On a question of Davenport
Let \(k\) be an algebraic number field and let \({\mathfrak o}_k\) be its ring of integers. Consider polynomials \(f,g\in {\mathfrak o}_k [X]\) such that \(V_{\mathfrak p} (f): = \{\overline f(u) \mid u \in{\mathfrak o}_k/{\mathfrak p}\} = V_{\mathfrak p} (g)\) for all but finitely many prime ideals \({\mathfrak p}\) of \({\mathfrak o}_k\). Does this imply \(f(X) = g(aX+b)\) for some \(a,b \in k\)? Davenport's original question concerns the case \(k= \mathbb{Q}\). Then we know that the answer is affirmative if \(f\) is an indecomposable polynomial or has odd prime degree. This result no longer remains true if we replace \(\mathbb{Q}\) by an arbitrary number field. The main result of the paper under review gives a large class of decomposable polynomials over number fields for which the answer is affirmative. The proof is based on the following known equivalence: \(V_{\mathfrak p} (f)= V_{\mathfrak p} (g)\) for all but finitely many prime ideals \({\mathfrak p}\) of \({\mathfrak o}_k\) if and only if \(f\) and \(g\) are Kronecker equivalent over \(k\). The latter property is a condition on the Galois group of \((f(X) - t)(g(X) -t)\) over \(k(t)\), where \(t\) is transcendental over \(k\). The monodromy group of \(f(X) \in E[X]\) is the Galois group of \(f(X) -t\) over \(\overline E(t)\). A polynomial \((aX+b)^n + c\) with \(a,b,c \in\mathbb{C}\), \(a\neq 0\) is called cyclic, since its monodromy group is cyclic. There is a similar definition of dihedral polynomials. We call a polynomial \(f\in \mathbb{C}[X]\) a Davenport polynomial if there is another polynomial \(g(X)\in \mathbb{C} [X]\) which is Kronecker equivalent to \(f(X)\) (over \(\mathbb{C})\), but \(f(X) \neq g(aX+b)\) for all \(a,b\in \mathbb{C}\). We are now able to state the main result: Let \(f,g\in \mathbb{C} [X]\) be Kronecker equivalent (over \(\mathbb{C})\), such that \(f\) can be written as a composition \(f= f_1 \circ \cdots \circ f_r\) of nonlinear indecomposable polynomials, none of which is cyclic, dihedral, or of degree 4. 1. If \(f,g\in \mathbb{Q} [X]\), then there exists \(a,b\in \mathbb{Q}\) with \(f(X) = g(aX +b)\). 2. If none of the \(f_i\) is a Davenport polynomial, then \(f(X) = g(aX+b)\) for some \(a,b\in \mathbb{Q}\). Since an indecomposable Davenport polynomial cannot have rational coefficients we easily deduce (1) from (2). The proof of (2) is based on a group-theoretic lemma, which gives conditions that certain subgroups are conjugate. This implies that \(g(X)-t\) has a root in \(\mathbb{C} (t)[X]/(f(X) -t)\). The proof further requires the classification of finite doubly transitive groups, which contains a regular cyclic subgroup. Hence it is based on the classification of finite simple groups.
decomposable polynomials over number fields
scientific article; zbMATH DE number 883830