If $\Omega_{X/Y}$ is locally free of rank $\mathrm{dim}\left(X\right)-\mathrm{dim}\left(Y\right)$, is $X\rightarrow Y$ smooth?

Question

Suppose I have a morphism $f:X\rightarrow Y$ such that the relative sheaf of differentials $\Omega_{X/Y}$ is locally free. Does it follow that $f$ is smooth?

The answer is no, but for a silly reason. You could have some non-reducedness $\mathrm{Spec}\left(k\left[e\right]/e^{2}\right)$ over $\mathrm{Spec}\left(k\right)$ has a free sheaf of differentials, but isn't smooth). But what if you add the hypothesis that the rank of $\Omega_{X/Y}$ is $\dim X-\dim Y$?

Edit: As Jonathan points out in his answer, I was careless with my counterexample. It only works if $\mathrm{char}\,k=2$.

Answer 1

I think Ishai's example is close, but one must be a little careful; the normalization of the node is a good example, but the normalization of the cusp is ramified, and the sheaf of relative differentials in that case is not even locally free.

The differential-wise condition you want is this: for the morphism morphism $f: X \to Y$ to be smooth, you need that the sequence

$$0 \to f^* \Omega_Y \to \Omega_X \to \Omega_{X/Y} \to 0$$

be exact and locally split (I can't find a reference that says this is sufficient, so it may not be). In the special case when $\dim X = \dim Y$, $\Omega_{X/Y}$ is 0 if and only if $f$ is unramified. But in this case $f^* \Omega_Y \to \Omega_X$ can still fail to be injective.

Answer 2

Let X = spec A be an affine integral scheme of dimension one which is not smooth. Smoothness may be checked smooth-locally on the source (given U --> V if there exists a W --> U which is smooth and surjective such that W --> V is smooth then U --> V was smooth). Thus, the normalization X~ = spec A~ cannot be smooth over X. But if d: A~ --> M is any derivation of A~ over A and a/b is an element of A~ then d(a/b)= (bda - adb)/(b^2)= 0; this shows that the rel. differentials are zero, hence in particular loc free of fin rank.

Answer 3

A variation on Ishai's example is a closed embedding: its sheaf of relative differentials is $0$, hence free of finite rank, even though it needn't be smooth.

However, $k[e] / e^2$ over $k$ is not actually a counterexample (except in characteristic $2$). The module of relative differentials of $\operatorname{Spec} k[e] / e^2$ over $\operatorname{Spec} k$ is not free if the characteristic of $k$ is not $2$. Let $A = k[e]$ and $B = k[e] / e^2$. Then

$$\Omega_B = \Omega_A (x) B / d(e^2) = k[e] / (e^2, 2e)$$

via the isomorphism $\Omega_A \to A : dt \to 1$. This is not isomorphic to $B$ unless $2 = 0$.

On the other hand, you can conclude that $B$ is smooth if its cotangent complex is a vector bundle in degree $0$. In the case of $k[e] / e^2$, the cotangent complex is

$$ [ I_{B/A} / I_{B/A}^2 \to \Omega_A (x) B ] = [ e^2 A / e^4 A \to B\ de ] $$

in degrees $[-1,0]$ and the differential is the universal derivation. (I write $I_{B/A}$ for the ideal of $B$ in $A$.) Even in characteristic $2$, the differential has a kernel, so the cotangent complex is not concentrated in degree $0$.