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If we have a "graded" sequence of polynomials $g_1\in {\mathbb C}(x_1),g_2\in {\mathbb C}(x_1,x_2),g_3\in {\mathbb C}(x_1,x_2,x_3) \ldots, g_n\in {\mathbb C}(x_1,x_2,\ldots,x_n)$ such that $g_i$ is non-constant in $x_i$ for each $i$, I call the map $f : {\mathbb C}^n \to {\mathbb C}^n$ defined by

$$ f(x_1,x_2,\ldots,x_n)=(g_1(x_1),g_2(x_1,x_2),\ldots,g_n(x_1,x_2,\ldots,x_n)) $$

a flag transformation. For any polynomial $P\in {\mathbb C}(x_1,x_2,\ldots,x_n)$, one can compose $P$ with $f$ to obtain another polynomial which I denote by $P\circ f$ :

$$ P\circ f(x_1,x_2,\ldots,x_n)= P(g_1(x_1),g_2(x_1,x_2),\ldots,g_n(x_1,x_2,\ldots,x_n)) $$

Given a nonzero polynomial $P$, can we always find a flag transformation $f$ such that $P\circ f$ has no zeroes on ${\mathbb Z}^n$ ?

This is clearly true for $n=1$, because in that case the zeroes of $P$ are bounded in some interval $[-M,M]$ and we can take $g_1(x_1)=M(1+x_1^2)$.

Ewan Delanoy
  • 61,600

1 Answers1

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We'll prove that it is always possible.

Take the polynomial $P$ and write it as

$$ \sum_r x_n^r P_r(x_1, \dots, x_{n-1}) $$

for polynomials $P_0, \dots, P_k$ not all zero. The linear dependences between the $P_r$ in $\mathbb{C}[x_1, \dots, x_{n-1}]$ form a linear subspace of $\mathbb{C}^{k+1}$, whose intersection with the set $\{(1, t, \dots, t^k): t \in \mathbb{R}\}$ is finite. In particular, we can choose $a_n \in \mathbb{R}$ such that no solution $t$ lies in $\mathbb{Z}+a_n$. Applying the flag transformation $x_n \mapsto x_n + a_n$, we get that $P(x_1, \dots, x_{n-1}, m)\neq 0$ in $\mathbb{C}[x_1, \dots, x_{n-1}]$ for all $m\in \mathbb{Z}$.

Now for each fixed value of $m$ we can apply the same argument to get a finite number of values of $t$ for which $P(x_1, \dots, x_{n-2}, t, m)=0$ in $\mathbb{C}[x_1, \dots, x_{n-2}]$. Taking the union over $m$, we have a countable collection of 'bad' $t$ values. We can now apply a translation in $x_{n-1}$ to make these bad values non-integers.

Keep going iteratively to get a flag transformation (in fact, a translation) which pushes all zeros of $P$ off the integer lattice.