%
\label{thm:spectral-determination}
Let $f, g \in \mathrm{Hol}(\mathbb{L})$.
If $f$ and $g$ have the same spectral coefficients
at every primorial stage ---
\[
    \forall\, k \geq 1:\;
    (a_k(f), b_k(f)) = (a_k(g), b_k(g))
\]
--- then $f = g$.
Equivalently:
\[
    \boxed{%
    \hat{f} = \hat{g}
    \;\;\Longrightarrow\;\;
    f = g.}
\]

\textbf{Step 1: Character basis at each stage.}
By the spectral decomposition
(Theorem~\ref{thm:spectral-decomposition}, I.T12),
$\chi_+$ and $\chi_-$ provide
a unique decomposition at each primorial stage.
If $(a_k(f), b_k(f)) = (a_k(g), b_k(g))$,
then:
\[
    f_k = a_k(f) \cdot e_+ + b_k(f) \cdot e_-
    = a_k(g) \cdot e_+ + b_k(g) \cdot e_-
    = g_k
\]
in $(\mathbb{Z}/M_k\mathbb{Z})[j]$.
So $f$ and $g$ agree at primorial depth $k$.

\textbf{Step 2: Tower coherence forces global agreement.}
Since spectral coefficients agree at \emph{all} stages,
$f$ and $g$ agree at every depth $k \geq 1$.
By the $\tau$-Identity Theorem
(Theorem~\ref{thm:tau-identity}, I.T21),
agreement at even a single depth suffices.
Therefore $f = g$.