Hilbert's Nullstellensatz

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. 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.

Going deeper

For a student

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: every radical ideal is an intersection of maximal ideals.