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$\begingroup$

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

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    It is my pleasant duty to thank you for the third time today, dear @Zev: I actually wanted to know how I could obtain such `elegant grey` ! And I am happy to inform you that I was made aware of your last comments through my newly discovered comment-inbox...2012-01-18

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Let $p\in Z(\mathfrak a) $. We know that $f^r(p)=0$ which implies $f(p)=0$ so $f$ annihilates everything in $Z(\mathfrak a)$.

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    @ZhenLin, composition does make sense for polynomials of one variable but not for several variables, though "multiple composition" $f(g_1,\dots,g_n)$ does (for lack of a better term).2012-01-18