On nilpotent Lie algebras of derivations of fraction fields

Let $K$ be an arbitrary field of characteristic zero and $A$ a commutative associative $K$-algebra which is an integral domain. Denote by $R$ the fraction field of $A$ and by $W(A)=RDer_{\mathbb K}A,$ the Lie algebra of $\mathbb K$-derivations of $R$ obtained from $Der_{\mathbb K}A$ via multiplication by elements of $R.$ If $L\subseteq W(A)$ is a subalgebra of $W(A)$ denote by $rk_{R}L$ the dimension of the vector space $RL$ over the field $R$ and by $F=R^{L}$ the field of constants of $L$ in $R.$ Let $L$ be a nilpotent subalgebra $L\subseteq W(A)$ with $rk_{R}L\leq 3$. It is proven that the Lie algebra $FL$ (as a Lie algebra over the field $F$) is isomorphic to a finite dimensional subalgebra of the triangular Lie subalgebra $u_{3}(F)$ of the Lie algebra $Der F[x_{1}, x_{2}, x_{3}],$ where $u_{3}(F)=\{f(x_{2}, x_{3})\frac{\partial}{\partial x_{1}}+g(x_{3})\frac{\partial}{\partial x_{2}}+c\frac{\partial}{\partial x_{3}}\}$ with $f\in F[x_{2}, x_{3}], g\in F[x_3]$, $c\in F.$ In particular, a characterization of nilpotent Lie algebras of vector fields with polynomial coefficients in three variables is obtained.Comment: Corrected typos. Revised formulation of Theorem 1, results unchange

On one-sided Lie nilpotent ideals of associative rings

We prove that a Lie nilpotent one-sided ideal of an associative ring $R$ is contained in a Lie solvable two-sided ideal of $R$. An estimation of derived length of such Lie solvable ideal is obtained depending on the class of Lie nilpotency of the Lie nilpotent one-sided ideal of $R.$ One-sided Lie nilpotent ideals contained in ideals generated by commutators of the form $[... [ [r_1, r_{2}], ... ], r_{n-1}], r_{n}]$ are also studied.Comment: 5 page

On closed rational functions in several variables

Let k be an algebraically closed field of characteristic zero. An element F from k(x_1,...,x_n) is called a closed rational function if the subfield k(F) is algebraically closed in the field k(x_1,...,x_n). We prove that a rational function F=f/g is closed if f and g are algebraically independent and at least one of them is irreducible. We also show that the rational function F=f/g is closed if and only if the pencil af+bg contains only finitely many reducible hypersurfaces. Some sufficient conditions for a polynomial to be irreducible are given.Comment: Added references, corrected some typo