Hilbert's fourteenth problem and field modifications (Q2178498)

As is often the case with Hilbert problems, there is some ambiguity in terminology. The original Hilbert's 14th problem asked if the ring of invariants of a subgroup \(\mathrm{GL}(V)\) of the general linear group acting on \(\mathrm{Sym}(V)\) for a vector space \(V\) of dimension \(n\) over a field \(k\) in the standard way is always a finitely generated \(k\)-algebra. A counterexample was found by \textit{M. Nagata} [Am. J. Math. 81, 766--772 (1959; Zbl 0192.13801)]. A generalization of this question, namely when \(A=\mathrm{Sym}(V)\cap L\) is a finitely generated \(k\)-algebra for a field extension \(k\subseteq L\), also became later known as Hilbert's 14th problem. We say that \(L\subseteq K\) (where \(K\) is the field of fractions of \(\mathrm{Sym}(V)\)) is \textit{minimal} if its transcendence degree \(r\) over \(k\) is the same as the transcendence degree of \(A\) over \(k\). (Then \(L\) is the field of fractions of \(A\).) Some interest appears to be in the question of what values of \((n,r)\) allow counterexamples. \textit{O. Zariski} [Bull. Sci. Math., II. Sér. 78, 155--168 (1954; Zbl 0056.39602)] showed that there are no counterexamples for \(r\leq 2\). There exists a counterexample for \((n,r)=(5,4)\) which is an invariant ring [\textit{D. Daigle} and \textit{G. Freudenburg}, J. Algebra 221, No. 2, 528--535 (1999; Zbl 0963.13024)], but no such example is known for \((n,r)=(4,3)\). The author previously gave examples for \((n,r)=(4,3)\) [J. Algebra 279, No. 1, 126--134 (2004; Zbl 1099.13035)] and for \((n,r)=(3,3)\) [Mich. Math. J. 53, No. 1, 123--132 (2005; Zbl 1108.13020)] which are not invariant rings. For \((n,r)=(3,3)\), the author previously also constructed examples with any \([K:L]=d\) for any \(d\geq 3\) [J. Algebra 309, No. 1, 282--291 (2007; Zbl 1116.12003)]. One major result of the present paper is finding a counterexample with \(d=2\). The method is to show that if \(V\) has basis \(x_1,\dots,x_n\) and \(S\subset k[x_1,\dots,x_{n-1}]\) such that \(k(S)\neq k(x_1,\dots,x_{n-1})\) and \(k(S)\) has transcendence degree \(\geq 2\) over \(k\), then there exis \(\sigma\in \mathrm{Aut}_k(K)\) such that \(L=\sigma(k(S\cup\{x_n\})\) is a counterexample to the generalized Hilbert 14th problem.