Let \(f \in \mathbb{C}[z]\) be a polynomial of degree \(m \geq 1\). A point \(z \in \mathbb{C}\) is a critical point of \(f\), if \(f'(z)=0\). The corresponding value \(t=f(z)\) is a critical value, and all other values are called regular values. If \(\Sigma\) denotes the set of critical values, then \(f : f^{-1}(\mathbb{C} \setminus \Sigma) \to \mathbb{C} \setminus \Sigma\) is a fibration with zero-dimensional fiber. For a regular value \(t\) of \(f\) the fiber \(f^{-1}(t)\) consists of \(m\) distinct points \(z_1(t),\dotsc,z_m(t)\) which can be continuously transported along any path in \(\mathbb{C} \setminus \Sigma\). Then \(z_1(t),\dotsc,z_m(t)\) define algebraic functions, and a zero-chain of \(f\) is given by \[ C(t) = \sum_{j=1}^m n_jz_j(t)\,, \] where \(n_1,\dotsc,n_m\) are integers. If, in addition, \[ \sum_{j=1}^m n_j = 0\,, \] then \(C(t)\) is called a zero-cycle. Now, the tangential center problem is to find all polynomials \(g \in \mathbb{C}[z]\) such that \[ \sum_{j=1}^m n_jg(z_j(t)) \equiv 0\,. \]
In this paper, the authors give a solution of this problem which is based on induction on the number of decomposition factors of \(f\). For that purpose, they first show the uniqueness of the decomposition \(f = f_1 \circ\dotsb\circ f_d\), where every \(f_k\) is \(2\)-transitive, a monomial or a Chebyshev polynomial. Here, a polynomial \(f\) is called \(2\)-transitive, if the monodromy group \(G_f\) (this group is isomorphic to the Galois group of \(f(z)-t\) seen as a polynomial over \(\mathbb{C}(t)\)) acts \(2\)-transitive on the finite set \(\{z_1(t),\dotsc,z_m(t)\}\). The action of a group \(G\) on a finite set \(X\) is called \(2\)-transitive if for any two pairs \((i,j)\), \((k,l) \in X \times X\) there is some \(\sigma \in G\) with \(\sigma(i)=k\) and \(\sigma(j)=l\). Then the authors give a complete solution of the tangential center problem under the assumption that there is no merging of critical values in the above decomposition. This assumption means the following. If \(\tilde{f}\) and \(h\) are two nonlinear polynomials and if \(f = \tilde{f} \circ h\), then the critical values of \(\tilde{f}\) and \(h\) do not merge, if \(\{\, f(z) : \tilde{f}'(h(z))=0 \,\} \cap \{\, f(z) : h'(z)=0 \,\} = \emptyset\) and if \(f\) is injective on the set of critical values of \(h\).