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Suppose $X$ is a variety of dimension $n$ over $k$ and there exists a dominant rational map $\mathbb{A}^N\dashrightarrow X$, where $N$ can be larger than $n$. Is it true that there is a dominant rational map $\mathbb{A}^n\dashrightarrow X$?

If $k$ is characteristic 0, this is true because we can apply generic smoothness and look at the map of tangent spaces to pick out an $n$-dimensional linear subspace $\mathbb{A}^n$ of $\mathbb{A}^N$ that dominates $X$.

In characteristic $p$, it seems unlikely, but one might worry that every $n$-dimensional linear subspace of $\mathbb{A}^N$ somehow fails to dominate $X$.

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It seems to me that the following arguement doses not depend on characteristic. Consider a general fiber of $\varphi:\mathbb{A}^N\dashrightarrow X$. Take its closure in $\mathbb{P}^N$. It is a subvariety of dimension $N-n$. Then a general $\mathbb{P}^n$ intersects it in a finite subscheme, so restriction of $f$ to this plane is still dominant.

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    $\begingroup$ What is a "general" $\mathbb{P}^n$ if $k$ is a finite field? $\endgroup$
    – Jason Starr
    Commented Sep 22, 2016 at 15:08
  • $\begingroup$ @JasonStarr Of course, you are right This arguement works only for algebraically closed filed since Hilbert's Nullstellensatz is substantially used. $\endgroup$
    – SashaP
    Commented Sep 22, 2016 at 18:58
  • $\begingroup$ I was thinking of the algebraically closed case (in particular infinite base field), but any insight into whether the map can actually be defined over a finite field is appreciated $\endgroup$
    – DCT
    Commented Sep 23, 2016 at 3:26
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The proof in the case when $k$ is a finite field is given in Lemma 11 in Unirationality of Cubic Hypersurfaces by Kollár, published in Journal of the Institute of Mathematics of Jussieu, Volume 1, Issue 3 (2002). But you can also find the article in

https://arxiv.org/abs/math/0005146

and Lemma 11 is given in page 5.

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