In algebraic geometry and commutative algebra, a ring homomorphism
is called formally smooth (from French: Formellement lisse) if it satisfies the following infinitesimal lifting property: Suppose B is given the structure of an A-algebra via the map f. Given a commutative A-algebra, C, and a nilpotent ideal
, any A-algebra homomorphism
may be lifted to an A-algebra map
If moreover any such lifting is unique, then f is said to be formally étale.
[1][2] Formally smooth maps were defined by Alexander Grothendieck in Éléments de géométrie algébrique IV.
For finitely presented morphisms, formal smoothness is equivalent to usual notion of smoothness.
All smooth morphisms
are equivalent to morphisms locally of finite presentation which are formally smooth.
Hence formal smoothness is a slight generalization of smooth morphisms.
One method for detecting formal smoothness of a scheme is using infinitesimal lifting criterion.
For example, using the truncation morphism
the infinitesimal lifting criterion can be described using the commutative square
Spec
Spec
{\displaystyle {\begin{matrix}X&\leftarrow &{\text{Spec}}\left({\frac {k[\varepsilon ]}{(\varepsilon ^{2})}}\right)\\\downarrow &&\downarrow \\S&\leftarrow &{\text{Spec}}\left({\frac {k[\varepsilon ]}{(\varepsilon ^{3})}}\right)\end{matrix}}}
Spec
{\displaystyle X={\text{Spec}}\left({\frac {k[x,y]}{(xy)}}\right)}
Spec
{\displaystyle Y={\text{Spec}}(k)}
then consider the tangent vector at the origin
given by the ring morphism
, this is a valid morphism of commutative rings.
Then, since a lifting of this morphism to
Spec
{\displaystyle {\text{Spec}}\left({\frac {k[\varepsilon ]}{(\varepsilon ^{3})}}\right)\to X}
, there cannot be an infinitesimal lift since this is non-zero, hence
is not formally smooth.
This also proves this morphism is not smooth from the equivalence between formally smooth morphisms locally of finite presentation and smooth morphisms.