Rank of normal sheaf in singular varieties Let $X$ be a $n$-dimensional normal, singular projective variety (over the field of complex numbers), and let $Y$ be a subvariety of $X$ of dimension $k$. I know that, if $X$ was non-singular, than the normal sheaf $N_{Y\mid X}$ would have been free of rank $n-k$.
In the singular case, can I still say something about the rank of the normal sheaf? Do we have at least $\text{rank}(N_{Y\mid X})\leq n-k$ (I think so, but I've never worked with singular varieties)?
I'm in particular interested in the case  $Y$ a divisor (that is, a codimension $1$ subvariety): is his normal sheaf a line bundle, or can strange things happen?
 A: If $i:Y\to X$ is a closed immersion so that $i(Y)$ is cut out by the sheaf of ideals $\mathcal{I}$, then $i^*(\mathcal{I}/\mathcal{I}^2)$ (often denoted just $\mathcal{I}/\mathcal{I}^2$ by abuse of notation) is the conormal sheaf. When $Y$ is a Cartier divisor, $\mathcal{I}$ is locally principal, and the conormal sheaf is a line bundle on $Y$. Here's why: $i^*(\mathcal{I})\cong i^*(\mathcal{I}/\mathcal{I}^2)$ and $\mathcal{I}$, being locally principal, is a line bundle. Since the pullback of a line bundle is a line bundle, the conormal sheaf is a line bundle, and its dual, the normal sheaf, is also a line bundle.
More generally, for any closed immersion which is locally a complete intersection, the normal/conormal sheaves are vector bundles by the same argument.

As for your more general question about the general properties of the normal sheaf, the appropriate tool is the conormal exact sequence: given an immersion of schemes $i:Y\to X$ over some base $S$, there is a canonical exact sequence $$ N^\vee_{Y/X}\to i^*\Omega_{X/S} \to \Omega_{Y/S}\to 0$$ where the first term is the conormal sheaf. If the composite $Y\to S$ is smooth or formally smooth, this is exact on the left as well (ref). This mostly invalidates your first paragraph. (If you're interested in more details, please leave a note - I am about to leave for a short time and figured it was better to get you most of your answer rather than make you wait.)
