# singular $m$-canonical divisors

[remark for v2] I began by considering curves in v1. I am convinced that the answer is positive. Thanks to Jason Starr and abx.

Let $$X$$ be a complex projective variety.
Let $$K_X$$ be its canonical bundle.

Let $$m$$ be an integer.
We assume that a generic element in the linear system $$|mK_X|$$ is a smooth divisor.
Let $$S\subseteq|mK_X|$$ be the set of singular divisors.

Is $$S$$ always a hypersurface, i.e., without irreducible component of codimension $$\geqslant 2$$ ?
If it is the case, could we calculate the degree of $$S$$ ?

• Regarding your general question: I believe that every "tangent defective" variety is uniruled. Every variety with nontrivial pluricanonical sections, i.e., with nonnegative Kodaira dimension, is non-uniruled. Thus, no such variety can be "tangent defective". – Jason Starr Dec 26 '18 at 13:33
• For your first question, you can just "count" the degree of this hypersurface to see that it is not codimension 2, cf. the following MO answer: mathoverflow.net/questions/165672/… – Jason Starr Dec 26 '18 at 13:37
• For cuves,this is all very classical. See for instance Arbarello et al., chap. VIII, §5 ("de Jonquières formula"). – abx Dec 26 '18 at 13:37