General form of Nullstellensatz

Question

At lots of places (some examples are 1, 2, 3), it is stated, that Hilbert's Nullstellensatz is well understood as theorem about more general Jacobson rings. Namely

(When $R$ is a Jacobson ring and $S$ finitely generated $R$-algebra, then $S$ is also Jacobson.) If $M$ is a maximal ideal of $S$, then the pullback ideal $N=M\cap R$ of $R$ is also maximal and the field $S/M$ is finite extension over the field $R/N$.

Why is this considered to be generalize Nullstellensatz? Isn't that mainly a statement about restrictions for existence of solutions, or more precise characterization of the Galois connection in question.

The general version statement is in my opinion analogous to Zariski's lemma, which is used in the first step of the usual Nullstellensatz' proof.

It's very well possible it's called Nullstellensatz even though the analogy is not a strong one, because the Zariski's lemma is the crucial ingredient in the proof. However, I'm not writing here to rant about the name. I want to make sure I'm not missing something: Are the other parts of Nullstellensatz deducible from the general form or is there more to the theory of Jacobson rings, that would generalize the usual case?

Answer 1

Note that the classical Nullstellensatz in $k[x_1, \dotsc, x_n]$ states $$\sqrt{\mathfrak a} = \mathcal I(\mathcal V(\mathfrak a)) = \bigcap\limits_{\mathfrak a \subset \mathfrak m} \mathfrak m,$$ with the intersection being taken over all maximal ideals containing $\mathfrak a$.

In any ring, we have a formal Nullstellensatz: $$\sqrt{\mathfrak a} = \bigcap\limits_{\mathfrak a \subset \mathfrak p} \mathfrak p,$$ with the intersection being taken over all prime ideals containing $\mathfrak a$. Note that the most proofs of this fact are actually modern versions of the Rabinowitsch trick (Rabinowitsch trick is just localizing, not so tricky from the modern point of view ;) ).

In a Jacobson ring, we have $\bigcap\limits_{\mathfrak a \subset \mathfrak p} \mathfrak p = \bigcap\limits_{\mathfrak a \subset \mathfrak m} \mathfrak m$, i.e. the formal Nullstellensatz boils down to the formulation of the classical Nullstellensatz. This is why a result about many rings being Jacobson is called a generalization of the Nullstellensatz.

Answer 2

$\def\bbA{\mathbb{A}} \def\fra{\mathfrak{a}} \def\jn{\operatorname{JN}} \def\an{\operatorname{AN}} \def\frm{\mathfrak{m}}$Let's call the statement in italics from the OP the Jacobson Nullstellensatz, and abbreviate it as $\jn(R,S)$, for a given Jacobson ring $R$ and a finitely generated $R$-algebra $S$.

Let $k$ be any field, and denote $\bbA_k^n=k^n$ to the $n$-dimensional classical affine space over $k$. For each subset $S\subset k[T_1,\dots,T_n]$, denote $V(S)=\{x\in\bbA_k^n\mid f(x)=0,\;\forall f\in S\}$ to the vanishing set of $S$. On the other hand, for each subset $X\subset\bbA_k^n$, denote $I(X)=\{f\in k[T_1,\dots,T_n]\mid f(x)=0,\;\forall x\in X\}$ to the vanishing polynomial ideal of $X$. Then, for any ideal $\fra\subset k[T_1,\dots,T_n]$, it is not difficult to verify that $I(V(\fra))\supset\sqrt{\fra}$.

Now suppose that $k$ is algebraically closed field (we will assume so for the rest of the post). Then it holds that $I(V(\fra))=\sqrt{\fra}$. Let's call algebraic Nullstellensatz to this assertion, and abbreviate it as $\an$.

I claim that $$ \jn(k,k[T_1,\dots,T_n])\iff\an. $$ Note: of course this equivalence is just a tautology, these assertions are true mathematical statements by their own right. What I actually mean with this logic equation is that there is a direct proof of each of the claims using the other one. This justifies why it makes sense to call $\jn(R,S)$ the “general form of the Nullstellensatz” (as they do in the links 2, 3 from the OP).

Let's give the argument. We start with an easy

Lemma. Let $K$ be any field (algebraically closed or not), and let $x\in\bbA_K^n$. Then $I(\{x\})=(T_1-x_1,\dots,T_n-x_n)$.

Proof. The ideal $(T_1-x_1,\dots,T_n-x_n)$ is maximal, for $K[T_1,\dots,T_n]/(T_1-x_1,\dots,T_n-x_n)\cong K$. On the other hand, this ideal it is contained in the proper ideal $I(\{x\})$, so they must be the same ideal. $\square$

$(\Leftarrow)$ Here it is proven that $k[T_1,\dots,T_n]$ is a Jacobson ring using $\an$. On the other hand, let $\frm\subset k[T_1,\dots,T_n]$ be a maximal ideal. Then $k\cap\frm=(0)$ is maximal in $k$, and it remains to show that $k[T_1,\dots,T_n]/\frm$ is a finite field extension of $k$. It suffices to show that there is $x\in\bbA_k^n$ with $\frm=(T_1-x_1,\dots,T_n-x_n)$. On the one hand, since $V(I(X))=\overline{X}$ for any subset $X\subset\bbA^n_k$ (true regardless of the algebraic closedness of $k$, see [GW], Prop. 1.12(2)), $\an$ implies that the operators $I(-),V(-)$ are the mutual inverses of a contravariant isomorphism between the posets of Zariski closed subsets of $\bbA_k^n$ and radical ideals of $k[T_1,\dots,T_n]$. In particular, the atoms in the Zariski closed subsets are the points. Thus, $\frm$, which is a coatom in the poset of radical ideals of $k[T_1,\dots,T_n]$, must be of the form $\frm=I(\{x\})=(T_1-x_1,\dots,T_n-x_n)$, for some $x\in\bbA_k^n$ (where the last equality is our lemma).

$(\Rightarrow)$ This is basically what [GW] does. There $\jn(k,A)$ is Theorem 1.7 and $\an$ is Prop. 1.12(1). With the objective to be self-contained, we reproduce the proof here. Assume $\jn(k,A=k[T_1,\dots,T_n])$. In particular, for a maximal ideal $\frm\subset A$, the field $A/\frm$ is a finite field extension of $k$ via the composite $k\to A\to A/\frm$. Since the finite field extensions of an algebraically closed field are all trivial, this map is an isomorphism.

(The following is taken from [GW], hoping this qualifies as fair use.)

Our lemma says that $\frm_x=I(\{x\})$, for $x\in\bbA_k^n$. Hence, for $Z\subset\bbA_k^n$, we have

In the middle equality we've used Corollary 1.11(2). The last equality is the fact that $k[T_1,\dots,T_n]$ is Jacobson. $\square$

---

References

[GW] U. Görtz, T. Wedhorn, Algebraic Geometry I: Schemes. Second ed. Springer Spektrum.