Let $P_1,\dots,P_k$ be polynomials over $\mathbf{C}$, no two of them being proportional.

Does there exist an integer $N$ such that $P_1^N,\dots,P_k^N$ are linearly independent?

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    $\begingroup$ Are these univariate or multivariate polynomials? $\endgroup$ – Zach Teitler Jun 19 at 1:31
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    $\begingroup$ I meant univariate, but of course the question is interesting as well for multivariate polynomials. $\endgroup$ – Guillaume Aubrun Jun 19 at 10:29
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    $\begingroup$ Lemma 4.4 of Katz's A conjecture in the arithmetic theory of differential equations has an elementary proof using Vandermonde determinants for the linear multivariate case. $\endgroup$ – AWO Jun 20 at 1:09

The answer is yes. In fact, an even stronger claim is true: there exists some $N$ such that for all $n \geq N, \ P_{1}^{n}, \dots, P_{k}^n$ are linearly independent over $\mathbb{C}$.

For this we will use a generalization of the Mason-Stother's theorem which appears on the Wikipedia page (though I have taken the special case of the curve $C = \mathbb{P}^{1} (\mathbb{C})$ and written it in slightly different language.):

Let $q_1, \dots, q_{k}$ be polynomials such that $q_1 + \cdots + q_{k} = 0$ and every proper subset of $q_1, \dots, q_{k}$ is linearly independent. Then, $$\max \left\{ \mathrm{deg} \left( q_1 \right), \dots, \mathrm{deg} \left( q_{k} \right) \right\} \leq \frac{(k - 1)(k - 2)}{2} \left( \mathrm{deg} \left( \mathrm{rad} \left( q_1 \cdots q_{k} \right) \right) - 1\right)$$

Now, we can prove the claim by induction on $k$. For $k = 2$ it is obvious. Now, by induction for all $n$ large enough every proper subset of $P_{1}^{n}, \dots, P_{k}^{n}$ is linearly independent. Suppose for contradiction that there exist constants $\lambda_{1}, \dots, \lambda_{k}$ such that $$\lambda_1 P_{1}^n + \cdots + \lambda_{k} P_{k}^n = 0$$ Letting $q_i = \lambda_i P_{i}^{n}$, notice that $q_1, \dots, q_k$ satisfy the requirements of the lemma (we have assumed that $\lambda_i \neq 0$), and therefore $$n \leq \max \left\{ \mathrm{deg} \left( q_1 \right), \dots, \mathrm{deg} \left( q_{k} \right) \right\} \leq \frac{(k - 1)(k - 2)}{2} \left( \mathrm{deg} \left( \mathrm{rad} \left( q_1 \cdots q_{k} \right) \right) - 1\right) = \frac{(k - 1)(k - 2)}{2} \left( \mathrm{deg} \left( \mathrm{rad} \left( P_1 \cdots P_k \right) \right) - 1 \right)$$ but the right hand side is constant, and so for $n$ large we get a contradiction.

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    $\begingroup$ I think a more elementary proof using Vandermonde matrices exists, if I have time later I might write it. $\endgroup$ – Random Jun 18 at 14:48
  • $\begingroup$ As a coarse bound, it looks like you can make this explicit with $N = (k-1)(k-2)/2$. $\endgroup$ – Kevin Ventullo Jun 18 at 15:26
  • $\begingroup$ Thanks! I am definitely interested in an elementary proof. $\endgroup$ – Guillaume Aubrun Jun 18 at 15:45
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    $\begingroup$ @Fedor Petrov Does the regular proof with Wronskians generalize to this case? $\endgroup$ – Random Jun 18 at 19:18
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    $\begingroup$ Wronskians? That’s fine for powers of linear polynomials $(at+b)^n$ but it will be messy for powers of higher degree polynomials. $\endgroup$ – Zach Teitler Jun 19 at 1:39

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