Fix a positive integer $n \ge 2$. Let $\pi:\mathcal{X} \to B$ be a family (flat, projective and surjective morphism) of projective subschemes of $\mathbb{P}^n$. Assume $B$ is reduced, irreducible. Suppose there is a hyperplane $H$ of $\mathbb{P}^n$ and a $b_0 \in B$ such that $H$ intersects $\mathcal{X}_{b_0}$ transversally. Can we then find an open neighbourhood $U \subset B$ containing $b_0$ such that $H$ intersects $\mathcal{X}_b$ transversally for all $b \in U$?
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$\begingroup$ What is your definition of "intersects transversally"? In the literature, there are many different properties that get called "transverse intersection". $\endgroup$– Jason StarrJan 24, 2014 at 22:18
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$\begingroup$ @Starr: I mean codimension of $H.\mathcal{X}_{b_0}$ is one more than the codimension of $\mathcal{X}_{b_0}$ in $\mathbb{P}^n$. $\endgroup$– user45397Jan 24, 2014 at 23:06
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1$\begingroup$ I think that what you call transversal is usually called "proper intersection". $\endgroup$– Damian RösslerJan 25, 2014 at 10:49
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I suppose your assumption on $\pi$ means that $\mathcal X\subseteq \mathbb P^n\times B=:Z$ and $\pi$ is the restriction of the projection to the second factor. Then $H_B:=H\times B$ is a $\pi$-ample Cartier divisor on $Z$. Your assumption on the special fiber implies that $\mathcal X\not\subseteq H_B$, so $\mathcal X\cdot H_B$ is a Cartier divisor on $\mathcal X$ that does not contain $\mathcal X_{b_0}$, but is also not disjoint from it. These two together clearly form an open condition which is also equivalent to the intersection being a Cartier divisor on the fiber, so your desired outcome is true.
answered Jan 25, 2014 at 1:16