# Are there more Nullstellensätze?

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

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

Are there further Nullstellensätze (say, for non-Henselian fields to rule out variations on 3)?

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.

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 $I \mapsto \overline{I}$.

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If I had a second vote, I would give it to you for using the correct plural of "Satz." –  Ben Webster Jul 19 '10 at 23:44
@Ben: In that case, please find an upvote to give to Georges Elencwajg, who informed me of this in an email some months ago. (I had used an incorrect pluralization in some online lecture notes.) Don't be fooled -- unfortunately, I speak no German. –  Pete L. Clark Jul 19 '10 at 23:47
Hi Pete - welcome back! I don't have time for a real answer... but have you seen the Ruckert Nullstellensatz? It's a nullstellensatz for germs of holomorphic functions - I got interested in it a bit, due to a model-theoretic proof by Weispfenning (there's a model-theoretic approach to all the Nullstellensatz that I know). –  Marty Jul 20 '10 at 2:48
@Marty: No, I don't think I've seen it, but I'll check it out. As for the model theory -- yes, of course that's part of where I'm coming from. In fact I am teaching a short course on model theory as we speak (well, if we were to speak tomorrow afternoon at 2:15 pm EST), and I just typed up notes on the real Nullstellensatz following from the model completeness of real-closed fields. –  Pete L. Clark Jul 20 '10 at 3:13
One reference for an analytic Nullstellensatz is this: archive.numdam.org/article/SHC_1960-1961__13_2_A6_0.pdf (starting on page 15). In this series of exposes, k is a nondiscretely valued complete field. –  Lars Jul 20 '10 at 7:18

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.

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This is a very nice observation! It came a long time after I originally posted the question and, though I apparently upvoted it at the time, I seem to have forgotten about it. Does it appear in the literature anywhere? –  Pete L. Clark Aug 2 '12 at 11:29
This is a common knowledge in algebraic proof complexity. I have no idea who could have observed it first, but see e.g. the introduction to dx.doi.org/10.1112/plms/s3-73.1.1 –  Emil Jeřábek Aug 2 '12 at 16:40

Pete, a Nullstellensatz-like result for finite fields is the "Combinatorial Nullstellensatz" formulated by Noga Alon, and it does imply the Chevalley-Warning theorem. Searching for CN will produce Alon's paper and several others on the first page of results.

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Thanks, T. I have seen the Combinatorial Nullstellensatz before, but I haven't really internalized it or appreciated its usefulness. For instance, the derivation of Warning's Theorem from the CN reminds me a lot of Warning's proof of Warning's theorem. I say this not to denigrate the result but just to be honest about my current level of understanding. Any insight you can provide would be welcome: in particular, does it give further information about the closure operator in general, or just in a particular case as in p.1 of Alon's paper? –  Pete L. Clark Jul 21 '10 at 4:08
Just to be clear: I was not thinking about CN when I posted the question, and, in that it is in the same circle of results as Chevalley-Warning, it is quite plausible to me that it has something to do with a Hilbert-style Nullstellensatz over finite fields. I appreciate the pointer. (+1.) But it thickens the plot rather than definitively answering my question (not that I have any guarantee that a definitive answer exists...). –  Pete L. Clark Jul 21 '10 at 4:26