$\newcommand\C{\mathbb C}\newcommand\syz{\operatorname{syz}} \newcommand\x{\mathbf x}\newcommand\a{\mathbf a}$Let $\a=a_1,\dots,a_m$ and $\x = x_1,\dots, x_n$ be indeterminates and consider a polynomial $F(\a, \x) \in \C[\a,\x]$ and having a constant term $1$ with respect to the indeterminates $a_i$'s i.e. $F(\mathbf 0,\x)\equiv 1$. I want to see if I can solve for $a_i\in \C[x_1,\dots,x_n]$ such that $F(\a,\x)$ is identical to $0$. Let $R=\C[\x]$ and consider now the coefficients of the monomials of $F$ regarded as a polynomial in $R[\a]$, lets call them $\alpha_0,\dots, \alpha_N$ where $\alpha_0=1$ is the constant term of $F$. I can now compute the first syzygy $S:=\syz(\alpha_0,\dots,\alpha_N) \subset R^{N+1}$ of these coefficients. $S$ is a finitely generated $R$-module and it can be generated by say $k$ elements in $R^{N+1}$. And now comes my question...
If the ideal generated by the $0$-coordinate of these $k$ elements above (these corresponds to $\alpha_0$) is not the whole ring $R$, then is it OK to conclude that I do not have a polynomial solution for the $a_i$'s such that $F(\a,\x) \equiv 0$?
I can imagine something going wrong like for instance some of the $\alpha_0,\dots,\alpha_N$ could be algebraically dependent (or even equal) but I don't think this matters here, right?