I'm trying to prove that $I(Z(\mathfrak{a})) = \sqrt{\mathfrak{a}}$, where $\mathfrak{a}$ is an ideal of $A = K[x_1, ... , x_n]$ and $K$ is an algebraically closed field. In case this notation is nonstandard:
if $T \subseteq A$ then $Z(T) = \{P \in \mathbb A^n \mid f(P) = 0 \ \forall f \in T\;\}$
and if $Y \subseteq \mathbb A^n$, then $I(Y) = \{f \in A \mid f(P) = 0 \ \forall P \in Y\;\}$.
Using Hilbert's Nullstellensatz I can get that $ I(Z(\mathfrak{a})) \subseteq \sqrt{\mathfrak{a}}$. I'm having trouble with the other direction though. Suppose $f \in \sqrt{\mathfrak{a}}$. Then $f^r \in \mathfrak{a}$ for some integral $r > 0 $. I want to show that $f$ annihilates everything in $Z(\mathfrak{a})$, given that $f^r$ annihilates everything in $Z(\mathfrak{a})$. I can't see how to make it happen, though.
Any help would be great. Thanks