Formally smooth map

All smooth morphisms ${\displaystyle f:X\to S}$ are equivalent to morphisms locally of finite presentation which are formally smooth. Hence formal smoothness is a slight generalization of smooth morphisms[3].

One method for detecting formal smoothness of a scheme is using infinitesimal lifting criterion. For example, using the truncation morphism ${\displaystyle k[t]/(t^{3})\to k[t]/(t^{2})}$ the infinitesimal lifting criterion can be described using the commutative square

${\displaystyle {\begin{matrix}X&\leftarrow &{\text{Spec}}\left({\frac {k[x]}{(x^{2})}}\right)\\\downarrow &&\downarrow \\S&\leftarrow &{\text{Spec}}\left({\frac {k[x]}{(x^{3})}}\right)\end{matrix}}}$

where ${\displaystyle X,S\in Sch/S}$. For example, if

${\displaystyle X={\text{Spec}}\left({\frac {k[x,y]}{(xy)}}\right)}$ and ${\displaystyle Y={\text{Spec}}(k)}$

then consider the tangent vector at the origin ${\displaystyle (0,0)\in X(k)}$ given by the ring morphism

${\displaystyle {\frac {k[x,y]}{(xy)}}\to {\frac {k[t]}{(t^{2})}}}$

sending

${\displaystyle {\begin{aligned}x&\mapsto \varepsilon \\y&\mapsto \varepsilon \end{aligned}}}$

Note because ${\displaystyle xy\mapsto \varepsilon ^{2}=0}$, this is a valid morphism of commutative rings. Then, since a lifting of this morphism to

${\displaystyle {\text{Spec}}\left({\frac {k[t]}{(t^{3})}}\right)\to X}$

is of the form

${\displaystyle {\begin{aligned}x&\mapsto \varepsilon +a\varepsilon ^{2}\\y&\mapsto \varepsilon +b\varepsilon ^{2}\end{aligned}}}$

and ${\displaystyle xy\mapsto \varepsilon ^{2}+(a+b)\varepsilon ^{3}=\varepsilon ^{2}}$, there cannot be an infinitesimal lift since this is non-zero, hence ${\displaystyle X\in Sch/k}$ 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.