# What's a more general nullstellensatz?

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Let $$k$$ be an algebraically closed field. For an ideal $$I$$ in $$k[x_1,\ldots, x_n]$$ we define $$\mc V(I)$$ as the set of all the points in $$k^n$$ on which each element of $$I$$ vanishes. For a subset $$S$$ of $$k^n$$ we define $$\mc I(S)$$ as the set of all the functions $$f\in k[x_1,\ldots, x_n]$$ which vanish on each point of $$S$$. The nullstellensatz states that $$\mc I(\mc V(I)) = \sqrt{I}$$ for any ideal $$I$$ in $$k[x_1,\ldots, x_n]$$.

Now suppose we were working over the ring $$R =k[x_1^{\pm},\ldots, x_n^{\pm}]$$, that is, the localization of $$k[x_1,\ldots, x_n]$$ with respect to the multiplicatively closed set $$\set{x_1^{m_1} \cdots x_n^{m_n}:\ m_i\geq 0}$$. Here we define, for an ideal $$I$$ in $$R$$, $$\mc V(I)$$ as $$\mc V(I) = \set{p\in (k^\times)^n:\ f(p) = 0 \text{ for all } f\in I}$$ For any subset $$S$$ of $$(k^\times)^n$$ we define $$\mc I(S)$$ as the set of all the elements of $$R$$ which vanish at every point of $$S$$. Then again we have $$\mc I(\mc V(I)) = \sqrt{I}$$ and this can be proved using the nullstellensatz stated before.

So we recover the nullstellensatz by replacing $$k$$ with $$k^\times$$. I believe the `reason' for this is that $$k[x_1^{\pm},\ldots, x_n^{\pm}]$$ is the coordinate ring of $$(k^\times)^n$$.

Question. Is there a more general version of the nullstellensatz which captures both the statements simultaneously (and also is there is a unified proof?)

Here are many such statements. We will assume $$R$$ is Noetherian. Let $$X=\operatorname{Spec} R$$ the set of all prime ideals of $$R$$ and let $$X_m\subset X$$ be the set of maximal ideals.
1. Let $$I\subset R$$ be any ideal and let $$V(I)\subset X$$ be the the set of all prime ideals containing $$I$$. Let $$J=\cap_{P\in V(I)} P$$. Then $$\sqrt{I}=J$$.
2. Assume $$R$$ is Jacobson. This means for any prime ideal $$P$$, $$P=\cap_{P\subset M\in X_m} M$$. Then, $$J=\cap_{M\in X_m,\, I\subset M} M$$.
3. If $$R$$ is a finitely generated algebra over a field $$k$$, then $$R$$ is Jacobson.
4. Finally, if $$k$$ is algebraically closed field, $$R=k[x_1,\ldots, x_n]$$ then any maximal ideal is of the form $$(x_1-a_1,\ldots, x_n-a_n)$$ for $$a_i\in k$$.