2017
DOI: 10.1112/s0010437x17007187
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The existence of Zariski dense orbits for polynomial endomorphisms of the affine plane

Abstract: Abstract. In this paper we prove the following theorem. Let f be a dominant polynomial endomorphism of the affine plane defined over an algebraically closed field of characteristic 0. If there is no nonconstant invariant rational function under f , then there exists a closed point in the plane whose orbit under f is Zariski dense.This result gives us a positive answer to a conjecture proposed by Medvedev and Scanlon, by Amerik, Bogomolov and Rovinsky, and by Zhang, for polynomial endomorphisms of the affine pl… Show more

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Cited by 16 publications
(7 citation statements)
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“…The proof of the Theorem 6.2 relies on the theory of valuative tree introduced by Favre and Jonsson in [8], which was developed by Favre, Jonsson and the author in [9,10,24,25,26]. 6.1.…”
Section: Endomorphisms Of Amentioning
confidence: 99%
“…The proof of the Theorem 6.2 relies on the theory of valuative tree introduced by Favre and Jonsson in [8], which was developed by Favre, Jonsson and the author in [9,10,24,25,26]. 6.1.…”
Section: Endomorphisms Of Amentioning
confidence: 99%
“…This is explained in [39, Theorem 6.2], and [75,Theorem 5.6]. In fact, this version of Amerik's theorem has been used throughout the literature; see for instance [7,Lemma 3.3] and [76].…”
Section: Dominant Self-maps Of Pseudo-mordellic Varietiesmentioning
confidence: 99%
“…In case (3), each point in the orbit of [3 : 2 : 1] under φ 6 is (D, S)-integral for S = {∞, 2, 3}, and it is easy to see from the 2-adic and the 3-adic valuations that an algebraic curve can only contain finitely many points in this orbit. For cases (1) and (2), it follows from a recent work of Xie [35] that there exists an algebraic point whose orbit under φ 6 is Zariski-dense (if there exists ℓ ≥ 1 such that the two eigenvalues of the tangent map at a fixed point of φ 6ℓ are multiplicatively independent, we can invoke [2, Corollary 2.7] instead). By enlarging k and S so that all coordinates of this point are S-units and that all the coefficients of F and G lie in O k,S , we see that each point in this orbit is (D, S)-integral.…”
Section: Integral Points In Orbitsmentioning
confidence: 99%