I was trying to understand completely the post of Terrence Tao on 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?

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    $\begingroup$ This was previously answered: mathoverflow.net/questions/30599/… $\endgroup$
    – fherzig
    Mar 14 '11 at 22:19
  • $\begingroup$ I still wonder where this little Lemma appears in the literature ... $\endgroup$ Oct 13 '13 at 8:52
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    $\begingroup$ According to Serre, there's a proof of this lemma in Bourbaki, N. Algèbre Commutative. Chapitre V. Entiers, Hermann, Paris, 1964. (p. 68, cor. 1) $\endgroup$ May 19 '18 at 6:39
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    $\begingroup$ @MikePierce Indeed I find it p64 (chap 5, §3, no.4). Statement is (translated and summarized) "Every finitely generated [commutative] algebra over $\mathbf{Z}$ is a Jacobson ring; a prime ideal $P$ is maximal iff the quotient ring is finite." $\endgroup$
    – YCor
    May 19 '18 at 8:50

To prove Nullstellensatz over $\mathbb{Z}$: as the morphism $f: \mathrm{Spec}(R)\to\mathrm{Spec}(\mathbb Z)$ 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 $\mathrm{Spec}(\mathbb Z)$, 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: 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 exits $f\in\mathbb Z$ non-zero and a finite injective homomorphism $\mathbb Z_f[X_1,\dots, X_d]\hookrightarrow R/\mathfrak m$. But then $\mathbb Z_f[X_1,\dots, X_d]$ must be a field. This is impossible because the units of this ring are $\pm f^k$, $k$ relative integers.

  • $\begingroup$ This first argument is quite nice! $\endgroup$ Mar 6 '11 at 2:49
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    $\begingroup$ The theorem of Chevalley can be proved by induction, using the second argument :). $\endgroup$
    – Qing Liu
    Mar 6 '11 at 10:17
  • $\begingroup$ @Qing Liu, thanks for the answer. I am lost when you say in lines 2-3 of Edit: "So there exits $f\in \mathbb Z$ non-zero and a finite injective homomorphism ...". Why is there such $f$? $\endgroup$
    – aglearner
    Mar 7 '11 at 0:16
  • $\begingroup$ Sorry, my punctions were not good. I should write "If a maximal ideal ..., so there exists f". The existence of such a $f$ is a form of Noether's normalization lemma that you can find at mathoverflow.net/questions/42276 $\endgroup$
    – Qing Liu
    Mar 7 '11 at 8:14
  • $\begingroup$ After one year and half I guess I understand the logic of both answers. Could you please say if there is a pedagogic explanation of Chevalet's theorem somewhere on the web? $\endgroup$
    – aglearner
    Oct 14 '12 at 13:34

Let $R$ be a finitely generated $\mathbb{Z}$-algebra, and $\mathfrak{m}\subset R$ are maximal ideal. We wish to show $R/\mathfrak{m}$ is a finite field.

Let $i: \mathbb{Z}\to R$ be the unique ring map; then $i^{-1}(\mathfrak{m})$ is a maximal ideal in $\mathbb{Z}$ (as $R$ is finitely generated over $\mathbb{Z})$, and thus $\mathbb{Z}/i^{-1}(\mathfrak{m})$ is a finite field $\mathbb{F}_p$ for some prime $p$. As $R$ is finitely generated over $\mathbb{Z}$, $R/\mathfrak{m}$ is finitely generated over $\mathbb{F}_p$. But all finite field extensions of $\mathbb{F}_p$ are still finite, completing the proof.

