The theorem $8.15$(p.177) from the Hartshorne's book "Algebraic Geometry" says:
" Let $X$ be an irreducible separated scheme of finite type over an algebraically closed field $k$. Then $\Omega_{X/k}$ is a locally free sheaf of rank $n= \dim \ X$ iff $X$ is nonsingular variety over $k$."
I can't understand where does we use the condition that $X$ is separated.
Hartshorne's definition of variety is an integral separated scheme over $k$. From the assumption on $\Omega_{X/k}$ he can deduce that $X$ is reduced, and since he assumes $X$ separable, he can then deduce that it is integral. But the assumption on $\Omega_{X/k}$ can't give that $X$ is separated. Thus he has to assume this in order to conclude that $X$ is a variety. (The fact that it is non-singular then follows from the assumption on $\Omega_{X/k}$.)
If he had worked with a more expansive definition of variety, one that allowed non-separated objects (e.g. what Mumford calls a pre-variety in the Red Book), then he wouldn't need the separated assumption.