Let $k$ be a field of characteristic zero and $R$ be the polynomial ring $k[x_1,...,x_n,t_1,...,t_n]$. Let $P_i = (a_{i1}:a_{i2}:a_{i3})$ be $n$ points in the projective plane over $k$, such that not all of them lie on a line. Now define $t = t_1\cdots t_n$, $w_i = x_i t t_i^{-1}$ for $1 \leq i \leq n$, and $z_j = \sum_{i=1}^n a_{ij} w_i$ for $1\leq j \leq 3$. I want to show that $z_1,z_2,z_3,$ and $t$ are algebraically independent over $k$. Note that since the points $P_i$ do not all lie on a line, we have that $z_1,z_2,z_3$ are linearly independent.
Given any relation $F(z_1,z_2,z_3,t) = 0$ with coefficients in $k$ (i.e. a polynomial $F \in k[X_1,X_2,X_3,X_4]$) we can obtain another relation $F'(x_1,...,x_n,t_1,...,t_n) = 0$ with coefficients in $k$, by using the above definitions of $z_1,z_2,z_3,$ and $t$ in terms of the variables in $R$. Since these variables are algebraically independent over $k$, the coefficients of $F'$ vanish. But these coefficients are polynomials in the coefficients of our original relation $F$. Can we use this to show that the coefficients of $F$ vanish? Or, more generally, do you think this is the right way to proceed?
It seems that in order to push this argument further I need to determine the specific relationship between the coefficients of $F'$ and those of $F$, but this would be messy.
