Hilbert's Nullstellensatz: intersection over maximal ideals? In Reid's commutative algebra, the author gives some exposition about the Nullstellensatz, $I(V(J))=\operatorname{rad}(J)$. But I can't understand it: The Nullstellensatz says that we can take the intersection just over maximal ideals of $k[X_1,\cdots,X_n]$. 
 A: Because the maximal ideals correspond to points in $k^n$. Then $f(a_1,\ldots,a_n)=0$ iff $f \in (X_1-a_1,\ldots,X_n-a_n)$.
A: In a commutative ring $A$, an ideal $I\subset A$ has two radicals:
a) The nilpotent radical $\sqrt I=Rad(I)=\cap_{I\subset P \:prime}$ which is the intersection of all prime ideals containing $I$.
b) The Jacobson  radical $J(I)=\cap_{I\subset M \: maximal}$ which is the intersection of all maximal ideals containing $I$.
Now, there is a class of rings, deservedly called Jacobson rings, for which the two radicals coincide. Any algebra of finite type over a field is a Jacobson ring, and so there is no contradiction in the statements you mention in your question.
Warning Many  rings  are not Jacobson rings. For example a local domain (A,M) which is not a field is never Jacobson because the nilpotent  ideal $(0)$ has $Rad((0))=(0)$ whereas $J((0))=M$.
So, beware: the local rings $\mathcal O_{X,x}$ of points $x\in X$ of the affine algebraic  $k$-variety $X=Spec(A)$ are not Jacobson rings even though the finitely generated $k$-algebra $A$ is a Jacobson ring! 
