18
$\begingroup$

The question was cross-posted from Math.SE: https://math.stackexchange.com/questions/4566017/strengthening-ax-grothendieck


The question is simple. The Ax-Grothendieck theorem says a polynomial map $p\colon\mathbb C^n\to\mathbb C^n$ that is injective is also surjective.

Is assuming $p$ has finite fibers enough?

I can't come up with any easy counter-examples, but the proof I know using finite fields does not work.

$\endgroup$
7
  • 3
    $\begingroup$ This sounds way too good to be true, but I am not an expert in complex analysis... (also, etiquette generally dictates a grace period of at least a few days before cross-posting) $\endgroup$
    – Alec Rhea
    Commented Nov 1, 2022 at 17:29
  • 1
    $\begingroup$ @Mohan I don't understand: that map has finite fibers and is also surjective, so it's not a counterexample (nor a proof). $\endgroup$ Commented Nov 1, 2022 at 18:14
  • 3
    $\begingroup$ @JasonStarr Maybe you could give some more details of the argument you have in mind? Already the Ax-Grothendieck theorem (the case when the fibers have size at most $1$) seems to require a more involved argument than what you are suggesting. $\endgroup$ Commented Nov 2, 2022 at 13:05
  • 3
    $\begingroup$ @AlexKruckman The Ax-Grothendieck Theorem is not the result that the OP cites above. Polynomial rings are much simpler than general quotient rings of polynomial rings. $\endgroup$ Commented Nov 2, 2022 at 15:11
  • 2
    $\begingroup$ @JasonStarr Indeed they are! But if you see Ax's model-theoretic proof worked out in the case of $\mathbb{C}^n$, it's essentially obvious how to generalize to arbitrary varieties over algebraically closed fields. Perhaps I was misled by this into thinking that in "pure" algebraic geometry proofs, the kernel of the difficulty would likewise already be present in the case of $\mathbb{C}^n$. Anyway, thank you for expanding on your "sketch". $\endgroup$ Commented Nov 2, 2022 at 16:12

3 Answers 3

17
$\begingroup$

A counterexample for $n = 2$ is the map $(x_1, x_2) \mapsto (x_1x_2 - 1, x_2(x_1x_2 - 1) + x_1)$. This example, which I learned from a paper of Zbigniew Jelonek, was mentioned in an earlier answer of mine.

$\endgroup$
4
  • 2
    $\begingroup$ We have $(x,0)\mapsto(-1,1)$. $\endgroup$ Commented Nov 2, 2022 at 16:15
  • 3
    $\begingroup$ @KentaSuzuki I think it's a typo for $(x_1x_2 - 1,x_2(x_1x_2 - 1)+x_1)$. This is the example in pinaki's linked answer. $\endgroup$ Commented Nov 2, 2022 at 16:17
  • $\begingroup$ @AlexKruckman: thanks! $\endgroup$
    – pinaki
    Commented Nov 2, 2022 at 16:18
  • 4
    $\begingroup$ Nice example! This polynomial misses (0,0), but fibers are all size at most $2$. $\endgroup$ Commented Nov 2, 2022 at 16:25
5
$\begingroup$

Edit. To try and make something productive out of the argument below, let me explain the proof of the Ax-Grothendieck Theorem for affine space using this argument. Also, the argument below includes the proof of the conjecture of the OP when $n$ equals $1$. As noted in the accepted answer, the conjecture is false when $n$ equals $2$. The argument below shows that when the result fails, the image of the morphism is an open subset of affine space whose closed complement has codimension $\geq 2$, containing the image of the singular locus of the "integral closure" variety $X$ and intersecting the singular locus of the branch divisor (this is why the Bertini result fails).

The proof strategy here is similar to that used by Jouanolou in his book on Bertini theorems to simplify the proof of the Quillen-Suslin Theorem.

Lemma 1. For a field $k$, every $k$-subalgebra $R$ of $k[x]$ that strictly contains $k$ is a finite type $k$-algebra, and $k[x]$ is a finitely generated $R$-module. Thus, every nonconstant, dominant $k$-morphism to an integral affine $k$-scheme from $\mathbb{A}^1_k$ is finite, hence surjective.

Proof. This is a variant of the proof of Noether's bound in invariant theory. Let $m(x)\in R$ be any nonconstant monic element. Then the $k$-subalgebra $S$ of $R$ generated by $m(x)$ is a finite type $k$-algebra. Moreover, the element $x\in k[x]$ satisfies a monic polynomial in $t$ with coefficients in $S$, namely $m(t) - m(x)$.

Thus $k[x]$ is a finitely generated $S$-module, and $R$ is an $S$-submodule. Since $S$ is a finite type $k$-algebra, hence Noetherian by the Hilbert Basis Theorem, the $S$-submodule $R$ is finitely generated. Hence $R$ is a finite type $k$-algebra, and the finitely generated $S$-module $k[x]$ is also finitely generated as an $R$-module. QED

Lemma 2. For a field $k$, for every affine, dominant, quasi-finite $k$-morphism $p:U\to Y$ between normal, integral, finite type $k$-schemes, there is a factorization of $p$ as $i:U\to X$ composed with $\overline{p}:X\to Y$, where $i$ is a dense open immersion, and where $\overline{p}$ is a finite $k$-morphism between normal, integral, finite type $k$-schemes.

