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[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 $S$ ?

Improve this question
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  • $\begingroup$ 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". $\endgroup$
    – Jason Starr
    Commented Dec 26, 2018 at 13:33
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    $\begingroup$ 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/… $\endgroup$
    – Jason Starr
    Commented Dec 26, 2018 at 13:37
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    $\begingroup$ For cuves,this is all very classical. See for instance Arbarello et al., chap. VIII, §5 ("de Jonquières formula"). $\endgroup$
    – abx
    Commented Dec 26, 2018 at 13:37

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