Preperiodic portraits for unicritical polynomials (Q2802109)

One may check that \(z\mapsto z^2 -z\) has no points of period two. I.\ N.\ Baker showed that, up to conjugacy, this is the only complex polynomial without a period two point [\textit{I. N. Baker}, J. Lond. Math. Soc. 39, 615--622 (1964; Zbl 0138.05503)].NEWLINENEWLINEGiven a polynomial \(\phi\), the author says a point \(x\) has \textit{preperiodic portrait} \((M,N)\) if \(M,N\) are minimal with \(\phi^{M+N}(x) = \phi^M(x)\). From Baker's result, it is easy to check that, up to the above class of counterexamples, for any \(M \geq0, N\geq1\), a polynomial of degree at least two always has a point with portrait \((M,N)\).NEWLINENEWLINEOne may mark a point \(x\) and ask a dual question, namely whether there exists a map for which the point has a given portrait. Naturally, one must rule out affine conjugacies. The author considers the family \(f_{d,c} \mapsto z^d + c\) and shows the following result:NEWLINENEWLINETheorem. Let \(K\) be an algebraically closed field of characteristic zero, and let \((x,M,N,d) \in K \times \mathbb Z^3\) with \(M \geq 0\), \(N \geq 1\), and \(d \geq 2\). Then there exists \(c \in K\) for which \(x\) has portrait \((M,N)\) under \(f_{d,c}\) if and only if NEWLINE\[NEWLINE (x,M) \neq (0,1) \text{ and } (x,M,N,d) \not \in \left\{ \left(-\frac{1}{2},0,2,2 \right), \left(\frac{1}{2}, 1, 2, 2 \right), \left( \pm 1, 2, 2, 2 \right) \right\}. NEWLINE\]NEWLINE This answers a question of [\textit{D. Ghioca} et al., Math. Proc. Camb. Phil. Soc. 159, No. 1, 165--186 (2015)], where ``simultaneous multi-portraits'' are examined.