Universal hyperplane section and nondegeneracy of general hyperplane section I have a question about Exercise 18.11 In Harris' book Algebraic Geometry, on page 231:
Give a proof of the nondegeneracy of the general hyperplane
section of an projective irredicible nondegenerated variety $X$ of degree $\ge 2$ (...that's just to avoid the case $X$ linear; see also EDIT below) over $k= \mathbb{C}$ without invoking the notion of degree (or Bezout's theorem), as follows. First, show that if the general hyperplane section of $X$ spans a $k$-plane, we have a rational map
$$  \varphi: (\mathbb{P}^{n})^* \dashrightarrow  \mathbb{G}(k,n)$$
defined by sending a general hyperplane $H \in  (\mathbb{P}^{n})^*$ to the span of $H \cap X$. Next, use the fact that the universal hyperplane section $\Omega_X$ (for definition see p 43 or below) of $X$ is irreducible to deduce that for any
$H \in  (\mathbb{P}^{n})^*$ and any point $\Lambda \in \varphi(H)= \Gamma_{\varphi} \cap (\{H\} \times \mathbb{G}(k,n))$ (that is, any point in the image of the fiber of the
graph $\Gamma_{\varphi}$, over $H$), the hyperplane section $H \cap X$ lies on the $k$-plane $\Lambda$. It follows that if the general hyperplane section of $X$ is degenerate, then all are - but any $n$
independent points of $X$ will span a hyperplane $H$ with $X \cap H $nondegenerate, contradiction.
Remark: The universal hyperplane section $\Omega_X \subset X \times (\mathbb{P}^{n})^*$ is defined as subvariety $\{(p, H) \ \vert \ p \in H \cap X \}$ und is irreducible (Theorem 5.8, page 53).
Question: I not understand how the irreducibility of $\Omega_X$ can be related to this problem, ie to use it here to conclude that $H \cap X \subset \Lambda$ for every $\Lambda \in \varphi(H)$.
Some thoughts: Let $U \subset (\mathbb{P}^{n})^*$ the open dense subset where $\varphi$ is regular. Since $\Gamma_{\varphi}$ is closure of the graph $\Gamma_U:=\{(H, \varphi(H) \ \vert \ H \in U \}$, it follows that $\Gamma_{\varphi}$ is irreducible and it suffice to find a closed subset of $\Gamma_{\varphi}$ which contains $\Gamma_U$.
A natural choice seems to take the closed subset $A:= \{(H, \Lambda) \ \vert H \cap X \subset \Lambda \}$ and intersect it with $\Gamma_{\varphi}$. By contruction it contains $\Gamma_U$.
But this approach nowhere makes use of the universal hyperplane section $\Omega_X$. So my question is not how to prove the claim somehow, but how to argue as Harris suggested using explicitly the irreducibility of the universal hyperplane section.
EDIT 1: as Libli noticed the exercise is flawed if we not exclude the case that $X$ is linear, let's do it and add the additional assumption that the degree of $X$ is $\ge 2$.
EDIT 2: actually I noticed that it could happen that that the set $A:= \{(H, \Lambda) \ \vert H \cap X \subset \Lambda \}$ might be not always closed inside $(\mathbb{P}^{n})^* \times \mathbb{G}(k,n)$; so my approach is flawed, but the concern of the question stays the same, how is it possible to deduce the exercise's Claim exploiting irreducibility of $\Omega_X$?
My guess is that presumably the argument I'm seeking for might work like that there exist certain dense subset $B \subset \Omega_X$ carrying informations about $\Gamma_{\varphi}$ and showing equality $B = \Omega_X$ would imply consequently the claim that $ H \cap X \subset \Lambda$ for all $ \Lambda \in \varphi(H)$. And the task becomes to show that this $B$ is closed...
Here is also the most naive candidate: set $B:= \{(p,H) \in \Omega_X \ \vert \ \forall \Lambda \in \varphi(H): p \in \Lambda \}$. As I said, that's a very naive guess :) But it is of course dense in $\Omega_X$ and the big question becomes if this subset is closed?
What I can show that the set $ \{(p,H) \in \Omega_X \ \vert \ \exists \Lambda \in \varphi(H): p \in \Lambda \}$ is closed as image under proper projection map $p_{12}: X \times (\mathbb{P}^{n})^* \times \mathbb{G}(k,n) \to X \times (\mathbb{P}^{n})^* $ of closed subset
$$ p_{12}^{-1}(\Omega_X) \cap 
p_{13}^{-1}(\{ (p,\Lambda) \ \vert \ p \in \Lambda \}) \cap p_{23}^{-1}(\Gamma_{\varphi})  $$
but the conclusion that this set indeed equals $\Omega_X$ implies just that for every $H$ there exist some $\Lambda \in \varphi(H)$ with $ H \cap X \subset \Lambda$, but that's not enough.
 A: The result of this exercice is correct (and well-known). I am however not sure that the intermediate result used in the exercice is true.
Take $X \subset \mathbb{P}^2$ be a line. Then for generic $H \in \left(\mathbb{P}^2\right)^*$, the intersection $X \cap H$ is degenerate (it is a point in a $\mathbb{P}^1$). But for $H = X$, the intersection $H \cap X$ is not degenerate in $H$. I guess the problem comes from the definition of $\{(H,\Lambda), \ H \cap X \subset \Lambda \}$ as a closed subset. You can always define it as the closure of some set, but then, at the boundary, things will get messy. In the example above, then all couples $(X,x)$ for $x \in X$ will be included in the closure, but of course, $X \cap X \not\subset x$.
I believe a better approach to prove the result of the exercice is the following. Let $J_X$ be the ideal sheaf of $X$ in $\mathbb{P}^n$. Then $h^0(\mathbb{P}^n, J_X(1))$ is the number of independant points in $\left(\mathbb{P}^n\right)^*$ which contains $X$. If $X \subset \mathbb{P}^n$ is not degenerate, then for all $H \in \left(\mathbb{P}^n\right)^*$ the intersection $X \cap H$ is proper (that is has the right codimension) and so we have $J_{X \cap H} = J_X \otimes \mathcal{O}_H$, where $J_{X \cap H}$ is the ideal sheaf of $X \cap H$ in $H$.
Assume now that for generic $H \in \left(\mathbb{P}^n \right)^*$, the section $X \cap H$ is degenerate, that is $h^0(H, J_X \otimes \mathcal{O}_H(1)) > 0$. By the semi-continuity Theorem for $h^0$, we get that for all $H \in \left(\mathbb{P}^n \right)^*, \ h^0(H, J_X \otimes \mathcal{O}_H(1)) > 0$, that is for all $H \in \left(\mathbb{P}^n \right)^*$, the section $X \cap H$ is degenerate. And now you can conclude as in the exercice.
