# Generic hyperplane and transversal intersection

I'm trying to solve an exercise I.7.7 in Hartshorne's Algebraic Geometry:

7.7. Let $Y$ be a variety of dimension $r$ and degree $d>1$ in $\mathbb{P}^n$. Let $P \in Y$ be a nonsingular point. Define $X$ to be the closure of the union of all lines $\overline{PQ}$, where $Q \in Y, > Q\neq P$.

(a) Show that $X$ is a variety of dimension $r+1$.

(b) Show that $deg X < d$.

I've proved (a); it follows from the fact that there is a dominant rational map from the projective cone over $Y$ to $X$.

However, to prove (b), I'm stuck.

I found that it suffices to prove the following:

Let us consider the set $(\mathbb{P}^n)^*$ of all hyperplanes in $\mathbb{P}^n$ as the projective $n$-space $\mathbb{P}^n$ with its Zariski topology.

Let $W$ be the set of all hyperplanes containing $P$. Then W is a hyperplane in $(\mathbb{P}^n)^*$.

Then, for any (closed) variety $Y \subset \mathbb{P}^n$ which contains $P$ and is nonsingular at $P$, the set

$$\{ H\in W: \mbox{For every irreducible component Z of } Y\cap H, i(Y,H;Z) = 1. \}$$

contains a nonempty open subset of $W$.

Now $i(Y,H;Z)$ is the intersection multiplicity of $Y$ and $H$ along $Z$, i.e., the length of the $\mathcal{O}_Z$-module $\mathcal{O}_Z / (I(Y)+I(H))\mathcal{O}_Z$, where $\mathcal{O}_Z$ is the local ring of $\mathbb{P}^n$ at $Z$.

How can I prove this? Thanks.

Edited: I found that if the ideal $I(Y)+I(H)$ is a radical ideal, then $i(Y,H;Z) = 1$ for all $Z$.

• do you know that the degree of a variety can be expressed as the generic number of intersection points that it has with a linear subspace if complementary dimension? – Jesko Hüttenhain Aug 5 '18 at 15:33
• Yes, and thus the claim is true for $n=2$. But I don't know how to prove it for higher dimension. – Hiro Wat Aug 7 '18 at 5:33