Ax-Grothendieck Theorem - Sheaf of Thoughts

Introduction¶

We all know from basic linear algebra that injective endomorphisms of a finite-dimensional vector space are also surjective. The same holds for finite sets. Over algebraically closed fields, there is an analogous result:

We will talk about an analogous and more general statement about schemes later on.

Let us comment briefly on the hypotheses. Injectivity is the right assumption for $F$ : the map $z\mapsto z^2$ is surjective but not injective on $\mathbb{C}$ . Furthermore, the result is generally false for non-algebraically closed fields; for example, $x \rightarrow x^3$ is injective but not surjective on $\mathbb{Q}$ .

As a sanity check, let’s verify the result for $n=1$ . Consider a polynomial $f\in k[x]$ such that the map $x\mapsto f(x)$ is injective, then $f$ can have at most 1 root. Since $k$ is algebraically closed, $f=ax+b$ with $a\neq 0$ (otherwise our map would not be injective). Hence $f$ is surjective.

Moreover, the theorem is easy to prove when $k=\overline{\mathbb{F}}_p$ .

Proof of the theorem¶

We start by proving two lemmas, and then proceed with the proof of the theorem.

Now we are ready to prove the theorem.

Proof 4 (proof of the Theorem 1)

Suppose $\underline{z}\in k^n$ is not in the image of $F$ . Consider $A$ to be the $\mathbb{Z}$ -algebra generated by:

the coefficients of the $f_i$ ’s defining $F$ ;
the $z_i$ ’s; the coefficients of the $q_{i,j}$ as in Lemma 1;
the coefficients of the $r_i$ as in Lemma 2.

Since finitely many coefficients are involved, the algebra $A$ is of finite type over $\mathbb{Z}$ . In addition, we can restrict $F:A^n \rightarrow A^n$ is still injective. Moreover, we have the following identities in $A[x_1,\dots,x_n,y_1,\dots,y_n]$ given respectively by Lemma 1 and Lemma 2:

\sum_{i=1}^ng_i\cdot q_{i,j}=(x_j-y_j)^{k_j}

(3)

\sum_{i=1}^n r_i\cdot (f_i-z_i)=1

(4)

Now pick a maximal ideal $m$ in $A$ . The field $A/m$ is a finite by the following claim:

CLAIM: Let $B$ be an algebra of finite type over $\mathbb{Z}$ . If $m$ is a maximal ideal of $B$ , then $B/m$ is a finite field

Assuming the claim, set $L=A/m$ and note that the reduction $\overline{F}:L^n\rightarrow L^n$ of $F$ modulo $m$ is injective. Indeed, consider $\underline x,\underline y\in L^n$ such that $\overline{F}(\underline x)=\overline{F}(\underline y)$ . By evaluating equation (3) in $(\underline x,\underline y)$ we get $\underline x=\underline y$ in $L^n$ (having $\overline g_i(\underline x, \underline y)=0$ for all $i$ ).

Since $L^n$ is finite and $\overline{F}$ is injective, $\overline F$ is surjective. However, reducing $\underline{z}$ modulo $m$ and evaluating equation (4) at a preimage of that reduction under $\overline F$ yields a contradiction. Hence, up to proving the claim, we have the theorem.

To prove the claim, consider two cases:

If $m\cap \mathbb{Z}=p \mathbb{Z}$ , then $B/m$ is an algebra of finite type over $\mathbb{F}_p$ ; by Zariski’s lemma, $B/m$ is a finite extension of $\mathbb{F}_p$ , hence finite.
If $m\cap \mathbb{Z}=0$ , then in the tower $\mathbb{Z} \subset \mathbb{Q} \subset B/m$ , we see that $B/m$ is of finite type over $\mathbb{Q}$ , hence is finite over $\mathbb{Q}$ by Zariski’s lemma. Artin-Tate’s lemma then implies that $\mathbb{Q}$ would be of finite type over $\mathbb{Z}$ , which is a contradiction.

Generalization to schemes¶

Note that the map $F$ considered in Theorem 1 can be seen as the the map induced on closed points by a morphism of $k$ -schemes $\mathbb{A}^n_k\rightarrow \mathbb{A}^n_k$ . This leads to the following question:

Let $k$ be an algebraically closed field and $X$ a scheme of finite type over $k$ . If $f:X\rightarrow X$ is a $k$ -endomorphism injective on $k$ points, is it surjective on all points?

The finite-type assumption is necessary: the map $k[X_1,\dots, X_n,\dots]\to k[X_1,\dots, X_n,\dots]$ sending $X_1\mapsto 0$ and $X_i\mapsto X_{i-1}$ , for $i>1$ , is injective but not surjective.

