Timeline for For which smooth varieties is the Jacobian conjecture known to be true?
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17 events
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| Jul 9, 2021 at 0:07 | comment | added | David E Speyer | @TakumiMurayama Thanks! | |
| Jul 8, 2021 at 23:53 | comment | added | Takumi Murayama | @DavidESpeyer I think that version of the Ax–Grothendieck theorem is [EGAIV$_3$, Proposition 10.4.11]. There are other versions in [EGAIV$_4$, Proposition 17.9.6] and [EGAIV$_4$, (Err$_{\text{IV}}$, 31)]. | |
| Jul 8, 2021 at 23:47 | comment | added | Qfwfq | @Angelo: you weren't thinking of Ax-Grothendieck as the easy step, were you? | |
| Jul 8, 2021 at 14:34 | comment | added | David E Speyer | The model theory wiki states Ax-Grothendieck in this generality modeltheory.fandom.com/wiki/Ax-Grothendieck_Theorem , and it seems to me that the standard proof adapts easily, although I couldn't quickly find an official source which says this. | |
| Jul 8, 2021 at 14:33 | comment | added | David E Speyer | I believe the Ax-Grothendieck theorem has the following generalization: If $V$ is a variety over an algebraically closed field $K$, and we have a morphism $f : V \to V$ which induces an injection $V(K) \to V(K)$, then $V(K) \to V(K)$ is a bijection. In particular, we cannot have a non-surjective open inclusion from $V$ into itself. | |
| Jul 8, 2021 at 13:59 | comment | added | Francesco Polizzi | @Qfwfq: sorry, I was thinking only about the projective case | |
| Jul 8, 2021 at 13:58 | comment | added | Qfwfq | I must be missing something. Are you saying $f(X)$ has to be a projective variety? But $X$ isn't nec. projective | |
| Jul 8, 2021 at 13:49 | comment | added | Francesco Polizzi | Well, the image of $f$ must be both open in $X$ (since $f$ is open) and projective (since $f$ is an embedding)... | |
| Jul 8, 2021 at 13:32 | comment | added | Qfwfq | I figured ZMT (stacks.math.columbia.edu/tag/05K0) implies $f$ is an open embedding. How can one easily see that $f$ is also surjective? | |
| Jul 8, 2021 at 11:31 | comment | added | Angelo | By Zariski's main theorem, an injective morphism of smooth complex varieties is an open embedding. From this it is easy to deduce that an injective endomorfism of a smooth complex variety is an isomorphism. Hence RP and JC are equivalent. | |
| Jul 8, 2021 at 10:10 | comment | added | Qfwfq | Yes, of course you're right: I meant $x\mapsto 2x$. | |
| Jul 8, 2021 at 10:06 | comment | added | Francesco Polizzi | $x \mapsto -x$ is an isomorphism. To have a counterexample, one should take (for instance) $x \mapsto 2x$. Or am I missing something? | |
| Jul 8, 2021 at 10:03 | comment | added | Qfwfq | I wouldn't count Abelian varieties as interesting counterexamples in this context, because they always admit the $x\mapsto -x$ isogeny, so they trivially do not satisfy (RP) or (JC). | |
| Jul 8, 2021 at 9:51 | comment | added | Francesco Polizzi | At any rate, the study of smooth projective varieties admitting a non-trivial étale endomorphism is an interesting problem in itself. See for instance projecteuclid.org/journals/osaka-journal-of-mathematics/… for the 3-fold case. | |
| Jul 8, 2021 at 9:42 | comment | added | Francesco Polizzi | Do you include Abelian varieties among the "interesting counterexamples"? | |
| Jul 8, 2021 at 9:40 | history | edited | Qfwfq | CC BY-SA 4.0 |
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| Jul 8, 2021 at 8:57 | history | asked | Qfwfq | CC BY-SA 4.0 |