Let $R = k[x_1,\dotsc,x_n]$ and $J = (f_1,\dotsc,f_m) \subseteq R$ be an ideal. I know that under the assumption that algebraically closed, $V(J)=\emptyset$ iff $J=(1)$. Namely, "$\Leftarrow$" is clear, and "$\Rightarrow$" we get by using that in general, $V(J)$ iff $(1)=I(V(J))$, which by the Nullstellensatz is equals $\sqrt{J}$.
Now, what can I do if $k$ is not algebraically closed? I can still play the same game in the closure $\bar{k}$, but then I don't know how to decide if $V_k(J)$ is empty, given that $V_{\bar{k}}(J)$ happens to be non-empty.
Is there, probably, an effective way to compute, e.g., a generating set (or even Gröbner basis) of $I(V(J))$ from $J$?
Does it help if I restrict which generators $f_i$ the ideal $J$ can have? In my problem, $J$ will be generated by a number of linear polynomials in $k[x_1,\dotsc,x_{n-1}]$, and a single non-linear one of the form $g x_n-1$, where $g \in k[x_1,\dotsc,x_{n-1}]$ is arbitrary.
I apologize for the vague question. Any hint that might be useful appreciated.
