AI 中文总结
本文证明秩二对称化 Nahm 和若为模函数，则其 Nahm 点生成域的次数不超过 24，并针对特殊情形给出更紧的界，方法结合牛顿多边形与极点阶分析。
AI 中文摘要
设 $F$ 为秩二对称化 Nahm 和，具有任意对称化子 $\operatorname{diag}(d_1,d_2)$（正定有理矩阵）及有理平移，并设 $K$ 为由其 Nahm 点生成的域；记 $m=d_2/d_1$。我们证明：若 $q^cF$ 是模的，则 $[K:\mathbb{Q}]\le 24$。模性迫使对数径向渐近展开的高阶系数消失；我们通过牛顿多边形和沿行列式圆锥的极点阶分析研究所得多项式方程，并对 $m=1$ 情形证明其公共分量仅出现在一个显式族上，该族的 Nahm 点为有理点。对于相等的行和，Bloch 群挠给出 $[K:\mathbb{Q}]\le 2$。对于互补 Nahm 坐标，当 $m\ne 1$ 时界为 $2$，当 $m=1$ 时界为 $4$（后者使用前两个渐近修正）。证明的若干步骤依赖于精确计算。
英文摘要
Let $F$ be a rank-two symmetrizable Nahm sum, with any symmetrizer $\operatorname{diag}(d_1,d_2)$, a positive-definite rational matrix and rational shifts, and let $K$ be the field generated by its Nahm point; write $m=d_2/d_1$. We prove that if $q^cF$ is modular, then $[K:\mathbb{Q}]\le 24$. Modularity forces the higher coefficients of the logarithmic radial asymptotic expansion to vanish; we study the resulting polynomial equations through their Newton polygons and a pole-order analysis along the determinant conic, and for $m=1$ show that their common components occur only on one explicit family, whose Nahm point is rational. For equal row sums, Bloch-group torsion gives $[K:\mathbb{Q}]\le 2$. For complementary Nahm coordinates, the bound is $2$ when $m\ne 1$ and $4$ when $m=1$, using the first two asymptotic corrections in the latter case. Several steps of the proofs rely on exact computation.
Comments31 pages, 1 table. Ancillary files: all verification scripts, independent checks and their logs; a one-command check (verify_all.sh quick, about 30 s; full, about 8 min) reproduces every computational claim in Appendix A
