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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 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.

The Nullstellensatz over p-adically closed fields

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 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.

http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.jmsj/1240234705

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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 $f \in R$, say that $(x,f)$ 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

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