Smoothness over a characteristic-zero field via free differentials

Regularity of the local ring. Fix $x\in X$ and choose an affine open $\mathrm{Spec}A\subseteq X$ containing $x$, with $A$ a finite-type $k$-algebra and $\mathfrak{q}\in \mathrm{Spec}A$ the prime corresponding to $x$; such a chart exists by [F6]. Local freeness of ${\mathrm{\Omega }}_{X\mathrm{/}k}$ means that on some neighbourhood of $x$ the sheaf restricts to a free module, so ${\mathrm{\Omega }}_{A\mathrm{/}k,\mathfrak{q}}$ is a free $A_{\mathfrak{q}}$-module; the hypothesis of [F2] is therefore satisfied and $A_{\mathfrak{q}}$ is a regular local ring.

A standard smooth chart. The field $k$ has characteristic $0$, hence is perfect by [F7]. As $A$ is a finite-type $k$-algebra and $A_{\mathfrak{q}}$ is regular, [F3] supplies $t\in A\setminus \mathfrak{q}$ with $A_t$ standard smooth over $k$.

Smoothness at $x$. The algebra $A_t$ is finite type over the Noetherian field $k$, hence finitely presented by [F5]. Applying the pointwise criterion [F4] to $k\to A_t$ at $\mathfrak{q}{A}_t$ shows that $k\to A_{\mathfrak{q}}$ is flat and that the fibre $A_t{\otimes }_k\kappa (0)=A_t$ is geometrically regular at $\mathfrak{q}{A}_t$. This is precisely geometric regularity of the fibre of the chart $D(t)\to \mathrm{Spec}k$ at $x$. Its finite presentation also gives local finite presentation at $x$ by [F9]. These three conditions make $X\to \mathrm{Spec}k$ smooth at $x$ by [F1], since smoothness is unchanged on restricting to an open neighbourhood. As $x$ was arbitrary, $X$ is smooth over $k$.

The converse and the equivalence. Conversely, if $X$ is smooth over $k$, then [F8] makes ${\mathrm{\Omega }}_{X\mathrm{/}k}$ locally free of finite rank near every point, hence locally free; this is the stated converse. Combining it with the implication from a locally free ${\mathrm{\Omega }}_{X\mathrm{/}k}$ to smoothness proved above, the two conditions are equivalent over a field of characteristic $0$, and the Axiom of Choice is inherited through [F2], [F3], [F4] and [F8].