  • $\begingroup$ Daniel, thanks for the proof of the lemma! $\endgroup$
    – aglearner
    Mar 6 '11 at 0:16
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    $\begingroup$ Perhaps it's worth emphasizing that what makes this argument work is the non-trivial fact (proved in Qing Liu's answer) that if $f:\mathbb{Z}\rightarrow\mathbb{R}$ is of finite type, then the pre-image of a maximal ideal of $R$ under $f$ is non-zero. This is a particular case of the general Nullstellensatz on Jacobson rings (found in, e.g., Eisenbud's book). A proof of the statement about fields finitely generated as rings using the usual Nullstellensatz is outlined in Exercise 6 of the Noetherian rings chapter of Atiyah and Macdonald. $\endgroup$ Mar 6 '11 at 1:07
  • $\begingroup$ @Keenan Kidwell: Of course; thanks for making that explicit. $\endgroup$ Mar 6 '11 at 1:13
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    $\begingroup$ An ignorant question: "$R/\mathfrak{m}$ is finitely generated over $\mathbb F_p$. But all finite field extensions of $\mathbb F_p$ are still finite, completing the proof". I'm missing something here. How did finitely generated extension get to be the same as finite field extension? (in particular isn't $\mathbb F_p(x)$ a finitely generated extension but not a finite extension)? $\endgroup$ Mar 6 '11 at 21:44
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    $\begingroup$ $\mathbb{F}_p(x)$ is not finitely generated as an $\mathbb{F}_p$-algebra; in particular, all the irreducibles need to be inverted. See e.g. mathreference.com/ag,fgaf.html $\endgroup$ Mar 6 '11 at 21:54

One can give a more elementary proof of the fact that $\mathfrak{m} \cap \mathbb{Z} \neq 0$ - By more elementary I mean a proof that only uses the Nullstellensatz over $\mathbb{Q}$.

Notice that it is enough to verify the claim for $R=\mathbb{Z}[x_1,..,x_n]$, and $\mathfrak{m} \in Max(R)$.

Suppose there is $\mathfrak{m} \in Max(R)$ such that $\mathfrak{m} \cap \mathbb{Z} =0$. Then, we may assume that $\mathbb{Z} \subseteq F :=\mathbb{Z}[x_1,..,x_n]/\mathfrak{m}$. If we denote by $\alpha_{i}=x_i+\mathfrak{m}$ we have that $F=\mathbb{Z}[\alpha_1,..,\alpha_n]$. Since $F$ is a field we conclude that $\mathbb{Z}[\alpha_1,..,\alpha_n]=\mathbb{Q}(\alpha_1,..,\alpha_n)$.

Claim: $F/\mathbb{Q}$ is an algebraic extension.

proof: $F/\mathbb{Q}$ is a finitely generated field extension- generated as an algebra- in particular $F$ is of the form $\mathbb{Q}[y_1,..,y_m]/M$ for some $M$ maximal ideal of $\mathbb{Q}[y_1,..,y_m].$ By the Nullstellensatz $M$ has a zero $(\beta_1,...,\beta_m)$ where each $b_i$ is algebraic over $\mathbb{Q}$, so $F=\mathbb{Q}(\beta_1,...,\beta_m)$ is algebraic over $\mathbb{Q}$.

Since each $\alpha_{i}$ is algebraic, there are integers $q_i$'s such that $q_{i}\alpha_{i}$ is integral over $\mathbb{Z}$ for all $i$. In particular $F=\mathbb{Z}[\alpha_1,..,\alpha_n]$ is an integral extension of $\displaystyle \mathbb{Z}[\frac{1}{q_1},..,\frac{1}{q_n}]$. Since $F$ is a field we have that $\displaystyle \mathbb{Z}[\frac{1}{q_1},..,\frac{1}{q_n}]$ is a field, which is a contradiction( $p$ is not invertible for any prime not dividing $q_{1}...q_{n}$).

  • $\begingroup$ Unfortunately, I can not understand when you write " By the Nullstellensatz we have that each $\alpha_i$ is algebraic over $\mathbb Q$". Could you please explain this point? Are you using Nullstelensatz over $\mathbb Q$ here? To which ring are you applying it? $\endgroup$
    – aglearner
    Mar 7 '11 at 0:12
  • $\begingroup$ @aglearner: I've added an explanation to what you are wondering. The point is that one version of the Nullstellensatz, which I learned by the name algebraic Nullstellensatz, is the following: A finitely generated extensions of fields $F/K$ is algebraic. $\endgroup$ Mar 7 '11 at 1:21

This is not an answer to your question, but let me point out that the Ax-Grothendieck theorem is now easy to prove using E-polynomials (Hodge-Deligne polynomials). If $f:X \to X $ is an injective endomorphism of a complex algebraic variety, then $E(X) = E(f(X))=E(X)-E(X\setminus f(X))$. So $E(X\setminus f(X))=0$ and $X\setminus f(X) = \emptyset$, because the degree of a constructible set is twice its dimension. Since one supposes the mixed Hodge theory, this proof is not trivial at all. But at least for me, this looks more natural.

