3
$\begingroup$

Let $X$ be an integral Noetherian scheme that has no automorphisms other than the identity. Is it possible to characterize (in a not completely tautological way) $X$ such that

  • there is no finite morphism $X\rightarrow X$ surjective on the underlying topological spaces (other than the identity)?
  • there is no finite flat morphism $X\rightarrow X$ surjective on the underlying topological spaces (other than the identity)?

If $X$ is regular, then there is no distinction between two bullet points.

Among spectra of fields, one can find examples satisfying the first (eq. second) bullet point ($\mathrm{Spec}\,\mathbb{Q}$) or not satisfying the first (eq. second) bullet point ($\mathrm{Spec}\,k$ for $k$ rigid non-perfect).

$\endgroup$
  • 3
    $\begingroup$ I think the answer is pretty clearly "no". The existence of such a morphism is a rare event. It's possible to give many obstructions to its existence, but the only way to guarantee its existence is to exhibit one. $\endgroup$ – Will Sawin Apr 24 '19 at 13:58
  • $\begingroup$ @WillSawin maybe you could give some necessary conditions then? $\endgroup$ – user138661 Apr 24 '19 at 14:13
  • 1
    $\begingroup$ Well for instance if the canonical bundle is very ample, because any finite map has to be injective on sections of the canonical bundle, the map would be isomorphic on sections of the canonical bundle, which makes it injective because the canonical bundle is very ample, hence an isomorphism, hence the identity by assumption. $\endgroup$ – Will Sawin Apr 24 '19 at 14:34
  • 1
    $\begingroup$ Just a remark -- Will's comment is for smooth projective schemes over a field of characteristic zero; in characteristic p every scheme has the Frobenius endomorphism, which will often be an example of what you're looking for. $\endgroup$ – Daniel Litt Apr 24 '19 at 14:49
  • $\begingroup$ Sort of similarly to the Frobenius, toric varieties have "toric Frobenius" endomorphisms $F_k$ for each $k$, obtained by extended the map $(x_1,\ldots,x_n) \mapsto (x_1^k, \ldots, x_n^k)$ from the torus to the whole variety. $\endgroup$ – Bort Apr 24 '19 at 14:58
2
$\begingroup$

The question as stated seems to me too broad. Nevertheless, in some cases it is actually possible to provide a complete characterization. For instance, ruled surfaces admitting non-trivial, surjective endomorphisms are classified in

N. Nakayama: Ruled surfaces with non-trivial surjective endomorphisms, Kyushu J. Math. 56, No. 2, 433-446 (2002). ZBL1049.14029.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ among surfaces with no finite surjective morphisms that are not automorphisms, are there surfaces with no automorphisms (other than the identity)? Quick scan through the paper did not tell me the answer and I am bad at surfaces. $\endgroup$ – user138661 Apr 24 '19 at 14:50
  • $\begingroup$ It seems to me (but I should check better) that all examples given by Nakayama have non-trivial automorphisms. $\endgroup$ – Francesco Polizzi Apr 24 '19 at 15:02
1
$\begingroup$

A related question studied quite extensively in the literature is the existence of polarized endomorphisms. Let $X$ be a variety over $k$, for simplicity smooth and projective. An endomorphism $f\colon X\to X$ is polarized if there exists an ample line bundle $L$ on $X$ an an integer $n>1$ such that $f^* L\simeq L^n$. There is a conjecture (I think due to Fakhruddin) that if $k=\mathbb{C}$ and $X$ admits a polarized endomorphism, then up to a finite etale cover, $X$ is a bundle of toric varieties over an abelian variety. I believe the same should hold in positive characteristic under a separability assumption (the Frobenius is polarized endomorphism).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy