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.
$\begingroup$
$\endgroup$
2
-
1$\begingroup$ You do not understand correctly. Even for irreducible varieties, $S_d$ may not be irreducible. For a simple example, take a nodal curve in the plane and $d=1$. $\endgroup$– MohanCommented May 16, 2018 at 19:35
-
$\begingroup$ I think your $S_d$ is the dual variety of, not $X \subset \mathbb{P}^n$, but rather the degree $d$ Veronese re-embedding $\nu_d(X) \subset \mathbb{P}^N$, $N = \binom{n+d}{n}-1$. $\endgroup$– Zach TeitlerCommented May 17, 2018 at 2:37
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
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.
