$\mathrm H_{p^m}^{n+1}(F)$中指数较低的类
Classes in $\mathrm H_{p^m}^{n+1}(F)$ of lower exponent
AI总结：
研究了特征$p>0$的域中，$H_{p^m}^{n+1}(F)$中符号的指数与符号长度的关系，证明了在特定条件下符号长度的上界。
摘要：
令$F$为特征$p>0$的域。我们证明，如果符号$A=ω⊗β_1⊗⋯⊗β_n$属于$H_{p^m}^{n+1}(F)$，且其指数能被$p^{m-1}$整除，则其在$H_{p^{m-1}}^{n+1}(F)$中的符号长度至多为$p^n$。在$n=2$的情况下，我们还证明，如果$A=ω_1⊗β_1+⋯+ω_r⊗β_r$属于$H_{p^{m}}^2(F)$，且$\exp(A)|p^{m-1}$，则$A$在$H_{p^{m-1}}^2(F)$中的符号长度至多为$p^r + r - 1$。最后，我们考虑$p=2$的情况，证明如果$A$是$H_{2^m}^{n+1}(F)$中两个符号的和，且$\exp A |2^{m-1}$，则$A$在$H_{2^{m-1}}^{n+1}(F)$中的符号长度至多为$(2n+1)2^n$。我们的结果在特征$p$中使用规范条件的方式与Matrzi在论文``On the symbol length of symbols''中相同。
英文摘要：
Let $F$ be a field of characteristic $p>0$. We prove that if a symbol $A=ω\otimes β_1 \otimes \dots \otimes β_n$ in $H_{p^m}^{n+1}(F)$ is of exponent dividing $p^{m-1}$, then its symbol length in $H_{p^{m-1}}^{n+1}(F)$ is at most $p^n$. In the case $n=2$ we also prove that if $A= ω_1\otimes β_1+\cdots+ω_r\otimes β_r$ in $H_{p^{m}}^2(F)$ satisfies $\exp(A)|p^{m-1}$, then the symbol length of $A$ in $H_{p^{m-1}}^2(F)$ is at most $p^r+r-1$. We conclude by looking at the case $p=2$ and proving that if $A$ is a sum of two symbols in $H_{2^m}^{n+1}(F)$ and $\exp A |2^{m-1}$, then the symbol length of $A$ in $H_{2^{m-1}}^{n+1}(F)$ is at most $(2n+1)2^n$. Our results use norm conditions in characteristic $p$ in the same manner as Matrzi in his paper ``On the symbol length of symbols''.
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。
