= Solution
Let $S=\operatorname{diag}(1,-1)$ be the base-<spin> insertion matrix. In the <eigenbasis> from <1/c/solution>, its diagonal matrix elements vanish: $\langle+|S|+\rangle=\langle-|S|-\rangle=0$. Hence
$$
\boxed{\langle\sigma_1\rangle=\frac{\operatorname{Tr}(ST^L)}{\operatorname{Tr}T^L}=0,}
$$
not only at large $L$ but for every finite ring at zero field.
This vanishing is not merely an averaging artifact hiding a finite-temperature ordered phase. The two-point <correlation function> is $\langle\sigma_1\sigma_{1+r}\rangle=(\lambda_-^r\lambda_+^{L-r}+\lambda_+^r\lambda_-^{L-r})/Z_L$. In the <thermodynamic limit> it becomes $(\lambda_-/\lambda_+)^r$, with finite <correlation length> $\xi^{-1}=\log(\lambda_+/\lambda_-)$ for every $T>0$ and $J>0$. Thermal <domain walls> destroy long-range alignment in this one-dimensional short-range system. At zero <temperature>, the two fully aligned <ground states> are degenerate; that limiting case must be distinguished from any fixed positive <temperature>.
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