  • $\begingroup$ I hadn't heard about this, and I don't know anything about E-polynomials. Do you know if the proof you describe uses anything in characteristic p in the background? Any "spreading" over Z? $\endgroup$
    – Marty
    Mar 6 '11 at 3:17
  • $\begingroup$ This proof uses only the mixed Hodge theory, and one does not need to switch the base field or ring. On the other hand, both proofs by E-polynomial and finite fields have the "motivic" nature. Namely the E-polynomial and the number of rational points are generalized Euler charqcteristics, that is, they have the additivity and multipicativity. $\endgroup$ Mar 6 '11 at 9:39
  • $\begingroup$ But I wonder, since much of mixed Hodge theory (after Deligne) lifts characteristic p results to get characteristic zero results. And perhaps even transcendental proofs in Hodge theory (like those of Saito) might hide some characteristic p aspects. Do you know if characteristic p is hidden in the background for the results on the E-polynomial? $\endgroup$
    – Marty
    Mar 10 '11 at 22:05

Let me add a self-contained answer which is completely elementary and avoids both the Nullstellensatz and Noether's normalization. It is a polished version of this article (already linked in Daniel Litt's comments). It partially overlaps with the answers of Daniel Litt and Guillermo Mantilla.

Lemma 1. Let $R$ be a UFD with infinitely many primes. Then an algebraic field extension $L$ of $F:=\operatorname{Frac}(R)$ cannot be finitely generated as an $R$-algebra.

Proof. Assume $L=R[y_1,\dots,y_n]$, with each $y_j$ being a root of a certain monic polynomial $p_j\in F[x]$. Taking the common denominator $d\in R$ of the coefficients of these polynomials $p_j$, we get that $y_j$ is integral over $R':=R[1/d]\subseteq F$. But then, given a prime $p\nmid d$ (here we use that $R$ has infinitely many primes), the same holds for $1/p\in L=R'[y_1,\dots,y_n]$. However, since $R'$ is still a UFD, this implies $1/p\in R'$ (as a UFD is integrally closed), contradiction. $\blacksquare$

Lemma 2. Given a field extension $K\subseteq L$, if $L$ is finitely generated as a $K$-algebra then the extension is algebraic, and in particular finite (meaning $\operatorname{dim}_K L<\infty$).

Proof. Assume $L=K[z_1,\dots,z_m]$. Since $L=K(z_1)[z_2,\dots,z_m]$, by induction $L$ is algebraic over $K(z_1)$. If $z_1$ is transcendental over $K$, then $R:=K[z_1]\cong K[x]$ satisfies the hypotheses of Lemma 1, which contradicts that $L$ is finitely generated as an $R$-algebra. So $z_1$ is algebraic over $K$, hence also $L$. (Note that the base case $m=1$ is obvious, as $1/z_1\in K[z_1]$ implies that $z_1$ is the root of some polynomial over $K$.) $\blacksquare$

Theorem. If $L$ is a field which is finitely generated, then $L$ is a finite field.

Proof. $L$ is (isomorphic to) the quotient of the ring $\mathbb{Z}[x_1,\dots,x_n]$ by a maximal ideal $M$. Observe that $M\cap\mathbb Z$ is a prime ideal of $\mathbb Z$. If $M\cap\mathbb Z=\{0\}$, then $\mathbb Z$ embeds into $L$; but this contradicts Lemma 1 (with $R:=\mathbb Z$)! Hence $\mathbb F_p$ embeds into $L$ for some prime number $p$ and Lemma 2 gives that $L$ is finite. $\blacksquare$


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