Quotient of $Z[x_1,...,x_n]$ by a maximal ideal is a finite field 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
 A: Use that ${\mathbb Z}$ is a Jacobson ring, so that according to the generalized Nullstellensatz, the inverse image of the maximal ideal $(0)$ in $k={\mathbb Z}[x_1,\ldots, x_n]/I$ is also a maximal ideal, i.e. an ideal of the form $(p)$  (with $p\neq 0$).  This means the image of ${\mathbb Z}$ in $k$ is finite, so $k$ is a finite extension of a finite field, hence finite.
A: 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. 
