Let $X \subset \mathbb{P}^n$ be a reducible, projective subscheme. Assume that $X$ is reduced (meaning that every local ring is reduced i.e., does not contain nilpotent element). Denote by $S\_d$ the locus of degree $d$ hypersurfaces $H$ in $\mathbb{P}^n$ such that $X \cap H$ is singular. Is $S\_d$ irreducible for any $d>0$?
If I understand correctly, this is true if $X$ is irreducible.



Bertini-type theorem for reducible schemes
Take $X = X\_1\cup\dots\cup X\_k\subset\mathbb{P}^n$ be the union of $k$ distinct smooth hypersurfaces with $\deg(X\_i)\geq 2$ for $i = 1,\dots,k$, and $d = 1$. For any point $p\in X\_i$ then intersection $T\_pX\cap X$ is singular at $p$. Therefore, the dual hypersurface $X\_i^{\*}$ is a component of $S\_1$.


Now, if $X\_{i}^{\*} = X\_{j}^{\*}$ then $X\_i = X\_i^{\*\*} = X\_j^{\*\*} = X\_j$. Then $X\_i^{\*} \neq X\_j^{\*}$ if $i\neq j$, and $S\_1$ has at least $k$ irreducible components. 


The irreducibility of $S\_1$ holds if $X$ is smooth since the dual variety of a smooth variety is irreducible.


