Nullstellensatz: varieties and radical ideals

Let $k$ be an algebraically closed field and let $A=k[x_1,\dots,x_n]$ . For an ideal $I\subseteq A$ , define its zero 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 ideal of functions vanishing on $X$ by

I(X)=\{f\in A : f(a)=0 \text{ for all } a\in X\}.

The sets $V(I)$ are exactly the closed sets of the Zariski topology on $k^n$ .

Theorem (Nullstellensatz, variety–ideal correspondence). With $k$ and $A$ as above, the following hold:

For every ideal $I\subseteq A$ , one has $I\bigl(V(I)\bigr)=\sqrt{I},$ the radical of $I$ .
For every Zariski-closed set $X\subseteq k^n$ , one has $V\bigl(I(X)\bigr)=X.$

Consequently, the assignments $I\mapsto V(I)$ and $X\mapsto I(X)$ restrict to mutually inverse, inclusion-reversing bijections between:

radical ideals of $A$ , and
Zariski-closed subsets of $k^n$ .

A useful special case is the "weak" form: maximal ideals of $A$ are exactly the ideals $(x_1-a_1,\dots,x_n-a_n)$ for points $a=(a_1,\dots,a_n)\in k^n$ . In other words, the maximal spectrum $\operatorname{MaxSpec}(A)$ can be identified with $k^n$ , and the corresponding residue field at such a maximal ideal is (canonically) $k$ .

Examples

A non-radical ideal with the same zero set as its radical. In $k[x]$ , let $I=(x^2)$ . Then $V(I)=\{0\}$ , but $I\bigl(V(I)\bigr)=(x)=\sqrt{(x^2)}.$ This illustrates that taking $V(-)$ forgets nilpotent structure, and the theorem recovers exactly the radical.

A point. In $k[x,y]$ , let $I=(x,y)$ . Then $V(I)=\{(0,0)\}$ . Conversely, for the closed set $X=\{(0,0)\}$ one has $I(X)=(x,y)$ , a maximal ideal, matching the identification of maximal ideals with points.

A reducible algebraic set. In $k[x,y]$ , let $I=(xy)$ . Then $V(I)$ is the union of the two coordinate axes. The ideal of this union is $I\bigl(V(I)\bigr)=(x)\cap (y)=\sqrt{(xy)},$ reflecting that $V(I)$ has two irreducible components.