Over which fields $k$ is there a reasonable analogue of Hilbert's Nullstellensatz?


Here is a more precise formulation: let $k$ be an arbitrary field, $n$ a positive integer, and $R = k[t\_1,..,t\_n]$. There is a natural relation between $k^n$ and $R$: for $x \in k^n$ and $f \in R$, $(x,f)$ lies in the relation if $f(x) = 0$.  

This relation induces a Galois connection between the power set of $k^n$ and the set of all ideals of $R$ (both partially ordered by inclusion). In more standard algebraic-geometric language, if $S$ is a subset of $k^n$ and $J$ is an ideal of $R$, put


$I(S) = \{f \in R \ | \ \forall x \in S, \ f(x) = 0\}$


and


$V(J) = \{x \in k^n \ | \ \forall f \in J, \ f(x) = 0\}$.


There are induced closure operators: for a subset $S$, $\overline{S} := V(I(S))$ and for an ideal $J$, $\overline{J} := I(V(S))$.


The closure operator on subsets is compatible with finite unions so is the closure operator for a topology on $k^n$, the Zariski topology.


The question is: what is the closure operator $I \mapsto \overline{I}$ on ideals of $R$? By a Nullstellensatz, I mean a nice description of this closure operator.


Some remarks and examples:


1. Over any field $k$, one sees that $\overline{I}$ is a radical ideal hence contains $\operatorname{rad}(I) = \{x \in R \ | \ \exists n \in \mathbb{Z}^+ \ | \ x^n \in I\}$.


If $k$ is algebraically closed, then Hilbert's Nullstellensatz says that $\overline{I} = \operatorname{rad}(I)$.


It is easy to see that if $\overline{I} = \operatorname{rad}(I)$ for all maximal ideals of $k[t]$, then $k$ is algebraically closed.


2. If $k$ is formally real, then for any ideal $I$ of $R$, $\overline{I}$ is a **real ideal**, i.e., $x\_1,\ldots,x\_n \in R, \ x\_1^2 + \ldots + x\_n^2 \in I \implies x\_1,\ldots,x\_n \in I$. Moreover, for any ideal $I$ in a commutative ring, there is a unique minimal real ideal containing $I$, its **real radical** $\mathbb{R}ad(I)$, which is the intersection of all real prime ideals $\mathfrak{p}$ containing $I$.


If $k$ is real-closed, then for any ideal $I$ in $k[t\_1,\ldots,t\_n]$, $\overline{I} = \mathbb{R}ad(I)$: this is Risler's Nullstellensatz.


3. There is also a Nullstellensatz for p-adically closed fields (in particular, for $p$-adic fields) due to Jarden and Roquette: see


[The Nullstellensatz over p-adically closed fields](https://projecteuclid.org/journals/journal-of-the-mathematical-society-of-japan/volume-32/issue-3/The-Nullstellensatz-over-p-adically-closed-fields/10.2969/jmsj/03230425.full)



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> Are there further Nullstellensätze (say, for non-Henselian fields to rule out variations on 3)?
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Although I haven't been precise on what a description of $\overline{I}$ means (I don't know how), it seems reasonable to guess that there is no good Nullstellensatz over a field like $\mathbb{Q}$ for which it is believed that [Hilbert's 10th problem](https://en.wikipedia.org/wiki/Hilbert%27s_tenth_problem) has a negative answer. Briefly: if you had a system of polynomial equations $P\_1,\ldots,P\_m$ with $\mathbb{Q}$-coefficients, then they have a simultaneous solution over $\mathbb{Q}$ iff
the closure of $\langle P\_1,\ldots,P\_m \rangle$ is a proper ideal, so if you had a sufficiently nice description of the closure operation, you could use it to answer H10 over $\mathbb{Q}$ affirmatively.


A case of persistent interest to me over recent years is that of a finite field. In some sense this is the worst case, since it is not hard to show that the zero ideal in $k[t\_1,\ldots,t\_n]$ is closed iff $k$ is infinite. Nevertheless, I vaguely feel like there should be something to say here, possibly something having to do with reduced polynomials -- i.e., for which each exponent of each variable is at most $\# k - 1$ -- as in one of the proofs of the [Chevalley-Warning theorem](https://en.wikipedia.org/wiki/Chevalley%E2%80%93Warning_theorem).


P.S.: I am aware of other algebraic results about $k[x\_1,\ldots,x\_n]$ over a general field $k$ which, when $k$ is algebraically closed, imply Hilbert's Nullstellensatz, e.g. that a finitely generated $k$-algebra which is a field is finite-dimensional over $k$, or that every prime ideal in $k[t\_1,\ldots,t\_n]$ is an intersection of maximal ideals. These are interesting and useful, but here I am really interested in $I \mapsto \overline{I}$.



Are there more Nullstellensätze?
If $k$ is a finite field with $q$ elements and $I$ an ideal of $k[x\_1,\dots,x\_n]$, then $\overline I=I+I\_0$, where $I\_0=(x\_1^q-x\_1,\dots,x\_n^q-x\_n)$. This follows immediately from Hilbert’s Nullstellensatz applied to the algebraic closure of $k$, and the observation that any ideal extending $I\_0$ is a radical ideal (as it contains all polynomials of the form $f^q-f$).


On an unrelated note, a more explicit description for the case of $k$ real-closed follows from Stengle’s (Positiv- and) Nullstellensatz: $f\in\overline I$ iff $-f^{2n}\in I+\Sigma$ for some $n\in\mathbb N$, where $\Sigma$ is the set of all sums of squares of polynomials.


