Let $f: X \rightarrow Y$ be a map of schemes. $f$ is formally smooth if maps $\text{Spec}(A/I) \rightarrow \text{Spec}(A)$ lift against it (not necessarily uniquely), for $A$ a commutative ring and $I$ a nilpotent ideal.
Let $f : X \rightarrow Y$ be a map of schemes. Suppose that maps $\text{Spec}(A/I) \rightarrow \text{Spec}(A)$ lift against it, for $A$ a commutative ring and $I$ a square zero ideal. I suspect this is like being differentiable once. Indeed, it seems we can phrase this in terms of derivations. My question is, under what conditions is $f$ also formally smooth?
This is equivalent to formal smoothness. Indeed, if you can always lift for square zero ideals, then for any ideal $I$ in a ring $A$ you can lift from $A/I$ to $A/I^2$, and then to $A/I^4$, and then to $A/I^8$, and so on. If $I$ is nilpotent, then $I^{2^n}$ will eventually be $0$, so you eventually lift to $A$ itself.