A Galois counterexample to Hilbert's fourteenth problem in dimension three with rational coefficients
Let \(k\) be a field, \(k\left[ X \right] = k[X_1,\dots ,X_n]\) the polynomial ring in \(n\) variables over \(k\) for some \(n \in \mathbb{N}\), and \(k(X)\) the field of fractions of \(k\left[ X \right]\). Assume that \(L\) is a subfield of \(k(X)\) containing k. Then, \textit{the Fourteenth Problem of Hilbert} asks whether the \(k\) subalgebra \(L/k[X] \) of \(k[X]\) is finitely generated. To date a variety of counterexamples to Hilbert Fourteenth Problem has been constructed by many mathematicians. For the case where \(n \geq\;3 \), and any field \( k \neq \mathbb{Q}\) of characteristic zero, and a finite group \(G\) with \(|G| \geq 2\), \textit{S. Kuroda} [``Hilbert's Fourteenth Problem and invariant fields of finite groups'', preprint] constructed a counterexample \(L\) such that \(k(x)/L\) is a Galois extension with Galois group isomorphic to \(G\). However for \(n =3 \) and \(k = Q \) he posed the following \textit{open problem} (2008):NEWLINENEWLINEAssume that \( n =3 \). Let \(L\) be a subfield of \(Q(x)\) such that \(Q(x)/L\) is a Galois extension. Is the \(Q\)-subalgebra \(L\bigcap Q[x] \) of \(Q[x] \) always finitely generated?NEWLINENEWLINE This problem was solved in \textit{the negative} by the first author of this paper, as part of his Master Thesis, by proving the following:NEWLINENEWLINETheorem. Assume that \(n = 3 \) and \(k\) is any field of characteristic zero. If \(\epsilon \geq 4\), then the \(k\)-subalgebra \(k(x)^{\Gamma_{\epsilon }}\bigcap k[x] \) is not finitely generated. Where \(\Gamma \) is a certain subgroup of \(GL(2, k) \) and \(\Gamma_{\epsilon}\) is a certain conjugate of \(\Gamma\).NEWLINENEWLINEFor the entire collection see [Zbl 1266.14004].
scientific article; zbMATH DE number 6212736