Exercise 11.2

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Prove the converse to Proposition 11.1.1.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{
}{\mathrm{N}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
If $N_s = \sum_{j=1}^e \beta_j^s - \sum_{i=1}^d \alpha_i^s,$ where $\alpha_i,\beta_j$ are complex numbers,  then
\begin{align*}
\sum_{s=1}^\infty \frac{N_s u^s}{s} &=  \sum_{j=1}^e \left( \sum_{s=1}^\infty \frac{(\beta_j u)^s}{s} ight) - \sum_{i=1}^d \left( \sum_{s=1}^\infty \frac{(\alpha_i u)^s}{s} ight)\
&=-\sum_{j=1}^e \ln(1 - \beta_ju) + \sum_{i=1}^d \ln(1 - \alpha_i u).
\end{align*}
Here $u$ is a variable, and both members are formal polynomials in $\C[[u]]$, so we don't study convergence. Nevertheless, the left member has a radius of convergence at least $q^{-n}$, and the right member $\min_{i,j}(1/|\beta_j|,1/|\alpha_i|)$.

Therefore,
$$Z_f(u) = \exp\left(\sum_{s=1}^\infty \frac{N_s u^s}{s} ight) = \prod_{j=1}^e(1 - \beta_j u)^{-1} \prod_{i=1}^d (1-\alpha_i u) = \frac{\prod_{i=1}^d (1 - \alpha_i u)}{\prod_{j=1}^e (1 - \beta_j u)}$$
is a rational fraction.
\end{proof}

Exercise 11.2

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Exercise 11.2

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