I was trying to understand completely the [post of Terrence Tao on Ax-Grothendieck theorem](https://terrytao.wordpress.com/2009/03/07/infinite-fields-finite-fields-and-the-ax-grothendieck-theorem/). This is very cute. Using finite fields you prove that every injective polynomial map $\mathbb C^n\to \mathbb C^n$ is bijective. It seems to me that the only point in the proof presented in the post that is not explained completely is the following lemma:



> 
> Take any finitely generated ring over $\mathbb Z$ and quotient it by a maximal ideal. Then the quotient is a finite field. 
> 
> 
> 


Is there some comprehensible reference for the proof of this lemma?


In slightly different wording, the question is the following: assuming Nullstellensatz, can one really give a complete proof of Ax-Grothendick theorem in two pages, so that it can be completely explained in one (2 hours) lecture of an undergraduate course on algebraic geometry? 



A finitely generated $\mathbb{Z}$-algebra that is a field has to be finite
$\DeclareMathOperator\Spec{Spec}$To prove Nullstellensatz over $\mathbb{Z}$: as the morphism $f: \Spec(R)\to\Spec(\mathbb Z)$ is of finite type, a theorem of Chevalley says that the image of any constructible subset is constructible. So the image of any closed point by $f$ is a point which is a constructible subset. This can not be the generic point of $\Spec(\mathbb Z)$, so it must be a closed point.


Note that this does not hold in general. For example, over the ring of $p$-adic integers, the ideal $(pX-1)\mathbb{Z}\_p[X]$ is maximal, but its preimage in $\mathbb{Z}\_p$ is $0$ and it not maximal.


[**EDIT**] Another proof using Noether's normalization lemma: [Noether's normalization lemma over a ring A](https://mathoverflow.net/questions/42276): if a maximal ideal $\mathfrak m$ of $R$ is such that $\mathfrak m\cap \mathbb Z=0$, then $R/\mathfrak m$ is finite type over (and contains) $\mathbb Z$. So there exists $f\in\mathbb Z$ non-zero and a finite injective homomorphism $\mathbb Z\_f[X\_1,\dotsc, X\_d]\hookrightarrow R/\mathfrak m$. But then $\mathbb Z\_f[X\_1,\dotsc, X\_d]$ must be a field. This is impossible because the units of this ring are $\pm f^k$, $k$ relative integers.


