Exercise 15.4.16

Question

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Prove that for each integer $k\geq 0$ there exists a polynomial $S_k(u) \in \Q[u]$ of degree $4k$ such that (15.59) holds for all $\beta \in \Z[i]$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

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{\U}{\mathbb{U}}

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

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ewcommand{\Gal}{\mathrm{Gal}}

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\begin{proof} 
By (15.56) and (15.57),
\begin{align*}
\varphi(\beta z) &= \sum_{j=0}^\infty c_j \beta^{4j+1} z^{4j+1}\
&= \sum_{k=0}^\infty b_k(\beta) \left( \sum_{j=0}^\infty c_j z^{4j+1}ight)^{4k+1}\
&=  \sum_{k=0}^\infty b_k(\beta) S_k(z),
\end{align*}
where
\begin{align*}
S_k(z) &= \left( \sum_{j=0}^\infty c_j z^{4j+1}ight)^{4k+1}\
&=z^{4k+1}\left(1 + \sum_{j=1}^\infty c_j z^{4j}ight)^{4k+1}
\end{align*}
Consider
$$g(u) = 1 + \sum_{j=1}^\infty c_ju^j, $$
which verifies $S_k(z) = z^{4k+1} g^{4k+1}(z^4)$.

Since $\varphi(z) = \sum_{j=0}^\infty c_jz^{4j+1}$ is analytic in the neighborhood of $0$, it is the same for $g(u)$, so that the radius of convergence of $ \sum_{j=0}^\infty c_ju^j$ doesn't vanish. Then
\begin{align*}
g^{4k+1}(u) &= \left(1 + \sum_{j=1}^\infty c_ju^jight)^{4k+1}\
&=\sum_{l=0}^\infty \delta_{k,l}u^l.
\end{align*}
We want to prove that $\delta_{k,l} \in \Q$, knowing that $c_1,c_2,\cdots $ are rational.
To consider only finite sums, we use an asymptotic expansion in the neighborhood of $0$, up to  degree $m$.
If $v =\sum_{j=1}^\infty c_ju^j$, then $v \sim c_1u = \frac{-1}{10}u$, and
$$v = \sum_{j=1}^m c_ju^j + o(u^m).$$
Therefore
\begin{align*}
g^{4k+1}(u) &= (1+v)^{4k+1}\
&=1 + \sum_{r=1}^{4k+1} \binom{4k+1}{r} v^k\
&=1 + \sum_{r=1}^{4k+1} \binom{4k+1}{r} \left(\sum_{j=1}^m c_ju^jight)^k  + o(u^m)\
&= 1 + \sum_{l=1}^m P_{k,l}(c_1,\ldots,c_m) u^l + o(u^m),
\end{align*}
where $P_{k,l}$ is a multivariate polynomial with coefficients in $\Z$.

The unicity of the asymptotic expansion shows that $\delta_{k,0} = 1$, and 
$$\delta_{k,l} = P_{k,l}(c_1,\cdots,c_m) \in \Q.$$
(Note that the unicity shows that $P_{k,l}(c_1,\cdots,c_m)$ depends only of $c_1,\ldots, c_l$, so we can write $P_{k,l}(c_1,\cdots,c_m) = P_{k,l}(c_1,\cdots,c_l)$.)

Therefore
$$S_k(z)  =  \sum_{l=0}^\infty \delta_{k,l} z^{4l+4k+1},\qquad \delta_{k,l} \in \Q,$$
and
\begin{align*}
\varphi(\beta z)  &= \sum_{k=0}^\infty b_k(\beta) S_k(z)\
&= \sum_{k=0}^\infty \sum_{l=0}^\infty  b_k(\beta)  \delta_{k,l} z^{4l+4k+1}\
&= \sum_{j=0}^\infty \left( \sum_{k+l = j}  b_k(\beta)  \delta_{k,l}ight) z^{4j+1}\
&= \sum_{j=0}^\infty \left( \sum_{k=0}^j  \delta_{k,j-k} b_k(\beta)  ight) z^{4j+1}\
&=\sum_{j=0}^\infty \left( \sum_{k=0}^j  a_{j,k} b_k(\beta)  ight) z^{4j+1}
\end{align*}
where $a_{j,k} = \delta_{k,j-k} \in \Q,$ and $a_{j,j} = \delta_{j,0} = 1$.
This gives the generalization of (15.58):
$$c_j \beta^{4j+1} = \sum_{k=0}^j  a_{j,k}\,  b_k(\beta) = a_{j,0}\, b_0(\beta) + \cdots + a_{j,j-1}\, b_{j-1}(\beta) + b_j(\beta), \qquad a_{j,k} \in \Q,$$

\bigskip

We prove by induction that for any $j$, there is a polynomial $S_j(u) \in \Q[u]$ of degree $4k$ such that $b_j(\beta) = \beta S_j(\beta)$.

Since $b_0(\beta) = \beta$, this is true for $k=0$ with $S_0(u) = 1, \deg(S_0) = 0$.

Now we suppose that $b_i(\beta) = \beta S_{i}(\beta)$ for $0\leq i \leq j-1$. Then
\begin{align*}
b_j(\beta) &= c_j \beta^{4j+1} - \sum_{k=0}^{j-1} a_{j,k} b_k(\beta)\
&=c_j \beta^{4j+1} - \sum_{k=0}^{j-1} a_{j,k}  \beta S_k(\beta)\
&= \beta \left(c_j \beta^{4j} - \sum_{k=0}^{j-1} a_{j,k}  S_k(\beta)ight)\
&=\beta S_j(\beta),
\end{align*}
where $S_j(u) = c_j u^{4j} - \sum_{k=0}^{j-1} a_{j,k}  S_k(u) \in \Q[u]$.

Since $c_j =\frac{ \varphi^{(4j+1)}(0)}{(4j+1)!} 
e 0$, and $\deg(S_k) = 4k < 4j$ for $k=0,\ldots,j-1$, we obtain $\deg(S_j )= 4j$, and the induction is done.

For each integer $k\geq 0$, there exists a polynomial $S_k(u) \in \Q[u]$ of degree $4k$ such that $b_k(\beta) = \beta S_k(\beta)$ holds for all odd $\beta \in \Z[i]$.
\end{proof}

Exercise 15.4.16

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Exercise 15.4.16

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