# For any ideal $\mathfrak{a}\subseteq A$, $I(Z(\mathfrak{a})) = \sqrt{\mathfrak{a}}$, the radical of $\mathfrak{a}$

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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\frak{a} is $\frak{a}$ – yoyo Jan 18 '12 at 17:45
You might want the symbol $\mathfrak{a}$, which is a Fraktur letter 'a' (Fraktur letters are often used for ideals of rings, among other things). You can get Fraktur characters with the command \mathfrak. – Calvin McPhail-Snyder Jan 18 '12 at 17:45
Thanks: edited. – Matt Jan 18 '12 at 17:48
Note: \mathfrak is the better command for this site; for example, $\frak{a} \subseteq A$ produces $$\frak{a} \subseteq A$$ while $\mathfrak{a} \subseteq A$ produces $$\mathfrak{a} \subseteq A$$ (i.e. \frak changes the current font to Fraktur, while \mathfrak applies it only to its argument). However one could use ${\frak a} \subseteq A$, which produces $${\frak a} \subseteq A$$ (the curly braces contain the effects of the \frak command) – Zev Chonoles Jan 18 '12 at 18:39
This is very useful, @Zev: I had always be bothered by the fact that \frak changes the current font and had to resort to clumsy curly braces to remedy that. So thanks a lot for solving my problems for the second time in a few minutes (and now, without my even having to ask!) – Georges Elencwajg Jan 18 '12 at 20:35

Let $p\in Z(\mathfrak a)$. We know that $f^r(p)=0$ which implies $f(p)=0$ so $f$ annihilates everything in $Z(\mathfrak a)$.
I'm obviously being very slow: why does $f^r(p) = 0$ imply $f(p) = 0$? – Matt Jan 18 '12 at 18:10
@Matt: Because $f(p) \in K$ and $K$ is a field, so a fortiori an integral domain, and thus $f(p)^r = 0$ implies $f(p) = 0$. – Zhen Lin Jan 18 '12 at 18:32
@ZhenLin Does the $f^r$ in Hilbert's Nullstellensatz refer to composition or multiplication? – Matt Jan 18 '12 at 18:37
@Matt, $f^r(p) = (f(p))^r$ – lhf Jan 18 '12 at 18:48
@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). – lhf Jan 18 '12 at 20:48