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Is every flat morphism of schemes $Xrightarrow S$ such that the fiber over any point is smooth necessarily formally smooth? There are formally smooth morphisms that are not flat so the converse fails. If we assume that the morphism is locally of finite presentation, then it is necessarily smooth so a fortiori formally smooth.

Welcome new contributor. This is too long for a comment. Let $R$ be a DVR with a uniformizing element $t$, e.g., $k[[t]]$ where $k$ is a field. Let $C$ be the $R$-algebra $$C=R[x_n:nin mathbbZ_geq 0].$$ Let $J$ be the ideal in $C$, $$J=langle x_n-tx_n+1 :nin mathbbZ_geq 0 rangle.$$ Let $A$ be the $R$-algebra $C/J$.

Since $A$ has no $t$-torsion, the $R$-module $A$ is $R$-flat. The fiber ring $Aotimes_R (R/t R)$ equals $R/tR$, which is smooth over $R/tR$. The fiber ring $Aotimes_R R[1/t]$ equals $R[1/t][x_0]$, which is smooth over $R[1/t]$.

Now consider the $R$-algebra $B=C/J^2$. Let $I$ be the image ideal in $B$, $$I=J/J^2.$$ This is a square-zero ideal whose quotient algebra equals $A$. Thus, if $A$ is formally smooth, there exists a retraction of $R$-algebras, $$s:Ato B.$$ In particular, $s(x_0)$ is a $t$-divsible element of $B$ that maps to $x_0$ in the quotient algebra $A$.

Consider the quotient of $B$ by the ideal $$K = langle x_mx_n :m,nin mathbbZ_geq 0 rangle.$$ The quotient $B$-algebra equals $Roplus M$ where $M$ is a square-zero ideal that is the free, countably-generated $R$-module with basis $$ x_n : nin mathbbZ_geq 0 .$$ In particular, the element $x_0$ in this quotient algebra is not $t$-divisible in $M$. Thus, there is no $t$-divisible element of $B$ that maps to $x_0$ in the quotient algebra $A$. Therefore, $A$ is not formally smooth over $R$.