Proof. This follows from Grothendieck's version of Zariski's Main Theorem and the Noether Normalization Theorem: $X$ is the integral closure of $Y$ in the function field of $U$, this is finite type over $Y$, and $U$ is an open subscheme of $X$. QED

Denote by $D$ the closed complement $X\setminus U$ with its reduced structure.

Lemma 3. The morphism $p$ is not finite if and only if $D$ is a nonempty divisor in $X$.

Proof. Of course the statement that $p$ is finite is equivalent to the statement that $D$ is empty.
Since $X$ is normal and affine over $Y$, the (reduced) closed complement $D=X\setminus U$ of the $Y$-affine scheme $U$ is either empty or has pure codimension $1$. QED

Hypothesis 4. The field $k$ is perfect, both $U$ and $Y$ are smooth $k$-schemes, and $p$ is generically étale.

By construction of $X$, also the non-smooth locus $X_{\text{sing}}$ of $X$ has codimension $\geq 2$. On the smooth open complement $X^o:=X\setminus X^{\text{sing}}$, the reduced Weil divisor is Cartier, i.e., the ideal sheaf is locally principal. Since $\overline{p}$, the ideal sheaf is even principal when restricted to $\overline{p}$-preimages of sufficiently small Zariski open subsets of $Y$, e.g., the inverse image $\overline{p}^{\text{pre}}(Y\setminus E)$ for a divisor $E$ in $Y$ that contains the image $\overline{p}(X_{\text{sing}})$. If $D$ is nonempty, then we can choose $E$ so that it does not contain $\overline{p}(D)$, i.e., so that the restriction of $D$ to the preimage open subset is nonempty and the generator of the principal ideal is a nonzero nonunit.

Problem 5. Assuming $D$ is nonempty, among irreducible divisors $E$ in $Y$ containing $\overline{p}(X_{\text{sing}})$ yet not contain $\overline{p}(D)$, and such that $D$ is principal on $\overline{p}^{\text{pre}}(Y\setminus E)$, is there one such that $\overline{p}^{\text{pre}}(E)$ is irreducible?

Lemma 6. If $p$ is birational, then there is such a divisor $E$.

Proof. If $p$ is birational, then by Zariski's Main Theorem, also $\overline{p}$ is an isomorphism. In particular, it is a homeomorphism for the Zariski topology, and inverse images of irreducible closed subsets are irreducible. QED

Proposition 7. For a dominant,quasi-finite $k$-morphism $p:U\to Y$ between affine spaces over $k$, if there exists a divisor $E$ as in the problem, then already $p$ is finite, hence surjective. In particular, if $p$ is birational, then $p$ is an isomorphism of $k$-schemes.

Proof. The morphism $p$ is finite if and only if the divisor $D$ is nonempty. By way of contradiction, assume that $D$ is nonempty and a divisor $E$ as in the problem exists. Since $Y$ is factorial (by Gauss's Lemma), the irreducible divisor $E$ is $\text{Zero}(f)$ for some irreducible $f\in k[Y] \cong k[y_1,\dots,y_n]$. By hypothesis, in the $k$-algebra $k[X][f^{-1}]$, the ideal of $D$ is principal, say $\langle g\rangle$. Thus, the fraction ring $k[X][\overline{p}^*f^{-1},g^{-1}]$ is already contained in the $k$-subalgebra $k[U][p^*f^{-1}] \cong k[x_1,\dots,x_n][p^*f^{-1}]$.

By hypothesis on $E$, the element $p^*f$ is irreducible in $k[x_1,\dots,x_n]$. Thus, the invertible elements of $k[x_1,\dots,x_n][p^*f^{-1}]$ are precisely the elements of the form $cf^d$ for $c\in k^\times$ and for $d\in \mathbb{Z}$. Since $g$ is invertible in $k[x_1,\dots,x_n][p^*f^{-1}]$, the element $g$ equals $cf^d$ for some $c$ and integer $d$. Thus, $g$ is already invertible in $k[X][\overline{p}^*f^{-1}]$. This contradicts the hypothesis that $E$ does not contain $\overline{p}(D)$. QED

Nota bene. In my original post, I attempted to use Bertini theorems to find $E$ as above: the inverse image of $E$ is connected by the Bertini connectedness theorem. If $E$ intersects transversally the branch divisor of $\overline{p}$ at all generic points of the intersection set, then the inverse image of $E$ is also locally irreducible, and hence $E$ is irreducible. However, as the example in the accepted answer shows, it can happen that $\overline{p}(X_{\text{sing}})$ contains an irreducible component of the singular locus of the branch divisor that has codimension $2$ in $Y$. Thus, the requirement that $E$ contains $\overline{p}(X_{\text{sing}})$ forces $E$ to intersect the branch divisor nontransverally at the generic point of $\overline{p}(X_{\text{sing}})$.

$\endgroup$
2
$\begingroup$

A version of OP's claim (with an additional hypothesis) is mentioned in a footnote on the first page of

where he says that M. Raynaud proved the following:

If $V$ is an algebraic variety, $f: V\to V$ has finite fibers, and is an immersion on a dense open subset of $V$, then $f$ is an isomorphism.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.