Monomials of linearly independent polynomials remains independent?

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?

• There is a sufficient condition for $P_{n - 1}$ to be a linearly independent set over $R$, assuming $R$ is a field. Let $A$ be an $R$-algebra which is also a unique factorisation domain. For $a \in A$, denote by $\pi(a)$ the set of prime elements of $A$ dividing $a$. Let $E = \{a_1, \dots, a_n\}$ be a subset of $A$ such that $\pi(a_i)$ is not contained in $\bigcup_{j \neq i} \pi(a_j)$ for every $i$ (e.g., the sets $\pi(a_i)$ are pairwise disjoint). Then the set $P_{n - 1}$ consisting of the products of $n - 1$ distinct elements of $E$ is linearly independent over $R$. Commented Jul 4, 2023 at 6:08
• Could you please clarify what $R$ is? It seems likely that your ring of coefficients is the field of real numbers $\mathbb{R}$, but I cannot exclude that you meant something different. Commented Jul 4, 2023 at 6:34
• @luc guyot: Just let R be either real or complex numbers, and thanks for the editing and comments. The sufficient condition looks nice to work on the context of Question 2 where prime factors are just linear polynomials. Are there any references on why $P_{n-1}$ is special? Commented Jul 4, 2023 at 8:36
• The case $P_{n - 1}$ is special in the sense that it relates linear dependence over $K$ to divisibility in the ring of polynomials in an obvious way. Similar criteria can be established for $P_k$ with $k < n - 1$, see my answer below, which is a more general version of my comment. However, this is only a remark with limited bearing and it doesn't refer to homogeneous polynomials specifically. Commented Jul 4, 2023 at 15:43

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 $$K$$.