Let $f: X \to Y$ be a morphism of schemes. Then $f$ is called (EGA IV.17) *formally smooth* if whenever $T$ is an affine $Y$-scheme and $T'$ a closed subscheme of $T$ defined by a nilpotent ideal (it's sufficient to consider ideals of square zero), any $Y$-morphism $T' \to X$ can be extended to $T \to X$. $f$ is called *formally étale* if such a lifting always exists and is unique. $f$ is called *formally unramified* if such a lifting is unique. I am trying to understand what this should mean geometrically. I don't have a specific question in mind, but I don't see much geometric intuition in the EGA definition, and would appreciate any explanation.

For the geometric intuition that I see, suppose $X,Y$ are algebraic varieties over an algebraically closed field $k$. Let $y \in Y$ be the image of $x \in X$ (consider only closed points for simplicity). Then, formal smoothness states that any tangent vector (i.e. a map $\mathrm{Spec} k[\epsilon]/\epsilon^2 \to Y$) to $y$ lifts to a tangent vector of $x$. Formal unramifiedness states that any tangent vector can lift in only one way to a tangent vector of $x$. Formal étaleness implies that the map on tangent spaces is an isomorphism.

I don't know whether these tangent space conditions are *equivalent* to the formal definitions for varieties. I believe it is true for formal unramifiedness, at least.

(For schemes of finite type over a noetherian scheme, formal unramifiedness is equivalent to the relative sheaf of differentials being zero. This can be checked on the fibers. So let $f: X \to Y$ be a morphism of $k$-varieties. Then $f$ is formally unramified if $\Omega_{X/Y} = 0$. The exact sequence $f^*{\Omega_{Y/k}} \to \Omega_{X/k} \to \Omega_{X/Y} \to 0$ shows that formal unramifiedness holds iff the map on the cotangent spaces is surjective, i.e. the map on the tangent spaces is injective.)

**Question:** Am I correct in thinking of a formally smooth map as a submersion, a formally unramified map as an immersion (in the sense of differential geometry), and a formally étale morphism as a local isomorphism (again, using neighborhoods smaller than the Zariski neighborhoods---I understand when the residue fields are the same étaleness implies that the map on the completions of the local rings are the same)? Is this geometric intuition reasonable?

(This intuition will, of course, ignore the distinction between plain unramifiedness/smoothness/étaleness and the formal analog, which only exists when one leaves noetherian and finite-type hypotheses.)

**Question'** Are the tangent space remarks sufficient for etaleness/smoothness/unramifiedness in the variety case?

smoothcase at $k$-points (as then complete local ring is power series ring over $k$). – BCnrd Nov 14 '10 at 18:57finitein order to "lift" (via Nakayama) the "primitive element thm" description at a point in the fiber to a description on a Zariski-neighborhood in the total space. See EGA IV$_4$, 18.4.6(ii) (beware (i) is false in formally etale case, as I noted elsewhere on MO with counterexample). – BCnrd Nov 14 '10 at 20:34usesthe local structure thm for etale maps. Stick to f. presented ZMT to avoid circularity. Cannot avoid ZMT. That is the engine. – BCnrd Nov 14 '10 at 23:59