$\newcommand{\set}[1]{\{#1\}}$ $\newcommand{\mr}{\mathscr}$ $\newcommand{\mc}{\mathcal}$

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?)


1 Answer 1


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 $a_i\in k$.

These give both your statements (and many more).


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