# Are large powers of polynomials linearly independent?

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?

• Are these univariate or multivariate polynomials? – Zach Teitler Jun 19 at 1:31
• I meant univariate, but of course the question is interesting as well for multivariate polynomials. – Guillaume Aubrun Jun 19 at 10:29
• 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. – 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.

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