Let $X$ be the cuspidal curve $y^2 = x^3$. This is a singular curve with a unique singularity in the origin. One can easily see this using the Jacobi criterion.
How does one show explicitly that the cotangent bundle $\Omega^1_{X/k}$ is not a locally free sheaf?
Of course, the stalk at the origin should have torsion, but I can't seem to locate it explicitly.
I've wondered about this for a while. The reason being that I never really understood via an example why a variety is smooth if and only if its cotangent bundle is locally free.
The module of Kahler differentials is spanned by $dx$ and $dy$, and differentiating the relation $y^2 = x^3$ gives the relation $2y \, dy = 3x^2 \, dx$. Multiplying by $y$ gives
$$2y^2 \, dy = 2x^3 \, dy = 3 x^2 y \, dx$$
from which it follows that $2x \, dy - 3y \, dx$ is torsion, since it is annihilated by multiplication by $x^2$. (Of course then we need to show that this element isn't just zero, which it is away from the origin.)