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I asked this question on math stack exchange but did not receive any reaction. I would like to try again here since I am sure this is known.

In the book An invitation to algebraic geometry by Smith et. al. (2000) the following definition is given for the degree of a projective variety:

The degree of the projective variety $V \subset \mathbb{P}^n$ is the greatest possible finite number of intersection points of $V$ with a linear subvariety $L \subset \mathbb{P}^n$ of dimension equal to the codimension of $V$.

Then it is added:

In fact, the maximal number of intersection points is almost always achieved: The degree of $V$ is the number of points common to $V$ and a $(\mathrm{codim} V)$-dimensional generic linear subvariety. One should interpret the word "generic" here to mean the intuitive idea of a typical, representative, or "sufficiently general" linear subvariety. To make this idea precise, the reader should prove that there is a dense open subset $U$ of the Grassmannian of all $(\mathrm{codim} V)$-dimensional subspaces of $\mathbb{P}^n$ such that for any $\Lambda$ in this open set, $V \cap \Lambda$ consists of precisely $d = \mathrm{deg} V$ distinct points. In this case, "generic" would mean simply "member of $U$."

I am having troubles proving this claim. There is a similar question here from which I got some ideas but I couldn't finish an argument. Here is my attempt.

Let $k = \mathrm{codim} V$. Let $\mathrm{Gr}(k+1,n+1)$ be the Grassmannian of $(k+1)$-dimensional planes in $\mathbb{A}^{n+1}$ (= $k$-planes in $\mathbb{P}^n$). First, one can show that always $V \cap L \neq \emptyset$ for any $k$-plane $L$ in $\mathbb{P}^n$ using the projective dimension theorem. I find this is somehow also missing in the book.

Now, consider the incidence correspondence $X = \lbrace (v,L) \in V \times \mathrm{Gr}(k+1,n+1) \mid v \in L \rbrace \subset V \times \mathrm{Gr}(k+1,n+1)$. This is a projective variety. The projection $\pi \colon X \to \mathrm{Gr}(k+1,n+1)$ is surjective by the observation above. Moreover, $\pi^{-1}(L) = V \cap L$. So, the degree is about fiber cardinality of $\pi$. It is now a general property of surjective morphisms $f \colon X \to Y$ of varieties that there is a non-empty open subset $U$ of $Y$ such that $\mathrm{dim} f^{-1}(y) = \mathrm{dim}X - \mathrm{dim} Y$. This fact is not covered in the book.

One can compute that $\mathrm{dim} X = \mathrm{dim} \mathrm{Gr}(k+1,n+1)$ here, so we conclude that there is a non-empty open (thus dense) subset $U$ of $\mathrm{Gr}(k+1,n+1)$ such that $\mathrm{dim}(V \cap L) = 0$ for all $L \in U$. This means $\pi$ is generically finite. I believe seeing a statement in the stacks project that one can take $U$ to be the set of points with finite fibers indeed, and then $\pi|_{\pi^{-1}(U)}$ is a finite morphism.

But now my problem is to show that on an open subset of $U$ the cardinality of $V \cap L$ is indeed the maximal cardinality of all finite intersections. I do not see this at all. If $\pi$ were flat, then it is a general property that fiber cardinality is always $\leq$ than the generic fiber cardinality. But I do not see why $\pi$ should be flat. And in general it is not true that the generic fiber cardinality is the maximal fiber cardinality. And somehow all this also goes beyond the material of the book.

So, how does one prove the claim?

Is the claim in the book even correct?

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  • $\begingroup$ One way to argue is as follows: show that $X$ is a variety (i.e. integral) since the other projection $X \to V$ is a 'Grassmannian bundle' and $V$ is integral. If $\operatorname{char} k = 0$, this implies that $\pi$ is generically smooth, so there exists a dense open $U$ such that $\pi|_{\pi^{-1}(U)}$ is finite étale, so in particular flat with reduced geometric fibres. Then the number of points in the fibre is constant. $\endgroup$ Commented May 3 at 8:27
  • $\begingroup$ How do you know that there is no point outside of U with finite fibers and possibly larger fiber cardinality? The claim in the book is really that the generic fiber cardinality is the maximal finite fiber cardinality. $\endgroup$
    – user527391
    Commented May 3 at 12:46
  • $\begingroup$ Ah, good point. I don't know actually, and it does make the statement sound a little fishy. I wonder if there's some other argument we're missing... $\endgroup$ Commented May 3 at 21:41
  • $\begingroup$ @R.vanDobbendeBruyn: how about the Hilbert polynomial argument that I sketched in a MSE comment (but never got around to fleshing out)? I’m a bit torn about it, since it seems to work – but it also seems too easy somehow? $\endgroup$
    – Aphelli
    Commented May 4 at 13:39
  • $\begingroup$ @Aphelli ah, I didn't see that. I think you're right that this should be a better strategy. This will prove that whenever the intersection is zero-dimensional, its length will be the leading term of the Hilbert polynomial of $V$. Then use the generic smoothness to conclude that on a nonempty open, the zero-dimensional scheme you get is reduced, so its length equals its number of points. $\endgroup$ Commented May 4 at 15:14

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