---
tag: D85VZWS9QK
kind: statement
role: theorem
name: Hilbert's Nullstellensatz
msc: [13A15, 14A05]
uses: [0PF6TM8TWT, 0SDC934R52, 096Z9MP2ME, 437YRCNYRJ]
about: [437YRCNYRJ]
sources:
  - url: https://en.wikipedia.org/wiki/Hilbert's_Nullstellensatz
same_as:
  wikidata: Q1068976
  wikipedia: Hilbert's Nullstellensatz
---

## Statement

Let $k$ be an [[algebraically closed field|0PF6TM8TWT]], let $n \ge 1$, and let $I$ be an [[ideal|096Z9MP2ME]] of the [[polynomial ring|0SDC934R52]] $k[x_1, \ldots, x_n]$.
Let $V(I) = \set{a \in k^n : f(a) = 0 \text{ for all } f \in I}$ be the zero set of $I$.
Let $I(V(I)) = \set{f \in k[x_1, \ldots, x_n] : f(a) = 0 \text{ for all } a \in V(I)}$ be the ideal of polynomials vanishing on it.
Then
$$I(V(I)) = \sqrt{I},$$
where $\sqrt{I} = \set{f : f^m \in I \text{ for some } m \ge 1}$ is the [[radical|437YRCNYRJ]] of $I$.

## Remarks

The inclusion $\sqrt{I} \subseteq I(V(I))$ is clear: if $f^m$ vanishes at a point, so does $f$.
For the other inclusion, let $f$ vanish on $V(I)$ and add a new variable $y$ (the Rabinowitsch trick).
The ideal generated by $I$ and $1 - y f$ in $k[x_1, \ldots, x_n, y]$ has no zeros, so it is the unit ideal by the [[weak Nullstellensatz|ZWS0YWGF17]].
Writing $1$ as a combination and substituting $y = 1/f$, then clearing denominators, gives $f^m \in I$ for some $m$.

Example: in $\C[x]$, let $I = (x^2)$; then $V(I) = \set{0}$ and $I(V(I)) = (x) = \sqrt{(x^2)}$.

The field must be algebraically closed: over $\R$, let $I = (x^2 + 1)$ in $\R[x]$.
Then $V(I)$ is empty, so $I(V(I)) = \R[x]$, but $\sqrt{I} = I$ is a proper ideal, since $I$ is prime.

## Explanations

### L1 human

A polynomial vanishes at all common zeros of some polynomials exactly when one of its powers is a combination of them.
So, over $\C$, systems of polynomial equations and ideals of polynomials carry the same information, up to taking radicals.

### L2 human

The theorem gives an inclusion-reversing bijection between radical ideals of $k[x_1, \ldots, x_n]$ and algebraic subsets of $k^n$.
Prime ideals correspond to irreducible algebraic sets and maximal ideals to points.
It is the bridge between commutative algebra and classical algebraic geometry.
In ring-theoretic form it says that $k[x_1, \ldots, x_n]$ is a [[Jacobson ring|7EB64ZBXAT]]: every radical ideal is an intersection of maximal ideals.