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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

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  • $\begingroup$ A very short proof due to Shalen can be found in Lemma 10.3 and Thm 10.4 of math.berkeley.edu/~stall/ch3oz.pdf $\endgroup$
    – Benjamin Steinberg
    Commented Oct 11, 2013 at 2:40
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    $\begingroup$ Let $K$ be a field finitely generated as a $\mathbf{Z}$-algebra. Assume $\mathbf{Q}\subset K$, so $K$ is finitely generated as a $\mathbf{Q}$-algebra, so $\mathbf{Q}$-finite by the Nullstellensatz over fields. A finite set of $\mathbf{Z}$-algebra generators of $K=\mathbf{Q}\otimes_{\mathbf{Z}}O_K$ lies in $O_K[1/N]$ for a sufficiently divisible integer $N>0$, yet $O_K[1/N]$ is a ring, so $K=O_K[1/N]$, absurd (Euclid!). Thus, $p:={\rm{char}}(K)>0$, so $K$ is a finitely generated $\mathbf{F}_p$-algebra, so $\mathbf{F}_p$-finite (Nullstellensatz again). No Jacobson rings, no first-order logic. :) $\endgroup$
    – Marguax
    Commented Oct 11, 2013 at 2:55
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    $\begingroup$ @Marguax, what you really seem to be using is Zariski's lemma en.wikipedia.org/wiki/Zariski%27s_lemma. It seems to me that this is not really any easier than first order logic, although it has the advantage of being part of Commutative Algebra. The logic proof just uses ultraproducts and that two algebraically closed fields of the same characteristic and cardinality are isomorphic. It takes a couple of pages in Marker's book. $\endgroup$
    – Benjamin Steinberg
    Commented Oct 11, 2013 at 15:22
  • $\begingroup$ @Benjamin Steinberg: Sorry, I don't know anything about mathematical logic. Also, perhaps it's a matter of convention for the names of results, but when working over a general field I always thought that the meaning of "Nullstellensatz" is exactly what is called "Zariski's Lemma" by you and Wikipedia (corporations are people, my friend). $\endgroup$
    – Marguax
    Commented Oct 11, 2013 at 22:03
  • $\begingroup$ @Margaux, what's in a name :) $\endgroup$
    – Benjamin Steinberg
    Commented Oct 11, 2013 at 23:16

2 Answers 2

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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.

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  • $\begingroup$ If we replace $\mathbb{Z}$ by the $p$-adic integers $\mathbb{Z}_p$ (which is the $p$-adic completion of $\mathbb{Z}$), then also the above result holds true. Am I correct ? $\endgroup$
    – MAS
    Commented Apr 7, 2021 at 18:35
  • $\begingroup$ I'm probably not the best person to ask on that. I dont think the logic approach world but maybe one of the others does $\endgroup$
    – Benjamin Steinberg
    Commented Apr 7, 2021 at 19:35
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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.

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    $\begingroup$ Actually the fact that the image of Z is finite just means that k is finitely generated over the prime field. But according to the Wikipedia article it is part of the generalized Nullstellensatz that k is a finite extension of Z/p. $\endgroup$
    – Benjamin Steinberg
    Commented Oct 11, 2013 at 2:43
  • $\begingroup$ @BenjaminSteinberg: Yes, there are two parts to the argument --- first, $k$ is finitely generated over the prime field, then this implies that $k$ is finite. Wikipedia uses the phrase "generalized Nullstellensatz" to mean the conjunction of these two statements; I'm not sure how standard that is. $\endgroup$
    – Steven Landsburg
    Commented Oct 11, 2013 at 4:18
  • $\begingroup$ @BenjaminSteinberg: Though if your point was that it takes additional argument to conclude that $k$ is finite, that's of course correct. $\endgroup$
    – Steven Landsburg
    Commented Oct 11, 2013 at 4:19

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