Differential rank alone does not prove smoothness

The vanishing, or the local freeness of constant rank, of ${\mathrm{\Omega }}_{X\mathrm{/}S}$ records first-order infinitesimal information about a morphism $f\text{\hspace{0.17em}⁣}:X\to S$. The differential-rank condition of Relative differential-rank condition is therefore not a smoothness criterion. Vanishing of ${\mathrm{\Omega }}_{X\mathrm{/}S}$ is equivalent to formal unramifiedness (Formal unramifiedness iff Omega vanishes), a uniqueness statement about square-zero lifts; it does not by itself imply existence of lifts or flatness. It does have further consequences under finiteness hypotheses: if $f$ is locally of finite type, vanishing of ${\mathrm{\Omega }}_{X\mathrm{/}S}$ makes $f$ unramified (Unramified morphism). The warning here is that differential rank alone does not establish smoothness, not that vanishing differentials carry no geometric information.

The source treatment makes the separation explicitly. Smoothness of a ring map is defined by finite presentation together with a condition on the naive cotangent complex, not by the module of differentials alone; and after defining the relative-dimension condition the Stacks text records that it is not enough to assume that $f$ is flat, of finite presentation, and ${\mathrm{\Omega }}_{X\mathrm{/}S}$ finite locally free of rank $d$: a counterexample is given by $\mathrm{Spec}({\mathbb{F}}_p[t])⟶\mathrm{Spec}({\mathbb{F}}_p[t^p])\mathrm{.}$ That morphism is flat of finite presentation with $\mathrm{\Omega }$ free of rank one, and it is precisely the Frobenius morphism discussed below; the rank of $\mathrm{\Omega }$ is not the fibre-dimension computation and no regularity of the fibres follows from it.

The same distinction appears at the level of tangent maps. A morphism $F\text{\hspace{0.17em}⁣}:{\mathbb{A}}_k^1\to {\mathbb{A}}_k^1$ over ${\mathbb{F}}_p$ can have $\mathrm{d}F=0$ as a map between the absolute modules $F^{\ast }{\mathrm{\Omega }}_{{\mathbb{A}}_k^1\mathrm{/}k}\to {\mathrm{\Omega }}_{{\mathbb{A}}_k^1\mathrm{/}k}$ (Differential of an S-morphism), so that its dual fibre map vanishes at every point (with the target cotangent space extended to the source residue field), while the relative module ${\mathrm{\Omega }}_{{\mathbb{A}}_k^1\mathrm{/}{\mathbb{A}}_k^1,F}$ of the morphism is nonzero, so that $F$ is not formally unramified and hence not formally etale. The companion examples page records this Frobenius witness; the moral is that a zero map on absolute differentials checks a different module from the one whose vanishing would give formal unramifiedness, and that a rank computation may not be substituted for the flatness, finiteness and fibre hypotheses that enter smoothness. In particular, no item on this page may conclude smoothness from a differential rank computation alone.