There's no surjective ring homomorphism from $\mathbb{Z}[x_1,\dots,x_n]$ onto $\mathbb{Q}$. I'm trying to prove that, if a field $A$ is also a finitely generated $\mathbb{Z}$-algebra, then $A$ is finite. The proof I found for this depends on the fact that $\mathbb{Q}$ cannot be a finitely generated $\mathbb{Z}$-algebra.
So, my question (which is equivalent to the title of the post, right?) is how to prove that $\mathbb{Q}$ cannot be a finitely generated $\mathbb{Z}$-algebra?
Thanks.
 A: You can prove explicitly that $\Bbb Q$ is not finitely generated. Say $q_1,\ldots, q_n$ generate $\Bbb Q$. Then set $q_i={r_i/s_i}$ with $r_i,s_i$ reduced fractions. Then if we look at a prime $p$ which does not divide any of the $s_i$--such a thing exists because only finitely many primes divide the finitely many $s_i$--Then take an arbitrary element in the algebra $\Bbb Z[q_1,q_2,\ldots, q_n]$
$$S=f(q_1, q_2,\ldots, q_n), \quad f(x)\in\Bbb Z[x_1,\ldots q_n]$$
If a highest degree term of $f$ is given by $x_1^{m_1}\ldots x_n^{m_n}$ and $m=m_1+\ldots +m_n$ then a common denominator for the sum is given by $(s_1\ldots s_n)^m$. Even when reducing fractions, we cannot have any other primes in algebraic (i.e. polynomial) combinations of such elements, so $f(x)\ne {1\over p}$ and our set of generators was insufficient to generate ${1\over p}$ and so $\Bbb Q$ is not finitely generated.

With respect to your question of "equivalent to the thing in the title," the answer is "yes." Because if you have a basis $q_1,\ldots, q_n$ then the surjective map would be $x_i\mapsto q_i$, all $n\in\Bbb Z$ map to themselves.
A: Hint  $\,\ \Bbb Q = \Bbb Z\left[\dfrac{a_1}{b_1},\cdots,\dfrac{a_n}{b_n}\right]\,\Rightarrow\,\Bbb Q = \Bbb Z\left[\dfrac{1}b\right],\ b = b_1\cdots b_n\,\Rightarrow\,$ every prime $\,p\mid b\ $ since
$$\frac{1}p \in \Bbb Z\left[\frac{1}b\right]\,\Rightarrow\, \frac{1}p =  \frac{a}{b^n}\,\Rightarrow\, ap = b^n\,\Rightarrow\, p\mid b^n\,\Rightarrow\, p\mid b$$
Remark $\ $ Kaplansky calls G-domains those domains whose quotient field is so finitely generated (equivalently, as above, singly-generated), see section $1$-$3$ of his Commutative Rings.
The above yields that a PID is a G-domain iff it has finitely many primes. More generally one can show that a Noetherian domain is a G-domain iff it has finitely many nonzero prime ideals all of which are maximal, i.e. the domain is $1$-dimensional and semilocal.
