth over $k$ by the same argument, but it is not finitely generated, so clause 1 fails.

Another non-example: $A = k[x, y]/(y^2 - x^3)$ is finitely presented over $k$ but not smooth. Here $\Omega_{A/k}$ is generated by $dx, dy$ with the one relation $2y \, dy - 3x^2 \, dx = 0$. At the maximal ideal $\mathfrak m = (x, y)$, this relation vanishes modulo $\mathfrak m$, so $\Omega_{A/k} \otimes_A A/\mathfrak m$ has dimension $2$. A smooth algebra over a field has differentials that are locally free of rank equal to the local Krull dimension, which is $1$ here.