On function compositions that are polynomials (Q901022)

Let \(k\) be a field. For positive integers \(m\) and \(n\), let \(g,f_1,\dots,f_m\in k[t_1,\dots,t_n]\). \(g\) is said to be determined by \(f=(f_1,\dots,f_m)\) if and only if there is a function \(h:k^m\to k\) such that \(g(a)=h(f_1(a),\dots,f_m(a)),\text{ for all }a\in k^n.\) Now suppose \(k\) is algebraically closed. By \(\mathrm{range}(f)\), we denote the image of the mapping \(f:k^n\to k^m\) that is induced by \(f\). \(f\) is said to be almost surjective on \(k\) if the dimension of the Zariski-closure of \(k^m\setminus\mathrm{range}(f)\) is at most \(m-2\). The set of all elements of \(k[t_1,\dots,t_{n}]\) that are determined by \(f=(f_1,\dots,f_m)\) is a \(k\)-subalgebra of \(k[t_1,\dots,t_n]\) and will be denoted by \(k\langle f\rangle\). The author of the paper under review describes \(k\langle f\rangle\) in the case where \(k\) is algebraically closed and f induces a map from \(k^n\) to \(k^m\) that is surjective or almost surjective. As a result, author deduces in the case that \(k\) is algebraically closed of characteristic 0 and \(f\) is surjective, then \(g=h\circ f\) implies that \(h\) is a polynomial.