Nullstellensatz Variety–Ideal Correspondence

Statement

Let $k$ be an algebraically closed field , and let

be the polynomial ring .

• For an ideal $I\subseteq R$, define the affine algebraic set $V(I) = \{a\in k^n : f(a)=0 \ \text{for all } f\in I\}.$
• For a subset $X\subseteq k^n$, define the vanishing ideal $I(X)=\{f\in R : f(a)=0 \ \text{for all } a\in X\}.$

Then Hilbert’s Nullstellensatz implies:

where $\sqrt{I}$ is the radical of I .

Consequences:

• The assignments $I \mapsto V(I)$ and $X \mapsto I(X)$ are inclusion-reversing.
• $I$ is radical iff $I=I(V(I))$.
• There is a bijection between radical ideals of $R$ and affine algebraic sets in $k^n$.
• Under this correspondence, prime ideals correspond to irreducible algebraic sets, and maximal ideals correspond to points.

This viewpoint is the algebra behind the Zariski topology on $k^n$.

Cross-links

• Topology side: Zariski topology
• Algebra side: radical ideal , prime ideal , maximal ideal
• Spectral viewpoint: prime spectrum
• Classical theorems: weak Nullstellensatz , strong Nullstellensatz

Examples

1. A point in the plane.
   In $k[x,y]$, the ideal $(x-a,\,y-b)$ is maximal, and

   $$V(x-a,y-b)=\{(a,b)\}.$$

2. A curve.
   In $k[x,y]$, the ideal $(y-x^2)$ defines the parabola:

   $$V(y-x^2)=\{(t,t^2): t\in k\}.$$

   Since $(y-x^2)$ is prime, the parabola is irreducible.

3. Radical matters: same variety, different ideal.
   In $k[x,y]$,

   $$V(x^2)=V(x)=\{(0,y): y\in k\},$$

   and indeed $I(V(x^2))=\sqrt{(x^2)}=(x)$.