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 $\mathrm{char}\,k=2$.



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


