Slick proof the intersection of maximal ideals of a finitely generated domain is zero? I have seen in several places the result that taking polynomial rings preserves (and reflects) the Jacobson-Hilbert property, which is also preserved by quotients. Thus, in any finitely generated commutative ring, the nilradical equals the Jacobson radical. In particular, in a finitely generated domain, the intersection of all maximal ideals is zero.
I don't know commutative algebra, but I was wondering whether this result about zero intersection of maximal ideals can be proved "directly" for a finitely generated domain using Noether normalization. By 'directly' I mean 'without developing the theory of Jacobson rings'.
 A: I'm not sure this is quite what you're looking for, since it's really just a proof that a finitely generated algebra over a Jacobson ring is Jacobson, specialized to the case of $\mathbb{Z}$.  But here goes.  The key tool is the Artin-Tate lemma (you can use Noether normalization instead, but Artin-Tate gives what I would consider a more elementary argument).
First, let us prove that any field $K$ which is finitely generated as a ring is a finite field.  Let $k$ be the prime subfield of $K$.  Since $K$ is finitely generated as a $k$-algebra, it suffices to show $K$ has positive characteristic and is finite over $k$.  Let $B\subset K$ be a transcendence basis and note that $K$ is algebraic and hence finite over $k(B)$.  By the Artin-Tate lemma applied to the $\mathbb{Z}$-algebras $k(B)\subseteq K$, $k(B)$ is finitely generated as a ring.  This implies $B=\emptyset$ (any finitely generated subring of $k(B)$ has only finitely many irreducible polynomials which are factors of denominators of its elements, but there are infinitely many different irreducible polynomials in $k[B]$ if $B$ is nonempty).  Moreover, since $\mathbb{Q}$ is not a finitely generated ring, $k$ must have positive characteristic.  Thus $K$ is algebraic over $k=\mathbb{F}_p$ for some $p$ and hence is a finite field.
Now suppose $R$ is a finitely generated domain and $f\in R$ is nonzero.  The ring $S=R[f^{-1}]$ is then also a finitely generated domain, and so it has a maximal ideal $M$.  I claim that $M\cap R$ is a maximal ideal of $R$, and thus there exists a maximal ideal of $R$ which does not contain $f$.
To prove this, observe that field $S/M$ is finitely generated as a ring, and hence finite.  But $R/(M\cap R)$ is naturally a subring of $S/M$, and any subring of a finite field is a field.  Thus $R/(M\cap R)$ is a field, so $M\cap R$ is maximal.
