Density of maximal spectrum It's well known that for algebraically closed field $k$ maximal spectrum of finitely generated $k$-algebra is everywhere dense in whole spectrum of this algebra.
What can be said in the case of non-algebraically closed field, or even if $k$ is not a field, $\mathbb{Z}$, for instance?
 A: Claim: For a ring $R$, its maximal spectrum is dense in its whole spectrum if and only if its Jacobson radical $J(R)$ is equal to its nilradical $N(R)$.
Proof: The max spectrum of $R$ is dense in the spectrum if and only if, for all $f \in R$, either $D(f) = \varnothing$ or there exists a maximal ideal $M$ such that $M \in D(f)$. That is, either $f$ is contained in all prime ideals of $R$ (so that $D(f) = \varnothing$) or there exists a maximal ideal $M$ with $f \notin M$, or equivalently, either $f \in N(R)$ or $f \notin J(R)$. Because $N(R) \subseteq J(R)$, this dichotomy is finally equivalent to $N(R) = J(R)$ as claimed.  $\square$

The well known case that you cite for a finitely generated algebra over an algebraically closed field follows from the fact that $k[x_1, \dots, x_n]$ is a Jacobson ring, so that every homomorphic image thereof is also Jacobson.  In fact, since polynomial rings over any Jacboson ring are Jacobson, and because fields and the ring of integers are Jacobson rings, we can see immediately that every finitely generated $k$-algebra has the property that you desire when $k$ is any field (not necessarily algebraically closed) or when $k = \mathbb{Z}$.
