Minimality implies algebraic independence?

Question

$\DeclareMathOperator\supp{supp}$Given $f_1,...,f_n \in \mathbb{F}[x_1,...,x_n]$ such that

$f_1 = x_1 + q_1$

$f_2 = x_2 + q_2$

$\cdot \cdot \cdot$

$f_{n-1} = x_{n-1} + q_{n-1}$

$f_{n} = q_n$

such that $deg(q_i) = 2$ and $q_i$ is homogenous, for all $i \in \{1,...,n \}$.

For $f = \sum_{l}a_l x_1^{l_1}\cdot \cdot x_n^{l_n}$, define $$\supp(f) = \{j \mid \exists l \in \mathbb{Z}_{\geq 0}^n \,,\; a_l \neq 0 \text{ and } l_j\neq 0 \}.$$

We say $f_1,...,f_n$ satisfy the minimality condition if for $1 \leq r \leq n-1$, any $r$ subset of distinct polynomials $f_{i_1},...,f_{i_r}$ we have $|\bigcup_j\supp(f_{i_j})| \geq r+1$.

My question is whether $f_1,...,f_n$ satisfying the minimality condition implies algebraic independence. I am mostly interested in the case $\mathbb{F} = \bar{\mathbb{F}}_2$ but would welcome insights about any general field.

Answer 1

Here's an example over any field, with five variables. Let $u=x_1+x_2+x_3+x_4$, and let $f_1=x_1+ux_5$, $f_2=x_2+ux_5$, $f_3=x_3-ux_5$, $f_4=x_4-ux_5$, $f_5=x_1x_2+x_2x_3+x_3x_4+x_4x_1$. Then $$f_5=f_1f_2+f_2f_3+f_3f_4+f_4f_1,$$ so $f_1,f_2,f_3,f_4,f_5$ are not algebraically independent. If I'm not mistaken, the supports satisfy your conditions.