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arXiv 0907.4913math.COmath.NT
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Davenport常数与群代数

On the Davenport constant and group algebras

Daniel Smertnig

摘要:

对于有限阿贝尔群$G$和$G$的分裂域$K$,令$d(G, K)$表示最大整数$l \in \N$,对于该最大整数,存在$G$上的序列$S = g_1 \cdot ... \cdot g_l$,使得对于所有$a_1, ..., a_l \in K^{\times}$,$(X^{g_1} - a_1) \cdot ... \cdot (X^{g_l} - a_l) \ne 0 \in K[G]$。如果$D(G)$表示$G$的达文波特常数,则存在简单的不等式$D(G)-1 \le d (G, K)$。平等适用于各种群体,以及W. Gao等人指出,平等适用于所有群体。我们提供了更多的群,其中等式成立,但我们也给出了第一个例子的群体$G$,其中$D(G) -1 < d(G, K)$持有。我们要以实际行动证明这一点。

英文摘要:

For a finite abelian group $G$ and a splitting field $K$ of $G$, let $d(G, K)$ denote the largest integer $l \in \N$ for which there is a sequence $S = g_1 \cdot ... \cdot g_l$ over $G$ such that $(X^{g_1} - a_1) \cdot ... \cdot (X^{g_l} - a_l) \ne 0 \in K[G]$ for all $a_1, ..., a_l \in K^{\times}$. If $D(G)$ denotes the Davenport constant of $G$, then there is the straightforward inequality $D(G)-1 \le d (G, K)$. Equality holds for a variety of groups, and a standing conjecture of W. Gao et.al. states that equality holds for all groups. We offer further groups for which equality holds, but we also give the first examples of groups $G$ for which $D(G) -1 < d(G, K)$ holds. Thus we disprove the conjecture.

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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