TITLE: 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 [9 upvotes]: 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 $\mathrm{char}\,k=2$.

REPLY [5 votes]: 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.