Smooth algebras always lift

Question

Let $k$ be a finite field, $A$ a smooth $k$-algebra.

Does there exists a smooth algebra $B$ over the Witt vectors $W(k)$, such that $B/p\simeq A$? How is it constructed?

Answer 1

This follows from a result of Elkik in [R. Elkik Solutions d’équations à coefficients dans un anneau hensélien Annales scientifiques de l’É.N.S. 4e série, tome 6, no 4 (1973), p. 553-603.].

Theorem (Theorem 6 in Section 4 on p. 580, changed notation to fit yours). Let $(R, J)$ be a noetherian henselian couple. Then every smooth $R/J$-algebra $A$ lifts to a smooth $R$-algebra $B$.

Of course if you only want to lift formally over $R=W(k)$, that is to find a system $B_n$ over $W_{n+1}(k)$ lifting $B_{n-1}$ with $B_0 = A$, then the result follows from elementary deformation theory: the obstruction to lifting $B_{n-1}$ over $W_{n+1}(k)$ lies in $H^2({\rm Spec}(A), T_{{\rm Spec}(A)})=0$. But why should the inverse limit of the $B_n$ be the $p$-adic completion of a smooth $W(k)$-algebra?

Similarly, if $A$ is a smooth complete intersection, then it is easy to lift $A$ over $W(k)$: just lift the equations, and then maybe localize if the lifting is not smooth.