I'm working on an exercise from Kunz' Into to Algebraic Geometry which I've paraphrased:
Let $L|K$ be a field extension where $L$ infinite. Given $f_1,\cdots,f_n \in K[t_1,\cdots,t_m]$, let $V$ denote the closure (in the Zariski topology over $k^n$) of the set $V_0=\{(f_1(t_1,\cdots,t_m),\cdots,f_n(t_1,\cdots,t_m)),(t_1,\cdots,t_m)\in L^m\}$. Show that $V$ is irreducible.
(hint: Consider the $K$-homomorphism $K[x_1,\cdots,x_n]\rightarrow K[t_1,\cdots,t_m]$ defined by sending $x_i\to f_i(t_1,\cdots,t_m)$, then recall that the complement of any nontrivial variety in $L^n$ is infinite.)
My approach is to show the ideal of $V$, $\mathbb{J}(V)=\{f:f(x_1,\cdots,x_n)=0,(x_1,\cdots,x_n)\in L^n\}$, is prime. Also note that $\mathbb{J}(V)$ is the kernel of the mapping \begin{equation} \Phi:K[x_1,\cdots,x_n]\rightarrow K[t_1,\cdots,t_m] \end{equation} defined in the hint. So by the first isomorphism theorem, $\mathbb{J}(V)$ is prime if and only if $Im(\Phi)$ is an integral domain. It would then be sufficient to show that $\Phi$ is surjective, but I don't know if this is the case, or how to prove it if it is. I'm also not sure where I'd be using the fact that the complement of any variety is infinite.