Hilbert's Nullstellensatz implies $I(Z(I)) = \sqrt{I}$ Let $k$ be an algebraically closed field.  Let $I$ be the map that takes algebraic sets in $k^n$ to the ideal generated by them: $I : \{$ algebraic sets $\} \to \{$ ideals of $k[x_1,\dots, x_n] \}$, $I(X) = \{ f \in k[x_1, \dots, x_n] : f(X) = \{0\}\}$.  Similarly let $Z$ be the map that takes ideals in $k[x_1,\dots, x_n]$ to algebraic sets.  Then $I(Z(I)) = \sqrt{I}$.
How does that follow from Hilbert's Nullstellensatz?
This version of Nullstellensatz:
If $k$ is an algebraically closed field then, the maximal ideals of $k[x_1,\dots, x_n]$ are of the form $(x_1 - a_1, \dots, x_n - a_n)$ where $a = (a_1, \dots, a_n)$ is a point in $k^n$.
 A: Let $I$ be an ideal of $k[x_1, \dots, x_n]$. It is easy to see that $\sqrt{I}\subseteq IZ(I)$ ($f^l \in I$ implies $f^l(P)=0$ for every $P \in Z(I)$, which implies $f(P)=0$ for every $P \in Z(I)$ ).
The other inclusion is a bit more tricky. Assume that $ f \in IZ(I)$ and take some generators $f_1, f_2, \dots, f_k$ of $I$. Then the set $Z(\{f_1, f_2, \dots, f_k,x_{n+1}f-1\})$ is empty, since $f(P)=0$ whenever $f_1(P)=f_2(P)=\dots =f_k(P)=0$ (note that here we operate "one dimension higher", i.e. in $k^{n+1}$; $x_{n+1}$ is a new indeterminate). Hence $$J:=(f_1, f_2, \dots, f_k,x_{n+1}f-1)=k[x_1,\dots , x_n,x_{n+1}]$$ (if $J$ were a proper ideal, it would be contained in some maximal ideal $m$, but $Z(m)$ is a point, hence nonempty, and $Z(m)\subseteq Z(J)$).
Therefore, one can write 
$$1=\sum_{i=1}^{k}f_i(x_1, \dots x_{n})h_i(x_1, \dots x_n,x_{n+1})+(x_{n+1}f(x_1, \dots x_{n})-1)h(x_1, \dots x_n,x_{n+1})$$
by using the substitution $x_{n+1}=\frac{1}{f}$ (formally, using an evaluation homomorphism), one has (in the localisation $k[x_1, \dots x_n]_f$)
$$1=\sum_{i=1}^{k}f_i(x_1, \dots x_{n})h_i(x_1, \dots x_n,\frac{1}{f})+(\frac{1}{f}f(x_1, \dots x_{n})-1)h(x_1, \dots x_n,\frac{1}{f}),$$
but the last term clearly vanishes, so
$$1=\sum_{i=1}^{k}f_i(x_1, \dots x_{n})h_i(x_1, \dots x_n,\frac{1}{f}).$$
One can then easily see that the terms $h_i(x_1, \dots x_n,\frac{1}{f})$ are of the form $\frac{g_i}{f^{m_i}}$, where $g_i$ are polynomials in indeterminates $x_1, \dots, x_n$. Take $M:=\max_i m_i$. Then 
$$f^M=\sum_{i=1}^{k}f_ig_if^{M-m_i},$$
which is an equality in $k[x_1, \dots x_n]$. Hence, $f \in \sqrt{(f_1, \dots, f_k)}=\sqrt{I}$.
