Coming up with an example What is an example of a finite type $\mathbb{Z}$-algebra $R$ which satisfies the following conditions:
(1) There is no ring map from $R \rightarrow \mathbb{Q}$
(2) For every positive prime $p \in \mathbb{Z}, \exists$ a maximal ideal $m \subset R$ such that $R/m \cong \mathbb{F}_{p}$
 A: It suffices to find an integer polynomial $p(x_1, ... x_n)$ such that $p = 0$ has no rational solutions but has solutions $\bmod p$ for all primes $p$; then $\mathbb{Z}[x_1, ... x_n]/p(x_1, ... x_n)$ is such an algebra. 
In fact we can take $p(x) = (x^2 - 2)(x^2 - 3)(x^2 - 6)$. The equation $p(x) \equiv 0 \bmod q$ has solutions modulo $q = 2, 3$ by inspection; for all larger primes, $\left( \frac{2}{q} \right) \left( \frac{3}{q} \right) = \left( \frac{6}{q} \right)$, so one of $2, 3, 6$ must be a quadratic residue $\bmod q$. 

Some additional comments.


*

*$p$ clearly cannot be linear. 

*The Hasse principle suggests we should not try to look for examples with $p$ quadratic. 

*$p$ cannot be cubic if $n = 1$. This is a consequence of the following more general fact:


Proposition: A monic irreducible integer polynomial in one variable has no roots modulo infinitely many primes $p$.
This is a nice exercise in group theory and Galois theory together with the Frobenius density theorem (nowadays usually subsumed under Chebotarev's density theorem). 
