Suppose $R$ is a Noetherian ring and $I$ a nontrivial ideal of $R$, and $A\_0\to B\_0$ a finite faithfully flat lci map of smooth $R\_0 := R/I$-algebras.


We fix a smooth $R$-algebra $A$ lifting $A\_0$ and assume $A$ and $A\_0$ integral.


Assume $R$ is $I$-adically complete.



> 
> Does there exist a smooth $R$-algebra $B$ with a **finite** $R$-map $A\to B$ lifting $A\_0\to B\_0$?
> 
> 
> 


If a finite $R$-map $A\to B$ exists, then it is finite lci and faithfully flat, because it is an integral extension and so it is a finite surjective lci on spectra.


I'm trying to use [this](https://stacks.math.columbia.edu/tag/07M8) and variations on the theme. I've been able to show that there is a quasi-finite syntomic map $A\to B$ with $B$ not necessarily smooth.



Lifting of flat lci maps
No, I don't think so.


Lemma: Given $R$, $I$, $R\_0 = R/I$, $A$, $A\_0 = A/I$. Assume $2$ is invertible in $R\_0$. If the answer to the question is "yes" then the image of the map $Pic(A) \to Pic(A\_0)$ contains all $2$-torsion of $Pic(A\_0)$.


Proof: Namely, suppose $L\_0$ is an invertible $A\_0$-module of order $2$ in the Picard group of $A\_0$. Choosing an isomorphism $\psi : L\_0 \otimes\_{A\_0} L\_0 \to A\_0$ we can set $B\_0 = A\_0 \oplus L\_0$ with multiplication defined by $\psi$. Then $A\_0 \to B\_0$ is finite etale because $2$ is invertible in $A\_0$ by the branched covering trick, whence $B\_0$ is smooth. If we can find $B$ as desired, then we see that $\det\_A(B) \in Pic(A)$ lifts $L\_0$. QED


So now we just have to find $R$ and $A$ where the $2$-torsion in $Pic(A\_0)$ does not lift. Read on if you want to see what I came up with. To understand it, you might have to recall some facts about the deformation to the normal cone. See example at the end for an explicit case.


Say $R$ is a complete discrete valuation ring with uniformizer $t$ and residue field $k$. Let $n \geq 2$ and let $C \subset \mathbf{A}^n\_k$ be an irreducible closed smooth subscheme such that $Pic(C)$ has lots of $2$-torsion. Denote
$$
b : T \longrightarrow \mathbf{A}^n\_R
$$
the blowing up of $C$ viewed as a closed subscheme by identifying $\mathbf{A}^n\_k$ with a closed subscheme of $\mathbf{A}^n\_R$. Then we have
$$
V(t) = E + X
$$
where $E$ is the exceptional fibre of $b$ and $X$ is the strict transform of $\mathbf{A}^n\_k$, i.e., $X$ is the blowing up of $C$ in $\mathbf{A}^n\_k$. Since $E$ is an anti-ample divisor, we see that $X$ is an ample divisor on $T$. Hence we see that
$$
U = T \setminus X = \text{Spec}(A)
$$
is an affine scheme (complement of an ample divisor on a scheme projective over an affine scheme is affine). It is smooth over $R$ as the non-smooth locus of $b$ is $E \cap X$. The special fibre $U\_0 = \text{Spec}(A\_0)$ is an $\mathbf{A}^c$-bundle over $C$ where $c$ is the codimension of $C$ in $\mathbf{A}^n\_k$. Thus we see that $Pic(A\_0) = Pic(U\_0) = Pic(C)$ has nonzero $2$-torsion elements.


Finally, the generic fibre of $\text{Spec}(A)$ is $\mathbf{A}^n\_K$ where $K$ is the fraction field of $R$. Since $t$ is a prime element of $A$ this implies that $\text{Pic}(A)$ is trivial.


Example: If $C$ is a hypersurface $C = V(\overline{f})$, then I think you just get $A = R[x\_1, \ldots, x\_n, y]/(ty - f)$ where $f$ is a lift of $\overline{f}$. You can verify all the properties of this directly, I guess.


