Monomials of linearly independent polynomials remains independent?

Question

I met with a very elementary question on linear algebra which turns out to be quite resistant when I work on it. Any comments or references on either solutions or the possible value of the question will be greatly appreciated.

Initial problem: Given $n$ homogeneous polynomials $ F_1,\dots, F_n $ in $R[x_1,\dots, x_m]$ such that $\deg(F_1)= \dots= \deg(F_n)=d$. For a fixed positive integer $k < n$, set $$P_k=\{F_{i_1}F_{i_2} \cdots F_{i_k} \,\vert\, 1\leq i_1 < i_2 < \cdots < i_k \leq n \}$$ If $ F_1,\dots, F_n $ are linearly independent, that is, $P_1$ is a linearly independent set, then is $P_k$ a linearly independent set for $k > 1$?

The question is trivial for $d=1$. In this case monomials in $F_1,\dots,F_n$ are always linearly independent as we can assume $n \le m$ and $F_i = x_i$ up to a coordinate change. On the other hand, the question has a negative answer when $d > 1$ if we do not impose any additional conditions. For instance, set $$F_1=x_1^2, F_2=x_1x_2,F_3=x_1x_3,F_4=x_2x_3,$$ then $F_1,F_2,F_3,F_4$ are independent. However, $P_2$ is dependent since $F_1F_4=F_2F_3$.

Now the question becomes:

Question 1. Assume $d > 1$. Under what condition will $P_k$ be linearly independent?

A first thought is that we restrict to homogeneous polynomials that are powers of linear polynomials. But this seems to be complicated as well. For instance, the question for $P_2$ and $d = 2$ goes as follows:

Question 2. Let $F_1,\dots, F_n \in R[x_1,\dots, x_m]$ be homogeneous polynomials of degree one such that $F_1^2, F_2^2,\dots, F_n^2$ are linearly independent ($F_1,\dots, F_n$ might be linearly dependent in this case). Under what condition will $F_1^2F_2^2, F_1^2F_3^2,\cdots, F_{n -1}^2F_n^2$ be linearly independent?

Answer 1

This is only a long comment with limited bearing. It consists of two ideas of a similar flavour. Each can be used to address specific instances of the problem.

Claim 1. Let $K$ be a field and let $A$ be $K$-algebra with no zero divisors. Let $E$ be subset of $A$ with $n > 1$ elements and let $k < n$ be such that for every subset $S \subset E$ with $n - k$ elements, there is a valuation $\nu = \nu_S$ defined on the field of fractions of $A$, and non-negative on $A$, such that $$\min_{a \in S} \nu(a) > \sum_{a \in E \setminus S} \nu(a).$$ Then the products of $k$ distinct elements of $E$ are linearly independent over $K$.

Example 1. Let $p$ be a prime element of an integral domain $A$ which satisfies the ascending chain conditions on principal ideals. The largest power $\nu_p(a)$ of $p$ which divides $a \in A$ defines a non-negative valuation on $A$ which extends to a discrete valuation on the field of fractions of $A$.

Example 2. Let $K$ be an algebraically closed field and let $\mathfrak{m}$ be a maximal ideal of $A = K[x_1, \dots, x_m]$. Then $\nu_{\mathfrak{m}}(a) := \sup \{n \ge 0 \, \vert \, a \in \mathfrak{m}^n \}$ defines a non-negative valuation of $A$ which extends to a discrete valuation of $K(x_1, \dots, x_m)$.

Proof of Claim 1. A non-trivial $K$-linear dependence relation between $k$-fold products yields a relation of the form $$a_1 \cdots a_k \in Aa_{k + 1} + \cdots + A a_n$$ for some suitable labelling $E = \{a_1, \dots, a_n\}$. Let $S = \{a_{k + 1}, \dots, a_n\}$ and let $\nu = \nu_S$ be the valuation afforded by hypothesis. As $\nu(\sum_{a \in S}Aa) \ge \min_{a \in S} \nu(a)$, we obtain a contradiction.

Claim 2. Let $K$ be field, $A = K[x_1, \dots, x_m]$ and let $\prec$ be a monomial order on $A$. For a polynomial $a \in A$, denote by $\operatorname{LT}(a)$ the leading term with respect to $\prec$ and by $\deg(M) \in \mathbb{N}^m$ the multi-degree of a monomial $M$ of $A$. Let $a_1, \dots, a_n$ be elements of $A$ such that the sums $\deg(\operatorname{LT}(a_i)) + \deg(\operatorname{LT}(a_j))$, with $i < j$, are distinct. Then the products $a_i a_j$, with $i < j$, are linearly independent over $K$.

Proof of Claim 2. By hypothesis, the leading terms $\operatorname{LT}(a_i a_i)$, with $i < j$, are linearly independent over $K$. This ensures that the products $a_i a_j$, with $i < j$, are linearly independent over $K$.