Degree of a projective variety

Question

Let $X \subset \mathbb{P}^n$ be a projective variety of dimension $k <n$. By an equivalent definition of dimension, $k$ is the smallest integer such that there exists an open set of $G(n-k-1,n)$, i.e. an open set of $(n-k-1)$-planes of $\mathbb{P}^n$ that are disjoint from $X$. In other words, the general $(n-k-1)$-plane does not meet $X$.

Question 1: How can we show that there exists an open set of $G(n-k,n)$, such that every $(n-k)$-plane in that open set meets $X$ in zero dimension?

Question 2: Taking a step further, how can we show that the number of points that the general $(n-k)$-plane (of Question 1) meets $X$ is constant? This is known as the degree of $X$.

PS: I am interested in arguments making use of basic principles rather than invoking results from intersection theory. I am aware that if we intersect $X$ with an $(n-1)$-plane that meets $X$ but does not contain it, then the dimension of the intersection is precisely $k-1$. Now, an $(n-k)$-plane is the intersection of $k$ $(n-1)$-planes and so i can intuitively see the existence of the required open set of Question 1. But how to establish this rigorously? Many thanks.

Answer 1

For simplicity I am assuming $X$ is irreducible, if not, just run this argument for each irreducible component. I also assume you are working over an algebraically closed field $\mathbb{k}$ of characteristic 0. There is an incidence correspondence $I_X \subset X \times G(n-k,n)$ defined as

$I_X = \{ (x,H) | x \in X \cap H\}$.

We want to count the dimension of $I_X$. For each point $x\in X$ there is a $G(n-k-1,n-1)$ of $n-k$-planes that go through $x$. The dimension of $G(n-k-1,n-1)$ is $(n-k) \cdot k$, therefore the dimension of $I_X$ is $(n-k) \cdot k + k = (n-k+1)\cdot k = $ dim $G(n-k,n)$. Moreover if $X$ is irreducible, this shows $I_X$ is irreducible.

The projection map from $\pi : I_X \rightarrow G(k,n)$ is surjective (this is because any $n-k$ plane must intersect $X$ at least at one point). A map between two algebraic varieties of the same dimension can only be surjective if the generic fiber contains finitely many points. That is there is an open set $U \subset G(k,n)$ over which $\pi^{-1} U$ is a finite branched cover.

Moreover, there is a discriminant locus $D \subset U$ which is a closed locus over which the map $\pi: \pi^{-1} U \rightarrow U$ branches. (This can be defined as the zero locus of the discriminant of the field extension of $\mathbb{k}(G(n-k,n)) \subset \mathbb{k}(I_X)$.) Over the complement $U - D$ (a nonempty open set in $G(k-n,n)$) the map $\pi: \pi^{-1} (U -D)\rightarrow U - D$ has finite fibers all of the same number.

For $H \in U-D$ the fiber $\pi^{-1} H$ is the intersection $X \cap H$. Thus we see the number of intersections of $n-k$-planes with $X$ is finite and constant on an open set of $G(n-k,n)$.