= Solution
The <Boltzmann factor> factorizes over bonds. Define the <spin-chain transfer matrix> by
$$
T_{ab}=e^{\beta J\delta_{ab}}
=1+(e^{\beta J}-1)\delta_{ab},\qquad a,b=1,\ldots,Q.
$$
Here $\delta_{ab}$ is the <Kronecker delta>. Summing the product of these <matrix elements> over all configurations, and closing the final bond by <periodic boundary conditions>, gives the <canonical partition function>
$$
Z=\sum_{\sigma_1,\ldots,\sigma_L}
T_{\sigma_1\sigma_2}\cdots T_{\sigma_L\sigma_1}
=\operatorname{Tr}T^L.
$$
Hence, writing $I$ for the <identity matrix> and $P$ for the <all-ones matrix>,
$$
\boxed{T=g(\beta J)I+P,\qquad g(\beta J)=e^{\beta J}-1.}
$$
Every element of $P^2$ is a sum of $Q$ ones, while every diagonal element of $P$ is one. Thus
$$
\boxed{P^2=QP,\qquad \operatorname{Tr}P=Q.}
$$
More explicitly, for $u=(1,\ldots,1)^T$, $Pu=Qu$. If $\sum_a v_a=0$, then $Pv=0$. These complementary <eigenspaces> have <dimensions> $1$ and $Q-1$. Equivalently, $P/Q$ is the <orthogonal projector> onto the <linear span> of $u$. The <eigenvalues> and their <eigenvalue multiplicities> are therefore
$$
\boxed{\begin{aligned}
P:&\quad Q\ (1),\qquad 0\ (Q-1),\\
T:&\quad \lambda_+=e^{\beta J}+Q-1\ (1),\qquad
\lambda_-=e^{\beta J}-1\ (Q-1).
\end{aligned}}
$$
Since the two <matrices> share these <eigenvectors>, the exact finite-ring <canonical partition function> is
$$
\boxed{Z=(e^{\beta J}+Q-1)^L+(Q-1)(e^{\beta J}-1)^L.}
$$
Both terms matter in the finite-ring low-<temperature> limit: discarding the second would lose the correct <ground state> <degeneracy factor> and introduce a spurious single-wall contribution.
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