# Minimality implies algebraic independence?

$$\DeclareMathOperator\supp{supp}$$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.

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.