Rückert Nullstellensatz

Remarks

The inclusion $\sqrt{I} \subseteq \mathcal{I}(V(I))$ is clear: if $f^m \in I$, then $f^m$, and hence $f$, vanishes on $V(I)$. For the other inclusion, write $\sqrt{I}$ as a finite intersection of prime ideals, which is possible because $\mathcal{O}_{n,a}$ is Noetherian, and reduce to the case of a prime ideal $\mathfrak{p}$. For $\mathfrak{p}$ prime, the local parametrization of the irreducible germ $V(\mathfrak{p})$ shows that a germ vanishing on $V(\mathfrak{p})$ lies in $\mathfrak{p}$. Example: for $I = (z_1^2) \subseteq \mathcal{O}_{2,0}$, the germ $V(I)$ is the line $\set{z_1 = 0}$, and $\mathcal{I}(V(I)) = (z_1) = \sqrt{I}$; here $z_1^2 \in I$. As a consequence, $I \mapsto V(I)$ and $A_a \mapsto \mathcal{I}(A_a)$ are mutually inverse bijections between radical ideals of $\mathcal{O}_{n,a}$ and germs of analytic sets at $a$, and prime ideals correspond to irreducible germs. As for Hilbert's Nullstellensatz, the base field matters: the statement fails for real analytic germs, since $x^2 + y^2$ vanishes only at $0 \in \R^2$ but $x \notin \sqrt{(x^2 + y^2)}$.

Going deeper

For a student

The Nullstellensatz is the dictionary of local analytic geometry. With it, the decomposition of an analytic germ into irreducible components corresponds to the primary decomposition of its ideal. The dimension of a germ $A_a$ equals the Krull dimension of $\mathcal{O}_{n,a}/\mathcal{I}(A_a)$. It is also an ingredient in Cartan's coherence theorem for the ideal sheaf of an analytic set.