$\mathbb Q$ is not finitely generated $\mathbb Z$-algebra I proved that $\mathbb Q$ is not finitely generated as $\mathbb Z$-module using Nakayama Lemma. Now I want to prove the following:

$\mathbb Q$ is not finitely generated as $\mathbb Z$- algebra.

I don't want to use the result that: A field which is finitely generated $\mathbb Z$-algebra is finite.
Any help will be appreciated. Many thanks.
 A: Suppose it were. What primes can appear as denominators in any algebra element of $\mathbb Q$? Can there be infinitely many? 
In fact, suppose we have a domain $D$ for which $F=Q(D)$ is a finitely generated $D$-algebra. By taking a common denominator, it follows that $F = D[s^{-1}]$ for some $s\in D$. It is evident from this that $\mathbb Q$ is not a finitely generated $\mathbb Z$-algebra, since $s$ can contain only finitely many primes, as suggested above. 
Domains $D$ for which $F=Q(D)$ is a finitely generated $D$-algebra are called Goldstein domains. They play an important role in a generalization of Hilbert's Nullstellensatz to what are known as Jacobson (or Hilbert) rings.
If $R$ is any (commutative) ring, a prime $\mathfrak p$ in $R$ is Goldstein if the domain $R/\mathfrak p$ is Goldstein. A Hilbert ring $R$ is a ring all whose Goldstein prime ideals are maximal -- this equivalent to $R$ having all its quotients with nilradical equal to its Jacobson radical.
If $R$ is a Hilbert ring and $S$ is a finitely generated $R$-algebra, then


*

*$S$ is also a Hilbert ring,

*If $\mathfrak n$ is a maximal ideal in $S$, its contraction $\mathfrak m$ to $R$ is maximal in $R$ and,

*The field extension $\dfrac{S}{\mathfrak n}  / \dfrac{R}{\mathfrak m}$ is finite. 


In particular taking $k=R$ a field, $k$ is trivially a Hilbert ring, and the above says that if $K$ is a finitely generated algebra that is a field then $[K:k]$ is finite -- this is the content of the Zariski lemma, crucial to proving Hilbert's Nullstellensatz.
Add. I should have noted that the integers are a Jacobson ring, and hence the fact you don't want to use is a consequence of this generalized Nullstellensatz. 
