I am seeing the proof of the Ax-Groethendieck theorem from commutative algebra and I have a problem. How can I prove that if $x\_1,...,x\_n$ are complex numbers and $I$ is a maximal ideal of $\mathbb{Z}[x\_1,...,x\_n]$, the quotient $\mathbb{Z}[x\_1,...,x\_n]/I$ is a finite field?


Thanks.


[Infinite fields, finite fields, and the Ax-Grothendieck theorem](http://terrytao.wordpress.com/2009/03/07/infinite-fields-finite-fields-and-the-ax-grothendieck-theorem/)



Quotient of $Z[x_1,...,x_n]$ by a maximal ideal is a finite field
Let $R$ be a finitely generated integral domain (over $\mathbb Z$) and let $I$ be a maximal ideal of $R$. We show $R/I$ is a finite field. Let $K$ be an algebraic closure of $R/I$. Let $p$ be the characteristic of $K$. Suppose $n$ elements generate $R$. Then we can write $R/I= \mathbb Z[x\_1,\ldots x\_n]/(f\_1,\ldots, f\_m)$. Therefore, the first order sentence $\phi=\exists y\_1,\ldots, y\_n[f\_1(y\_1,\ldots,y\_n)=0\wedge\cdots \wedge f\_m(y\_1,\ldots,y\_n)=0]$ 


is true in $K$. There are two cases.


If $p>0$, then since the first order theory of algebraically closed fields of characteristic $p$ is complete we have $\overline {\mathbb F\_p}\models \phi$. It follows that $R/I$ embeds in $\overline {\mathbb F\_p}$ and hence is finite being finitely generated. 


Next suppose $p=0$. By completeness the theory of an algebraically closed field of characteristic $0$ models $\phi$. It is a standard consequence of the compactness theorem of first order logic that there is an algebraically closed field of prime characteristic that models $\phi$. The previous paragraph now shows $R/I$ is a finite field. 


