Lemma 35.29.10. Let $k$ be a field. Let $n \geq 2$. For $1 \leq i, j \leq n$ with $i \not= j$ and $d \geq 0$ denote $T_{i, j, d}$ the automorphism of $\mathbf{A}^ n_ k$ given in coordinates by

\[ (x_1, \ldots , x_ n) \longmapsto (x_1, \ldots , x_{i - 1}, x_ i + x_ j^ d, x_{i + 1}, \ldots , x_ n) \]

Let $W \subset \mathbf{A}^ n_ k$ be a nonempty open subscheme such that $T_{i, j, d}(W) = W$ for all $i, j, d$ as above. Then either $W = \mathbf{A}^ n_ k$ or the characteristic of $k$ is $p > 0$ and $\mathbf{A}^ n_ k \setminus W$ is a finite set of closed points whose coordinates are algebraic over $\mathbf{F}_ p$.

**Proof.**
We may replace $k$ by any extension field in order to prove this. Let $Z$ be an irreducible component of $\mathbf{A}^ n_ k \setminus W$. Assume $\dim (Z) \geq 1$, to get a contradiction. Then there exists an extension field $k'/k$ and a $k'$-valued point $\xi = (\xi _1, \ldots , \xi _ n) \in (k')^ n$ of $Z_{k'} \subset \mathbf{A}^ n_{k'}$ such that at least one of $x_1, \ldots , x_ n$ is transcendental over the prime field. Claim: the orbit of $\xi $ under the group generated by the transformations $T_{i, j, d}$ is Zariski dense in $\mathbf{A}^ n_{k'}$. The claim will give the desired contradiction.

If the characteristic of $k'$ is zero, then already the operators $T_{i, j, 0}$ will be enough since these transform $\xi $ into the points

\[ (\xi _1 + a_1, \ldots , \xi _ n + a_ n) \]

for arbitrary $(a_1, \ldots , a_ n) \in \mathbf{Z}_{\geq 0}^ n$. If the characteristic is $p > 0$, we may assume after renumbering that $\xi _ n$ is transcendental over $\mathbf{F}_ p$. By successively applying the operators $T_{i, n, d}$ for $i < n$ we see the orbit of $\xi $ contains the elements

\[ (\xi _1 + P_1(\xi _ n), \ldots , \xi _{n - 1} + P_{n - 1}(\xi _ n), \xi _ n) \]

for arbitrary $(P_1, \ldots , P_{n - 1}) \in \mathbf{F}_ p[t]$. Thus the Zariski closure of the orbit contains the coordinate hyperplane $x_ n = \xi _ n$. Repeating the argument with a different coordinate, we conclude that the Zariski closure contains $x_ i = \xi _ i + P(\xi _ n)$ for any $P \in \mathbf{F}_ p[t]$ such that $\xi _ i + P(\xi _ n)$ is transcendental over $\mathbf{F}_ p$. Since there are infinitely many such $P$ the claim follows.

Of course the argument in the preceding paragraph also applies if $Z = \{ z\} $ has dimension $0$ and the coordinates of $z$ in $\kappa (z)$ are not algebraic over $\mathbf{F}_ p$. The lemma follows.
$\square$

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