Some problems related to unirational varieties

Question

I am trying to understand the following contents from the notes and have several questions:

If $X$ is unirational, that is, there is a dominant rational map $\mathbf{P}^N \dashrightarrow X$, then there is a rational dominant map $\mathbf{P}^n \dashrightarrow X$, where $n = \dim X$.

Proof when k is infinite. The idea is to keep replacing $\mathbf{P}^N$ by a general hyperplane. If $\mathbf{P}^N \dashrightarrow X$ is a rational dominant map, then there exists a commutative diagram

where $N \geq n$.

If $N > n$, then consider a general fiber $f^{−1}(x)$ of $f$, which is of dimension $N − n$. Now choose a general hyperplane $H \subset \mathbf{P}_k^N$, which satisfies $H\cap U \neq \emptyset$ and $$\dim H\cap f^{−1}(x)< \dim f^{−1}(x).$$ The map $H \dashrightarrow X$ is still rational dominant, for otherwise all fibers would have dimension greater than $(N −1)−(n−1)$, which is not the case for $H \cap f^{−1}(x)$. Now repeat.

This statement is still true if $k$ is finite.

I got the idea of the proof. But I have some questions about details in the proof:

1. Why the $k$ is infinite or finite matters?

2. Why $N\geq n$? I know it is right intuitively, but how to prove it rigorously?

3. Why is the dimension of every fiber $N-n$?

4. Why the $\dim H\cap f^{−1}(x)$ is strictly less than $\dim f^{−1}(x)$?

5. Why is the $H \cap f^{−1}(x)$ still rational dominant? I cannot understand the verification of the author.

6. How to prove the existence of $H$?

Any hints and reference recommendations are welcomed!

Answer 1

1. It matters because when $k$ is finite, then in general you cannot chose a hyperplane $H$ such that $H\cap U \ne \emptyset$. If $k$ is finite, then you can consider the closed set defined by the (finite) union of all hyperplanes and take its complement.
2. I think you're asking why having a dominant rational map $\mathbb{P}^N \dashrightarrow X$ implies that $N\ge n$, where $n = \dim X$. To see this, note that this map gives a morphism of fields $k(X) \hookrightarrow k(\mathbb{P}^N)$, then look at their transcendence degrees.
3. The dimension of a general fiber is $N-n$. See for example Vakil Theorem 11.4.1.
4. Follows from 6.
5. Suppose $H\dashrightarrow X$ is not dominant, then the closure of its image is a proper closed subset of $X$, which necessarily has smaller dimension since $X$ is irreducible. So you have a dominant rational map $H \dashrightarrow Y \subsetneq X$, where $\dim(Y) < \dim(X) = n$. Then this follows from a general result on dimension of fibers, e.g. an application of Exercise 11.4.A, also from Vakil's notes, tells you that the dimension of the fiber is necessarily greater than or equal to $N-n$. But by construction, $\dim H\cap f^{-1}(x) < N-n$.
6. Choose a point in $f(U)$ such that $\dim f^{-1}(x) = N-n$. Now choose a hyperplane $H$ such that $H$ does not contain any irreducible component of $f^{-1}(x)$ (this also uses $k$ is infinite). Then $\dim H \cap f^{-1}(x)$ is necessarily smaller that $\dim f^{-1}(x) = N-n$.