The answer to the question is positive, and it follows not trivially from the following theorem.

Moreover, if $X$ is reduced and $k$ is an algebraically closed field of characteristic 0, then an injective endomorphism on $k$ -points is in fact an isomorphism (not merely a bijection). This is not trivial since we have bijective morphisms of schemes that are not isomorphisms (for example the one induced by $k[t^2,t^3]\rightarrow k[t]$ ). A reference for this is Lemma 1 of this article.

We will not give a proof of Theorem 2; see J. Ax paper Injective endomorphisms of varieties and schemes, or Grothendieck’s version in EGA-IV $_3$ , Proposition 10.4.11.

Let us explain why Theorem 2 implies that injectivity on $k$ -points gives surjectivity on all points.

The first ingredient is the Chevalley’s theorem on constructible sets (see Hartshorne exercise II.3.19):

The following lemma will explain why Theorem 2 implies Theorem 1.

Assuming this lemma, consider a morphism $\mathbb{A}^n_k\to \mathbb{A}^n_k$ over $k$ which is injective on closed points (which are exactly $k$ -points, since $k$ is algebraically closed). Then in particular $f$ results to be injective on all points. From Theorem 2 we get the surjectivity of $f$ on all points (answering to the question raised before) and in particular also surjectivity on closed points, since only closed points are sent to closed points. In particular, we get Theorem 1.

Proof 5 (proof of Lemma 3)

Observe that the diagonal $\Delta_f:X\to X\times_Y X$ is of finite type. Indeed, it is locally of finite type because $X\to \mathrm{Spec}(k)$ is locally of finite type (and $f$ can be seen as the first factor of this map), and is quasi compact because $X$ is noetherian (being of finite type over $k$ ).

Moreover the diagonal $\Delta_f$ is surjective on $k$ -points. Indeed for any $k$ -point of $X\times_Y X$ given by the morphism $\psi: \mathrm{Spec}(k)\to X\times_Y X$ we can consider the diagram

and by injectivity of the morphism $X\to Y$ we get that $\phi_1=\phi_2$ as map of topological space. They are also the same on the structure sheaf, because $k$ is algebraically closed. Therefore, $\phi_1=\phi_2$ , which means that $\psi$ factorizes through the diagonal $\Delta_f$ .

Furhtermore, $X\times _Y X$ is a jacobson scheme (i.e. its underlying topological space is jacobson) because it is of finite type over $k$ which is jacobson. By Theorem 3 the set $\Delta_f(X)$ is constructible. Consider that $Z:=(X\times_Y X)\smallsetminus \Delta_f(X)$ is constructible and that every closed point of $X\times_Y X$ is inside $\Delta_f(X)$ (by surjectivity of $\Delta_f$ on closed points). Since a constructible in a jacobson scheme always contains a closed point, we get that $Z=\emptyset$ . Therefore, $\Delta_f$ is surjective on all points.

Now we prove that the surjectivity of $\Delta_f$ implies the universal injectivity of $f$ . Indeed, consider any $g: Z\to Y$ and the diagonal $\Delta: X\times_Y Z\to (X\times_Y Z)\times_Z(X\times_Y Z)$ of the base change. This diagonal map $\Delta$ is surjective because it is the base change of $\Delta_f$ , which is surjective. Thus, we reduce to prove that $f$ is injective.

Consider $x_1,x_2\in X$ with $f(x_1)=f(x_2)=:y$ . We can find a field $L$ such that the following commutes:

We get the following commutative diagram:

By surjectivity of $\Delta_f$ we can find $x\in X$ such that $\Delta_f(x)$ is the image of $\epsilon$ , hence we get $x_1=x_2$ .

Note that the complex-geometric picture doesn’t match the algebraic one. For example, there are holomorphic maps $\mathbb{C}^n\rightarrow \mathbb{C}^n$ that are injective but not surjective. However, this cannot happen for $n=1$ , as it can be shown that all holomorphic injective $f: \mathbb{C}\rightarrow \mathbb{C}$ are of the form $f=ax+b$ , by the Casorati-Weierstrass Theorem.

Conclusion¶

To conclude we point your attention to the fact that this theorem is linked to a famous conjecture, namely the Jacobian’s conjecture:

Over an algebraically closed field of char 0, the fact that the bijection in Theorem 2 becomes an isomorphism shows that to prove the jacobian conjecture for $\mathbb{A}^n_k$ it is enough to show that a map $F:\mathbb{A}^n_k\rightarrow \mathbb{A}^n_k$ which has invertible jacobian determinant is injective.

I want to thank the friend of mine who suggested me this topic and helped me writing this post.

References¶

This post was mainly inspired by this post from Terence Tao’s blog. Other references are cited in the text when used.