When can I say that $\mathcal{I}(V)=\{0\}$ iff $V=\mathbb{F}^k$? In algebraic geometry, for which fields $\mathbb{F}$ can I say that $\mathcal{I}(V)=\{0\}$ iff $V=\mathbb{F}^k$, whenever $V \subseteq \mathbb{F}^k$ is an affine algebraic set?
The forward direction always holds: if $\mathcal{I}(V)=\{0\}$, then for all $a \in \mathbb{F}^k$, we have $\mathcal{I}(V) \subseteq m_a$ which is equivalent to $a \in \mathcal{V}(\mathcal{I}(V))=V$. Therefore $V = \mathbb{F}^k$.
For the converse, "if $V=\mathbb{F}^k$ then $\mathcal{I}(V)=\{0\}$", we at least require that $\mathbb{F}$ must be infinite. Otherwise, write $\mathbb{F}=F_{p^n}$; the nonzero polynomial $x^{p^n}-x$ vanishes on the whole of $\mathbb{F}$, so $x^{p^n}-x \in \mathcal{I}(\mathbb{F}^k)$. (As an aside, can anybody say precisely what $\mathcal{I}(\mathbb{F}^k)$ is? Is it just generated by $x^{p^n}-x$?)
I also suspect that if $\mathbb{F}$ is algebraically closed, then this suffices for $\mathcal{I}(\mathbb{F}^k) = \{0\}$. Does this use that maximal ideals of $R=F[t_1,\ldots,t_k]$ are all of the form $m_a$ for $a \in \mathbb{F}^k$, for algebraically closed $\mathbb{F}$?
Summary:


*

*How to show that if $\mathbb{F}$ is algebraically closed, then $\mathcal{I}(\mathbb{F}^k) = \{0\}$?

*Is there some not necessarily algebraically closed (but of course infinite) $\mathbb{F}$ such that $\mathcal{I}(\mathbb{F}^k) \neq \{0\}$?

*Aside: what is $\mathcal{I}(\mathbb{F}^k)$, when $\mathbb{F}=F_{p^n}$? Is it just generated by $x^{p^n}-x$?
 A: I'll give a quick sketch to show that over an infinite field $\mathbb{F}$, $I(\mathbb{F}^k)=\{0\}$. In particular the result holds for algebraically closed fields. It suffices to show that any nonzero polynomial $f\in \mathbb{F}[x_1,\dots,x_k]$ does not vanish on $\mathbb{F}^k$; we can do this by induction on $k$. Clearly the result is true for $k=1$, since $\mathbb{F}[x]$ admits a division algorithm and our field is infinite.
Now suppose we have a nonzero polynomial $f\in \mathbb{F}[x_1,\dots,x_k]$; the nontrivial case to consider is when $f$ depends on $x_k$. (Ruling out the case of, for instance, $f(x,y)=x^2\in \mathbb{F}[x,y]$). Write 
$$
f(x_1,\dots,x_k)=\sum_{j=0}^sc_j(x_1,\dots,x_{k-1})x_k^j
$$
where $c_j\in \mathbb{F}[x_1,\dots,x_{k-1}]$, $s>1$, and $c_s$ is not identically zero.
By our induction hypothesis, we may find $(a_1,\dots,a_{k-1})\in \mathbb{F}^{k-1}$ for which $c_s(a_1,\dots,a_{k-1})\neq 0$. Now consider the one variate polynomial 
$$
f_s(x):=f(a_1,\dots,a_{k-1},x_k).
$$
By construction this is a nonzero, nonconstant polynomial of one variable. Hence it does not vanish entirely on $\mathbb{F}$, so we may find an $a_k\in \mathbb{F}$ for which $f_s(a_k)\neq 0$, which is exactly what we want! Note that really the only place we used that $\mathbb{F}$ is infinite is in the one variate case (which of course then spills over to the multivariate setting).
