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带有2次分次恒等式的分次代数

The Graded Algebras with a Graded Identity of Degree 2

Federal Univ. of Campina Grande（坎皮纳格兰德联邦大学）

Antonio de França

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AI总结：

本文研究了满足2次分次多项式恒等式的分次结合代数，证明了代数在代数闭域上的结构分类，并探讨了有限循环群下中性分量为中心时换位子理想的幂零性及李代数的可解性。

摘要：

本文致力于研究满足2次分次多项式恒等式的分次结合代数。设$\mathsf{G}$为有限阿贝尔群，$\mathbb{F}$为特征零的域，$\mathfrak{A}$为$\mathsf{G}$-分次$\mathbb{F}$-代数。我们证明，当$\mathbb{F}$代数闭时，若$\mathfrak{A}_e$满足2次多项式恒等式$g=g(x_1^{(e)}, \dots, x_n^{(e)})\in\mathbb{F}\langle X^\mathsf{G} \rangle$，则$\mathfrak{A}$要么是幂零的，要么具有交换的中性分量，并且我们确保由$g$确定的$\mathsf{G}$-分次簇$\mathfrak{W}^\mathsf{G}$等于$\mathsf{var}^\mathsf{G}([x^{(e)},y^{(e)}])$或某个幂零$\mathsf{G}$-分次代数$N$的$\mathsf{var}^\mathsf{G}(N)$。随后，我们研究了$\mathfrak{A}_e$在$\mathfrak{A}$中为中心的含义。所得结果使我们能够证明，当$\mathsf{G}$为有限循环群时，若$\mathfrak{A}$是有限生成的且$\mathfrak{A}_e$在$\mathfrak{A}$中为中心，则$\mathfrak{A}$的换位子理想是幂零的，且代数$\mathfrak{A}^{(-)}=(\mathfrak{A},[\ ,\ ])$是可解李代数；并且，若$\mathsf{G}$的阶为奇数，则在$\mathfrak{A}$中$[x_1,x_2][x_3,x_4]\cdots[x_{2d-1},x_{2d}]\equiv0$，对于某个$d\in\mathbb{N}$。

英文摘要：

This paper is devoted to the study of graded associative algebras that satisfy a graded polynomial identity of degree $2$. % Let $\mathsf{G}$ be a finite abelian group, $\mathbb{F}$ a field of characteristic zero and $\mathfrak{A}$ a $\mathsf{G}$-graded $\mathbb{F}$-algebra. % We prove that, for $\mathbb{F}$ algebraically closed, if $\mathfrak{A}_e$ satisfies a polynomial identity $g=g(x_1^{(e)}, \dots, x_n^{(e)})\in\mathbb{F}\langle X^\mathsf{G} \rangle$ of degree $2$, then $\mathfrak{A}$ is either nilpotent or has commutative neutral component, % and we ensure that the $\mathsf{G}$-graded variety $\mathfrak{W}^\mathsf{G}$ determined by $g$ is equal to either $\mathsf{var}^\mathsf{G}([x^{(e)},y^{(e)}])$ or $\mathsf{var}^\mathsf{G}(N)$ for some nilpotent $\mathsf{G}$-graded algebra $N$. % Posteriorly, we investigate the implications of $\mathfrak{A}_e$ being central in $\mathfrak{A}$. The results obtained allow us to prove that, when $\mathsf{G}$ is finite cyclic, if $\mathfrak{A}$ is finitely generated and $\mathfrak{A}_e$ is central in $\mathfrak{A}$, then the commutator ideal of $\mathfrak{A}$ is nilpotent, and the algebra $\mathfrak{A}^{(-)}=(\mathfrak{A},[\ ,\ ])$ is a solvable Lie algebra, % and, if $\mathsf{G}$ has odd order, then $[x_1,x_2][x_3,x_4]\cdots[x_{2d-1},x_{2d}]\equiv0$ in $\mathfrak{A}$, for some $d\in\mathbb{N}$.

机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。

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