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

For a subset $X\subseteq k^n$, define the ideal of functions vanishing on $X$ by

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:

1. For every ideal $I\subseteq A$, one has
   $$I\bigl(V(I)\bigr)=\sqrt{I},$$
   the radical of $I$.
2. 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

1. 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.

1. 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.

1. 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.