Smooth morphism of schemes

Let $f:X\to S$ be a morphism of schemes, let $x\in X$ and put $s=f(x)$. The morphism $f$ is smooth at $x$ when the following three conditions hold at $x$:

1. $f$ is locally of finite presentation at $x$ (Locally finite presentation morphisms);
2. $f$ is flat at $x$ (Flat morphism of schemes);
3. the scheme-theoretic fibre $X_s$ (Geometric fibres and geometric points) is geometrically regular at $x$, that is, for every field extension $K\mathrm{/}\kappa (s)$ the local ring of $X_s{\times }_{\mathrm{Spec}\kappa (s)}\mathrm{Spec}K$ at every point over $x$ is regular (Geometrically regular algebras and geometrically regular fibres).

The morphism $f$ is smooth if it is smooth at every point of $X$. A morphism with empty source is smooth vacuously, and no condition is imposed by an empty fibre: the pointwise clause of the definition is quantified over primes of the fibre, of which there are none.

Smoothness is a condition on the germ of $f$ at $x$, so it is preserved by restricting the source to an open neighbourhood of $x$ and by shrinking the target to an open neighbourhood of $s$; affine-locally it becomes a condition on a finitely presented ring map at a prime, and the equivalence with the formal lifting and Jacobian formulations is proved on this page, not assumed here. Relative dimension is defined separately, and in relative dimension zero smoothness is equivalent to étaleness later on this page.