AI 中文总结
本文利用矩阵理论与史密斯标准形，推导有限环Z_{p^k}上四元数范数纤维大小公式，完全解决了该环上的局部四平方问题，给出元素表示为四个平方和的精确个数。
AI 中文摘要
本文研究有限环\boldsymbol{\text{Z}}_{p^k}上的四元数环\boldsymbol{\text{H}}_{\boldsymbol{\text{Z}}_{p^k}}上的范数映射N:\boldsymbol{\text{H}}_{\boldsymbol{\text{Z}}_{p^k}}\to \boldsymbol{\text{Z}}_{p^k}，其中p为奇素数，k\boldsymbol{\text{\textgreater =}}1为整数。利用同构\boldsymbol{\text{H}}_{\boldsymbol{\text{Z}}_{p^k}}\boldsymbol{\text{\textcong}}M_2(\boldsymbol{\text{Z}}_{p^k})，通过矩阵方法对四元数展开研究。证明纤维大小a_{p^k}(m)=|\boldsymbol{\text{\braceleft}}q\boldsymbol{\text{\textin}} \boldsymbol{\text{H}}_{\boldsymbol{\text{Z}}_{p^k}}:N(q)=m\boldsymbol{\text{\braceright}}仅依赖于m的p进赋值v_p(m)。对任意m\boldsymbol{\text{\textin}}\boldsymbol{\text{Z}}_{p^k}推导得到纤维大小的显式公式：当t=0时，a_{p^k}(m)=p^{3k-2}(p^2-1)；当0\boldsymbol{\text{\textless}}t\boldsymbol{\text{\textless}}k时，a_{p^k}(m)=p^{3k-2-t}(p+1)(p^{t+1}-1)；当t=k时，a_{p^k}(m)=p^{2k-1}(p^{k+1}+p^k-1)，其中t=v_p(m)（约定v_p(0)=k）。本文的主要结果是完全解决了环\boldsymbol{\text{Z}}_{p^k}上的局部四平方问题：a_{p^k}(m)给出了任意元素m\boldsymbol{\text{\textin}}\boldsymbol{\text{Z}}_{p^k}表示为四个平方和x_1^2+x_2^2+x_3^2+x_4^2=m的精确个数。证明是纯代数的，仅依赖矩阵理论与史密斯标准形，因此避免了数论的抽象工具。本预印本尚未经过同行评审（若适用）或任何投稿后的改进与修正。
英文摘要
The norm map \(N:\mathcal{H}_{\mathbb{Z}_{p^k}}\to \mathbb{Z}_{p^k}\) is studied on the quaternion ring over \(\mathbb{Z}_{p^k}\), where \(p\) is an odd prime and $k\ge 1$ an integer. By means of the isomorphism \(\mathcal{H}_{\mathbb{Z}_{p^k}}\cong M_2(\mathbb{Z}_{p^k})\), quaternions are investigated using matrix methods. It is shown that the fibre size \[ a_{p^k}(m)=|\{q\in \mathcal{H}_{\mathbb{Z}_{p^k}}:N(q)=m\}| \] depends only on the \(p\)-adic valuation \(v_p(m)\) of \(m\). Explicit formulas for the fibre sizes are derived for every \(m\in\mathbb{Z}_{p^k}\): \[ a_{p^k}(m)= \begin{cases} p^{3k-2}(p^2-1), & t=0,\\[6pt] p^{3k-2-t}(p+1)(p^{t+1}-1), & 0<t<k,\\[6pt] p^{2k-1}(p^{k+1}+p^k-1), & t=k, \end{cases} \] where \(t=v_p(m)\) (with the convention \(v_p(0)=k\)). The main result of the paper is a complete solution to the \emph{local four-square problem} over the ring \(\mathbb{Z}_{p^k}\): the number \(a_{p^k}(m)\) gives the exact number of representations of an arbitrary element \(m\in \mathbb{Z}_{p^k}\) as a sum of four squares, \[ x_1^2+x_2^2+x_3^2+x_4^2=m. \] The proof is purely algebraic; it relies only on matrix theory and Smith normal form, thus avoiding the abstract machinery of number theory. This preprint has not undergone peer review (when applicable) or any post-submission improvements or corrections
CommentsThis preprint has not undergone peer review (when applicable) or any post-submission improvements or corrections
