Are deformations of quotients of local rings embedded?

Question

In Hartshorne's book "Deformation Theory" one can find a statement (inside the proof of Theorem 10.1) that every deformation $X' \to Spec(C)$ of an affine scheme $$X = Spec(k[x_1,\ldots,x_n]/(f_1,\ldots,f_i)) \to Spec(k)$$ for $C \to k$ is in fact isomorphic to an embedded deformation, i.e., it arises from lifting of the elements $f_j$ to $C[x_1,\ldots,x_n]$.

Is this also true for closed subschemes of regular local rings? Namely, if one fixes a regular local rings $R/k$ together with a flat $C$-lifting $R'/C$ of $R$. Does any flat lifting of $Spec(R/(f_1,\ldots,f_i))$ necessarily come from a lifting of $f_j$'s to $R'$? Maybe, there are some special cases concerning $R$ being essentially of finite type over some field $k$? Is there any additional condition one can impose on the sequence $(f_1,\ldots,f_i)$ (regular sequence? for example) to make such a statement true?

Answer 1

Not true in general. For example, say $k$ is a non perfect field and $C = k[\epsilon]$ (ring of dual numbers) and $R = k'$ is a finite extension of $k$ and $R' = k'[\epsilon]$ with obvious structure of $C$-algebra. Then your statement would mean that every deformation of $k'$ would be trivial which isn't true in general. For example if $k' = k[x]/(x^p - \lambda)$ for a non-$p$th power $\lambda$ of $k$, then the deformation $C[x]/(x^p - \lambda - \epsilon x)$ is not a trival deformation.

What is sufficient is to assume that $R$ is formally smooth over $k$ as defined in Definition Tag 07EB. I am not going to explain in full why this is sufficient. There are some criteria that allows one to check this (Proposition Tag 07NQ and Proposition Tag 07PM) but the easiest thing is to remember that it is true if (1) if $R$ is the localization of a smooth $k$-algebra, or if (2) $R$ is a regular local ring and the residue field of $R$ is a separable extension of $k$.

Conclusion: in characteristic zero it works. Good luck!