My question is as in the title:
Suppose $f_1,...,f_{n+1}\in \mathbb{C}[x_1,....,x_n]$. Is there polynomial $g$ (or a system of polynomials) with variables given by the coefficients of the $f_i$ with the property that $g$ vanishes iff there is a common root of the $f_i$?
For my particular application, I don't need the explicit form of $g$ - I simply need its existence.
What Google has told me is the following: There is something similar to what I want called the Macaulay resultant $R(f_1,...,f_{n+1})$ (where one must first homogenize the $f_i$). Then the Macaluay resultant vanishes iff the $f_i$ have a common root in $\mathbb{C}P^n$.
What I'm asking is something slightly stronger, I want a polynomial whose vanishing is necessary and sufficient for the $f_i$ to have a common root in $\mathbb{C}$, that is, I disallow points at infinity in $\mathbb{C}P^n$.
Further, Googling tells me that in the case that $n=1$, there is something called the resultant which does exactly what I want.
If I nest resultants (thinking of $\mathbb{C}[x_1,...,x_n]$ as $\mathbb{C}[x_1,...,x_{n-1}][x_n]$), does that work? I'm thinking of something like $Res_{x_{n-1}}(Res_{x_n}(f_1,f_2),Res_{x_n}(f_1,f_3))$ where the subscript indicates which thing we view as the variable.
In case these questions are too broad, here's the specific example I'm interested in. I have 8 variables and 9 polynomials. Writing $\mathbb{C}[a_1,a_2,a_3,a_4,b_1,b_2,b_3,c]$, 8 of the polynomials have the form $$f_i(\vec{a},\vec{b},c) = g_i(\vec{a}) + \left(\sum a_i^2\right)( h_i(\vec{a}) + j_i(\vec{b}))$$ where $g_i$, $h_i$, and $j_i$ are all linear functions (with trivial constant coefficient). The $9$th equation is $$f_9(\vec{a},\vec{b},c) = \left(\sum a_i^2\right)c -1.$$
If one homogenizes by inserting the variable $z$, then one find the point $[\vec{a}:\vec{b}:c:z] = [\vec{0}: \vec{b}:c:0]$ is a simultaneous solution for all $\vec{b}$ and $c$. In particular, I think the Macaulay resultant vanishes in this situation.
Thanks in advance!
No: $f_1=ax_1+x_2, f_2=x_2-1, f_3=0$.
The condition for solvability is $a\neq0$, which is not closed in the parameter space $\mathbb C$ for $a$.